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ERROR FREE DETAILED SOLUTION

RPSC AE MAINS 2018 CE P1(TECH) FULL SYLLABUS

ENGINEERS PRIDE TEST DATE 24TH OCT 2019 (SET A)

PART-A

Marks: 40

Note- Attempt all the twenty questions. Each question

carries 2 marks. Answer should not exceed 15 words

Q.1 What is the relation between modules of elasticity E, bulk modules K and Poisson’s

ratio?

Solution:

3 (1 2 )E K = −

Where,

E→ Modulus of elasticity

K→ bulk modulus

→ Poisson’s ratio

Q.2 What is the significance of Mohr's circle?

Solution:




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Mohr's circle is a graphical method to represent the stresses on any plane at a point. It is

used for evaluation of principal stresses, maximum shear stress; normal and tangential

stresses on any given plane.

Q.3 A triangular section of a beam has width 200 mm and depth 100 mm. then for shear

force = 100 kN, what will be the maximum shear stress anywhere in the section?

Solution:

Maximum shear stress in triangular section = 3

2avg

33 3 100 10

12 2200 100

2

V

Area

= =

215 N/mmavg =

Q.4 A shaft turns at 250 rpm under a torque of 1500 N-m. Calculate power transmitted by

shaft?

Solution:

Power transmitted by shaft (P) = Torque (T) × rotational speed ( )

2

150060

NP

=

2 250

150060

P

= W

Power = 39.269 kW

Q.5 If the values of 60D , 30D and 10D , for a soil are 0.71, 0.34 and 0.18 respectively.

Determine the values of coefficient of uniformity and coefficient of curvature?

Solution:

Coefficient of uniformity ( ) 60

10

0.713.944

0.18

DCu

D= = =




P a g e 3 | 31

Coefficient of curvature ( )2 2

30

60 10

0.340.9045

0.71 0.18

DCc

D D= = =

Q.6 Define Phreatic line?

Solution:

The top flow line of a saturated soil mass below which seepage takes place is called the

phreatic line.

Note- Hydrostatic pressure acts below the phreatic line whereas atmospheric pressure

exists above the phreatic line. This line separates a saturated soil mass from an

unsaturated soil mass. It is not an equipotential line, but a flow line. For an earthen

dam, the phreatic line approximately assumes the shape of a parabola.

Q.7 List the parameters, apart from Terzaghi's bearing capacity factors which determine the

bearing capacity of shallow foundation?

Solution:

a) Cohesion of soil

b) Surcharge or overburden

c) Bulk unit weight of soil in shearing zone

d) Width of foundation

e) Depth of water tube.

Q.8 What is primary consolidation?

Solution:

When a saturated soil is subjected to a pressure, initially all the applied pressure is taken

up by water as an excess pore water pressure. A hydraulic gradient will develop and the

water starts flowing out and a decrease in volume occurs. This reduction in volume due

to expulsion of water is called as the primary consolidation of soil.

It is a time dependent phenomenon.

Q.9 Define influence line?

Solution:




P a g e 4 | 31

An influence line represents the variation of either the reaction, shear, moment or

deflection at a specific point in a member due to unity concentrated force moves over

the member.

Q.10 What is kinematic indeterminacy?

Solution:

The number of joint displacements/Rotations are required to analyze the structures is

called kinematic indeterminacy (or) degrees of freedom.

Q.11 What is Muller Breslau's principle?

Solution:

It states that the influence line for any force response function in a structure is given by

the deflected shape of the structure resulting from a unit displacement corresponding

to the force under consideration.

Q.12 Write importance of providing "Distribution Reinforcement", in reinforced concrete

structures?

Solution:

The distribution reinforcement is provided to resist stress due to shrinkage and temp. It

also distributes the concentrated loads uniformly. It tie and keep the main

reinforcement in position.

Q.13 What is ideal shape of Cable-profile in a pre-stressed concrete beam carrying u.d.l ‘W’

and why?

Solution:

For U.D.L. ideal shape of cable profile in a pre-stress concrete is parabolic. A parabolic

shape of the profile will exert a uniformly distributed upward load on the beam to

counteract a part or whole of the external downward loading.

Q.14 What are the different limit states of design?

Solution:

The different limit states of design are as follows:




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(i) Limit state of collapse:- This state corresponds to maximum load carrying capacity in

flexure, tension, compression, torsion etc.

