Serial : Gh1_W_ME_Strength of Material_250318

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CLASS TEST MECHANICAL

2018-19 ENGINEERING

Subject : Strength of Material

Date of test : 25/03/2018

Answer Key

1. (b) 7. (c) 13. (c) 19. (b) 25. (d)

2. (d) 8. (a) 14. (b) 20. (a) 26. (a)

3. (b) 9. (c) 15. (b) 21. (c) 27. (b)

4. (b) 10. (b) 16. (c) 22. (a) 28. (a)

5. (c) 11. (b) 17. (b) 23. (a) 29. (c)

6. (b) 12. (a) 18. (b) 24. (a) 30. (c)

 CT-2018 | ME • Strength of Material 7

Detailed Explanations

1. (b)

4 kN

P Q R S

1m 1m 1m
RP RR

∑Fy = 0
RP + R R = 4 kN
∑MR = 0
–4 × 2 + RR × 2 = 0
⇒ RR = 4 kN
RP = 0
MP = 0
MQ = 0

1
MR = −2 × 1 × = −1 kN-m
2
MS = –4 × 1 + 4 × 1 = 0

P Q R S

–1
BMD

2. (d)

Q S R S T
A. P R T 4. P Q

P
Q R S
B. 2. P Q R S T
T

P Q R S T P Q R S T
C. 3.

P Q R S T
P Q R
D. 5. S T

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 8 Mechanical Engineering

3. (b)
7 2
Gd 4 10 N/cm × 2 cm
=
4 4
( ) ( )
( )
Stiffness of spring =
8D 3n 8 × 203 cm3 × 25
= 100 N/cm
4. (b)
I 0.6 × 13
z1 = = = 0.1 m3
y max 12 × 0.5
I 1 × 0.63
and z2 = = = 0.06 m3
y max 12 × 0.3
(z = section modulus)
z2 0.06
∴ = 0.6 times
z1 = 0.1
5. (c)
As per maximum shear stress theory,
 σ − σ 2 σ1 σ 2 
Absolute τmax = Max of  1 , , 
 2 2 2 
σ yt 80 σ yt
=
2
( σ yt = yield point stress = ) 2
=
2
∴ σyt = 80 MPa

6. (b)
T τ J
= , Here T and τ are same, so should be same i.e. polar section modulus will be same.
J R R

7. (c)
1
(σ x − σy )
2
Radius of Mohr’s circle = + 4τ 2xy
2

(σ x − σ y )
2
Diameter of Mohr’s circle = + 4τ 2xy = 67.082 MPa

8. (a)
It is a case of pure shear stress,
τ MPa

σ MPa

σ2 σ1

σ1 = +400 MPa
σ2 = –400 MPa
τ = 0

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 CT-2018 | ME • Strength of Material 9

9. (c)
 α∆T l − ∆ 
Stress induced in the bar =   × E
l
 20 × 10−6 × 100 × 25 − 0.01
 × 10 × 0.5 = 80 MPa
5
= 
 25

10. (b)
Load (P ) = 5 kN = 5 × 103 N
stress = 100 MPa = 100 N/mm2
We know that
σimpact = σstatic × Impact factor
For suddenly applied load, Impact factor = 2
Here for safety σimpact should not exceed 100 MPa.
σImpact 100
⇒ σstatic = =
Impact factor 2
= 50 N/mm2
P 5 × 103 6366.1977
⇒ 50 = = =
A π × d2 d2
4
⇒ d = 11.28 mm

11. (b)
FBD of BC PBC 20 kN

FBD of AB PAB 80 kN (30 + 30 + 20)

Here on AB, 30 kN is on upper half and 30kN is on lower half

PBC = 20 kN, PAB = 80 kN
AAB = 0.785 (0.1)2 = 7.85 × 10–3 m2
(complete c/s area of AB will be taken because of additional 20 kN)
ABC = 0.785(0.075)2 = 4.4156 × 10–3 m2
2
PAB L AB P2 L
U = + BC BC
2A AB E st 2ABCEbr

=
(80 × 10 ) (1.5)
3 2

+
(20 × 10 )
3 2
(0.5)
−3
2 × 7.85 × 10 × 200 × 10 9
2 × 4.4156 × 101 × 109
= 3.28 J
The above result is valid only if σ < σy

