Quiz: Civil Engineering
Exam: Punjab PSC

Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes

Q1. While estimating for plastering, usually no deduction is made for –
(a) Ends of beams
(b) Small opening up to 0.5 sq. m.
(c) End of rafters
(d) All the above

Q2. A beam is structural member predominantly subjected to
(a) Transverse loads
(b) Axial forces
(c) Twisting moment
(d) None of these

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) anywhere in the span

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

Q5. The brick work is measured in square meter, in case of
(a) Honey comb brick work
(b) Brick flat Soling
(c) Half brick walls or the partition
(d) All options are correct

Q6. For a beam, the term M/EI is:
(a) Stress
(b) Rigidity
(c) Curvature
(d) Shear force

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

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

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

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


S1. Ans.(d)
Plastering is measured in terms of m²If,
(a) opening area < 0.5 m² (No deduction is applied) (b) opening area is between 0.5 m² – 3m², then (single side deduction is applied) (c) opening area > 3 m², then (double side deduction is applied)

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 its 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.(d)
Generally, the brick work is measured in terms of m³. but in case of honey comb brick work, brick flat soling, half brick wall or partition wall then, it is measured in m².

S6. Ans.(c)
Sol. By Bending equation
Curvature = 1/R

S7. Ans.(b)
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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