(ii) Limit state of serviceability:- This state corresponds to excessive deformation due to

deflection, cracking, vibration etc.

Q.15 What is the maximum strain value in tension reinforcement and compression concrete

in the section at the failure?

Solution:

In case of concrete = 0.0035

In case of tension reinforcement

0.0021.15

y

s

f

E= +

Where yf → yield stress

sE → Modulus of elasticity of steel

Q.16 Define post-tensioning in pre-stressed concrete.

Solution:

Post-tensioning is a method of reinforcing (strengthening) concrete or other materials

with high-strength steel strands or bars, typically referred to as tendons, after the

concrete has set.

Q.17 What is the basis of design of a steel tension member?

Solution:

I. Gross section yielding

II. Net section yielding

III. Block shear failure

Note- Design strength of steel tension member is least of its strength in-

I. Yielding of gross section

II. Rupture of critical section




P a g e 6 | 31

III. Block shear

Q.18 What is buckling failure and crushing failure?

Solution:

Crushing or compression failure occurs when short columns experience heavy loads that

exceeds the material limits.

Buckling failure occurs when tall, slender columns experience loads that cause them to

buckle or shift outward to relieve stress.

Q.19 What are the different elements of plate girder?

Solution:

A plate girder consists of following components:

1. Web plate 2. Flange plates

3. Flange angles 4. Web splice plates

5. Flange splice plates 6. Transverse or vertical stiffeners

7. Bearing stiffeners 8. Lontidudinal or horizontal stiffeners

9. End bearings stiffeners

Q.20 What is the formula to calculate thickness of slab base [both WSM and LSM ]

Solution:

For LSM:-

( )2 2

min

2.50.3

bs

wt a b

= −

mint → the slab thickness, in mm;

bs →Permissible bending stress in base plate=1.1

yf

a,b→Greater and smaller projections of base plate from column edge in mm

w→Upward pressure exerted by concrete on steel plate in Mpa




P a g e 7 | 31

For WSM:-

22

min

3

4bs

w bt a

= −

mint → the slab thickness, in mm;

bs →Permissible bending stress in base plate (for all steels, shall be assumed

as 185 Mpa)

a,b→Greater and smaller projections of base plate from column edge in mm

w→Upward pressure exerted by concrete on steel plate in Mpa

PART-B

Marks: 60

Note- Attempt all the twelve questions. Each question

carries 5 marks. Answer should not exceed 50 words

Q.21 Write Five Theories of Elastic Failure. Explain maximum principal stress theory.

Solution:

The five theories of failure are as follows:

1. Maximum principal stress theory or Rankine's Theory

2. Maximum principal strain theory or Venant’s Theory

3. Maximum shear stress theory or Guest’s theory




P a g e 8 | 31

4. Maximum strain energy theory or Haigh’s Theory

5. Maximum shear stress energy theory or Mises-Henky Theory

Maximum principal stress theory or Rankine's Theory: According to this theory for no

failure, maximum principal stress developed in a strained body must not exceed the

yield stress in uniaxial loading.

For no failure, 1 y

For design a suitable factor of safety is applied.

For design, 1

.

y

F S

This theory has certain limitations such as:

(i) It is not suitable for ductile metals.

(ii) It is not suitable for hydrostatic loading.

(iii) It is not suitable for the case of pure shear.

Q.22 A 20 mm diameter brass rod was subjected to a tensile load of 40 kN. The extension of

the rod was found to be 254 divisions in the 200 mm extension meter. If each division is

equal to 0.001 mm, find the elastic modulus of brass.

Solution:

Area of the rod = ( )2 2 220 mm 314.1568 mm

4A

= =

Pull = P = 40 kN = 40,000N

Stress = 40,000

314.568

Pf

A= = = 127.325 N/mm2

Length of specimen = 𝑙 = 200mm

Extension = 𝑑𝑙 = 254 × 0.001 = 0.254 mm

Strain 0.254

0.00127200

dle

l= = =




P a g e 9 | 31

Young's Modulus 2 5 2127.325/ 1.0025 10 /

0.00127

fE N mm N mm

e= = =

Q.23 Find the length from A, where shear force is zero?