PAB 80 × 103
σAB = = = 10.19 MPa <(σy)st = 250 MPa (O.K.)
A AB 7.85 × 10 −3

PBC 20 × 103
σBC = = = 4.537 MPa <(σy)br = 410 MPa (O.K.)
ABC 4.4156 × 10 −3

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 10 Mechanical Engineering

12. (a)
E = 2G(1 + ν)
105 = 2 × 39.7 (1 + ν)
ν = 0.3224
δd δl 0.025 l − 74.96
ν = =
d l 20 l
74.96
1− = 3.87 × 10–3
l
⇒ l = 75.2517 mm

13. (c)

P(N) P(N)
w(N/m)

A C D B
x l x

From the symmetry of the figure,

wl
RC = RD = P +
2
Bending moment at mid point,
wl l l  l
M = − × + RC × − P  x +  = 0
2 4 2  2
wl 2
gives x =
8P
14. (b)

A B C D
50 kN 80 kN 20 kN 10 kN

600 mm 1000 mm 1200 mm

Part AB: The section of the bar in this part is subjected to a tension of 50 kN.

P1 l1 50 × 1000 × 600
Extension of AB = = mm = 0.2857 mm (extension)
AE 1000 × 1.05 × 105
Part BC: The section of the bar in this part is subjected to a compression of 80 – 50 = 30 kN.

P2 l2 30 × 1000 × 1000
Contraction of BC = = mm = 0.2857 mm (contraction)
AE 1000 × 1.05 × 10 5
Part CD: The section of the bar in this part is subjected to a compression of 10 kN.

P3 l3 10 × 1000 × 1200
Contraction of BC = = mm = 0.1143 mm (contraction)
AE 1000 × 1.05 × 10 5
∴ Change in length of the bar
= 0.2857 – 0.2857 – 0.1143 = –0.1143
i.e., decrease in length of the bar is 0.1143 mm.

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 CT-2018 | ME • Strength of Material 11

15. (b)
d = 30 mm, f = 12 Hz, P = 25 kW
2 πf T
P =
1000
2 π × 12T
⇒ 25 =
1000
∴ T = 331.57 Nm

16T 16 × 331.57 × 10 3
Maximum shear stress, τs = = N/mm 2
πd 3 π × 30 3
= 62.5439 N/mm2 = 62.5439 MPa

16. (c)
According to the given conditions
50 MPa

60 MPa 60 MPa

50 MPa

1
(σ x + σ y ) ± (σ x − σy ) + 4τ xy 
2
σ1, 2 =
2  

1
(σ x + σy ) + (σ x − σy )2 + 4τ xy 
2
σ1 = 
2  

1
125 = (60 − 50) + (60 + 50)2 + 4τ xy 2 
2 
⇒ τxy = 106.65 MPa

17. (b)
Critical Point T = 500 Nm
τs

P
σa + σb σa + σb

M = Pe

P 50 × 10 3
σa = = = 25.46MPa
A π × 502
4
My 50 × 103 × 10 × 25 × 64
σb = = = 40.74 MPa
INA π × (50)4

16T 16 × 500 × 103

τs = = = 20.37MPa
πd 3 π × (50)3

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 12 Mechanical Engineering

66.2 MPa 66.2 MPa

20.37 MPa
According to MDET,
Syt
σ x + 3 τ xy
2 2
=
N
N = 2.67

18. (b)

w
1
A C
B l/2
l

By use of superposition principle

w
2
C

l
w

l/2
l
δ1 = δ2 – δ3
 wl 4 
δ2 = 
 8EI
We know that

For case (3)

B.M.D
l l
2 2

 
 l l 
3 4

w  2  l
w  
 2
δ3 =  × + 
 6E I 2 8E I 
 
wl4 wl4 wl4
⇒ δ 1 = (δ 2 − δ 3 ) = − −
8 E I 96 E I 128 E I
48 w l 4 − 4 w l 4 − 3 w l 4 41w l 4
= =
384 E I 384 E I

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 CT-2018 | ME • Strength of Material 13

19. (b)
Cross-section remains plane and undistorted for circular shaft only but not for non-circular shaft.

