RCC & STEEL STRUCTURES

RCC &

STEEL STRUCTURES

For CIVIL ENGINEERING

SYLLABUS RCC: Working stress, Limit state and Ultimate load design concepts; Design of beams, slabs, columns; Bond and development length; Prestressed concrete; Analysis of beam sections at transfer and service loads. Steel Structures: Working stress and Limit state design concepts; Design of tension and compression members, beams and beam- columns, column bases; Connections - simple and eccentric, beam- column connections, plate girders and trusses; Plastic analysis of beams and frames.

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RCC & STEEL STRUCTURES

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Topics Page No 1. BASICS

1.1 Plain and Reinforced Concrete 01 1.2 Compressive Strength of Concrete in Structure 02 1.3 Flexural Strength of Concrete (Modules of Rupture) 02 1.4 Tensile Strength of Concrete 02 1.5 Stress-Strain Curve of Concrete 03 1.6 Concrete Strain at Ultimate Strength 04 1.7 Design Stress-Strain Curve 04 1.8 Shrinkage and Creep in Concrete 04 1.9 Design Stress Strain Curve for Steel 06

2. DESIGN OF RCC STRUCTERES

2.1 Introduction 07 2.2 What is Limit State Method 07 2.3 Characteristic Strength of Method 07 2.4 Characteristic Load 08 2.5 Partial Safety of Factor 08 2.6 Assumption Limit State of Collapse: Flexure 09 2.7 Working Stress Method 12 2.8 Deficiency in Working Stress Method 12 2.9 Permissible Stresses 13 2.10 Permissible Stresses in Concrete 13 2.11 Permissible Stress in Steel Reinforcement 13

3. CONCRETE TECHNOLOGY

3.1 Introduction 15 3.2 Chemical Composition of Raw Materials 16 3.3 Type of Cements 16 3.4 Admixture 18 3.5 Different Type of Admixture 19 3.6 Aggregates 20 3.7 Permissible Limits for Impurities in Water 21 3.8 Concrete 21 3.9 Methods of Proportional Concrete 213.10 Water Cement Ratio 22

4. REINFORCED SECTION

4.1 Balanced, Under-Reinforced And Over Reinforced Section 23

CONTENTS (RCC)


4.2 Moment of Resistance Calculation 24 4.3 Doubly Reinforced Section 26 4.4 Analysis Doubly of Reinforced Section 26 4.5 Moment of Resistance Doubly of Reinforced Section 27 4.6 Design of Doubly of Reinforced Section 27 4.7 Flanged Beam 28 4.8 Effective Flange Width 29 4.9 IS Code Specification 32 4.10 Design of Flanged Section 33 4.11 Limit State of Collapse in Shear 34 4.12 Diagonal Tension and Diagonal Compression 34 4.13 Mechanism of Shear Resistance 36 4.14 Shear Stress 36 4.15 Design Shear Strength of Concrete in Beams 37 4.16 Shear Reinforcement in Beam 39 4.17 Minimum Shear Reinforcement 42 4.18 Design Steps 42 4.19 Bond and Anchorage 43 4.20 Torsion 45 4.21 Effect of Torsional Moment 46

5. BEAM COLUMN

5.1 Design of Beam and Slab 48 5.2 I.S 456 Provisions 48 5.3 Control of Deflection 49 5.4 One-way Slab 50 5.5 Introduction 50 5.6 One-way and Two-way Slabs 51 5.7 Code Requirements on Reinforcement and Detailing 51 5.8 Column Subjected to Axial Compression and Uniaxial bending

(clause 39.5 ) 55 5.9 Column Subjected to Axial Compression and Biaxial bending

(clause 39.6) 56 5.10 Code Procedure for Design of biaxilly Loaded Columns 56 5.11 Depth of Foundation 58 5.12 Design Consideration 58

6. BASIC ELEMENT OF PRESTRESS CONCRETE

6.1 Losses in Priestess 62 6.2 Loss of Prestress Due to Friction 63 6.3 Loss of Prestress Due Anchorage Slip 63 6.4 Loss of Prestress due to Creep of Concrete 64 6.5 Loss due to Shrinkage of Concrete 65 6.6 Loss of Prestress due to Relaxation of Steel 65 6.7 Deflection of Restressed Beam 66


6.8 Prestressed Concrete 68 6.9 Requirement of High Strength Steel & Concrete in Prestressing 69

7. GATE QUESTIONS 71

Topics Page No

1. GENERAL DESIGN SPECIFICATION

1.1 Design Objective 981.2 Methods of Design 981.3 Loads and Forces 981.4 Load Combinations 981.5 Geometrical Properties 981.6 Classification of Cross Sections 981.7 Limit State Designs of Steel Structures 991.8 Connections: Riveted Connection 991.9 Terminology in Bolted Connection 1011.10 IS 800-2007 Specification for Spacing and Edge Distances

of Bolt Holes 1011.11 Types of Joints 1021.12 Design Strength of Plates in a Joint 1031.13 Block Shear Strength 1041.14 Design Strength of Bearing Blots 1041.15 Bearing Capacity or Bolts �Vdpb� 1041.16 Tensile Capacity of Bolts 1051.17 Welded Connection 1051.18 Types of Welded Joints 1051.19 Specification for Welding 1061.20 Fillet Weld 1071.21 Reduction in Design Stresses for Long Joints 108 1.22 Eccentric Connection - Plane of Moment and The Plane of

Welds is the same 108

2. DESIGN OF TENSION MEMBERS

2.1 Tension Member 110

CONTENTS (STEEL STRUCTURES)


2.2 Design Strength Due to Yielding of Gross Section 110 2.3 Design Strength Due to Rapture of Critical Section 110 2.4 Design Strength Due to Block Shear 112 2.5 Lug Angles 112

3. DESIGN OF COMPRESSION MEMBERS

3.1 Introduction 114 3.2 Slenderness Ration 1143.3 Design Compressive Stress and Strength 115 3.4 How to Select the Shapes of Compression Members 115 3.5 Steps for Design of Compression Members 116 3.6 Battens 116 3.7 Design of Battened Columns 116 3.8 Design of Slab Base 117

4. DESIGN OF BEAMS

4.1 Introduction 119 4.2 Bending Strength of Laterally Supported Beam 119 4.3 Design Procedure 120 4.4 Shear Strength of Laterally Supported Beam 120 4.5 Web Buckling Strength 121 4.6 Web Crippling 121

5. Plastic Analysis 122 5.1 Introduction 122 5.2 Fully Plastic Moment of a Section 122 5.3 Bending of Beams Symmetrical About Both Axes 123 5.4 Shape Factor 124 5.5 Load Factor 124 5.6 Fundamental Conditions for Plastic Analysis 1245.7 Mechanism 124 5.8 Basic Theorems of Plastic Analysis 125

6. GATE QUESTIONS 126

7. ASSIGNMENT QUESTIONS (RCC & STEEL STRUCTURE) 151


RCC


1.1 PLAIN & REINFORCED CONCRETE

Plain Concrete: It is a mixture of sand, gravel, cement, and water which results in a solid mass. Concrete is strong in compression but weak in tension. Its tensile strength is approx, one tenth of compressive strength. Plain concrete is mostly used in mass concrete work. (As in dams)

1.1.1 REINFORCED CONCRETE

1) It is a concrete with reinforcementembedded in it. The embeddedreinforcement makes it capable ofresisting tension also.

2) Steel bars embedded in the tensionzone of concrete, relieves concrete ofany tension and takes all tensionwithout separating from concrete.

3) The bond between steel andsurrounding concrete ensures straincompatibility i.e., the strain at any pointin the steel is equal to that in theadjoining concrete.

4) Reinforcing steel imparts ductility toconcrete which is otherwise brittlematerial.

(5) Here ductility means large deflection owing to yielding of steel, thereby giving ample warning of impending collapse.

6) Tensile stress in concrete arises onaccount of direct tension, flexuraltension, diagonal tension (due to shear),temperature and shrinkage effect,restraint to deformation.

7) Under these conditions, reinforcementsmust be provided across potentialtensile crack.

1.1.2 GRADE OF CONCRETE

•

• Compressive strength of concrete isthe most important property ofconcrete. Because other properties like

tensile strength, shear strength, bond strength, density, impermeability, durability etc. can be inferred from the compressive strength using established correlations.

• Compressive strength can bemeasured by standard test on concretecube. (or cylinder) specimen.

• Strength of concrete in uniaxialcompression is determined by loadingstandard test cube (150 mm size) tofailure in compression testing machine.

• The test specimens are generally tested28 days after casting (and continuouscuring)

• Cube is always tested on sides i.e., facein touch with mould.

• Strength of cube is expressed to benearest of 0.5N/mm2.

• As per IS 456: 2000, three testspecimens of a sample are taken.

1 BASICS


• Additional samples may be required for various purposes such as to determine the strength of concrete at 7 days or for finding out the strength for striking the formwork, or to determine the duration of curing, or to check the testing error. Additional specimen may also be required for testing samples cured by accelerated methods.

• To report, strength of cube, we take average of the strength of three specimen of a sample.

• Individual variation should not be more than ±15% of average if variation is more, test results of the sample are invalid.

1.2 COMPRESSIVE STRENGTH OF CONCRETE IN STRUCTURE Strength of concrete is found to decrease with increase in the size of specimen. However, beyond 450mm size, there is no decrease in the compressive strength of concrete. Thus, compressive strength of concrete in structure is taken as 0.67 fck. 1.3 FLEXURAL STRENGTH OF CONCRETE (MODULES OF RUPTURE) Tensile strength of concrete in flexure is called flexural strength.

( )3

2M W×100= ×10Z 6×100× 100

6

N/mm2 [if W is in

kN] [Assuming linear stress strain curve and contribution of steel area to be negligible]

fcr = 0.4W N/mm2 [W is in kN for onset of cracking.] However, stress strain variation is not linear hence as per IS code fcr = 0.7 ckf N/mm2 Flexural strength is used to determine the onset of cracking or the loading at which cracking starts in a structure. 1.4 TENSILE STRENGTH OF CONCRETE Tensile strength of plain concrete is obtained by the splitting test.

Splitting tensile strength ct2Pf

πdL=

• fct = splitting tensile strength = 0.66 fcr, where fcr is Modulus of rupture

• Direct tensile strength = [0.5 – 0.625] fcr.

1.5 STRESS-STRAIN CURVE OF CONCRETE

• It is found by testing cylinder under

compression. The max. Strength obtained is the cylinder strength. Cylinder strength = 0.8 × cube strength (150 mmφ , 300 long)


• Cylinder is tested to obtain stress strain curve, because curve, we have to obtain condition for uniaxial stress condition. In case of cubes, due to friction between the concrete surface and the steel plate of testing machine, lateral restraints occurs.

• The effect of this lateral restraint is to compressive strength in longitudinal direction. This effect dies down with increasing distance from the friction surface [called platen restraint]. Thus as the distance from the friction surface increases (i.e., as height/width ratio increases), compressive strength decreases.

From the above stress-strain curve,

following points must be noted. 1) Max compressive stress occurs at a

strain value of 0.002. i.e., 0.2%. The value of stress at 0.002 strain is called compressive strength of concrete.

2) Lower strength concrete has great deformability i.e., ductility than high strength concrete.

3) Descending part of high strength concrete is sleeper.

4) High strength concrete gets crushed at smaller strain.

5) The point where curve ends is called crushing strain.

6) High strength concrete is more brittle as compared to low strength concrete.

(Crushing strain is 0.3%-0.5%) 7) Curves are generally linear up to a

stress of 0.6 times the peak stress. 8) Modulus of elasticity of concrete for

all practical purpose is taken as secant modulus at a stress of around 0.33 ckf . This Ec is generally found acceptable in representing an average value of Ec under service load condition (static loading).

• Modulus of elasticity is primarily

influenced by the elastic properties of aggregate and to a lesser extent by the conditions of curving, age of concrete, mix proportion and type of cement.

• As per IS code: c ckE 5000 f= ( cE and ckf are in

N/mm2) Short term modulus of elasticity of concrete

• Long term modulus of elasticity

including creep cce

EE1+θ

=

Where, θ = creep coefficient cE = short term modulus of elasticity

θ = creep coefficient ultimate creep strain

elastic strain at the age of loading=

Age at loading Creep coefficient 7 days 2.2 28 days 1.6 1 year 1.1

1.6 CONCRETESTRAIN AT ULTIMATE STRENGTH

• If a concrete cylinder is axially loaded the ultimate strength is at 0.2% strain.


• For flexure, crushing is assumed to occur at 0.35% strain.

Note:- Actually it is seen that if the stress strain distribution is

a) Rectangular, strain = 0.2% b) Triangular, strain = 0.35% c) Trapezoidal, strain = 0.2 – 0.35% 1.7 DESIGN STRESS-STRAIN CURVE • Ascending part is taken as 2nd degree

parabola. • 0.67 fck = Strength of concrete in

structure • 𝛾𝛾m = partial safety factor for material

strength. 𝛾𝛾m = 1.5 for limit state of collapse 𝛾𝛾m = 1.0 for limit state of serviceability

1.8 SHRINKAGE & CREEP IN CONCRETE 1.8.1 Creep • When concrete is subjected to sustained

loading, its deformation keeps on increasing with time, even though the stress level is not altered.

• Time dependent component of total strain is called creep.

• Creep is thought to occur due to:

1) Internal movement of adsorbed water

2) Viscous flow or sliding between gel particles

3) Moisture loss 4) Growth in micro cracks

• Effect of creep are 1) Increase in deflection of beams and

slabs 2) Increased deflection of slender

column that may lead to buckling 3) Gradual transfer of load from

concrete to reinforcing steel in comp. members

4) Loss of prestress in prestressed concrete

• Beneficial effects of creep are:

1) Reduction in stress induced by non-uniform load or restrained shrinkage

2) In indeterminate structure, stress induced due to settlement of support is reduced due to creep.

• Factors influencing creep Creep increase when:- (a) Cement content is high (b) W/c ratio is high (c) Aggregate content is low (d) Air entrainment is high (e) Relative humidity is low (f) Temperature (causing moisture loss)

is high (g) Size/thickness of member is small (h) Loading occurs at early age


(i) Loading sustained over a long period Creep coefficient for design

As long as stress in concrete does not exceed one-third of its characteristic strength, creep may be assumed to be proportional to the applied stress. Thus under service load condition, creep will be proportional to stress. This concept can be used to compute total deflection (initial + creep) by usual linear elastic analysis with reduced modulus of elasticity. The reduced modulus of elasticity Ece = Ec/(1+θ)

Where Ece = reduced modulus of elasticity taking into account long term effect of creep

cE = short term modulus of elasticity

θ = creep coefficient

Age at loading θ

7 days 28 days 1 Year

2.2 1.6 1.1

• Intermediate value of creep coefficient

may be interpolated by assuming that the creep coefficient decreases linearly with the log of time in days. Thus, creep coefficient for age of loading at 15 days in

[ ][ ]

10 1015

10 10

0.6 log 28-log 15θ = 1.6

log 28-log 7+

i,e., 0θ = C - θ log t

• Effect of creep can be reduced by 1) Using high strength concrete 2) Delaying the application of finishes,

partitions walls etc. 3) Adding reinforcement 4) Steam curing under pressure.

Note:

Steam curing under pressure reduces drying shrinkage and moisture movement

1.8.2 SHRINKAGE The shortening in length of a member or contraction of the concrete per unit length due to drying when concrete sets are known as shrinkage. Shrinkage can be classified as a) Plastic shrinkage b) Drying shrinkage 1.8.2.1Plastic shrinkage This type of shrinkage manifests itself soon after the concrete is placed in the form while the concrete is still in plastic state. Loss of water by evaporation from the surface of concrete or by absorption by aggregate or sub grade is believed to be the reason for plastic-shrinkage. As aggregate & steel restrain this effect, cracks appear at the surface or internally around aggregate or reinforcement. Thus shrinkage is prevented by using aluminium powder and expansive cement or shrinkage compensating cement. 1.8.2.2Drying shrinkage • It is an everlasting process and occurs

mainly due to loss of water held in the gel pores when concrete is kept in drying condition.

• The finer the gel, the more is the shrinkage.

• Harder aggregate leads to lower shrinkage but higher shrinkage stress. Reverse is true for softer aggregate.

• Shrinkage decreases with increase in size of member.

• Concrete made with smaller size aggregate shrinkage more than concrete made with larger size aggregate.

• Shrinkage produces tensile cracks in any member which is restrained.


• For a given environment, the total shrinkage depends more on the total amount of water present in concrete at the time of mixing and to a lesser extent on the cement content.

• In the absence of test data, an apex value of total shrinkage strain for design may be taken as 0.0003 shrinkage decreases with increase in relative humidity.

• ½ of total shrinkage assumed in 1st month and 3/4th in 1st 6 months.

1.9 DESIGN STRESS STRAIN CURVE FOR STEEL

mγ = Factor of safety

mγ = 1 for limit state of serviceability = 1.15 for limit state of collapse • Increase of stress in strain hardening

region is neglected • In RCC steel never reaches its ultimate

strength because strain at which ultimate strength is reaches will never be reached.

• Concrete will get crushed before steel reaches its ultimate strength.

Note:-Failure of RCC beam always occurs due to crushing of concrete. It never occurs due to failure of steel. (i.e., shaping of steel) • We may use same curve for tension and

compression for both grades of steel (Hot rolled or cold worked.) This may lead to insignificant error. (in case of compression)


2.1 INTRODUCTION

The philosophy of the limit state method of design represents a definite advancement over the traditional design philosophies. Unlike working stress method, which is based on calculations at service load conditions, alone, and unlike ultimate load method, which is based on calculations on ultimate load conditions alone, limit state method aims for a comprehensive and rational solution to the design problem, by considering safety at ultimate load and serviceability at working loads. The limit state philosophy uses a partial safety factor format which attempts to provide adequate safety at ultimate loads as well as adequate serviceability at service loads by considering all possible ‘limit state’.

2.2 WHAT IS LIMIT STATE METHOD?

• The acceptable limit for the safetyserviceability requirement of astructure or structural element beforefailure occurs is called limit state.

• In this structure is so designed that itcarry the loads with sufficient degree ofsafety and serviceability and structurewill not become unfit for which it is tobe designed.

2.3 CHARACTERISTIC STRENGTH OF MATERIAL

The strength of material below which not more than 5% of the test results are expected to fall is known as the characteristic strength of the material and denoted by ckf . fck= fm-1.65δ 2

x-xδ=n-1

∑

mf = Mean of strength δ = Standard deviation fck = Characteristics strength of material x = Concrete cube test result x = mean of concrete cube test result N = No. of samples.

2 DESIGN OF RCC STRUCTURE


Table1.1: Specified characteristic compressive strength of concrete at 28 days.

Grade

Designation Specified characteristic

compressive strength at 28 days N/mm2

M10 M15 M20 M25 M30 M35 M40

10 15 20 25 30 35 40

Note:- 1. In the designation of a concrete mix,

letter M refers to the mix and the number to the specified characteristic compressive strength of 15-cm cube at 28 days, expressed in N/mm2.

2. M5 and M7.5 grades of concrete may be used for lean concrete bases and simple foundation for masonry walls. These mixes need not be designed.

3. Grade of concrete lower than M20 shall not be used in reinforced concrete. (As per IS 456 : 2000)

4. For sea water grade of concrete lower thanM30 shall not be used in reinforced concrete. (As per IS 456: 2000)

2.4 CHARACTERISTIC LOAD Value of load which has a 95% probability of not being exceeded during the life of the structure is known as characteristic load and is denoted by F

mF F 1.65δ= +

mF = mean value of load

(Fm-1.65δ) and (Fm+1.65δ) are two important limit within which ‘probability of lying test result’ is maximum. These limit called confidence limit.

Definition Curve For Characteristic Load 2.5 PARTIAL SAFETY OF FACTOR In limit state method of design two factors of ms safety are used one to account for uncertainty in material property and other for uncertainty in loading. Hence the factors are called partial factor of safety γ = Partial safety factor for material strength. γmc for concrete = 1.5 (because of greater variation is excepted in the strength of concrete than that in the steel reinforcement) msγ for steel = 1.15 2.5.1 Materials The value of partial safety factor for material strength should account for the following parameters. a) Possibility of deviation of the strength

of material. b) Deviation of the sectional dimensions. c) Accuracy of the calculation procedure,

and d) Risk to life and economic consequences. 2.5.2 Loads:- The value of partial safety factor for load should account for the following parameters:- a) Unusual increase in loads beyond that

used for deriving characteristic values, b) Unforeseen stress redistribution,


c) Inaccurate assessment of the effects of loading and

d) Importance of the limit state considered Table 1.2: Value partial safety for loads under various load combination

No Description Collapse Service

ability 1 D.L + L.L 1.5 1 2 D.L + (W.L) or

(E.L) combination

(i) for normal case D.L. + W.L (or E.L)

(ii) For checking stability against

overturning/stress reversal D.L +

W.L (or E.L)

1.5

0.9

1

1

3 (D.L) + L.L + W.L (orE.L)

combination D.L L.L

W.L (or E.L)

1.2 1.2 1.2

1 0.8 0.8

Design values are obtained when partial safety factor are applied to characteristic load and materials. 1. Materials:-The design strength of the

materials fd is given by fd = fck/γmc

ckf = characteristic strength of the material

γmc = partial safety of factor for material 2. Loads:- The design load Fd is given by

Fd = F γf

F = Characteristic load

fγ = Partial safety of factor for load Table1.3: Indian standard specifications for loads.

Characteristic Loads Indian Standard Specification

Dead loads Imposed loads Wind loads Seismic loads

IS : 875 (Part 1) : 1987 IS : 875 (Part 2) : 1987 IS : 875 (Part 2) : 1987 IS : 1893 : 1984

2.6 ASSUMPTION LIMIT STATE OF

COLLAPSE: FLEXURE 1. Plane Section before Bending

Remains Plane even after the Bending This assumption mean that strain at any point on the cross-section is directly proportional to its distance from its neutral axis, it mean strain diagram is linear.

2. The maximum strain in concrete at the outermost compression fiber is taken as 0.0035 in bending. The maximum strain in concrete in compression will be fiber AB, (1.1) i.e., in the Topmost fiber, and according to this assumption its value will be limited to 0.0035. Concrete has very low ductility and so it gets crushed in compression at such low strain only.

3. Relationship between compressive

stress distribution in concrete and strain in concrete may be assumed to be rectangular, Trapezoidal, parabolic or any other shape which results in prediction of strength in substantial agreement with the result of test. An acceptable strain-stress curve is as shown below note that Characteristic strength of concrete fck is reduced to 2/3fck = 0.67 fck because of size effect. Design compressive stress in concrete may be taken as fd; where

fd = ck ckck

ms

0.67f 0.67f= =0.45fγ 1.5


• For design purpose, the compressive

strength of concrete shall be assumed to be0.67 times the characteristic strength and partial factor of safety γmc = 1.5 shall be applied in addition to this.

• It may be noted that for the design of flexural members the characteristic strength of concrete is taken as 0.67 ckf instead of ckf . This is on account for the fact that in actual structure size of concrete member may be more that the cube size tested in laboratory. Larger size leads to more variability in strength. Hence, strength of the concrete member will be lesser than that in cube. However beyond 450 mm size cube the variation of strength is not much.

• The variation of strain-stress curve shall be parabolic upto 0.002 strain and, thereafter, the stress remains constant up to the maximum permissible strain of0.0035. At limit state, this is an idealized curve for concrete in compression and is valid for all grades of concrete irrespective of percentage of tensile reinforcement.

• Design stress block parameter for the

above curve (Fig. 1.2) is calculated as follows.

From strain diagram (similar triangle) 0.0035 0.002=

AC AB

u0.002 2AB = AC = x0.0035 3.5

u4AB = x7

BC = AC – AB = u u4x - x7

BC = u3 x7

Compression force of rectangular portion = Area of stress diagram × width of the section

1 ck u3C 0.45f . x b7

=

1 ck uC 0.193 f x b=

This force will act at a distance y1 =

u3 x

14 from the top.

Compression force of parabolic portion = Area of stress diagram × Width of the section

2 base Height Width of the section3

= × ×

ck u2 4×0.45f . x b3 7

2 ck uC 0.171f .x .b= This will act at a distance

2 u u u u3 3 4 3 12y = x + × x = x + x =7 8 7 7 56

u uu

24x +12x 36= x56 56

u9y2= x

14 from top

This combined compressive force will act at a y distance from top. C.G. location of total force from top (y)

1 1 2 2

1 2

C y +C yC +C

=


ck u u ck u u

ck u ck u

3 90.193f .x .b× x +0.171f .x . x14 14

0.193f .x .b+0.171f .x .b

uy 0.42x= ∴ Lever arm distance between centroid of compressive force to centroid of tensile force (z) is given by z d y= −

uz d – 0.42x=

Total compressive force 1 2 ck u ck uC C C 0.193f .x .b 0.171f .x .b= + = +

ck uC 0.36f .x .b= 4. The tensile strength of concrete is

ignored. The cracked concrete does not contribute towards enhancement of moment of Resistance of the section, except providing bond between concrete and steel for developing tensile strain and thus stress in steel.

5. For design purpose partial factor of safety for steel msγ 1.15= and the stress in steel are derived from stress-strain curve. • A nominal strain-stress curve for

mild steel bars from test may be obtained a shown in Fig 1.6(a). The strain is obtained by dividing change in length by actual length; whereas stress is determined as load divided by original cross sectional area*.

• This strain-stress curve is simplified to a curve as shown in Fig. 1.5(a).

• Similarly the strain-stress curve for cold worked deformed bars of Grade Fe 415 and Fe 500 which are, generally, in use in India is as shown in Fig. 1.6(b). This is idealized to a curve of Fig. 1.5(b).

• For all type of steel, modulus of elasticity, E = 2 × 105 N/mm2.

• The strain-stress relationship for steel in tension and in compression is assumed to be the same. Original cross section area is defined as the area of cross section of steel at zero stress.

Note:- 1. Design stress for steel=

Permessible stressPartial safety factor

= yf1.15

= 0.87 fy

For Fe 250 = 0.87 × 250 = 257.0 N/mm2.

2. Total strain at yield ( )∈ for mild steel

will be y5

s

f 250E 2 10

=×

= 0.00125; whereas

total strain at design yield stress, y

ydms s

f= =

γ E∈ 5

2501.15 2 10× ×

= 0.00109.

• In a similar manner total strains '∈ and ε’d’(design strain) at any stress, say 0.9 fy and 0.9 fyd’ cold-worked deformed bars of grade Fe 415 can be evaluated as under(from the fig 1.3b) :

y

s

0.9f' 0.0003

E∈ = +

5

0.9×415 +0.0003.=0.00222×10

=

Where 0.0003 is the inelastic strain at 0.9 fy (figure 1.3b) and

y5

ms s

0.9f 0.9 415'd= +0.0003.=γ E 1.15 2 10

0.0003 0.00192

×∈

× ×+ =

• In a flexural member, where compression reinforcement is provided, the strain at the c.g. of compression reinforcement is calculated from strain diagram and the corresponding stress is determined as explained above.

• For ready reference, typical points on design strain-stress curves for all types of steel bars are given in Table 1.4.


Table 1.4:- Typical points on the design strain-stress curves.

Stress level

For fy = 250 MPa

For fy = 415 MPa For fy = 500 MPa

Strain Stress Strain Stress Strain Stress 0.80fyd

0.85fyd 0.90fyd 0.95fyd

0.975fyd 1.0 fyd

0.00087 0.00093 0.00098 0.00104 0.00106 0.00109

173.9 184.8 195.7 206.5 212.0 217.4

0.00144 0.00163 0.00192 0.00241 0.00276 0.00380

288.7 306.7 324.8 342.8 351.8 360.9

0.00174 0.00195 0.00226 0.00277 0.00312 0.00417

347.8 369.6 391.3 413.0 423.9 434.8

6. Maximum strain in Tension

reinforcement in the section at failure shall not

be less than

y y

sts s

f 0.87f+0.002= +0.0020

1.15E E∈ =

The above value of strains for different grades of steel as follows: i) For Fe250,

εst = [250/(1.15⨉2⨉105)]+0.002 or εst = 0.0031

ii) For Fe 415 εst = [415/(1.15⨉2⨉105)]+0.002 or εst = 0.0038 and iii) For Fe 500,

εst = [500/ (1.15⨉2⨉105)] +0.002 or εst = 0.0042

iv) Minimum strains at collapse

ensures that the design stress in tensile steel at collapse will always be fyd. (0.87fy)

v) Note that for Fe 250, the strain is 0.0031 is much larger than that required for development of stress equal to 0.87fy. In mild steel the strain at which stress is

0.87 fy is y

s

0.87fE

= 0.00109 but

the code does not give separate recommendation for mild steel

(Fe 250). Thus a strain of 0.0031 will ensure sufficient ductility in mild steel reinforcement besides causing the stress equal to 0.87 fy. This assumption thus, restricts the depth of neutral axis.

2.7 WORKING STRESS METHOD • This was the traditional method of

design. • Both concrete and steel are assumed to

behave in linearly elastic manner. • Stresses within the materials are not

allowed to exceed the permissible stresses.

• Working stress method of design used to be the basis of design for all RCC structures in the past. But these days it finds application in calculating serviceability requirement like deflection and crack width under service load condition.

• It is also used in the design of few structures like liquid retaining structures and highway bridges and chimney.

2.8 DEFICIENCY IN WORKING STRESS METHOD • It may not be possible to keep the stress

with in permissible stress. This is because of a) Term effect of shrinkage and creep b) Effect of stress concentration and

other secondary effect. All such effects result in significant local increase in stresses into inelastic range and redistribution of the calculated stress.

• In working stress method actual margin of safety is not equal to the factor of safety used in WSM because the stress strain curve is not linear upto collapse. Actual margin of safety here is given in


term of factor lie collapse loadworking load

the F.O.S.

on the other hand is characterstic stresspermissible stress

• WSM fails to discriminate between different types of loads that act simultaneously, but have different degrees of uncertainty. This may sometimes lead to significantly conservative design particularly when two different loads have counteracting effect. Example, if dead load and wind load produce counteracting stress but if they are simply added, the design load would be much larger.

2.9 PERMISSIBLE STRESSES • In working stress method, the stresses

in materials are not exceeded beyond their permissible values. The permissible stress in a material is given by

Permissible stress = Limiting strenghFactor of safety

• In case of steel reinforcement, the limiting strength is either the yield stress or 0.2% proof stress, as the case may be. For concrete, the limiting strength is the crushing strength in compression.

• The factor of safety in the case of tensile steel reinforcement is approximately = 1.82. Hence, the permissible tensile stress in steel is st yσ =0.55f .

• For concrete, the factor of safety is higher than for steel. This is so because concrete suffers from higher degree of

variability regarding its strength and properties than steel which is produced under well controlled conditions.

• The factor of safety for flexural compressive in concrete is = 3. Thus, the permissible compressive stress in concrete in flexural compression is

abc ckσ =0.333f .

2.10 PERMISSIBLE STRESSES IN CONCRETE • The permissible stress of concrete in

direct tension is denoted by tdσ . The values of tdσ for member in direct tension for different grades of concrete are given in the table below.

• It may be worth nothing that the factor of safety of concrete in direct tension is from 8.5 to 9.5.

• The permissible stresses of concrete in bending compression cbc'σ in direct compression cc'σ and the permissible stress in bond for plain bars in tension

bdT are given in table 21 of IS 456 for different grades of concrete which is presented in table below.

• For plain bars in compression, the values of bond stress are obtained by increasing the respective value in tension by 25 percent, as given in the table below and

• For deformed bars in tension the values bond stress given in table are to be increased by 60%.

2.11 PERMISSIBLE STRESS IN STEEL REINFORCEMENT • Permissible stresses in steel

reinforcement for different grades of steel, diameters of bars and the types of stress in steel reinforcement are given in table IS 456.

• Selective values of permissible stress of steel of grade Fe 250 (mild steel) and Fe

Grade of concrete

Direct tension

tdσ

(N/mm2)

Bending compression

abcσ

(N/mm2)

Direct compression

ccσ

(N/mm2)

Permissible bond

stress in bdT for

plain bars in tension (N/mm2)

M 20 2.8 7.0 5.0 0.8 M 25 3.2 8.5 6.0 0.9 M 30 3.6 10.0 8.0 1.0 N 35 4.0 11.5 9.0 1.1 M 40 4.4 13.0 10.0 1.2


415. (high yield) strength deformed bars) in tension ( )st ssσ or σ and compression in column ( scσ ) are furnished in Table 1.6 as a ready reference.

Table:- 1.6:- Permissible stress in steel reinforcement Type of stress in steel reinforcement

Mild steel bars, Fe 250, (N/mm2)

High yield strength deformed bars Fe 415, (N/mm2)

Tension stσ or ssσ

(a) Up to and including 20 mm diameter

(b) Over 20 mm diameter Compression in column

bars scσ

140

130

130

230

230

190

Note:-

It can be observed that for a given grade of concrete abc cbcσ <σ i.e. a greater F.O.S. is adopted for direct stress than for a bending stress. This is because when a c/s is subjected to bending stress, the stress induced on it is variable being maximum at extreme fibre and zero at N.A. When the maximum stress exceeds the permissible value, the extreme fibre will not fail actually, but will transfer the additional force to the inner fibre which has a lower stress. However, when the section is subjected to a direct stress, all points of the section have uniform stress having no scope for such a transfer of the force. It is for this reason larger F.O.S. is adopted for direct stress than bending stress. Note:-

In case of steel reinforcement of small diameter, the stress will be uniform for direct stress as well as for bending stress. Therefore in steel bars, the permissible stresses in bending and direct stresses re the same for lower dia bars up to 20mmφ . For more than20 mm, permissible tensile stress is usually reduced. Note:- The value of stσ is given at the centroid of tension reinforcement subjected to the condition that when more than one layer of tensile reinforcement is provided, stress at the centroid of outer most layer shall not exceed by more than10% that given in the above table.

Effect Wind and Earthquake on the permissible stress:-

• When the effect of wind or earthquake

is taken into account, the permissible values for stresses in steel and concrete

are increased by 133 %3


3.1 INTRODUCTION

Concrete a composite man-made material, is the most widely used building material in the construction industry. It consists of a rationally chosen mixture of binding material such as lime or cement, well grade fine and coarse aggregates, water and admixtures (to produce concrete with special properties.) • The good concrete are which fulfills two

criteria :-a) The concrete has to be satisfactory

in its hardened state, andb) Also in its fresh state while being

transported from the mixer andplaced in the formwork.

• The requirements in the fresh state arethat the consistency of the mix be suchthat it can be compacted by the meansdesired without excessive effort, andalso that the mix be cohesive enough forthe methods of transporting and placingused so as not produce segregation witha consequent lack of homogeneity of thefinished product.

• The primary requirements of goodconcrete in its hardened state are asatisfactory compressive strength andan adequate durability.

The flow of this chapter is as per the flow diagram described below

In this chapter we will discuss the constituent material of concrete as discussed in above flow diagram of concrete one by one.

3.1.1 Cement

Cement is a material which has cohesive and adhesive properties in the presence of water.

3.1.2 Manufacturing Process:

1. Portland cement is manufactured bygrinding together 1 part of calcareous(CaCO3) and one part of argillaceous(clay and shales). (Dry process)

2. The mixture is burnt in a kiln at atemperature of about 1300°C to 1500°Cwhere material use and form a smallClinkers of 3 mm to 20 mm in size.

3. Then the clinker is cooled down usingmoderate cooling condition (It isobserved that slow or quick coolingcondition reduced the strength of thecement.)

4. In moderate cooling conditiontemperature of clinker is brought down

3 CONCRETE TECHNOLOGY


to 500°C in 15 min and in further 10 minute the temperature is brought, down to atmospheric.

5. Cooled, clinker is then mix with gypsum.

6. The mixture is ground to required fineness in ball mills to get the final product known as cement.

7. Gypsum is required to retard the setting time.

3.2 CHEMICAL COMPOSITION OF RAW MATERIALS

• The three constituents of hydraulic cements are lime, silica and alumina.

• In addition, most cements contain small proportions of iron oxide, magnesia, sulphur trioxide and alkalis.

• There has been a change in the composition of Portland cement over the years, mainly reflected in the increase in lime cement and in a slight decrease in silica content.

• An increase in lime content beyond a certain value makes it difficult to combine completely with other compounds.

• Consequently, free lime will exist in the clinker and will result in an unsound cement. An increase in silica content at the expense of alumina and ferric oxide makes the cement difficult to fuse and form clinker. The approximate limits of chemical composition in cement are given in table.

Constituents of Portland Cement (Raw Material)

Oxide Function Composition (%)

CaO Controls strength and soundness. Its deficiency reduces strength and setting time.

60-65

SiO2 Gives strength. Excess of it causes slow setting.

17 – 25

Al2O5 Responsible for quick setting, if In excess, it lowers the strength

3 – 8

Fe2O5 Gives colour and helps in fusion of different ingredients

0.5 – 6

MgO Imparts colour and hardness. If in excess, it causes cracks in mortar and concrete and unsoundness.

0.5 – 4

Na2O + K2O TiO2 P2O5

These are residues, and if in excess cause efflorescence and cracking

0.5 – 1.3 0.1 – 0.4 0.1 - 0.2

SO3 Makes cement sound 1– 2

Notes:- 1. The rate of setting of cement paste is

controlled by regulating the ratio ( )2 2 3 2 3SiO / Al O Fe O .+

2. Where development of much heat of hydration is undesirable, the silica content is increased to about 21 percent, and the alumina and iron oxide contents are limited to 6 percent each.

3. Resistance to the action of sulphate waters is increased by raising further the silica content to 24 percent and reducing the alumina and iron contents in 4 percent each.

4. The variation in composition depends largely on the ration of CaO to SiO2 in the raw materials.

3.3 TYPES OF CEMENTS 1. Ordinary Portland Cements:-

• It is widely used cement for most of the work. It is suitable for general construction works when there is no exposure to sulphates in the soil or in the ground water.

• Three different grade of ordinary Portland cements to produce high strength cement is 33 grade of O.P.C. 43 grade of O.P.C. 53 grade of O.P.C.

2. Rapid hardening Cements:-

• (Name itself shows that it hardened and attains its strength earlier than ordinary Portland cement).


• Rapid hardening cement is finer than ordinary Portland cement and it contains more C3S and less C2S than the OPC

• 3 days strength of RHC is same of 7 days strength of OPC

• The main advantage of rapid hardening cement is that shuttering may be removed much earlier, thus saving considerable time and expenses.

• Rapid hardening cement is also used for road work where it is necessary to open the road traffic with the minimum delay.

3. Extra Rapid Hardening Cement:-

• In this cement 2% Cacl2 is mixed with rapid hardening cements.

• This Cacl2 imparts quick setting properties.

• While using this cement maximum time of 20 minute is available for mixing, transporting and placing the concrete and also this cement is used within one month.

• Extra rapid hardening cement used in cold weather because of large heat of evolution.

4. Portland Blast Furnace Cement:-

• It is manufactured by mixing Portland cement clinker with granulated blast furnace slag (This slag is waste product obtained from blast furnace which contains oxide of lime, silica and alumina) and gypsum in suitable proportions and grinding the mixture to the required fineness.

• The proportion of the slag should not less than 25% and not more than 65% of the total mass of the mixture.

• It contains approximately 45% CaO and 35% silica.

• It is similar and cheaper than O.P.C (and it can be replace it.)

• It has low heat of hydration and is better resistant to soil and water containing excessive amounts of sulphates, alkalies.

• Because of its low heat evolution it can be used in mass concrete structure such as dams & foundations.

5. Pozzolana Cements:-

• This is manufactured by grinding Portland cement clinker with Pozzolana and required quantity of gypsum. (The pozzolana are materials which at ordinary temperature react with lime in presence of water, resulting in cementing materials.) Fly ash, burnt clay are used as Pozzolana.

• The proportion of Pozzolana may be 10-25% by weight of Pozzolana cement.

• This cement has higher resistance to chemical agencies and to sea water because of absence of lime.

• Advantage of this cement are reduction in cost, increased impermeability.

6. High Alumina Cement:-

• It is non-Portland cement, it is manufacture by Melting mixture of aluminous and Calcareous Materials in Suitable proportion and grinding the resulting clinker to fine powder (which is black in color).

• The raw materials used for its manufacturing are chalk and bauxite which is special clay of extremely high alumina content.

• It is used where early removal of form work is required.

• Its rapid hardening properties arise from the presence of Calcium aluminate like, Calcium silicates in Portland cement.

• It’s one day strength is equal to 28 days strength of ordinary Portland cement.


• It is recommended for sea and under water work.

Table 10.3 Composition of typical high alumina cement

Composition (%) Al2O3, TiO2

Fe2O3, FeO, Fe3O4

CaO SiO2

MgO SO3

Insoluble material Loss on ignition

43.5 13.1 37.5 3.8 0.3 0.4 1.2 0.2

7. Low Heat Portland Cement:-

• It is manufactured by reducing the % of C3S and C3A of ordinary Portland cement, because of this cement gets the strength at a slower rate and the heat of hydration is less. This will require long time curing and keeping forms for a long time.