Solution:

Taking moment about B

02 3

A

wl lR l − =

6

A

wlR =

0yf =

1

2A BR R w l+ =

2

A B

wlR R+ =

2 6 3

B

wl wl wlR = − =

Let at distance x from A, S.F. = 0

1

. . . 06 2

wl wS f x x

l= − =

2

2

3

lx =

A

𝑙

B

w kN/m

A

𝑅𝐴 =𝑤𝑙

6

𝑤

𝑤𝑙

3

x




P a g e 10 | 31

3

lx =

Q.24 The soil at a site consists of two layers of thickness H each. The coefficient of

permeability of the Soil of 1st layer is 1k in both horizontal and vertical directions,

whereas for the 2nd layer, it is 1 / 2k . What will be the equivalent permeability of the

two-layered soil in horizontal and vertical directions?

Solution:

Equivalent horizontal permeability,

1 12

H

Hk H k

kHk

H H H

+

= = +

1

3

4Hk k=

Equivalent vertical permeability,

1

1 1

2

2/ 3V

H H Hk k

H HH K

k k

+= = =

+

Q.25 Explain the factors which affect the bearing capacity of soils.

Solution:

Factors affecting bearing capacity:

1. Size and Shape of footing:

By increasing size of footing the bearing capacity of soils generally increases. For

cohesive soils bearing capacity is independent of size. Bearing capacity also depends on

shape of footings like strip footing rectangular, square and circular footings.

2. Type of Soil:

Types of soil is an important factor for determining bearing capacity of soils. Bearing

capacity of cohesive soils is generally less than that cohesionless soils of same void ratio.

𝑘1

𝑘1/2 𝐻

𝐻




P a g e 11 | 31

3. Type of footing:

Bearing capacity also depends on the fact that the footing is rigid or flexible. Because

pressure distribution is different for rigid and flexible footing.

4. Ground water table:

Bearing capacity of soil below the ground water table is less as compared to that of

above ground water table

Q.26 Find the total degree of statical indeterminacy (both internal and external) for the

bridge truss shown in the figure.

Solution:

Total number of members (m) = 20

Total number of joints (j) = 10

Total number of external reactions (r) = 3

Total degree of external indeterminacy (E) = r − 3 = 3 − 3 = 0

Total degree of internal indeterminacy (𝐼) = m − (2j − 3)

= 20 − (2 × 10 − 3) = 20 −17 = 3

Total indeterminacy = 𝐸 + 𝐼

= 0 + 3 = 3

Alternatively,

Total indeterminacy = (r − 3) + (m − (2j − 3))




P a g e 12 | 31

= m + r − 2j

= 20 + 3 − 2(10) = 3

Q.27 Write Muller-Breslau principle and its application?

Solution:

Muller − It states that the influence line for any force response function in a structure is

given by the deflected shape of the structure resulting from a unit displacement

corresponding to the force under consideration.

It is used to find influence line of any stress function (reaction, shear, moment) for both

statically determinate and indeterminate structure.

Q.28 Write down the losses in prestress concrete. Explain loss due to shrinkage.

Solution:

Following are the losses in pre-tensioned and post tensioned prestress concrete.

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

Loss of prestress due to shrinkage: Shrinkage means contraction of concrete due to

chemical changes and drying. The shrinkage of concrete in prestressed members result

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

shrinkage of concrete is influenced by the type of cement and aggregates and the

method of curing used. Use of high strength concrete with low water cement ratios

result in reduction in shrinkage and consequent loss of prestress. The primary cause of

drying shrinkage is the progressive loss of water from concrete. The differential

shrinkage between the interior and surface of large members may result in strain




P a g e 13 | 31

gradients leading to surface cracking. Hence, proper curing is essential to prevent

shrinkage cracks in prestressed members.

In the case of pre-tensioned members, generally moist curing is resorted to in order to

prevent shrinkage until the time of transfer. Consequently, the total residual shrinkage

strain will be larger in pre-tensioned members after transfer of prestress in comparison

with post-tensioned members, where a portion of shrinkage will have already taken

place by the time of transfer of stress.

The loss of stress in steel due to the shrinkage of concrete is estimated as

Loss of stress = ∈𝑐𝑠 × 𝐸𝑠

where ∈𝑐𝑠 is total residual shrinkage strain having values of 43 10− for pre-tensioning

and ( )

4

10

2 10

log 2t

−

+ for post-tensioning where t is age of concrete at transfer in days.

This value may be increased by 50 per cent in dry atmospheric conditions, subject to a

maximum value of 43 10− units. sE is modulus of elasticity.

Q.29 What do you mean by a two-way slab? How is it economical over one-way slab?