20. (a)
ε1 = 0.0013
ε2 = –0.0013
E = 2 × 105; µ = 0.3
E
σ1 = ( ε1 + µε2 )
(1− µ2 )
2 × 105
= [0.0013 − 0.3 × 0.0013]
1 − 0.09
σ1 = 200 MPa
σ2 = – 200 MPa
σ1 − σ2
τmax = = 200MPa
2

21. (c)
Extension of a tapering circular bar is given by

4PL
∆l = πd d E
1 2

4 × 25 × 103 × 0.5
=
π × 1.25 × 2.5 × 10−4 × 210 × 109
= 0.2425 mm
For bar of uniform diameter
4PL
∆l =
πd2E
4 × 25 × 103 × 0.5
= = 0.2156 mm
π × (1.875)2 × 10 −4 × 210 × 10−9
 0.2425 − 0.2156 
% error =   × 100 = 11.10%
0.2425

22. (a)

π 3
For solid shaft, T = D τ
16

π (D04 − Di4 )
For hollow shaft, T = τ
16 D0
1
  D  3 4
⇒ D i = 0 1−   
D
  D0  
1
 45  4
3
= 50 1−    = 36.07 mm
  50 

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 14 Mechanical Engineering

23. (a)
∆l = 0.3 mm l = 200 mm

∆ l 0.3
⇒ ∈ = = P P = 23.5
l 200
P
⇒ σ = = ∈E Dia. = 10 mm
A
23.5 × 103 0.3
⇒ = ×E
π 200
× 102
4
235 200
⇒ E = × = 200 × 103 N/mm2
0.785 0.3
⇒ E = 200 kN/mm2

24. (a)
Force diagram
W W

R1 R2

+ +
_
_

SFD
2nd degree curve
+

_ _

1st degree curve

BMD

25. (d)
P P

Pb Pb

C A E B D

RA = P RB = – P

VE = RA – P = 0
 L  L

ME = −P  b +  +  P ×  + Pb = 0
2 2

26. (a)
π
Weight of water = ρVg = 1000 × × d 2 × l × 9.81
4
π
× ( 0.5) × 10 × 9.81 = 19261.89 N
2
= 1000 ×
4
And this is uniformly distributed,
w l 2 19261.89 × 10 × 1000
⇒ Mmaximum = = N-mm
8 8

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 CT-2018 | ME • Strength of Material 15

M max 19261.89 × 10 × 1000

σmaximum = = ×4
Z 8 × π ×d2 ×t
19261.89 × 10 × 1000 × 4
= = 4.91 MPa
8 × π × ( 500) × 25
2

27. (b)
We know that,
∆V 3σ
= [1 − 2µ ]
V E
∆V 3 × 15
∴ = [1 − (2 × 0.3)]
200 × 100 × 50 200 × 1000
∆V = 90 mm3

28. (a)
x

wx w0 Loading diagram

l/2 l/2

We know that
dV x πx
= −w 0 sin
dx l
w 0l πx
⇒ Vx = cos + C1
π l
dM x w 0l πx
= Vx = cos + C1
dx π l

w 0l 2 πx
Mx = sin + C1x + C 2 ...(i)
π 2 l
Using boundary conditions,
When, x = 0, Mx = 0
x = l, Mx = 0
We get, C1 = 0 and C2 = 0
w 0l 2 πx
⇒ Mx = sin
π2 l
l
At, x =
2
 w 0l 2 
(M ) x =
l =  2 
 π 
2

29. (c)

σ x sin 2φ
= σ x sin φ cos φ = τ ...(i)
2

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 16 Mechanical Engineering

σ x cos2 φ = σn ...(ii)
Dividing (i) by (ii), we get
τ  100 
tanφ = = 
σ n  300 
⇒ φ = 18.435°
⇒ Putting in equation (ii), we get

σ x cos2 18.435 = 300

⇒ σx = 333.33 MPa

30. (c)
ΣMB = 0
RA × 4.5 = 48 × 3 + 60 × 1.5 – 60 × 0.9
RA = 40 kN (upward)
RB = 48 – 40 = 8 kN (upward)
Making shear force diagram
40 kN

C D E B
A
– 8 kN – 8 kN
– Shear force diagram

– 68 kN
Maximum shear force = 68 kN

„„„„

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