• It is used in Mass Concrete Works, such as dams and retaining walls. (For mass construction work when OPC is used temperature may rises at its highest level when time passes outer layer will cool and contract while the inner mass is still a higher temp. This may leads to serious cracking).

• Heat generated in OPC at the end of 3 days –80cal/gm. While in low heat cement it is about 50 cal/gm of cement.

8. Sulphate resisting cement:-

• It is similar to ordinary Portland cement except that it contains more silicates and less quantity of aluminates.

• It is used for under water structure particularly exposed to alkali actions.

• Soluble sulphates like MgSO4, CaSO4 and Na2SO4 when present in ground reacts with cement and form

Sulpho-aluminates which have expansive properties and so causes disintegration of concrete (Due to this nature it expand and this larger volume will create pressure on concrete which will result in cracks and finally disintegration of concrete.)

• It is strongly recommended for structure in sea water, coastal areas and Marshy lands.

• This is used in canal.

9. Super Sulphated Cement:- • It is manufactured by grinding

together a mixture of 80 – 85% of granulated blast furnace slag and 10 – 15% of calcium sulphate and about 5% of Portland cement clinker.

• The cement is highly resistant to sulphate attack.

• Because of low heat of evolution it is useful for mass concrete works. It is also useful for foundation works where aggressive chemical conditions exist.

10. Quick Setting Portland Cement:-

• The quantity of gypsum is reduced and small percentage of aluminium sulphate is added. It is ground much finer that ordinary Portland cement.

Properties :- Initial setting time – 5 minutes Final setting time = 30 minutes

Use :- It is used when concrete is to be laid under water or in running water.

3.4 ADMIXTURE Suitable materials known as admixtures which may be added to concrete mix, Just before or during the mixing to modify one or more properties of the concrete in the plastic or hardened states as desired. The objective of admixture are :-


• To increase the rate of strength development at early ages.

• To retard the initial setting times. • To increase the workability without

changing the water content. • To increase the resistance to freezing

and thawing, vinsol resins and air entertainment admixture which is used for this purpose which cause air to be incorporated in the form of minute tiny bubbles in concrete during mixing to increase the workability and resistance to freezing and thawing.

3.5 DIFFERENT TYPE OF ADMIXTURE 1. Mineral Admixtures a) Fly ash

• This increase impermeability because it is finer than the cement particles.

• This is by product of wooden thermal power plant and is produced daily in large quantities.

• These require very less water to wet their surface.

• Same work which is obtained by OPC is achieved by lower water/cement ratio.

2. Chemical Admixtures a) Accelerating Admixture:-

A substance which increases the rate of strength development or reduces the setting time. CaCl2, Sodium silicate, Sodium chloride, Calcium nitrite and calcium nitrate are used as accelerator. • Barium chloride acts as an

accelerator only under warm condition

• Most frequently used accelerator is calcium chloride (CaCl2)

b) Retarding Admixture:- A substance which delays the setting time of cement paste. Gypsum, Tartaric acid, sugar are used as retarding admixture.

• A small quantity of sugar about 0.05 percent of the mass of cement will act as an acceptable retarder: the delay in setting of concrete is about 4 hours.

• A large quantity of sugar, say 0.2 to 1 percent of the mass of cement, will virtually prevent the setting of cement.

• When sugar is used as a controlled set retarder, the early strength of concrete is severely reduced but beyond about 7 days, there is an increase in strength of several percent compared with a non-retarded mix. This is probably due to the fact that delayed setting produces a denser hydrated cement gel.

Note:- Retarders tend to increase the plastic shrinkage because the duration of the plastic stage is extended, but drying shrinkage is not affected. c) Water Reducing Admixture:

A substance which either increases workability of freshly mixed mortar or concrete without increasing water-cement ratio or maintains workability with reduced water-cement ratio. These are: • Lignosulphonic acid and its salts. • Formaldehyde derivatives • Hydroxylated carboxylic acids • Calcium lignosulphonate

d) Air-entraining Admixture :-

A substance which causes air to be entrapped in the form of tiny bubbles in motor or concrete, during mixing to increasing its workability and resistance to freezing and thawing. Example: Vinsol resin, using of Aluminum powder.

e) Super Plasticizing Admixture :-


• A substance which imports very high workability with a large decrease in water content (at least 20%) for a given workability. The resulting improvement in workability can be exploited in two ways. By a) By producing concrete with a very

high workability or concrete with a very high strength.

b) Concrete of normal workability but with an extremely high strength owing to a very substantial reduction in the water/cement ratio.

• A high range water reducing admixture is also referred to as a super plasticizer and these are sulphonated melamine formaldehyde condensates; sulfonated naphthalene-formaldehyde condensates; modified lignosulfonates; and other such as sulfonic-acid esters and carbohydrate esters.

• The role of super plasticizer is to disperse, the particles remove air bubbles, example; Sulphonated melanin formaldehyde and to retard setting.

3.6 AGGREGATES

The aggregates consist of about 75% of volume of concrete and they greatly influence the properties of concrete.

• Aggregate give body to the concrete reduce the shrinkage effect of cement and make the concrete durable.

• Generally rounded shape aggregate is used because shape of aggregate affects the workability of concrete.

• To achieve the best possible strength, concrete should be as dense as possible i.e. it should contain minimum void. Voids are greatly influenced by the shape of aggregate.

• The rounded particles can be packed to produce a concrete with 33% void means 67% of the volume of concrete is occupied by the aggregate.

• The Rounded particle produce smoother mix for a given water/cement ratio.

• On the other hand, angular or flaky particles reduce the workability and demand more cement and water to give the specified strength of concrete mix.

• Not more than 10 – 15% of flaky

particles should be used in concrete. 3.6.1 Bulking of Sand Due to the presence of moisture content, aggregates bulk in volume. The moisture particles form a thin film around the aggregates and exert surface tension. This keeps the particles away from each other and thus aggregates bulk in volume. • This bulking in volume generally

negligible in the case of course aggregates. It has great importance in case of fine aggregates or sand.

• For sand the volume goes on increasing until the moisture content is about 8% by the mass of sand.

• Bulking increases with fineness of the aggregates.

• The bulking of sand may be as large as 30 – 40%

• With further addition of moisture content the thin film of water coated round the sand start disappearing and the volume of sand begins to decrease till finally at 25 – 30% of moisture content. The volume of sand returns to its original volume when it is dry.


3.6.2 Fineness Modulus In order to ensure the presence of all sizes of particles, the property of aggregate called fineness modulus is defined. The fineness modulus of an aggregate is an index number which is roughly proportional to the average size of the particles in the aggregate. 3.6.3 Fineness Modulus for Different Type of Sand Fine sand = 2.2 – 2.6 Medium sand = 2.6 – 2.9 Coarse sand = 2.9 – 3.2 Sand having fineness modulus more than 3.2 will be unsuitable for making satisfactory concrete. The fineness modulus is the sum divided by 100 of the cumulative percentage mass which is retained on each of the ten sieves specified by I.S. code. Sieves are 150mμ, 300mµ , 600mµ , 1.18mm, 2.36 mm, 4.75 mm, 10 mm, 20 mm, 40 mm, 80 mm and larger is required increasing in the ration of 2 : 1 Note:- Fineness modules of 4.00 can be interpreted to mean that the fourth sieve, i.e. the average size of particle is 1.18 mm. Water Water is required for proper chemical action and amount of water required is about (25 – 35%) of the weight of cement used. Water/cement = 0.4 3.7 PERMISSIBLEL LIMITS FOR IMPURITIES IN WATER Impurity Permissible Limits Organic 200 mg/l Inorganic 3000 mg/l Sulphates 400 mg/l Chlorides 2000mg/l for plain

concrete work 500

mg/l for reinforced concrete work

Suspended matter 2000 mg/l 3.8 CONCRETE Concrete is a carefully proportioned mixture of cement, fine aggregate, coarse aggregate and water. Sometimes to modify the physical properties of concrete a variety of admixture may be added. The preparation of concrete consists of the following operations:- 1. Proportioning of ingredients 2. Measurement of materials 3. Mixing and placing of concrete 4. Compaction 5. Curing. 1) Proportioning of Ingredient

Proportioning of ingredients means determining the relative amounts of ingredients (cement, FA, CA to get the required strength of concrete.) This can be done by two way: i) Design Mix. ii) Nominal Mix.

i) Design Mix:-in design mix the

proportion of ingredients of concrete to obtain a desired strength can be found out by laboratory method. Concrete grades > M 20 are design mix concrete.

ii) Nominal Mix:- In this method

proportion of ingredients of concrete can be choose on the arbitrary method M 5, M7.5 M 10 and M 20 are the nominal mix

3.9 METHODS OF PROPORTIONAL CONCRETE 1) Maximum Density Method

It is used as a theoretical approach to determine the grading of aggregate of obtained maximum density. It is given by Fuller’s


ddP =100D

d = Maximum size of fine aggregate. D = Maximum size of Coarse aggregate.

dP =% by weight of particle size finer than d in total mixture.

3.10 WATER CEMENT RATIO • Water cement ratio is the ratio of

volume of water mixed in concrete to volume of cement used.

• The strength and workability of concrete depends to a great extent on the amount of water used.

• For a given proportion of the materials, there is an amount of water which gives the greatest strength.

• Amount of water less than this optimum water decreases the strength and about 10percent less be insufficient to ensure complete setting of cement. Similarly, more than the optimum water increases the workability but decreases the strength.

• The use of an excessive amount of water not only produces low strength but increases shrinking and decreases density and durability.

• According to Abram’s water cement ratio law, for any given conditions of test, the strength of a workable concrete mix is dependent only on the water-cement ratio.

• Lesser the water-cement ratio in a workable mix, greater will be its strength.

• From Abram’s law, it follows that provided the concrete is fully compacted, the strength is not affected by aggregate shape, type or surface texture, or the aggregate grading, the workability and the richness of the mix.

• Concrete vibrated by efficient mechanical vibrators requires less water-cement ratio, and hence have more strength. Sometimes, plasticising

agents may be mixed to increase the workability of the mix.

• For such concrete, therefore, water-cement ratio is reduced; resulting is an increase in the strength.


4.1 BALANCED, UNDER-REINFORCED AND OVER-REINFORCED SECTION

4.1.1 Balanced Section

It is a section in which both steel and concrete reaches their maximum permissible value simultaneously. i.e., when st stf σ= (permissible tensile stress in steel), cbc cbcf σ= (permissible compression stress in concrete) Under this condition

01N=n = σst1+

mσcbc

0st

1n = 3σ1+280

Note Neutral axis depth coefficient under balanced condition depends only upon stress in tension steel. If Ast bal = Area of tension steel under balanced condition

Then stbal cbc0 0

st

A σp = = ηbd 2σ

00

nJ =1-3

4.1.2 Under reinforced and over reinforce section

4.1.3 When the section is under reinforced

a) crx < xb) Tension steel reaches the max.

Permissible value prior to concrete.c) Area of tension steel is less than that in

the balanced section

d) In under reinforced section failure istensile failure. Under failure condition,in under reinforced section, neutral axisis above balanced section N.A. Hencestrain in steel increase leading to widerand deeper tension cracks andincreased beam curvature anddeflections. The process continues untilthe max. strain in concrete in comp. sidereaches the ultimate compressive strainof concrete resulting in crushing ofconcrete. This type of failure is calledtension failure. Note that moment ofresistance beyond moment ofresistance corresponding to yielding oftension steel (i.e., Point A) does notincrease significantly, but curvatureincrease significantly.• Under reinforced condition, section

suffers large deflection and crackingprior to failure. Thus it is a ductilefailure.

• Ductile failure gives sufficientwarning before failure henceadvisable.

4 REINFORCED SECTION


4.1.4 When section is over reinforced a) crx>x b) Concrete reaches its max. permissible

value prior to steel. c) st st balA >A d) Over reinforced section fails to utilize

the full strength of costlier material i.e., steel. In the over reinforced section, N.A. depth is large hence strain across the section remains low. Consequently, the curvature, deflection and crack width also remain low. Thus failure is sudden (without any sign of warning). This type of failure is called brittle failure.

• As there is no significant warning before failure. IS code not permit over-reinforced section design.

Note:- Other types of failure besides tension and brittle failure are • Failure due to snapping of tension

reinforcement which occurs when there is extremely low amount of reinforcement provided. But this situation is rather rare, we safeguard against this by providing minimum tension reinforcement.

• Failure due to dynamic loading • Minimum tensile steel in beams is given

by stmin

y

A 0.85=bd f

• Maximum tensile steel in beams is given by

stmaxA =0.04bD

4.2 MOMENT OF RESISTANCE CALCULATION For under reinforcement [ ]st st cbc cbcf =σ ,f <σ :

( )wR st st

xM = σ .A d-3

WR cbc1 xM = f ×x.b d-2 3

Where, stcbc

σ xf =m d-x

x can be calculated from bx2/2 = mAst (d – x) For balanced section:- fst = σst ; fcbc=σcbc

0 0

RWO 0x xσcbc σcbcM = x b d- =

2 3 2 d

20x1 .bd3d

−

2rwo 0 0 cbc

1M = n J σ .bd2

2RWO wM =R bd ,

where w 0 0 cbc1R = n J σ2

For over reinforced section: [ ]cbc cbc st stf =σ ,f <σ

RW RW cbc1 xM =M = σ bx d-2 3

( )RW st stxM = f A d-3

Where stcbc

f d-x= .σm x

The following table must be remembered as a ready reference

Grade of Steel Fe 250 Fe 415

stσ N/mm2 [forφ < 20 mm] [for φ > 20 mm]*

140 130*

230 230*

n0 0.4 0.4179*

0.2887 0.2887*

J0 0.867 0.860*

0.904 0.904*

cbc 00

st

P2σ η

= ×σ

Percentage tensile reinforcement in fractions

0.143

cbcσ 0.160

cbcσ *

0.063

cbcσ 0.063

cbcσ *


w 0 0 cbc1R n J2

= σ

Rw =

0.1734

cbcσ 0.179

cbcσ *

0.130 cbcσ

0.130 cbcσ *

4.2.1 Relation between percentage reinforcement & moment of resistance

RWOM = Moment of resistance for balanced section • For 0p p< , MOR increases rapidly and

nearly proportional to p. Hence OA line can be approximated to be a straight line

• Rate of gain of MOR increases with use of higher grade steel

• Balanced section is reached at smaller percentage steel with use of higher grade steel. For p>po, the rate of gain of MOR with increase in percentage steel drops off rapidly

• Most economical utilization of steel (in

WSM) occurs in under reinforced design or balanced design. In over reinforced design, steel is not economically utilized.

Note:- A section which is over reinforced in WSM may be under reinforced in LSM. Except when the percentage reinforcement is large. 4.2.2 Design of section for a given moment a) If section dimension, are not given, fix the depth and width of section such that

( )balanced sectionM MOR ,≤ Where M = maximum applied BM i.e., 2

wM R bd≤

w

MdR b

≥

b) (width of section) could be chosen as 115 mm, 230mm, 250mm, 300mm, 350 for economy of form work

( )

st st

2

st

xM = A .σ d-3

bx = mA d-x2

Simultaneous solution of these two will yield the value of stA

Note: -

stA can be approximately calculated as

stst 0

M =Aσ J d

Where = 0J lever arm coefficient b) If the size of beam is given, find out moment of resistance of balanced section i.e., (i) If applied moment RWOM M ,< area of

steel can be found out by the simultaneous solution of equation (a) and (b) as given below

st stxM=A .σ d-3

. . . . . . . . . (a)

( )2

stbx mA d x2

= − . . . . . . . . . . (b)

(ii) If applied moment > (MOR) balanced section simultaneous solution of (a) and (b) will lead to over-reinforced design, which is not recommended. Hence we will design doubly reinforced section. • Provision of doubly reinforced

section ensures ductile failure. • Doubly reinforced section is the one

in which compression reinforced is provided in addition to tension reinforcement.

Note:- Under reinforced or balanced design is ensured by checking that percentage steel of the designed section


st0

A ×100 Pbd

≤

(percentage tensile

reinforcement) Where,

cbc0 0

st

σP = ×n ×1002σ

4.3 DOUBLY REINFORCED SECTION Whenever the size of beam is restricted and beam has to bear higher value of B.M. than the moment of Resistance of the ( )balanceM.R. balance section of given beam, then to resist this higher moment doubly reinforced section is to be provided. There are two methods to design such beams: 1) Increase the concrete mix to increase

the capacity of the section 2) Reinforcement are provided in

compression zone to give additional strength to the concrete in compression. Such beam are called double reinforced beams.

• A concrete structure when loaded undergoes elastic as well as plastic deformation.

• The plastic deformation is termed as creep.

• Elastic deformation is an instantaneous process while creep is a long process.

• The creep deformation of concrete produces additional strain in compression steel and gradually raises the level of stress.

• To account for this increases in stress, the modular ratio is increased.

• IS 456 states that the permissible stress for compression in bars account, shall be taken as the calculated compressive stress in the surrounding concrete multiplied by 1.5 times the modular ratio or scσ whichever is lower where

scσ is the permissible stress in compression in column bars.

• Generally the use of compression steel is considered uneconomical. However,

in some special cases it can be advantageous.

• Advantage of compression steel a) It permits smaller size which look

aesthetic b) It reduces the long term deflection

and increases ductility of the beam c) It can be used as anchor bar for

positioning the shear reinforcement d) As the compression reinforcement

increases ductility of beam, they are provided (even when not required for strength) in the seismic zone to with stand repeated reversals produces.

Note:- In tension steel, shrinkage reduces the tensile stress and creep produces additional tensile stress but in compressive steel both shrinkage and creep add additional stress so we have to use different value of modular ratio (m). In case of Doubly reinforced section stress in steel on tension side = m × stress in surrounding concrete. But as per IS456, for steel in compression zone, Stress in steel = 1.5m × stress in surrounding concrete = 1.5m × cbcf .

4.4 ANALYSIS OF DOUBLY REINFORCED SECTION: (RECTANGULAR)

scA = Area of compression steel

stA = Area of tension steel m’ = 1.5 m = modular ratio of compression steel. m = modular ratio of tension steel

scf = Stress in compression steel

stf = Stress in tension steel M = applied moment • When compression reinforcement is

provided in addition to tension


reinforcement in beam, such beams are termed as doubly reinforced beams.

• Hanger bars of nominal dia used for the purpose of holding stirrups, do not normally qualify as compression reinforcement, unless the area of such bars is significant (greater than 0.2%)

a) To find depth of neutral axis

( ) ( ) ( )2

sc stbx + m'-1 A x-d' =mA d-x2

( ) ( ) ( )2

sc stbx + 1.5m-1 A x-d' =mA d-x2

. . . . (A) b) To find out stress in concrete and

steel for a given applied moment (i) By flexure formula:-

32

sc

2st

bxI (1.5m 1)A (x d ')3

mA (d x)

= + − −

= −

;

I = M.O.I. about N.A.

cbcMxf =

I⇒Find cbcf from this, cbcf =

stress in concrete in bending compression fsc/1.5m = M(x-d’)/I Find scf from this; scf = stress in compression steel

stf M(d-x)=M I

⇒Find stf from this; stf

= stress in tension steel (ii) By internal couple method

scf = Stress in compression steel

1C = Compression carried by concrete = (1/2)fcbc xb

2C = Additional force carried by compression steel

2 SC SCC (1.5m 1)A f= − ×

SC cbc(x d ')(1.5m 1)A .f

x−

= −

M = C1{d-(x/3)} + C2(d-d’)

From this equation ( )α find fcbc .... ( )α fsc/1.5m = fcbc(x-d’)/x Find scf from this (fst/m) = fcbc{(d-x)/x} Find stf from this

4.5 MOMENT OF RESISTANCE OF DOUBLY REINFORCED SECTION I) Find depth of N.A. (i.e., x)

II) Find n = xd

III) Find 0n = Neutral axis coefficient for singly reinforced balanced section

VI) If 0n < n Tension steel reaches permissible stress

st stf = σ⇒

stcbc

σxf =d-x m

M.O.R.

= ( )1 2xC d- +C d-d'3

( ) ( )1 cbc 2 sc cbc

x-d'1C = f xb,C = 1.5m-1 A f2 x

V) If 0n n≥ , concrete, reaches its permissible stress

st cbcf =σ⇒

1 2 1 cbcx 1MOR = C d- +C (d-d')C = σ bx3 2

⇒

( )

2 sc cbc

x-d'C = (1.5m-1)A σ

x 4.6 DESIGN OF DOUBLY REINFORCED SECTION (RECTANGULAR) • When the size of the beam is restricted

and applied moment is more than the moment of resistance of balanced


section, we would design the section as doubly reinforced section. Compression steel provided in doubly reinforced section is useful even otherwise because

• It permits smaller size beam • Reduces long term deflection and

increases ductility of beam • Can be used as anchor bar for

positioning shear reinforcement • As compression steel increase ductility,

they are provided (even if not required for strength) in seismic zone to withstand repeated reversals produced. A doubly reinforced section is designed as a combination of (a) A balanced section (singly reinforced) and (b) A section containing only tension and compression reinforcement.

Applied moment M = Rwbd2 + (M-Rwbd2)

i.e., 1 2M M M= +

1 2st st stA =A +A 2 2

w wst

st 0 st

R bd (M-R bd )A = +σ (J cl) σ (d-d')

2w scM - R bd = [(1.5m-1)A ]

cbcx-d'σ . (d - d')x

⇒

[ ]

2w

sc

cbc

M-R bdA =x-d'(1.5m-1) σ d-d'x

Example Find out the moment of resistance of the section given below? Grade of concrete is M

20 and Fe 415 grade of steel is used. Take, m = 13, 2 3σcbc = 7N/m , σst = 230 N/m . Solution: d = 600 – 40 = 560 mm Critical depth of N.A.

cbcm

st cbc

mx dm

σ= σ + σ

= 13 7 560230 13 7

××

+ × mx = 158.75 mm

Actual depth of N.A. 2

sc c stbx +(1.5m-1)A (x-d ) = mA (d-x)2

22300x +(1.5×13-1)×π×12 (x-40)

2=

213×π25 (560-x) 150x2 + 8369.20(x – 40) = 25525.4(560 – x)150x2 + 8369.20x –334768 = 14.29 × 106 –25525.4x 105x2 + 33894.6x – 14.62 × 106= 0 X = 219.03 mm x > xm section is over reinforced For over reinforced section x > xm fcbc = cbcσ fst< stσ

c cbc1 xMR = bxσ d- +2 3

sc cbc c(1.5m-1)A f '(d-d )

cbcf ' can be find similar triangle

cbc cbc

c

σ f=x x-d

cbcf '= 7 (219.03 40)219.03

× − = 5.72 N/mm2.

MRc= 1 219.33300 219.33 7 5602 3

× × × −

+

2(1.5 13 1) 12 5.72(560 40)π× − × × × − = 111.99 × 106 + 24.89 × 106

MRc = 136.9 kN.m

4.7 FLANGED BEAM


4.7.1 INTRODUCTION • In monolithic construction slab and

beams are cast together. If slab in such cases is in compression zone they become effective (partially or wholly) in adding significantly to the area of concrete in compression in beam. However if flanges (slab) are located in tension zone, concrete in the flange (slab) becomes ineffective in cracked section analysis.

• In support region of a continues beam,

B.M. is negative (-ve) i.e., hogging and slab therefore is in tension. Thus beam in that region even if cast monolithically, is designed as rectangular beam.

• Away for the support region, the slab

will be in compression (BM being sagging). Hence in this region, beam is designed as flanged beam in which part of the slab is taking as a part of beam.

4.7.2 Inverted beam Inverted beams are recommended for architectural requirement only (i.e., to provide greater overhead clearance) such beam are designed as rectangular beam, because slab is a tension zone and does not resist any compression. 4.8 EFFECTIVE FLANGE WIDTH

• The compressive stress variation due to flexure is non-uniform over the width of the slab.

• For ease in calculation, we assume the stress distribution across the width to be uniform.

• The width of flange with constant compressive stress equal to the peak actual flexural compressive stress (α ) which leads to the same longitudinal compressive force as due the original stress distribution is called effective width of flange.

Note:- Effective flange width increases with a) Increased span b) Increased web width c) Increase flange thickness d) The actual variation of flexural compressive stress with max value near web and decreasing away from web location is attributed to shear lag phenomenon. Effective flange width also depends on type of loading (concentrated or distributed) and on the support conditions. 4.9 IS CODE SPECIFICATIONS 4.9.1 Effective flange with for T and L beams L-beam

0

f w flb =b +3D +12


fb 1

2wlb +

T-beam

066f w flb b D= + +

1 2

f wl lb b

2+

≥ +

Where bf = effective width of the flange, bw = breadth of the web. Df = thickness of the flange,

0l = distance between point of zero moment in the beam

0l = effective span beam for simply supported (SS) beam and = 0.7 × (effective span of beam) for continuous beam.

Effective width for flange in T or L beam (Isolated):-

Note:-

For fixed condition also we can take 0l to be equal to that corresponding to continuous case. • Examples of isolated T beams are

stringer beam of strain case • For T-or L-beam action, i.e., slab will act

as compression flange of a T-beam or L-beam only when a) Concrete of slab and web is

monolithic b) Web reinforcement is adequately

anchored in slab c) Adequate transverse reinforcement

must be provided near the top of the flange such transverse reinforcement is usually present in the form of (-ve) moment reinforcement in the continuous slab which spans across and form the flanges of the T-beam. When this is not the case (as in slabs where the main bars run parallel to beam), transverse reinforcement must be provided in the flange of the T-beam or L-beam. The area of such transverse reinforcement is 60% of the main area of steel provided at mid span of the slab and should extend on either side of beam to a distance not less than 1

4 th of the

span of the beam. Analysis of flanged section:-

1) 2

f fst f

b D mA (d-D )2

≥ then x will be less

than fD i.e., neutral axis lies in the flange

2) If (bfDf2/2)< mAst(d-Df) then x will be

more than fD i.e., neutral axis lies in the web.


Cases – I: - When neutral axis lies in the flange, analysis and design will be same as that of a singly reinforced rectangular section of width bf. Cases – II: - When neutral axis lies in web

C1 = compression carried by (+ ve) area [bf x] C2 = Compression carried by (− ve) area [bf – bw) (x – Df)] (i) To find out N.A.

( ) 22ff

f w st

x-Db x -(b -b ) =mA (d-x)2 2

..(i)

Solve the above equation to find out x. (ii)To determine stress in concrete and steel: (a)By internal couple method:

f1 2 f

x x-DC d- -C d-D - =M3 3

. . . . .(ii)

Where C1 = (1/2)fcbc xbf

2 ck f f w1C = f (x-D )×(b -b )2

[ ]cbcf f f w

f1= × (x-D ) x-D (b -b )2 x

by solving equation (ii) we can find out fcbc

cbcst

mff = (d-x)x

⇒

(b) Using flexure formula:

3f f

f wb x (x-D )3I= -(b -b ) +

3 3

2

Astm (d-x)

cbcMxf =

I

M(d-x)fst=mI

(ii)To find out moment of resistance of the section:

a) Find x and hence n = xd

(check if x > Df)

• If n 0 st stn ,f =σ< and fcbc = σst xm d-x

• If n 0 cbc cbc>n ,f =σ • Moment of resistance MOR = C1[d-(x/3)]-C2[d-Df -{(x-Df)/3}]

cbc f

1C1= f b x2

C2 = ( )( )f

cbc f f w1 x-Df × x-D b -b2 x

Note:T-beams hardly requires comp. steel Doubly reinforced flanged beam:

)2

f1 f sc f

Dx =b +1.5m-1 A (D -d')2 2x =mAst (d

– Df) a) x1> x2, Neutral axis lies in flange b) x1< x2, Neutral axis lies in web


Case:- 1 when the neutral axis lies in flange, the beam will be analysed designed in the same way as doubly reinforced rectangular section of width bf

Case: - 2 When neutral axis lies in web

1) To find neutral axis: (bf x2/2)- (bf-bw)⨉{(x-Df)2/2} + (1.5m-1)Asc(x – d’) = mAst (d – x) …….(A)

Solve this equation (A) to find x . . . . . (A) 2) To find out stresses:

f1 2 f

x x-DM=C d- -C d-D - +3 3

( )x-d'(1.5m-1)Ast fcbc d-d'x

C2 = (1/2)[fcbc{(x-Df)/x}](bf-bw) Solve this equation (B) to find out cbcf

st cbcd-xf =m fx

sc cbcx-d'f =1.5mfx

3) Calculation of moment of resistance: i) Find x and hence

xn=d

(check if x > Df)

ii) If n < n0, fst = stσ ,

stcbc

σ xf =m d-x

n > n, fcbc = cbcσ Moment of resistance is given by

M.O.R.=

( )

f1 2 f

cbc

x x-DC d- -C d-D -3 3

x-d'+(1.5m-1)Asc f d-x'x

cbc f1C1= f b x2

( )f2 cbc f w

x-DC = f b -bx

4.10 DESIGN OF FLANGED SECTION a)Fixed the width of beam from architectural requirement and also to accommodate the reinforcement fulfilling the requirement of spacing of bars. b)Normally, bw is taken as less than (l/25) and should never be < 230 mm (l = span of beam)

c)Adopt overall depth as l

10 to

l12

d) Assume the lever arm to be fDd-2

and

MAst=Dfσst d-2

Note:- MOR is calculated after providing Ast and it should be greater than Mapplied. Note that T-beam hardly requires compression steel. Example Find the moment of the resistance of isolated T-beam shown in figure below is simply supported over a span of 6 m use M


20 grade of concrete and Fe 415 grade of steel. Take cbcσ = 7 N/mm2, stσ =230 N/mm2, m = 13. Solution : Effective width of flange is lesser of (bf) (1)

= 0w

0

l +bl +4b

. . . . . (i)

=6000 3006000 4

1200

++

= 966.67 mm O.K.

1) b = 1200 mm . . . . . (ii) bf = 966.67 mm Critical depth of N.A. xm = [mσcbc/(σst+mσcbc)]d

= 13 7230 13 7

× + ×

× 500

xm= 141.74 mm Actual depth of N.A. Assume N.A. is lies in flange portion

2

f stxb . =mA (d-x)2

966.672x

2 = 13 × 1256.64 (500-x)

48.3x2 = 8.168 ×106 – 16336.28x X =114.19 mm less than 150 mm so our assumption in correct. And x < xm section is under reinforced For under reinforced section x < xm

cbc cbcf ,σ

st stf =σ

M.R. = st stxσ A d- =230×π×3

202 114.205003

−

= 133.51 kN-m

OR

M.R. = f cbc1 x b xf d-2 3

cbcf114.20

= 23013

500 114.20−

cbcf = 5.24 N/mm2.

M.R.= 12×966.67×114.20×5.2 114.20500-

3

M.R. = 133.60 kN-m 4.11Limit State of Collapse in Shear The section of a structural member may be subjected to shear force due to flexure, punching or torsion. Accordingly, the shear may be flexural shear, punching shear or torsional shear. 1) Flexural shear: The shear associated with change of bending moment along the span is known as flexural shear, or simply shear. Flexural shear force present in beam is

given by dMdx

= V

The horizontal and vertical shear stresses are to be accounted for in the designs of beams. Exact analysis of shear in a reinforced concrete beam is quite complex, several experimental studies have been conducted to understand the various modes of failure which could occur due to possible combination of shear and bending moment acting at a given section.

2) Punching shear: • The shear associated with the

possibility of punching a thin member by a concentrated load is called punching shear.

• A slab carrying a concentrated wall load, a beamless floor slab supported directly by columns (called flat slab)or a footing slab carrying a concentrated column load are subjected to punching shear.

For the member subjected to both the above types of shear, the flexural shear is referred to as one-way shear,


whereas punching shear is called two way shear. A footing slab carrying concentrated column load is subjected to both these shears.

3) Torsion shear: When a member is subjected to torsion, it is subjected to torsional shear. • The beams are usually subjected to

flexural shear, and sometimes to torsion shear also.

• The slab is the plate elements and usually subjected to flexural shear. However, sometimes they are subjected to all the types of shears as in case of restrained two- way slabs and the flat slabs.

• Usually the shear failures of shallow RCC beams may not lead to immediate failure, however it considerably reduces its flexural strength and thus there is a state of impending shear failure. Hence the shear design is considered as limit state of collapse.

• If the shear failures take place before flexural failures, they are brittle and occur without warning. lf the flexural failure takes place prior to shear failure, the ductile failure of the beam is ensured.

Note: Concrete is very strong in compression and also quite strong in shear, however the combination of vertical and horizontal shear stress along with tension due to bending produces diagonal tension which is quite and will be now examined. The complementary diagonal compression should also be taken into account. 4.12DIAGONAL TENSION AND DIAGONAL COMPRESSION Consider a small element along the length of the beam. Subjected to pure shear. The principal stress on this element are given

by ( )2

2σ σσ1 or σ2= ± + τ2 2

the inclination of principal planes is given

by 2τtan2θ=σ

The major principal stress is tensile and is equal to

( )2

2σ σσ1= + + τ2 2

The minor principal stress is compressive and equal to

( )2

2σ σσ2= - + τ2 2

Two important cases are discussed below (i)If B.M. = 0, i.e., σ =0, then 1σ = τ and 2σ = -τ tan2θ = ∞ and θ = 450 or 1350 • This means that near the support for

simply supported beam, where B.M. is zero, and at N.A. of any section the principal stress is equal to shear stress and is inclined at 450.

• This is known as diagonal tension. As the concrete is weak in tension, the concrete near the support cracks at 450 with horizontal (i.e., perpendicular to diagonal tension) as shown in fig below.

• To avoid the shear cracks, the beam should be reinforced across the cracks (i.e., along the principal tension)

• The other principal stress is inclined at 1350 and is compressive.

• This is known as diagonal compression and is of the same value as the shear stress.

• The concrete is strong in compression and for usual cases diagonal


compression is below the permissible value.

• However, if the shear stress is very high, precautions to avoid the diagonal compression failure also have to be taken.

• The diagonal tension and compression near the support are shown in fig. The shear cracks are shown in fig.

(ii)When bending stress (σ ) is maximum, while shear stress is zero (i.e. τ = 0), we get, σ1=σ2=σ andθ = 900 i.e. principal plane is perpendicular to the beam axis. This means that principal tensile stress acts in horizontal direction and cracks will be vertical as shown in Fig. below

Thus at mid-section, where bending stresses are predominant, the cracks will start developing vertically. These cracks are called flexural cracks. From above discussion, it is clear that between the two limits, the cracks will change from a vertical direction at a point of zero shear to a direction inclined at an angle of 450 at a point where bending stress is zero.

From above the discussion, different modes of failure are: 1. Diagonal Tension failure: Which occur under large shear force and less bending moment? Such cracks are normally at 450 with the longitudinal axis of beam

2. Flexural shear failure: Which occurs under large bending moment and less shear force? Which occurs normally at closer to 900 with horizontal? When flexural crack occurs in combination with a diagonal tension crack (as in usually the case), the crack is sometimes called flexural shear crack.

It should be noted that it is due flexure crack that usually forms Ist and due to


increased shear stress at the tip of the crack this flexural crack extends into diagonal tension crack. 3.Diagonal compression failure: Which occurs under large shear force? It is characterized by the crushing of concrete. Normally it occurs in beams which are reinforced against heavy shear.

Note: For deep beam and thin walled sections subjected to large concentrated loads, maximum tensile stress is located near N.A. resulting in crack which generally starts at N.A. and propagate to support. This crack is also called web shear crack or diagonal tension crack tension crack. Such type of failure occurs when area of shear reinforcement exceeds a certain limit such that concrete section becomes stronger in diagonal tension compared to diagonal compression. Thus diagonal compression failure may occurs even before the shear reinforcement has yielded. 4.13 MECHANISM OF SHEAR RESISTANCE

The relative proportion of various mechanisms depends on the loading stage, the extent of cracking and material and geometric properties of the beam. • Prior to flexural crack All shear resisted

by czV • At the commencement of flexural

cracking aV and dV develop. • At the diagonal tension cracking sV

develops (all the four mechanism exist at this stage)

• Increase in longitudinal reinforcement not only increases dV but also controls the propagation of flexural crack and contribute to increase the depth of N.A and thereby the depth of uncracked concrete due to increased % tensile reinforcement crack formed as are smaller which improves the aggregate interlocking.

4.14 SHEAR STRESS

Distribution of shear stress in reinforced concrete is shown as below.


However for the solve of simplification we use nominal shear stress as the design stress for shear

Nominal shear stress = uv

V =τbd

For uniform depth uV = shear force due to design load

b = breadth of member (which for flanged section shall be taken as the breadth of web bw) d = effective depth For beam of varying depth

𝜏𝜏𝑣𝑣=

𝑉𝑉𝑢𝑢 ± �𝑀𝑀𝑢𝑢 tan𝛽𝛽𝑑𝑑 �

𝑏𝑏𝑑𝑑

uM =bending moment at the section β = angle between the top and bottom

edges of the beam. (-) sign = when BM (numerically increases in the direction in which depth increases) i.e., when BM decreases in the direction of effective depth increase (+) other Wise β

Note:

u u

u uM MC TZ d

= = =

tanu net u uV V T β= +

u net u u

uu net u

V V T tan

MV V tand

= − β

= = − β

uu

MV tand

= + β

4.15 DESIGN SHEAR STRENGTH OF CONCRETE IN BEAMS The design shear strength of concrete in beams shall be taken as follows: (1) Without shear reinforcement: From table 19 of IS: 456. The values given in table are based on the following equation:

( )ck

c

0.85 0.85f 1+5β-1τ =

6β

ck

t

0.8fβ=6.89p

, but not less than 1.0.

st

100Ap =bd

; for flanged section b shall be

taken as wb , the width of rib. Table: Design shears strength of concrete

cτ N/mm2. 100 sA

bd

(1)

Concrete grade M20 (2)

M25 (3)

M30 (4)

≤0.15 0.28 0.29 0.29 0.25 0.36 0.36 0.37 0.50 0.48 0.49 0.50 0.75 0.56 0.57 0.59


1.00 0.62 0.64 0.66 1.25 0.67 0.70 0.71 1.50 0.72 0.74 0.76

Note: The term sA is the area of longitudinal tension reinforcement which continues at least one effective depth beyond the section being considered except at support where the full area of tension reinforcement may be used provided the detailing conforms to the code requirements. • From above it is clear that the design

shear strength cτ in concrete depends on the percentage of tension steel. This is because: (i) When the amount of tension steel

increase, the depth of neutral axis increases and thus, the depth of uncracked concrete increases. This increases the capacity of concrete in shear.

(ii) When the amount of tension steel increases,the cracks formed are smaller, which improves the aggregate inter-lock. Also because of larger steel area the dowel action is improved. This further improves the capacity of section in shear.

• Further it is indicates that shear strength of concrete is also related to the compressive strength of concrete, In fact the shear strength of concrete is a function of ckf where fck is the characteristic compressive strength of concrete.

• When the longitudinal bars are not required to resist moment, they are sometimes curtailed or bent up. lf the bars are curtailed, they create complicated stresses at the point of curtailment, thereby reducing the shear capacity of the section. Therefore while using table any longitudinal bars which are terminated within a distance of d of the section under consideration, shall

not be considered for calculation of s100A

bd

For solid slabs (resisting on beam) design shear strength of concrete shall be K cτ where the value of K has given as below

overall depth of slab mm

≥ 300 275 250 225 200 175 ≤ 150

K 1 1.05 1.1 1.15 1.20 1.25 1.30

Note: Experiment studies have shown that slabs and shallow beams fail at loads corresponding to nominal stress that is higher than that applicable for beam of usual proportion • Thinner is the slab greater is the

increase in shear strength. • Slabs subjected to normal distributed

loads satisfy vτ < k cτ , and hence do not need.

Shear reinforcement. This is because thickness of slab (controlled by Limiting deflection criteria is usually adequate in term of shear capacity) • lf the nominal shear stress does not

exceed the value of cτ , the section is safe for shear and shear reinforcements are not required theoretically. However, some minimum shear reinforcement shall be provided. lf the nominal shear stress exceeds the design shear strength cτ , the section shall be suitably reinforced with shear reinforcements.

Max shear stress with shear reinforcement: • Under no circumstances, even with

shear reinforcement shall the nominal

shear stress in beams exceed maxcτ

maxcτ Depends on grade of concrete


Grade

M15

M20

M25

M30

M35

M40 and above

maxcτN/mm2

2.5 2.8 3.1 3.5 3.7 4.0

• By this provision failure of beam by

diagonal compression is prevented. • If vτ exceeds maxcτ section shall be

redesigned either by using higher concrete mix or by increasing the size of section.