Solution:

Two-way slab: A two-way slab is a slab when all the four sides are supported and ratio

of long span to short span is small 2y

x

L

L

. In such condition, bending takes place

along the both spans instead to shorter span as in the case of one-way slab. It is

economical over one-way slab due to following reasons.

a) Maximum moments and deflection are smaller than one-way slab, so less slab

thickness is required.

b) As the slab is supported at four sides, it provides good restraining effect.

c) In case of two way slab, Redistribution of moment occurs in both direction (X & Y).So

its MOR is very high and it is very economical.

Q.30 Differentiate between under-reinforced and over-reinforced.

Solution:




P a g e 14 | 31

Under reinforced section: Under reinforced section is section in which tension steel

reaches its maximum permissible stress prior to concrete area of tension steel in this

section is less than balanced section and NA is above the NA of balanced section. Failure

is tensile or ductile failure in these section which gives sufficient warning before failure.

Over reinforced section: Over reinforced section is section in which concrete reaches its

maximum permissible stress value prior to steel. Area of tension steel is more than that

of balanced section and NA is below the NA of balanced section. Failure is compression

failure or brittle failure which is sudden failure.

Q.31 State the assumptions made for designing riveted connections in steel.

Solution:

Assumptions in riveted connections:

1. Friction between the plates is neglected.

2. Shearing stresses are assumed to be distributed uniformly (but actually shear stress

varies parabolically over the cross-section).

3. Bearing stress distribution is also assumed uniform and contact area is d × t, where

'd’ is the diameter and 't’ is the thickness of the plate.

4. Bending stresses in rivets are neglected.

5. The rivets fill the hole completely.

6. The distribution of direct stresses on the portion between the rivets hole is assumed

uniform.

c/s of rivet

(Actual)

(Assumed)

𝜏𝑚𝑎𝑥 = 4/3 𝜏𝑎𝑣𝑔

𝜏𝑎𝑣𝑔 = F/A




P a g e 15 | 31

7. Rivets in group subjected to direct load, share the direct load equally.

Q.32 Give the design steps for compression members.

Solution:

Step -1. On the basis of actual length and the end conditions, find the effective length 𝑙.

Step -2. Assume suitable value of slenderness ratio of the member, as under:

(a) For single struts : Assume between 120 to 150

(b) For double angle struts : Assume between 100 to 120

(c) For channel section struts : Assume between 80 to 100

(d) For 𝐼-section stanchions : Assume between 60 to 90

(e) For built-up sections : Assume between 30 to 50

Step -3. Select the value of ac , corresponding to the assumed value of .

Step -4. Compute the gross-area by the relation:

ac

PA

=

Step -5. From the steel tables, select a suitable section having the above area. Find the

minimum radius of gyration minr for this section.

Step -6. Compute /l r = . If this matches with the assumed value of , the design is OK.

Otherwise repeat steps 3 to 6.

PART-C

Marks: 100

(Actual) (Assumed)




P a g e 16 | 31

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

carries 20 mark. Answer should not exceed 200 words

Q.33 A solid circular shaft of diameter 50 mm is subjected to pure bending of 3.5 kNm. Find

the maximum twisting moment that can be applied on this shaft such that the material

of the shaft does not yield. Use Tresca's theory (maximum shear stress theory) of

failure. The yield stress of the material in uniaxial tension is 400 N/mm2.

Solution:

Diameter, D = 50 mm, M = 3.5 kNm = 63.5 10 Nmm

Let maximum twisting moment is T kNm

Bending stress due to bending moment at top or bottom of shaft is given by

62

3 3

32 32 3.5 10285.21 N/mm

50

M M

Z D

= = = =

Shear stress due to torque T,

62

3 3

16 16 1040.74 N/mm

50P

T T TT

Z D

= = = =

Due to the combined bending and torsion at the surface of shaft, principal stresses will

be

2

2

1 2/2 2

x y x y

xy

+ − = +

( )2

2

1 2

285.21 285.21/ 40.74

2 2T

= +

( )2 2142.605 142.605 1660.04T= +

2

1 142.605 20336.19 1660.04T = + +

and 2

2 142.605 20336.19 1660.04T = − +




P a g e 17 | 31

As per Tresca theory,

1 2

2 2

y −

2 2 142.605 20336.19 1660.04 142.605 20336.19 1660.04 400T T + + − + +

22 20336.19 1660.04 400T +

2 19663.81T

2 11.8453T

3.44 kNmT

Hence maximum twisting moment T is 3.44 kNm.

