QUIZ : CIVIL ENGINEERING (13-03-2021)

Quiz: Civil Engineering
Exam: PUNJAB PSC
Topic: Miscellaneous

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

Q1. For a beam, the term M/EI is:
(a) Stress
(b) Rigidity
(c) Curvature
(d) Shear force

Q2. 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

Q3. 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

Q4. 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

Q5. Rankines Gorden formula used to find out buckling load of column
(a) Long column
(b) Medium column
(c) Short column
(d) None of these

Q6. The minimum tension reinforcement in beam should not be less than __.
(a) A_st/bd=0.85/(fy )
(b) 0.47/A_st =fy/(√3 f_ck )
(c) 0.85/A_st =fy/(√3 f_ck )
(d) 0.45/A_st =(fy )/(√3 f_ck )

Q7. A T-beam roof section has the following particulars:
Thickness of slab = 100 mm
Width of rib = 300 mm
Depth of beam = 500 mm
C/c distance of beams = 3.0 m
Distance between points of contraflexure is 3.6 m.
The effective width of flange of the beam is:
(a) 1600 mm
(b) 1900 mm
(c) 1500 mm
(d) 3000 mm

Q8. The allowable stress in a long column can be increased by increasing the
(a) radius of gyration
(b) eccentricity
(c) slenderness ration
(d) length of the column

Q9. Which of the following is the correct sequence of plasticity of minerals in soil in an increasing order?
(a) Kaolinite, silica, montmorillonite, illite
(b) Kaolinite, silica, illite, montmorillonite
(c) Silica, kaolinite, illite, montmorillonite
(d) Silica, kaolinite, montmorillonite, illite

Q10. What is the power (in kW) transmitted by a rotating shaft? (Where : P = power, N= Speed in rpm and T = Torque transmitted)
(a) 2πNT/60
(b) 2πNT²/60000
(c) NT/60000π
(d) 2πNT/60000

SOLUTION
S1. Ans.(c)
Sol. By Bending equation
M/I=σ/y=E/R
M/I=E/R
M/EI=1/R
Curvature = 1/R

S2. 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

S3. Ans.(a)
Sol. Diameter = 18 mm
Ultimate load = 9.0 ton
Tensile stress = (9.0 ton)/(π/4×18²)
= 0.035 ton/mm²

S4. 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

S5. 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.
S6. Ans.(a)
Sol. Minimum reinforcement in beam (〖Ast〗_min)
▭(〖Ast〗_min/Bd=0.85/fy)

S7. Ans.(c)
Sol.

c/c distance b/w beam = 3000 mm
= 300 + 3000/2+3000/2
▭(b=3300 mm)
Effective width = (lo/6+bw+6df)
=((3.6×1000)/6+300+6×100)
=( 600+300+600)
=1500 mm.

S8. Ans.(a)
Sol. The allowable stress in a long column can be increased by increasing it’s radius of gyration and radius of gyration depends on moment of inertia.
R=√(I/A)
R- radius of gyration
I- Moment of Inertia

S9. Ans.(c)
Sol. increasing order of plasticity in clay mineral –
Silica → kaolinite → illite → montmorillonite

S10. Ans.(d)
Sol. Power (P) = 2πNT/60 (watts)
P= 2πNT/60×1/10³ (Kilo watt)
P=2πNT/60000

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