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P a g e 1 | 27

ERROR FREE DETAILED SOLUTION

RPSC AE MAINS 2018 CE P1(TECH) FULL SYLLABUS

ENGINEERS PRIDE TEST DATE 7TH NOV 2019 (SET B)

PART-A

Marks: 40

Note- Attempt all the twenty questions. Each question

carries 2 marks. Answer should not exceed 15 words

Q.1 If principal stresses in a two-dimensional case are −10Mpa and 20Mpa respectively, then

maximum shear stress at the point is

Solution:

21 2max

10 2015 N/mm

2 2

− − −= = =

Q.2 A cantilever beam of span L is loaded with a concentrated load P at the free end. Find the

deflection at the free end?

Solution:

P

A B

L




P a g e 2 | 27

Deflection at B=3

3B

PL

EI =

Q.3 What is the modulus of resilience?

Solution:

Modulus of resilience:

Maximum strain energy per unit volume which can be stored in a material up to elastic

limit is known as modulus of resilience.

I.e. proof resilience per unit volume is termed as modulus of resilience.

Proof resilience

MORVolume

=

1 1

2 2

P l

A l

= =

21

2 2E

= =

Q.4 A solid shaft having diameter 80 mm is subjected to a torque of 6kN-m. Calculate

maximum shear stress developed?

Solution:

( )

62

4

. 6 1059.68 N/mm

8032

T r

J

= = =

Q.5 Define void ratio and porosity?

Solution:

Void ratio; v

s

Ve

V= and Porosity; vV

V =

Where;




P a g e 3 | 27

vV →Volume of void

sV →Volume of solid

V →Total volume i.e. v sV V+

Q.6 Write the equation for the zero air void line.

Solution:

Equation for the zero air voids line 1

wd

G

wG

=

+;

Where;

G → Specific gravity of soil

w →Unit wt. of water

w→Water content of soil

Q.7 Define coefficient of compressibility of soil.

Solution:

Coefficient of compressibility is defined as the ratio of change in void ratio to change in

effective stress.

Coefficient of compressibility v

ea

−=

Note- Coefficient of compressibility is always –ve as the void ratio decreases with increase

in stress.

Q.8 For a saturated sand deposit, the void ratio and the sp. gravity of solids are 0.70 and 2.67

respectively. Find the critical hydraulic gradient.

Solution:

Critical hydraulic gradient 1 2.67 1

0.981 1 0.7

c

Gi

e

− −= = =

+ +

Q.9 What is static indeterminacy?

Solution:




P a g e 4 | 27

Static indeterminacy of a structure is defined as

static equilibrium

No.of unknown forces equations available

.s

No ofD

= −

Q.10 Write the method of analysis of determinate truss.

Solution:

There are two method of analysis of determinate truss.

1. Method of joint

2. Method of section

Q.11 What is stiffness of a member?

Solution:

The load required to produce a unit deflection is called stiffness (K).

P MK

= =

Q.12 What are the parameters include for partial safety factor used for material strength?

Solution:

Partial safety factors are used for material strength allowing uncertainties of element

behaviour, possible strength reduction due to manufacturing or costing tolerance and

imperfections in the materials.

Q.13 The flexural strength of M30 concrete as per IS 456:2000 is-

Solution:

As per IS456:2000, Modulus of elasticity= 5000 ckf

= 5000 30

=27386.127 Mpa

Q.14 What are the criteria to provide doubly reinforced beam?

Solution:




P a g e 5 | 27

Following are the criteria to provide doubly reinforced beams:

• When size of beam is restricted either due to architectural point of view or headroom

consideration.

• When the beam has to take bending moment more than moment of resistance of

singly reinforced balance section.

Q.15 What is concordant profile?

Solution:

If the cable is provided in such a way that there is no change in support reaction due to

pre stressing force, it is called concordant profile. The shape of concordant profile is

similar to bending moment diagram. If concordant profile is used, secondary moments

will not be develop in the span.

Q.16 Define ‘principal rafter’?

Solution:

One of the upper diagonal members of a roof truss supporting the purlins on which the

common rafter rest is called principal rafter.

Q.17 What are the different types of tension members are used?

Solution:




P a g e 6 | 27

Different types of tension members are:-

(1) Rods and bars

(2) Single structured shapes and plates

(3) Wires and cables

(4) Built up members

Q.18 What are the different modes of failure of column?