Q.34 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:

Taking member BC

A B

C D

40 kN

2 m 2 m 2 m

𝐸𝐼 = constant

A B

C D

40 kN

2 m 2 m 2 m

𝐸𝐼 = constant

Hinge




P a g e 18 | 31

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

= − = −

Deflection under the load

/ 0BCC B B C BBC = + + =

160 1 40

4 4 2 03 2BCBEI EI

− + + =

20 kN

B A C

D

40 kN

20 kN 20 kN

B

C

𝑅𝐵

D

40 kN

2 m 2 m

𝑅𝐶

A B C D

40

𝐸𝐼 kN-m 2 m 2 m

2 m 40

𝐸𝐼 kN-m




P a g e 19 | 31

160 160

43BCBEI EI

= −

40 40 80

3 3BCBEI EI EI

= − = −

/BCD B B D BBD = + +

160 80 1 40 2 160 160 80

23 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.35 A deposit of sand has porosity of 35% and G = 2.7. The soil is dry in top 1.5 m depth, it

has 15% moisture content in next 1.8 m depth and it is submerged below it.

(i) Find effective pressure at a depth of 8 m below the ground level.

(ii) Find change in effective pressure if the water table suddenly drops to a level of 6 m

below the ground level.

(iii) Find shear strength of the soil on horizontal plane at 8 m depth for both the

positions of ground water table if = 30°.

Solution:

2° curve

2° curve




P a g e 20 | 31

For 1.5 m depth:

We know that 0.35

0.541 1 0.35

ne

n= = =

− −

For depth up to 1.5m, S = 0, since sand is dry

( )

11

wG Se

e

+=

+

( )

1

2.7 0 0.54 9.81

1 0.54

+ =+

3

1

2.7 9.8117.20 kN/m

1.54

= =

For a depth of 1.8 m below 1.5 m:

( )2

1

wG Se

e

+=

+

( )

21

wG Gw

e

+=

+ Se Gw=

( )

2

1

1

ww G

e

+=

+

( ) 31 0.15 2.7 9.81

19.78 kN/m1 0.54

+ = =

+

For a depth below 3.3 m:

( )

31

wG Se

e

+=

+ 1S =

3.3 m

8 m

4.7 m

1.8 m

1.5 m

n = 35%

G = 2.7

S = 1

n = 35% G = 2.7

S = 0

w = 15% n = 35%

G = 2.7 Water Table

Ground level




P a g e 21 | 31

( )

3

3

2.7 1 0.54 9.8120.64 kN/m

1 0.54

+ = =+

(i) Total pressure at 8 m below GL, ( ) ( ) ( )1.5 17.20 + 1.8 19.78 + 4.7 20.64 =

2158.41 kN/m =

Pore water pressure at 8 m below GL, u = 4.7 × 9.81 = 46.11 kN /m2

Effective pressure at 8 m below GL, 2158.41 46.11 112.30 kN/mu = − = − =

(ii) When the water table drops to 6 m from ground level, the soil will remain saturated

for sometime; Thus, the effective pressure will be reduced only due to reduction in pore

water pressure.

For 1.5 m depth from GL, 1 = 17.20 kN/m3

For next 1.8 m after it, 2 = 19.78 kN/m3

For next 4.7 m after it, 3 = 20.64 kN/m3

Hence total pressure at 8 m below GL, ( ) ( ) ( )1.5 17.20 + 1.8 19.78 + 4.7 20.64 =

2158.41 kN/m =

Pore water pressure, u = 2 × 9.81 = 19.62 kN/m2

Effective pressure at 8 m depth, u = − = 158.41 −19.62 = 138.79 kN/m2

(iii) Shear strength, tanc = +

Given soil strata is sand, hence c = 0, 30 =

tan =

Now for effective pressure, = 112.30kN/m2 and 30 =

= 112.30 tan 30° = 64.84 kN/m2

and for effective pressure, = 138.79 kN/m2 and 30 =

= 138.79 tan 30° = 80.13 kN/m2




P a g e 22 | 31

A B C

D

4 m 4 m 3 m

2.5 𝐼 2.5 𝐼 2 𝐼

A C D

5 mm

Q.36 Analyse the continuous beam shown in the figure below by slope deflection method.

Support B settles down by 5 mm.

