Simplified Reinforced Concrete Design 2010 NSCP

7/13/2019 Simplified Reinforced Concrete Design 2010 NSCP

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CHAPTER 1

Introduction

Concrete

Concrete is a mixture of water, cement, sand, gravel crushed rock, or other aggregates.The aggregates (sand, gravel, crushed rock) are held together in a rocklike mass with a

paste of cement and water.

REINFORCED CONCRETE

As with most rocklike mass, concrete has very high compressive strength but have avery low tensile strength. As a structural member, concrete can be made to carry tensilestresses (as in beam in flexure). In this regard, it is necessary to provide steel bars to

provide the tensile strength lacking in concrete. The composite member is calledreinforced concrete.

AGGREGATES

Aggregates used in concrete may be fine aggregates (usually sand) and coarseaggregates (usually gravel or crushed stone). Fine aggregates are those that passesthrough a No. 4 sieve (about 6 mm in size). Materials retained are coarse aggregates.The nominal maximum sizes of coarse aggregate are specified in Section 5.3.3 ofNSCP. These are follows: 1/5 the narrowest dimension between sides of forms, 1/3 thedepth of slabs, or 3/4 the minimum clear spacing between individual reinforcing bars orwires, bundles of bars, or prestressing tendons or ducts. These limitations may not beapplied if, in the judgment the Engineer, workability and methods of consolidation are

such that concrete can be placed without honeycomb or voids.

WATER

According to Section 5.3.4, water used in mixing concrete shall be clean and free frominjurious of oils, acids, alkalis, salts organic materials or other substances that may bedeleterious to concrete or reinforcement. Mixing water for prestressed concrete or forconcrete that will contain aluminum embedments, including that portion of mixing watercontributed in the form of free moisture on aggregates, shall not be used in concreteunless the following are satisfied: (a) Selection of concrete proportions shall be basedon concrete mixes using water from the same source and (b) mortar test cubes madewith non-portable mixing water shall have 7-days and 28 day strengths equal to at least

90


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MODULUS OF ELASTICITYUnlike steel and other materials, concrete has no definite modulus of elasticity. Its value is

dependent on the characteristics of cement and aggregates used, age of concrete and

strengths.

According to NSCP (Section 5.8.5), modulus of elasticity Ec for concrete for values of wc,between 1500 and 2500 kg/ m3 may be taken as.

Eq. 1-1 Whereis the day 28-day compressive strength of concrete in MPa is the unit weight

onconcrete in

. For normal weight concrete,

Modulus of elasticity Es

for nonprestressed reinforced may be taken as 200,000 MPa.

DETAILS OF REINFORCEMENT

STANDARD HOOKS

Standard hooks refer to one of the following:

1. 180-degree bend plus extension but not less than 60 mm at free end of bar.2. 90-degree bed plus

extension at free end of bar.

3. For stirrups and tie hooks:a) 61 mm diameter bar and smaller, 90-degree bend plus extension atfree end bar, orb) 20 and 25 mm diameter bar, 90-degree bend, plus extension at free

end of bar, or

c) 25mm diameter bar and smaller, 135-degree bend d plus extension atfree end of bar.


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MINIMUM BEND DIAMETERS (SECTION 407.3)

Diameter of bend measured on the inside of the bar, other than for stirrups and ties insizes 10mm through 15 mm, shall not be less than the values in Table 1.1.

Inside diameter of bend for stirrups and ties shall not be less than

16 mm bar and

smaller. For bars larger than 16 mm, diameter of bend shall be in accordance withTable 1.1

Inside diameter of bend in welded wire fabric /9plain or deformed) for stirrups and ties

shall not be less than for deformed wire larger than D56 and for all other wires.Bends with inside diameter of less than 8db shall not be less than from nearestwelded intersection.

Table 1.1- Minimum Diameters of Bend

Bar Size Minimum Diameter

10 mm to25 mm 28 mm, 32 mm, and 36 mm

PLAIN REINFORCEMENT (407.6)

Reinforcement, prestressing tendons, and ducts shall not be accurately placed andadequately before concrete is placed, and shall be secured against displacement withintolerance permitted.

Unless otherwise specified by the Engineer, reinforcement prestressing tendons, andprestressing ducts shall be placed within the following tolerances:

Tolerance for depth d, and minimum concrete over a flexural members walls andcompression members shall be as follows:

Effect ive depth, d Tolerance on d Tolerance on minim um

con crete cover

d -10 mmd

-12 mm


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Except that tolerance for the clear distance to formed soffits shall be minus 6 mm andtolerance for cover shall not exceed minus 1/3 the minimum concrete cover required inthe design drawings or specifications.

Tolerance for longitudinal location of bends and ends of reinforcement shall be 50mm except at discontinuous ends of members where tolerance shall be mm.SPACING L IMITS FOR REINFORCEMENT

According for Section 5.7.6 of NSCP, the minimum clear spacing between parallel barsin a layer should be db but not less than 25 mm. Where parallel reinforcement is placedin two or more layers, bars in the upper layers should be placed directly above bars inthe bottom layer with clear distance between layers not less than 25mm. In spirallyreinforced or tied reinforced compression members, clear distance between longitudinal

bars shall be not less than 1.5 db nor 40mm.

In walls and slabs other than concrete joist construction, primary flexural reinforced shallbe spaced not for farther apart than three times the wall or slab thickness, nor 450 mm.

BUNDLED BARS

Groups of parallel reinforcing bars bundled in contact to act as unit shall be limited tofour in any one bundle. Bundled bars shall be enclosed within stirrups or ties and barslarger than 32 mm shall not be bundle in beams. The individual bars within the span offlexural members should terminate at different points with at least 40 dbstagger. Sincespacing limitations and minimum concrete cover of most members are based on a

single diameter db, bundled bars shall be treated as a single bar of a diameter derivedfrom the equivalent total area.


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Figure 1.1Bundled-bar arrangement

Diameter of single bar equivalent to bundled bars according to NSCP to be used for

spacing limitation and concrete cover.

=

3-25mm Equivalent diameter, D

(25)2x 3 D2

CONCRETE PROTECTION FOR REINFORCEMENT (SECTION 407.8.1)

Cast- in place Concrete (nonp restressed). The following minimum concrete covershall be provided for reinforcement:

Minimumcover, mm

(a) Concrete cast against permanently exposed to earth 75

(b) Concrete exposed to earth or weather:20 mm through 36 mm bars16 mm bar, W31 or D31 wire, and smaller

5040


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(C) Concrete not exposed to weather or in contact with ground:slabs, walls, joists:

32 mm bar and smallerBeams, columns

Primary reinforcement, ties, stirrups, spirals

Shells, folded place members:20 mm bar and larger16 mm, Wr1 or D31 wire, and smaller

20

40

20

15

Precast conc rete (Manufactured Under Plant Condi t ions ).The Following minimumconcrete shall be provided for reinforcement

Minimumcover, mm

(a) Concrete exposed to earth or weather:Wall panels:32 mm bar and smaller

Other members:20 mm through 32 mm bars16 mm bar, W31 wire, and smaller

20

4030

(b) Concrete not exposed to weather or in contact withground:

slabs, walls, joists:32 mm bar and smaller

Beams, columns

Primary reinforcement

Ties, stirrups, spiralsShells, folded plate members:

20 mm bar and larger16 mm, Wr1 or D31 wire, and smaller

15

db but not less 15, &need not exceed 4010

1510


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Prestressed Concrete

The following minimum concrete cover shall be provided for prestressed andnonprestressed reinforcement, ducts and end fittings.

Minimum

cover, mm

(a) Concrete cast against permanently exposed to earth 75

(b) Concrete exposed to earth or weather:Wall panels, slabs joistsother members

2540

(C) Concrete not exposed to weather or in contact withground:slabs, walls, joists:Beams, columns:

Primary reinforcement,

Ties, stirrups, spiralsShells, folded plate members:16 mm, Wr1 or D31 wire, and smaller

Other Reinforcement

20

40

25

10

db but not less than

20

Bundled Bars

For bundled bars, the minimum concrete cover shall be equal to the equivalent diameterof the bundle, but need to be greater than 50 mm, except for concrete cast against and

permanently exposed to earth, the minimum cover shall be 75 mm.


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SHRINKAGE AND TEMPERATURE REINFORCEMENT (2010 NSCP)

Shrinkage and temperature reinforcement is required at right angles to the principles

reinforcement to minimize cracking and to tie the structure together to ensure its acting

as assumed in the design. The provisions of this section are intended for structural

slabs only; they are not intended for soil-supported slabs on grade.