Solution:

Following are the modes of failure of column.

• Crushing

• Buckling

• Mixed mode of buckling and crushing

Q.19 What you understand by web crippling?

Solution:

Web crippling is the localized failure of a beam web due to introduction of an excessive

load over a small length of the beam. It occurs at point of application of concentrated

load or at point of support in a beam.

Q.20 What is plasticity index?

Solution:

(b) At support

(a) Below load

Applied load

Reaction

(b)




P a g e 7 | 27

Plasticity Index: It indicates the range of water content over which the soil remains in

plastic state. It is equal to the difference of liquid limit and plastic limit i.e.

Plasticity index ( ) ( )p L pI w w= −

Where;

Lw = Liquid limit of the soil

pw = Plastic limit of the soil

PART-B

Marks: 60

Note- Attempt all the twelve questions. Each question


Q.21 The maximum shear stress in a solid shaft of circular cross-section having diameter `d’

and objected to a torque T is . If the torque is increased by four times and the diameter

of shaft is increased by two times, then find the maximum shear stress in the shaft.

Solution:

We know that

Pl

T

R

=

2

1

P1 1 1

2 2 2 P

l

l

R T

R T

=

4

1 1 1 1

2 1 1 1

2

2 4

R T R

R T R

=

1

2

1 116

2 4

=




P a g e 8 | 27

12

2

=

22

=

Q.22 The principal stresses at a point in an elastic material are given as →

1 2 31.5 , , & 0.3, 250 MPayf = = = − = =

For no failure,find the value of as per maximum strain energy theory?

Solution:

As per maximum strain energy theory -

2

2 2 2

1 2 3 1 2 2 3 3 1

12

2 2

yf

E E + + − + +

( ) ( ) ( )2 2 2 2 2 2 21.5 2 1.5 1.5 yf + + − − − −

( )22 2 2 22.25 2 0.3 250 + + − −

2 250 250

4.85

2 113.51 MPa

Q.23 Write between homogeneous, isotropic and anisotropic?

Solution:

Homogeneous material: A material is said to be homogeneous when it has identical

properties at all the points in space or which has a uniform composition throughout the

body.

Isotropic material: A material is said to be isotropic when it has identical properties in all

the directions at a particular point.

Anisotropic material: A material is said to be anisotropic when its one or more properties

get changed with direction at a point within the material.




P a g e 9 | 27

Q.24 The porosity (n) and the degree of saturation (S) of a soil sample are 0.7 and 40%,

respectively. In a 3100 m volume of soil, find the volume (in 3m ) of air is ____.

Solution:

30.7, 40%, 100 , ?aS V m V = = = =

vV

V =

0.7100

vV=

370 mvV =

Now, 1w v a a

v v v

V V V VS

V V V

−= = = −

0.40 1 a

v

V

V= −

30.60 0.60 70 42 maa

v

VV

V= = =

Q.25 Write the assumption Terzaghi's theory for ultimate bearing capacity.

Solution:

Terzaghi's Bearing Capacity Theory: Terzaghi (1943) gave a general theory for the bearing

capacity of soils under a strip footing, making the following assumptions:

• The base of footing is rough.

• The footing is laid at a shallow depth, (fD B ) i.e. shallow foundation.

Air

Solids

Water V

𝑉𝑣 = 𝑉 𝑎 + 𝑉𝑤




P a g e 10 | 27

• The shear strength of the soil above the base of the footing is neglected. The soil

above the base is replaced by a uniform surcharge fD

• The load on the footing is vertical and symmetric. i.e. (moment=0)

• Footing is a strip footing (L>>B).

• The shear strength of the soil is governed by the Mohr-Coulomb equation.

• Failure is general shear failure.

• Ground is horizontal.

Q.26 Write difference between compaction and consolidation

Compaction

• Almost an instantaneous phenomenon.

• Soil is always unsaturated.

• Densification is due to a reduction in the volume of air voids at given water

content.

• Specified compaction techniques are used in this process.

Consolidation

• It is a time dependent phenomenon.

• Soil is completely saturated.

• Volume reduction is due to expulsion of pore water from voids.

• Consolidation occurs on account of a load placed on the soil.

Q.27 What do you mean by Influence Line of a stress-resultant in a structure at a given

point/section? Draw ILD for BM at quarter span in a simply supported beam?