• In solid slabs, vτ ≯ 0.5 maxcτ 4.16 SHEAR REINFOCEMENT IN BEAM Due to diagonal tension concrete cracks and if this crack is not intercepted, concreted section will split under overloading condition. To avoid it, we provide shear reinforcement in the form of (a) Vertical stirrups (b) Inclined stirrups (c) Bent up bars along with stirrups, Shear reinforcement restrains the growth of shear crack and resisting part of shear not resisted by the concrete. (1)Vertical stirrups: Vertical Stirrups are the commonly used shear reinforcement. These consist of a series of vertical open closed bars spaced along the beam span. The various types of vertical stirrups are

The first stirrups shall be provided within 50 mm from the face of the support The primary functions of the stirrups are: (i) to resist a part of the shear (ii) to resist the growth of the inclined

cracks and improve aggregate interlock, and

(iii)to tie the longitudinal bars in place, thereby increased the dowel action.

Closed stirrups are better because it resist torsion and helps in confining the compression reinforcement (in doubly reinforced beam)

Hook of the closed stirrups should be located in compression zone because in that case it helps in improving the anchorage and avoid crack initiation. The shear resisted through shear reinforcement is explained as follows: When the applied shear less than the concrete capacity, the stirrups carry practically negligible shear because the concrete is not cracked in diagonal tension. When the concrete capacity is exhausted, a shear crack forms. Simultaneously, the stirrups crossing the potential crack are


put into action. The additional shear, i.e., the shear in excess to the concrete capacity is now carried by the stirrups which cross this potential crack. This also means that any stirrup not crossing the crack essentially remains unstressed. Thus, the stirrups will be designed observing that they cross a crack, say the first crack at 450 near the support as shown in fig. The stirrups are spaced at a distance sv apart and they cross the crack. Let svA = area of legs of one stirrup

yf = characteristic strength of stirrup (or bent up bar) which shall not be greater than 415 N/mm2 0.87 yf = design tensile strength of shear reinforcement

usV = strength of shear reinforcement. Now, The horizontal length of crack = d –d’≈ d (d’ neglected),

'0 'd dtan 45 , x d d d

x−

= = − =

No of stirrups crossing the crack n = v

ds

Shear taken by stirrups

= sv yv

d A 0.87fs× ×

This shall be equal to the required strength

usV , of the shear reinforcement,

us sv yv

dV = ×A ×0.87fs

y svus

v

0.87f .A .dV =

S …. (a)

This formula is given in IS: 456:2000 and will be used to design stirrups. Grade Fe 250 (mild steel) and Fe 415 (HYSD) bars are used as stirrups. For design, usually we fix the grade of reinforcement (i.e., yf ), and the diameter of stirrups (i.e., svA ). Then we use above equation (a) to determine the spacing of stirrups as:

y sv y svv

us v c

y sv

v c

0.87f .A .d 0.87f .A .ds

V ( ) bd0.87f .Ab( )

= =τ − τ

=τ − τ

…. (b)

• Note that the value of yf is limited to

415 N/mm2 by the code. This is because the width of the diagonal crack is directly related to the strain in the stirrups. Limiting the stress in the stirrups will limit the strain and consequently the width of the shear crack. Limiting the width of crack also enables more aggregate interlock to develop. If higher grade steel, say Fe 500 is used for shear reinforcement, then also use yf = 415 N/mm2 for shear reinforcement.

• The stirrups are usually two-legged.

However stirrups of four-legged, six-legged, etc., also can be formed as shown in fig. More than two-legged stirrups are used for heavy shear. Unless otherwise specified, stirrups are always two-legged.

Maximum spacing: The shear reinforcements are provided to prevent the shear cracks in the beam. The horizontal distance between two successive cracks is approximately equal to the effective depth d. Stirrups shall be provided such that they cross the crack and also no crack shall remain unreinforced. To ensure this, the maximum spacing of stirrups as per code shall not exceed:


(i) 0.75 d for vertical stirrups (ii)d for inclined stirrups maximum spacing shall be limited to 300 mm. • It may be emphasized that closely

spaced stirrups prevent the failure caused by tearing of main reinforcement through concrete cover and slipping relative to the concrete.

(2) Inclined stirrups: A series of inclined stirrups near the support crossing a 450 crack is shown in fig, lf svA is the area of legs of inclined stirrups and 0.87 yf is the design tensile strength in the shear reinforcements, the tensile force in an inclined stirrup = y sv0.87f .A .

Let inclination of stirrups is α with horizontal. and crack is inclined at 450 (normally assumed) with the horizontal. from ∆ obs

d-d'tanα=x

x=(d-d')cotα Neglecting d', No. of stirrups crossing the

crack, ( )d+x d+dcotα dn= = = 1+cotαSv Sv Sv

Tensile force in an inclined stirrups = y sv0.87f .A Shear taken by stirrups = (d/Sv)(1+ cotα ) 0.87 fy Asv Vertical component of this force effective in shear

= ( ) y svv

d1+cotα 0.87f AS

sinα

This shall equal to usV , the strength of the stirrups required. Thus for a series of inclined stirrups

us y svv

dV =0.87f A sinα(1+cotα)S

y svv

cosα d=0.87f A sinα+ .sinαsinα S

( )us svv

dV =0.87fyA sinα+cosαS

The inclined stirrups are very effective in carrying flexural shear and restrain the inclined crack very effectively. However, in case of stress reversal (due to earthquake loads for example), the direction of crack will be reversed and inclined stirrups become ineffective α ≮ 450 because for α <450 there is a possibility of inclined stirrups slipping along the longitudinal bars. So that at least one reinforcement intercept the crack Max. Spacing ( vS ) of included stirrups (with α = 450) ≯ d also, vS ≯ 300 mm

yf ≯ 415 N/mm2

(3) Bent up bars When a single bar or a group of bars, all bent up at the same section

us y svV =0.87.f .A ….. (11)

svA = No. of bars bent 2π× d4

D= dia. of bent-up bars.

For series of bars bent up at different cross section,

( )us y svv

dV =0.87f A sinα+cosαS

The bent-up bars alone are not satisfactory for the shear reinforcements. This is because the exact behavior of bent bars in


resisting shear is not clearly understood. Also, the bent bars do not resist the reversal of shear force, IS: 456 states that when bent bars are provided, their contribution towards shear resistance shall not be taken as more than half of the total shear to be resisted by shear reinforcement, The remaining half of the shear to be resisted through shear reinforcement shall be resisted by vertical/inclined stirrup which are provided along with the bent up bars. 4.17 MINIMUM SHEAR REINFORCEMENT Minimum shear reinforcement in the form of vertical stirrups shall be provided such that

sv

sv y

A 0.4b 0.87f

≥

Where svA = total cross-sectional area of stirrup legs effective in shear

vS = stirrup spacing along the length of the member in mm b = width of the beam or width of rib of flanged (i.e., wb for flanged beam)

yf = characteristic strength of the stirrup reinforcement in N/mm2 which shall not be taken greater than 415 N/mm2. • However, where the maximum shear

stress calculated is less than half the permissible value and in the members of minor structural importance such as lintels, this provision need not be complied with. The above provision of minimum shear reinforcement provides a shear resistance of 0.4 N/mm2. For minimum shear reinforcement, the spacing of stirrup shall not exceed

y svv

0.87.f . As

0.4b≤ ….. (12 b)

It can be seen that for given type of steel and selected diameter of stirrups, the spacing that provides minimum shear resistance is inversely proportional to the Width ‘b’ of the member. In case of

‘T’ or ‘L’ beams the b shall be equal to

wb , the width of web. The minimum shear reinforcement is provided for the following: (i)Any sudden failure of beams is

prevented if concrete cover bursts and the bond to the tension steel is lost.

(ii)Brittle shear failure is arrested which would have occurred without shear reinforcement.

(iii)Tension failure is prevented which would have occurred due to shrinkage, thermal stresses and internal cracking in beams.

(iv)To hold the reinforcement in place when concrete is poured.

(v) Section becomes effective with the tie effect of the compression steel.

4.18 DESIGN STEPS

(a) Find uv

V =τbd

(b) Find cmaxτ (depends on the grade of concrete)

(c) Check v cmaxτ >τ if satisfied, redesign the section.

(d) Find cτ from the steel and grade of concrete

(e) If v cτ <0.5τ , no shear reinforcement provided

(f) If C V C0.5τ τ τ≤ ≤ ,provided min shear reinforcement

(g) If V Cτ >τ , provide shear reinforcement for ( )V Cτ -τ .This reinforcement not be less than the min shear reinforcement.

Example: A ‘T’ beam section having 230 mm with of web and 460 mm effective depth is reinforced with 5 no. of 16 mm dia bar as tension reinforcement, which continues for distance greater than effective depth , past the section, the section is subjected to a factored shear force of 52.5 kN check in


shear stress and design the Shear stress, the material M20, HYSD Fe415 for stirrups use 6 mm mild steel bar. Solution : i) τv=

Vubwd

3

v52.5×10τ =230×460

2vτ =0.4962N/mm

ii) Hence τc value corresponding to tP = cτ = 0.608N/mm2

t s wP =A /b d

( )2

t

π5× × 164P =

230×460

tP =0.95% iii)Check: v cmaxτ >τ 0.4962>2.81 ( cmaxτ =2.81 forM20) …..(O.K.) v cτ (0.4962)<τ (0.608) iv) So provide minimum reinforcement in

form of stirrup.

sv

v y

A 0.4bS 0.87f

≥

Use 2 logged 6 mm φ mild steel. 4.19 BOND AND ANCHORAGE 4.19.1 INTRODUCTION • While analyzing and designing the

reinforced concrete structures, the basic assumption is that there is a perfect bond between concrete and steel, i.e., there is absolutely no slippage between the concrete and steel.

• It is this bond which is responsible for the transfer of axial force from a reinforcing bar to the surrounding concrete, thereby providing strain compatibility and ‘Composite action’ of concrete and steel.

• If this bond is inadequate, ‘slipping’ of the reinforcing bar will occur, destroying full ‘composite action’. Hence, the fundamental assumption of the theory of flexure, viz. plane sections

remain plane even after bending, becomes valid in reinforced concrete only if the mechanism of bond is fully effective.

• The term ‘bond' describes the means by which the relative movement between concrete and steel is prevented and the intensity of adhesive force is called bond stress.

• Thus, bond stress is defined as longitudinal shear stress acting on the surface between steel and concrete.

• Bond between steel and concrete is due to combined effect of adhesive resistance, frictional resistance and mechanical resistance (for deformed bars).

• The adhesive resistance is provided by ‘chemical gum' produced by concrete during setting.

• The bond due to friction is provided by gripping of bars due to shrinkage. The friction gives considerable bond resistance. With increasing force in bar, the adhesion is lost first than the friction between concrete and steel.

• The mechanical resistance is provided by deformed bars only (not by the plain bars). The deformed bars have lugs, or corrugations and give higher bond resistance by providing an interlock between steel and concrete.

• ln deformed bar, adhesion and friction become minor elements, and the bond strength is primarily dependent on bearing of concrete against the lugs or corrugations.

Note:If bond is absent, the stress at all points on a straight bar would be constant, as in a string or a straight cable.

4.19.2 TYPES OF BOND STRERSS The bond stress in reinforced concrete members arises due to two distinct situations.


a) The change in the bar force along its length due to variation in bending moment in this length, This type of bond stress is called flexural band stress.

b) From the anchorage of bar in case of tension or compression. This type of bond force is known as Anchorage bond stress. Anchorage bond is also called development bond. • lt may be noted that the actual bond

stress will be influenced by flexural cracking, local slip, splitting and other secondary effects. In particular, flexural cracking has a major influence in governing the magnitude and distribution of local bond stresses.

4.19.3 Anchorage Bond Stress and Development Length

• Anchorage bond stress arises when a

bar is carrying certain force. ln anchorage bond, it is necessary to transfer this force in the bar to the surrounding concrete over a certain length. The length of bar ‘Ld’ is required to transfer the force in the bar.

• This condition arises near the extreme end (or cutoff point) of a bar subjected to tension (or Compression).

• Following figure shows a steel bar embedded in concrete and subjected to a tensile force T. Due to this force, there will be a tendency of the bar to slip out and this tendency is resisted by the bond stress developed over the perimeter of the bar, along its length of embedment.

• This required length ‘Ld' is called anchorage length in case of axial tension (or compression) & development length in case of flexural tension.

• The development length is an embedded length of the bar required to develop the design strength of reinforcement at the critical section.

If φ is the nominal diameter of a bar, Then Tension, y stT=(0.87f )A

2πT=(0.87fy)4φ

This force must be transferred from steel to concrete through bond acting over the perimeter of the bar along its length of embedment Ld.

lf bdτ is the average bond stress, than Force bd d=τ ×(π ).Lφ For equilibrium 2

yπ(0.87f )4φ

= bd dτ ×(π ).Lφ

y

dbd

0.87fL =

4τφ

The concept underlying ‘development length’ is that a certain minimum length of the bar is required on either side of a point of maximum steel stress, to prevent the bar from pulling out under tension (or pushing in, under compression).

The value of design bond stress for plain bars in tension prescribed by IS code are reproduced in table below :

Table: Permissible bond stress in tension

Grade of concrete

M20

M25

M30

M35

M40 and above


Design bond

stress bdτ , (N/mm2)

1.2

1.4

1.5

1.7

1.9

• In the above discussion φ is defined as

the nominal diameter of the bar. For the plain bars, nominal diameter and actual diameter are the same.

• However, for deformed bars, the cross-section is not perfectly circular.

• The actual area of the cross-section of a deformed bar is equated with an area of a circle and corresponding diameter of the circle is known as the normal diameter.

• Thus, although the shapes of cross-sections of plain and deformed bars are not equal but for a particular diameter, their cross-sectional areas and mass per metre length are the same.

Note: 1. bdτ for deformed bar is increased by

60%. This is because for deformed bars, the actual contact area of a bar with concrete is taken into account which is much more than contact area based on nominal diameter.

2. For bars in compression, the above values may be increased by 25 percent. This is because the end bearing of the bar helps in resisting compression. Also the possibility of cracking the concrete is nil in compression and hence, the allowable bond stresses are increased.

4.20 TORSION 4.20.1 INTRODUCTION Torsion produced in a beam is classified as a) Primary torsion or equilibrium torsion b) Secondary torsion or compatibility

torsion Torsion which can be determined only by using the static equilibrium condition is

called primary equilibrium torsion such torsion is induced in beams curved in plan and subjected to gravity loads and also in beams where the transverse loads are eccentric with respect to the shear centre of the x- section. For example a slab cantilevered from a beam which is assumed to be fixed at supporting columns. Slab load W induces a torsional moment of Wx in beam, or a small beam BC is cantilevered from beam B1. Beam B1 is considered fixed at columns. Here negative moment of beam BC will be the torsional moment of beam B1.

Secondary torsion is induced in a structural member by rotation (twist) applied at one or more points along the length of member through inter connected members, instead of by directly applied load. Twisting moment induced in this case is


proportional to the torsional stiffness of the member. These moments are generally statically indeterminate and their value is obtained using compatibility equation. This is the reason these torsion are called compatibility torsion.

As per IS code where torsion can be eliminated by releasing redundant restraints, no specific design for torsion is necessary provided torsional stiffness is neglected in the calculation of internal forces. The maximum torsional shear stress occurs at the middle of the wider face and has a value given by

t; max 2

Tb D

τ =α

4.21 EFFECT OF TORSIONAL MOMENT

• Torsional moment induces shear

stresses in the beam because of the torsion a beam fails in diagonal tension forming spiral cracks around the beam as shown in Fig. 4.2.

• Hence the ideal Way of providing reinforcement against torsion is by providing the steel in the form of a spiral along the direction of principal tensile stress. But this is not practical and hence torsional reinforcement is provided in the forms of longitudinal & transverse reinforcement, longitudinal

reinforcement is provided in the form of bars distributed around, the x-section, close to periphery and transverse reinforcement in the form of closed rectangular stirrups placed perpendicular to the beam axis.

• The longitudinal reinforcement helps in reducing the crack width through dowel action and stirrups crossing the cracks resist shear due to vertical loads and torsion,

As a simplification, the effect of torsional moment is split up into

a) Equivalent shear, and b) Equivalent bending moment

Code Provisions: (Clause 41) • If in the analysis of structure, the

torsional resistance or stiffness of members has not been taken into account, no specific calculation for torsion is necessary, adequate control of any torsional cracking being provided by the required nominal shear reinforcement.

• Where the torsional resistance or stiffness of members is taken into account in the analysis, the members shall be designed for torsion.

• For design of torsion, section located at a distance less than ‘d' from the face of the support may be designed for the same torsion as computed at a distance ‘d’, where d is the effective depth.

The design rules for torsion are based on equivalent moment. They are explained as per IS code 456: 2000

(a) Shear and Torsion (Clause 41.3.1): The combined effect of torsion and shear is considered and called equivalent shear.

Equivalent shear Ve is calculated from the formula

ue u

TV V 1.6b

= + −

Where, Ve = Equivalent shear at the section (in kN or N) Vu = Ultimate vertical shear at the section (in kN or N)


Tu = Ultimate torsional moment (in kNm or Nmm) And b = Breadth of beam. (in m or mm) The equivalent nominal shear stress veτ is given by,

eve

Vτ =b.d

The equivalent nominal shear stress veτ shall not exceed the values of c,maxτ as given in table 20 of lS-456. If veτ does not exceed cτ as given in table 19 of IS 456, only minimum shear reinforcement shall be provided min shear reinforcement is given

y sv

v

0.87f (A )d=0.4bd

S

lt veτ exceeds cτ as given in table 19 of IS 456, both longitudinal and traverse reinforcement shall be provided (clause 41.3.3). (b) Longitudinal Reinforcement (Clause

41.4.2): The longitudinal reinforcement shall be designed to resist an equivalent bending moment Me1 , given by

e1 u tM =M +M Where, uM = bending moment at the cross-section (factored)

Mt = Tu �1 + D

b1.7

�

Where uT = Torsional moment D = Overall depth of beam b = breadth of the beam When Mt<Mu there is resultant torsion due to ( tM - uM ). Therefore, the longitudinal reinforcement shall be provided on the flexural compression face, to resist the equivalent moment Me2 = Mt - Mu When t uM <M , no additional steel is required on bending compression side,

(c) Transverse reinforcement (Clause

41,4.3): Two-legged closed hoops enclosing the corner longitudinal bars shall have an area of cross-section Asv , given by

( )u v u v

sv1 y1 1 y

T .S V .SA = +2.5d (0.87f )b d 0.87f

(τve - τc)bd = (0.87fyAsvd/Sv) But the total transverse reinforcement shall not be less than Asv ≥ [{(τve-τcb)bSv}/(0.87 fy)] Where,

uT = Ultimate torsional moment

uV = Ultimate shear force

vS = Spacing of the stirrup reinforcement

1b = c/c distance between corner bars in the direction of width

1d = c/c distance between corner bars in the direction of depth b = Breadth of the member

yf = Characteristics strength of shear reinforcement ≤ 415N/mm2

eve

Vτ =b.d

= Equivalent shear stress

cτ = Shear strength of concrete, as in table 19 of IS 456.


5.1 DESIGN OF BEAM AND SLAB

A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own Weight, span and external reactions to these loads is called a bending moment.

5.2 I.S 456 PROVISIONS

1. Effective Span (Clause 22.2)(a) Simply supported Beam or Slab

The effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre of supports, whichever is less.

eff 0l (Effective span)=l +dOR

eff 0w wl l2 2

= + +

Whichever is less d = effective depth; w = width of support

0l = clear span;

effl = effective span

(b) Continuous Beams or Slab in case of continuous beam or slab, if the width of support is less than (1/12) of the clear span, the effective span shall be as in (a) above. lf the supports are wider than 1

12 of the clear span or 600 mm,

whichever is less, the effective span shall be taken as under:

Case (1):

If width of support 0lspan< w<12 12

Then effective span is calculated same as for simply supported case,

eff 0l =l +d Or eff 0w wl =l + +2 2

Whichever in less

Case(2): If width of support > [(span/12) or 600 mm] (a) (i) For one end fixed other continuous

(ii) Both end continuous (Intermediate span)

eff 0l =l = clear span (b) One end discontinuous other continuous

(simply supported) 0eff=l +d/2

Or eff 0wl =l +2

Whichever is less? Case(3): Cantilever

eff 01 =l +d/2 leff = lo+(w/2)

5 BEAM COLUMN


Case (4) : For frame its centre to centre distance between members.

effl = Centre to Centre distance between

members. Case (5) : In the case of spans with roller or rocket bearings, the effective span shall always be the distance between the centres of bearings. 2. Check for Deflection (cl 23.2/P-37) The deflection shall generally be limited to the following 1. The final deflection due to all loads

including the effect of temperature, creep and shrinkage and measured from at cast level of support of floor, roof and all horizontal. Deflection should not normally exceed span/250.

2. The deflection including the effect of creep, temperature and shrinkage

occurring after erection of partition and

application of finishes. span350

or 20 mm

whichever is less. Note: 1. Deflection can be reduced by providing

more tensile steel than required for flexure, the service stress in steel is reduced and hence the deflection. However care must be taken so that section will not become over reinforced.

2. Calculation of Deflection are required. (a) When designer wishes to exceed

span/ depth ratio (b) Where specific deflection control is

required. (c) Where the structure is abnormal

due to loading or behavior. 5.3 CONTROL OF DEFLECTION The deflection of structure or part thereof shall not adversely affect the appearance or efficiency the structure or finishes of partitions. • For beam and slabs, the vertical

deflection limits may generally be assumed to be satisfied, provided that the span to depth ratio are not greater than the values obtained as below a) Basic values of span to effective

depth ratios for spans upto 10m. Cantilever 7 Simply supported 20 Continuous 26 b) For spans above 10m, the values in

(a) may be multiplied by 10/span in meters, except for cantilever in which case deflection calculations should be made.

c) Depending on the area the stress of steel for tension reinforcement, the values in (a), or (b) shall be modified by multiplying with the modification factor obtained as per Fig. 5.7 depending on reinforcement percentage and type.

• In above formula of finding out the modification factor for tension


reinforcement indicates that if the designer wishes to use shallow member he or she increase the area of reinforcement there by reduced the service stress hence the deflection, however it should be remember that by increased Ast, section should not become over reinforced.

• The total depth D can be determined by adding nominal cover and half of the diameter of the bar used to the effective depth.

Selection of Diameters of Bar of Tension Reinforcement • Reinforcement bars are available in

different diameters such as 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 36 and 40 mm.

• Some of these bars are less available. • The selection of the diameter of bars

depends on its availability, minimum stiffness to resist while persons walk over them during construction, bond requirement etc.

• Normally, the diameters of main tensile bars are chosen from 12, 16, 20, 22, 25 and 32 mm.

Selection of Grade of Concrete Besides strength and deflection, durability is a major factor to decide on the grade of concrete. Table of lS 456 recommends M 20 as the minimum grade under mild environmental exposure and other grades of concrete under different environmental exposures also. Selection of Grade of Steel Normally, Fe 250, 415 and 500 are in used in reinforced concrete work. Mild steel (Fe 250) is more ductile and is preferred for structures in earthquake zones where there are possibilities of vibration, impact, blast etc. 5.4 ONEWAY SLAB Slabs are plate elements having depth much smaller than its other two

dimensions. So slab is a two dimensional element. Slabs form roof or floor of the building. Slabs are designed same as beams with unit width. 5.5 INTRODUCTION • Slabs, used in floors and roofs of

buildings mostly integrated with the supporting beams, carry the distributed loads primarily by bending.

• lt has been mentioned that a part of the integrated slab is considered as flange of T - or L – beams because of monolithic construction.

• However, the remaining part of the slab needs design considerations.

• These slabs are either single span or continuous having different support conditions like fixed, hinged or free along the edges. Though normally these slabs are horizontal, inclined slabs are also used in ramps, stair cases and inclined roofs.

• While square or rectangular plan forms are normally used, triangular, circular and other plan forms are also needed for different functional requirements.

• This lesson takes up horizontal and rectangular/square slabs of buildings supported by beams in one or both directions and subjected to uniformly distributed vertical loadings.


Note: 1. One – way slab if y xl >2l 2. Indicates that no support is needed if

y xl >2l and is needed if y xl 2l≤ 3. End supports may be simply supported

or clamped.

5.6 ONE-WAY AND TWO-WAY SLABS

• Figures 5.13 (a) and (b) explain the share of loads on beams supporting solid slabs along four edges when vertical loads are uniformly distributed.

• lt is evident from the figures that the share of loads on beams in two perpendicular directions depends upon the aspect ratio /y xl l of the slab, xl being the shorter span.

• For large values of yl , the triangular area is much less than the trapezoidal area. Hence, the share of loads on beams along shorter span will gradually reduce with increasing ratio of /y xl l .

• In such case, it may be said that the loads are primarily taken by beams of longer span. The deflection profiles of the slab along both directions are also shown in the figure.

• The deflection profile is found to be constant along the longer span except

near the edges for the slab panel of Fig. 5.13 (a).

5.7 CODE REQUIREMENTS ON REINFORCEMENT AND DETAILING Longitudinal Reinforcement

5.7.1 Minimum Reinforcement

The longitudinal bars must, in general, have a cross-sectional area not less than 0.8% of the gross area of the column section. Such a minimum limit is specified by the Code: • To ensure nominal flexural resistance

under unforeseen eccentricities in loading; and to prevent the yielding of the bars due to creep and shrinkage effects, which result in a transfer of load from the concrete to the steel.

• However, in the case of pedestals which are designed as plain concrete columns, the minimum requirement of longitudinal bars may be taken as 0.15 per cent of the gross area of cross-section.

• In the case of reinforcement concrete walls, the Code (Cl.32.5) has introduced detailed provisions regarding minimum reinforcement requirements for vertical (and horizontal) steel.

• The vertical reinforcement should not be less than 0.15% of the gross area in general.

• This may be reduced to 0. 12% if Welded wire fabric or deformed bars (Fe 415/Fe 500 grade steel) is used, provided the bar diameter does not exceed 16 mm.

• This reinforcement should be placed in two layers if the wall is more than 200 mm thick.

• ln all cases, the bar spacing should not exceed three times the wall thickness 450 mm, whichever is less.

5.7.2 Maximum Reinforcement: • The maximum cross-sectional area of

longitudinal bars should not exceed 6


per cent of the gross area of the column section.

• However, a reduced maximum limit of 4 per cent is recommended in general in the interest of better placement and compaction of concrete and, in particular, at lapped splice locations.

• In tall buildings, columns locate in the lowermost storeys generally carry heavy reinforcement (~ 4 per cent). The bars are progressively curtailed in stages at higher levels.

5.7.3 Minimum diameter/number of bars and their location

• Longitudinal bars in column (and pedestals) should not be less than 12 mm in diameter and should not be spaced more than 300mm apart (centre-to-centre) along the periphery of the column [Fig. 6.3(a)].

• At least 4 bars (one at each corner) should be provided in a column with rectangular cross-section, and at least 6 bars (equally spaced near the periphery) in a circular column.

• ln ‘spiral columns' (including noncircular shapes), the longitudinal bars should be placed in contact with the spiral reinforcement, & equidistant around its inner circumference [Fig, 6.3(b)]. In columns with T-, L-, other cross sectional shapes, at least one bar should be located at each corner or apex [Fig 6.3(c)].

• Longitudinal bars are usually located close to the periphery (for better flexural resistance), but may be placed in the interior of the column when eccentricities in loading are minimal, When a large number of bars need to be accommodated, they may be bundled, or, alternatively, grouped, as shown in [Fig 6.3(d)].

5.7.4 Arrangement of transverse reinforcement lf the longitudinal bars are not spaced more than 75 mm on either side, transverse reinforcement need only to go round corner and alternate bars for the purpose of providing effective lateral supports. If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of the tie are effectively tied in two direction, additional longitudinal bars in between these bars need to be tied in one direction by open ties.


5.7.5 Cover to Reinforcement: A minimum clear cover of 40mm or bar diameter (whichever is greater), to the column bars is recommended by the Code (Cl. 26.4.2.1) for columns in general; a reduced clear cover of 25 mm is permitted in small-sized columns (D ≤ 200 mm and whose reinforcing bars do not exceed 12 mm) and a minimum clear cover of 15 mm (or bar diameter, whichever is greater) is specified for walls. however, in aggressive environments, it is desirable, in the interest of durability, to provide increased cover but preferably not greater than 75 mm. 5.7.6 Transverse Reinforcement (Cl. 26.5.3.2 of the Code) • General All longitudinal reinforcement

in a compression member must be enclosed with in transverse reinforcement comprising either lateral ties (with infernal angles not exceeding 1350) or spirals. This is required:

To prevent the premature buckling of individual bars;

to confine the concrete in the ‘core’, thus improving ductility and strength;

to hold the longitudinal bars in position during construction; and

to provide resistance against shear and torsion, if required.

5.7.7 Lateral Ties • The arrangement of lateral ties should

be effective in fulfilling the above requirements.

• They should provide adequate lateral support to each longitudinal bar, thereby preventing the outward movement of the bar.

• The diameter of the tie φ , is governed by requirements of stiffness, rather than strength, and so is independent of the grade of steel.

• The pitch st (centre-to centre spacing along the longitudinal axis of the

column) of the ties should be small enough to reduce adequately the unsupported length (and hence, slenderness ratio) of each longitudinal bar. The code recommendations are as follows:

tie diameter t long,max / 4φ ≥ φ or 6mm whichever is maximum

tie spacing St ,min16 longφ≤ or D or 300 mm whichever is minimum

• Minimum tie diameter was specified as 5 mm, instead of 6 mm in earlier version of code.

• Maximum tie spacing was specified as 48 tφ instead of 300 mm in earlier version of code.

Where longφ denotes the diameter of longitudinal bar to be tied and D denotes the least lateral dimension of the column.

• When the spacing of longitudinal bars is less than 75mm, lateral support need only be provided for the corner and alternate bars.

• If the longitudinal bars spaced at a distance not exceeding 48φ are effectively tied in two directions, then the additional longitudinal bars in between these bars need be tied only in one direction by open ties.

5.7.8 Spirals

Helical reinforcement provides very good confinement to the concrete in the ‘core’ and enhances significantly the ductility of the column at ultimate loads. The diameter and pitch of the spiral may be computed as in the case of ties except when the column is designed to carry a 5 per cent overload (as permitted by the Code), in which case Pitch st < { 75mm or (core diameter/6) St> {25mm or 3фt

The ends of the spiral should be anchored properly by providing one and a half extra turns.


5.7.9 Design of Short Columns under Axial Compression It load P is applied on a column of width B and Depth D.

Then u c scP =P +P …. (i)

cP = load taken by concrete

scP =load taken by steel

uP = total load taken up by the column (ultimate load)

We know that stress = LoadArea

ccc

c

PfA

= and scsc

sc

PfA

=

where ccf = Stress in concrete

scf =Stress in steel

cA =Area of concrete

scA =Area of steel A = Gross area

u cc c sc sc

u cc g sc cc sc

P f A f A orP f A (f f )A

= += + −

…(ii)

Maximum compressive strain in concrete under axial loading at the limit state of collapse in compression is specified of εc = 0.002 by the code. Corresponding to this limiting strain of 0.002 stress in the

concrete is ck0.67 f1.5

= 0.45fck and the design

strength of a short column is u ck g sc ck scP =0.45f A +(f -0.45f )A sc yf =0.87f for Fe250

y0.790f for Fe415 y0.746f for Fe500

The design stress in concrete is ck

ck0.67f 0.447f

1.5=

And design stress in steel is ck0.87f in case of Fe 250 under ‘pure’ axial loading

conditions, the design strength of a short column is

u ck c y scP =0.447f A +0.87f A … (iii) However the code requires all columns to be designed for minimum eccentricities in loading hence equation (iii) cannot be applied directly. Nevertheless, where the calculated minimum eccentricity (in any plane) does not exceeds 0.05 time the least lateral dimension (in the plane considered), the code (Cl. 39.3) permits the use of the following simplified formula, obtained by reducing the uP by 10%. Now, u ck c y scP =0.4f A +0.67f A Example Design A R-Column of size 450 mm × 600 mm subject to an axial total of 2000 kN under service load condition. Unsupported length of column is 3m, use M20 concrete and Fe 415 steel. Assume effl l= . Solution: 1) Check whether long or short column =

3000 6.67 12450

= < so column is short

2) Check mine along 450 mm mine is greater of

(i) 3000 450500 30

+ =21mm

(ii) 20 mm

mine = 21

mine must be less than 0.05×450 = 22.5mm

mine (21) < 0.05B (22.5) O.K

mine along 600mm

mine is greater of (1) 3000 600 26500 30

+ =

(1) 20 mm

mine = 26 < 30 (0.05 D) So we can use u ck c y scP =0.4f A +0.67f A

( )sc sc=0.4 20 A A 0.67 415 A× × − + × ×


3sc

sc

2000 10 1.5 0.4 20 (450 600 A )0.67 415A

× × = × × × −+ ×

4sc sc=216×10 -8A +278.05A

2

scA =3110.5mm Provide 4 nos. 25 mm φ at corner

24×491=1964mm

An additional 4 nos. 20φ

2=4×314=1256mm

2 2

scA =3220mm >3110.5mm

scA 3220p'= ×100=

bd 450×600 1.197 0.8> Minimum reinforcement Lateral ties is greater of

i) 25/4 = 6.25 ii) 6 mm use 8 mmφ Spacing is mini. of

i) 450 mm ii) 16 ×20 = 320 mm iii) 300 mm

Hence use 8 mm φ @ 300 mm c/c 5.8 COLUMN SUBJECTED TO AXIAL COMPRESSION AND UNIAXIAL BENDING (CLAUSE 39.5 ) 5.8.1 INTRODUCTION • The load on the column is rarely axial.

There is always some minimum inherent eccentricity on account of inaccuracies in loading, bad workmanship etc.

• ln such cases, the column is subjected to axial compression uP and bending moment uM . This loading system can be reduced to a single resultant load uP

acting at an eccentricity u ue=M /P as shown in Fig. 6.5.

• But the design of member subjected to combined axial load and uniaxial bending involves lengthy calculations by trial and error.

• ln order to overcome these difficulties, interaction diagrams may be used. These have been prepared and published by BIS in “SP: 16 design aids for reinforced concrete to IS 456".

5.8.2 Interaction Curve

• The ‘interaction curve' is a complete

graphical representation of the design strength of a uniaxially eccentrically loaded column of given proportions.

• Each point on the curve corresponds to the design strength values of uP and

uM associated with a specific eccentricity (e) of loading.

• That is to say, if load ‘P’ is applied on a short column with an eccentricity ‘e’ and if this load is gradually increased till the ultimate limit state is reached and that ultimate load at failure is given by uP=P and the corresponding moment by u uM=M =P .e, then the co-ordinates

u u(M ,P ) form a unique point on the diagram.

• The interaction curves define the different u u(M ,P ) combinations for all possible eccentricities of loading 0≤ e <∞

• For design purposes, the calculations of uM and uP are based on the design

stress strain curves (including partial safety factors) and the resulting


interaction curve is referred to as the ‘design interaction curve' (which is different from the characteristic interaction curve).

• Using the design interaction curve for a given column section, it is possible to make a quick judgment as to whether or not the section is safe under a specified factored load effect combination

u u(P ,M ) . • lf the point given by the co-ordinates

u u(P ,M ) falls within the design interaction curve, the column is ‘safe', otherwise it is not.

5.8.3 Salient Point on the Interaction Curve The salient points, marked 1 to 5 on the interaction curve corresponds to the failure strain profiles, marked 1 to 5 in Fig. 6.6. a) The point 1 corresponds to the

condition of axial loading with e=0 For this case of ‘pure’ axial

compression, uM =0 And

u u ck c y scP =P =0.45f A +0.75f A b) The point (1) corresponds to the

condition of axial loading with the minimum eccentricity. mine = 0.05×D [clause 25.4 & 39.3] The corresponding ultimate resistance is written as

u ck c y scP =0.4f A +0.67f A c) The point (3) corresponds to the

condition e = eD (i.e. xu = D). For e< or eD, the entire section is in under compression and neutral axis is located outside the section (xu> D)with 0.002 <

uε < 0.0035. For e > eD, the natural axis is located

within the section (xu< D) & uε =0.0035 at the “highly compressed edge”. Point(2) represents a general case, with the neutral axis outside the section (e <eD). d) The point (4) corresponds to the balanced failure condition, with e = eb and xu = xu. lim. The design strength

values for this ‘balanced failure’ condition are denoted as ubP and ubM.For u ub bP <P (i.e e>e ) the mode of failure is called tension failure.

e) The point (5) corresponds to a ‘pure’ bending condition u(e= ,P =0)∞ . The resulting ultimate moment of resistance is denoted by ubM and the corresponding neutral axis depth takes on a minimum value u,minX . This case is the same as the under- reinforced section of beam.

5.9 COLUMN SUBJECTED TO AXIAL COMPRESSION AND BIAXIAL BENDING (CLAUSE 39.6) 5.9.1 INTRODUCTION The condition of biaxial bending along with axial compression is of common occurrence for design of R.C. columns due to its monolithic construction with beams in two different directions. This is especially true for corner columns in frames. The strength of the section under axial compression and biaxial bending is a function of uP , uxM and

uyM . The ultimate moment uxM about the major axis of bending (i.e. X-axis) and uyM about the minor axis of bending (i.e. Y-axis) can be expressed in terms of axial compression uP acting at eccentricities

uxx

u

Me =P

and uyy

u

Me =

P

5.10 CODE PROCEDURE FOR DESIGN OF BIAXILLY LOADED COLUMNS

The column subjected to axial compression and biaxial bending may be designed by the following equation

nn

uyux

ux1 uy1

MM 1.0M M

αα + ≤


Where uxM and uyM = Moment about X and Y axes due to design loads.

uxlM and uylM = Maximum uniaxial moment capacity for an axial load of

uP , bending about X and Y axis respectively nα is related to u

uz

PP

Where Puz = 0.45fckAc + 0.75fyAsc = 0.45fckAg + (0.75fy-0.45fck)Asc

For values of u

uz

PP

= 0.2 to 0.8, the values of

nα vary lineraly from 1.0 to 2.0 For values less than 0.2, nα is 1.0; for values greater than 0.8, nα is 2.0 as shwon in Fig. 6.7

Accordingly,

nα = 1.0 for u

uz

P <0.2P

nα = 2.0 for u

uz

P >0.8P

nα =0.667 + 1.667 u

uz

PP

otherwise

5.10.1 Slender Columns

An essential step in the design of a column is to determine whether the proposed dimensions will make it a short column or a slender column. A short compression

member, whose l/D ratio is less than 12, is not in danger of buckling prior to achieving its ultimate strength based on the properties of the cross-section. Moreover, the lateral deflection short compression members subjected to bending moments are small, thus, contributing little secondary bending moment P- ∆ as shown in Fig. 6.8. These buckling and additional deflection effects are more pronounced in slender compression members and reduce their ultimate strength as compared to that of a short column having the same cross-section and amount of steel. 5.10.2 P - ∆ Effect in a column The CEB-FIP recommendations for buckling in compression and bending advise a check for columns with effective slenderness ratio more than 35. The effective slenderness ratio should not exceed 140 for normal aggregate concrete and 80 for light weight aggregate. However, clause 25.3.1 of the Code restricts maximum slenderness ratio of a column to 60. If in any given plane, one end of a column is unrestrained, its effective length l should not exceed.

2b100D

Where, b = width of cross-section D = depth of cross-section measured in the plane under consideration In the code, the effect of slenderness is approximated by using a moment magnifier approach, whereby the sum of the primary and secondary moment is used as total design moment. The additional or secondary moments are given by: May = (Pb/2000){(ley/b)2} And Max= (PD/2000){(lex/D)2} Where, lex= effective length in respect of major axis, that is, bending occurs about the major axis ley= effective length in respect of minor axis, that is, bending occurs about the minor axis


D = depth of cross-section at right angles to the major axis b = breadth of cross-section These expressions are applicable to a balance design of a slender column subjected to uniaxial loading as well as biaxial bending. As the axial load increases from zero, the tensile stress in the steel decreases to zero and changes to a compressive stress. As this occurs, the curvature and deflection decreases. Clause 39.7.1.1 of the code permits a reduction in the additional moments by a factor k given by:

z

z b

P Pk 1P P

−= ≤

−

zP = capacity of cross section under pure axial load

zP = z ck c y scP =0.45f A +0.75f A

bP = balance axial load corresponding to the condition of maximum compressive strain of 0.0035 in concrete and tensile strain of 0.002 in the outermost layer of tension steel.The value of bP will depend on arrangement of reinforcement and the cover ratio d’/D, in addition to the grades of concrete and steel. The values of bP can be determined using the following equations: For rectangular sections,

2b 1 ck

ck

q pP q f bD

= + σ For Circular sections,

22b 1 ck

ck

q pP q f D

= + σ

The coefficients 1q & 2q are given in tables.

5.11 DEPTH OF FOUNDATION • All type of foundation should have a

minimum depth of 50 cm as per IS 1080-1962. This minimum depth is required to ensure the availability of soil having the safe bearing capacity assumed in the design.

