Quiz: Civil Engineering

Exam: RSMSSB-JEn

Topic: Strength of Material

Each question carries 1 mark

Negative marking: 1/3 mark

Time: 10 Minutes

Q1. Which one of the following assumptions in the theory of pure torsion is false?

(a) The twist is uniform along the length

(b) The shaft is of uniform circular section throughout

(c) Cross-section plane before torsion remain plane after torsion

(d) all radii get twisted due to torsion

Q2. A circular shaft can transmit a torque of 13 kN-m. if the torque is reduced to 12 kN-m, then the maximum value of bending moment that can be applied to the shaft is-

(a) 1 kN – m

(b) 3 kN – m

(c) 5 kN – m

(d) 7 kN – m

Q3. The spindle of a machine tool is subjected to the following type of load-

(a) Torsional load

(b) Bending load

(c) Axial compressive load

(d) Axial tensile load

Q4. For any given power and permissible shear stress, the rotational speed of shaft and its diameter are correlated by the expression:

(a) ND^3=constant

(b) ND^2=constant

(c) ND = constant

(d) √ND = constant

Q5. A solid shaft transmits 44 kW power at 700 rps. Calculate the torque produced (in Nm)

(a) 60

(b) 100

(c) 600

(d) 10

Q6. The outside diameter of a hollow shaft is twice it’s inside diameter. The ratio of its torque carrying capacity to that of a that of a solid shaft of the same material and the same outside diameter is _____________.

(a) 15/16

(b) 3/4

(c) 1/2

(d) 1/16

Q7. The safe twisting moment for a compound shaft is equal to the ____________.

(a) maximum calculated value

(b) minimum calculated value

(c) mean value

(d) extreme value

Q8. Two solid shafts are made of same material and have their diameter D and D/2. The ratio of strength of bigger shaft to smaller one in torsion is

(a) 4

(b) 2

(c) 8

(d) 16

Q9. Two shafts A and B are made of the same material. The diameter of shaft B is twice that of shaft A. the ratio of power which can be transmitted by shaft A to that of shaft B is _________.

(a) 1/2

(b) 1/4

(c) 1/8

(d) 1/16

Q10. If a body is transmitting torque T Nm at N rpm, then horsepower (watts) transmitted will be

(a) TN

(b) TN/75

(c) TN/4500

(d) 2πNT/ (750×60)

SOLUTION

S1. Ans.(a)

Sol. we know that, the torsion equation

T/J=τ/r=Gθ/L

θ=TL/GJ

θ α L

Twist is varying with the length linearly so it is not uniform throughout the length.

S2. Ans.(c)

Sol. for compensate the torque addition of bending moment is required

T_e=√(T^2+m^2 )

13=√((12)^2+m^2 )

169=12^2+m^2

169-144=m^2

m=5.kNm

S3. Ans.(a)

Sol. The spindle of a machine tool is subjected to torsional load. the spindle undergoes rotation, loading to torsional stress.

S4. Ans.(a)

Sol. we know that, the power is

P = T.W = 2πNT/60,000 Kw——(1)

And torsional shear stress τ=16T/(πD^3 )

T=(πD^3)/16 τ—–(2)

From (1) and (2)

P=2πN/60,000×((πD^3)/16.τ)

Everything is constant except N & d for given power (P) and shear stress (τ)

∵ND^3=constant

S5. Ans.(d)

Sol. we know that, power is

P=2πNT if N=rps

P=2×22/7×700×T

44000=2×22×100×T

T=10 N.M

S6. Ans.(a)

Sol.

T_S/T_H =(τ_max×π/16 d^3 )_S/(τ_max×π/16 d^3 (1-k^4 ))_H {∵K=d_i/d_O }

=1/(1±k^4 )

Given that K=4

=1/(1-1/16)=16/15

T_S/T_H =(16/15)

S7. Ans.(b)

Sol. the safe twisting moment for a compound shaft is equal t the minimum calculated value let there are two torques T_1 & T_2 (T_1> T_2 ) here, we are assuming T_2 is lesser than T_1. So, if the material will fail at T_2 then, there is no mean of T_1 because material will not able to reach at T_1 because failure occurs at T_2. So, the safe limit is T_2 i.e. minimum calculated value.

S8. Ans.(c)

Sol. Given that shafts are diameter D and D/2

Let bigger shaft strength T_b

τ_b=(16T_b)/πDׇ ——-(1)

Smaller shaft strength ‘τ_s^’

τ_s=(16T_S)/(π(D/2)^3 )=8×(16T_S)/(πD^3 )—–(2)

τ_b/τ_s =(16T_b \/πD^3)/(8×16T_S \/πD^3 )

Material is same b (∵τ_b=τ_s )

8.(16T_S)/(πD^3 )=(16T_b)/(πD^3 )

▭(T_b=8.T_S )

S9. Ans.(c)

Sol. Torque transmitted torque shaft

T=π/16 D^3 τ

T α D^3

Power,P= Tω

P α T α D^3

P_A/P_B =(D_a^3)/(D_B^3 )=[1/2]^3=1/8

S10. Ans.(d)

Sol. power = Tω

P=T(2πN/60)=2πNT/60 Watt

in horse power=2πNT/(750×60)