Solution:

Water

Water

Solids Solids

𝑉2 𝑉1

∆𝑉

(a) Before consolidation (b) After consolidation




P a g e 11 | 27

An influence line represents the variation of reaction, shear, moments or deflection at a

specified point in a member as unit load moves across the member.

Influence line for BM at quarter span in a simple supported beam

By Muller Breslau Principal release moment at C. Introduce unit reaction at C

1A B + = , 13

4 4

y y

L L+ =

44

3y y L+ = ,

16

3y L=

3

16

Ly =

Q.28 Explain about principle of superposition?

Solution:

Total displacements (or) stresses at a point in a structure subjected to several external

loading can be determined by adding the displacement (or) stresses caused by each of

the external loads acting separately.

Requirements for principal of Superposition are

A B

C

L

L/4

L/4 3L/4

D

𝜃𝐴 𝜃𝐵

L/4

L




P a g e 12 | 27

1. Linear elastic response i.e., Hooke's law is valid. (I.e. Load ∝ displacement)

2. Small displacement theory applies i.e., geometry of the structure must not undergo

significant change when loads are applied.

Q.29 What is plate girder and gantry girders?

Solution:

Plate girder

When the span of beam increases (more than 15m) so that available rolled steel sections

are not able to provide the required section modulus then one has to go for plate girders.

Plate girder is an arrangement of plates that provides the required section modulus when

the span is large. These are particularly useful for spans ranging from 15 m to 50 m.

Gantry girders

Gantry girders are laterally unsupported beam to carry heavy loads from place to place at

the construction sites, mostly these are of steel material. A girder is a support beam used

in construction. It is the main horizontal support of a structure which supports smaller

beams.

Q.30 Write the principal step in the design of a steel member subjected to axial tension?

Solution:

Steps of designing: -

(1) Permissible stress in tension member = 0.6 yf = at

End carriage

Crab

Gantry girder

Gantry girder

Load

Crane bridge




P a g e 13 | 27

(2) Calculate net area of tension member = 0.6

net

y

PA

f=

(3) Increase netA by 40% to get and required after deduction for rivet holes.

(4) Choose appropriate section from steel table

(5) Check for netA

(6) netA (Provided) netA (calculated)

(7) Provide connections.

Q.31 Differentiate between Limit state method and Working state method.

Solution:

Working state method Limit state method

1 It is deterministic approach. 1 It is probabilistic approach.

2 In this method safety & serviceability

are checked at working load

2 In this method safety is checked at

ultimate load & serviceability is

checked at working load

3 In WSM, We underestimate the

material strength & don’t

overestimate the load

3 In LSM, We underestimate the

material strength & overestimate

the load

Q.32 What are the assumptions made for the limit state collapse in compression and write an

expression for factored axial load for a short column?

Solution:

Assumption: -

(1) Plane section before bending remains plane after bending

(2) Strain diagram is linear.

(3) Maximum permissible stress in concrete is 0.45fck and in steel 0.67fy.

(4) Maximum strain in concrete in direct compression is 0.002 and in case of bending is

0.0035.




P a g e 14 | 27

(5) The maximum compressive strain in highly compressed extreme fibre in concrete

subjected to axial compression and bending but when there is no tension on the section,

is taken as 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.

Expression for factored axial load for a short column:-

0.4 0.67 u ck c y scP f A f A= +

uP = Factored load for short column

ckf = characteristic strength of concrete

yf = characteristic strength of steel

cA = area of concrete

scA = Area of longitudinal reinforcement.

PART-C

Marks: 100

Note- Attempt any 5 out of 7 questions. Each question


Q.33 A beam fixed at one end and simply supported at the other end is having a hinge at B as

shown in the figure. Determine the deflections (a) under the load and (b) at the hinge B.

Use moment area method.