• Moreover, the foundation should be placed well below the level which will

not be affected by seasonal change of weather to cause swelling & shrinking of the soil. Further, frost also may endanger the foundation if placed at a very shallow depth. Rankine formula gives a preliminary estimate of the minimum depth of foundation and is expressed as

2cq 1-sind=γ 1+sin

φφ

Where d = minimum depth of foundation qc = gross bearing capacity of soil γ = density of soil φ = angle of repose of soil

• Though Rankine formula considers three major soil properties qc, γ and φ . It does not consider the load applied to the foundation. However, this may be a guideline for an initial estimate of the minimum depth which shall be checked subsequently for other requirements of the design.

5.12 DESIGN CONSIDERATION

(a) Minimum Nominal Cover (cl. 26.4.2.2 of IS 456)

The minimum nominal cover for the footings should be more than that of other structural elements of the superstructure as the footing are in direct contact with the soil. Clause 26.4.2.2 of IS 456 prescribes a minimum cover of 50 mm for footings. However, the actual cover may be ever more depending on the presence of harmful chemicals or minerals, water table etc.

(b)Thickness at the Edge of Footing (cls. 34.1.2 and 34.1.3 of IS 456)

The minimum thickness at the edge of reinforced and plain concrete footings shall be at least 150 mm for footings on soils and at least 300 mm above the top of piles for footing on piles, as pr the stipulation in cl. 34.1.2 of IS 456.

For plain concrete pedestals, the angle α between the plane passing through


the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal & the horizontal plane shall be determined from the following expression (cl. 34.1.3 of IS 456)

tanα ≥ 0.9 √{(100qa/fck)+1}

. . . . . (11.3) Where aq = calculated maximum bearing pressure at the base of pedestal in N/mm2, & ckf = characteristic strength of concrete at 28 days in N/mm2.

(c) Bending Moments (cl. 34.2 Of IS 456) 1. It may be necessary to compute the

bending moment at several section of the footing depending on the type of footing, nature of loads and the distribution of pressure at the base of the footing. However, bending moment at any section shall be determined taking all forces acting over the entire area on one side of the section of the footing which is obtained by passing vertical plane at that section extending across the footing (cl. 34.2.3.1 of IS 456.)

2. The critical section of maximum bending moment for the purpose of designing an isolated concrete footing which supports a column, pedestal or wall shall be: i) At the face of the column, pedestal

or wall for footing supporting a concrete column, pedestal or reinforced concrete wall, Figs (a_), (b) and (i), and

ii) Halfway between the centre-line and the edge of the wall, for footing under masonry wall (Fig. (i). this is stipulated in cl. 34.2.3.2. of IS 456.

The maximum moment at the critical section shall be determined as mentioned in 1 above.

For round or octagonal concrete column or pedestal, the face of the column or pedestal shall be taken as the side of a square inscribed within the perimeter of the round or octagonal column or pedestal (see cl. 34.2.2 of IS 456 and Figs. (a) and (b).

d) Shear Force (cl. 31.6 & 364.2.4 of IS 456) Footing slabs shall be checked in one-way or two-way shears depending on the nature of bending. If the slab bends primarily in one-way, the footing slab shall be checked in one-way vertical shear. On the other hand, when the bending is primarily two-way, the footing slab shall be checked in two-way shear or punching shear. The respective critical sections and design shear strength are given below: 1. One – way Shear (cl. 34.2.4 of IS 456) One-way shear has to be checked across the full width of the base slab on a vertical section located from the face of the column, pedestal or wall at a distance equal to (Figs. (a), (b) and (i). i) Effective depth of the footing slab in

case of footing slab on soil, and ii) Half the effective depth of the footing

slab if the footing slab is on piles Fig. (k). The design shear strength of concrete without shear reinforcement is given in Table 19 of Cl. 40.2 of IS 456.

2. Two – way or Punching Shear (cls. 31.6 and 34.2.4) Two-way or punching shear shall be checked around the column on a perimeter half the effective depth of the footing slab away from the face of the column or pedestal (Figs. (a) and (b). The permissible shear stress, when shear reinforcement is not provided, shall not


exceed ksTc, where ks = ( )c0.5+β , but not greater than one, cβ being the ratio of short side to long side of the column, and Tc =

0.25 ( )12

ckf

in limit state method of design,

as stipulated in cl. 31.6.3 of IS 456. Normally, the thickness of the base slab is governed by shear. Hence, the necessary thickness of the slab has to be provided to avoid shear reinforcement. e) Bond (cl. 34.2.4.3 of IS 456) The critical section for checking the development length in a footing slab shall be the same planes as those of bending moments in part (c) of this section. Moreover, development length shall be checked at all other sections where they change abruptly. The critical sections for checking the development length are given in cl. 34.2.4.3 of IS 456, which further recommends to check the anchorage requirements if the reinforcement is curtailed, which shall be done in accordance with cl. 26.2.3 of IS 456. f) Tensile Reinforcement (cl.34.3 of IS 456) The distribution of the total tensile reinforcement, calculated in accordance with the moment at critical sections, as specified in part (c) of this section, shall be done as given below for one-way and two-way footing slabs separately. i) In one-way reinforced footing slabs like

wall footings, the reinforcement shall be distributed uniformly across the full width of the footing i.e., perpendicular to the direction of wall. Nominal distribution reinforcement shall be provided as per cl. 34.5 of IS 456 along the length of the wall to take care of the secondary moment, differential settlement, shrinkage and temperature effects.

ii) In two-way reinforced square footing slabs, the reinforcement extending in each direction shall be distributed

uniformly across the full width/length of the footing.

iii) In two-way reinforced rectangular footing slabs, the reinforcement in the long direction shall be distributed uniformly long direction shall be distributed uniformly across the full width of the footing slab. In the short direction a central band equal to the width of the footing shall be marked along the length of the footing, where the portion of the reinforcement shall be determined as given in the equation below. This portion of the reinforcement shall be distributed across the central band:

Reinforcement in the band = {2/β +1} (Total reinforcement in the short direction) where β is the ratio of longer dimension to shorter dimension of the footing slab. Each of the two end bands shall be provided with half of the remaining reinforcement, distributed uniformly across the respective end band.

g)Transfer of Load at the base of Column (cl. 34.4 of IS 456) All forces and moments acting at the base of the column must be transferred to the pedestal, if any, and then from the base of the pedestal to the footing, (or directly from the base of the column to the footing if there is no pedestal) by compression in concrete and tension in steel. Compression forces are transferred through direct bearing while tension forces are transferred through developed reinforcement. The permissible bearing stresses on full area of concrete shall be taken as given below from cl. 34.4 of IS 456 br ck:σ =0.25f , in working stress method, and . . . . . (i)

br ckσ =0.45f , in limit state method . . . . (ii) The permissible bearing stress of concrete in footing is given by (cl. 34.4 of IS 456):

12

br ck 1 2σ =0.45f (A /A )


With a condition that ( )1

1 2 2A / A 2.0≤

Where 1A = maximum supporting area of footing for bearing which is geometrically similar to and concentric with the loaded area 2A , as shown in Fig.

2A = loaded area at the base of the column. The above clause further stipulates that in sloped or stepped footings, 1A may be taken as the area of the lower base of the largest frustum of a pyramid or cone contained wholly within the footing and having for its upper base, the area actually loaded and having side slope of one vertical to two horizontal, as shown below: i) Sufficient development length of the

reinforcement shall be Provided to transfer the compression or tension to the supporting member in accordance with cl. 26.2 of IS 456, when transfer of force is accomplished by reinforcement of column (cl. 34.4.2 of IS 456.)

ii) Minimum area of extended longitudinal bars or dowels shall be 0.5 percent of the cross sectional area of the supported column or pedestal (cl. 34.4.3 of IS 456.)

iii) A minimum of four bars shall be provided (cl. 34.4.3 of IS 456.)

iv) The diameter of dowels shall not exceed the diameter of column bars by more than 3 mm.

v) Column bars of diameter larger than 36 mm, in compression only can be doweled at the footings with bars of smaller size of the necessary area. The dowel shall extend into the column, a distance equal to the development

length of the column bar and into the footing, a distance equal to the development length of the dowel, as stipulated in cl. 34.4.4 of IS 456 and as shown in Fig.


6.1 LOSSES IN PRESTRESS

• In prestressed concrete application, themost important variable is theprestressing force. In the early days, itwas observed that the prestressingforce does not stay constant, butreduces with time.

• Even during prestressing of the tendonsand the transfer of prestress to theconcrete member, there is a drop of theprestressing force from the recordedvalue in the jack guage.

• The various reductions of theprestressing force are termed as thelosses in prestress.

• The losses are broadly classified intotwo groups, immediate & time-dependent.

• The immediate losses occur duringprestressing of the tendons and thetransfer of prestress to the concretemember.

• The time-dependent losses occurduring the service life of prestressedmember.

• The losses due to elastic shortening ofthe member, friction at the tendon-concrete interface and slip of theanchorage are the immediate losses.

• The losses due to the shrinkage andcreep of the concrete and relaxation ofthe steel are the time dependent losses.

• The various losses in prestress areshown in the following chart.

The different types of losses encountered in pretensioning and post tensioning are as given below.

Pre-tensioning Post-tensioning 1. Elastic deformation of concrete

1. No loss due to elasticshortening when all bars are simultaneously tensioned. If however, wires are successively tensioned there would be loss of prestress due to elastic deformation of concrete

2. Relaxation ofstress in steel

2. Relaxation of stress insteel

3. Shrinkage ofConcrete

3. Shrinkage of Concrete

4. Creep of concrete

4. Creep of concrete

5. Frictional losses5. Anchorage slip

Note: The total losses are around 15 to 20% of the initial prestressing. • In addition to the above, there may be

losses due to sudden changes intemperature, especially in steam curingof pretensioned units.

• The rise in temperature causes a partialtransfer of prestress (due to elongationof the tendons between adjacent unitsin the long – line process) which may

6 BASIC ELEMENT OF PRESTRESS CONCRETE


cause a large amount of creep if the concrete is not properly cured.

• If there is a possibility of a change of temperature between the times of tensioning and transfer, the corresponding loss should be allowed for in the design.

6.2LOSS OF PRESTRESS DUE TO FRICTION • The friction generated at the interface

of concrete and steel during the stretching of a curved tendon in a post-tensioned member, leads to a drop in the prestress along the member from the stretching end.

• The loss due to friction does not occur in pre-tensioned members because there is no concrete during the stretching of the tendons.

• The friction is generated due to the curvature of the tendon and the vertical component of the prestressing force.

• In addition to friction, the stretching has to overcome the wobble of the tendon. The wobble refers to the change in position of the tendon along the duct. The losses due to friction and wobble are grouped together under friction.

• Force in cable at a distance x from jacking end, after frictional loss – Px

Px = Poe-(µα+kx)

Where Px = Prestressing force at a distance x from jacking end.

P0 = Prestressing force at jacking end. k = coefficient called wobble correction

factor μ = Coefficient for friction in curve α = Cumulative angle in radian through

which the tangent to the cable profile turned between any two point under consideration.

• For small values ofμα+kx , the above expression can be simplified by the Taylor series expansion. Px = Po [1-(µα+kx)]

• Thus, for a tendon with single curvature, the variation of the prestressing force is linear with the distance from the stretching end. The following figure shows the variation of prestressing force.

6.3 LOSS OF PRESTRESSE DUE ANCHORAGE SLIP • In a post-tensioned member, when the

prestress is transferred to the concrete, the wedges slip through a little distance before they get properly seated in the conical space.

• The anchorage block also moves before it settles on the concrete.

• There is loss of prestress due to the consequent reduction in the length of the tendon.

• The total anchorage slip depends on the type of anchorage system. In absence of manufacture’s data, the following typical values for some systems can be used.

This loss due to anchorage slip = sE ΔL

= sΔ EL

sE = Young modulus of steel in N/mm2

∆ = Anchorage slip in mm L = Length of cable in mm

Table:- Typical values of anchorage slip Anchorage system Anchorage

slip (∆ )

Freyssinet 4 mm 12-5mm φ strands 6 mm 12-8 mm φ strands 8 mm Magnel 1 mm Dywidag system

• Since the loss of stress is caused by a definite total amount of shortening. The


percentage loss is higher for short members than for long ones.

• With the long line pretensioning system, the slip at the anchorage is normally very small in comparison with the length of the tensioned wire and hence is generally ignored.

• While prestressing a short member, due care should be taken to allow for the loss of stress due to anchorage slip, which forms a major portion of the total loss.

EXAMPLE: In a post tensioned beam the cable is subjected to stress of 1200 N/mm2. If the slip at the jacking end is found to be 4mm, find the percentage loss of stress due to this case if the beam is 25 m long. Take sE = 210 kN/mm2. Solution: Loss of stress due to anchorage slip

sΔ 4= E =L 25×1000

⨉210 ×103

= 33.6 N/mm2

%loss = 33.6 1001200

× = 2.8%

6.4LOSS OF PRESTRESS DUE TO CREEP OF CONCRETE Creep is the property of concrete by which it continues to deform with time under sustained loading. Creep coefficient is defined as

φ = creep strainelastic strain

= cp

c

εε

cp cε = εφ ……(1)

εc = fc/Ec Loss of stress cp s=ε ×E

= c ss

c c

f E. ×E m=E E

φ

Loss of stress = mφ fc

• Note that elastic shorting loss multiplied by creep co-efficient is equal to loss due to creep. Age at loading Creep co-efficient 7 days 2.2 28 days 1.6 1 year 1.1

• Intermediate values of creep coefficient may be interpolated by assuming that creep coefficient decreases linearly with the log of time in days.

For example creep coefficient for 15 days as age of loading is

= 1.6 + 10 10

10 10

(2.52 1.6) (log 28 log 15)log 28 log 7

− −−

• Creep losses are generally 2-3% of initial prestressing force.

EXAMPLE A prestressed concrete beam of rectangular section 120 mm wide and 300 mm deep is prestressed by 6 wires of 6 mm diameter, provided at an eccentricity of 55 mm. The initial stress in the wires is 1150 N/mm2. Find the loss of stress in steel due to creep of concrete. Take 5 2

sE =2×10 N/mm , 4 2

cE =3×10 N/mm , =1.50φ Solution : Modular ratio an

= s

c

EE

= 5

42 10 203 10 3×

=×

Area of the beam section A = 120 × 300 = 36000 mm2. Moment of Inertia of the beam section,

I = 3120 300

12× = 2.70 × 108 mm4.

Prestressing force = P = 6 ×4π× 62× 1150

= 195093 N Stress in concrete at the level of steel,

( )22

c 8

195093× 55Pe 195093f = = +I 36000 2.70×10

= 7.60 N/mm2. • Loss of stress in steel due to creep of

concrete=φ m fc = 1.50 203

×7.60 N/mm2.


= 76 N/mm2. 6.5LOSS DUE TO SHRINKAGE OF CONCRETE The shrinkage of concrete in prestressed members results in a shortening of tensioned wires and hence contributes to the loss of stress. • In the case of pre-tensioned members,

generally moist curing is resorted to in order to prevent shrinkage until the time of transfer.

• Consequently, the total residual shrinkage strain will be larger in pretensioned members after transfer of prestress in comparision with post-tensioned members, where a portion of shrinkage will have already taken place by the time of transfer of stress.

• This aspect has been considered in the recommendations made by the Indian standard code (IS: 1343) for the loss of prestress due to the shrinkages of concrete and as detailed below.

csε = total residual shrinkage strain having values of 3 × 10−4 for pre-tensioning and

εcs = [ (2⨉10-4)/log10(t+2)] for post-tensioning Where, t = age of concrete at transfer in

days. • The value may be increased by 50

percent in dry atmospheric conditions, subject to a maximum value of 3 × 10−4 units.

The loss of stress in steel due to the shrinkage of concrete is estimated as,

Loss of stress = cs sε ×E Where Es = modulus of elasticity of steel.

EXAMPLE A concrete beam is prestressed by a cable carrying an initial prestressing force of 300 KN. The cross-sectional area of the wires in the cable is 300 mm2. Calculate the percentage loss of stress in the cable only

due to shrinkage of concrete using IS: 1343 recommendations assuming the beam to be, (a) pre-tensioned and (b) post-tensioned. Assume Es = 210 KN/mm2 and age of concrete at transfer = 8 days. Solution:

Initial stress in wires = 3300 10

300 ×

= 1000 N/mm2.

a) If the beam is pre-tensioned, the total residual shrinkage strain = 300 × 10–6 units.

∴ Loss of stress = (300 × 10–6) (210 × 103) = 63 N/mm2. ∴ Percentage loss of stress

= 63 1001000 ×

= 6.3%

b) If the beam is post-tensioned, the total shrinkage strain

= 6

10

200 10log (8 2)

− × +

= 200 × 10–6 units

∴ Loss of stress = (200 × 10–6) (210 × 103) = 42 N/mm2. ∴ Percentage loss of stress = 42 100

1000 ×

= 4.2%

6.6LOSSS OF PRESTRESS DUE TO RELAXATION OF STEEL • Relaxation of steel is defined as the

decrease in stress with time under constant strain.

• Due to the relaxation of steel, the prestress in the tendon is reduced with time.

• The relaxation depends on the type of steel, initial prestress (fpi) and the temperature.

• To calculate the drop (or loss) in prestress (∆ fp), the recommendations of IS: 1343 – 1980 can be followed in absence of test data.


Table: Relaxationlosses for prestressing steel at 1000 H at 270 C (as per IS 1343 – 1980

Initial Stress (1) Relaxation Loss N/mm2 0.5 fp 0 0.6 fp 35 0.7fp 70 0.8 fp 90

Note fp is the characteristic strength of prestressing steel. Final conclusion of above discussions:-

Sr. No.

Type of loss Equation

1 Wobble & curvature effect

(µα + kx)P0

2 Anchorage slip Es∆ /L 3 Shrinkage loss ε Sc. Es 4 Creep of concrete mφ fc 5 Elastic shortening of

concrete mfc

6 Relaxation in steel 2 to 5% for initial stress in steel.

Type of loss Pretensione

d (%) Post tensioned (%)

Elastic shorting of conc.

3 1

Shrinkage 7 6 Creep 6 5 Relaxation 2 3 Total Loss 18% 15%

Losses Pretensioning Post

tensioning Length effect No Yes

Curvature effect No Yes Anchorage slip No Yes

Shrinkage of concrete

Yes Yes

Creep of concrete Yes Yes Elastic

deformation or shortening of

concrete

Yes No (If all wires are

simultaneously tensioned) Yes

(If wires are

successively tensioned)

6.7 DEFLECTION OF RESTRESSED BEAM

• Short term deflection under uncracked condition in a prestressed beam can be computed using the elastic theory. In this we account for dead load, live load (under service load condition) and moment due to eccentricity of prestressing cable.

• BM at any section due to all such loading can be taken and BMD can be made. With the help of this, deflection can be computed as done in elastic theory by talking complete x-section to be effective.

• Computation of deflection can be conveniently done using area moment theorem (Mohr’s theorem.)

6.7.1 Effect of tendon profile on deflection • In most of the cases of prestressed

beams, tendons are located with eccentricities towards the soffit of beams to counteract the sagging bending moments due to transverse loads.

• Consequently, the concrete beam deflect upwards (camber) on the application or transfer of prestress. Since the bending moment at every section is the product of the prestressing force and eccentricity, the tendon profile itself will represent the shape of the BMD.

• The method of computing deflection of beams with different cable profiles is outlined below.

1. Straight tendons : Figure below shows a beam with a straight tendon at a uniform eccentricity below the centroidal axis,if upward deflection are considered as negative and, P = effective prestressing force E = eccentricity L = length of the beam a = -(Pe/2)L(L/4)/(EI) = -(PeL2/8EI)


2. Trapezoidal tendons : A draped tendon with a trapezoidal profile is shown in Fig. Considering the B.M.D., the deflection at the centre of the beam is obtained by taking the moment of area of the B.M.D. over one – half of the span. Thus, a = - (𝑃𝑃𝑃𝑃

𝐸𝐸𝐸𝐸)[ L2 (L1+ L2/2)+ (L1/2)(2/3L1)]

= -(Pe/6EI)(2L12+6L1L2+3L2

2)

3. Parabolic tendons (central anchors) : The deflection of a beam with parabolic tendons having an eccentricity e at the centre and zero at the support is given by,

2Pe 2 L 5 L 5PeLα=- × × × =-EI 3 2 8 2 48EI

4. Parabolic tendons (eccentric anchors): Fig, below shows a beam, with a parabolic tendons having an eccentricity e1 at the centre of span and e2 at the support section. • The resultant deflection at the centre is

obtained as the sum of the upward deflection of a beam with a parabolic tendon of eccentricity (e1 + e2) at the centre and zero at the supports and the downward deflection of a beam subjected to a uniform sagging bending moment of intensity Pe2 throughout the length.

• Consequently, the resultant deflection becomes,

( )2 2

1 2-5 PL PeLa= e +e +48 EI 8EI

( )2

1 2PLa= -5e +e48EI

5. Sloping tendons (eccentric anchors) : From fig below the deflection is computed in a way similar to method 4 above. Thus


( )22

21 2

Pe LPLA= - e +e +12EI 8EI

= ( )2

1 2PL -2e +e24EI

Deflections due to self – weight and imposed loads. • At the time of transfer of prestress, the

beam hogs up due to the effect of prestressing.

• At this stage, the self-weight of the beam induces downwards deflections, which further increase due to the effect of imposed loads on the beam.

If g = self-weight of the beam/m q = imposed load/m (uniformly

distributed). The downward deflection is computed

as,

45(g+q)La=

384EI Deflection due to concentrated live loads can be directly computed by using Mohr’s theorems.

EXAMPLE 1 A concrete beam with a rectangular section 300 mm wide and 500 mm deep is prestressed by 2 post tensioned cables of area 600 mm2 each. Initially stressed to 1600 N/mm2. The cables are located at a constant eccentricity of 100 mm throughout the length of the beam having a span of 10m. The modulus of elasticity of steel and concrete is 210 and 38 kN/mm2. Respectively.

a) Neglecting all losses, find the deflection at the centre of span when it is supporting its own weight.

b) Allowing for 20 percent loss in prestress, find the final deflection at the centre of span when it carries an imposed load of 18 kN/m. Dc = 24 kN/m3.

Solution: Self-weight of the beam, g = (0.3 ×0.5×24) = 3.6 kN/m = 0.0036 kN/mm

Second moment of area, (I) 3300×500=

12

= 3125 × 106 mm4

Prestressing force, P = (2 × 600 × 1600) = 1920 × 103 N = 1920 kN Downward deflection due to self-weight =

45gL384EI

= 4

6

5 0.0036 (10 1000)384 8 3125 10

× × × × × × 3

= 3.95 mm Upward deflection due to the prestressing force

= 2PeL

8EI

= 2

6

1920 100 (10 1000)8 8 3125 10

× × × × × × 3

= 20.2 mm Net upward deflection of the beam, when it support its own wt = (20.2 – 3.9) = 16.3 mm Downward deflection due to live load

= 3.9 183.6×

= 19.5 mm

Net upward deflection due to prestress after losses = (0.8 × 20.2) = 16.16 mm. Final downward deflection of the beam due to (self – weight + prestress + live load) = (3.95 – 16.16 + 19.50) = 7.29 mm.

6.8 PRESTRESSED CONCRETE

• Prestressed concrete is basically a concrete in which internal stress of suitable magnitude and distribution are introduced so that the stresses resulting from external load are counteracted to a desired degree.

• A prestressed concrete is different from a conventional RCC structure due to the


application of an initial load on the structure prior to its use.

• For concrete internal stress are induced (usually by mean of tensioned steel) due to the following reasons.

1. Tensile strength of concrete is less and cracks may develop at early stages of loading in flexural members such as beams and slabs. Hence if the members are compressed prior to application of flexural load, the cracking will not occur.

2. Prestressing enhances the bending, shear and torsional capabilities of the flexural members.

3. In pipes and liquid storage tanks the hoop tensile stresses can be effectively counteracted by circular prestressing.

How do we apply prestress by means of tensioned wire?

There are two ways (1) Pretensioning (2) Post tensioning. 6.8.1 Pre-tensioning

• In pre-tensioning system, the high-

strength steel tendons are pulled between two end abutments (also called bulkheads) prior to the casting of concrete.

• The abutments are fixed at the ends of a prestressing bed.

• Once the concrete attains the desired strength for prestressing, the tendons are cut loose from the abutments.

• The prestress is transferred to the concrete from the tendons, due to the bond between them.

• The stages of prestensioning are shown schematically in the following figures.

6.9 REQUIRMENT OF HIGH STRENGTH STEEL & CONCRETE IN PRESTRESSING 6.9.1 Need for high strength steel

Prestressed concrete requires high strength steel and high strength concrete. This can be explained as follows:-

Strain in cable initially = s

Δl

Stress in cable initially = Ess

Δl

(Force in cable initially ss

s

E Δ .A =Pl

= prestressing force)

(Loss of strain in cable due s

δ=l

to

shortening of concrete over a period of time)

Loss of stress in cable= ss

δE .l

Loss of prestressing force s ss

δ=E . .Al

Normally due to creep and shrinkage in long term the strain lost is approximately 0.0008 Stress lost = 0.0008 × 2 × 105 = 160 N/mm2.If we use Fe250 or F415, all of the initial stress in it will be lost in due course. Hence there would not be any pressuressing force remaining in concrete, thus the beam will fail.


Hence we use high strength steel such that the initial prestress in it would be 1200 – 2000 N/mm2 in which the loss would be around 200 N/mm2. 6.9.2 Need for High Strength Concrete in Pre-stress Concrete • High strength concrete offers high

resistance to tension, shear, bond and bearing.

• In case of pre-tensioned members, tensile stress in steel of very high magnitude should be transferred to concrete as prestress through bonding between steel and surrounding concrete.

• In post-tensioned members transfer of stress is through bearing at end sector. Hence concrete of appreciable bond and bearing strength is quite essential for pre-stressed concrete.

• In addition to above, and bearing strength is quite essential for pre-stressed concrete. 1) High strength concrete is less liable

to shrinkage cracks and has higher modulus of elasticity and smaller ultimate creep strain. As a result loss of prestress in steel is reduced.

2) Use of high strength concrete results in reduction of cross sectional dimensions of prestressed concrete structural elements, with reduced dead weight longer spans becomes economically and practically viable.


Topics Page No

1. WORKING SRESS & LIMIT STATE METHOD 11

2. SHEAR, TORSION, BOND, ANCHORAGE & DEVELOPMENT LENGHTH 656

3. FOOTING, COLUMNS, BEAMS AND SLABS 44

4. PRESTRESSED CONCRETE 11

GATE QUESTIONS


Q.1 The following two statements are made with reference to a simply supported under reinforced RCC beam: I. Failure takes place by crushing

of concrete before the steel has yielded.

II. The neutral axis moves up as theload is increased.

With reference to the above statements, which of the following applies? a) Both the statements are falseb) I is true but II is falsec) Both the statements are trued) I is false but II is true

[GATE-2001]

Q.2 Maximum strains in an extreme fibre in concrete and in the tension reinforcement (Fe-415 grade and Es-200kN/mm2) in a balanced section at limit state of flexure are respectively a) 0.0033 and 0.0038b) 0.002 and 0.0018c) 0.0035 and 0.0041d) 0.002 and 0.0031

[GATE-2003]

Q.3 List-I contains some properties of concrete cement and List-II contains list of some tests on concrete/cement Match the property with the corresponding test. List-I A. Workability of commie B.Direct tensile strength of coma-etc C. Bond between concrete and steel D. Fineness of cement List-II 1. Cylinder splitting lest2. Tiee-Bee test3. Surface area test

4. Fineness modulus test5. Pull out. Lest

Codes: A B C D

a) 2 l 5 3 b) 4 5 1 3 c) 2 1 5 4 d) 2 5 1 4

[GATE-2003]

Q.4 For avoiding the limit state of collapse, the safety of RC structures is checked for appropriate combinations of Dead Load (DL), imposed Load or Live Load (M), Wind Load (WL,) and Earthquake Load (EL). Which of the following load combinations is NOT considered? a) 0.9 DL + 1.5 WLb) 1.5 DL + 1.5 WLc) 1.5 DL + 1.5 WL + 1.5 ELd) 1.2 DL + 1.2 IL + 1.2 WL

[GATE-2004]

Q.5 In the limit state design method of concrete structure, the recommended partial material safely factor (yn) for steel according to IS: 436-2000 is a)1.5 b)1.13 c)1.00 d)0.87

[GATE-2004]

Q.6 The partial factor of safety for concrete as per IS : 456-2000 is a) 1.50 b) 1.15c) 0.87 d) 0.446

[GATE-2005]

Q.7 The flexural strength of M30 concrete as per IS: 456-2000 is a) 3.83 MPa b) 5.47 MPa

1 WORKING SRESS & LIMIT STATE METHOD


c) 21.23 MPa d) 30.0 MPa [GATE-2005]

Q.8 In a random sampling procedure for

cube strength of concrete, one sample consists of X number of specimens. These specimens are tested at 28 days and average Strength of these specimens is considered as test result of the sample, provided the individual variation in the strength of specimens is not more than ∓ Y 5 percent of the average strength. The values of X and Y as per IS : 456 – 2000 are

a) 4 and 10 respectively b) 3 and 10 respectively c) 4 and 15 respectively d) 3 and 15 respectively

[GATE-2005]

Q.9 As per IS : 456 – 2000, consider the following statements: 1. The modular ratio considered in

the working stress method depends on the type of steel used.

2. There is an upper limit on the nominal shear stress in beams (even with shear reinforcement) due to the possibility of crushing of concrete in diagonal compression.

3. A rectangular slab whose length is equal to its width may not be a two way slab for some support conditions.

The TRUE statements are a) 1 and 2 b) 2 and 3 c) 1 and 3 d) 1, 2 and 3

[GATE-2006]

Q.10 Consider the following statements: 1. The compressive strength of

concrete decreases with increase in water-cement ratio of the mix.

2. Water is added to the for hydration of workability.

3. Creep and shrinkage are independent of the water –cement ratio in the concrete mix.

The TRUE statements are a) 1 and 2 b) 1, 2 and 3 c) 2 and 3 d) only 2

[GATE-2007]

Q.11 A reinforced concrete structure has to be constructed along a sea coast. The minimum grade of concrete to be used as per IS : 456-2000 is

a) M 15 b) M 20 c) M 25 d) M 30

[GATE-2008] Q.12 Un-factored maximum bending

moments at a section of a reinforced concrete beam resulting from a flame analysis are 50, 80, 120 and 180 kN-m under dead, live, wind and earthquake loads respectively. The design moment (kNm) as per IS: 456-2000 for the limit state of collapse (flexure) is

a) 195 b) 250 c) 345 d) 372

[GATE-2008]

Q.13 The modulus of rapture of concrete terms of its characteristic cube compressive strength (fck) in MPa according to IS 45 6:2000 is

a) 50000 fck b) 0.7 fck

c) 50000 ckF d) 0.7 ckF [GATE-2009]

Q.14 For limit state of collapse, the partial safety- factors recommended By IS 456:2000 for estimating the design strength of concrete and reinforcing steel are respectively

a) 1.15 and 1.5 b) 1.0 and 1.0 c) 1.5 and 1.15 d) 1.5 and 1.0

[GATE-2009]

Q.15 Match List-I (List of test methods for evaluating properties of concrete) with


List-II (List of properties) and select the correct answer using the codes given below the lists:

List-I A. Resonant frequency test B. Rebound hammer test C. Split cylinder teat D. Compacting factor lest List - II 1. Tensile strength 2. Dynamic modulus of elasticity 3. Workability 4. Compressive strength Codes:

A B C D a) 2 4 1 3

b) 2 1 4 3 c) 2 4 3 1 d) 4 3 1 2

[GATE-2009]

Q.16 As per IS 456:2000, in the Limit Suite Design of a flexural member, the swain in reinforcing bars under tension at ultimate state should not be haw than

a) f/E b) f/E + 0.002 c) f/1.15E d) f/1.15E + 0.002

[GATE-2012]

Q.17 Maximum possible value of Compacting Factor for fresh (green) concrete is:

a) 0.5 b) 1.0 c) 1.5 d) 2.0

[GATE-2013]

Q18. Match the information given in Group – I with those in Group II.

Group - I Group - II (p)

Factor to decrease ultimate strength to design strength

(1) Upper bound on ultimate load

(q)

Factor to increase working load to ultimate load for design

(2) Lower bound on ultimate load

(r)

Statical method of ultimate load analysis

(3) Material partial safety factor

(s)

Kinematical mechanism method of

(4)Load factor

ultimate load analysis

a) P – 1; Q – 2; R – 3; S – 4 b) P – 2; Q – 1; R – 4; S – 3 c) P – 3; Q – 4; R – 2; S – 1 d) P – 4; Q – 3; R – 2; S – 1

[GATE-2014] Q19. While designing, for a steel column

of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/mm2) in limit state method of design as per IS:456-2000 is _________

[GATE-2014]

Q.20. The first moment of area about the axis of bending for a beam cross-section is a) moment of inertia b) section modulus c) shape factor d) polar moment of inertia

[GATE-2014]

Q.21. Group I contains representative stress-strain curves as shown in the figure, while Group II gives the list of materials. Match the stress-strain curves with the corresponding materials.

Group I Group II (p) Curve J (1) Cement paste (q) Curve K (2) Coarse aggregate (r) Curve L (3) Concrete a) P-1; Q-3; R–2 b) P-2; Q-3; R-1 c) P-3; Q-1; R-2 d) P-3; Q-2; R-1

[GATE-2014]

Q.22. The target mean strength fcm for concrete mix design obtained from the characteristic strength fck and standard deviation σ, as defined in IS:456-2000, is a) ckF +1.35σ b) ckF +1.45σ


c) ckF +1.55σ d) ckF +1.65σ [GATE-2014]

Q.23 The flexural tensile strength of M25

grade of concrete, in N/mm2, per IS:456-2000 is ________

[GATE-2014]

Q.24 The modulus of elasticity, E = 5000 ck F where ck F is the

characteristic compressive strength of concrete, specified in IS:456-2000 is based on a) tangent modulus b) initial tangent modulus c) secant modulus d) chord modulus

[GATE-2014]

Q.25 In a reinforced concrete section, the stress at the extreme fibre in compression is 5.80 MPa. The depth of neutral axis in the section is 58 mm and the grade of concrete is M25. Assuming linear elastic behavior of the concrete, the effective curvature of the section (in per mm) is a) 2.0×10-6 b) 3.0×10-6

c) 4.0×10-6 d) 5.0×10-6 [GATE-2014]

Q.26 Workability of concrete can be

measured using slump, compaction factor and Vebe time. Consider the following statements for workability of concrete: (i) As the slump increases, the Vebe

time increases (ii) As the slum increases, the

compaction factor increases Which of the following is TRUE? a) Both (i) and (ii) are True b) Both (i) and (ii) are False c) (i) is True and (ii) is False d) (i) is False and (ii) is True

[GATE-2015]

Q.27 Consider the singly reinforced beam shown in the figure below:

At cross-section XX, which of the following statement is TRUE at the limit state? a) The variation of stress is linear

and that of strain is non-linear b) The variation of strain is linear

and that of stress is no-linear c) The variation of both stress and

strain is linear d) The variation of both stress and

strain is non-linear [GATE-2015]

Q.28 Consider the following statements

for air-entrained concrete: (i) Air-entrainment reduces the

water demand for a given level of workability

(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected.

Which of the following is TRUE? a) Both (i) and (ii) are True b) Both (i) and (ii) are False c) (i) is True and (ii) is False d) (i) is False and (ii) is True

[GATE-2015]

Q.29 The composition of an air-entrained concrete is given below: Water : 184kg/m3 Ordinary Portland Cement(OPC) : 368 kg/ m3 Sand : 606kg / m3 Coarse aggregate : 1155 kg/m3 Assume the specific gravity of OPC, sand and coarse aggregate to be


3.14, 2.67 and 2.74, respectively. The air content is _________liters/ m3.

[GATE-2015]

Q.30 Consider the singly reinforced beam section given below (left figure). The stress block parameters for the cross-section from IS:456-2000 are also given below (right figure). The moment of resistance for the given section by the limit state method is __________kN-m.

[GATE-2015]

Q.31 According to the concept of Limit State Design as per IS456: 2000, the probability of failure of a structure is ________

[GATE-2015]

Q.32 A reinforced concrete (RC) beam with width of 250 mm and effective depth of 400 mm is reinforced with Fe415 steel. As per the provisions of IS 456-2000, the minimum and maximum amount of tensile reinforcement (expressed in mm2) for the section are, respectively a) 250 and 3500 b) 205 and 4000 c) 270 and 2000 d)300 and 2500

[GATE-2016] Q.33 For M25 concrete with creep

coefficient of 1.5, the long-term static modulus of elasticity (expressed in M Pa) as per the provisions of IS:456-2000 is _____

[GATE-2016]

Q.34 Let the characteristic strength be

defined as that value, below which not more than 50% of the results are expected to fall. Assuming a standard deviation of 4 MPa, the target mean strength (in MPa ) to be considered in the mix design of a M25 concrete would be

a) 18.42 b) 21.00

c) 25.00 d) 31.58

[GATE-2017]

Q.35 Group I gives a list of test methods and test apparatus for evaluating some of the properties of Ordinary Cement (OPC) and concrete. Group II gives the list of these properties

Group I Group II

P. Le Chatelier test 1. Soundness of OPC

Q. Vee- Bee test 2. Consistency and setting time of OPC

R. Blaine air permeability test

3. Consistency or workability of concrete

S. The Vicat apparatus 4. Fineness of OPC

The correct match of the items in Group I with the items in Group II is

A) P-1, Q-3, R-4, S-2 C) P-4,Q-2, R-4, S-1 B) P-2, Q-3, R-1, S-4 D) P-1,Q-4, R-2, S-3

[GATE-2017]

Q.36 According to IS 456:2000, which one of the following statements about the depth of neutral axis xu,bal for a balanced reinforced concrete section is correct?


A) xu,bal depends on the grade of concrete only.

B) xu,bal depends on the grade of steel only. C) xu,bal depends on both the grade of

concrete and grade of steel . D) xu,bal does not depend on both the

grade of concrete and grade of steel . [GATE-2017]

Q.37 The Le Chatelier apparatus is used to determine a) compressive strength of cement b) fineness of cement

c) setting time of cement d) soundness of cement

[GATE-2018] Q.38 The setting time of cement is

determined using a) Le Chatelier apparatus

b) Briquette testing apparatus c) Vicat apparatus

d) Casagrande’s apparatus [GATE-2018]

Q.39 The deformation in concrete due to

sustained loading is a) creep

b) hydration c) segregation

d) shrinkage [GATE-2018]

Q.40 As per IS 456:2000, the minimum

percentage of tension reinforcement (up to two decimal places ) required in reinforced- concrete beams of rectangular cross-section ( considering effective depth in the calculation of area ) using Fe500 grade steel is _______________

[GATE-2018]

Q.41 A reinforced-concrete slab with

effective depth of 80 mm is simply

supported at two opposite ends on 230 mm thick masonry walls. The centre-to-centre distance between the walls is 3.3 m. As per IS 456:2000, the effective span of the slab (in m, up to two decimal places) is __________

[GATE-2018]

Q.42 A singly-reinforced rectangular

concrete beam of width 300 mm and effective depth 400 mm is to be designed using M25 grade concrete and Fe500 grade reinforcing steel. For the beam to be under-reinforced, the maximum number of 16 mm diameter reinforcing bars that can be provided is

a) 3 b) 4 c) 5 d) 6

[GATE-2018] Q.43 Two rectangular under-reinforced

concrete beam sections X and Y are similar in all aspects except that the longitudinal compression reinforcement in section Y is 10% more. Which one of the following is the correct statement?

a) Section X has less flexural strength and is less ductile than section Y

b) Section X has less flexural strength but is more ductile than section Y

c) Section X and Y have equal flexural strength but different ductility

d) Section X and Y have equal flexural strength and ductility

[GATE-2018]

Q.44 The frequency distribution of the compressive strength of 20 concrete cube specimens is given in the table


f(Mpa) Number of specimens with

compressive strength equal to f

23 4 28 2

22.5 5 31 5 29 4

If μ is the mean strength of the specimens and σ is the standard deviation, the number of Specimens (out of 20) with compressive strength less than μ-3σ is ________

[GATE-2018]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (b) (a) (a) (c) (c) (b) (d) (d) (d) (a) (d) (d) (a) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (c) (b) (b) (c) 9 (b) (b) (d) 3.5 (b) (c) (d) (b) (a) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 51 42.82 0.097 (b) 10000 (c) (a) (b) (d) (c) (a) 0.17 3.15m (c) 43 44

(a) zero

ANSWER KEY:


Q.1 (b)

Q.2 (a) c 0.0035ε =

steelE = 5

fy1.15E

+ 0.002

3

415 0.0021.15 200 10

= +× ×

= 0.0038

Q.3 (a)

Q.4 (c)

Q.5 (c)

Q.6 (b)

Q.7 (d)

Q.8 (d) Refer : IS : 456 – 2000 clause 15.3

Q.9 (d)

Q.10 (a)

Q.11 (d)

Q.12 (d) Load combinations (1) 1.5 (DL + LL) = 1.5 (50 + 80) = 195 kN-m (2) 1.5 (DL + EQ) = 1.5 (50 + 180) = 345 kN-m (3) 1.2 (DL + LL + EQ)

= 1.2 (50 + 80 + 180) = 372 kN-m Use maximum of above combination.