Reinforcement for shrinkage and temperature stresses normal to flexural reinforcement

shall be provided in structural slabs where the flexural reinforcement extends in one

direction only.

Shrinkage and temperature reinforcement shall be provided in accordance with either of

the following:

a) Where shrinkage and temperature movements are significantly restrained,the requirements of 408.3.4 and 408.3.3 shall be considered.

b) Deformed reinforcement conforming to 43.6.3 used for shrinkage and

temperature reinforcement shall be provided in accordance with the

following:

Areas of shrinkage and temperature reinforced shall be provided at least the followingrations of reinforcement area to gross concrete area, but no less than 0.014:

a) Slabs where Grade 280 or 350 deformed bars are used.0.0020b) Slabs where Grade 420 deformed bars or welded wire reinforcement are

used...0.0018

c) Slabs where reinforcement with stress exceeding 420 MPa measured at ayield strain of 0.35 percent is

used.......0.0018x415/Shrinkage and temperature reinforcement shall be spaced not farther apart than fivetimes the slab thickness, nor farther apart than 450 mm.

LOADS


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The most important and most critical task of an engineer is the determination of theloads that can be applied to a structure during its life, and the worst possiblecombination of these loads that might occur simultaneously. Loads on structure may beclassified as dead loads or live loads.

DEAD LOADDead loads are loads of constant magnitude that remain in one position. This consistsmainly of the weight of the structure and other permanent attachments to the frame .

LIVE LOAD

Live loads are loads that may change in magnitude and position. Live loads that moveunder their own power called moving loads. Other Live loads are those caused by wind,rain, earthquakes, soils, and temperature changes. Wind and earthquake loads arecalled lateral loads.

ARRENGMENTS OF LIVE LOAD

Live loads may be applied only to the floor or roof under consideration, and the far endsof columns built integrally with the structure may be considered fixed. It is permitted bythe code to assume the following arrangement of live loads:

(a) Factored dead load on all spans with full factored live load on two adjacentspans, and(b) Factored dead load on all spans with full factored live load on alternativespans.

REQUIRED STRENGHT (FACTIRED LOAD), U

Required strength U to resist dead load (D) and live load (L) shall be at least equal to:

Eq. 1-2 U=1.4D + 1.7L

If resistances to structural effects of a specified wind load W are included in design, thefollowing combination of D, L, and W shall be investigated to determine the greatestrequired strength U:

Eq. 1-3 U=0.75(1.4D + 1.7L + 1.7W)

Where load combinations shall be include both full value and zero value of L todetermine the more severe condition, and

Eq. 1-4 U=0.9D + 1.3W


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But for any combination of D, L, and W, required strength U shall not be less than Eq.1-2

If resistance to specified earthquake loads of forces E is included in design, the

following combinations of D, L and E shall be investigated to determine the greatestrequired strength U:

Eq. 1-5 U=1.1D + 1.3L + 1.1E

Where load combinations shall included both full value and zero value of L to determinethe more severe condition, and

Eq. 1-6 U=0.9D + 1.1E

But for any combination of D, L, and E, required strength U shall not be less than Eq. 1-2

If resistance to earth pressure H is included in design, required strength U shall be atleast equal to:

Eq. 1-7 U=1.4D + 1.7L + 1.7 H

Except where D or L reduces the effect of H, 0.9D shall be substituted for 1.4D and zerovalue of L shall be used to determine the greatest required strength U. For anycombination of D, L and H, required strength U shall not be less than.

If resistance to loadings due to weight and pressure of fluids with well defined densitiesand controllable maximum heights F is included in design, such loading shall have afactor of 1.4 and to be added to all loading combinations that include live load.

If resistance to impact effects is taken into account in design, such effects shall beincluded with live load L.


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Where structural effects T of differential settlement, creep, and shrinkage expansion ofshrinkage-compensating concrete or temperature change may be significant in design,required strength U shall be equal to

Eq. 1-8 U=1.75(1.4D +1.4T + 1.7L)

But required strength U shall not be less than

Eq. 1-9 U=1.4(D + T)

Estimations of differential settlement, creep, and shrinkage expansion of shrinkagecompensating concrete or temperature change shall be based on a realistic assessmentof such effects occurring in service.

STRENGTH REDUCTIONS FACTORS, (PHI)

The design strength provided by a concrete member, its connections to other members,

and its cross sections, in terms of flexure, axial load, shear, and torsion shall be taken

as the nominal strength multiplied by a strength reduction factor having following

values.

(a) Flexure without axial load 0.90

(b) Axial tension, and axial tension with flexure 0.90

(c)Axial tension and axial tension with flexure:

1. Spiral reinforcement.. 0.752. The reinforcement & other reinforced members... 0.75

(d) Shear and torsion.. 0.85(e) Bearing on concrete.. 0.70(f)Post-tensioned anchorage zones 0.85


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ACI-318-05 (NSCP C101-10-210)

Notations

gross of concrete sections . For a hollow section, is the area of theconcrete only and does not include the area of the void(s) area of shear reinforcement spacing, web width, or diameter of circular section, mmD = dead loads, or related internal moments and forces

d = distance from extreme compression fiber to centroid of longitudinal tensionreinforcement, mm

E = load effects of earthquake, or related internal moments and forces

specified yield strength of transverse reinforcement, MPaF = loads due to weight and pressures of fluids with well-defined densities andcontrollable maximum heights, or related internal moments and forces.

h = overall thickness or height of member, mm

H = loads due to weight and pressure of soil water in soil, or other materials, or related

internal moments and forces.

L = live loads or related internal moments and forces.

roof live loads or related internal moments and forces. factored moment at section, N-mm factored axial force normal to cross section occurring simultaneously with or ;to be taken as positive for compression and negative for tension, N

R = rain load, or related internal moments and forces.


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T = cumulative effect of temperature, creep, shrinkage , differential settlement, andshrinkage-compensating concrete.

U = required strength to resist factored loads or related internal moments and forces,

= nominal shear strength provided by concrete, N = nominal shear strength. = nominal shear strength provided by shear reinforcement N = factored shear force at section, NW = wind load, related internal moments and forces

= net tensile strain in extreme layer of longitudinal tension steel at nominal strength,excluding strains due to effective prestress, creep, shrinkage, and temperature = strength reduction factor = ratio of to CHAPTER 9 STRENGTH AND SERVVICEAB ILITY REQUIREMENTS

9.1- GENERAL

9.1.1 Structures and structural members shall be designed to have designstrengths at all sections at least equal to the required strengths calculated for thefactored loads and forces in such combinations as are stipulated in this code.

9.1.2 Members also shall meet all other requirements of this code to ensureadequate performance at service load levels.

9.1.3 Design of structures and structural members using the load factorcombinations and strength reduction factors of Appendix C shall be permitted.Use of load factor combinations from this chapter in conjunction with strengthreduction factors of appendix C shall be permitted.


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9.2 Requ ired strength

9.2.1 Required strength U shall be at least to the effects of factored loads in Eq.(9-1) through (9-7). The effect of one or more loads not acting simultaneouslyshall be investigated.

U = 1.4 (D+F) (9-1)

U = 1.2(D+F+T) + 1.6(L+H) + 0.5(or R) (9-2)U = 1.2D + 1.6( or R) + (1.0L or 0.8W) (9-3)U = 1.2D + 1.6W + 1.0L + 0.5(

or R) (9-4)

U = 1.2D + 1.0E+ 1.0L (9-5)U = 0.9D + 1.6W+ 1.6H (9-6)U = 0.9D + 1.0E+ 1.6H (9-8)

Except as follows:

a) The load factor on the live load L in Eq. (9-3) to (9-5) shall be permitted to bereduced to 0.5 except for garages, areas occupied as places of public assembly,

and all where L is greater than 4.8N/.b) Where wind load W has not been reduced by a directionality factor, it shall be

permitted to use 1.3 W in Eq. (9-4) and (9-6).c) Where E, the load effects of earthquake, is based on service-level seismic

forces, 1.4E shall be used in place of 1.0E Eq. (9-5) and (9-7).d) The load factor on H, loads due to weight and pressure of soil, water in soil or

other materials, shall be set equal to zero in Eq. (9-6) and (9-7) if the structuralaction due to H counteracts that due to W or E. Where lateral earth pressure

provides resistance to structural actions from other forces, it shall be not beincluded in H but shall be included in the design resistance.

9.2.2 If resistance to impact effects is taken into account id design, such effectsshall be included with L.


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9.2.3 Estimations of differential settlement, creep, shrinkage, expansion ofshrinkage-compensating concrete. or temperature change shall be based on arealistic assessment of such effects occurring in service.

