Quiz: Civil Engineering 26 March 2020

Quiz: Civil Engineering
Exam: UPPSC-AE
Topic: Strength of Material

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 10 Minutes

Q1. The stress level, below which a material has a high probability of not failing under reversal of stress, is known as
(a) elastic limit
(b) endurance limit
(c) proportional limit
(d) tolerance limit
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q2. A beam is structural member predominantly subjected to
(a) Transverse loads
(b) Axial forces
(c) Twisting moment
(d) None of these
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q3. The load at the free end of a uniform cantilever beam is increased, the failure will occur
(a) at the middle
(b) at the fixed end
(c) at the point of application of load
(d) any where in the span
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q4. Principal planes are planes of
(a) maximum shearing stress
(b) zero shearing stress
(c) shearing stress having a magnitude of 50% of principal stress
(d) will be maximum
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q5. Maximum shear strain occurs on
(a) Principal planes
(b) 45 deg. With principal planes
(c) 90 deg. With principal planes
(d) All of above
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q6. For a beam, the term M/EI is:
(a) Stress
(b) Rigidity
(c) Curvature
(d) Shear force
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q7. At any point of a beam, the section modulus may be obtained by dividing the moment of inertia of the section by …………
(a) depth of the section
(b) depth of the neutral axis
(c) maximum tensile stress at the section
(d) maximum compressive stress at the section
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q8. Find the tensile stress of a mild steel rod of 18mm diameter. Given: ultimate load = 9.0 ton.
(a) 0.03 Ton/sq.mm
(b) 0.02 Ton/sq.mm
(c) 0.05 Ton/sq.mm
(d) 0.06 Ton/sq.mm
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q9. The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimension and material, but with pinned ends is equal to
(a) 0.5
(b) 4.0
(c) 2.0
(d) 1.0
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

Q10. Rankines Gorden formula used to find out buckling load of column
(a) Long column
(b) Medium column
(c) Short column
(d) None of these
L1Difficulty 2
QTags Strength of Material
QCereator Anubhav Gupta

 

 

SOLUTION

S1. Ans.(b)
Sol. The stress which can be withstand for some infinite number of cycles without failure is called endurance limit. Endurance limit magnitude less then yield strength.

S2. Ans.(a)
Sol. The beam is a structure element that primarily resists loads applied laterally (transverse) to the beam axis.
S3. Ans.(b)
Sol. If the load at the free end of a cantilever beam is increased the failure will occure at the fixed end because due to increase of load moment resisted by fixed support is increased and when support moment reaction increased beyond it’s limit then failure occure.

S4. Ans.(b)
Sol. Infinite plane’s passes through a single point. Out of infinite plane’s three planes are mutually perpendicular to each other and having zero shear stress, that planes are called principal plane.

S5. Ans.(b)
Sol. Maxm shear strain occure on 45° with the principal planes.

S6. Ans.(c)
Sol. By Bending equation
M/I=σ/y=E/R
M/I=E/R
M/EI=1/R
Curvature = 1/R

S7. Ans.(b)
Sol.
Section Modulus (Z) = (Moment of Inertia (I))/(Depth of the neutral axis (y))

If we consider Rectangular section

Z= I/y=(bd^3/12)/(d/12)
Z= bd²/6

S8. Ans.(a)
Sol. Diameter = 18 mm
Ultimate load = 9.0 ton
Tensile stress = (9.0 ton)/(π/4×18²)
= 0.035 ton/mm²

S9. Ans.(b)
Sol. (i) Pcr=π²EI/Le²
Both end’s fixed
Le = L/2
Pcr_1=(π^2 EJ×4)/L² -(i)
(ii) both end’s pinned
Le= L
Pcr_2=π²EI/L² -(ii)
(Pcr_1)/(Pcr_2 )=((π^2 EI×4)/L^2 )/((π^2 EI)/L^2 )
= 4

S10. Ans.(b)
Sol. → Rankines Gorden formula used to find out buckling load of medium column.
→ Euler formula used for calculating buckling load of long column.

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