Quiz: CIVIL ENGINEERING
Exam: TELANGANA-PSCM
Topic: MISCELLANEOUS
Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes
Q1. When the liquid limit of a soil is less than the plastic limit, then the plasticity index is reported as
(a) Negative
(b) Zero
(c) 1
(d) Non-plastic (NP)
Q2. The hydraulic head that would produce a quick sand condition in a sand stratum of thickness 1.8m, specific gravity 2.65 and void ratio 0.65 is equal to:
(a) 1.0 meter
(b) 1.2 meter
(c) 1.6 meter
(d) 1.8 meter
Q3. In a soil deposit of two layers of equal thickness, the permeability first and second layers are in ratio 1 : 2. The ratio of average permeability of horizontal direction to the vertical direction will be:
(a) 9/8
(b) 8/9
(c) 12/8
(d) 8/12
Q4. For routine consolidation test in laboratory, the thickness of the specimen is
(a) 10 mm
(b) 20 mm
(c) 40 mm
(d) 60 mm
Q5. The consolidation time for soils
(a) Increases with increasing compressibility
(b) Decreases with increasing permeability
(c) Is independent of the stress change
(d) All of the above
Q6. The assumption that the plane sections normal before bending remains normal after bending is used_______:
(a) only in the working stress method of design
(b) only in the limit-state method of design
(c) in both working stress and limit state methods of design
(d) only in the ultimate load methods of design
Q7. For concrete of grade M-50, short-term modulus of elasticity will be nearly
(a) 20000 N/mm²
(b) 35000 N/mm²
(c) 50000 N/ mm²
(d) 75000 N/mm²
Q8. The stress in punching shear is checked _________ in an isolated reinforced column footing of effective depth ‘d’.
(a) at the centre of the column
(b) at a distance d/2 away from the face of the column
(c) at a distance d/2 away from the centre of the column
(d) at the face of the column
Q9. As per IS 456, the minimum nominal cover specified for footings is:
(a) 25 mm
(b) 40 mm
(c) 50 mm
(d) 75 mm
Q10. The reduction coefficient of a reinforced concrete column with an effective length of 4.5 m and size 250×300 mm² us _______:
(a) 0.8
(b) 0.85
(c) 0.9
(d) 0.95
SOLUTION
S1. Ans.(b)
Sol. Plasticity index (I.P.) = Liquid limit – Plastic limit
▭(I.P.=W_L-W_P )
→ When W_L is less then W_p then plastic index equal to zero.
S2. Ans.(d)
Sol. thickness (t) = 1.8 m
G = 2.65
Void ratio (e) = 0.65
For quick sand condition –
Critical hydraulic gradient (i_c ) = (G-1)/(1+e)
(G-1)/(1+e)=∆h/t
(2.65-1)/(1+0.65)=∆h/1.8
1.65/1.65=∆h/1.8
▭(∆h=1.8 meter)
S3. Ans.(a)
Sol. H_1=H_2=H
K_1/K_2 =1/2
K_1=K
K_2=2K
Average permeability in horizontal direction =
K_avg=(Σk_i H_i)/(ΣH_i )
K_avg=(k×H+2k×H)/(H+H)
=3KH/2H
=3/2 K
Average permeability in vertical direction
k_(avg_2 )=(ΣH_i)/((ΣH_i)/k_i )
=(H+H)/(H/K+H/2K)
=2H/(3H/2K)
=4/3 K
k_(avg_1 )/k_(avg_2 ) =(3⁄2 k)/(4⁄3 k)=9/8
S4. Ans.(b)
Sol. the thickness of the specimen should be as small as possible to reduce side friction. But a minimum thickness of 20 mm is usually required.
S5. Ans.(d)
Sol. T_v=C_v t/d^2 ————– (1)
▭(C_v=K/(m_v r_ω )) put in (1)
T_v=(K/m_v r_ω ) (t/d^2 ) ———— (2)
▭(m_v=a_v/(1+e_o )) Put in (2)
▭(T_v=(K(1+e_o )/(a_V (r_ω ) )) (t/d^2 ) )
▭(t α a_v )
▭(t α 1/K)
S6. Ans.(c)
Sol. plane section normal before bending remains normal after bending is valid for working stress method & Limit state method.
S7. Ans.(b)
Sol. short term modulus of elasticity (E_c)
E_c=5000 √fck
for M50
▭(fck=50)
E_c=5000 √50
E_c 35355.33 N\/mm^2
Most expected option is (b)
S8. Ans.(b)
Sol. → punching shear checked at a distance d/2 away from the face of the column.
→ In one way shear checked at ‘d’ distance from face of the column.
S9. Ans.(c)
Sol.
Component Nominal cover
SLAB 20 mm
BEAM 25 mm
COLUMN 40 mm
FOOTING 50 mm
S10. Ans.(c)
Sol. reduction coefficient (Cr) = 1.25 – l_eff/(48.b)
Cr=1.25-4500/(48×250)
▭(Cr=0.875≈0.9)