9.2.4 If structure is in a flood zone, or is subjected to forces from atmospheric ice

loads, the flood or ice loads and the appropriate load combinations ofSEI/ASCE7 shall be used.

9.2.5 For post-tensioned anchorage zone design, a load factor of 1.2 shall beapplied to the maximum prestressing steel jacking force.

9.3 Design st rength

9.3.1Design strength provided by a member, its connections to other members,and its cross sections, in terms of flexure, axial load, shear and torsion, shall be

taken as the nominal strength calculated in accordance with requirements andassumptions of this code, multiplied by the strength reduction factors in9.3.2,9.3.4, and 9.3.5.9.3.2Strength reduction factor shall be as given in 9.3.2.1 through 9.3.2.7:

9.3.2.1 Tension-controlled sections as defined in 10.3.4.0.90(See also 9.3.2.7)

9.3.2.2 Compression-controlled sections, as defined 10.3.3:a) Members with spiral reinforcement conforming to 10.9.3..0.70b) Other reinforced members..0.65

For sections in which the net tensile strain in the extreme tension steel at nominal

strength is between the limits for compression-controlled and tension-controlledsections, shall be permitted to be linearly increase from that for compression-limit to0.005.


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Alternatively, when Appendix B is used, for members in which does not exceed 415MPa, with symmetric reinforcement, and with (d-d)/h not less than 0.70,shall be

permitted to be increased linearly to 0.90 as decreases from 0.10 to zero. Forother reinforced members,

shall be permitted to be increased from 0.10

or

,

whichever is smaller, to zero.

9.3.2.3Shear and torsion0.759.3.2.4 Bearing on concrete (except for post-tensioned and anchoragezones and struct-and-tie models).0.65

CHAPTER 1

Analys is and Design of BeamNOTAIONS AND SYMBOLS USED = depth of equivalent stress block, mm = area of tension reinforcement, mm2 = area of skin reinforcement per unit height in one side face, mm2/ m = width of compression face of member, mm = distance from extreme compression fiber to neutral axis, mm = distance from extreme compression fiber to centroid of tension reinforcement,

mm =distance from extreme compression fiber to centroid of compressionreinforcement, mm

= thickness of concrete cover measured from extreme tension fiber to center ofbar or wire, mm =modulus of elasticity of concrete, MPa = modulus of elasticity of steel 200,000 MPa =specified compressive stress of concrete, MPa =calculated stress in reinforcement at service loads, MPa =specified yield strength of steel, MPa =overall thickness of member, mm =moment of inertia of gross concrete section about centroidal axis, neglectingreinforcement

=moment of inertia of reinforcement about centroidal axis of member cross-

section


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=nominal moment, N-mm =factored moment at section, N-mm =factor defined in Section 410.4 in Page 16 =strain in concrete (maximum = 0.003)

=strain in steel below yield point =

=strain in steel at yield point =ration of tension reinforcement =balance steel ratio=strength reduction factor

ASSUMPTION IN STRENGTH DESIGN IN FLEXURE

(CODE SECTION 5.10.2)1. Strain in reinforcement and concrete shall be based assumed directly proportional tothe distance from the neutral axis. Expect for deep flexural members with overall depth

to clear span to ratio, h/L> 2/5 for continuous spans and h/L >4/5 for simple spans, anonlinear distribution of strain shall be considered (See Sec. 5.10.7).

2. Maximum usable strain at extreme concrete compression fiber, shall be assumedequal to 0.003

3. For below shall,be taken as x for >,= .4. Tensile strength of concrete shall be neglected in axial and flexural calculations.

5. Relationships between compressive stress distribution and concrete strain may be

assumed rectangular, trapezoidal, parabolic, or any other from that result in predictionof strength in substantial agreement with results of comprehensive tests.

6. For rectangular distribution of stress:

a) Concrete stress of 0.85 shall be assumed uniformly distributed over anequivalent compression zone bounded by edges of the cross-section and astraight line located parallel to the maximum compressive strain.

b) Distance c from fiber of maximum strain to the neutral axis hall is measured inthe direction perpendicular to N.A.


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c) Factor shall be taken as 0.85 for 30 MPa and shall be reducedcontinuously at rate of 0.008 for each 1 MPa of strength in excess of 30 MPa, but shall not be taken less than 0.65. i.e

i. For

30 MPa,

= 0.85

ii. For > 30 MPa, =0.85-0.008(-30) but not shall be less than 0.65RECTANGULAR BEAM REINFORCED FOR TENSION ONLY

(SINGLY REINFORCED)

b 0.85 0.003c a c

d d-a/2NA

T= Stress Diagram Strain Diagram

Figu re 2.1:Stress and strain diagram for singly reinforced and rectangular beam

Eq. 2-1 For For ( but shall not be less than 0.65 C=T

0.85 Eq. 2-2

As


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Multiplying Eq. 2-2 by d/d:

The termis called the ratio of steel reinforcement and is denoted as.

Eq. 2-3

and

Eq. 2-4 Let Eq. 2-5

Nominal Moment Capacity:

From the stress diagram in Figure 2.1:


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Eq.2-6

Ultimate Momen t Capacity (Design Strength ): Eq.2-7 Coefficient of Resistance

Eq.2-8 Eq.2-9

Solving for an in Eq. 2-8 and replacing it with, , yields the following formula thesteel ratio :Eq.2-10 [ ]

BALANCE DESIGNBalance design refers to a design so proportioned that the maximum stresses in concrete

(with strain of 0.003) and steel and (with strain of ) are reached simultaneously once heultimate load is reached, causing them to fail simultaneously.UNDERREINFORCED DESIGN

Underreinforced design is a design in which the steel reinforced is lesser than what isrequiredfor balance condition . If the ultimate load is approached, the steel will begin to yield althoughthe compression concrete is still understressed. If the load is further increased, the steel willcontinue to elongate, resulting in appreciable deflections and large visible crack in the tensileconcrete. Failure under this condition is ductile and will give warning to the user of the

structure to decrease the load.


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OVERREINFORCED DESIGN

Overreinforced design is a design in which the steel reinforcement is more than what isrequired for balanced condition. If the beam is overreinforced, the steel will not before failure.

As the load is increased, deflections are not noticeable although the compression concrete is

highly stressed, and failure occurs suddenly without warning to the user of the structure.

Overreinforced as well as balanced design should be avoided in concrete because of its brittleproperty, that is why the Code limits the tensile steel percentage (Pmax=0.75pb) to ensureunderreinforced beam with ductile type of failure to give occupants warning before occurs.

BALANCED STEEL RATIO :In balanced condition, the concrete and steel yield simultaneously, In this condition, the strain

in concrete reached is maximum usable value of and the strain in steel is

where

=200,000 MPa.

By ratio and proportion in the triangle shown in Figure2.2:

Note:


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Eq.2-11

But a =

c=c Eq. 2-12 Note: Eq. 2-12 is for singly reinforced rectangular sections only. Eq. 2-11 is applicableto nay shape.

MAXIMUM STEEL REINFORCEMENT

Sectio n 410.4.3: For flexural and for subject to combined flexure and compressive axial

load when the design axial load strength is less than the smaller of or ,the ratio of reinforcement that would produce balance strain condition for the sectionunder flexure without axial; load. For members with compression reinforcement, the

portion of equalized by compression reinforcement need not be reduced by the0.75factor.

Eq. 2-13 and

Eq. 2-14


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This limitation is to ensure that the steel reinforcement will yield first to ensure ductilefailure.

MINIMUM REINFORCEMENT OF FLEXURAL MEMBERS

410.61 At very section of flexural members where tensile reinforcement is required by

analysis, the area provided shall not be less than that given by:Eq. 2-15 Eq.2-16 and not less than

410.62 For statically determinate T-section with flange in tension, the area shallbe equal to or greater than the smaller value given either by:Eq. 2-17

or Eq. 2-15 with set equal to the width of the flange.410.6.3 The requirements of Sections 410.6.1 and 410.6.2 need to be applied if at everysection the area of the tensile reinforcement is at least one-third greater than thatrequired by analysis.

410.6.4 For structural slabs and footings of uniform thickness, the minimum area oftensile reinforcement in the direction of span shall be the same as that required bySection 407.13 (Shr inking and Temperature Reinforcement). Maximum spacing ofthis reinforcement shall not exceed three times the thickness and 450 mm.

The provision for minimum amount of reinforcement applies to beams, which forarchitectural and other reasons are much larger in cross-section than required bystrength consideration. With a very small amount of tensile reinforcement, the computedmoment strength as a reinforced concrete section computed from its modulus ofrapture. Failure in such a case can be quite sudden.


