**Quiz: Civil Engineering **

**Exam: GATE**

**Topic: Miscellaneous**

**Each question carries 2 marks**

**Negative marking: 1/3 mark**

**Time: 12 Minutes**

Q1. A fine – grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when water content is between 15% and 28%. The soil behaves fluid-like when the water content is more than 40%. The ‘Activity’ of the soil is

(a) 3.33

(b) 0.42

(c) 0.30

(d) 0.20

Q2. A mild steel specimen is under uniaxial tensile stress. Young’s modulus and yield stress for mild steel are 2 × 10^5 MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is

(a) 156 N-mm/mm³

(b) 15.6 N-mm/mm³

(c) 1.56 N-mm/mm³

(d) 0.156 N-mm/mm³

Q3. The maximum super elevation to be provided on a road curve 1 in 15. If the rate of change of super-elevation is specified as 1 in 120 and the road width is 8m, then the minimum length of the transition curve on each end will be –

(a) 64 m.

(b) 80 m.

(c) 100 m.

(d) 120 m.

Q4. The water flow fully through the rectangular channel of latrel dimensions 5m × 4m. what is the velocity of flow (m/sec) through the channel, if the slope of energy line and Chezy’s constant as 0.0006 and 90 respectively?

(a) 4.6

(b) 2.7

(c) 2.2

(d) 1.5

Q5. The appropriate expression in assessing development length is

(a) L_d=(ϕσ_bc)/τ_bd

(b) L_d=σ_S/(4 τ_bd )

(c) L_d=(ϕσ_s)/〖8 τ〗_bd

(d) L_d=(ϕσ_s)/〖4 τ〗_bd

Solutions

S1. Ans.(c)

Sol. Given,

w_L=40%

w_P=28%

Activity (A_T )= ?

% Clay fraction finer than 2μ = 40%

〖Plasticity index (I〗_P) = w_L-w_P

= 40 – 28

= 12%

A_T=I_P/(% Clay fraction finer than 2 μ)

▭(A_T=12/40=0.3)

S2. Ans.(d)

Sol. Given;

E = 2 × 10^5 MPa

σ_y = 250 MPa

U = ?

Strain energy per unit volume is given by →

U=1/2 (σ_y )^2/E

=1/2×(250)^2/(2×10^5 )

=0.156 N-mm\/mm^3.

S3. Ans.(a)

Sol. Given,

Super elevation (e) = 1 in 15

Rate of change of super elevation (1 in N) = (1 in 120)

Width of road (w) = 8m.

Now,

Length of transition curve based on the rate of change of super elevation is

L=eNw

=1/15×120×8

▭(L=64m.)

S4. Ans.(b)

Sol. velocity of flow (v) = C√RS

C = Chezy’s constant = 90

R = hydraulic radius = A/P

=((5×4))/((5+2×4) )

▭(R=20/13)

S=Slope=0.0006

V=90√(20/13×0.0006)

▭(V=2.7 m\/sec.)

S5. Ans.(d)

Sol. As per IS : 456 – 2000, the development length (L_d) is given by

▭(L_d=(ϕσ_s)/(4τ_bd ))

Where ϕ = diameter of bar in mm.

τ_bd = Bond stress in N/mm²

σ_s = stress in bar