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# Quiz: Civil Engineering 20 May 2021

Quiz: Civil Engineering
Exam: DFCCIL
Topic: Solid Mechanics

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. In a closed helical spring subjected to an axial load, other quantities remaining the same, if the wire diameter is doubled and mean radius of the wire diameter is doubled and mean radius of the coil is also doubled, then stiffness of spring when compared to original one will become________.
(a) Twice
(b) Four times
(c) Eight times
(d) Sixteen times

Q2. What will be the effect on Euler’s crippling load for a column with both ends hinged, if the length of the column is halved?
(a) 0.25
(b) 0.5
(c) 2
(d) 4

Q3. The thin wall cylindrical vessel has the wall thickness t and diameter d is subjected to the gauge pressure of p. if the diameter of the vessel is doubled, then what is the ratio of hoop stress as compared to initial value?
(a) 1: 1
(b) 2: 1
(c) 1: 2
(d) 1: 4

Q4. Calculate the value of shear stress (MPa) in the solid circular shaft of diameter 0.1 m which is subjected to the torque of 10 kN-m.
(a) 40.5
(b) 50.93
(c) 60.5
(d) 70.5

Q5. Calculate the maximum shear strain at the point where principal strains are 100 × 10^(-6) and -200×10^(-6)
(a) 100×10^(-6)
(b) 200×10^(-6)
(c) 300×10^(-6)
(d) 400×10^(-6)

Q6. A downward vertical load of 10 kN net: at a distance of 40 cm from the left end on a 1 m long beam. This beam is simply supported at both ends. The vertical reaction at the left end is:
(a) 0.25 kN
(b) 4 kN
(c) 6 kN
(d) 5 kN

Q7. Modulus of rigidity is defined as the ratio of __.
(a) longitudinal stress and longitudinal strain
(b) volumetric stress and volumetric strain
(c) lateral stress and lateral strain
(d) shear stress and shear strain

Q8. A rod of dimension 20 mm × 20 mm is carrying an axial tensile load of 10 kN. If the modulus of elasticity is 250 MPa. Then the strain induced due to this load would be __.
(a) 0.1
(b) 0.25
(c) 0.2
(d) 10

Q9. Calculate the value of modulus of rigidity (N/mm²) if the Poisson’s ratio is 0.25 and modulus of elasticity for the material is 200 N/mm²
(a) 50
(b) 80
(c) 100
(d) 150

Q10. A cross sectional bar of area 700 mm² is subjected to an axial load shown in the figure below. What is the value of stress (MPa) in the section RS?

(a) 30
(b) 40
(c) 50
(d) 60

Solutions

S1. Ans.(a)
Sol. Spring stiffness, K=(Gd^4)/(8D^3 N)
K^’=(G×2^4.d^4)/(8×2^3×D^3 N)=2[(Gd^4)/(8D^3 N)]
=2K

S2. Ans.(d)
P_C=(π^2 EI)/(le^2 )
For both End hinged (le = l)
P_c=(π^2 EI)/l^2
If length column is halved [le=l\/2]
Pc^’=(π^2 EI)/(le)^2 =(4π^2 EI)/l^2
Pc^’=4(Pc)

S3. Ans.(b)
Sol.
σ_h1=pd/2t
σ_h2=P(2d)/2t
σ_h1/σ_h2 =(P(2d)\/2t)/(pd\/2t)=2/1
σ_h1:σ_h2=2∶1

S4. Ans.(b)
Sol. Shear stress in solid circular shaft
τ=16T/(πd^3 )=(16×10×10^3)/(π(0.1)^3 )
=50.93 MPa

S5. Ans.(c)
Sol. Maximum shear strain (γ_xy )
γ_xy/2=(∈_1-∈_1)/2
γ_xy=100×10^(-6)-(-200×10^(-6) )
=300×10^(-6)

S6. Ans.(c)
Sol. Taking moment about point B

For equilibrium moment about any point is zero.
∑▒M_B =R_A×1-10×(0.60)
0=R_A-6
R_A = 6.0 km

S7. Ans.(d)
Sol. Modules of rigidity is defined as the ratio of shear stress and shear strain.
As per Hook’s law
Shear Stress ∝ shear strain
Shear stress = G × Shear strain
G = (Shear Stress)/(Shear Strain)
Where G – modules of rigidity.

S8. Ans.(a)
Sol. Given that
P = 10 kN
A = 20 × 20 mm²
E = 250 MPa
We know that
SL=PL/AE
δL/L=P/AE
Strain =(10×10^3)/(20×20×10^(-6)×250×10^6 )
Strain =0.1

S9. Ans.(b)
Sol. Given; E = 200 N/mm², G = ?
μ = 0.25
we know that,
E=2G(1+μ)
200=2G(1+0.25)
100=G(1.25)
G=80 N\/mm^2

S10. Ans.(a)
Sol. FBD of PQ, QR and RS

Maximum load in RS = 21 kN
Maximum stress in RS =
σ_RS=P_RS/A_RS =(21×10^3)/(700×10^(-6) )
=30 MPa

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