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STEPS IN DESIGNING A SINGLY REINFORCED

RECTANGULAR BEAM FOR FLEXURE:

Note: The assumptions made in steps II, V,and VIII are the authors recommendationbased on his experience.

I. Identify the values of the dead load and live load to be carried by thebeam. (DL & LL)

II. Approximate the weight of beam (DL) between 20% to 25% of(DL+LL).This weight is added to the de load.

III. Compute the factored load and factored moment:Ex: factored Load =1.4 DL+1.7L

IV. Compute the factored moment to be resisted by the beam, V. Try a value of steel ratio from 0.5 but must not be less than . This

value will provided enough alloance in the actual value of due torounding-off of the number of bars to be used, for it not to exceed the

maximum of 0.05b. ( ) VI. Compute the value of VII. Solve for

:

VIII. Try ratio ( from d=15b to d=2b), and solve for d, (round-off this valueto reasonable dimension). Check also the minimum thickness of beamrequired by the Code a given in Table 2.1 in page 36.

After solving for d, substitute its value to Step VII, and solve for b.Compute the weight of the beam and compare it to the assumption madein Step II.

IX. Solve for the required steel area and number of bars.

Number of bars(diameter = D)


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x number of bars =STEPS IN COMPUTING THE REQUIRED TENSION STEEL AREA OF A BEMWITH KNOWN MOMENT NT

AND OTHER PROPERTIES:

I. Solve for (1-0.59

if design as singly reinforced (Step II)if

design as doubly reinforced (Step III)

II. Solve for :

III. Compression reinforcement is necessary. (See Chapter 3)

STEPS IN COPUTING OF A BEAM WITH KNOWN TENSION STEEL AREAAND OTHER BEAM PROPERTIES:


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I. Solve for : II. Check if steel yields by computing

III. ` if ,steel yields, proceed to IIIif ,steel does not yield, proceed to step IV.Note: if ,the givenis not adequate for the beam dimension.

IV.


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Solve forfrom the strain diagram: [Note: =200,000MPa]

Eq. 2-18 [ T=C but a=

Solve c by quadratic formula and solve for and a: or

MINIMUM THICKNESS OF FLEXURAL MEMBERS

According to Section 5.9.5 of NACP, minimum thickness stipulated in Table 2.1 shallapply for one-way construction not supporting are attached to portions or otherconstruction likely to be damaged by large deflections, unless computation of deflectionindicates a lesser thickness can be used without adverse effects.

Table 2.1 MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS OR ONE-WAYSLABS UNLESS DEFLECTIONS ARE COMPUTED *


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Minimum thickness, h

Simplysupported

One endcontinuous

Both endscontinuous

Cantilever

MemberMembers not supporting or attached to partitions or other constructionlikely to be damaged by large deflections

Solid one-wayslabs

L/20 L/24 L/28 L/10

Beams orribbed one-wayslabs

L/16 L/18.5 L/21 L/8

Span length L is in millimetersValues given shall be used directly for members with normal density concrete

( ) and grade 415 reinforcement. For other conditions, the values shallbe modified as follows:

(a) For structural lightweight concrete having weights in the range 1500-2000values shall be multiplied by (1.65-0.005) but not less than 1.09, where isthe unit mass in .

(b) For other than 415 MPa, the values shall be multiplied by (0.4 + BEAM DEFLECTION (SECTION 5.9.5

Sect. 5.9.5.2.2 Where deflections are to be computed, deflections that occurimmediately on application of load shall be computed by usual methods or formulas forelastic deflections, considering effects of cracking and reinforcement on member

stiffness.

Sec t. 5.9.5.2.3 Unless stiffness values are obtained by a more comprehensive analysis,

immediate deflection shall be computed with the modulus of elasticity for concreteand with the effective moment of inertia as follows, but not greater than .

Eq.2-19 * + * +3]


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Where = = modulus of rapture of concrete, MPa, for normal weightConcrete

= maximum moment in member at stage deflections is computed.= moment of inertia of gross concrete section about centroidal axis, neglectingreinforcement.= moment of inertia of cracked section transformed to concrete= distance from centroidal axis of gross section, neglecting reinforcement, to extremefiber in tension.

When Lightweight aggregate is used, one of the following modifications shall apply:

(a) When

is specified and concrete is proportioned in accordance with Sec. 5.5.2,

shall be modified by substituting 1.8 for but the value of 1.8 shall notexceed.(b) When is not specified,shall not be multiplied by 0.75 for all lightweightconcrete, and 0.85 for sand-lightweight concrete. Linear interpolation is permitted if

partial sand replacement is used.

Sec t. 5.9.5.2.4: For continuous members, effective moment of inertia may be taken asthe average of values obtained from Eq. 2-19 for the critical positive and negativemoment sections. For prismatic members, effective moment of inertia may be taken as

the value obtained from Eq. 2-19 at midspan for simple and continuous spans, and atthe support cantilevers.

Sect.5.9.5.2.5: Unless values are obtained by a more comprehensive analysis,additional long-term deflection resulting from creep and shrinkage of flexural members(normal weight or lightweight concrete) shall be determined by multiplying theimmediate caused by the sustained load considered, by the factor.

Eq. 2-10


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Where shall be taken the value of reinforcement ratio for non-prestressedcompression reinforcement at midspan for simple and continuous spans,a nd at support

for cantilevers. It is permitted to assume the time-dependent factor for sustained loadsto be equal to:5 years or more2.0

12 months...1.46 months..1.23 months1.0

Deflection computed in accordance with Sec. 5.9.5.2.2 through Sec.5.9.5.2.5 shall notexceed limits stipulated in Table 2.2.

Table 2.2: Maximum Permissible Computed Deflections

Type of member Deflection to be considered Deflection limitation

Flat roofs not supporting orattached to nonstructuralelements likely to be damage bylarge deflections

Immediate deflection due tolive load LL

L/180*

Floors not supporting orattached to nonstructuralelements likely to be damagedby large deflections

Immediate deflection due tolive load LL

L/360*

Roof or floor constructionsupporting, or attached tononstructural elements not likelyto be damaged by largedeflections

That part of the totaldeflection occurring afterattachment of non structuralelements (sum of the long-time deflection due to allsustained loads and theimmediate deflection due toany additional live load)****

L/480**

Roof or floor constructionsupporting, or attached tononstructural elements not likelyto be damaged by largedeflections

L/20****


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Limit not intended to safeguard against ponding. Ponding should be cheated bysuitable calculations of deflections, including added deflections due to pondedwater and considering long-term effects of all sustained loads, camber,construction tolerances, and reliability of provisions for damage.

Limit may be exceeded if adequate measures are taken to prevent damage to

supported or attached elements. Long=time deflections shall be determined in accordance with Sec.5.9.5.2.5 or

Sec. Attachment of nonstructural elements. This amount shall be determined onbasis of accepted engineering, data relating to time-deflection characteristics ofmembers similar to those being considered.

But not greater than tolerance provided for nonstructural elements. Limit may beexceeded if camber is provided so that deflection minus camber does notexceeded limit.

NSCP COEFFICICIENTS FOR CONTINUOUS BEAMS AND SLASBS

Section 5.8.3.3 of NSCP states that in lieu of frame analysis, the following approximatemoment and shears are permitted for design of continuous beams and one-way slabs(slabs reinforced to resist flexural stresses in only one direction), provided:

a) There are two or more spans,b) Spans are approximately equal, with the larger of two adjacent spans not greater

than uniformly than the shorter by more than 20 percent,

c) Loads are uniformly distributed,d) Unit live does not exceeded three times unit dead load, ande) Members are prismatic.

Positive momentEnd spans

Discontinuous end unrestrainedDiscontinuous end integral with support..

Interior spansNegative moment at exterior face of first interior support

Two spans .....


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More than two spans.....Negative moment at other faces of interior supports.Negative moment at face of all supports for:

Slabs with spans not exceeding 3 m; and beamsWhere ratio of sum of column stiffness to beams

Stiffness exceeds eight at each end of the spanNegative moment at interior face of exteriorSupport members built integrally with

Where support is a spandrel beamWhen support is a column....

Shear in end members at face of

First interior support.....................................Shear at face of all other supports..When =clear span positive moment or shear and average of adjacent clear spans fornegative moment.


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Figure 2.3: Shear and moment for continuous beam or slab with spans and

discontinuous end integral with support


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Figu re 2.5 Shear and moment for continuous beam or slab with more than two spansand discontinuous end unrestrained

ACI-318-05 (NSCP C101-10-2010)

10.2 Design ass um ption s (410.3)

10.2.1 Strength design of members for flexure and axial loads shall be based onassumptions given in 10.2.2 through 10.2.7, and on satisfaction of applicable conditionsof equilibrium and compatibility of strains.