Solution:

A B

C D

40 kN

2 m 2 m 2 m

𝐸𝐼 = constant

Hinge




P a g e 15 | 27

Taking member BC

By symmetry 20 kNB CR R= =

Thus, FBD of the given beam

Thus, BMD

EI of the given beam

According to moment area method, deflection

at the hinge point B ( )B is

/B A A B AAB = + +

( )1 40 2 160

2 22 3 3EI EI

= − = −

A B

C D

40 kN

2 m 2 m 2 m

𝐸𝐼 = constant

C

𝑅𝐵

D

40 kN

2 m 2 m

𝑅𝐶

20 kN

B A C

D

40 kN

20 kN 20 kN

B

A B C D

40

𝐸𝐼 kN-m 2 m 2 m

2 m 40

𝐸𝐼 kN-m




P a g e 16 | 27

Deflection under the load

/ 0BCC B B C BBC = + + =

160 1 40

4 4 2 03 2BCBEI EI

− + + =

160 1604

3BCBEI EI

= −

40 40 80

3 3BCBEI EI EI

= − = −

/BCD B B D BBD = + +

160 80 1 40 2 160 160 802

3 3 2 3 3 3 3EI EI EI EI EI EI

= − − + = − − +

( )320 80 240 80

3 3EI EI EI

− += = − = −

Deflection shape of the beam will be

Q.34 A vertical cylindrical steel storage tank has 30 m diameter and the same is filled up to a

depth of 15 m with the gasoline of relative density 0.74. If the yield stress for steel is 250

MPa, find the thickness required for the wall plate. Adopt a factor of safety of 2.5 and

neglect localized bending effects, if any.

Solution:

2° curve 2° curve




P a g e 17 | 27

Density of gasoline = 0.74 × 1000 = 740 kg/m3

Hoop stress ( )2

h

pd

t =

where 740 9.81 15p gh= =

5 21.089 10 N/m=

20.1089 N/mm=

( ) 2

0.1089 30000 250 N/mm

2h

t

=

6.534 mmt …(i)

Longitudinal stress ( )( ) 2

0.1089 30000 250 N/mm

4 4t

pd

t t = =

t 3.267 mm …(ii)

From (i) and (ii), t 6. 534 mm

Using a FOS of 2.5, t 6.534 × 2.5 = 16.335 mm

Adopt t = 18 mm

Q.35 A 3.0 m square footing is located in a dense soil and at a depth of 2.0 m. Determine the

ultimate bearing capacity for the following water table positions:

(i) at ground surface,

(ii) at footing level and

(i) at 1m below the footing.

30 m

15 m




P a g e 18 | 27

The moist unit weight of sand above the water table is 18 kN/m3 and the saturated unit

weight is 3

q20 kN/m , 35°, c 0, N 33 and N 34.0. = = = =

Solution:

Given data:

Width of square footing, B = 3.0m

Depth of footing, 2.0 mfD =

Bulk unit weight of sand, 318 kN/mt =

Saturated unit weight of sand, 320 kN/msat =

q35°, c 0, N 33 and N 34.0 = = = =

We know that for a square footing, ultimate bearing capacity is

1.3 0.4 u c f qq cN D N BN = + +

For the given soil which is sand c = 0

0.4 u f qq D N BN = +

The value of in the above equation is susceptible to the location of water table. Thus

for each water table position, the ultimate bearing capacity will be different. So, the

ultimate bearing capacity will be given as

1 20.4 u f qq D N BN = +

(i) When water table is at the ground surface




P a g e 19 | 27

Now for the given condition, soil above the footing and below the footing is submerged,

so

1 220 9.81 and 20 9.81 = − = −

3 3

1 210.19 kN/m and 10.19 kN/m = =

1 20.4u f qq D N BN = +

210.19 2 33 0.4 10.19 3 34 1088.29 kN/m= + =

(ii) When the water table is at the footing level

Now for the given condition, soil above the footing is moist soil and below the footing is

submerged, so

1 2 and sat w = = −

3 3

1 218 kN/m and 20 9.81 10.19 kN/m = = − =

1 20.4u f qq D N BN = +

18 2 33 0.4 10.19 3 34= +

21603.75 kN/m=

(iii) When the water table is at 1 m below the footing

Now for the given condition, soil above the footing is moist and below the footing it is

moist up to 1m and below this, it is submerged, so

𝐷𝑓 𝐷𝑤1

B 𝐷𝑤2




P a g e 20 | 27

3

1 218 kN/m and B 1 18 2 (20 9.81) 38.38 = = + − = KN/m2

1 20.4u f qq D N B N = +

218 2 33 0.4 34 1709.968 kN/m= + =

Q.36 A retaining wall 6 m high supports earth with its face vertical. The earth is cohesionless

with particle specific gravity 2.69, angle of internal friction 35° and porosity 40.5%. The

earth surface is horizontal and level with the top of the wall. Determine the earth thrust

and its line of action on the wall if the earth is water logged to level 2.5 m below the top

surface. Neglect wall friction. Draw the pressure diagrams.