E = 2 × 105 N/mm2, 𝐼 = 36 × 106 N/mm4

Solution:

Given:

Support B settles down by 5 mm

E = 2 × 105 N/mm2

𝐼 = 36 × 106 N/mm4

Using slope-deflection equations:

For AB ( )( ) ( ) ( )( )2 2 2 2 2 6 2 0.005

3 3 9

A B

AB

E I E I E IM

= + −

3 2.67 1.33 6.67 10A BEI EI EI −= + −

Since 0A =

31.33 6.67 10AB BM EI EI −= −

For BA ( )( ) ( ) ( )( )2 2 2 2 2 6 2 0.005

3 3 9

B A

BA

E I E I E IM

= + −

3 2.67 6.67 10BEI EI −= −

For BC ( ) ( )( )2 2.5 2 6 2.5 0.0052 2.5

4 4 16

B CBC

E I E IE IM

= + +

3 2.5 1.25 4.6875 10B CEI EI EI −= + +

A B C

D

4 m 4 m 3 m

2.5 𝐼 2.5 𝐼 2 𝐼




P a g e 23 | 31

B BA BCM M M = +

3 5.17 1.25 1.9825 10 0 ....(i)B CEI EI EI −= + − =

Applying, compatibility condition

For CB ( ) ( ) 3

2 2.5 2 2 2.54.6875 10

4 4

C B

CB

E I E IM EI

−

= + +

3 2.5 1.25 4.6875 10C BEI EI EI −= + +

For CD ( )( )2 2.5 2

2.5 4

C

CD C

E IM EI

= =

( )2 2.5

1.25 4

C

DC C

E IM EI

= =

0C CB CDM M M = + =

35 1.25 4.6875 10 0 ....(ii)C C BM EI EI EI − = + + =

By solving equation 1 and 2

46.49 10B−=

31.099 10C−= −

By putting the value of B and C in the equation of moments we get

35.8 10ABM EI−= −

3 5 6 4

25.8 10 2 10 36 10 mm /m

N

mm

−= −

3 5 6 65.8 10 2 10 36 10 10 N-m− −= −

41.76 kN-m= −

3 5 6

24.93 10 2 10 36 10 mm/mBA

NM

mm= −

3 5 6 64.93 10 2 10 36 10 10 N-m− −= −




P a g e 24 | 31

35.49 kN-m= −

35.49 kN-mBCM =

19.8 kN-mCBM =

19.8 kN-mCDM = −

9.864 kN-mDCM = −

Q.37 A 5 m effective span simply supported beam is subjected to a load of 40 kN/m including

its self weight. The size of the beam is 250 mm × 500 mm. The beam is reinforced with

4-20 at bottom (out of which two bars are curtailed) and 2-12 at top. Design the

beam against shear force and show the reinforcement details. Use M 20 and Fe 415. Use

limit state method of design.

% tension

steel

0.15 0.25 0.5 0.75 1.0

𝜏𝑐 (MPa) 0.28 0.36 0.48 0.56 0.62

% tension

steel

1.25 1.5 2.0 2.5 3 & above

𝜏𝑐 (MPa) 0.67 0.72 0.79 0.82 0.82

Solution:

𝑊 = 40 kN/m

Factored load uW = 1.5 × 40 60 kN/m

Max. SF, 60 5

150 kN2 2

uu

W lV

= = =

Effective depth, 20

500 25 465 mm2

d = − − =

40 kN/m

5 m

500

250




P a g e 25 | 31

Nominal shear stress, 3150 10

250 465

uv

V

bd

= =

21.29 N/mm=

2

max 0.63 2.8 N/mmc ckf = = ; maxv c (ok)

Design shear strength of concrete:

at critical section

% 100stt

AP

Bd=

At supports only 2-20 bars are present on tension side

( )

22 20

4% 100 0.54250 465

tP

= =

( ) 20.56 0.480.48 0.54 0.5 0.4928 N/mm

0.75 0.5c

−= + − =

−

SF to be resisted by shear stirrups,

us u cV V bd= −

3150 0.4928 250 465 10 92.712 kN−= − =

Using 2-legged 8 mm bars as shear stirrups

2 22 8 10.53 mm4

svA

= =

Spacing of 2-legged 8 mm stirrups

( )

3

0.87180 mm

92.712 10

sv y

v

us

A f dS

V

= = =

Max. spacing corresponding to min. shear reinforcement

0.4

. 0.87

sv

v y

A

B S f=




P a g e 26 | 31

0.87 415 100.53

3600.4 250

vS

=

Max. spacing should not exceed.