Q.13 (a)

= 0.7 ckf

Q.14 (a)

Q.15 (c)

Q.16 (b)

Q.17 (b) Compaction factor = weight of partially compacted concrete/ weight of fully compacted concrete For full workability weight of partially compacted concrete will be equal to weight of fully compacted concrete, therefore maximum possible value of compaction factor is equal to 1

Q.18. (c) Static method → Upper bound method of ultimate load analysis Kinematic method → Lower bound on ultimate load QDesign = Qw × load factor M0 = Mu × Ym' Ym = material partial safety factor

Q.19 (9 to 9) Permissible bearing strength = 0.45 fck 20.45 20 9N / mm= × =

Q.20 (b)

Section modulus, z =2I A.r = =A.r

r r,

i.e. Moment of Area.

Q.21 (b) So, P = 2,

Q = 3 R = 3

EXPLANATIONS


Q.22 (d)

m ckf = f + 1.65σ (As per IS : 456.2000)

Q.23 (3.5 to 3.5) Flexural tensile strength,

2crf 0.7 fck N/mm=

0.7 25= × 2= 3.5 N/ mm

Q.24 (b) Initial tangent modulus. Q.25 (c)

2

ckE = 5000 f 5000 25 = 25000N/mm= ×

M F EI Y R= =

Curvature, 65.8 4 1 0

58×25001 F= = per mmR y 0E

= ×

65.8 4 1 058×25000

F per mmyE

= = ×

Q.26. (d)

As the slump increases, the Vebe time decreases

Q.27 (b) At

Q.28 (a)

(i) Air-entrainment reduces the water demand for a given level of workability-True

(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected. -True

Q.29 (51)

Mc Ms Ma+ + +Vw+Va = 1Pc Ps Pa

v

368 606 11553.14×1000 2.67×1000 2

184 1.01000

.74×1000

V+

+

=

⇒ +

+

V0.117 0.227 0.421 0.184 V 1.0+ + + + =⇒ VV 0.051⇒ =

30.051 1000 51 50.32 1/ m= × = ;

Q.30 (42.82)

( )2 2st

πA = 4× × 12 = 453mm4

ck u y st0.36f b.x 0.87f A=

y stv

ck

0.87f A 0.87 415 453X0.36f .b 0.36 25 200

× ×⇒ =

× ×

= 90.86mm v,maxX 0.48d=

0.4 300 120mm= × = vX < Xv , max so U.R. section

v y st vM = 0.87 × f × A × (d - 0.42X )

( )0.87 415 453 300 – 42 90.86 42.82kNm= × × × × =

Q.31 0.097 Q.32 (b) Q.33 (10000)


Q.34 (c) 25MPa If fck is the value below which not more than 50% of test results are expected then fm = fck (Mean strength=Characteristics strength) So target mean strength of concrete to be considered in design mix = Mean strength fm = fck = 25MPa Q.35 (A) Q.36 (b) Q.37 (d) Q.38 (c) Q.39 (a) Q.40 Ast/bd = (0.85/fy) Ast/bd = (0.85/500)

Minimum Percentage of tension reinforcement = (Ast/bd)⨉100

= (0.85⨉100)/500 = 0.17 Q.41 effective depth (d) = 80mm lc/c = 3.3 m Width of support = 230 mm lclear = 3.3-(0.230/2)-(0.230/2) = 3.07 m

leff = min �𝑙𝑙𝑜𝑜 + 𝑤𝑤𝑙𝑙𝑜𝑜 + 𝑑𝑑

leff = min �3.07 + 0.230 = 3.30

3.07 + 0.08 = 3.15 = 3.15 m Q.42 (c) xu,lim for Fe500 = 0.46⨉400 = 184 mm For section to be under-reinforced Ast ≤ Ast,lim

Ast,lim = 0.36fckbxu,lim

0.87fy

= 0.36×25×300×1.84

0.87×500

= 1142.06 𝑚𝑚𝑚𝑚2 No.of bars for limited area of steel

= 1142.06𝜋𝜋4 162

= 5.68 So number of bars will be 5. Q.43 (a) Q.44 Mean Strength µ = (23×4)+(28×2)+(22.5×5)+(31×5)+(29×4)

20

µ = 26.575MPa n= 20

Standard deviation = σ = �∑(f−μ)2

(n−1)

𝜎𝜎 = �4(23−26.575)2+2(28−26.575)2+5(22.5−26.575)2+5(31−26.575)2+4(29−26.575)2

19

= �51.1225+4.06125+83.028+97.903+23.522519

= 3.697 MPa

𝜇𝜇 − 3𝜎𝜎 = 26.575 − 3(3.697) = 15.485MPa Hence No. of specimen with compressive strength < 15.485 MPa are zero.


Q.1 Consider the following two statement.' related to reinforced concrete design and identify whether they are TRUE/FALSE: I. Curtailment of bars in the flexural tension zone in beams reduces the shear strength at the cut-off locations II. When a rectangular columnsection is subject to will be eccentric compression. The neutral axis will be parallel to the resultant axis of banding. a) Both statements I and II are

TRUE.b) Statement I is TRUE and

Statement ll is FALSEc) Statement I is FALSE and

statement II is TRUEd) Both Statements I and II an

FALSE[GATE-2001]

Common data for Questions 2 and 3 At the limit state of collapse, an RC beam is subjected to flexural moment 200 KN-m, shear force 20 KN and torque 9 KN-m. The beam is 300 mm wide and has a gross depth of 425mm with an effective cover of 25 mm. The equivalent nominal shear stress ve( )τ as calculated by using the design code turns out to the lesser than the design shear strength c( )τ of the concrete.

Q.2 The equivalent shear force ( )cV is a) 20 kN b) 54 kNc) 56 kN d) 68 kN

[GATE-2004]

Q.3 The equivalent flexural moment el(M ) for designing the longitudinal

tension steel is a) 187 kN-m b) 200 kN-m

c) 209 kN-m d) 213 kN-m[GATE-2004]

Common data for Q.4 & Q.5 A reinforced con to beam of rectangular

crass section of breadth 230 mm and effective depth 400 mm is subjected to a maximum factored shear force of 120 kN. The grades of concrete, main steel and stirrup steel are M20, Fe415 and Fe250 respectively. For the area of main steel provided, the design shear strength 𝜏𝜏e as per IS: 456-2000 is 0.48 N/mm2. The beam is designed for collapse limit state.

Q.4 The spacing (nun) of 2 – legged 8 mm stirrups 10 be provided Is a)40 b)115 c)250 d)400

[GATE-2008]

Q.5 In addition, the beam is subjected to a torque whose factored value is 10.90 kN-m. The stirrups have to be provided to carry a shear (kN) equal to a)50.42 b)130.36 c)151.67 d)200.23

[GATE-2008]

Q.6 In the design of a reinforced concrete beam the requirement for bond is not getting satisfied. The economical option to satisfy the requirement for bond is by a) Bundling of barsb) Providing smaller diameter bars

more in numberc) Providing larger diameter bars

less in numberd) Providing same diameter bars

more in number[GATE-2008]

2 SHEAR, TORSION, BOND, ANCHORAGE & DEVELOPMENT LENGHTH


Q.7 Consider a reinforcing bar embedded in concrete. In a marine environment this her undergoes uniform corrosion which leads to the deposition of corrosion products on its surface and an increase In the apparent volume of the bar. 'This subjects the surrounding concrete to expansive pressure. As a result, corrosion induced cracks appear at the surface of concrete. Which of the following statements is 'TRUE? a) Corrosion onuses circumferential

tensile stresses in concrete and the creeks will be parallel to the corroded reinforcing bar.

b) Corrosion causes radial tensile stresses in concrete and the cracks will be parallel to the corroded reinforcing bar.

c) Corrosion causes circumferential tensile stresses in concrete and the cracks will he perpendicular to the direction of the corroded reinforcing bar.

d) Corrosion causes radial tensile anemia in concrete and the cracks will be perpendicular to the direction of the corroded reinforcing bar.

[GATE-2009]

Q.8 A rectangular beam of width (b) 230 mm and effective depth (d) 450mm is reinforced with four bars of 12 mm diameter. The grade of concrete is M20 and grade of steel is Fe500. Given that for M20 grade of concrete the ultimate shear strength,

20.36N / mm∞τ = for steel percentage, p=0.25, and 2

0.48N / mm∞τ = p = 0.50. For a factored shear force of 45kN, the diameter (in mm) of Fe500 steel two legged stirrups to be used at spacing of 375 mm, should be

a) 8 b) 10 c) 12 d) 16

[GATE-2014]

Q.9 The development length of a deformed reinforcement bar can be expressed as s bd(1/k)( / ).φσ τ From the IS:456-2000, the value of k can be calculated as___ .

[GATE-2015]

Q.10 In shear design of an RC beam, other than the allowable shear strength of concrete c(τ ), there is also an additional check suggested in IS 456-2000 with respect to the maximum permissible shear stress

c max(τ ) . The check for c maxτ max is required to take care of a) additional shear resistance from

reinforcing steel b) additional shear stress that

comes from accidental loading c) possibility of failure of concrete

by diagonal tension d) possibility of crushing of

concrete by diagonal compression

[GATE-2016]

Q.11 As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress

cmax( )τ depends on the a) grade of concrete and grade of

steel b) grade of steel only d) grade of concrete & percentage

of reinforcement [GATE-2016]

Q.12 An RCC beam of rectangular cross section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strength τc of concrete as 0.62 N/mm2 and maximum allowable shear stress τc, max in concrete as 2.8 N/mm2. If


two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per limit state method will be _________

[GATE-2018]

1 2 3 4 5 6 7 8 9 10 11 12

(b) (d) (b) (b) (c) (b) (c) (a) 6.4 (d) (b) 8.2 cm

ANSWER KEY:


Q.1 (b)

Q.2 (d)

Ve = Vu + 1.6 Tub

= 20 + ( )-3

1.6 9300×10

= 68 kN

Q.3 (b) Equivalent bending moment for the design of tension steel Me1=200 kN-m

Q.4 (b) Design shear force for vertical Stirrups Vus = VU— cτ bd

3120 10 0.48 230 400 44.16 kN= × − × × =Spacing for vertical stirrups

( ) ( )2

US 3

π0.87 250 2 8 2004V

44.16 10

× × =

×= 115 mm c/c.

Q.5 (c)

e u1.6TuV V

b= +

33 1.6 10.9 10120 10

230× ×

= × + =196 KN

Design shear force, eVus V cbd= −3196 10 0.48 330 400= × − × ×

151.67 kN m= −

Q.6 (b) If bond is not safe it is better provide smaller diameter bars more in no.

Q.7 (c)

Q.8 (a)

2uv

V 45×1000τ = 0.434 N/mmbd 230×450

= =

% tensile reinforcement (p) 2π4× ×(12)

4 100 0.437%230×450

= × =

c0.12τ 0.36 (0.437 0.25)0.25

= + × −

= 0.45 N/mm2 v cτ τ<

So, minimum shear reinforcement is required Minimum shear reinforcement

svsv

v y y

A 0.4 0.4×230×375= Ab×S 0.87f 0.80×f

⇒ =

2π 0.4×230×3752× ×f =4 0.80×500

⇒

7.10mm⇒φ =So, adopt 8 mmφ =

Q.9 (6.4) s

dbd

σL4τ

=φ But for deformed bars

bdτ is increased by 60%. So,

st sd

bd bd

σ σL = =4×1.6×τ 6.4τ

φ φ

So, k = 6.4

Q.10 (d)

Q.11 (b) By IS 456:2000

EXPLANATIONS


max ckτc 0.62 f=

maxτa depends on grade of concrete only.

Q.12 Given, Vu = 200kN, 𝜏𝜏c = 0.62 MPa, 𝜏𝜏𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚 = 2.8 MPa

τv = Vubd

= 200×103

250×350 = 2.286 MPa

As 𝜏𝜏𝑣𝑣 < 𝜏𝜏𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚 And design shear force = (𝜏𝜏𝑣𝑣 − 𝜏𝜏𝑐𝑐) bd = (2.286-0.62) × 250 × 350 = 145.775 kN Spacing of shear reinforcement = 𝑆𝑆𝑣𝑣 Vus = 145775 = 0.87 × 250 × 2 × π

4× 102 × 350

Sv

Hence Sv = 82.03 mm Spacing for minimum shear reinforcement

AsvbSv

≥ 0.40.87fy

Sv ≤ 341.65 mm Spacing should be minimum of

(i) 0.75d = 26.25 cm (ii) 8.2 cm (iii) 34.16 cm (iv) 30 cm

So spacing will be 8.2 cm.


Q.1 A concrete column carries axial load or 450 kN and a bending moment of 60 kN-m its base. An isolated footing of size 2m × 3m side along the plane of the bending moment is provided 'under the column Centers of gravity of column and footing coincide. The net maximum and the minimum pressures in kN/m2 on soil under the Coating are respectively. a) 95 and 55 b) 95 and 75c) 75 and 55 d) 75 and 75

[GATE-2003]

Common data for question 2 and 3: A reinforced concrete beam, size 200 mm wide and 300 mm deep overall is simply supported over a span of 3 m. It is subjected to two point loads P of equal magnitude placed at middle third points. The two leads are gradually increased simultaneously. Beam is reinforced with 2 HYDC bars of 16mm diameter placed at an effective cover of 40 mm on bottom face and nominal shear reinforcement. The characteristic compressive strength and the bending tensile strength of the concrete are 20.0 N/mm2 and d2.2 N/mm2 respectively.

Q.2 Ignoring the presence of tension reinforcement, the value of load P in kN when the first flexure crack will develop in the beam is a) 4.5 b) 5.0c) 6.6 d) 7.5

[GATE-2003]

Q.3 The theoretical failure load of the beam for attainment of limit state of collapse in flexure is a) 23.7 EN b) 25.6 KRc) 28.7 kN d) 31.6 kN

[GATE-2003]

Q.4 An RC short column with 300mm x 300mm square cross-section is made of M 20 grade concrete and has 4 numbers. 20 mm diameter longitudinal bars of Fe-415 steel. It is under the action of a concentric axial compressive load. Ignoring the reduction in the area of concrete due to steel bars, the ultimate axial load carrying capacity of the column is a) 1659 kN b) 1548 kNc) 1198 kN d) 1069 kN

[GATE-2004]

Q.5 A rectangular column section of 250 mm x 400 mm is reinforced with five stool bars of grade Fe-500, each of 20 mm Diameter. Concrete mix is M30. Axial load on the column section with minimum eccentricity as per IS:456-2000 using limit state method can be applied up to a) 1707.37 b) 1805.30c) 1806.40 d) 1903.7

[GATE-2005]

Common data for question 6 and 7: A single reinforced rectangular concrete beam has a width of 150 mm and an effective depth of 330 mm. The characteristic compressive strength of concrete is 20 MPa and the characteristic tensile strength of steel is 415 MPa. Adopt the stress block for concrete as given in LS: 456-2000 and take limiting value of depth of neutral axis as 0.48 times the effective depth of the beam.

Q.6 The limiting value of the moment of resistance of the beam in kN-m is a)0.14 b)0.43 c)45.08 d)156.82

[GATE-2007]

3 FOOTING, COLUMNS, BEAMS AND SLABS


Q.7 The limiting area of tension steel in mm2 is

a) 473.9 b) 412.3 c) 373.9 d) 312.3

[GATE-2007] Q.8 A reinforced concrete column

contains longitudinal steel equal to 1 percent of net cross-sectional area of the column. Assume modular ratio as 10. The loads carried (using the elastic theory) by the longitudinal steel and the net area of concrete. are Ps and Pc respectively. The ratio Ps/Pc expressed as per cent is

a)0.1 b)1 c)1.1 d)10

[GATE-2008]

Statements for linked answer question 9 and 10: A doubly reinforced rectangular concrete beam has a width of 300mm and an effective depth of 500mm. The beam is reinforced with 2200 mm2 of steel in tension and 628 mm2 of steel in compression. The effective cover for compression steel is 50mm. Assume that both tension and compression steel yield. The grades of concrete and steel used are M20 and re250 respectively. The stress block parameters (rounded off to first two decimal places) for concrete shall be as per IS 456:2000 Q.9 The depth of neutral axis is a) 205.30 man b) 184.56 mm c) 160.91 mm d) 145.30 mm

[GATE-2010]

Q.10 The moment of resistance of the section is

a) 206.00 kN-m b) 209.20 kN-m c) 237.80 kN-m d) 231.90 kN-m

[GATE-2010]

Q.11 The value of xu (in mm) computed per the Limit State Method of IS 456:2000 is

a) 200.0 b) 223.3

c) 236.3 d) 273.6 0.004 m-1 along the span, the

maximum deflection (in m) of the beam at mid-span is _______.

[GATE-2015]

Q.13. A haunched (varying depth) reinforced concrete beam is simply supported at both ends, as shown in the figure. The beam is subjected to a uniformly distributed factored load of intensity 10 kN/m. The design shear force (expressed in kN) at the section X-X of the beam is _______.

[GATE-2016]

Q.14 An RCC short column (with lateral ties) of rectangular cross section of 250 mm ⨉ 300 mm is reinforced with four numbers of 16 mm diameter longitudinal bars. The grades of steel and concrete are Fe 415 and M20, respectively. Neglect eccentricity effect. Considering limit state of collapse in compression (IS 456: 2000), the axial load carrying capacity of the column (in kN, up to one decimal place), is _________

[GATE-2018]

Q.15 A structure member subjected to compression has both translation and rotation restrained at one end, while only translation is restrained at the other end. As per IS 456:2000, the effective length factor recommended for design is a) 0.50 b) 0.65

c) 0.70 d) 0.80 [GATE-2018]


Q.12 A simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (d) (c) (d) (d) (a) (c) (b) (d) (c) (b) (c) 0.0005 65 918.08 (d)

ANSWER KEY:


Q.1 (d)

Q.2 (c) F M=Y l

2200×300M = f.Z = (2.2)6

M = P (1m) P = 6.6 kN

Q.3 (d) ux max 0.48 d 124.8 mm= =

C = T ck u y0.36 f b x 0.87 f Ast=

( )( ) u0.36 20 200 (x )

( ) 2π0.87 415 2 164

= × ×

ux 100.8 mm=

u uX < x max , under reinforced section

uMu 0.87 fy Ast(d 0.42x )= − = 31.6 kN -m Mu P 1m= × P 31.6 kN=

Q.4 (d) Pu = 0.4 fck Ac + 0.67 fy Asc

( )( ) ( )0.4 20 300 300 0.67 415= × +

2π4× ×20 = 1069kN4

Q.5 (a) Pu 0.4 fck Ac + 0.67 fy Asc= Ac Ag Asc= −

2250 400 5 204π

= × − × ×

298489.2 mm=

Pu 0.4 30 98429.2 0.67 500 1570.8 1707.3716= × × +

× × =

Q.6 (c) u lim ck u,max u,maxM 0.36 f bx (d-0.42 x )=

u,maxx 0.48d=

( )( )Mu lim 0.36 20 150 0.48 330

330 0.42 0.48 330

= × × × ×

− × ×= 45 kN -m

Q.7 (b)

Q.8 (d) Longitudinal steel, As = 1% of Ac Where Ac = Area of concrete Modular ratio, m = 10

c c s sc s

c c s s

P l P l= = (here l =l )A E A E

s s s

c c c

P E A= .P E A

1%Ac= m .Ac

= 10 × 1 s

c

P 10P

∴ =

Q.9 (c) ck u sc y St0.36f bx f Asc 0.87f A+ =

u0.36 20 300 x 0.87 250628 0.87 250 2200= × × × ×× × × ×

ux 160.91mm=

Q.10 (b) ck u u

sc 1

Mu 0.36f bx (d 0.42x )f Asc (d-d )= −

+

( )( )0.36 20 300 160.91(500 0.42

160.91) 0.87 250 628 500 50= × × × −

× + × −= 209.21 kN-m.

EXPLANATIONS


Q.11 (c) Q.12 (0.0005)

( )212- =249.9995m

2OA 250

AA’ 0.0005m∆ = Q.13 (65)

x

v

MV ± .tanβdτ =bd

xd v x

MV τ .bd=V ± tanβdx

⇒ =

xV =100-10×5=50kN;dx=500mm

x5×5M =100×5-10× =375kN-m

2

d375V = 50 + × tanβ0.5

600-400 200tan β = =10×1000 10,000

d375 200V = 50+ × = 50+15 = 65kN0.5 10,000

Q.14 918.08 kN Considering limit state of collapse in compression and the permissible axial strain as 0.002, the stress in concrete = 0.45fck and stress in steel = 0.87fy for Fe250 =0.79fy for Fe415 = 0.75fy for Fe500 The above stresses in steel are derived from stress strain curve of different grade of steel. The axial load capacity of column

neglecting eccentricity effect = 0.45fckAc + 0.75fyAsc

Ac = 250 × 300- 4× π4

× 162 = 74195.75 mm2

Asc = 4× π4

× 162 = 804.25 mm2

Axial load capacity = (0.45⨉20⨉74195.75)+ (0.75⨉415⨉804.25) = 918084 N = 918.08 kN Q.15 (d) Recommended value of effective length = 0.8ℓ


Q.1 A simply supported prestressed concrete beam is 6 m long and 300 mm wide. Its gross depth is 600 mm. It is prestressed by horizontal cable tendons at a uniform eccentricity of 100 mm. The prestressing tensile force in the cable tendons is 1000 kN. Neglect the self-weight of beam. The maximum normal compressive stress in the beam at transfer is a)12.5 N/mm2 b)5.55N/mm2

c)11.11 N/mm2 d)15.68 N/mm2

[GATE-2004]

Q.2 A concrete beam of rectangular cross section of 200 mm × 400 mm is pre stressed with a force 400 kN at eccentricity 100 mm. The maximum compressive stress in the concrete a) 12.5 N/mm2 b) 7.51N/mm2

c) 5.0 N/mm2 d) 2.5 N/mm2

[GATE-2005]

Q.3 The percentage loss of prestress due to anchorage slip of 3 mm in a concrete beam of length 30 m which is post-tensioned by a tendon with an initial stress of 1200 N/mm2 and modulus of elasticity equal to

5 22.1 10 N / mm× a) 0.0175 b) 0.175c) 1.75 d) 17.5

[GATE-2007]

Q.4 A concrete beam of rectangular cross-section of size 120 mm (width) and 200 mm(depth) is prestressed by a Straight tendon to an effective force of 150 kN at an eccentricity of 20 mm (below the centroidal axis in the depth direction) The stresses at the top and bottom fibres of the section are

a) 2.5 N/mm2 (compression), 10N/mm2 (compression)

b) 10 N/mm2 (tension), 2.5 N/mm2

(compression)c) 3.75 N/mm2 (tension), N/mm2

(compression)d) 2.75N/mm2 (compression), 3.75

(compression)[GATE-2007]

Q.5 A pre-tensioned concrete member of section 200 mm × 250 mm contains tendon of area 500 mm2 at centre of gravity of the section. The pre stress in the tendons is 1000 N/mm2. Assuming modular ratio as 10, the stress (N/mm2) in concrete a)11 b) 9c) 7 d)5

[GATE-2008]

Q.6 A rectangular concrete beam of width 120 mm and depth 200 mm is prestressed by pretension in to a force of 150 kN at an eccentricity: of 20 mm. The cross-sectional area of the prestressing steel is 87.5 mm2.Take modulus of elasticity steel and concrete as 2.1 × 105 MPa and 3.0 × 104 MPa respectively The percentage loss of stress in the prestressing steel due to elastic deformation of concrete is a) 8.75 b) 6.125c) 4.81 d) 2.19

[GATE-2009]

Q.7 A concrete beam prestressed with a parabolic tendon is shown in the sketch. The eccentricity of the tendon is measured from the centroid of the cross-section. The applied prestressing force at service is 1620 kN. The uniformly distributed load of 45 kN/m

4 PRESTRESSED CONCRETE


includes the self-weight. The stress (in N/mm2) in the bottom fibre at mid span is

a)tensile 2.90 b)compressive 2.90 c)tensile 4.32 d)compressive 4.32

[GATE-2012]

Q.8 In a pre-stressed concrete beam section shown in the figure, the net loss is 10% and the final pre-stressing force applied at X is 750 kN. The initial fibre stresses (in N/mm2) at the top and bottom of the beam were:

a)4.166 and 20.833 b)-4.166 and -20.833 c)4.166 and -20.833 d)-4.166 and 20.833

[GATE-2015]

Q.9 A pre-tensioned rectangular concrete beam 150 mm wide and 300 mm depth is prestressed with three straight tendons, each having a cross-sectional area of 50 mm2, to an initial stress of 1200N/mm2. The tendons are located at 100 mm from the soffit of the beam. If the modular ratio is 6, the loss of prestressing force (in kN, up to one decimal place) due to the elastic deformation of concrete only is ______________

[GATE-2017]

Q.10 A simply supported rectangular concrete beam of span 8 m has to be prestressed with a force of 1600 kN. The tendon is of parabolic profile having zero eccentricity at the supports. The beam has to carry an external uniformly distributed load of intensity 30 kN/m. Neglecting the self-weight of the beam, the maximum dip (in meters, up to two decimal places) of the tendon at the mid-span to balance the external load should be _____________

[GATE-2017]

Q.11 A 6 m long simply-supported beam is prestressed as shown in the figure.

The beam carries a uniformly distributed load of 6 kN/m over its entire span. If the effective flexural rigidity EI = 2⨉104 kNm2 and the effective prestressing force is 200 kN, the net increase in length of the prestressing cable (in mm, up to two decimal places) is_____________

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11

(a) (a) (c) (a) (c) (d) (b) (d) 4.8 kN 0.15m 0.120 mm

ANSWER KEY:


Q.1 (a)

minP

Zσ Pe-

A=

=3 3

2400 10 400 10 100( )

200 400200 4006

× ×= −

××

= 12.5 MPa

Q.2 (a)

Q.3 (c)

Q.4 (a) 3

bot top 2

P 150×10 ×20σ ,σ =A 120×200

6

m

( )bot 10 Mσ Pa C= +

( )top 2.5 Mσ Pa C= +

Q.5 (c) Stress in concrete at the level of steel is

P Pefc = + (e)A l

3 3 2

3

150 10 15010 (20)120 200 120 200

12

× ×+

× ×

= 7 MPa

Q.6 (d) Sec IS : 456 = clause 2.19

Q.7 (b) a Dl+LL

bP MPσ = + -

A Z Z

3 3

2

6

2

1620 10 1620 10 145500 750 500 750

6

299.75 10299.75750

6

× × ×+ −

× ×

×

= 2.9N/mm2 (comp)

Q.8 (d) Loss = 10% Find force = 750 kN

Initial force = 750 = 833.33 kN0.9

=

3

23

P mTop & Bottom stress = ±A z

833.33 10250 400

833.33×10 ×100×6250×400

= × ±×

8.33 12.5= ± Top = -4.166 (T) Bottom = 20.833 (C)

Q.9 4.8 kN

Initial prestressing force =3⨉50⨉1200 = 180000 N

Eccentricity (e) = ( D2 - 100)

= ( 3002

- 100) = 50 mm

Stress in concrete at the location of

steel = �𝑃𝑃𝐴𝐴� + �Pe

I� e

EXPLANATIONS


= 180000150×300

+ 180000 × 502

150×300312

= 4.0 + 1.33 = 5.333 N/mm2

Loss of stress = m ⨉ fc = 6 × 5.3333 = 32 N/mm2

Loss of prestressing force

= 3⨉50⨉321000

kN = 4.8 kN

Q.10 (0.15)

Central dip , e = wl2

8P

= 30×82

8×1600 = 0.15 m

Q.11 (0.120mm)

Downward UDL,

w = 6kN/m

Eccentricity, e = 50 mm

Prestressing force, P = 200 kN

EI = 2 ⨉ 104 kN m2

L = 6 m

Rotation due to prestress = PeL2EI

= 200×50×6×10−3

2×2×104 = 1.5⨉ 10−3

Rotation due to UDL = wL3

24EI

= 6×63

24×2×104 = 2.7 ⨉ 10−3

Net rotation = 2.7 ⨉ 10−3 - 1.5 ⨉ 10−3

= 1.2 ⨉ 10−3 radian

Elongation of the cable = 2⨉50⨉1.2⨉10−3

= 0.120 mm


STEEL STRUCTURE


1.1 DESIGN OBJECTIVE

The objective of design is the achievement of an acceptable probability that structures will perform satisfactorily for the intended purpose during the design life. With an appropriate degree of safety, they should sustain all the loads and deformation during construction and use and have adequate resistance to accidental loads and fire.

1.2 METHODS OF DESIGN

Structure and its elements shall normally be designed by limit state method as per IS 800-2007. Where the limit state method cannot be conveniently adopted; working stress method shall be used.

1.3 LOADS AND FORCES

For the purpose of designing any element, member or structure, the following loads and their effects shall be taken into account, where applicable, with partial safety factors and combinations : a) Dead loads;b) Imposed loads; (Live load, crane load,

snow load etc.)c) Wind loadsd) Earthquake loadse) Erection loadsf) Accidental loads such as those due to blastg) Secondary effects due to contraction or

expansion resulting from temperaturechanges, differential settlements of thestructure as a whole or of itscomponents, eccentric connections.

1.4 LOAD COMBINATIONS

The following load combinations with appropriate load factors may be considered in designing. a) Dead load + Imposed loadb) Dead load + Imposed load + Wind or

Earthquake loadc) Dead load + Wind or Earthquake loadd) Dead load + Erection load

1.5 GEOMETRICAL PROPERTIES

IS 800-2007 gives the concept of the gross and effective cross -sections of a member. • The properties of the gross cross-

section shall be calculated from thespecified size of the member or readfrom appropriate table.

• The effective cross -section of a memberis that portion of the gross cross –section that is effective in resisting thestresses.Highlighting this concept of effectivecross - section, IS 800-2007 hasclassified the members cross-section asfollows.

1.6 CLASSIFICATION OF CROSS SECTIONS

Basis of classification • The plate elements of a cross -section

may buckle locally due to compressivestresses.

• When plastic analysis is used, themembers shall be capable of formingplastic hinges with sufficient rotationcapacity (ductility) without localbuckling to enable the redistribution ofbending moment required beforeformation of failure mechanism.

• When elastic analysis is used, themember shall be capable of developingthe yield stress under compressionwithout local buckling.

1 GENERAL DESIGN SPECIFICATION


On the above basis, four classes of section are defined as follows : a) Plastic : Cross-sections, which can

develop plastic hinges and have the rotation capacity required for failure of the structure by formation of a plastic mechanism.

b) Compact : Cross -sections, which can develop plastic moment of resistance, but have inadequate plastic hinge rotation capacity for formation of a plastic mechanism.

c) Semi-compact : Cross -Sections, in which the extreme fibre in compression can reach yield stress, but cannot, develop the plastic moment of resistance, due to local buckling.

d) Slender : Cross -sections in which the elements buckle locally even before reaching yield stress.

1.7LIMIT STATE DESIGNS OF STEEL STRUCTURES Basis for Limit State Design In the limit state design method, the structure shall be designed to withstand safely all loads likely to act on it throughout its life. It shall also satisfy the serviceability requirements, such as limitations of deflection and vibrations and shall not collapse under accidental loads such as

from explosions or impact or due to consequences of human error to an extent not originally expected to occur. The acceptable limit for the safety and serviceability requirements before failure occurs is called a limit state. The objective of design is to achieve a structure that will not become unfit for use with acceptable target reliability. In other words, the probability of a limit state being reached during its lifetime should be very low. In general, the structure shall be designed on the basis of the most critical limit state and shall be checked for other limit states. In limit state design, structures are designed on the basis of safety against failure and are checked for serviceability requirements. 1.8 CONNECTIONS:RIVETED CONNECTION

• Riveting is a method of joining together pieces of metal by inserting ductile metal pins called rivets into holes of pieces to be connected and forming a head at the end of the rivet to prevent ach metal piece from coming out.

• Rivet holes are made in the structural members to be connected by punching or by drilling. The size of rivet hole is kept slightly more (1.5 to 2.0 mm) than the size of rivet.

• After the rivet holes in the members are matched, a red hot rivet is inserted which has a shop made head on one side and the length of which is slightly more than the combined thicknesses of the members to be connected.

• Then holding red hot rivet at shop head end, hammering is made.

• It results in to expansion of the rivet to completely fill up the rivet hole and also into formation of driven head.

• Desired shapes can be given to the driven head.

• The riveting is done may be in the workshops or in the field.


• Riveting has the following disadvantages. a) It is associated with high level of

noise pollution. b) It needs heating the rivet to red hot. c) Inspection of connection is a skilled

work. d) Removing poorly installed rivets is

costly. e) Labour cost in high.

• Production of wieldable quality steel and introduction of high strength frication grip bolts have replaced use of rivets.

• Design procedure for riveted connections is same as that for bolted connection except that the effective diameter of rivets may be taken as rivet hole diameter instead of nominal diameter of rivet.

• IS 800-2007 do not discuss riveted connection in it hence it is not discussed further here.

1.8.1 BOLTED CONNECTIONS

• A bolt is a metal pin with a head formed at one end and shank threaded at the other in order to receive a nut. Bolts are used for joining together pieces of metals by inserting them through holes in the metal and tightening the nut at the thread ends.

Bolts are classified as : a) Unfinished (black) bolts b) Finished (turned) bolts c) High strength friction grip (HSFG) bolts a) Unfinished/Black Bolts

• These bolts are made from MILD steel rods with square or hexagonal head.

The shank is left unfinished i.e. rough as rolled.

• In structural elements to be connected holes are made larger than nominal diameter of bolts.

• As shank of black bolts is unfinished, the bolts may not establish contact with structural member at entire zone of contact surface.

• Joints remain quite loose resulting into large deflections.

• These bolts are used for light structures under static loads such as trusses, bracings and also for temporary connections required during erections.

• It is not recommended for connection subjected to impact, fatigue or dynamic loading.

• Bolt of property class 4.6 means, ultimate strength is 400 N/mm2 and yield strength is 400 × 0.6 = 240 N/mm2

• If a bolt is designated as M16, M20, M24, M30, it means shank diameter of 16 mm, 20 mm, 24 mm, and 30 mm respectively.

b) Finished/Turned Bolts

• These bolts are also made from mild steel, but they are formed from hexagonal rods, which are finished by turning to circular shape.

• Tolerance available for fitting is quite small (0.15 mm to 0.5 mm).

• It needs special methods to align bolt holes before bolting.

• As connection is tighter, it results in to much better bearing contact between the bolts and holes. These bolts are used in special jobs like connecting machine parts subjected to dynamic loadings.

c) High strength Friction Grip (HSFG) Bolts High Strength bolts

a) Made from bars of medium carbon steel.

b) Normally class 8.8 and 10.9 are commonly used


c) Less ductile than black bolts d) Material of the bolt does not have well-

defined yield point. Instead of yield stress, proof load is used

e) As per IS 8000: 2007 proof load is taken as 0.7 × ultimate tensile stress of bolt

f) M16, M20, M24, M30, are generally used

g) Designated like 8.8S, 10.9S denotes high strength bolt.

h) Percentage elongation of these bolts at failure is approx 12%

i) Special techniques are used for tightening the nuts to induce specified initial tension in the bolts, which caused sufficient friction between the flaying forces.

j) These bolts with induced initial tension called as high strength friction grip (HSFG) bolts.

k) Due to friction, the sleep in the joint is eliminated hence; connection in this case is called nonslip connection or friction type connections.

1.9 TERMINOLOGY IN BOLTED CONNECTION 1. Pitch of the bolts (P)

It is the centre to centre spacing of bolts in a row, measured in the direction of load.

2. Gauge (g) It is the distance between the two consecutive bolts of adjacent rows and is measured at right angle to the direction of load.

3. Staggered Pitch (Ps) It is the centre to centre distance of staggered bolts measured in the direction of load.

4. Diameter of Bolt Hole

Diameter of bolt hole is larger than the nominal diameter (shank diameter) of the bolt to facilitate erection and to allow for in assurances in fabrication. Holes are a) Standard clearance hole normal b) Oversized holes (i.e. holes of size

larger than standard clearance hole) used in slip resistant connection.

c) Short and long slot used in slip resistant connection following table gives the diameter of holes for bolts.

Clearances for fastener Holes

Sl No.

Nominal size of Fastener, d

mm

Size of the Hole = Nominal Diameter of the Fastener + Clearances

mm Standard

Clearance in Diameter and Width of Slot

Over size Clearance in

Diameter

Clearance in the Length of

the slot Short Slot

Long Slot

(1) (2) (3) (4) (5) (6) i) 12 - 14 1.0 3.0 4.0 2.5 d ii) 16 - 22 2.0 4.0 6.0 2.5 d iii) 24 2.0 6.0 8.0 2.5 d iv) Larger than 24 3.0 8.0 10.0 2.5 d

From the above table: Diameter of Normal Bolt Holes is: Nominal size of Bolts in mm

12 14 16 20 22 24 30 36

Diameter of Bolt hole in mm

13 15 18 22 24 26 33 39

5. Area of Bolt at Root (Anb)

Area of bolt at root of the thread is less than that at shank of the Bolt. It is taken approximately equal to 0.78 times the shank area i.e. Anb = 0.78 × Asb Where Asb = Area of bolt at shank d=Nominal diameter of Bolt(shank diameter) Anb = Area of bolt at root 1.10 IS 800-2007 SPECIFICATION FOR SPACING AND EDGE DISTANCES OF BOLT HOLES

1. Pitch 'P' shall not be less than 2.5 d, where d is the nominal diameter of bolt.

2. Pitch 'P' shall not be more than


a) 16 t or 200 mm, whichever is less, in case of tension members.

b) 12 t or 200 mm, whichever is less, in case of compression members where t is the thickness of thinner plate.

c) In case of staggered pitch, pitch may be increased by 50 percent of values specified above provided gauge distance is less than 75 mm.

3. In case of compression member where forces are transferred through butting faces, i.e., (butt joints), maximum pitch is to be restricted to 4.5 d for a distance of 1.5 times the width of plate from the butting surface. (Refer Figure Below).

4. The gauge length 'g' should not be more

than 100 mm + 4 t or 200 mm whichever is less in compression and tension member where t is the thickness of thinner outside plate.

5. Minimum edge and end distance shall not be a) Less than 1.7 × hole diameter in case

of sheared or hand flame out edges. b) Less than 1.5 × hole diameter in case

of rolled, machine flame cut, sawn and planed edges.

6. Maximum edge distance (e) should not exceed

a) 12 tε, where y

250εf

= and t is the

thickness of thinner outer plate. This recommendation does not apply to fasteners interconnecting the components of back to back tension members.

b) Where the members are exposed to corrosive environment max edge

distance 40 mm + 4 t, where t is the thickness of thinner connected plate.

7. Apart from the required bolt from the consideration of design forces, additional bolts called tacking fasteners should be provided as specified below. a) Tacking rivets should be providing.

i) At 32 t or 300 mm, whichever is less, if plates are not exposed to weather.

ii) At 16 t or 200 mm, whichever is less, if plates are exposed to weather?

8. In case of a tension member made up of

two flats or angles or tees or channels, tacking rivets are to be provided along the length to connect its components as specified below: a) Not exceeding 1000 mm, if it is

tension member. b) Not exceeding 600 mm, if it is

compression member. Countersunk heads

• is neglected in calculating length of

fastener in bearing • For Fastener in tension having

countersunk heads, tensile strength is reduced by 33.3% and no reduction in shear strength calculation.

1.11 TYPES OF JOINTS

Types of joints may be grouped into the following two: a) Lap joint b) Butt joint a) Lap Joint: It is the simplest type of joint. In this the

plates to be connected overlap one another.


b) Butt Joint

In this type of connections, the two main plates butt against each other and the connection is made by providing a single cover plate connected to main plate or by double cover plates, one on either side connected to the main plates.

1.12DESIGN STRENGTH OF PLATES IN A JOINT

Plates in a joint made with bearing bolts may fail due to any one of the following 1. Shearing or bursting of the edge. 2. Crushing of plates. 3. Rupture of Plates. 4. Block shears failure of plates in tension.

The brushing or shearing and crushing failures are avoided if the minimum edge/end distance as per IS 800-2007 recommendations are provided. As per IS 800-2007: The minimum edge distance and end distances from the centre of any hole to the nearest edge of a plate shall not be less than. 1.7 times the hole diameter in case of sheared or hand – flame cut edges; and 1.5

times the hole diameter in case of rolled, machine-flame cut, sawn and planed edges. The maximum edge distance to the nearest line of fasteners from and edge of any un-stiffened part should not exceed 12 tε,

where ε = �250fy

and t is the thickness of the

thinner outer plate. This would not apply to fasteners interconnecting the components of back to back tension members. Where the members are exposed to corrosive influences, the maximum edge distance shall not exceed 40 mm plus 4t, where t is the thickness of thinner connected plate.