Solution:

Given data:

Height of retaining wall, H = 6 m

Specific gravity of soil particles, G = 2.69

Angle of internal friction, 35 =

Porosity, n = 40.5% = 0.405

Depth of water table below top surface = 2.5 m

We know that 1

en

e=

+

0.405

0.6811 1 0.405

ne

n= = =

− −

32.69 0.6819.81 19.67 kN/m

1 1 0.681sat w

G e

e

+ + = = =

+ +

32.69 9.8115.7 kN/m

1 1 0.681

wdry

G

e

= = =

+ +

3' 19.67 9.81 9.86 kN/msat w = − = − =




P a g e 21 | 27

1 sin 1 sin 350.271

1 sin 1 sin 35aK

− − = = =

+ +

for z = 2.5 m

20.271 15.7 2.5 10.64 kN/mv aP K z= = =

Total active thrust for z = 2.5 m is given by

1

1 12.5 10.64 2.5 13.3 kN/m

2 2vP P= = =

For z = 3.5 m

1

2' 0.271 9.86 3.5 9.35 kN/mv aP K z= = =

2

29.81 3.5 34.335 kN/mv wP z= = =

1 2

29.35 34.335 43.685 kN/mv vP P+ = + =

Total active thrust for z = 3.5 m is given by

( )2 3

143.685 3.5 10.64 3.5

2P P+ = +

76.45 37.24 113.69 kN/m= + =

Total active thrust on the retaining wall =

1 2 3P P P+ + = 13.3 + 76.45 + 37.24 = 126.99 kN/m

Location of total thrust from base =

H = 6 m

2.5 m

3.5 m

10.64 9.35 34.335

𝑃1

𝑃2

𝑃3

𝛾𝑑𝑟𝑦 = 15.7 kN/m3

𝛾′ = 9.86 kN/m3




P a g e 22 | 27

2.5 3.5 3.513.3 3.5 37.24 76.45

3 2 31.67m

126.99

+ + + =

Q.37 A continuous beam ABCD consists of three spans, and is loaded as shown in Figure. Ends

A and D are fixed. Determine the bending moments at the supports and plot the bending

moment diagram.

Solution:

(a) Fixed end moments (kN-m units)

22 66

12FABM

= − = −

22 66

12FBAM

= + = +

2

2

5 3 22.4

5FBCM

= − = −

2

2

5 2 33.6

5FCBM

= + = +

8 55

8FCDM

= − = −

8 55

8FDCM

= + = +

(b) Distribution factors:-

The relative stiffiness and distribution factors are calculated in Table

Joint Member Relative stiffness Sum

Distribution

6 m 5 m 5 m

3 m 2 m 2.5 m 2.5 m

A B C 2 kN/m

5 kN 8 kN

D

I 2I I




P a g e 23 | 27

B BA

BC

𝐼

6

2𝐼

5

17𝐼

30

5

17

12

17

C CB

CD

2𝐼

5

𝐼

5

3𝐼

5

2

3

1

3

(c) Moments distribution:- The moment distribution is carried out in Table. The final

bending moment diagram and the deflected shape of the beam are shown in figure (b)

and (c) respectively.

-6.00 +6.00

-1.06

-2.40 +3.60

-2.54 +0.93

-5.00 +5.00

+0.47

F.E.M

Balance

-0.53

-0.14

0.46 -1.27

-0.32 +0.85

+0.23

+0.42

Carry over

Balance

-0.07

-0.12

+0.42 -0.16

-0.30 +0.11

+0.21

+0.05

C

B

-0.06

-0.014

+0.05 -0.15

-0.03 +0.10

+0.03

+0.05

C

Balance

-6.66 +4.66 -4.66 +4.01 -4.01 +5.47 Final moments

A B

C D

5

17

12

17

2

3

1

3




P a g e 24 | 27

Q.38 Mention the various types of losses in prestressing members in pretensioning and post

tensioning. Explain them very briefly

Solution:

The losses in prestressed members are:

pre-tensioned concrete post tensioned concrete

1. Elastic deformation of concrete 1. No loss due to elastic deformation if all

the tendons are tensioned simultaneously.