(i) 0.75 d = 0.75 × 465 = 348.75 mm

(ii) 300 mm

Hence max. spacing = 300 mm

Location of min. shear reinforcement (Max. spacing)

2.5

92.712 150

x=

2.5 92.7121.545 m from mid

150x

= =

Location of max spacing, 2𝑥 = 3.09 m




P a g e 27 | 31

Q.38 Write the steps for designing a RCC column?

Solution:

(1) Depending upon the grade of concrete to be used, determine stress in concrete,

longitudinal bars and ties.

(2) Find the superimposed load the column is required to carry. To this, add dead

weight of column to get total load the column has to carry at its base.

(3) Assume suitable value of reinforcement scA B/W 0.8 to 2% of gross area of a

column.

. .cc c sc scP A A = +

( ) .cc g sc sc scP A A A = − +

𝑤𝑢

5 m

150 kN

150 kN

92.712 kN

92.712 kN

2𝑥 = 3.09

2-Legged 8𝜙 180 c/c

2-Legged 8𝜙 300 c/c

3.09 m

5 m

Shear reinforcement detailing

𝐶𝐿

𝐶𝐿




P a g e 28 | 31

( ) . cc g g sc gP A p A p A = − +

( )1

g

cc sc

PA

p p =

− +

Ag - Gross area of column

p - Ratio of steel to total area sc

g

AP

A=

(4) After having known the area Ag, determine the dimension of the column. It is a

square of side gP A=

If it is a circular column; D will be equal to 4 /gA

(5) For a given end conditions, determine the effective length Le of the column and

hence calculate Le/b ratio to find whether it is short column or long column.

(6) It Le/b < 12, it will be designed as a short column for which dimensions have

already been found in step −3 and step – 4 .So, determine the Asc and distribute the

bars suitably around the column keeping suitable cover.

(7) It Le/b > 12, it will be designed as long column for which the reduction Cr is

determined

1.2548

eff

r

LC

b= −

Calculate the design load P for an equivalent short column by reduce the permissible

stress of concrete and steel with a factor Cr. Calculate the area Ag and hence determine

the size of column and also calculate the area of steel Asc = P.Ag

(8) Find the diameter of bars used as ties and determine its pitch.

Q.39 Explain Euler's Buckling failure theory of compression member?

Solution:

Buckling Failure: Euler's theory:




P a g e 29 | 31

Let us take a column which is very long in proportion to its cross-sectional dimensions.

The column is assumed to be perfectly straight and homogenous in quality, and the

compressive load is perfectly axially applied. The strength to resist buckling is greatly

affected by conditions of the ends, whether fixed or free. To start with, we will take the

standard case of a column, hinged at both the ends.

At critical load causing neutral equilibrium, stated in the previous article, a compression

member may buckle in any direction, if its moment of inertia 𝐼 is same about the axes.

Generally, a compression member does not posses equal flexural rigidity in all

directions. The significant flexural rigidity 𝐸𝐼 of a column depends on the minimum 𝐼

and at the critical load, a column buckles either to one side or to the other, in that plane

about which 𝐼 is minimum.

Consider a column with its ends free to rotate around frictionless pins. The buckled

shape shown in Fig. is possible only at a critical load called buckling load, crippling load

or Euler load, as prior to this load, the column remains straight. The smallest force at

which a buckled shape is possible is called critical force.

Consider a section at a distance x from end A, and let y be its deflection from the centre

line. We have therefore, the following equation for the elastic curve:

2

2.x

d yEI M P y

dx= = −

The negative sign has been taken because the column has been bent concave to its

original centre line.

Rearranging the above equation, we get

P B

P

A

L Y

𝑥

Buckled

shape Initial

axis




P a g e 30 | 31

2

20

d y Py

dx EI+ =

This is a standard differential equation, the solution of which is

1 2cos sinP P

y C x C xEI EI

= +

where 1C and 2C are constants of integration and can be found by applying end

conditions.

At A, x = 0 and y = 0

1C = 0

At B, x = L and y = 0

20 sinP

C LEI

=

This is possible if 2C is zero, in which case, the column has not bent at all, or

sin 0P

LEI

=

0, ,2 ,......P

LEI

=

Taking the least significant value, we get

2

2

EIP

L

=

We shall call this critical load as Euler's load, crP

Hence, 2

2cr

EIP

L

=

Where L = Effective length of the column.




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