It the maximum distances are ensured in a joint, the design tensile strength of plate in the joint is the strength of the thinnest member against rupture. This strength is

given by. 1

n udn

m

0.9A fTγ

=

Where γmt= Partial safety factor for failure at ultimate stress = 1.25 fu = Ultimate stress of the material An = Net effective are of the plate at critical section, which is given by 𝐴𝐴𝑛𝑛 = �𝑏𝑏 − 𝑛𝑛𝑑𝑑ℎ + ∑ 𝑃𝑃2𝑠𝑠𝑠𝑠

4𝑔𝑔𝑠𝑠𝑖𝑖=𝑡𝑡 � 𝑡𝑡 b = Width of plate t = Thickness of thinner plate in joint


dh = Diameter of the bolt hole (2 mm in addition to the diameter of the hole, in case of directly punched holes) g = Gauge length between the bolt holes Ps = Staggered pitch length between lines of bolt holes n = Number of bolt holes in the critical section It may be noted that, if there is no staggering Psi = 0 and hence An = (b – ndh)t, which is the net area at critical section.

1.13 BLOCK SHEAR STRENGTH

Block Shear failure of plate occurs as shown below. It's a combination of yielding and rupture. Block shear failure of a plate occurs along a path involving tension on one plane and shear on a perpendicular plane along the fasteners.

Block Shear strength at an end connection is calculated as given below. It should be taken as the smaller of

( )0

1

u tndb vg y m

m

0.9f AT = A f 3 γ +γ

Shear yielding + tensile rupture OR

1

y bgu tndb

m m0

f A0.9f AT =γ γ

+

Shear rupture + tensile yielding Where Avg, Avn = Minimum gross and net area in shear along a line of transmitted force respectively (1-2 and 4-3 as shown in figure above.) Atg, Atn = Minimum gross and net area in tension from the hole to the toe of the angle

or next last row of bolt in plates, perpendicular to the line of force respectively (2-3). fu, fy = Ultimate and yield stress of the material respectively.

0 1m mγ , γ = Partial factor of safety in yielding and rupture respectively

0 1m m(γ 1.1, γ 1.25).= = 1.14 DESIGN STRENGTH OF BEARING BLOTS The design strength of bearing bolts under shear is the least of the following : a) Shear capacity (strength). b) Bearing capacity (strength) a) Shear Capacity (Strength) of Bearing Bolts in a joint Design strength of the bolt, Vdsb is given by

nsbdsb

mb

VV =

γ

Where Vnsb, nominal shear capacity of bolt and

mbγ = partial safety factor of material of bolt= 1.25. In the above expression Vnsb is

given by ( )ubdsb n nb s sb

fV = η A +η A

3

Where, fub=Ultimate tensile strength of a bolt nη = Number of shear planes with threads

intercepting the shear plane. sη = Number of shear planes without

threads intercepting the shear plane Asb = Nominal shank area of bolt

2d=4

π

Anb = Net shear area of the bolt at threads, may be taken as the area corresponding to root diameter of the thread.

Anb = 0.78π4

d2 or π4

(d − 0.9382p)2

Where p is the pitch of thread, d = nominal dia. of bolt i.e. shank dia. 1.15 BEARING CAPACITY OF BOLTS(Vdpb)

IS 800-2007 Suggest the following procedure to find bearing strength of bolts.


npbdpb

mb

VV =

γ

Where, Vdpb = Design bearing strength Vnpb = Nominal bearing strength

mbγ = Partial safety factor of material = 1.25 Nominal bearing strength may be found from the following relation.

npb b uV = 2.5 K dt f where Kb is smaller of

ub

h h u

fe P, – 0.25, , 1.03d 3d f

In which e, p are end and pitch distances respectively dh = diameter of hole fub, fu = ultimate tensile strength of bolt and plate respectively d = nominal diameter of bolt t = summation of the thickness of the connected plates experiencing bearing stress in the same direction. If bolts are countersunk, it is to be reduced by the half depth of countersinking. In plates the bearing strength is a linear function of end distances both bolts and plates are subjected to significant triaxial containment. Due to this, bearing behavior of plates is influenced by the proximity of neighbouring holes or boundary (edge distance)

1.16 TENSILE CAPACITY OF BOLTS

A bolt subjected to a factored tensile force Tb shall satisfy:

b dbT T≤ where, Tdb = nb mbT γ Tnb = Nominal tensile capacity of the bolt

mbnb ub n yb sb

m0

γT = 0.9 f A < f A

γ

where, fub = Ultimate tensile stress of the bolt fyb = Yield stress of the bolt An = Net tensile stress area, shall be taken as the area at the bottom of the threads Asb = Shank area of the bolt.

mb m0γ , γ = Partial safety factors = 1.25, 1.1 respectively It means that tensile capacity of bolts is minimum of

yb sbtb ndb

mb m0

f A0.9f AT Min ,γ γ

=

tb n

mb

0.9f Aγ

= tensile rupture

yb sb

m0

f Aγ

= tensile yielding

1.17 WELDED CONNECTIONS

Welding consists of joining two pieces of metal by establishing a metallurgical bond between them. The elements to be connected are brought closer and the metal is melted by means of electric arc or oxyacetylene flame along with weld rod which adds metal to the joint. After cooling the bond is established between the two elements.

1.18 TYPES OF WELDED JOINTS

There are three types of welded joints: 1. Butt weld 2. Fillet weld 3. Slot weld and Plug weld

1. Butt Weld Butt weld is also known as groove weld. Depending upon the shape of groove made for welding butt welds. Types of Butt Welds


2. Fillet Weld

Fillet weld is a weld of approximately triangular cross-section joining two surfaces approximately at right angles to each other in lap joint, tee joint or corner joint.

When the cross-section of fillet weld is isosceles triangle with face at 45º. It is known as standard fillet weld. In special circumstances 60º and 30º angel are also used. A fillet weld is known as concave fillet weld, convex fillet weld or as mitre fillet weld depending upon the shape of weld face.

3. Slot Weld and Plug Weld Figure below shows a typical slot weld in which a plate with circular hole is kept with another plate to be joined and then fillet welding is made along the periphery of the hole.

Figure below shows typical plug welds in which small holes are made in one plate and is kept over another plate to be connected and then the entire hole is filled with filler material.

1.19 SPECIFICATION FOR WELDING

Important specification regarding butt weld, fillet weld and plug and slot weld as per IS 800 – 2007 is:

Butt Weld

1. The size of butt weld shall be specified by the effective throat thickness. In case of a complete penetration butt weld it shall be taken as thickness of the thinner part joined. Double U, double V, double J and double level butt welds may be generally regarded as complete penetration butt welds. The effective throat thickness in case of incomplete penetration butt weld shall be taken as the minimum thickness of the weld metal common to the parts joined excluding reinforcement. In the absence of actual data it may be taken as 5/8th of thickness of thinner material.

2. The effective length of butt weld shall be taken as the length of full size weld.

3. The minimum length of butt weld shall be four times the size of the weld.

4. If intermittent butt welding is used, it shall have an effective length of not less than four times the weld size and space between the two welds shall not be more than 16 times the thickness of the thinner part joined.


1.20 FILLET WELD

1. Size of Fillet Weld a) The size of normal fillet weld shall be

taken as the minimum weld leg size. b) For deep penetration welds with

penetration not less than 2.4 mm, size of weld is minimum leg size + 2.4 mm.

c) For fillet welds made by semi-automatic or automatic processes with deep penetration more than 2.4 mm, if purchaser and contractor agree.

S = minimum leg size + Actual penetration

2. Minimum size of fillet weld specified in 3 mm. To avoid the risk of cracking in the absence of preheating the minimum size specified are: For less than 10 mm thickness plate – 3 mm For 10 to 20 mm thickness plate – 5 mm For 20 to 32 mm thickness plate – 6 mm For 32 to 50 mm thickness plate – 8 mm 3. Effective throat thickness: It shall not be less than 3 mm and shall not generally exceed 0.7 t (or t under special circumstances) where t is the thickness of the thinner plate of the elements being welded. If the face of plates being welded is inclined to each other, the effective throat thickness shall be taken as K times the fillet size where K is as given in table below:

4. Effective Length: The effective length of the weld is the length of the weld for which specified size and throat thickness exist. In drawings only effective length is shown. While welding length made is equal to effective length plus twice the size of the weld. Effective length should not be less than 4 times the size of the weld.

5. Lap joint: The minimum lap should be four times the thickness of thinner part joined or 40 mm whichever is more. The length of weld along either edge should not be less than the transverse spacing of welds. 6. Intermittent welds: Length shall not be less than 4 times the weld size or 40 mm whichever is more. The minimum clear spacing of intermittent weld shall be 12 t for compression joints and 16 t for tensile joints, where t is the thickness of thinner plate joined. The intermittent welds shall not be used in positions subject to dynamic repetitive and alternating stresses. 1.20.2 DESIGN STRESSES IN WELDS

Butt Welds Butt welds shall be treated as parent metal with a thickness equal to the throat thickness, and the stresses shall not exceed those permitted in the parent metal. It means that the strength of Butt weld is equal to the strength of the parent metal.

Fillet Weld, Slot or Plug Welds Design strength shall be based on its throat area and shall be given by

wnwd

mw

ff =

γ

Where, uwn

ff =

3

fu = Smaller of the ultimate stress of the weld or of the parent metal

mwγ = 1.25 for shop welds= 1.5 for field welds IS 800-2007 gives the following provisions for the fillet welds:

1. If a fillet welds is applied to the square edge of a part, the specified size of the weld should generally be at least 1.5 mm less than the edge thickness.


2. If fillet weld is applied to the rounded toe of a rolled section, the specified size of the weld should generally not exceed 34th of the thickness of the section at the

toe.

3. In members subjected to dynamic loading, the fillet weld shall be of full size with its leg length equal to thickness of plate.

4. End fillet weld, normal to the direction of force shall be of unequal size with throat thickness not less than 0.5 t. The difference in the thickness of weld shall be negotiated at a uniform slope.

1.21 REDUCTION IN DESIGN STRESSES FOR LONG JOINTS

If the length of the welded joint lj is greater than 150 t , where t is throat thickness, the design capacity of weld fwd shall be reduced by the factor.

jlw

t

0.2l=1.2 1.0

150 tβ − ≤

lj = length of the joint in the direction of the force transfer and t = throat thickness of the weld

The reduction in design strength of long joint is to cater for non uniform mobilization of stress in a long joint causing ineffectiveness of certain length of joint. 1.22ECCENTRIC CONNECTION – PLANE OF MOMENT AND THE PLANE OF WELDS IS THE SAME

Figure above shows a typical case. The eccentric load P is equivalent to i) a direct load P at the centre of gravity of

the group of weld and ii) a twisting moment P × e

Let a weld of uniform size be applied throughout and 't' be the effective throat thickness. It’d’ is the depth of weld and 'b' is the width as shown in the figure, the direct shear stress in the weld is

( )1Pq =

2b+d t

The stress in the weld due to twisting moment is the maximum in the weld at the extreme distance from the centre of gravity of the group of weld and acts in


the direction perpendicular to the radius vector. The maximum stress due to the moment.

max2

zz

P×e×rq =

I

where rmax is the distance of the extreme weld from the c.g. of the group. Izz = Ixx + Iyy, the polar moment of inertia. The vector sum of the stress is

2 21 2 1 2q = q + q + 2q q cosθ.

For the safe design it should be less than the resistance per unit area.


2.1TENSION MEMBER

Tension members are linear members in which axial forces act causing elongation (stretch). Such members can sustain loads upto ultimate load, at which stage they may fail by rupture at a critical section. However, if the gross area of the member yields over a major portion of its length before the rupture load is reached, the member may become non functional due to excessive elongation, plates and other rolled sections in tension may also fail by block shear of end bolted regions, under combined shear along longitudinal section along bolt lines and normal tensile stresses on a transverse section across a bolt line. The factored design tension T, in the members shall satisfy the following requirement

dT < TWhere Td = Design strength of member. The design strength of member under axial tension Td is the lowest of the design strength due to yielding of gross section, Tdg, rupture of critical section, Tdn and block shear, Tdb.

2.2DESIGN STRENGTH DUE TO YIELDING OF GROSS SECTION

The design strength of members under axial tension Tdg as governed by yielding of gross-section is given by

dg y g m0T = f A γ Where fy = Yield strength of the material in MPa Ag = Gross area of cross -section in mm2

= Partial safety factor for failure in tension by yielding = 1.1 2.3DESIGN STRENGTH DUE TO RUPTURE OF CRITICAL SECTION

Before going in to this discussion, lets first understand few terms.

2.3.1Net Effective Area It is defined as the gross area (Ag) of the section minus the deductions (Ad) made for any loss of material. Thus, An = Ag – Ad.

2.3.2 Diameter of Bolt Hole According to IS 800-2007 diameter of Bolt hole is to be taken as larger than nominal dia. of Bolt. Following table gives the diameter of holes for Bolts.

2.3.3Strength of Plate in Rupture The design strength in tension of a plate Tdn as governed by rupture of net cross-sectional area An, at the hole is given by

Tdn = 0.9Anfu ∕ γm1 γm1 = Partial safety factor for failure in tension by rupture = 1.25 fu = Ultimate stress of the material in MPa An= Net effective area of the member 2.3.4 Calculation of Net Effective Area (An)

2 DESIGN OF TENSION MEMBERS


* Rupture can occur along 1-2-3-4-5-6-7

An = �b − ndh + ��P2

4g�� t

In this case,n = 5

Hence 2

n h2PA in this case = b – 5d + t4g

Where, dh = Diameter of Bolt hole b, t = Width and thickness of plate respectively n = Number of Bolt holes P=Staggered pitch length (parallel to pull) g=Gauge length between bolt holes (perpendicular to pull) 2.3.5 Strength of threaded Rods in Rupture

The design of threaded rods in tension, Tdn , as governed by rupture is Given by

1dn u n mT = 0.9f A γ Where An = net root area at the threaded

section 20.784

dπ= × ×

d = Nominal diameter of Bolt

2.3.6 Strength of single Angles in Rupture

The tearing/rupture strength of an angle connected through one leg is affected by shear lag. Shear lag : It is a term used for non-uniform transfer of stresses near the connection. the connected leg will have higher stresses than the unconnected leg. Shear lag effect reduces with increase in connection length.

The design strength of single angle, Tdn, as governed by tearing at net section is given by Tdn = 0.9 fuAnc / ϒm1 + βAgo fy / ϒm0 where

β = 1.40 − 0.076 �wt� �fy

fu� �bs

L� ≤ �fu

fy

γm0γm1

� ≥ 0.7

sb=1.4 – 0.54

L

Where w & bs are as shown in figure below. L = Length of end connection i.e. distance between the outermost bolt in the joint along the force direction or length of the weld along the force direction.


Alternatively the tearing strength of net section may be taken as

1

n udn

m

A fT = α

γ

Where, α = 0.6 for one or two bolts, 0.7 for three bolts and 0.8 for four or more bolts in the end connection or equivalent weld length. An= Net area of the total cross-section Anc=Net area of the connected leg. Ag0=Gross area of the outstanding leg t=Thickness of the leg 2.4DESIGN STRENGTH DUE TO BLOCK SHEAR In this failure made, the failure of the member occurs along a path involving tension on one plane and shear on a perpendicular plane along the fasteners as shown in the figure below.

Block shear failure is a combination of tensile and shear failure under tensile load

Block shear strength at an end connection is calculated as given below. It should be taken as the smaller of Tdn = (Avgfy/√3 ϒmo) + (0.9fuAtn/ϒm1) Shear yielding + tensile rupture

OR

( )1 0db u vn m y tg mT = 0.9f A 3 γ f A /γ+ Shear

rupture + tensile yielding where, Avg, Avn = Minimum gross and net area in

shear along a line of transmitted force, respectively (1-2 and 4-3 as shown in figure(1) and (1-2) as shown in figure (2). Atg, Atn = Minimum gross and net area in tension from the hole to the toe of the angle or next last row of bolt in plates, perpendicular to the line of force respectively (2-3) as shown in figure (1) and figure (2). fu, fy=Ultimate and yield stress of the material respectively.

0 1m mγ , γ =Partial factor of safety in yielding and rupture respectively

0 1m m(γ 1.1, γ 1.25)= =

2.5LUG ANGLES


Length of the end connection of a heavily loaded tension member may be reduced by using lug angles. By using Lug angles there will be saving in gusset plate, but it is upset by additional fasteners and angle required. Hence, now-a-days it is not preferred, IS 800-2007 specification for Lug angles are:

1. The effective connection of the Lug

angles shall be as far as possible terminate at the end of the member connected.

2. The connection of Lug angle to main member shall preferably start in advance of the member to the gusset plate.

3. Minimum of two bolts, rivets or equivalent welds be used for attaching Lug angles to the gusset.

4. If the main member is angle: a) The whole area of the member shall

be taken as the effective rather than net effective section (i.e. with reduction for outstanding Leg area.) The whole area of the member is the gross area less deduction for bolt holes.

b) The strength of Lug angles and fastener connecting Lug angle to gusset plate should be at least 20% more than the force in outstanding Leg.

c) The strength of the fastener connecting Lug angle and main member shall be at least 40% more than the force carried by the outstanding leg.

5. In case the main member is a channel and like: a) As far as possible Lug angles should

be placed symmetrically.

b) The strength of fasteners connecting Lug angle to the gusset should be at least 10% more than the force not accounted for by the direct connection of the member and the attachment of the Lug angles to the member shall be capable of developing 20% in excess of that force.


3.1 INTRODUCTION While designing as structure there are member that are in compression. Vertical compression members in buildings are called column, posts or sanctions. Compression members in trusses are called struts. The jib of crane which carries compression is called boom. Whatever care is taken by the engineers to transfer load axially unexpected eccentricity of load is unavoidable due to imperfection. This eccentricity causes lateral bending moment which results in to bending compression also. As the axial compression increase the lateral deflection increases resulting into additional bending stresses. A stage of instability is reached at load much below crushing strength of compression members. This phenomenon is called buckling of columns. Because of buckling tendency the load carrying capacity of columns is reduced considerably. The load carrying capacity depends upon the end conditions and also on slenderness ratio of the column sections 3.2 SLENDERNESS RATIO Slenderness ratio of a column is defined as the ratio of effective length to corresponding radius of gyration of section. Thus

Slenderness ratio e KL=r rl

=

Where L = Actual length of compression Member

el = KL, effective length r = Appropriate radius of gyration

3.2.1 ACTUAL LENGTH

It is centre to centre distance of compression member between the

restrained ends. In figure below 6 m column is restrained at ends A and B in both y-y and z-z direction. At C it is restrained in z – z direction only. Hence its actual length in -y direction is 6 m while in z-z direction it is equal to AC = 3m only.

3.2.2 EFFECTIVE LENGTH The effective length KL is calculated from the actual length L, of the member considering the rotational and relative translation boundary conditions at the end. IS800-2007 recommends the following. a) If End conditions can be assessed

Where the boundary conditions in planeof buckling can be assessed the effectivelength KL can be calculated on the basisof table given below.

b) Compression Member in Trussesi) In the case of bolted riveted or

welded trusses and braced frames.The effective length KL shall betaken as 0.7 to 1.0 times the actuallength, depending upon the degreeof end restrained provided.

ii) For buckling in the planeperpendicular to the plane of truss,the effective length may be takenas actual length.

c) In FramesIn frame analysis, if deformed shape isnot considered (second order analysis

3 DESIGN OF COMPRESSION MEMBERS


is not used) the effective length depends upon stiffness's of the members connecting at the joint. The method of finding effective length factor K are shown in annex D of IS 800-2007. One can use the graphs given in annexure.

d) In Case of Stepped Columns Expressions for finding effective length

factor for various stepped columns are presented in IS 800-2007 annexure D2 and D3.

Effective length of Prismatic Compression Members

3.3 DESIGN COMPRESSIVE STRESS AND STRENGTH The design compressive stress fcd of axially loaded compression members shall be calculated using the following equation.

( )0 0

y ycd 0.52 2

m m

f f1f = ×γ γ+ – λφ φ

≤

0 0

y ycd

m m

f ff =

γ γχ

≤

Where, 20.5[1 ( 0.2) ]φ α λ λ= + − +

λ =Non-dimensional effective slenderness ratio

2y 2

ycc

f KL= f Erf

= ∏

Fcc = Euler buckling stress 2

2

EKLr

∏=

α = Imperfection factor given in table below χ = Stress reduction factor for different buckling class, slenderness ratio n and yields stress

2 2 0.5

1( )φ φ λ

= + −

0mγ = Partial safety factor for material strength in yielding = 1.1 for steel The design compressive strength Pd of a members is given by Pd = Ae fcd Where Ae is the effective cross -sectional area, which is the same as gross area if bolt holes are filled with bolts. Deductions for bolt holes may be made only if the holes are not fitted with bolts. Imperfection Factor,

Buckling Class

a B c d

α 0.21 0.34 0.49 0.76 3.4 HOW TO SELECT THE SHAPES OF COMPRESSION MEMBERS Since the design stress in compression member decreases with the least radius of gyration, the section should be proportioned to have maximum moment of inertial for the same sectional area. As far


as possible the section should have approximately the same radius of gyration about any axis. This requirement is fulfilled by circular tubes. Next best shape may be square tubing. Moreover it is preferable to use equal angles instead of unequal angles as compression members since such angles have higher rmin values for the same cross –sectional areas. Various shapes of commonly used compression members are shown below.

3.5 STEPS FOR DESIGN OF COMPRESSION MEMBERS The following are the usual steps in the design of compression members: 1. Design stress in compression member

is too assumed. For rolled beam section the slenderness

ratio varies from 70 to 90. Hence design stress may be assumed as 135 N/mm2. For angle struts, the slenderness ratio varies from 110 to 130. Hence, design stress for such members may be assumed as 90 N/mm2. For compression members carrying large loads, the slenderness ration is comparatively small. For such members

design stress may be assumed as 200 N/mm2.

2. Effective sectional area required, d

ecd

PA =

f

Where Pd = Factored load in N fcd = Design stress in N/mm2

3. Select a section to give effective are required and calculate rmin.

4. Determine the effective length and slenderness ration.

5. Calculate design stress and use it to calculate capacity of the section Pc.

6. Revise the section of Pc< Pd (factored load) or if Pc>> Pd.

3.6 BATTENS

Instead of lacing one can use battens to keep members of columns at required distances. Figure below shows the use of batten plates.

3.7 DESIGN OF BATTENED COLUMNS

IS 800-2007 specifies the following rules for the design of battened columns: 1. Batten plates should be provided

symmetrically. 2. The battens shall be placed opposite to

each other at each end of the member and at points where the member is stayed in its length and as far as


practicable, be spaced and proportioned uniformly throughout. The number of battens shall be such that the member is divided into not less than three bays within its actual length from centre to centre of end connections.

3. Battens shall be of plates, angles, channels or I-sections and at their ends shall be riveted, bolted or welded.

4. By providing battens distance between the members of columns is so maintained that radius of gyration about the axis perpendicular to the plane of battens is not less than the radius of gyration about the axis parallel to the plane of battens

yy zz(r > r )

5. The effective slenderness ration of battened columns shall be taken as 1.1 times the maximum actual slenderness ration of the column, to account for shear deformation.

6. The vertical spacing of battens, measured as centre to centre distance between end connection of consecutive battens shall be such that the slenderness ration of any component of column over that distance shall be neither greater than 50 nor greater than 0.7 times the slenderness ration of the member as a whole about its z-z axis.

7. Battens shall be designed to carry the bending moments and shear forces arising from the transverse shear force Vt equal to 2.5% of the total axial force.

8. In case columns are subjected to moments also, the resulting shear should be found and then the design

shear is sum of this shear and 2.5% of axial load.

9. The design shear and moments for batten plates is given by t

bV C

V =NS

and

tV C

M =2N

at each connection.

Where Vt = Transverse shear force as defined

in 7 and 8 C = Distance between centre to centre

of battens longitudinally N = Number of parallel planes S=Minimum transverse distance

between the centroid of the fasteners connecting batten to the main members.

10. The effective depth of end battens (longitudinally) shall not be less than the perpendicular distance between the centroids of the main members. The effective depth of the intermediate battens shall not be less than 3

4 of above

distance. 11. In no case the effective depth of battens

shall be less than twice the width of one member in the plane of the batten. It is to be noted that the effective depth of a batten shall be taken as the longitudinal distance between the outermost fasteners.

12. The thickness of battens shall be not less than 1

50 of the distance between the

innermost connecting lines of rivets, bolts or welds.

13. The length of the weld connecting batten plate to the member shall not be less than half the depth of batten plate. At least one third of the weld shall be placed at each end of this edge.

3.8 DESIGN OF SLAB BASE

3.8.1 When the Column is subjected to Pure Axial Load

In this case pressure will be uniformly distributed under the slab base.


3.8.2 Size of base slab

1. Find the bearing strength of concrete

which given by 0.45 fck. 2. Find the required area of base plate

where Pu is factored axial load on column.

3. Select the size of base plate. For economy as far as possible keep the projections a and b equal.

3.8.3 Thickness of base plate 1. Find the intensity of pressure

PW =

Actualarea of base plate 2. Find the minimum thickness required

using the following formula given by IS 800 – 2007

ts = �2.5w(a2−0.3b2)γmo

fy > tf

where γmo

= 1.1 fy = yield strength of base plate tf = thickness of flange of column The above mentioned formula can be derived by taking Poisson’s ratio 0.3µ = and using elastic bending theory.

Moment at section (1) - (1) for 1 mm width

2a wa=1× a×w× =2 2

2

x-x1×tz =

6

2 2

1 2 2

wa ×6 3waBending stress = f = =2 × t t

Similarly, for bending about (2)-(2) 2

z 2

3wbf =t

Maximum in direction of, 1 21

f ff

E EE µ = −

Now this stress should be less than equal to

0

y

m

fγ

Hence 0

y 2 22

m

f 3w a –μbγ t

≥

0

2 2

my

3w (a – 0.3b )t γf

≥

Where, 0.3µ = Now IS code has modified the above derived formula to account for elastic-plastic behavior of the plate, hence it has taken minimum thickness

0

2 2m

y

2.5w (a – 0.3b ) γt

f=

IS code has taken 21

5×

=tz in between

value of 2

elastic1 tz

6×

= and zplastic = 1×t2

4


4.1 INTRODUCTION

Beam is structural member with length considerably larger than corss sectional dimension subject to lateral loads which give rise to bending moment, shear forces in the member. There are mainly two types of beams based on the lateral supports to compression flanges a) Laterally supported beamsb) Laterally unsupported beamsIf the compression flanges are laterally supported by flooring it is mainly subjected to bending and shear. If the compression flange of beam is not laterally supported the lateral buckling of the compression flange reduces the load carrying capacity of the beam. 4.2BENDING STRENGTH OF LATERALLY SUPPORTED BEAM a) According to IS 800-2007 the design

bending strength of a section which isnot susceptible to web bucklingunder shear before yielding (where

/ wd t ≤ 67 e) shall be determinedas under:Here,d = depth of web;tw = thickness of web

y250 fε =

Case I : If factored design shear force u(V ) 0.6≤times design shear strength of the cross= section (Vd). i.e. u dV 0.6 V≤

where 0

yw

m

Area of web fVd

3 γ×

=×

γm0 = 1.1, area of web = dtw fyw= yield stress of web

The design bending strength Md shall be taken as

0d b p y mM = β z f γTo avoid irreversible deformation under (serviceability loads)

0

e yd

m

1.2 z fM <

γFor simply supported beams

0

e yd

m

1.5z fM <

γ For cantilever beams Where,

bβ = 1.0 for plastic and compact sections

bβ = e

p

zz

for semi-compact sections

p ez , z = plastic and elastic section moduleof the cross-section respectively fy = yield stress of the material and

0mγ = partial safety factor = 1.1 IS 800 – 2007 does not give the design of members belonging to class 4 (slender sections).

4 DESIGN OF BEAMS


4.3 DESIGN PROCEDURE 1. A trial section is selected assuming it is

going to be plastic section (class 1). 2. Then it is checked for the class it

belongs. 3. Check for bending strength. 4. Check for shear strength. 5. Check for the deflection. Case II :If factored design shear force (Vu) > £ 0.6 times design shear strength of the cross – section i.e. Vu> 0.6 Vd The design bending strength shall be taken as Md = Mdv Where, Mdv=design bending strength under high shear Mdv is calculated as given below a) Plastic or Compact Section

( )0

e ydv d d fd

m

1.2 z fM = M –β M –M

γ≤

Where, 2

u

d

2Vβ = –1

V

Md = Plastic design moment of the whole section as calculated in case I.

Vu = Factored applied force. Vd = Design shear Strength. Mfd = Plastic design strength of the area

of the cross-section excluding the shear area, considering partial safety factor 𝛾𝛾𝑚𝑚𝑜𝑜 , Ze = Elastic section modulus of the whole-section.

b) Semi-compact Section

0

e ydv

m

z fM =

γ

Design bending strength of a section which is susceptible to web buckling under shear before yielding

( )67 ,wd t ε> the design bending strength shall be calculated using one of the following methods. a) The bending moment and axial force

acting on the section may be assumed to be resisted by flanges only and the web is designed only to resist shear.

b) The bending moment and axial force acting on the section may be assumed to be resisted by the whole section. In such a case the web shall be designed for combined shear and normal stresses using simple elastic theory in case of semi-compact webs and simple plastic theory in the case of compact and plastic webs.

4.4 SHEAR STRENGTH OF LATERALLY SUPPORTED BEAM

The design shear strength of a section is given by

Vd = Avfyw√3γmo

Where, Av = Shear area ; fyw = Yield strength of the web The shear area may be calculated as given below:

For I and Channel Sections i) Major axis bending

Hot rolled - htw Welded - dtw

ii) Minor axis bending Hot rolled or welded - 2btf

For Rectangular Hollow Sections of Uniform Thickness

i) Loaded parallel to depth (h) Ah–

(b+h)

ii) Loaded parallel to width (b) Ab–(b+h)

iii) Circular hollow tubes of uniform

thickness 2A−

∏


iv) Plates and solid bars A Where, A = Cross-section area: b = Overall breadth of tubular section, breadth of I-section flanges d = Clear depth of web between flanges h = Overall depth of the section tf = Thickness of the flange and tw = Thickness of the web 4.5 WEB BUCKLING STRENGTH Certain portion of beam at supports acts as column to transfer the load from beam to the support. Hence under this compressive force the web may buckle. This may happen under a concentrated load on the beam also. The load dispersion angle may be taken as 45º. Hence, there is need to check for web buckling. However, the rolled section is provided with suitable thickness for web so that web buckling is avoided. In case of built-up sections it is necessary to check for buckling of web and provide web stiffeners.

Hence as per IS 800-2007, effective web buckling strength is to be found based on the cross -section of web

At support, 1 whA = b + t2

Web buckling strength

= cdw 1 w chF = b + t f2

And fc is the allowable compressive stress corresponding to the assumed web column of effective length = 0.7 d where, d is the web height. At concentrated load

( )

cdw 1 w c

cdw 1 w c

hF = b + 2× t ×f2

F = b + h t × f

4.6 WEB CRIPPLING Near the support web of the beam may cripple due to lack of bearing capacity as shown in figure below. The crippling occurs at the root of the radius. IS 800-2007 has accepted the following formula to find crippling of web.

( )0

yww 1 c w

m

fF = b + n t ×

γ

Where, b1 = Stiff bearing length nc = Length obtained by dispersion through the flange to the web junction at a slope 1 : 2.5 to the plane of flange ⇒ nc = 2.5 tf fyw = Yield stress of the web

The care is taken in fixing the web thickness of rolled steel sections to avoid such failure. Hence rolled steel section is selected as a beam section there is no need to check for this failure. However, when built-up sections are selected the web should be checked for this local failure.


5.1 INTRODUCTION Steel has unique physical property, ductility, because of which it is able to absorb large deformations beyond the elastic limit without fracture. Due to this property, steel possesses a reserve of strength beyond its yield, which engineers have tried to utilize in the plastic method of design. Plastic design is an aspect of the limit state design that extends the structural usefulness up to the plastic strength or ultimate load carrying capacity. The term plastic has occurred due to the fact that the ultimate load is found from the strength of steel in the plastic range. Plastic analysis is based on idealised stress-strain curve. The effect of strain hardening is neglected. This method is rapid and provides a rational approach for the analysis of the structure. It also provides striking economy as regards the weight of steel since the sections required by this method are smaller in size than those required by the method of elastic analysis. Plastic analysis and design has its main application in the analysis and design of statically indeterminate framed structures.

5.2 FULLY PLASTIC MOMENT OF A SECTION The fully plastic moment Mp, of a section is defined as the maximum moment of resistance of a fully plasticized or yielded cross-section.

In order to find out the fully plastic moment of a yielded section of a beam as shown in Fig.1, we employ the force equilibrium equation, namely the total force in compression and the total force in tension over that section are equal. Total compression , C = Total tension , T

fy. A1 = fy. A2 ∴ A1 = A2

A = A1 + A2 A1 = A2 = A/2

Plastic Moment of resistance, Mp = fyA1y1� + fyA2y2��

= fyA2

(y1� + y2��) = fyZp where 𝑍𝑍𝑝𝑝, the plastic modulus of the section= 𝐴𝐴

2(𝑦𝑦1�� + 𝑦𝑦2��)

5.3 BENDING OF BEAMS SYMMETRICAL ABOUT BOTH AXES The fibres of the beam across the cross section are stressed in tension or compression according to their position relative to the neutral axis and are strained in accordance with Fig. 2

Fig. 2 Elastic stresses in beams

While the beam remains entirely elastic the stress in every fibre is proportional to its strain and to its distance from the neutral axis. The stress (f) in the extreme fibres cannot exceed fy. (see Fig. 2) When the beam is subjected to a moment

5 PLASTIC ANALYSIS


slightly greater than that, which first produces yield in the extreme fibres, it does not fail. Instead the outer fibres yield at constant stress(fy) while the fibres nearer to the neutral axis sustain increased elastic stresses. Fig. 3 shows the stress distribution for beams subjected to such moments. Such beams are said to be 'partially plastic' and those portions of their cross-sections, which have reached the yield stress, are described as 'plastic zones'.

Fig. 3 Stresses in partially plastic beams The depths of the plastic zones depend upon the magnitude of the applied moment. As the moment is increased, the plastic zones increase in depth, and, it is assumed that plastic yielding can occur at yield stress (fy) resulting in two stress blocks, one zone yielding in tension and one in compression. Fig. 4 represents the stress distribution in beams stressed to this stage. The plastic zones occupy the whole of the cross section, and are described as being 'fully plastic'. When the cross section of a member is fully plastic under a bending moment, any attempt to increase this moment will cause the member to act as if hinged at the neutral axis. This is referred to as a plastic hinge. The bending moment producing a plastic hinge is called the full plastic moment and is denoted by 'Mp'. Note that a plastic hinge carries a constant moment, MP.

Fig. 4 Stresses in fully plastic beams

5.4 SHAPE FACTOR For equilibrium of the cross section, the areas in compression and tension must be equal. For a rectangular cross section, the elastic moment is given by, M = bd2

6fy

The plastic moment is obtained from, Mp = 2. b. d

2. d4

fy = bd2

4fy

Here the plastic moment Mp is about 1.5 times greater than the elastic moment capacity. The ratio of the plastic modulus (Zp) to the elastic modulus (Z) is known as the shape factor (S). Shape factor signifies the reserve strength of the beam section beyond yield point. More shape factor implies, reserve strength is more beyond yielding. 5.4.1 SHAPE FACTOR FOR DIFFERENT CROSS SECTION (a) Rectangular Section S =1.5 (b) Circle S = 1.7 (c) I Section S = 1.12 to 1.14 (d) H Section S = 1.5 (e) Triangle S = 2.34 (f) Diamond Section S = 2 (g) Thin Hollow Circular Section S = 4

𝜋𝜋 = 1.27


5.5 LOAD FACTOR The load factor is the ratio of the collapse load to the working load: Load Factor= Ultimate Load

Working Load= Plastic Moment

Working Moment

Load Factor= 𝜆𝜆 = 𝑀𝑀𝑝𝑝

𝑀𝑀= 𝑓𝑓𝑦𝑦𝑍𝑍𝑝𝑝

𝑓𝑓𝑓𝑓

Load Factor= Shape Factor × Factor of Safety 5.6 FUNDAMENTAL CONDITIONS FOR PLASTIC ANALYSIS (i) Mechanism condition: The ultimate or collapse load is reached when a mechanism is formed. The number of plastic hinges developed should be just sufficient to form a mechanism. (ii) Equilibrium condition: ∑𝐹𝐹𝑥𝑥 = 0, ∑𝐹𝐹𝑦𝑦 = 0, ∑𝑀𝑀𝑥𝑥𝑦𝑦 = 0 (iii) Plastic moment condition: The bending moment at any section of the structure should not be more than the fully plastic moment of the section. 5.7 MECHANISM When a system of loads is applied to an elastic body, it will deform and will show a resistance against deformation. Such a body is known as a structure. On the other hand if no resistance is set up against deformation in the body, then it is known as a mechanism. Various types of independent mechanisms are 5.7.1 Beam Mechanism (a) A simply supported beam has to form one plastic hinge at the point of maximum bending moment. Redundancy, r = 0

(b) A propped cantilever requires two hinges to form a mechanism. Redundancy, r = 1 No. of plastic hinges formed, = r + 1 = 2

(c) A fixed beam requires three hinges to form a mechanism. Redundancy, r = 2 No. of plastic hinges = 2 + 1 = 3

Hence the number of hinges needed to form a mechanism equals the statical redundancy of the structure plus one. 5.7.2 Panel or Sway Mechanism Fig. 5 shows a panel or sway mechanism for a portal frame fixed at both ends.

Fig.5 Panel Mechanism 5.7.3 Gable Mechanism Fig.6 shows the gable mechanism for a gable structure fixed at both the supports.

Fig.6 Gable Mechanism 5.7.4 Joint Mechanism Fig.7 shows a joint mechanism. It occurs at a joint where more than two structural members meet.


Fig.7 Joint Mechanism 5.7.5 Combined Mechanism Various combinations of independent mechanisms can be made depending upon whether the frame is made of strong beam and weak column combination or strong column and weak beam combination. The one shown in Fig.8 is a combination of a beam and sway mechanism. Failure is triggered by formation of hinges at the bases of the columns and the weak beam developing two hinges. This is illustrated by the right hinge being shown on the beam, in a position slightly away from the joint.

Fig.8 Combined Mechanism Note: The number of independent mechanisms (n) is related to the number of possible plastic hinge locations (h) and the number of degree of redundancy (r) of the frame by the equation. n = h – r 5.8 BASIC THEOREMS OF PLASTIC ANALYSIS 1. Lower Bound or Static Theorem A load factor (λs) computed on the basis of an arbitrarily assumed bending moment diagram which is in equilibrium with the applied loads and where the fully plastic moment of resistance is nowhere exceeded will always be less than or at best equal to the load factor at rigid plastic collapse, (λp). λp is the highest value of λswhich can be found.

2. Upper Bound or Kinematic Theorem A load factor (𝜆𝜆𝑘𝑘) computed on the basis of an arbitrarily assumed mechanism will always be greater than, or at best equal to the load factor at rigid plastic collapse (𝜆𝜆𝑝𝑝) 𝜆𝜆𝑝𝑝 is the lowest value of 𝜆𝜆𝑘𝑘which can be found. 3. Uniqueness Theorem If both the above criteria are satisfied, then the resulting load factor corresponds to its value at rigid plastic collapse (𝜆𝜆𝑝𝑝).


