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# Quiz: Mechanical Engineering 29th April

Quiz: Mechanical Engineering
Exam: DFCCIL
Topic: Strength of Material

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 15 Minutes

(a) Ductile materials
(b) Brittle materials
(c) Equally serious in both cases
(d) Depends on other factors

Q2. A rod with a length of 100 cm and diameter of 4 cm undergoes and axial pull of 50 kN. What is the stress (in N/mm²)?
(a) 0.04
(b) 0.4
(c) 4
(d) 40

Q3. A steel bar of dimension 10 mm × 1mm × 1mm is free to expand is heated from 15°C to 25°C what stress shall be developed?
(a) Tensile stress
(b) No stress
(c) Shear stress
(d) Compressive stress

Q4. The strain energy stored in a cantilever beam loaded as shown, will be:

(a) (P^2 l^3)/4E
(b) (P^2 l^3)/3EI
(c) (P^2 l^3)/2EI
(d) (P^2 l^3)/6EI

Q5. Mohr’s circle can be used to determine following stress on inclined surface:
Principal stress
Normal stress
Tangential stress
Maximum shear stress
(a) only A
(b) only B
(c) only C
(d) A, B, C and D

Q6. A circular shaft can transmit a torque of 13 kN-m. if the torque is reduced to 12 kN-m, then the maximum value of bending moment that can be applied to the shaft is-
(a) 1 kN – m
(b) 3 kN – m
(c) 5 kN – m
(d) 7 kN – m

Q7. If ‘L’ is the span of a light suspension bridge, whose, each cable carries total weight (W) and the central dip is ‘Y’, the horizontal pull at each support is __.
(a) wL/4y
(b) wL/8y
(c) wL/2y
(d) wL/y

Q8. What is the ratio of maximum tangential stress (σ_t ) and maximum radial stress (σ_r) of a solid disk flywheel?
(a) σ_t/σ_r =2.0
(b) σ_t/σ_r =0.5
(c) σ_t/σ_r =1.5
(d) σ_t/σ_r =1.0

Q9. What is the volumetric strain in the thin cylinder subjected to internal pressure having hoop stress of 200 MPa, modulus of elasticity, E = 200 GPa and Poisson’s ratio = 0.25?
(a) 20/1000
(b) 2/1000
(c) 0.2/1000
(d) 0.02/1000

Q10. The equivalent length of the column when one end is fixed and other end is pinned is __.
(a) l
(b) l/2
(c) l/√2
(d) 2l

Solution

S1. Ans.(b)
Sol. Stress concentration in static loading is more serious in Brittle material because in case of ductile material local yielding will take place so chances of stress concentration is less as yielding is missing in brittle material.

S2. Ans.(d)
Sol. Given,
L = 100 cm, d = 4 cm, P = 50 kN
σ=P/Q=P/(π/4 d^2 )
=4P/(πd^2 ) =(4×50×10^3)/(π×[0.04]^2 )
=39.789 MPa
=40 N\/mm^2

S3. Ans.(b)
Sol. No stress will develop because it is already mention in the question that bar is free to expand.

S4. Ans.(d)
Sol.

In this beam strain energy is because of shear force and bending moment but due to shear force is not considerable in amount so can be neglected than, Strain energy due to bending only
U=∫0toL((M_xy )^2 dx)/2EI
U_AB=∫0toL((Px)^2 dx)/2EI
U_AB=[(P^2 x^3)/(3×2EI)]_0^2
(U_AB=(P^2 L^3)/6EI)

S5. Ans.(c)
Sol. the Mohr’s circle is used to determine the stress components and i.e. the co-ordinates of any point on the circle when stresses are on inclined surface, then tangential stress can be determined because the normal stress and tangential stress are to be already known.

S6. Ans.(c)
Sol. for compensate the torque addition of bending moment is required
T_e=√(T^2+m^2 )
13=√((12)^2+m^2 )
169=12^2+m^2
169-144=m^2
m=5kNm

S7. Ans.(b)
Sol. if L is the span of a light suspension bridge, where, each cable carries total weight (W) and the central dip is ‘y’ the horizontal pull at each support is
= Wl⁄8y

S8. Ans.(d)
Sol. in case of solid disk flywheel,
Maximum tangential stress (σ_t ) = maximum radial stress (σ_t )
σ_t=σ_r
∵σ_t/σ_r =1

S9. Ans.(b)
Sol. we know that, the volumetric strain is
∈_V=δV/V=Pd/4tE (5-4μ)
Given that, hoop stress = pd/2t=200 MPa
Now,
∈_v=1/2×hoop stress×1/E=(5-4μ)
=1/2×200×1/(200×10^3 )×(5-4×0.025)
=2/1000

S10. Ans.(c)
Sol. we know that, the crippling load
P_Cr=(π^2 EI)/(le^2 )
le=l/2 (When one end fixed and one end hinged)

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