2. Shrinkage of concrete 2. Shrinkage of concrete

3. creep of concrete 3. creep of concrete

4. Relaxation of steel 4. Relaxation of steel

5.Frictional and wobbling losses

6. Anchorage slip

1. Loss due to elastic deformation of concrete: This loss depends on the modular ratio and

average stress in concrete at the level of steel. In case of post tensioned beams if all the

wires are simultaneously tensioned no loss due to elastic deformation occurs. But if wires

are successively tensioned, there will be loss of prestress due to elastic deformation.

Loss = m.fc

6.66

6 m

4.01kN-m 5.47

5 m 5 m

3 m 2 m 2.5 m 2.5 m A

B (a) C 2𝐼 2 kN/m

5 kN 4.66kN-m 8 kN

D 𝐼

6 10 9

6.66 kN-m

(b)

A B

(c) C D

4.66kN-m 5.47kN-m 4.01kN-m

- + -

+




P a g e 25 | 27

2. Loss due to shrinkage of concrete: This shrinkage of concrete in prestressed members

results in shortening of tensioned wires and hence contributes to the loss of stress.

Loss = Shrinkage strain × sE

3. Loss due to creep of concrete: The sustained prestress in the concrete of a prestressed

member results in creep of concrete which effectively reduces the stress in high tensile

steel.

Loss . . cm f=

4. Loss due to relaxation of stress in steel: It is generally equal to 1 to 5% of prestress.

5. Loss of stress due to friction: In case of post tensioned members, the tendons are housed

in ducts preformed in concrete. The ducts are either straight or follow curved profile

depending upon the design requirements. Consequently on tensioning the curved

tendons, loss of stress occurs in post-tensioned members due to friction between the

tendons and the surrounding concrete ducts.

Prestressing force at a distance x from jacking end ( )

0

kx

xP Pe − +=

6. Loss due to anchorage slip: In most post-tensioning systems, when the cable is tensioned

and the jack is released to transfer prestress to concrete, the friction wedges, employed

to grip the wires, slip over a small distance before the wires are firmly housed between

the wedges. The magnitude of slip depends upon the type of wedge and the stress in

wires

Loss = s

LE

L

Q.39 Two wheels, placed at a distance of 2.5 m part, with a load of 200 kN on each of them,

are moving on a simply supported girder (𝐼-section) of span 6.0 m from left to right. The

top and bottom flanges of the 𝐼-section are 200 × 20 mm and the size of web plate 800

× 6 mm. If the allowable bending compressive, bending tensile and average shear

stresses are 110 MPa, 165 MPa and 100 MPa respectively, check the adequacy of the

section against bending and shear stress. Self weight of the girder may be neglected.

Solution:

Maximum S.F. will occur when left 200 kN load is at A or right 200 KN load is at B.




P a g e 26 | 27

( )6 200 6 200 6 2.5AR = + −

AR = 316.67 kN

Maximum S.F = 316.67 kN

Maximum B.M. will occur when one of 200 kN load and resultant of both 200 kN loads

are equidistant from center of span as shown.

Max. BM under left 200 kN load = AR (3−0.625)

= 158.33 (2.375) = 376.034 kNm

A B

6 m

1

1

+

−

1.25 m 1.25 m

2.5 m 200 kN 200 kN

ILD of S.F. at A

ILD of S.F. at B

ILD for Maximum BM

A B

0.79167

0.625 m 1.875 m

6 m




P a g e 27 | 27

3 3840 800200 194

12 12xxI = −

6 41601.067 10 mm=

Maximum bending compressive stress 6

6

376.034 10420

1601.067 10

=

2 2 98.64 N/mm < 110 N/mm= (OK)

Maximum bending tensile stress = 6

6

376.034 10400

1601.067 10

2 2 98.64 N/mm < 165 N/mm= (OK)

Maximum shear stress ( )2 2 2

8 w

F BD d d

I t

= − +

( )3

2 2 2

6

316.67 10 200840 800 800

8 1601.067 10 6

= − +

2 2 69.88 N/mm < 100 N/mm= (OK)

Section is safe against bending and shear

200

200

800

6

20

20

X X

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ERROR FREE DETAILED SOLUTION · (200 Meter from Riddhi Siddhi Tiraha), Gopal Pura Mode (between Gandhi Nagar Railway Station and Durga Pura Railway Station), Jaipur, Rajasthan, 9660807149,