10.2.2 Strain in reinforcement and concrete shall be assumed directlyproportional to the distance from the neutral axis, except that, for deep beams asdefined in 10.7.1, an analysis that considers a nonlinear distribution of strain shall be

used alternatively, it shall be permitted to use a struct-and tie model. See 10.7,118, andAppendix A.

10.2.3 Maximum usable strain at extreme concrete compression fiber shall beassumed equal to 0.003.

10.2.4 Stress in reinforcement below shall be taken as times steel strain.For strains greater than that corresponding to, stress in reinforcement shall beconsidered independent of strain and equal to.

10.2.5 Tensile strength of concrete shall be neglected in axial and flexural

calculations of reinforced concrete, except when meeting requirements of 18.4.

10.2.6 The relationship between concrete compressive stress distribution andconcrete strain shall be assumed to be rectangular, trapezoidal, parabolic, or any othershape that results in prediction of strength in substantial agreement with results ofcomprehensive tests.

10.2.7 Requirements of 10.2.6 are satisfied by an equivalent rectangularconcrete stress distribution defined by the following:

10.2.7.1 Concrete stress of 0.85

shall be assumed uniformly distributed

over an equivalent compression zone bounded by edges of the cross section and a


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straight line located parallel to the neutral axis at distance a= form the fiber ofmaximum compressive strain.

10.2.7.2 Distance from the fiber of maximum strain to the neutral axis, c ,shall be measured in direction perpendicular to the neutral axis.

10.2.7.3 For between 17 and 18 MPa, shall be taken as 0.85. Forabove 28 MPa, shall not be taken less than 0.6510.3General princip les and requ irements (410.4)

10.3.1 Design of cross sections subject to flexure or axial loads, or to combinedflexure and axial loads, shall be based on stress and strain compatibility usingassumptions in10.2.

10.3.2 Balanced strain conditions exist at a cross section when tension

reinforcement reaches the strain corresponding to just as concrete in compressionreaches its assumed ultimate strain of 0.003.

10.3.3 Sections are compression-controlled if the next tensile strain in the

extreme tension steel, , is equal to or less than the compression-controlled strain limitwhen the concrete in reaches its assumed strain limit of 0.003. The compression-controlled strain limit is the net tensile strain in the reinforcement at balanced strainconditions. For Grade 415 reinforcement, and for all prestressed reinforcement, it shall

be permitted to set the compression-controlled strain limit equal to 0.002.

10.3.4 Sections are tension-controlled if the net tensile strain in the extreme

tension steel is equal to greater than 0.005 when the concrete in compressionreaches its assumed strain limit of 0.003. Sections with between the compression-controlled strain limit and 0.005 constitute a transition region between compression-controlled and tension-controlled sections.

Derivation: for E = 200 GPaThe beam is tension-controlled

When

= 0.005 (or

=1000MPa)


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Eq. 2-21

For rectangular beam:

Eq. 2-22

10.3.5 For nonprestressed flexural members and nonprestressed members with

factored axial compressive load less than 0.10 steel strain at nominal strengthshall not be less than 0.004.

10.3.5.1 Use of compression reinforcement shall be permitted in conjunction withadditional tension reinforcement to increase the strength of flexural members.

Derivation: for E =200 GPa


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Maximum steel area and when beam is singly reinforced:

Eq. 2-23 For rectangular section:

T= ) b

Eq. 2-14


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Eq. 2-25 )Eq. 2-26

10.3.6 Design axial strength

of compression members shall not be taken greater

than computed by Eq. (10-1) or (10-2).10.3.6.1 For nonprestressed members with spiral reinforcement conforming to

7.10.4 or composite members conforming to 10.16:

(10-1)10.3.6.2 For non nonprestressed members with spiral reinforcement conforming

to 7.10.5: (10-2)10.3.6.3 For prestressed members, design axial shall not be taken greaterthan 0.85 (for members with spiral reinforcement) or 0.80 (for members with tie

reinforcement) of the design axial strength at zero capacity .10.3.7 Members subject to compressive axial load shall be designed for the maximum

moment that can accompany the axial load. The factored axial force at giveneccentricity shall not exceed that given in 10.3.6. The maximum factored momentshall be magnified for slenderness effects in accordance with 10.1010.4 Distance between lateral sup po rts of flexural members


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10.4.1 Spacing of lateral supports for a beam shall not exceed 50 times b, theleast width of compression flange or face.

10.4.2 Effects of lateral eccentricity of load shall be taken into account in

determining spacing of lateral supports.

10.5.1 Minimum reinforcement of f lexural m embers

10.5.1 At every section of flexural members where tensile reinforcement isrequired by analysis, except as provided in 10.5.2, 10.5.3, and 10.5.4, as provided shallnot be less than that given by

(10-3)and not less than

(10-3)

10.5.2 For statically determinate members with a flange in tension, shall not beless than the value given by eq. (10-3), except that is replaced by either or thewidth of the flange, whichever is smaller.

STEPS IN THE DESIGN OF SINGLY REINFORCEDRECTANGULAR BEAM FOR FLEXURE

Note: The assumption made in steps II, V, and VIII are the authors recommendationbased on his experience.

I. Determine the values of loads, Dl, LL and other loadsII. Approximate the weight of beam (DL) as follows:

Small beams: 2kN/mMedium-sized beams: 3.5kN/mLarge-sixed beams: 7kN/m

or Weight of beam in kN/m=24kN/x beam area in III. Compute the factored load on different load combinations

Example: Factored Load =1.2 DL + 1.6 LL


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IV. Compute the factored moment to be resisted by the beam, V. Try a value of steel ratio from 0.7 to 0.8but must not be less

than . This value of will provided enough allowance in the actualvalue of due to rounding-off the numbers bars to be used so that it willnot exceed the maximum

.

()

( )

VI. Compute the value of VII. Solve for the reduction factor

Solve for c:

Note: For singly reinforced rectangular beam, is directlyproportional to c:c=(assumed factor) x The assumed factor may range from 0.7 to 0.8 as suggested instep V.

if


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if VIII. Solve for :

IX. Try a ratio d/b (from d= 1.5b to d=2b), and solve for d. (round-off this valueto reasonable dimension)

Check also the minimum thickness of beam required by the code as givenin Table 2.1 in Page 26.

After solving for d, substitute its value to Step VII, and solve for b.

Compute the weight of the beam and it to the assumption made in Step II.

X. Solve for the required steel area and number of bars. Number of bars (diameter=D) x number of bars =

STEPS IN FINDING THE REQUIRED TENSION STEEL AREA

OF A BEAM WITH

KNOW REQUIRE MOMENT AND OTHER BEAM PROPERTIESGiven b, d, and :I. Solve for and .


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if design as Singly Reinforced (Step II)if design as Doubly Reinforced (Step V)

II. Determine if the section in tension-controlled or transition

From Eq. 2-11:

if ,proceed to step IIIif region, proceed to step IVIII.

IV.


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Solve for c and

* +

if V. Compression reinforcement is necessary.(See chapter 2)

STEPS IN FINDING OF A BEAM WITH KNOWN TENSION STEEL AREAAND OTHER BEAM PROPERTIES:Given: b, d, ,,:I. Solve for

II. Check if steel yields by computing () if steel yields, proceed to step IIIif steel dos not yield, proceed to step IV.Note: if the givenis not adequate for the beam dimension.


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III. Solve for : ;c==_________

if if

IV. Compression-controlled b 0.85

a c=

. dd-a/2

T=C but a= T=

c=__________ ; =__________a=

or


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ILLUSTRATIVE PROBLEMS

DESIGN PROBLEMS

PROBLEM 2.1A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 mm

and is reinforced for tension only. Assuming and , determinethe balance steel area in sq.mm.

SOLUTION

PROBLEM 2.2

A rectangular beam has b = 300 mm and d =490 mm. Concrete compressive strength and steel yield strength . Calculate the required tensionsteel area if the factored moment is (a) 20 kN-m,(b)140 kN-m,(c) 485 kN-m, and(d)620 kN-m.

SOLUTION

Solve for


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a) 20 x

b) (singly reinforced) 140 x 1


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c) (singly reinforced) 485 x

d) The beam will be doubly reinforced. See Chapter 3.

PROBLEM 2.3 (CE MAY 2012)


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A reinforced concrete beam has a width of 300 mm and an overall depth of 480 mm.

