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DFCCIL’21 CE: Daily Practice Quiz. 31-July-2021

Quiz: Civil Engineering
Exam: DFCCIL-JE
Topic: STRENGTH OF MATERIAL

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. Which formula CORRECTLY depicts the elongation in a composite body?
(a) δ=P/E [l_1/A_1 +l_2/A_2 +l_3/A_3 +l_4/A_4 +⋯.]
(b) δ=P/AE [l_1+l_2+l_3+l_4….]
(c) δ=P/AE [l_1-l_2+l_3-l_4+⋯.]
(d) δ=Pl/E [1/A_1 +1/A_1 +1/A_1 +1/A_1 +⋯]

Q2. What will be the thermal stress developed in a rod having a diameter of 4 cm and length of 2m. it experiences heating from temperature 50°C to 200°C. the coefficient of thermal expansion is:
α=10×10^(-6) \/°C and young modulus is 250 GPa
(a) 300
(b) 325
(c) 350
(d) 375

Q3. 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

Q4. Determine the value of shear force at point ‘A’ in the figure shown below:
DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_30.1
(a) (wl^2)/6
(b) (wl^2)/2
(c) wl^2
(d) (wl^2)/3

Q5. The diagram shown below shows the shearing force diagram for a beam.
Which of the following is the correct bending moment diagram for the above shear force diagram?
DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_40.1
DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_50.1

Q6. Maximum principal stress theory was postulated by __________.
(a) St. Venant
(b) Mohr
(c) Rankine
(d) Tresca

SOLUTION
S1. Ans.(a)
Sol. Bar is composite so axial force is same throughout the bars.
DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_60.1
Total elongation is the sum of each bars elongation.
Total elongation δ=δ_1+ δ_2+δ_3+⋯
δ=(Pl_1)/(A_1 E)+(Pl_2)/(A_2 E)+(Pl_3)/(A_3 E)……
δ=P/E [l_1/A_1 +l_2/A_2 +l_3/A_3 ]

S2. Ans.(d)
Sol. Thermal stress in rod
σ_thermal=α∆TE
=10×10^(-6)×(200-50)×250×10^9
=375 MPa

S3. 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
S4. Ans.(a)
Sol. Distributed load
DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_70.1
f=∫_0^L▒〖W_(x-x) dx〗
=∫_0^L▒〖wx/l.dx〗
=[(wx^2)/2L]_0^l
=wL/2 – it is at L/3 from A then bending moment at A
M_A=f.x
=WL/2.L/3
=(WL^2)/6

S5. Ans.(a)
Sol. In SFD value of shear force is decreasing from A to B. So, slope of bending moment diagram also has decreasing slope.

DFCCIL'21 CE: Daily Practice Quiz. 31-July-2021_80.1
S6. Ans.(c)
Sol. maximum principal stress theory also known as Rankine’s theory because it is postulated by Rankine.

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