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# UPSSSC-JE’21 CE: Daily Practice Quiz. 25-October-2021

Know your strengths and practice your concepts with this quiz on UPSSSC-JE Recruitment 2021. This quiz for UPSSSC-JE Recruitment 2021 is designed specially according to UPSSSC-JE Syllabus 2021.

Quiz: Civil Engineering
Exam: UPSSSC-JE
Topic: Strength of material

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

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

Q2. The stress produced by a suddenly applied load as compared to that produced by the same load when applied gradually is __ times.
(a) 1.5
(b) 2
(c) 3
(d) 4

Q3. The elongation (mm) in a steel bar having a square cross section of diameter 40 mm × 40 mm is 2.5 mm and is subjected to an axial compressive load of P (kN). If the length of the bar is 4 m and modulus of elasticity is E = 250 GPa. What is the value of P(kN)?
(a) 100
(b) 150
(c) 200
(d) 250

Q4. Calculate the value of modulus of elasticity (N/mm²), if the Poisson’s ratio is 0.25 and modulus of rigidity of the material is 80 N/mm²?
(a) 100
(b) 200
(c) 250
(d) 300

Q5. For simply supported beam having load at the Center the bending moment will be:
(a) minimum at the support
(b) minimum at the center
(c) maximum at the support
(d) None of the above

Q6. In case of a circular section the section modulus is given as–
(a) (πd^2)/16
(b) (πd^3)/32
(c) (πd^3)/16
(d) (πd^4)/64

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.(b)
σ=σ_static×I.F
Where I.F – impact factor
I.F.=1+√(1+2h/δ_static ) ;let h=0
=1+√(1+(2×0)/δ_static )=2
So, the stress produced by a suddenly applied load as compared that produced by the same based when applied gradually is 2 times.
S3. Ans.(d)
Sol. Given
A = 40 × 40 mm², L = 4 m. δl = 2.5 mm
E = 250 GPa
δl=PL/AE
P=(δl.A.E)/l
P=(2.5×10^(-3)×40×40×10^(-6)×250×10^9)/4
P=250 kN

S4. Ans.(b)
Sol. Given E = 2, G = 80 N/mm², μ = 0.25
We know that,
E=2G(1+μ)
E=2×80×(1+0.25)
=160×1.25
E=200 N\/mm^2
S5. Ans.(a)
Sol. From figure we can see that for simply supported beam having lead at the center the bending moment will be minimum at the support.

S6. Ans.(b)
Sol. We know that the section modules Z=I/y
I=π/64 d^4
y=d/2
z=π/y=(π/64 d^4)/(d/2)
▭(Z=π/32 d^3 )

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