The beam is simply supported over span of 5 m. Steel strength MPa andconcrete . Concrete cover is 70 mm from the centroid of the steel area. Unitweight concrete is 23.5kN/.Other than the weight of the beam , the beam carries asuperimposed dead of 18 kN/m and a live load of 14 kN/m. Use the strength design

method.a) Determine the maximum factored moment on the beam.b) If the design ultimate moment capacity of the beam is 280 kN-m,

determine the required number of 20 mm tension bars.c) If the beam will carry a factored load of 240 kN at midsoan, determine the

required number of 20 mm tension bars.

SOLUTION

Given: b=300m d=480-70=410 mm

Bar diameter , Weight of beam,

a) Maximum factored moment on the beam.Factored load, Factored load, Maximum factored moment:

b)

Solve for to determine whether compression steel is needed


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Required


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2498 = N PROBLEM 2.4 (CE MAY 1993)

A reinforced concrete beam has a width of 300 mm and an effective depth totension bars of 600 mm. compression reinforcement if needed will be placed at a

depth of 60 mm below the top. If and , determine thetension steel area if the beam is to resist an ultimate moment of 650 kN-m.

SOLUTION

Solve for and :


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[1-0.59(0.309) Since , the beam may be designed as singly reinforced. 650 x 1 Solve for :

PROBLEM 2.5 (CE Novem ber 2000)

A rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm.The beam is simply supported over a span 6 m and is used to carry a uniform dead load

of 25 kN/m and a uniform live load of 40 kN/m. Assume and . Compression reinforcement if necessary shall be placed at a depth 80 mmfrom the outermost compression concrete.

a) Determine 80 mm from the outermost compression concrete.

b) Determine the required tension steel area.


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c) Determine the required number of 25-mm tension bars.

SOLUTIONa) Maximum steel area:

9

b) Required tension steel area:

Factored load: Required strength:

=463.5kN-mSolve for


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c) Number of 25 mm bars:

Number of 25-mm bars= Number of 25-mm bars=

PROBLEM 2.6 (CE MAY 2009)

A reinforced concrete beam has a width of 300 mm and total depth of 600 mm.The beam will be design to carry a factored moment of 540kN-m. Concrete

strength and steel yield strength . Solve using thestrength design method.

a) Determine the balanced steel ratio in percent.

b) Determine the minimum effective depth of the beam using a steel ratio equal to 0.5 of balanced steel ratio.

c) Determine the minimum effective depth of the beam using the maximumallowable steel ratio.


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SOLUTION

Given:

b=300 mm h=600 mm

a) Balanced steel ratio:

( )

b) Effective depth using

540 x 1


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PROBLEM 2.7

A concrete one-way slab has a total thickness of 120 mm. The slab will be reinforced

with 12-mm-diameter bars with .Concrete strength .Determine the required spacing 12 mm main bar if the total factored moment acting on1-m width of slab is 23 kN-m width of slab is 23 kN-m. Clear concrete cover is 20 mm.

SOLUTION

Note: Slabs are practically singly reinforced because of its small depths.

. 12mm bars d h=120

. mm

s s cover=20 mm

b = 1000 mm

Effective depth, d= 120 -20-1/2(12)=94 mmWidth, b = 1000 mm

23 x


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Spacing of bars (for walls and slabs using unit width):

Eq. 2-17

PROBLEM 2.8

A 2.8 m square column fooring has a total thickness of 47 mm. The factored moment atcritical section for moment is 640 kN-m. Assume and . Clearconcrete cover is 75 mm. Determine the required number of 20 mm tension bars. SOLUTION

Effective depth, d=470-75-1/2(20)=385 mmWidth, b =2800 mm

Design strength, Maximum and minimum requirements:


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(Procedure is not shown anymore see Problem 2.2)

Singly reinforced:

Number of 20 mm bars:


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PROBLEM 2.9

Design a rectangular beam reinforced for tension only to carry a dead load moment of

60 kN-m (including its own weight) and a live load moment of 48 kN- m. Use and SOLUTION

Required strength:

(Note: this already includes the weight of beam)

Try Note: this is the authors suggestion

165.6 x


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Try d = 1.75 b b=228 mm say 230 mmd=399 say 30 mm

Summ ary: b = 230 mmd = 400 mm

PROBLEM 2.10

Design a singly reinforced rectangular beam for a 6-m simple span to support asuperimposed dead load of 29 kN/m and a live load of 44 kN/m. Assume normal weigth

oncrete with

. Use

SOLUTION

Weight of beam: (this is the authors assumption)

Assuming a 300 mm x 600 mm, .7(44) .


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Assume d = 1.75 b (this is the authors assumption)

546.516 x Use b = 280 mm, d = 490 mm

Minimum beam the thickness (Section 409.6.2.1)

Using 32 mm bars (#100):


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280 mm

.

. h

Beam weight = 24 (0.28)(0.5545)

Beam weight = 3.73 kN/m < 4.32(OK)

PROBLEM 2.11

A propped cantilever beam shown in Figure 2.6 is made of reinforced concrete having awidth of 290 mm overall depth of 490 mm. The beam is loaded with uniform dead loadof 35 kN/m (including its own weight), and a uniform live load of 55 kN/m. Given

Concrete cover is 60 mm from the centroid of the bars.

Determine the required tension steel area for maximum positive moment. AssumeEI=constant.

290mm

490 mm

A 6m B 2m C

6 - 10


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Figure 2.6

SOLUTION

Given: O A B 2m C x

R

Moment Diagram

Solve for moment reactions using the three-moment equation:

Mo Lo +

D


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-489.75 = R(6)- 142.5(8)(4)R=676.875 kN

Maximum positive moment:

142.5(2 + x) - 676.875 = 0x = 2.75 m

Solve for :


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At a point of maximum positive moment:

(Singly reinforced) 253.828 x


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ANALYSIS OF RECTANGULAR BEAMS WHERE

STEEL YIELDS ( PROBLEM 2.12(CE MAY 1999)A reinforced concrete rectangular beam with b = 400 mm and d= 720 mm is reinforced

for tension only with 6-25 mm diameter bars. If and a) The coefficient of resistance of the beam.b) The ultimate moment capacity of the beam.

SOLUTION

Answer

Answer


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PROBLEM 2.13

A rectangular beam reinforced for tension only has b= 300 m, d = 490 mm. The tensionsteel area provided is 4,500 sq. mm. Determine the ultimate moment capcity of the

beam in kN-m. Assume

,

SOLUTION

PROBLEM 2.14

A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate the design moment


68/200

SOLUTION

Check if the beam satisfies the minimum requirement:

PROBLEM 2.15

A 130-mm-thick-one-way slab is reinforced with 12-mm-diameter tension bars spaced at

110 on centers. Concrete cover is 20 mm, concrete strength MPa and steelyield strength . Unit weight of concrete is 23.5 kN/.

a) What is the ultimate moment capacity of the slab?b) If the slab is simply supported over a span of 4 m, what safe uniform live

load pressure can the slab carry?


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SOLUTION

a) Consider 1 m width of slab, b = 1000 mm

Effective depth: d = hcover- 1/2 d = 130-20-1/2(12)=104 mm

Check if the beam satisfies the minimum steel requirement on flexures:


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b) Dead load pressure,

x thickness of concrete.

Dead load pressure, PROBLEM 2.16

A rectangular beam with b = 250 mm and d = 460 m is reinforced for tension only with3-25 mm bars. The beam is simply supported over a span of 6 m and carries a uniformdead load of 680 N/m including its own weight. Calculate the uniform live load that the

beam can carry. Assume and .SOLUTION

)Check if the beam satisfies the minimum steel requirement on flexure:


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PROBLEM 2.17 (CE JANUARY 2008)

A reinforced concrete rectangular beam has a width of 300 mm and an effective depth

of 55 mm. The beam is reinforced with six 25-mm-diameter tension bars. Steel yield

is 415 MPa and concrete strength is 28 MPa.a) What is the balanced steel ratio?b) What is the maximum steel area for singly reinforced?c) What is the nominal moment capacity of the beam?

SOLUTION


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a) Balanced steel ratio:

b) Maximum steel area

c) Nominal moment capacity

Using 6-25 mm bars:


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PROBLEM 2.18

A 350 mm x 500 mm rectangular is reinforced for tension only with 5-28 mmbars. The beam has an effective depth of 446 mm. The beam carries a uniformdead load of 4.5 kN/m (including its own weight), a uniform live load of 3 kN/m,and concentrated dead load of P and 2P as shown in Figure 2.7. Assume . Calculate the following:

a) The ultimate moment capacity of the section in kN-m, andb) The maximum value of P in kN.

2P P

2m 2m 2m

Figu re 2.7

SOLUTION


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Check if the beam satisfies the minimum requirement:

1.4(2P) 1.4P


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A

B C D2m 2m 2m

Figure 2.8Beam with factored loads

For the given loads, the maximum moment can occur at B or C:

At point C: Set 440.18 = 1.4P(2) + 11.4(2)(1)

At point B: (First solve for

Set Thus the maximum value of P such that will not exceed 440.18 kN-m is 149 kN.