Topics Page No

1. STRUCTURAL FASTENERS 111

2. TENSION MEMBER 11

3. COMPRESSION MEMBER 111

4. BEAMS 111

5. PLATE GIRDERS & INDUSTRIAL ROOTS 111

6. PLASTIC ANALYSIS 111

GATE QUESTIONS


Q.1 Identify the most efficient but joint (with double cover plates) for a plate in tension from the patterns (plan views) shown below, each comprising 6 identical bolts with the same pitch and gauge.

a)

b)

c)

d)

[GATE –2001]

Q.2 ISA 100 x 100 x 10mm (Cross sectional area = 1908mm2) is welded along A and B (Refer to figure given below) such that the lengths of the weld along A and B are l1 and l2 respectively. Which of the following is a possibly acceptable combination of l1 and l2?

a) I1 = 60 mm and I2 = 150 mmb) I1 = 150mm and I2 = 60mmc) I1= 150mm and I2 = 150 mmd) Any of the above, depending on

the size of the weld[GATE – 2002]

Q.3 ISA 100 x100 x 10mm (Cross sectional area = 1908mm2) serves as tensile member. This angle is welded to a gusset plate along A and B appropriately as shown. Assuming the yield strength of the steel to be 260 N/mm2 the tensile strength of the member can be taken to be approximately

a) 500kN b) 300kNc) 225kN d) 375kN

[GATE – 2002]

Q.4 When designing steel structures, one must ensure that local buckling in webs does not take place. This check may not be very critical when using rolled steel sections because a) Quality control at the time of

manufacture of rolled sections is very good

b) Web depths available are smallc) Web stiffeners are in-built in

rolled sectionsd) Depth to thickness ratios (of the

web) are appropriately adjusted[GATE – 2002]

Q.5 An ISMB 500 is used as a beam in a multi-storey construction. From the view point of structural design, it can be considered to be 'laterally restrained' when, a) The tension flange is' laterally

restrained'

1 STRUCTURAL FASTENERS


b) The compression flange is `laterally restrained'

c) The web is adequately stiffened d) The conditions in (a) and (c) are

met [GATE – 2002]

Q.6 Rivet value is defined as a) Lesser of the bearing strength of

rivet and the shearing strength of the rivet

b) Lesser of the bearing strength of rivet and the tearing strength of thinner plate

c) Greater of the bearing strength of rivet and the shearing strength of the rivet

d) Lesser of the shearing strength of the rivet and the tearing strength of thinner plate

[GATE –2004]

Q.7 A moment M of magnitude 50 kN-m is transmitted to a column flange through a bracket by using four 20 mm diameter rivets as shown in the figure. The shear force induced in the rivet A is

a) 250 kN b) 176.8 kN c) 125 kN d) 88.4 kN

[GATE – 2004]

Q.8 Which one of the following is NOT correct for steel sections as per IS: 800-1984? a) The maximum bending stress in

tension or in compression in extreme fibre calculated on the effective section of a beam shall not exceed 0.66fy.

b) The bearing stress in any part of a beam when calculated on the area shall not exceed 0.75fy.

c) The direct stress in compression on the gross sectional area of axially loaded compression member shall not exceed 0.6fy.

d) None of the above. [GATE -2005]

Q.9 A fillet-welded joint of 6 mm size is shown in the figure. The welded surfaces meet at 60-90 degree and permissible stress in the fillet weld is 108 MPa. The safe load that can be transmitted by the joint is

a)162.7 kN b)151.6 kN c)113.4 kN d)109.5 kN

[GATE – 2005]

Q.10 In the design of welded tension members, consider the following statements: I. The entire cross-sectional area

of the connected leg is assumed to contribute to the effective area in case of angles.

II. Two angles back-to-back and tack-welded as per code requirements may be assumed to behave as a tee section

III. A check on slenderness ratio may be necessary in some cases. The TRUE statements are

a) only I and II b)onlyII and III c) only I and III d) I, II and III

[GATE – 2006]

Q.11 A steel flat of rectangular section of size 70⨉6 mm is connected to a gusset plate by three bolts each having a shear capacity of 15 kN in holes having diameter 11.5 mm. If the allowable tensile stress in the flat is 150 MPa, the maximum tension that can be applied to the


flat is

a) 42.3 kN b) 52.65 kN c) 59.5 kN d) 63.0 kN [GATE – 2007] Q.12 A bracket connection is made with

four bolts of 10mm diameter and supports a load of 10 kN at an eccentricity of 100mm. The maximum force to be resisted by any bolt will be

a)5 kN b)6.5 kN c)6.8 kN d)7.16 kN

[GATE –2007] Q.13 Rivets and bolts subjected to both

shear stress �τvf,cal� and axial tensile stress �σtf,cal� shall be so proportioned that the stresses do not exceed the respective allowable stresses vfτ , and tf ,σ and the value

of �τvf,calτvf

+ σtf,calσtf

� does not

exceed a)1.0 b)1.2 c)1.4 d)1.8

[GATE – 2008]

Q.14 A 12 mm thick plate is connected to two 8 mm plates, on either side through a 16 mm diameter power driven field rivet as shown in the figure below. Assuming permissible shear stress as 90 MPa and

permissible bearing stress as 270 MPa in the rivet, the rivet value of the joint is

a) 56.70 kN b) 43.29 kN c) 36.19 kN d) 21.65 kN

[GATE – 2009]

Q.15 Two plates, subjected to direct tension, each of 10 mm thickness and having widths of 100 mm and 175 mm, respectively are to be fillet welded with an overlap of 200 mm. Given that the permissible weld stress is 110 MPa and the permissible stress in steel is 150 Mpa, then length of the weld required using the maximum permissible weld size as per IS : 800 — 1984 is

a) 245.3 mm b) 229.2 mm c) 205.5 mm d) 194.8 mm

[GATE – 2010]

Q.16 For the fillet weld of size's' shown in the adjoining figure the effective throat thickness is

a) 0.61 s b) 0.65 s

c) 0.70 s d) 0.75 s [GATE – 2011]


Q.17 In a steel plate with bolted connections, the rupture of the net section is a mode of failure under

a)tension b)compression c)flexure d)shear

[GATE – 2012] Q.18 Two plates are connected by fillet

welds of size 10 mm and subjected to tension, as shown in the sketch. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa respectively. The welding is done in the working shop (γmw = 1.25). As per the Limit State Method of IS 800-2007. The minimum length (rounded off to the nearest higher multiple of 5mm) of each weld to transmit force P equal to 270 kN is

a)100 mm b)105mm c)110 mm d)115 mm

[GATE – 2012] Q.19 A steel section is subjected to a

combination of shear and bending actions. The applied shear force is V and the shear capacity of the section is Vs. For such a section, high shear force (as per IS:800-2007) is defined as

a) V > 0.6Vs b)V > 0.7Vs

c) V > 0.8Vs d) V > 0.9 Vs

[GATE – 2013]

Q.20 The tension and shear force (both in kN) in each bolt of the joint, as shown below, respectively are ___________

a) 30.33 and 20.00 b) 30.33 and 25.00 c) 33.33 and 20.00 d) 33.33 and 25.00

[GATE – 2014]

Q.21 A bracket plate connected to a column flange transmits a load of 100 kN as shown in the following figure. The maximum force for which the bolts should be designed is _______ kN.

[GATE – 2015]

Q.22 Prying forces are a) shearing forces on the bolts

because of the joints b) tensile forces due to the

flexibility of connected parts c) bending forces on the bolts

because of the joints d) forces due the friction between

connected parts [GATE – 2015]

Q.23 Two plates are connected by fillet

welds of size 10 mm and subjected to tension, as shown in the figure. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 Mpa, respectively.


The welding is done in the workshop (γmw =1.25).

As per the Limit State Method of IS 800: 2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN (factored) is (a) 90 mm (b) 105 mm (c) 110 mm (d) 115 mm

[GATE – 2016]

Q.24 Two bolted plates under tension with alternative arrangement of bolt holes are shown in figures 1 and 2. The hole diameter, pitch, and gauge length are d, p and g, respectively.

a) 2p > 2gd b) 2p 4gd< c) 2p 4gd> d) p > 4gd

[GATE – 16]

Q.25 A column is subjected to a load through a bracket as shown in the figure

The resultant force (in kN, up to one decimal place) in bolt 1 is ___________ [GATE – 17]

Q.26 A fillet weld is simultaneously subjected to factored normal and shear stress of 120 MPa and 50 MPa, respectively. As per IS 800:2007, the equivalent stress (in MPa, up to two decimal place) is_____________ [GATE – 18] Q.27 Four bolts P, Q, R and S of equal diameter are used for a bracket subjected to a load of 130 kN as shown in the figure.

The force in bolt P is

(a) 32.50 kN (b) 69.32 kN (c) 82.50 kN (d) 119.32 kN

[GATE – 18] Q.28 In a fillet weld, the direct shear stress and bending tensile stress are 50 MPa and 150 MPa, respectively. As per IS 800:2007, the equivalent stress (in MPa, up to two decimal places) will be ________________ [GATE – 18]


Q.1 (a) For main plate critical most section is outermost section for that a outer most section net area should be more to resist more load in all four case we will get more netA in case (a)

Q.2 (a) The C.G of the angle is close to B compared to A. Therefore, the length of the weld, 2l shall be more than 1l so that the algebraic sum of moments of the two welds 1l & 2l about C.G of the angle will be zero and there is no eccentricity.

Q.3 (a) Tensile strength of the member

= Ag⨉fy = 1908⨉260 = 500 kN

Q.4 (d)

Q.5 (b )

Q.6 (a)

Q.7 (b)

Force due to moment effect,

m 2

M.rFr

=∑

2 2r 50 50= + 70.71mm 0.0707m= =

m 2

50 0.0707F4 0.0707×

=×

176.8kN=

Q.8 (d)

Q.9 (c)

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(a) (a) (a) (d) (b) (a) (b) (d) (c) (d) (a) (d) (c) (b)

15 16 17 18 19 20 21 22 23 24 25 26 27 28

(b) (b) (a) (b) (a) (d) 156.20 (b) (b) (c) 6 147.99 69.32 173.2

ANSWER KEY:

EXPLANATIONS


Total length of weld,

l = 100+50+100 = 250 mm Throat thickness of fillet, t = 0.7⨉ size

= 0.7⨉ 6 = 4.2 mm Strength, P = l × t × τvf = 250⨉4.2⨉108 = 113.4 kN Q.10 (d) Q.11 (a) Steel flat 70⨉6 mm Capacity of each bolt = 15 kN, No. of bolt = 3 So, maximum load transferred = 3⨉15 = 45 kN Tearing Strength of flat = (70-2⨉11.5)⨉6⨉150 = 42300 N = 42.3 kN Maximum tension that can be applied to the flat = minimum of above two = 42.3 kN Q.12 (d)

Force due to axial load effect,

aw 10F = = 2.5 kNN 4

=

Force due to moment effect,

m 2

M.rF = ;r∑

2 2r = 30 +40 = 50mm = 0.05m M = W e = 10 0.1 = 1 kNm× ×

m 2

1 0.05F =4 0.05××

mF = 5 kN Resultant force,

2 2r a m a mP F +F +2F F cos θ=

40Cos θ 0.850

= =

2 2r P = 2.5 5 (2 2.5 5 0.8)+ + × × ×

Pr = 7.16 kN Q.13 (c) Q.14 (b) Gross diameter of rivet, D = 16 + 1.5 = 17.5 mm Strength of rivet in shearing,

Ps = 2 × π4

× d2 × τvf (Double shear for butt joints)

( )2π=2 17.5 904

× × ×

sP 43.29 kN= Strength of rivet in bearing,

Pb = d.t.σpf = 17.5⨉12⨉270 bP 56.7 kN= Rivet value vR is smaller of s bP & P

vR 43.29 kN= Q15 (b)

Max. size of weld as per code

maxS = T - 1.5 = 10 - 1.5 = 8.5 mm Size, S = 8.5 mm

Throat thickness ‘t’ = 0.7⨉ 8.5 = 5.95 mm Strength of the plate ‘P’ is

stP = (b.T) σ = (100 10) 150× ×

3P = 150 ×10 N: Let l = total length of the weld required. Strength of the weld,

s vfP = lt × τ = l × 5.95 × 110 Equating sP = P


150⨉103 = l⨉5.95⨉110 l = 229.2 mm

Q.16 (b) Throat thickness, t = k.S

Angle between fusion faces

k

60° - 90° 0.70 91° - 100° 0.65 101° - 106° 0.6 107° - 113° 0.55 114° - 120° 0.50

Q.17 (a) Q.18 (b) Design strength of fillet weld,

Pd = Lwtfu

√3γmw

270⨉103 = 2Lw(0.7 × 10) ×410

√3 × 1.25

wL 101.84mm= Rounded off to 105 mm Q.19 (a)

As per clause 9.2.1 (IS: 800-2007) for combined shear and bending: Factored value of applied shear force is greater than or equal to shear strength for high shear. i.e., V > 0.6 Vs

Q.20 (d)

3 4 4sin θ = , cos θ = Pu cos θ = .Pu5 5 5

Pu sin θ = 3 .Pu5

Tension in each bolt = u uP cosθ 4P 4 250 33.33kN

6 5×6 30= = × =

Shear in each bolt =

u uP sinθ P3 3 250 25 kN6 5 6 5×6

= = × =

Q.21 (156.20)

D100F 20 kN

5P n

== =

Ft = (P. d)r ∑ r2 =

100 × 600 × 75√2

4 × �75√2�2 = 141.42 kN

2 2

R D t D tF = F +F +2×F ×Fcosθ

( ) ( )2 2 120 + 141.42 +2×20×141.42×2

= 156.20kN

1Cos θ =2

θ = 45°⇒ Q.22 (b) Q.23 (b) Maximum force carried by plates,

0

g y

m

A f 100 12 250P = 272.73kNγ 1.1

× ×= =

Load carried by each weld = P2

= 136.36 kN For minimum length of weld, Strength of weld=Load carried by weld

𝑙𝑙𝑤𝑤 × 𝑠𝑠 × 𝑓𝑓𝑢𝑢

√3𝛾𝛾𝑚𝑚1= 136.36 × 103

3

w410l (10 0.7) 136.36 103×1.2

⇒ × × × = ×


lw = 102.9mm next multiple of 5 is ≈ 105mm

Q.24 (c) 2p 4gd>

This question can be solved by trick, Option (b) and (d) are not dimensionally correct.

Q.25 (6.0 kN) F1 = P

n = 10

4= 2.5 kN

F2 = Pe

∑ri2 r1 = 10×15

4×52× 5 = 7.5 kN

𝐹𝐹𝑟𝑟 = �𝐹𝐹12 + 𝐹𝐹22 + 2𝐹𝐹1𝐹𝐹2 cos 𝜃𝜃 = √2.52 + 7.52 + 2 × 2.5 × 7.5 × cos 135° = 6.0 kN Q.26 (147.99 MPa) Equivalent stress = √𝜎𝜎2 + 3𝜏𝜏2 = 147.99 MPa Q.27 (b) 69.32 kN Direct shear in bolt P, F1 = 130

4= 32.5 kN

Shear due to twisting moment F2 = 130×200×√1202+502

4×�√1202+502�2

𝐹𝐹2 = 50 𝑘𝑘𝑘𝑘

𝜃𝜃 = tan−112050

= 67.38° Resultant force in bolt P

= �𝐹𝐹12 + 𝐹𝐹22 + 2𝐹𝐹1𝐹𝐹2 cos𝜃𝜃

=√32.52 + 502 + 2 × 32.5 × 50 × cos 67.38°

= 69.32 kN Q.28 (173.20MPa) Direct shear stress (τ) = 50 MPa

Bending tensile stress (σ) = 150 MPa fR = �σ2 + 3τ2 = √1502 + 3 × 1502 = 173.20 MPa


Data for Q.1 and Q.2 is given below: A truss tie consisting of 2 ISA75 75 8mm× ×carries a pull of 150 kN. At ends the two angles are connected, one each on either side of a 10 mm thick gusset plate, by 18 mm diameter rivets arranged in one row. The allowable stresses in rivet are

2sf 90N / mm= and 2

brf 250 N / mm .=

Q.1 Maximum tensile stress in the tie in N/mm2 is a) 93.6 b) 87.5c) 77.2 d) 66.0

[GATE – 03]

Q.2 Minimum number of rivets required at each end is a) 2 b) 3c) 4 d) 5

[GATE – 03]

Q.3 Two equal angles ISA 100mm ⨉ 100mm of thickness 10mm are placed back-to-back and connected to the either side of a gusset plate through a single row of 16 mm diameter rivets in double shear. The effective areas of the connected and unconnected legs of each of these angles are 775 mm2 and 950 mm2, respectively. If these angles are NOT tack riveted, the net effective area of this pair of angle is

a) 3650 mm2 b) 3450 mm2

c) 3076 mm2 d) 2899 mm2

[GATE – 04]

Q.4 The permissible stress in axial tension in steel member on the net effective area of the section shall not exceed the following value (fy is the yield stress) a) 0.80fy b) 0.75fy

c) 0.60fy d) 0.50fy

[GATE – 05]

1 2 3 4 (a) (c) (d) (c)

ANSWER KEY:

2 TENSION MEMBER


Q.1 (a) If angles are not tack riveted they behave individually. Gross diameter of rivet, d = 18 + 1.5 = 19.5 mm

Only connected to gusset plate (Not tack riveted) 𝐴𝐴𝑛𝑛𝑛𝑛𝑛𝑛 = 2 × (𝐴𝐴1 + 𝑘𝑘𝐴𝐴2)

1A = Net Area of connected leg

2A = Gross c/s area of unconnected leg

18A = 75 19.5 82

− − × Q

2= 412 mm [ ]2A = 75 4 8 568− × =

11

1 2

3AK = 0.6853A +A

=Q

net A = 2[412 + 0.685 568]×Q 21602.2 mm=

Q maximum tensile stress, 3

2tcal

net

P 150 10σ 93.61 N / mmA 1602.2

×= = =

Q.2 (c) Strength of rivet in shear;

2S vf

πP = 2× d ×τ (Rivets are under4

double shear)2 3

SπP = 2× (19.5) × 90 = 53.76 104

× N

= 53.76 kN Strength of rivet in bearing

b vfP = d.t.τ3=19.5 10 250 = 48.75 10 N× × ×

= 48.75 kN Rivet value (Rv) is minimum of above two value Rv = 48.75 kN Number of rivets; N = load/Rv Provide 4 rivets.

Q.3 (d) If two angles are not tack riveted they will be considered as single angle connected on one side of gusset plate.

2 21 2A = 775 mm ; A = 950 mm

11

1 2

3A 3×775K = = = 0.713A +A (3×775)+950

1K = 0.71

net 1 1 2A = 2(A + k A )= 2 (775+0.71 950)× ×

2 Anet = 2899 mm∴

Q.4 (c) Permissible stress in axial tension is 0.60fy as per the code of practice, IS : 800

EXPLANATIONS


Q.1 Consider the following two statements related to structural steel design, and identify whether they are true or FALSE. I. The Euler buckling load of a

slender steel column depends on the yield strength of steel.

II. In the design of laced column,the maximum spacing of the lacing does not depend on the slenderness of column as a whole.

a) Both statements I and II areTRUE

b) Statement I is TRUE, andStatement II is FALSE

c) Statement I is FALSE, andStatement II is TRUE

d) Both Statements I & II areFALSE

[GATE – 2001]

Q.2 In the design of lacing system for a built-up steel column, the maximum allowable slenderness ratio of a lacing bar is a) 120 b) 145c) 180 d) 250

[GATE – 2003]

Q.3 A strut in a steel truss is composed of two equal angles ISA 150 mm x 150 mm of thickness 100mm connected back-to-back to the same side of a gusset plate. The cross sectional area of each angle is 22921 mm and moment of inertia ( )xx yyI I= is 6335000 mm4. The distance of the centroid of the angle from its surface( )x yC C= is 40.8 mm. The minimum radius of gyration of the strut is

a) 93.2 mm b) 62.7 mmc) 46.6 mm d) 29.8 mm[GATE – 2004]

Q.4 Consider the following statements I. Effective length of a battened

column is usually increased to account for the additional load on battens due to the lateral expansion of columns.

II. As per IS: 800-1984,permissible stress in bending compression depends on both Euler buckling stress & the yield stress of steel.

III. As per IS: 800-1984, theeffective length of a column effectively held in position at both ends but not restrained against rotation, is taken to be greater than that in the ideal end conditions.

The TRUE statements are a)only I and II b)only II and III c)only I and III d)I, II and III

[GATE – 2006]

Q.5 The square root of the ratio of moment of inertia of the cross-section to its cross-sectional area is called a)second moment of area b)slenderness ratio c)section modulus d)radius of gyration

[GATE –2009]

Q.6 Consider the following statements for a compression member:

I. The elastic critical stress in compression increases with decrease in slenderness ratio.

3 COMPRESSION MEMBER


II. The effective length depends on the boundary conditions at its ends.

III. The elastic critical stress in compression is independent of the slenderness ratio.

IV. The ratio of the effective length to its radius of gyration is called as slenderness ratio.

The TRUE statements are (a) II and III (b) III and IV (c) II,III and IV (d) I,II and IV [GATE – 2009] Q.7 The ratio of the theoretical critical

buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is equal to

(A) 0.5 (B) 1.0 (C) 2.0 (D) 4.0 [GATE – 2012] Q.8 Two steel columns P (length L and

yield strength fy = 250 MPa) and Q (length 2L and yield strength fy = 500 MPa) have the same cross-sections and end conditions. The ratio of buckling load of column P to that of column Q is:

(A) 0.5 (B) 1.0 (C) 2.0 (D) 4.0

[GATE – 2013]

Q.9 Consider two axially loaded columns, namely, 1 and 2, made of a linear elastic material with Young’s modulus 2⨉105 MPa, square cross-section with side 10 mm, and length 1 m. For column 1, one end is fixed and the other end is free. For Column 2, one end is fixed and the other end is pinned. Based on the Euler’s theory, the ratio (up to one decimal place) of the buckling load of column 2 to the buckling load of column 1 is________

[GATE – 2017] Q.10 A column of height h with

rectangular cross section of size of a × 2a has a buckling load of P. If the cross-section is changed to 0.5a × 3a and its height changed to 1.5h, the buckling load of the redesigned column will be

(a) P/12 (b)P/4 (c) P/2 (d)3P/4

[GATE – 2018]

1 2 3 4 5 6 7 8 9 10

(b) (b) (c) (a) (d) (d) (d) (d) 8 (a)

ANSWER KEY:


Q.1 (b)

Euler buckling load 2

2EI

lπ

=

E = σyε

Q Euler buckling load depends on yield strength for riveted or welded lacing system, L

rmin = 50

or 0.7 times 'λ' (slenderness ratio) of member as a whole whichever is less.

Q.2 (b) The slenderness ratio of a lacing bar should not exceed 145 as per the code (IS : 800)

Q.3 (c)

Total area 2A 2 a 2 2921 5842mm= = × = × =

Since X-X axis of the built-up section passes through the C.G of individual Angle, minimum moment of inertia occurs about axis XX. Minimum moment of inertia as about X – X is

XX XXI 2 I of each angle= × 2= 2 6335000 mm×

XXmin

I r = A

∴

2 6335000 46.6 mm5842

×= =

Q.4 (a) For two ends hinged condition (statement No. III), eI L;= hence III is wrong.

Q.5 (d) Radius of Gyration

Moment of inertiaCross Sectional Area

=

Q.6 (d)

fcc = π2E

�𝑳𝑳𝒓𝒓�𝟐𝟐

The elastic critical stress in compression increases with decrease in slenderness ratio.

Q.7 (d) Theoretical critical buckling load

Pcr = π2EI(L)2

Pcr ∝1L2

PcrfixedPcrpinned

= �LpinnedLfixed

�2

PcrfixedPcrpinned

= � L0.5 L

�2

PcrfixedPcrpinned

= 4

Q.8 (d)

PcrPPcrQ

= �LQLP�2

PcrfixedPcrpinned

= �2LL�2

PcrfixedPcrpinned

= 4

Q.9

EXPLANATIONS


𝑃𝑃𝑐𝑐𝑐𝑐2𝑃𝑃𝑐𝑐𝑐𝑐1

= �𝐿𝐿1𝐿𝐿2�2

𝑃𝑃𝑐𝑐𝑐𝑐2𝑃𝑃𝑐𝑐𝑐𝑐1

= � 2𝐿𝐿1√2𝐿𝐿�2

𝑃𝑃𝑐𝑐𝑐𝑐𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

𝑃𝑃𝑐𝑐𝑐𝑐𝑝𝑝𝑓𝑓𝑝𝑝𝑝𝑝𝑓𝑓𝑓𝑓= 8

Q.10 (a) Case- I

Pcr = π2EImin

leff2

Imin = 2a(a)3

12= a4

6

leff

2 = h2

∴ Pcr 1 = P =π2Ea

4

6h2

= 16π2Ea4

h2

π

2Ea4

h2= 6P

Case – II

Pcr = π2EImin

leff2

Imin = 3a(0.5a)3

12= a4

32

leff

2 = (1.5h)2 = 94

h2

∴ Pcr2 =π2E 1

32a4

94h

2 = 172

π2Ea4

h2

Pcr2 = 1

72× 6P = P

12

Pcr2 = P

12


Q.1 Group-I contains some elements in design of a simply supported plate girder and Group-II gives some qualitative locations on the girder. Match the items of two lists as per good design practice and relevant codal provisions. Group I P. flange splice Q. web splice R. bearing stiffeners S. horizontal stiffener Group II 1. at supports (minimum)2. away from centre of span3. away from support4. in the middle of span5. longitudinally somewhere in the

compression flangeCodes:

P Q R S a) 2 3 1 5 b) 4 2 1 3 c) 3 4 2 1 d) 1 5 2 3

[GATE – 2003]

Q.2 A square steel slab base of area 1 m2 is provided for a column made of two rolled channel sections. The 300 mm x 300 mm column carries an axial compressive load of 2000 kN. The line of action of the load passes through the centroid of the column section as well as of the slab base. The permissible bending stress in the slab base is 185 MPa. The required minimum thickness of the slab base is a) 110 mm b) 89 mmc) 63 mm d) 55 mm

[GATE –2004]

Q.3 An unstiffened web I section is fabricated from a 10mm thick plate by fillet welding as shown in the figure. If yield stress of steel is 25OMPa, the maximum shear load that section can take is

a) 750kN b) 350kNc) 337.5 kW d) 300kN

[GATE – 2005]

Q.4 The adjoining figure shows a schematic representation of a steel plate girder to be used as a simply supported beam with a concentrated load. For stiffeners, PQ (running along the beam axis) and RS (running between the top and bottom flanges) which of the following pairs of statements will be TRUE?

a) i) RS should be provided under the concentrated load only ii) PQ should be placed in thetension side of the flange.

b) i) RS helps to prevent local buckling of the web. ii) PQ should be placed in thecompression side of the flange.

c) i) RS should be provided at supports ii) PQ should be placed along theneutral axis

4 BEAMS


d) i) RS should be provided away from points of action of concentrated loads. ii) PQ should be provided on the compression side of the flange.

[GATE – 2011] Q.5 The semi-compact section of a

laterally unsupported steel beam has an elastic section modulus, plastic section modulus and design bending compressive stress of

3500 cm , 3650 cm and 200MPa,

respectively. The design flexural capacity (expressed in kN-m) of the section is ____.

[GATE – 2016]

1 2 3 4 5

(a) (d) (d) (b) 100

ANSWER KEY:


Q.1 (a) Flange splice: Flange resists moment. Since moment will be generally maximum at centre, there should not be flange joint at the centre. Web spice: Web resists shear force. Maximum shear force occurs at support. Hence avoid the web splice at the support. Bearing stiffener: It is to avoid compression buckling. Hence provided near the compressive flange.

Q.2 (d) a b 1000 300 / 2 350mm= = − =

32

2P 2000 10w 2N / mm

Actual area (1000)×

= = =

Thickness ‘t’

22

bs

3w bt a4

= − σ

223 2 350t 350

185 4 ×

= −

t = 55mm∴

Q.3 (d) τva = V

d×tww v.aV d.t . = τ

2v.aτ = 0.4fy = 0.4 × 250 = 100 N/mm

3V 300 10 100 300 10 N= × × = × = 300 kN

Q.4 (b) Vertical stiffeners are provided to prevent buckling of web. Longitudinal stiffener (PQ) is provided to prevent bucking of web due to compression. Hence it should be provided on compression side.

Q.5 (100) As per IS 800, the design bending strength of laterally unsupported beam as governed by lateral torsional buckling is:

d b p bdM .Z .f= β

eb

p

Zβ =Z

for semi compact section

ed p bd

p

ZSo M .Z .fZ

=

e. bd3 6500 10 200 10 100kNZ f -m−= × × × =

EXPLANATIONS


Q.1 Which of the following elements of a pitched roof industrial steel building primarily resists lateral load parallel to the ridge? a) Bracings b) purloinsc) Truss d) columns

[GATE – 2003]

Q.2 In a plate girder, the web plate is connected to the flange plates by fillet welding. The size of the fillet welds is designed to safely resist

a) the bending stresses in theflanges

b) the vertical shear force at thesection

c) the horizontal shear forcebetween the flanges and the webplate

d) the forces causing buckling inthe web

[GATE – 2004]

1 2 (a) (c)

ANSWER KEY:

5 PLATE GIRDERS & INDUSTRIAL ROOTS


Q.1 A prismatic beam (as shown below) has plastic moment capacity of Mp, then the collapse load P of the beam is

a) p2ML

b) p4ML

c) p6.ML

d) p8ML

[GATE-2014]

Q.2 For formation of collapse mechanism in the following figure, the minimum value of Pu is cMP/L. Mp and 3Mp denote the plastic moment capacities of beam sections as shown in this figure. The value of c is _______

[GATE-2015]

Q.3 A fixed end beam is subjected to a load, W at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is

a)16.5 MP/L b)15.5 MP/L c)15.0 MP/L d)16.0 MP/L

[GATE-2015]

Q.4 A propped cantilever of span L carries a vertical concentrated load at the mid-span. If the plastic moment capacity of the section is Mp, the magnitude of the collapse load is

a) P8ML

b) P6ML

c) P4ML

d) P2ML

[GATE-2016]

Q.5 A fixed-end beam is subjected to a concentrated load (P) as shown in the figure. The beam has two different segments having different plastic moment capacities (Mp, 2Mp) as shown.

The minimum value of load (P) at which the beam would collapse (ultimate load) is a) p7.5M / L b) p5.0M / L c) p4.5M / L d) p2.5M / L

[GATE-2016]

6 PLASTIC ANALYSIS


Q.6 The prismatic propped cantilever beam of span L and plastic moment capacity Mp is subjected to a concentrated load at its mid-span. If the collapse load of

the beam is αMp

L , the value of α is__________

[GATE-2018]

1 2 3 4 5 6 (c) 13.33 (c) (b) (a) 6

ANSWER KEY:


Q.1 (c) Degree of static indeterminacy DS =0 ∴ Number of plastic hinges=Ds+1 = 1 From principal of virtual work

p PL P L-M .θ - M .θ + P. .θ - . .θ =02 2 3

P

LP = 6M

⇒

Q.2 (13.33) Mechanism-I

( ) uP P3M . M 2 M LP ×P. ×θ4

θ+ θ + θ =

p uL6M .θ = P . .θ4

⇒

pu

MP 24

L⇒ =

Mechanism-II 1. 3. 3φ = θ⇒ φ = θ

( )P P P uL3M . M M . P4

θ+ θ+ φ + φ = × φ

( )P P P3M . M 3 M .3⇒ θ+ θ+ θ + θ

u 3 LP θ×4

= ×

P uLP10M . 3 θ×4

⇒ θ = ×

P PM M40 . 13.333 L L

⇒ =

So, C = 13.33

Q.3 (c)

Plastic hinges formed = 3 (1)

θα2

=

2θ = α

θ3

VL/

=

L= θ3

∆

α2L3

∆ =

2θ = α

P P P P2M 2M 2M M Wθ+ θ+ α + α = ∆

P P P P2M θ + 2M θ +M θ2

M θ+

LW θ3

= ×

P PWL 16.55.5M W M

3 L= ⇒ =

(2)

P P P P2M M M M wθ+ θ+ ∆ + θ = ×∆

P5 M θ θW L3

=

EXPLANATIONS


pM15 =W

L

Lowest is collapse load P15ML

Q.4 (b)

p p pM θ M θ M L θθ P 02

− − − + × =

pPLθ3M θ =

2

p6MP =

L

Q.5 (a)

Mechanism -I

p p p p2LθM θ - M θ - M θ - -2 M θ +P = 0

3

p2P- 5 LM θ+ θ=0

3

p2PL =5M

3

p15MP 7.5Mp / L

2L=

Mechanism -II

21 41θ = 3 3

φ

θ 2= φ

p p p p2L-2M θ-2M θ-2M f-M f+P θ = 0-43

p p2PLM θ-3M f+ θ=0

3

p pθ 2PL-4M θ-3M + θ=02 3

p11 2PLM θ= θ2 3

p p33P M 8.25M4

= =

So the minimum value of load = 7.5 Mp/L

Q.6 (6)


−3𝑀𝑀𝑝𝑝𝜃𝜃 = −𝑊𝑊𝐿𝐿2𝜃𝜃

6 𝑀𝑀𝑝𝑝

𝐿𝐿= 𝑊𝑊

∴ 𝛼𝛼 = 6


Q.1 A singly reinforced balanced section designed by limit state method will have 1. Lesser depth compared to designed

by working stress method2. Higher value of steel compared to

designed by working stress method3. Lower cost compared to designed

by working stress methodWhich of these statements is/are correct? a) 1 and 2 b) 2 and 3c) 3 and 1 d) Only 3

Q.2 Which one represents correct deflected profile corresponding to section 1-1for a two way simply supported RCC slab subjected to UDL as shown in the figure below?

a) b)

c) d)

Q.3 Which expression represents effective flange width of isolated reinforced concrete T- beam? Symbols have their usual meaning.

a) 0w f

l +b +6d6

Or `b’ whichever is lesser

b) 0w f

l +b +3D12


c) 0w

0

0.5l +bl +4b


d) 0w

0

l +bl +4b


Q.4 A deep continuous beam has effective depth of 500 mm and effective cover of 50 mm. Its maximum effective length is a) 1375 mmb) 1100 mmc) 1275 mmd) cannot be determined from the

given data

Q.5 If a beam of cross-section 350 mm × 600 mm is subjected to torsional moment of T = 50 KN-m then equivalent shear will be a) 129 kN b) 229 kNc) 329 kN d) Zero

Q.6 For a continuous RC beam, match list-1(condition) with list-2 (Placement of live load) and select the correct answer using the codes given below the lists: List-1 A. For maximum sagging moment

in the span B. For maximum hogging moment

at a support C. For maximum hogging moment

in a span List- 2 1. The spans adjoining the span as

well as alternate span2. The same span as well as

alternate span3. The adjacent span on both side

of this support as well as spansalternate to these

ASSIGNMENT QUESTIONS


4. Span next to the adjacent spans of the support plus alternate spans

A B C a) 2 3 1 b) 1 4 2 c) 2 4 1 d) 1 3 2

Q.7 For a continuous slab of 3 m × 3.5 m size the minimum overall depth of slab to satisfy vertical deflection limits is a)5 cm b) 7.5 cm c) 10 cm d) 12 cm

Q.8 Which one of the following

statements about the percentage of tensile steel required to produce a balanced reinforced concrete section is correct? The required percentage of steel a) Reduces as the yield strength of

steel increases b) Remains unchanged irrespective

of the yield strength of steel c) Is the same for a given quality of

steel irrespective of whether working stress method is followed or ultimate load method used

d) Is only function of the modulus of elasticity of steel

Q.9 The effective depth of the singly

reinforced rectangular beam is 30 cm. The section is over reinforced and the neutral axis is 12 cm below the top. If the maximum stress attained by concrete is 50 kg/cm 2 and modular ratio is 18, then stress developed in the steel would be a) 1800 kg/cm 2 b) 1600 kg/cm 2 c) 1350 kg/cm 2 d) 1300 kg/cm 2

Q.10 If permissible stress in steel in tension is 140 N/mm 2 , then the depth of neutral balance section using working stress method is a) 0.35d

b) 0.40d c) 0.45 d d) Dependent on grade of concrete

also Q.11 As per the provision of IS: 456-2000,

in the limit state method for design of beams, the limiting value of the depth of neutral axis in a reinforced concrete beam of effective depth `d’ is given as a) 0.53 d b) 0.48 d c) 0.46 d d) Any of the above depending on

the different grade of steel

Q.12 The span to depth ratio limit is specified in IS: 456-2000 for the reinforced concrete beam, in order to ensure that the a) Tensile crack width is below a

limit b) Shear failure is avoided c) Stress in the tension

reinforcement is less than the allowable value

d) Deflection of the beam is below a limiting value

Q.13 A floor slab of thickness t is cast monolithically transverse to a rectangular continuous beam of span L, and width, B. If the distance between two consecutive points of contra flexure is, L 0 , the effective width of compression flange at a continuous support is a)B b)

3L

c)B+ 12t d) B+ 6t+ 06

l

Q.14 The total compressive force at the

time of failure of a concrete beam section of a width `b’ without considering the partial safety factor of the material is a) 0.36 ckf b uX b) 0.54 ckf b uX


c) 0.66 ckf b uX d) 0.8 ckf b uX 4

Q.15 What is the modular ratio to be used in the analysis of RC beam using working stress method if the grade of concrete is M 20? a)18.6 b)13.3 c)9.9 d)6.5

Q.16 A simply supported RC beam having clear span 5 m and support width 300 mm has the cross-section as shown in the figure below?

What is the effective span of the

beam as per IS: 456? a) 5300 mm b) 5400 mm c) 5200 mm d) 5150 mm

Q.17 What is the adoptable maximum

spacing between vertical stirrups in a RCC beam of a rectangular cross-section having an effective depth of 300 mm a) 300 mm b) 275 mm c) 250 mm d) 225 mm

Q.18 In limit state design method, the

moment of resistance for a balanced section using M20 grade concrete and HYSD steel of grade Fe 415 is given by M ,limu = Kbd 2 , What is the value of K? a)2.98 b)2.76 c) 1.19 d) 0.89

Q.19 The maximum strain in the tension reinforcement in the section at failure when designed for the limit state of collapse should be

a) y

s

f> +0.002

1.15E

b) y

s

f< +0.002

1.15E

c) Exactly equal to y

s

f+0.002

1.15E

d) < 0.002 Where, F y = Characteristic strength of steel, & E S = Modulus of elasticity of steel

Q.20 Consider the following statements. In an under-reinforced concrete beam, 1) Actual depth of neutral axis is

less than the critical depth of neutral axis

2) Concrete reaches ultimate stress prior to steel reaching the ultimate stress

3) Moment of resistance is less than that of balanced sections.

4) Lever arm of resisting couple is less than of balanced section

Which of these statements is/are correct? a) 1 and 2 b)1 and 3 c) 2, 3 and 4 d)1 and 4

Q.21 A doubly reinforced concrete beam

ahs effective cover `d’ to the centre of compression reinforcement. ` ux ’ is the depth of neutral axis and `d’ is the effective depth to centre of tension reinforcement. What is the maximum strain in concrete at the level of compression reinforcement? a) 0.0035 (1- d’/d) b) 0.0035 (1-d’/ ux ) c) 0.002 (1-d’/ ux ) d) 0.002(1- d’/d)

Q.22 A reinforced concrete beam is subjected to the following bending moments: Dead load – 20 kN-m

Live load - 30 kN-m Seismic load – 10 kN-m The design bending moment for

limit state of collapse is


a) 60 kN-m b) 75 kN-m c) 72 kN-m d) 80 kN-m

Q.23 The maximum depth of neutral axis

for a beam width `d’ as the effective depth, in limit state method of design for Fe 415 steel is a) 0.46d b) 0.48d c) 0.50d d) 0.53d

Q.24 The main reinforcement of a RC slab

consists of 10mm bars at 10 cm spacing. If it is desired to replace 10 mm bars by 12 mm bars, then the spacing of 12 mm bars should be a)12 cm b)14 cm c)14.40 cm d)16 cm

Q.25 In case of 2-way slab, the limiting deflection of the slab is a) Primarily a function of the long

span b) Primarily a function of the short

span c) Independent of long or short

spans d) Dependent on both long and

short spans

Q.26 When shear reinforcement is not provided in flat slab, the calculated shear stress at the critical section shall not exceed s cK τ , where sK is a) (1 + β C ) b) (0.6 + β C ) c) (0.5 + β C ) d) (1.5 + β C )

Q.27 A reinforced concrete slab is 75 mm thick. The maximum size of reinforcement bar that can be used is of a)12 mm diameter b)10 mm diameter c)8 mm diameter d)6 mm diameter

Q.28 In the design of two-way slab restrained at all edges, torsional reinforcement required is

a) 0.75 times the area of steel provided at the mid span in the same direction

b) 0.375 times the area of steel provided at the mid span in the same direction

c) 0.375 times the area of steel provided in the shorter span

d) nil

Q.29 A reinforced concrete member is subjected to combined action of compressive axial force and bending moment. If ε c is the least compressive strain in the member, fy , the yield stress of steel and, E S ,

the modulus of elasticity of steel, the maximum permissible compressive strain in concrete member will be a) 0.002 b) 0.002 + f y / (1.15 E S ) c) 0.0035 – 0.75 ε c d) 0.0035

Q.30 In doubly reinforced rectangular

beam, the allowable stress in compression steel is a) Equal to the permissible stress in

tension in steel b) More than permissible stress in

tension in steel c) Less than permissible stress in

tension in steel d) Not related to the permissible

concrete compression stress Q.31 The splicing of reinforcement bars

in RCC beam can be done at a section where

a) Bending moment is zero b) Bending moment is less than the

maximum bending moment c) Bending moment is less than the

one half of the maximum bending moment

d) Shear force is zero


Q.32 The side face reinforcement, if required, in a T-beam will be

a) 0.1% of the web area b) 0.15% of the web area c) 0.2% to 0.3% of the web area

depending upon the breadth of the web

d) Half the longitudinal reinforcement Q.33 What shall be the maximum area of

reinforcement (i) in compression and (ii) in tension to be provided in an RC beam, respectively, as per IS: 456? a) 0.08% and 2% b) 2% and 4% c) 4% and 2% d) 4% and 4%

Q.34 In the reinforced concrete slab, the spacing between main reinforcement should not exceed a) Three times its effective depth b) Four times its effective depth c) Five times its effective depth d) Six times its effective depth

Q.35 The maximum percentage of

moment redistribution allowed in RCC beams is a) 10% b) 20% c) 30% d) 40%

Q.36 Ratio of permissible shear stress in

limit state method of design and working stress method of design is a) 25 : 16 b) 5 : 4 c) 16 : 25 d) 4 : 5

Q.37 The contribution of bent up bars towards shear resistance shall not be more than a) 50% of total shear

reinforcement b) 40% of total shear

reinforcement c) 55% of total shear

reinforcement

d) 30% of total shear reinforcement

Q.38 If the nominal shear stress ( vτ ) at a

section does not exceed the permissible shear stress ( cτ ) a) Minimum shear reinforcement is

still provided b) Shear reinforcement is provided

to resist the nominal shear stress c) No shear reinforcement is

provided to resist the nominal shear stress

d) Shear reinforcement is provided for the difference of the two

Q.39 Abeam of rectangular cross-section

(b × d) is subjected to a torque. What is the maximum torsional stress induced in the beam (b < d and α is a constant)? a)

2

Tb dα

b) 2

Tbdα

c) Tbdα

d) Tbd

Q.40 Minimum shear reinforcement in

beams is provided in the form of stirrups a) To resist extra shear force due to

live load b) To resist the effect of shrinkage

of concrete c) To resist principal tension d) To resist shear cracks at the

bottom of beam

Q.41 The development length in compression for a 20 mm diameter deformed bar of grade Fe-415 embedded in concrete of grade M-25, whose design bond stress is 1.40 N/mm 2 is a) 20 0.87 415

4 1.40× ××

b) 20 0.87 4154 1.25 1.40× ×× ×

c) 20 0.87 4154 1.6 1.40× ×× ×

d) 20 0.87 4154 1.25 1.6 1.40

× ×× × ×


Q.42 Consider the following statements: 1. Development length includes anchorage value of hooks in tension reinforcement. 2. The development length of each bar of bundled bars shall be that for the individual bar, increased by 33% for three bars in contact. 3. Deformed bars may be used without end anchorage provided development length Requirement is satisfied.

