Quiz: Civil Engineering

Exam: DFCCIL-JE

Topic: MISCELLANEOUS

Each question carries 1 mark

Negative marking: 1/4 mark

Time: 10 Minutes

Q1. 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

Q2. 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

Q3. For routine consolidation test in laboratory, the thickness of the specimen is

(a) 10 mm

(b) 20 mm

(c) 40 mm

(d) 60 mm

Q4. 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

Q5. 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

Q6. As per IS 456, the minimum nominal cover specified for footings is:

(a) 25 mm

(b) 40 mm

(c) 50 mm

(d) 75 mm

Q7. 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

Q8. The diameter of rivet hole in tension member using rivet of diameter more than 25mm will be equal to

(a) rivet diameter +3mm

(b) rivet diameter +2 mm

(c) rivet diameter +1.5 mm

(d) rivet diameter

Q9. The permissible longitudinal pitch in a riveted joint in tension is

(a) 12t or 200 mm

(b) 20t or 200 mm

(c) 16t or 200 mm

(d) 24t or 200 mm

Q10. Bearing strength of single riveted lap joint is equal to:

(a) d×t×f_b

(b) 2d×t×f_b

(c) 2×d²×t×f_b

(d) None of the above

` SOLUTION`

S1. 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)

S2. 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

S3. 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.

S4. 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)

S5. 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)

S6. Ans.(c)

Sol.

Component Nominal cover

SLAB 20 mm

BEAM 25 mm

COLUMN 40 mm

FOOTING 50 mm

S7. Ans.(c)

Sol. reduction coefficient (Cr) = 1.25 – l_eff/(48.b)

Cr=1.25-4500/(48×250)

▭(Cr=0.875≈0.9)

S8. Ans.(b)

Sol. in case of rivet

→ if rivet diameter is less than 25 mm then diameter of hole equal to diameter of rivet plus 1.5 mm.

→ if rivet diameter is more than 25 mm then diameter of hole equal to diameter of rivet plus 2 mm.

S9. Ans.(c)

Sol.

→ Permissible longitudinal pitch in tension is (16 t, 200 mm) whichever is less

→ Permissible longitudinal pitch in compression is (12 t, 200 mm) whichever is less

Where, t = thickness of thinner outside plate

S10. Ans.(a)

Sol. Bearing strength of rivet joint

(d×t×σ_bp )

σ_bp→ permissible bearing stress in rivet.