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ANALYSIS OF RECTANGULAR BEAMS WHERE

STEEL DOES NOT YIELDS ( )PROBLEM 2.19

A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate the ultimate moment capacity of thebeam.

SOLUTION


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From Eq. 2-18


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PROBLEM 2.20

A rectangular beam reinforced for tension only has b=300 mm, d = 490 mm. The

tension steel area provided is 7-25 mm diameter bars with .Calculate the ultimate moment capacity of the beam.SOLUTION


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From Eq.2-18:


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ANALYSIS & DESIGN OF SINGLY REINFORCED

NON-RECTANGULAR BEAMS

PROBLEM 2.21

Compute the ultimate moment capacity of the beam shown in Figure 2.9. Assume and .

SOLUTION

Note: This is not a rectangular beam. Some formulas derived above (such

as ,) may not be applicable. The moment can be computed using theassumptions in the Code and the conditions of equilibrium.


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Solve for the balanced to determine whether the given steel yield or not.From Eq. 2-11

Since


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PROBLEM 2.22

Compute the ultimate moment capacity of the beam shown in Figure 2.10. Assume and .


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SOLUTION

Solve for

Since , tension steel does not yield ( solve for c:

0.85


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* +

PROBLEM 2.23

A hallow beam is shown in Figure 2.11. Assume and .a) Calculate the required tension steel area when .b) What is the balanced moment capacity of the beam?c) What is the maximum steel area under singly reinforced condition?d) What is the maximum design moment strength under singly reinforced condition?

e) Calculate the required tension steel area when

.


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Figu re 2.11-Hallow beam

SOLUTION

To guide us whether a: will exceed 150 mm or not, let us solve the design

moment when a=150 mm.

d = 80075 = 725 mm

a)

Since the required Assuming tension steel yields:

Check is steel yields:


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b) Balanced condition (See Figure 2.12)

)


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Figu re 2.12

c) Maximum steel area,

d) Maximum moment ,

Refer to Figure 2.12:

e) Refer to Figure 2.12


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BEAM DEFLECTION PROBLEM

PROBLEM 2.24

A reinforced concrete beam is 350 mm wide and 600 mm deep. The beam is simplysupported over a span of 8 m and carries a uniform dead load of 11 kN/m including itsown weight and a uniform live load of 15 kN/m. The beam is reinforced tension bars of

530 mm. Modulus of elasticity ofconcrete

and

a) Calculate the maximum instantaneous deflection due to service loads.b) Calculate the deflection for the same loads after five years assuming that 40% of

the live load is sustained.

SOLUTION


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Figure 2.13

Effective moment of inertia,

Eq. 2-19

Moment of inertia of cracked section with steel transformed to concrete FromFigure 2.13:

Modular ratio,


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Solve for c:Moment of area above N.A. = Moment of area below N.A.

350 x c x c/2 = 27,208(350-c)c = 219.7 mm

a) Instantaneous Deflection:

b) Long-term Deflection

Since only 40% of the live load was sustained:

w = 11 + 0.4(15) = 17 kN/m


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Instantaneous deflection

Note: Since deflections are directly proportional to the load, the instantaneous deflectiondue to sustained load can be found by ratio and proportion using the result in Parta. Long-term deflection =

Long-term deflection = 16.36 + 2(10.7)Long-term deflection = 37.76 mm


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PROBLEM 2.25 (CE NOVEMBER 2002)

The continuous reinforced concrete beam shown in Figure 2.14 is subjected to auniform service dead load of 16 k/m and a service live load of 32 kN/m,resulting in thebending moment diagram shown. Twenty percent of the live load will be sustained in

nature, while 80% will be applied only intermittently. The concrete strength

The modulus of elasticity of concrete is given by the expression and the modulus of rapture is given by the expression . Determine the following:a) The effective moment of inertia at the supports (maximum negative moment).b) The effective moment of inertia for the continuous member.c) The additional deflection (in addition to the initial deflection) after 5 years, under

the sustained loading if the instantaneous deflection due to the combined servicedead and live load is 5 mm.

Figu re 2.14


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SOLUTION a) Effective moment of inertia at the supports

Maximum moment, Distance from NA of gross section to extreme tension fiber,


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Moment of inertia of gross section, Moment of inertia of cracked section,

b) Effective moment of inertia for the continuous member

At maximum negative moment (at support)

Solving for at maximum positive moment (at midspan)


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c) Additional long term deflection= long term deflection x

Solving for the instantaneous deflection under sustained loading:

Instantaneous deflection = 5mm (given)Instantaneous loading = 16 kN/m + 32 kN/mInstantaneous loading = 48 kN/m

Sustained loading = 16 + 20%(32)


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Sustained loading = 22.4 kN/m

Sine deflection is directly proportional to the load:

Additional long term deflection = 2.333 x

=2.333 x 2Additional long term deflection = 4.67 mm

ONE-WAY SLA B

Reinforced concrete design slabs are large flat plates that are supported at its sides byreinforced concrete beams, walls, columns, steel beams, or by the ground. If a slab issupported on two opposite sides only, they are referred to a one-way slabs since thebending occurs in one direction only. If the slab is supported on all four sides, it is calledtwo-way slab since the bending occurs in both direction.

If a rectangular slab is supported in all four sides but the long is two or more times theshort side, the slab will, for all practical purposes, act as one way slab, with bendingoccurring in the short direction.


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A one-way slab is considered as a wide, swallow, rectangular beam. The reinforcingsteel is usually spaced uniformly over its width. One way-way slabs are analyzed byconsidering one-meter strip, which is assumed independent of the adjacent strips. Thismethod of analysis is somewhat conservative because we neglect the lateral restraint

provided by the adjacent strips.

MAXIMUM SPACING OF REINFORCEMENT

According to Section 407.7.5, the flexural reinforcement shall not be spaced fartherapart than 3 times the slab thickness, nor 450 mm.

SHRINKAGE AND TEMPERATURE REINFORCEMENT,Concrete shrinks as it hardens. In addition, temperature changes occur that causesexpansion and construction of concrete. In this effect, the code (407.13) requires thatone-way slab, where flexural reinforcement extends in one direction only, should bereinforced for shrinkage and temperature stresses perpendicular to flexuralreinforcement. According to Section 407.132.2.1, the area of shrinkage reinforcementshall provide at least the following ratios of gross concrete area bh, (where h is the slabthickness) but not less than 0.0014.

a) Where Grades 230 & 275 deformed bars are used..0.0020b) Where Grade 415 deformed bars or welded wire

fabric (plain or deformed ) are used..0.0018

c) Where reinforcement with measured atyield strain of 0.35% are used.


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Shrinkage and temperature reinforcement may not be spaced not farther apart than 5times the slab thickness, nor 450 mm (Section 407.13.2.2).

STEPS IN THE DESIGN OF ONE-WAY SLABS (FLEXURE)

I. Identify the uniform floor pressure (Pa) to be carried by the slab. This load mayconsist of:1) Live load pressure2) Dead load pressure3) Ceiling load and other attachments below the slab

II. Determine the minimum slab thickness h from Table 2.1. If necessary adjustthis value depending on your judgment.

III. Compute the weight of slab (Pa)

Weight = IV. Calculate the factored moment (to be carried by the slab.

Uniform load,

.

V. Compute the effective depth, d:d=h-covering (usually 20 mm)-1/2 (main bar diameter)

VI. Compute the required steel ratio :Solve for from Solve for If

is less than

and greater than

, use

If is greater than, increase the depth of slab to ensure ductile failureIf is less than VII. Compute the required main bar spacing.

Spacing, Use the smallest of the following for the main bar spacing:

a) b)

c) 450 mm


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VIII. Temperature bars: See Page 81 for the required steel ratio, Use the smallest of the following for temperature bar spacing:

a) b) c) 450 mm

ILLUSTRATIVE PROBLEMS

Prob lem 2.36

Design a one-way slab having a simple span 3 m. The slab is to carry a uniform live

load of 7,500 Pa. Assume and for main and temperaturebars. The slab is not exposed to earth or weather. Use unit weight of concrete .SOLUTION

Consider 1 m strip of slab, b= 1000 m

Uniform live load,


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Minimum slab thickness from Table 2.1:

Effective depth:

d = 120-20 mm (covering)-1/2 bar diameter (12mm)d=94 mm

Weight of slab: Factored floor pressure load:


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Check for and :

per meter width of slabUsing 12-mm main bars:

Spacing s =


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Maximum spacing required by the Code:

a) b)

Thus, use 12 mm main bars at 135 mm o.c.