Which of these statements is/are correct? a) Only 2 b) both 1 and 3 c) Both 1 and 2 d) 1, 2 and 3

Q.43 An RC structural member

rectangular in cross section of width b and depth d is subjected to a combined action of bending moment M and torsional moment. The longitudinal reinforcement shall be designed for a moment M e given by

a) M e = M + (1 / )1.7

T D b+

b) M e = M + (1 / )1.7

T D b−

c) M e = (1 / )1.7

T D b+

d) M e = (1 / )1.7

T b D−

Q.44 Which one of the following is the

correct expression to estimate the development length of deformed reinforcing bar as per IS code in limit state design?

a)4.5

bd

sφστ

b)5

bd

sφστ

c)6.4

bd

sφστ

d)8

bd

sφστ

Q.45 Bond strength can be enhanced

when

i) High grade concrete is used ii) Mechanical anchorages are

employed iii) Smaller diameter bars are used iv) Deformed bars are used

Which of these statements are correct? a) (i), (ii) and (iii) b) (i), (ii) and (iv) c) (i), (iii) and (iv) d) (i), (ii), (iii) and (iv)

Q.46 The effective length of a column in a reinforced concrete building frame, as per IS: 456-2000, is independent of the a) Frame type i.e., braced (no sway)

or un-braced (with sway) b) span of the beam c) Height of the column d) loads acting on the frame

Q.47 The reduction coefficient of a

reinforced concrete column with an effective length of 4.8 m and size of 250 mm × 300 mm is a) 0.80 b) 0.85 c) 0.90 d) 0.95

Q.48) The ratio of the maximum and minimum percentage of longitudinal reinforcement that can be provided in a column is a) 6 b) 7.5 c) 5 d) 8

Q.49 In the pedestal, the factor by which

the effective length should not exceed the least lateral dimension is a) 2 b) 3 c) 4 d) 5

Q.50 A reinforced concrete beam of 10 m

effective spam and 1 m effective depth is supported on 500 mm × 500 mm columns. If the total uniformly distributed load on the beam is 10 kN/m, the design shear force for the beam is


a) 50 kN b) 47.5 kN c) 37.5 kN d) 43 kN

Q.51 Which of the following are the

additional moments considered for design of slender compression member in lieu of deflection in x and y direction

a) 2

2000u exP l

D And

2

2000u eyP l

D

b) 2000

u exP lD

And 2000

u eyP lD

c) 2

2000u exP l

D And

2

2000u eyP l

b

d) 2

200u exP l

D And

2

200u eyP l

D

Where P u is axial load; l ex and l ey are effective length in respective directions; D is depth of section perpendicular to major axis; b is width of the member

Q.52 What is the minimum number of longitudinal bars provided in reinforced concrete column of circular cross-section? a) 4 b) 5 c) 6 d) 8

Q.53 An axially loaded column is of 300 mm × 300 mm size. Effective length of column is 3 m. What is the minimum eccentricity of the axial load for the column? a) 0 b) 10 mm c) 16 mm d) 20 mm

Q.54 The load carrying capacity of a

column designed by working stress method is 500 kN. The ultimate collapse load of the column is a) 500kN b) 662.5kN c) 750kN d) 1100kN

Q.55 Consider the following statements:

1.In a helical reinforced concrete column, the concrete core is subjected to tri-axial state of stress. 2.Helically reinforced column are very much suitable for earthquake resistant structures. Which of these statements is/are correct? a) 1 only b) 2 only c) Both 1 and 2 d)None of these

Q.56 According to IS: 456-2000, the column or the strut is the member whose effective length is greater than a) The least lateral dimension b) 2 times the least lateral

dimension c) 3 times the least lateral

dimension d) 4 times the least lateral

dimension Q.57 According to IS 456, the slenderness

ratio for a short column should not exceed a) 12 b) 18 c) 24 d)None of these

Q.58) Consider the following statements:

Bars that extend into a simple support must be able to develop their full strength at a designated point L so that their moment capacity is more than the bending moment at that point. The clauses of the code require that ( sσ = 0.85 syσ )

1 L d ≤ 10

1.3M LV

+

2 4

bd

sφστ

≤ 10

1.3M LV

+

3 ∅≤ 10

4 1.3bd

s

M LV

τσ

+

Which of these statements is/are correct? a) 1 and 2 b) 2 and 3 c) 1 and 3 d) 1, 2 and 3


Q.59 In a combined footing for two

columns carrying unequal loads, the maximum hogging moment occurs at a) inside face of the heavier column b) a section equidistant from both

the column c) a section having maximum shear

force d) a section heaving zero shear force

Q.60 A square column section of size

350mm × 350mm is reinforced with four bars of 25mm diameter and four bars of 16mm diameter. Then the transverse steel should be a) 5 mm dia at 240 mm c/c b) 6 mm dia at 250 mm c/c c) 8 mm dia at 250 mm c/c d) 8mm dia at 350 mm c/c

Q.61 A trapezoidal combined footing for two axially loaded columns is provided when i) Width of the footing near the

heavier column is restricted. ii) Length of the footing is

restricted. iii) Projections of the footing beyond

the heavier column are restricted.

Select the correct answer using the codes given below: a) 1 and 2 b) 1 and 3 c) 2 and 3 d) 1, 2 and 3

Q.62 In RC footing on soil, the thickness

at the edge should not be less than a) 10 cm b) 15 cm c) 20 cm d) 25 cm

Q.63 A strap footing is a special type of a) Strip footing b) raft footing c) Combined footing d) spread footing

Q.64 If the foundations of all the columns of a structure are designed on the total live and dead load basis, then a) There will be no settlement of

columns b) There will be no differential

settlement c) The settlement of exterior

columns will be more than interior columns

d) The settlement of interior columns will be more than exterior columns

Q.65 In the case of isolated square concrete footing, match the locations at which the stress resultants are to be checked, where d is effective depth of footing and select the correct answer using the codes given below the lists:

Stress Resultant Location A. Bending moment 1.At face of Column B. One way shear 2.At d/2 from face of column C. Punching shear 3. At d from face of column

Codes: A B C a) 1 2 3 b) 3 1 2 c) 1 1 3 d) 1 3 2

Q.66 IS 1343-1980 limits the minimum characteristic strength of prestressed concrete for post-tensioned work and pretension work as a) 25 MPa, 30 MPa respectively b) 25 MPa, 35 MPa respectively c) 30 MPa, 35 MPa respectively d) 30 MPa, 40 MPa respectively

Q.67 A simply supported post-tensioned prestressed concrete beam of span L is prestressed by a straight tendon at a uniform eccentricity è’ below the centroidal axis. If he magnitude of prestressing force is P and


flexural rigidity of beam is EI, the maximum central deflection of the beam is

a)2

8PeL

EI(Downwards)

b)2

48PeL

EI(Upwards)

c)3

8PeL

EI(Upwards)

d)2

8PeL

EI(Upwards)

Q.68 Concordant cable profile is

a) A cable profile that produces no support reactions due to prestressing

b) A cable profile which is parabolic in nature

c) A cable profile which produces no bending moment at the supports of a beam

d) A cable profile laid corresponding to axial stress diagram

Q.69 Match List-1 (post-tensioning

system) with List-2( type of anchorage) and select the correct answer using the codes given bellow the lists:

List-1 A. Freyssinet B. Giford-udall C. Lee-McCall D. Magnel Blaton List-2 1. Flat steel wedges in sandwich

plates 2. High strength nuts 3. Split conical wedges 4. Conical serrated concrete wedges

Code: A B C D a) 2 1 4 3 b) 4 3 2 1 c) 2 3 4 1 d) 4 1 2 3

Q.70 Match List-1 (post-tensioning system) with List-2 (type of anchorage) and select the correct answer using the codes given bellow the lists: List-1 A Freyssinet B Giford-udall C Lee-McCall D Magnel Blaton List-2 1. Single bars 2. Wires evenly spaced by

perforated spacers 3. Horizontal rows of four wired

spaced by metal grills 4. Wires spaced by helical wire

core in annular spacer Code: A B C D a) 4 1 2 3 b) 3 2 1 4 c) 4 2 1 3 d) 3 1 2 4

Q.71 An ordinary mild steel bar has been

prestressed to a working stress of 200MPa. Young’s modulus of steel is 200 GPa. Permanent negative strain due to shrinkage and creep is 0.0008. How much is the effective stress left in steel? a) 184 MPa b) 160 MPa c) 40 MPa d) 16 MPa

Q.72 What is the allowable upward

deflection in a prestress concrete member under serviceability limit state condition? a) Span/250 b) Span/300 c) Span/350 d) Span/500

Q.73 At the time of initial tensioning, the

maximum tensile stress immediately behind the anchorage should not exceed which one of the following? a) 0.50 × ultimate tensile stress b) 0.60 × ultimate tensile stress


c) 0.70 × ultimate tensile stress d) 0.80 × ultimate tensile stress

Q.74 If a simply supported concrete beam

, prestressed with a force of 2500 kN, is designed by load balancing concept for an effective span of 10 m and to carry a total load of 40 kN/m, the central dip of the cable profile should be a) 100 mm b) 200 mm c) 300 mm d) 400 mm

Q.75 A prestressed concrete beam has a

cross-section with the following properties:

A = 46400 mm 2 , l = 75.8 × 10 7 mm4 , Y bottom = 244 mm , Y top = 156 mm It is subjected to a prestressing force at an eccentricity è’ so as to have a zero stress at the top fibre. The value of è’ is given by a) 66.66 mm b) 66.95 mm c) 104.72 mm d) 133.33 mm

Q.76 In prestressed concrete

a) Forces of tension and compression change but lever arm remains unchanged

b) Forces of tension and compressions remain unchanged but lever arm changes with the moment

c) Both forces of tension and compression as well as lever arm change

d) both forces of tension and compression as well as lever arm remain unchanged

Q.77 If `P’ is the prestressing force

applied at a maximum eccentricity `g’ at mid-span, figure, to balance the concentrated load `W’, the balancing load will be

a) 2.5 Pg/L b) 3.0 Pg/L

c) 3.5 Pg/L d) 4.0 Pg/L

Q.78 If percentage gain in stress due to UDL in tendon is zero, then eccentricity will be a) 2

12S

C

EWlp E

b) 2 1

12 C

Wlp E

c) 2

12C

S

EWlp E

d) 2

12Wl

p

Where P is prestressing force in tendon, W is UDL, E c - modulus of elasticity of concrete and E s - modulus of elasticity of steel.

Q.79 A concentrated live load at the centre of span of prestressed concrete beam can be counter balanced by selecting a) Straight cable profile b) Parabolic cable profile c) Linearly varying profile with zero eccentricity at centre of span d) concentric cable profile

Q.80 Deflection of prestressed concrete beam is excessive in the a) Pre-cracking stage b) Plastic stage c) Post-cracking stage d) none of the above

Q.81 Combination of partial safety factors for loads under limit state of collapse and limit state of serviceability will be a) 1.5 (DL + LL) or 1.5 (DL + WL) or

1.2 (DL + LL + WL) and DL + 0.8 (LL + WL)

b) 1.5(DL+LL) and Dl+0.8(LL + WL) c) 1.5 (DL + LL) or 1.5 (DL + WL) or

1.2 (DL + LL + WL) and 1.0 (DL + LL) or 1.0 (Dl + WL) or 0.8 (LL + WL)


d) 1.2 (DL + LL + WL) and 1.0 (DL + LL) or 1.0 (Dl + WL) or 0.8 (LL + WL)

Q.82 If the depth of neutral axis for a singly reinforced balanced rectangular section is represented by kd in working stress design, then the value of k for a balanced section, a) Depends on stσ only b) Depends on cbcσ only c) Depends on both stσ and cbcσ d) is independent of both stσ and

cbcσ Q.83 The expression for modular ratio m

= 2803 cbcσ

, where cbcσ is the

permissible compressive stress due to bending in concrete in N/mm 2 , a) Fully takes into account the long-

term effect such as creep b) Partially takes into account the

long-term effect such as creep c) Does not take into account the

long-term effect such as creep d) is the same as the modular ratio

based on the value of modulus of elasticity of structural concrete E c

Q.84 Examine the following statements:

1. Factor of safety for steel should be based on its yield stress.

2. Factor of safety for steel should be based on its ultimate stress.

3. Factor of safety for concrete should be based on its yield stress.

4. Factor of safety for concrete should be based on its ultimate stress. The correct statements are:

a) 1 and 3 b) 1 and 4 c) 2 and 3 d) 2 and 4

Q.85 What should be the minimum grade of reinforced concrete in and around sea coast construction?

a) M 35 b) M 30 c) M 25 d) M 20

Q.86 On which one of the following

concepts is the basic principle of structural design based? a) Weak column strong beam b) Strong column and weak beam c) Equally strong column and beam d) Partial weak column- beam

Q.87 Match List-1 with List-2 and select

the correct answer using codes given below the lists:

List-1

A. IS-875 B. IS-1343 C. IS-1893 D. IS-3370 List-2 1. Earthquake resistant design 2. Loads 3. Liquid structure 4. Prestressed Concrete A B C D a) 3 1 4 2 b) 2 1 4 3 c) 3 4 1 2 d) 2 4 1 3

Q.88) MatchList-1 with List-2and select

the correct answer using codes given below the lists:

List-1 A. Moment and shear coefficients B. Fire resistance C. Sliding D. Span to depth ratio of beam List-2 1. Durability 2. Stability 3. Analysis of structure 4. Deflection limits A B C D a) 4 2 1 3 b) 3 2 1 4 c) 4 1 2 3 d) 3 1 2 4


Q.89 Consider the following statements:

1. Maximum strain in concrete at the outermost compression fibre is taken to be 0.0035 in bending. 2. The maximum compressive strain in concrete in axial compression is taken as 0.002. Keeping the provision of IS: 456-2000 on limit state design in mind, which of the following is true?

a) Statement 1 is true but 2 is false b) Statement 1 is false but 2 is true

c) Both statements 1 and 2 are true d) Both statements 1 and 2 are false

Q.90 To determine the modulus of rupture, the size of test specimen used is a) 150 × 150 × 500 mm b) 100 × 100 × 700 mm c) 150 × 150 × 700 mm d) 100 × 100 × 500 mm

Q.91 For the design of retaining walls, the minimum factor of safety against overturning is taken as a) 2 b) 1.4

c) 1.5 d) 3.0

Q.92 The centroid of compressive force, from the extreme compression fibre, in limit state design lies at a distance of a) 0.367 X U b) 0.416 X U c) 0.446 X U d) 0.573 X U

Q.93 Maximum strain at the level of compression steel for a rectangular section having effective cover to compression steel as `d’ and neutral axis depth from compression face as X U is

a)0.0035U

d1X

−

b)0.002

U

d1X

−

c)0.0035 UX1d

−

d)0.002 UX1d

−

Q.94 Design strength of concrete for limit state of collapse is a) ckf b) 0.67 ckf /γ c) 0.67 ckf d) γ ckf Where,

ckf = Characteristic compressive strength of concrete

γ = Partial factor of safety Q.95 The diagonal tension failure in a

concrete beam occurs due to a) Large shear force and less

bending moment b) Large bending moment and less

shear force c) Equal shear force and bending

moment d) None of these

Q.96 If the cross sectional areas of the three basic structural elements viz. beam slab and column are related to the amount of steel reinforcement then which of the following statements is correct 1. Percentage steel is usually

maximum in a column 2. Percentage steel is least in a slab a) only 1 b) only 2 c) Both 1 and 2 d) neither 1 nor 2

Q.97 Consider the following statements in

relation to stress-strain curve for concrete and steel: 1. Factor of safety (FOS)is applied for whole curve for steel 2. Factor of safety (FOS) is applied only for a part of steel for which hook’s law is not applicable. 3. Factor of safety (FOS) is applied for whole curve for concrete. 4. Factor of safety (FOS) is applied only for a part of concrete for which hook’s law is not applicable.

Which of these statements are correct? a) 1 and 3 b) 1 and 4 c) 2 and 3 d) 2 and 4


Q.98 Compressive strength of concrete is

taken as 0.67 ckf and not ckf because of 1. Size factor which is constant for

a cube of size greater than 450 mm.

2. Size factor which is constant for a cube of size greater than 550 mm.

3. End friction zone which acts throughout the length of cube.

4. End friction zone which acts throughout the length of cylinder.

Which of these statements are correct? a) 1 and 3 b) 1 and 4 c) 2 and 3 d) 2 and 4

Q.99 The rise and tread of staircase are

150 mm and 250 mm and weight of slab on slope is 5 kN/m 2 . The dead weight of horizontal area is

a) 4.5 kN/m 2 b) 5.0 kN/m 2 c) 5.83 kN/m 2 d) 6.8 kN/m 2

Q.100 If the creep coefficient for concrete at 7 days is 1K and at 28 days is 2K then a) 1K > 2K b) 1K < 2K c) 1K = 2K d) 1K ≤ 2K

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (a) (d) (a) (b) (a) (b) (a) (c) (b) (d) (d) (d) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (b) (a) (d) (b) (a) (b) (b) (b) (b) (c) (b) (c) (c) (a) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (c) (c) (c) (a) (d) (a) (c) (a) (a) (a) (a) (c) (d) (b) 43 44 45 46 47 48 49 50 51 52 53 54 55 56 (a) (c) (d) (d) (b) (b) (b) (c) (c) (c) (d) (c) (c) (c) 57 58 59 60 61 62 63 64 65 66 67 68 69 70 (a) (d) (d) (c) (c) (b) (c) (c) (d) (d) (d) (a) (b) (c) 71 72 73 74 75 76 77 78 79 80 81 82 83 84 (c) (b) (d) (b) (c) (b) (d) (d) (c) (c) (c) (a) (b) (b) 85 86 87 88 89 90 91 92 93 94 95 96 97 98 (b) (b) (d) (d) (c) (c) (b) (b) (a) (b) (a) (c) (c) (a) 99 100 (c) (a)

ANSWER KEY:


Q. 1 (a)

Q. 2 (a)

Q. 3 (d) See clause 23.1.2 of IS 456 : 2000

Q. 4 (a) LD

<2.5 for continuous beam

So L < 2.5×D So Maximum effective length = 2.5× (500+50)=1375 mm

Q. 5 (b) u

e uTV V 1.6b

= +

Here uV 0= 3

e50 10V 1.6 229kN

350×

= × =

Q. 6 (a)

Q. 7 (b) In case of two way slab, for mild steel, (span/Overall Depth) = 40

D =(3⨉100/40)=7.5cm

Q. 8 (a) For a balanced section

ck y st0.39f *b*X 0.87f A=

ckst

y

0.36f *b*XA0.87f

⇒ =

As yf increases stA decreases.

Q. 9 (c) The limiting value l/D for simply supported and continuous span deep beams are as follows:

For simple span deep beam, l/D ≤2.0 For Continuous span deep beam, l/D ≤.5 Where, l = Effective span taken as centre to centre distance between the support of 1.15 time the clear span, whichever is smaller. D = Overall depth D = 30 cm xu= 12 cm c = 50 kg/ cm 2

U

U

XCt d Xm

⇒ =−

⇒ t = 18 5012× × (30-12)

18 50t (30 12)12×

⇒ = × −

= 1350kg/ cm 2

Q. 10 (b) Depth of the balanced neutral section using working stress method is given by

U minSR

280X d280 3

⇒ =+ σ

280 d 0.4d280 3 140

= =+ +

Q. 11 (d) The limiting value of depth of neutral axis depending on the different grades of steel is as follows:

yf U minX / d 250 0.53 415 0.48 500 0.46

Q. 12 (d) Basic value of span to effective depth ratios for spans up to 10 m to satisfy vertical deflection limits are for

EXPLANATIONS


Cantilever beam 7 Simply supported 20 Continuous 26

Q. 13 (d)

For T-beams effective width of compression flange

0f

Lb B 6t6

⇒ = + +

For L- beam bf = (Lo/12) + bw +3 Df

Q. 14 (b)

Total compressive force without considering the partial safety factor of material ⇒ 0.36 ck Uf b X∗ ∗ ×1.5 = 0.54 ck Uf b X∗ ∗

Q. 15 (b)

As per IS: 456-2000 modular ratio is given by

M = 2803 cbcσ

For M20 concrete cbcσ =7N/mm 2

∴ m= 2803 7×

= 13.3

Q. 16 (a)

As per the clause 22.2 of IS: 456-2000, for simply supported beam and slab, the effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre of supports, whichever is less. ∴ Effective span = 5 × 1000 + 40

= 5400 mm And effective span

= 5 × 1000 + 300

2 +

3002

= 5300 mm

Q. 17 (d)

As per clause 26.5.1.5 of IS: 456-2000, the maximum spacing of shear

reinforcement measured along the axis of member shall not exceed 0.75 d for vertical stirrups and 1d’ for inclined stirrups at 45 ° , where 1d’ is the effective depth of the section under consideration. In n case shall the spacing exceed 300 mm. ∴ Maximum spacing = 0.75 d = 0.75 × 300 = 225 mm

Q. 18 (b)

Moment of resistance for a balanced section is given by M ,limU = 0.36 f ck × ,limUX × (1- 0.42 ×

,limUXd

) × bd 2

But for FE 415 ,limUX = 0.48 d

M ,limU = 0.36 f ck × 0.48 × (1- 0.42 ×

0.48) × bd 2 ⇒M ,limU = 0.13796 f ck × bd 2 On comparison, we get K = 0.13796 f ck

= 0.13796 × 20 = 2.76

Q. 19 (a)

To ensure ductility the maximum strain in tension reinforcement in the section at failure shall not be

less than 1.15

y

S

fE

+ 0.002

Q. 20 (b)

For under- reinforced beam

i) UX < ,limUX or CX

ii) sc∈ Steel reaches ultimate

stress before concrete reaching the ultimate stress

iii) uM < ,limuM


iv) Lever arm = d-0.42 UX will be more in the case of under- reinforced section

Q. 21 (b)

sc

uX d∈

′− = cu

uX∈

∴ sc∈ = 0.0035(1-u

dX′

)

Q. 22 (b)

The various load combination are as follows: i) For the dead load and live load

the ultimate bending moment is given by,

M u = 1.5 (DL + LL)

= 1.5 × (20+30) = 75 kN-m ii) For dead load and earthquake

(seismic) load the ultimate bending moment is given by,

M u = 1.5 (DL + EL) = 1.5 × (20 + 10) = 45 kN-m iii) For dead load, live load and

earthquake (seismic) load the ultimate bending moment is given by

M u = 1.2 (DL + EL + LL) = 1.2 (20 + 30 + 10) = 72 kN - m

Q. 23 (b)

X ,maxU = 0.0035

0.870.0055 y

s

dfE

+

For f y = 415 N/mm 2 And E s = 2 × 10 5 N/mm 2

X ,maxU = 0.48d and

for f y = 250 N/mm 2 X ,maxU = 0.53d Q. 24 (c)

The horizontal distance between parallel main reinforcement bars shall not be more than three times he effective depth of solid slab or 300 mm whichever is smaller. The total reinforcement in the slab should remain same. By replacing 10 mm bars by 12 mm bars, the spacing will increase as

2

2

1

dd

× S1 =

21210

× 10

= 14.4 cm Q. 25 (b)

The strip of two way slab may be checked against shorter span to effective depth ratios. Span to effective depth ratio

Q. 26 (c) k S = 0.5 + 𝛃𝛃 C > 1 𝛃𝛃 C = Shorter side/Longer side Q. 27 (c)

The diameter of the bars shall not exceed one eighth of the total thickness of the slab.

∴ Maximum size = 758

= 9.375 mm So the diameter will be 8 mm.

Q. 28 (a)

For restrained slab, the area of reinforcement in each of the four corner layers shall be three-quarters


of the area required for the maximum mid span moment in the slab simply supported on both edges meeting at that corner. If the corner contained by edges over only one of which the slab is continuous, torsion reinforcement equal to 0.375 times the area reinforcement provided at the mid-span in the same direction shall be provided. If both edges are continuous, no torsion reinforcement shall be provided.

Q. 29 (c)

Maximum permissible compressive strain in concrete under axial compression is taken as 0.002. The maximum compressive strain at the highly compressed extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.

Q. 30 (c)

The stress in the compression steel depends on the strain, ϵ sc , the level of the compression steel.The strain at the level of compression steel under loading does not reach the yield strain value of steel. Therefore, allowable stress is less than the permissible stress in tension in steel. The maximum permissible compressive stress in steel is 130MPa and 190MPa for Fe 250, Fe 415 and Fe 500 grades respectively.

Q. 31 (c)

It is recommended that splices in flexural members should not be provided at sections where the bending moment is more than 50% of the moment of resistance; and not

more than half the bars shall be spliced at a section.

Q. 32 (a)

For beams exceeding overall depth of 750 mm side face reinforcement is provided. Such reinforcement shall be not less than 0.1 percent of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness whichever is less.

Q. 33 (d) As per the clause 26.5.1.1 of IS: 456-2000, the maximum area of tension reinforcement in beams shall not exceed 0.04 bD. As per clause 26.5.1.2 of IS: 456-2000, the maximum area of compression reinforcement in beams shall not exceed 0.04 bD.

Q. 34 (a)

Maximum diameter of reinforcing bars in slabs is limited to one-eight of the total thickness of slab and the maximum spacing of main bars is limited to 3d or 300 mm (whichever is less).

Q. 35 (b)

As per clause 37.1.1 of IS: 456-2000 the ultimate moment of resistance at any section should not be less than 70% of the factored moment at that section as obtained from the elastic moment envelope (considering all loading combinations).

Q. 36 (a) Permissible shear stress, In limit state method of design, τ C = 0.25 ckf In working state method of design τ C = 0.16 ckf


∴ Ratio = 2516

Q. 37 (a)

Total shear strength = Shear resistance of effective concrete area as a function of longitudinal bars + shear resistance of vertical shear stirrups + shear resistance of inclined shear stirrups.Tests have shown that inclined bars alone do not provide a satisfactory solution and their contribution is limited to 50% of net shear strength after deducting the contribution of concrete. The remaining shear resistance is provided by vertical stirrups.

Q. 38 (a) If the calculated shear stress ( vτ ) is

i) Less than allowable shear stress (cτ ) but more than 0.5 cτ , the

minimum shear reinforcement in the form of stirrups shall be provided such that

SV

V

AbS ≥

0.40.87 yf

ii) More than cτ , shear reinforcement shall be provided in the form of vertical stirrups or bent up bars with stirrups or inclined stirrups to resist

usV = uV - cτ bd. Here vτ < ,maxcτ iii) If vτ > ,maxcτ redesign the section.

Q. 39 (a)

Torsion constant of a rectangular section of width b and depth d (b < d) may be expressed as J = b 3 d For T, L and I sections torsion constant

J = 313 i ib d∑

Where b i and d i are the dimensions of each of the component rectangles

into which the section may be divided. Torsional shear stress for rectangular section

iT ′ = T 3

3i i

i i

b db d

∑

Q. 40 (c)

The shear, at which the inclined crack in beam without shear reinforcement is formed first, is taken as the shear strength of concrete as the difference between the loads corresponding to the first crack and the ultimate failure is very less. Formations of such crack occur when the principal tensile stress reaches the tensile strength of concrete. At the mid-span of a simply supported beam subjected to uniformly distributed load, where shear is small and bending stress is large, the direction of principal tensile stress is flat and is nearly equal to the flexural tensile stress. This will cause flexural cracks nearly vertical to the axis of the beam. These are initiated even, when 0.5 cτ< vτ < cτ .Thus minimum reinforcement is needed to prevent flexural crack due to principal tension.

Q. 41 (d) For bars in compression the value of

bond stress is increased by 25% For deformed bars value of bond

stress is increased by 60% Q. 42 (b)

In case of bundled bars in contact the development length is increased than that for individual bar by 1 10% for two bars in contact

2 20% for three bars in contact


3 33% for four bars in contact Q. 43 (a)

The longitudinal reinforcement shall be designed to resist an equivalent bending moment

eM = M + tM

Where, tM = T1 /

1.7d B+

∴ eM = M + T1 /

1.7D b+

Equivalent shear

V e = V + 1.6 T/b Q. 44 (c) As per clause 26.2.1 of IS: 456-2000,

the development length L d is given by

L d = 4bd

sφστ

Where,∅ = nominal diameter of bar σ s = stress in the bars at the section considered at design load

bdτ = design bond stress for plain bars

The value of bd

τ should be increased by 60% for deformed bars.

∴ L d = 604100 bd

s

bd

φσ

τ τ +

L d = 6.4bd

sφστ

Q. 45 (d) Q. 46 (d)

Effective length of a column depends upon: i)Flexural stiffness (E/L) of beams joining at a point. ii) Flexural stiffness (E/L) of columns joining a point.

iii) Type of frame (sway or non sway)

Q. 47 (b)

In the working stress method the design of a long column is made by considering reduced stresses obtained by multiplying the permissible stresses by a coefficient,

C r = 1.25 - 48

efflD

or 1.25 - 48

efflB

Whichever is smaller

C r = 1.25 - 4800

48 250× or

1.25 - 480048 300×

= 0.85 or 0.92

Q. 48 (b) Maximum reinforcement = 6% Minimum reinforcement = 0.8% Ratio = 6

0.8 = 7.5

Q. 49 (b)

Very short columns with effective length less than three least lateral dimensions are called pedestal columns.

Q. 50 (c)

The shear force should be calculated at critical section i.e. 1m away from the face of the column. Effective span of the beam is to be taken as the clear span plus effective depth of beam or centre to centre spacing of supports whichever is less. The location for shear calculation will be

X = 0.25+ 1= 1.25 ∴ design shear force = 10 10 10 1.25

2×

− × = 37.5kN

Q. 51 (c)

M ax = 2

2000u exP D l

D

=2

2000u exP l

D


M ay =

2

2000eyu lP bb

=2

2000u eyP l

b

Q. 52 (c) As per IS: 456- 2000 clause

26.5.3.1(a) the minimum number of longitudinal bars provided in a column shall be four in rectangular columns and six in circular columns.

Q. 53 (d)

Minimum eccentricity equals to unsupported length of column/ 500 plus lateral dimension/30, subjected to a minimum of 20 mm.

3000 300 16500 30 500 30L B

+ = + =

So eccentricity = 20mm Q. 54 (c) The ultimate collapse load on the

column is given by W u = W s × 1.5 = 500 × 1.5 = 750 kN Q. 55 (c)

Confinement of concrete by providing transverse reinforcement in the form of steel hoops and spirals increases the strength and ductility of concrete. This reduces the tendency for internal cracking and volume increase prior to failure.

Q. 56 (c)

The code defines the column as a compression member, the effective length of which exceeds three times the least lateral dimension. The term `pedestal’ is used to describe a vertical compression member whose` effective length’ is less than three times its least lateral dimension.

Q. 57 (a)

According to the IS (CI. 25.1.2), a compression member may be classified as a `short column’ if its slenderness ratios with respect to the `major principal axis’ (l ex /D x ) as well as the `minor principal axis (l ey /D y ) are both less than 12.

Q. 58 (d) Q. 59 (d)

The combined footing resists the load by bending in two directions. The footing in the longitudinal direction is subjected to a sagging moment in the cantilever portion and under the columns; and hogging moment in the central portion between the columns. Footing in the transverse direction develops sagging moments. The maximum hogging moment occurs in the central portion where shear force is zero.

Q. 60 (c)

The diameter of transverse reinforcement shall not be less than one fourth of the diameters of the largest longitudinal bar and in no case less than 6 mm. So the diameter of the bar 25

4 = 6.25

mm. Choose 8 mm diameter bar. The pitch of the transverse

reinforcement shall not be more than the least of the following. i) The least lateral dimension of the compression member i.e. 350 mm. ii) Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied i.e. 16×6 = 256 mm iii) 300 mm

So pitch will be 250 mm c/c. Q. 61 (c)


Combined footing supporting two columns may be used for column on property lines or sewer line. In this case load will be eccentric.

Q. 62 (b)

According to the code (CI. 34.1.2) restricts the minimum thickness at the edge of the footing to 150 mm for footings in general (and to 300 mm in the case of pile caps).

Q. 63 (c)

A strap footing consists of a spread footing of two columns connected by a strap beam. This type of footing is useful when the

external column is very near to the property line so that its footing cannot be spread beyond the property line

Q. 64 (c)

The ratio of live load to dead load is greater in the case of interior column as compared to that in exterior column.

Q. 65 (d) For footings on piles the critical

section for one way shear is d/2 from the face of the wall.

Q. 66 (d) Q. 67 (d) Maximum central deflection,

δ = 2

8LR

1R

= PeEI

∴ δ = 2

8PeL

EI (Upwards)

Q. 68 (a)

Concordant cable profile does not produce secondary moments and

there will be no reactions at the supports due to prestressing action.

Q. 69 (b)

In the freyssinet system anchorage consists of cylinder of good quality concrete and is provided with corrugations on the outside. It has a central conical hole and is provided with heavy top reinforcement.

In Gifford-Udall system, the wire are stressed and anchored one by one in a separate cylinder using small wedding grip called Udall grips. Each grip consists of two split cones.

In Lee Mc Call system anchoring of bars is done by screwing special threaded nuts. The nuts bear against a distribution plate provided at the end of the beam.

In the Magnel Blaton system, the wires are anchored by wedging, two at a time into sandwich plates. The sandwich plates are provided with two wedge shaped grooves on its two faces. The wires are taken into each groove and tightened. Then a steel wedge is driven between the tightened wires to anchor them against the plate.

Q. 70 (c) In Freyssinet system, high tension

steel wires 5 mm to 8 mm diameter about 12 in number are arranged to form a group into a cable with spiral spring inside.

In the Gifford Udall system the wires are stressed and anchored one by one in a separate cylinder using small wedging grips called Udall grips.

In the Lee- Mc Call system, High tensile alloy steel bars are used as the prestressing tends. In the Magnel Blaton system cable of rectangular section is provided. The wires are arranged with four wires


per layer. The geometric pattern of the wires is maintained in the same form throughout the length of the cable by providing grills or spacers at regular intervals.

Q. 71 (c)

Loss of stress due to shrinkage and creep = 200 × 10 3 × 0.0008 = 160 MPa

Effective stress left = 200 – 160 = 40 MPa

Q. 72 (b)

As per IS: 1343-1980 (codes of practice for prestressed concrete), if finishes are to be applied to the prestressed concrete members, the total upward deflection should not exceed span/300, unless uniformity of camber between adjacent units can be insured.

Q. 73 (d)

As per clause 18.5.1 of IS: 1343-1980, at the time of initial tensioning, the maximum tensile stress immediately behind the anchorages shall not exceed 80% of the ultimate tensile strength of the wire or bar or strand.

Q. 74 (b) Let the dip of the cable be h. Upward pressure provided by the

parabolic cable = 2

8PhL

In order this upward pressure may fully balance the external loading.

2

8PhL

= W

⇒ 2

8 250010

h× × = 40

⇒h = 0.2 m = 200 mm Q. 75 (c) For zero stress at top fibre

PA

= PeZ

⇒PA

= / top

PeI Y

⇒e = 775.8 10

156 46400×

× = 104.72 mm

Q. 76 (b) Q. 77 (d) Q. 78 (d)

Net percentage gain due to load = 2 net se E

lθ

If % gain is zero, then 2 net se E

lθ

= 0 [e≠0, E s ≠ 0] ∴θ net = 0

⇒3

24 c

WlE l

- 2 c

PelE l

= 0

⇒e = 2

12Wl

P

Q. 79 (c) Q. 80 (c) Q. 81 (c) Q. 82 (a) The value of K is given by,

K = mc

t mc+

For a balanced section,

m = 2803 cbcσ

c = cbc

σ , t = stσ

∴ K =

2803

2803

cbccbc

st cbccbc

σσ

σ σσ

×

+ ×


=

2803

2803stσ +

Q. 83 (b) Q. 84 (a) A beam collapses when stress is

equal to its yield strength and stress in concrete reaches its ultimate strength.

Q. 85 (b) Exposure Minimum

grade of plain concrete

Minimum grade of reinforced concrete

i. Mild - M20 ii. Moderate M15 M25 iii. Severe M20 M30 iv. Very severe M20 M30 v. Extreme M25 M40

Concrete in sea water or exposed directly along the sea coast shall be at least M20 grade in he case of plain concrete and M30 in case of reinforced concrete. The use of slag or pozzolana cement is advantageous under such condition.

Q. 86 (b) In tall buildings, it is important to

control lateral displacements within the serviceability limit state. A structural system may be classified as follows: i) Building frame system ii) Moment resisting frame system iii) Dual frame system iv) Tube system

In a moment resistant frame, The relative stiffness of beam and columns is very important. A frame may be designed using weak column strong beam proportions or strong column weak beam proportions. A frame with weak beam-strong column is more stable and therefore highly desirable.

Q. 87 (d) Q. 88 (d) Q. 89 (c) Q. 90 (c) The modulus of rupture test

employs a 150×150×700 mm plain concrete beam. The beam is simply supported and subjected to third-points loading until failure. Assuming a linear stress distribution across the cross section, the theoretical maximum tensile stress reached in the extreme fibre is termed by applying the flexure formula

f cr = MZ

Modulus of rupture is given by f cr = 0.7 ckf The factor of safety against

overturning and sliding should not be less than 1.4i.e.

FOS = Stabilizing force or moment / Destabilizing force or moment

In case of retaining walls the stabilizing force is due to dead load, thus these stabilizing forces are factored by a value of 0.9 (IS 456 Cl. 20)

∴FOS = 0.9 Stabilizing force or moment / Destabilizing force or moment ≥ 1.4

Q. 92 (b) The stress diagram for concrete

compression is shown below

Q. 93 (a) Maximum strain in concrete under

compression is 0.0035


Using similar triangle

0.0035

UX =

( )sc

U dX −

∈

sc∈ = 1u

dX

−

0.0035

Q. 94 (b) Design strength of concrete is

obtained is obtained after applying adequate factor of safety (FOS = 1.5 for concrete) to the characteristic strength of concrete. Here characteristic strength of concrete in the actual structure is taken as 0.67fck and hence the design strength of concrete is 0.67fck/γ

∴ Design strength =0.67fck/γ

= 0.67 fck/1.5 = 0.447fck Q. 95 (a)

For such a failure to happen τ>> σ Q. 96 (c) Q. 97 (c) 1.In case of concrete, the partial

safety factor γc is applicable at all stress levels because the stress - strain curve is directly affected by changes in the compressive strength of concrete.

2. In case of steel, the partial safety factor γs is applicable only for the

inelastic region because the modulus of elasticity is independent of the variations in the yield strength.

Q. 98 (a) Q. 99 (c)

W 1 = 2 2R TWT+′

= 5 ×2 2(150) (250)250+

= 5.83 kN/ m 2 Q.100 (a) See clause 6.2.5 of IS 456: 2000

strain that develops due o constant sustain loading is called creep strain but in the initial age of concrete, creep strain of concrete is higher than later age. However, elastic strain remains constant throughout. So creep coefficient = Creep strain/Elastic strain, decrease with time.


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RCC & STEEL STRUCTURES