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Strength of material part -2

Strength of material part -2
Question:-
Option (A)
10%
Option (B)
20%
Option (C)
30%
Option (D)
more than 30%
Correct Option:
D
Question Solution:
Euler’s load = P α d4

=> P1 = kd4

When diameter is reduced by 10% P2 = k(0.9 d)4

Reduction in crippling load = (P1 - P2)/P1 X 100 = (1 - 0.94)/1 X 100 = 34.39%
Question:-
Option (A)
3/4
Option (B)
4/3
Option (C)
3/2
Option (D)
2/3
Correct Option:
D
Question Solution:
Let torque in AC and CB be TA and TB. Compatibility condition: θAC = θCB


Question:-
Option (A)
Option (B)
Option (C)
Option (D)
Correct Option:
C
Question Solution:
Maximum shear stress in case (c) = [σ1 - (-σ1)]/2 = σ1
which is equal to maximum tensile stress; σ1
Question:-
Option (A)
4 t
Option (B)
6.67 t
Option (C)
4.45 t
Option (D)
2.5 t
Correct Option:
B
Question Solution:
Shear force at end A = Reaction at end A.

ΣMB = 0
RA × 4 =

RA = 6.67 t
Question:-
Option (A)
7.21 t-m
Option (B)
5.2 t-m
Option (C)
6.5 t-m
Option (D)
3.2 t-m
Correct Option:
A
Question Solution:
Te =

= 7.21 t-m
Question:-
Option (A)
μc < μg < μr < μm
Option (B)
μc < μg < μm < μr
Option (C)
μg < μc < μm < μr
Option (D)
μg < μc < μr < μm
Correct Option:
C
Question Solution:
μg = 0.01 to 0.05
μc = 0.1 to 0.2
μm = 0.25 to 0.42 and μr = 0.5.