Temperature bars: (Grade 275)

Spacing =


a) b) 450 mm OK

Thus, use 10 mm temperature bars at 325 mm o.c.


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PROBLEM 2.27

Design a one-way slab to carry a service live load of 4000 Pa. The slab has a length of

4m with both ends continuous. Assume and for main barsand for temperature bars. Steel cover is 20 mm. Unit weight of concreteis 23.5 kN/.SOLUTION

Consider 1 m strip, b = 1000 mm

Uniform live load, Minimum slab thickness from Table 2.1:

Weight of beam (DL):

Maximum factored moment, Section 408.4 (See Page 29)LL < 3 DL


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Effective depth, d = 1.50201/2 (12)Effective depth, d = 124 mm

At midspan:


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Use

Spacing, s =


a) b) 450 mmThus, use 12 mm bottom bars at 270 mm o .c. at midspan

At support:


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Use

Spacing,

Thus, use 12 mm top bars @ 265 mm o.c. at suppo rt

Temperature bars (10 mm): ( Spacing, s =


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a)

b) 450 mm

Thus, use 10 mm temperature bars @ 260 mm o.c.

PROBLEM 2.28

A one-way slab having a simple span of 3 m is 160 mm thick. The slab is reinforced with

12 mm tension bars spaced at 140 mm o.c. Steel covering is 20 mm.Calculate the uniform live load pressure that a slab can carry. Use . Unitweight of concrete is 23.5 kN/.SOLUTION

Consider 1 m strip of slab, b = 1000 m


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Dead load: Effective depth: d = 160201/2(12)

d = 134 mm

Steel area,


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= 22.696 kN/m

10.25 = Uniform pressure x 1Uniform live load pressure = 10.25 kPa

Solved Prob lems Usin g 2010 NSCP

PROBLEM 2.29

A reinforced concrete beam has width of 310 mm and an effective depth of 490 mm. , . Determine the following:a) The balanced steel areab) The maximum steel area for singly reinforced conditionc) The maximum design strength if the beam is singly reinforcedd) The required steel area if the beam is subjected to dead load moment of 120 kN-

m and live load moment of 170 kN-m.

SOLUTION

Since ( )

a) Balanced steel area:

()


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b) Maximum steel area when beam is singly reinforced:

From Eq. 2-24:

c) Maximum design strength, From Eq. 2-25 :

From Eq. 2-26:


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d)

Thus, the beam is singly reinforced.

Determine if the beam is tension-controlled:

From Eq. 2-22:

Since the required the section is tension controlled.

Check if it is really tension-controlled:

PROBLEM 2.30


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Given the following data for a rectangular beam: width , effective depth , , . Dead load moment Liveload moment SOLUTION

Solve for to determine if compression steel area is required.

=0.8237 Solve for to determine if the section is tension-controlled. Since , the section is within transition region, i.e 0.65


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PROBLEM 2.31

Given the following properties of a rectangular concrete beam: b = 280 mm, d = 480

mm, , . The beam is reinforced for tension only.Determine the design strength under the following conditions.

a) When the beam is reinforced with three 25 mm diameter bars.b) When the beam is reinforced with four 25 mm diameter bars.c) When the beam is reinforced with seven 25 mm diameter bars.


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SOLUTION

()

a)

The section is tension-controlled,


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b)

)

The section within transition region, i. e 0.65


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c)

The section is compression-controlled,

PROBLEM 2.32


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A hallow beam is shown in Figure 2.16. Assume and .a) Calculate the required tension steel area when b) What is the balanced moment capacity of the beam?c) What is the maximum steel area under singly reinforced condition?d) What is the maximum design moment strength under singly reinforced condition?

e) Calculate the required tension steel area when .

SOLUTION

This problem is the same as Problem 2.23.


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To guide us whether a will exceed 150 mm or not, let us solve the design momentwhen a =150 mm.

a) Since the required

Stress in steel


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b) Balanced condition:

241.27) = 454.37


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c) Maximum steel area,

(114.11)=517.95

d) Maximum moment, :


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e)

Refer to Figure 2.17

=575-1/2(0.85c-150)


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PROBLEM 2.33

Design a singly reinforced rectangular beam to carry dead load moment of 110 kN-m

(including self weight) and live load moment of180 kN-m. Use steel ratio andtake Assume and SOLUTION

Note: For singly reinforced rectangular beam,

is directly proportional to c.


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Thus,

PROBLEM 2.34

Repeat Problem 2.33 using a steel ratio SOLUTION


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SUPPLEMENTARY PROBLEMS

PROBLEM 2.35

A rectangular beam has .Determine (a) the maximum design moment if the beam is singly reinforced and (b) therequired steel area if the beam is required to carry a dead load moment of 50 kN-m anda live load moment of 30 kN-m. Use the 2001 NSCP.

PROBLEM 2.36

Repeat Problem 2.35 using the 2010 NSCP.


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PROBLEM 2.37

Design a rectangular beam reinforced for tension only carry dead load moment of 85

kN-m (including its estimated weight) and a live load of 102 kN-m. Use anduse d= 1.75b. Assume and . Use the 2001 NSCP

PROBLEM 2.38


PROBLEM 2.39

A reinforced concrete beam has the following properties: Use 2001 NSCP)

beam with, effective depth, concrete strength, reinforcing steel, reinforcing steel modulus, service dead load moment

a) If the beam is to be designed for a balanced condition, find the required area ofsteel area reinforcement, design balanced moment, and the correspondingservice live load moment.

b) Find the maximum steel area, the maximum design moment, and the


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corresponding service live load moment if the beam is to be designed as singlyreinforced.

PROBLEM 2.40


PROBLEM 2.41

Calculate the ultimate moment capacity of a rectangular beam with , , . Assume . . Use 2001 NSCP

PROBLEM 2.42


PROBLEM 2.43

Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NSCP.

PROBLEM 2.44


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PROBLEM 2.45

Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NSCP

PROBLEM 2.46


CHAPTER 3

Analysis and Design of T-Beams and

Doubly Reinforced Beams


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T-Beams

Reinforced concrete floors usually consist of slab and beams, which are placed orpoured monolithically. In this effect, the beam will have extra width on top (which isusually under compression) called flangers, and the resulting section is called a T-

beam. The beam may also be L-shaped if it is located at the end of slab.

ANALYSIS AND DESIGN OF T-BEAMS

WITH FLANGE IN COMPRESSION

Because of the huge amount of compression concrete when the flange of a T-beams iscompression, the section is usually tension-controlled (extreme tension yields).

The compression block of T-beam may fall within the flange only or partly in the web. Ifit falls within the flange as shown in Figure 3.1 (a), the rectangular beam formulas in

Chapter 2 applies since the concrete below neutral axis is assumed to be cracked andits shape has no effect on the flexure calculations. If however it cover part of the web asshown in Figure 3.1 (b), the compression concrete no longer consist of a singlerectangle and thus the rectangular formulas do not apply.

BALANCED AND MAXIMUM STEEL AREA AND MOMENTThe balanced value of c for any beam shape, as discussed in Chapter 2 is given by:


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and If a is less than the slab thickness, the formulas for rectangular beam may be used, or

However, if a is greater than the slab thickness, the following formula will be used.

Eq. 3-1 Eq. 3-2


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DESIGN OF T-BEAMS WITH NEGATIVE MOMENTS

When T-beams are resisting negative moments so that far their flangers are in tensionand the bottom of their stems in compression, the formulas for rectangular beams canbe applied. The following code requirements shall be applied for this case:410.7.6: Where flangers of T-beam construction are in tension, part of the flexuraltension reinforcement shall be distributed over an effective flange width as defined inSec. 408.11, or width equal to 1/10 the span, whichever is smaller. If the effective flangewidth exceeds 1/10 the span, some longitudinal reinforcement shall be provided in theouter portions of the flange.

The intention of this section is to minimize the possibilities of flexural cracks that willoccur at the top face of the flange due to negative moments.

MINIMUM STEEL RA TIO

For statically determinate T-section with flange in tension, the minimum steel area isequal to or greater than the smaller value of Eq. 3-3 and Eq. 3-4:


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Eq. 3-3

Eq. 3-4 CODE REQUIREMENTS ON T-BEAMS (SECTION 408.11)

NOTE: THESE REQUIREMENTS ARE THE SAME WITH 2010 NSCP

1. In T-beam construction, the flange and web shall be built integrally or otherwiseeffectively bonded together.

2. The width of s

Documents

Simplified Reinforced Concrete Design 2010 NSCP