Torsion of Shafts

Question 1
A cylindrical rod of diameter 10 mm and length 1.0 m is fixed at one end. The other end is twisted by an angle of 10^{\circ} by applying a torque. If the maximum shear strain in the rod is p \times 10^{-3}, then p is equal to ______ (round off to two decimal places).
A
0.25
B
0.5
C
0.8
D
0.6
GATE ME 2019 SET-1   Strength of Materials
Question 1 Explanation: 


\begin{aligned} &\mathrm{d}=10 \mathrm{mm}\\ &\theta=10^{\circ}=10^{\circ} \times \frac{\pi}{180} \text { radian }\\ &\mathrm{R}=\frac{\mathrm{d}}{2}=5 \mathrm{mm}\\ &\mathrm{L}=1 \mathrm{m}\\ &\phi_{\max }=P \times 10^{-3}\\ &\text { from torsion equation, } \\ \frac{\mathrm{T}}{\mathrm{J}}&=\frac{\tau_{\max }}{\mathrm{R}}=\frac{\mathrm{G} \theta}{\ell}\\ &\Rightarrow \frac{\tau_{\max }}{G}=\frac{R \theta}{\ell}\\ &\phi_{\max }=\frac{R \theta}{\ell} \\ P \times 10^{-3}&=\frac{5 \times 10^{\circ} \times \frac{\pi}{180}}{1000}=0.8726 \times 10^{-3} \\ \therefore \;\;P&=0.8726\end{aligned}
Question 2
A bar of circular cross section is clamped at ends P and Q as shown in the figure. A torsional moment T=150 Nm is applied at a distance of 100 mm from end P. The torsional reactions \left ( T_{P}, T_{Q} \right ) in Nm at the ends P and Q respectively are
A
\left (50,100\right)
B
\left (75,75\right)
C
\left (100,50 \right )
D
\left (120,30 \right )
GATE ME 2018 SET-2   Strength of Materials
Question 2 Explanation: 


\begin{aligned} T_{P}+T_{Q} &=T \ldots(i)\\ Q_{P R}+Q_{R Q} &=0 \\ \frac{T_{P} \cdot L}{G I_{P}}+\frac{\left(T_{P}-T\right) 2 L}{G I_{P}} &=0 \\ T_{P}+2 T_{P} &=2 T \\ T_{P} &=\frac{2 T}{3}=100 \mathrm{Nm}\\ \text{From equation (i)}\\ T_{Q}&=\frac{T}{3}=50 \mathrm{N}-\mathrm{m} \end{aligned}
Question 3
A hollow circular shaft of inner radius 10 mm, outer radius 20 mm and length 1 m is to be used as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is ________ (correct to two decimal places).
A
12.36
B
18.26
C
35.34
D
48.22
GATE ME 2018 SET-2   Strength of Materials
Question 3 Explanation: 
\qquad \begin{aligned} \text { Torsional stiffness }&=\frac{G I_{P}}{L} \\ &=\frac{150 \times 10^{9} \times \frac{\pi}{32}\left[0.04^{4}-0.02^{4}\right]}{1} \\ &=35343 \mathrm{Nm} / \mathrm{rad}=35.343 \mathrm{kNm} / \mathrm{rad} \end{aligned}
Question 4
For an Oldham coupling used between two shafts, which among the following statements are correct?
I. Torsional load is transferred along shaft axis.
II. A velocity ratio of 1:2 between shafts is obtained without using gears.
III. Bending load is transferred transverse to shaft axis.
IV. Rotation is transferred along shaft axis.
A
I and III
B
I and IV
C
II and III
D
II and IV
GATE ME 2018 SET-1   Strength of Materials
Question 4 Explanation: 
Oldham coupling is used to connect two shafts which are not on the same axis means they are not aligned to the same axis
So,
(1) Torsional load is transferred along shaft axis as both shafts are rotating member.
So, that statement is correct.
(2) A velocity ratio 1:1 between shafts is obtained using gears.

So, this statement is wrong
(3) Bending load is not transferred transverse to shaft axis as there is no transverse load.
(4) Rotation is transferred along shaft axis
So, this statement is correct
Question 5
A motor is driving a solid circular steel shaft transmits 40kW of power at 500 rpm. If the diameter of the shaft is 40 mm, The maximum shear stress in the shaft is ____MPa.
A
60.7
B
50.5
C
55.25
D
51.36
GATE ME 2017 SET-1   Strength of Materials
Question 5 Explanation: 
\begin{aligned} T &=\frac{P \times 60 \times 10^{6}}{2 \pi N}=\frac{40 \times 60 \times 10^{6}}{2 \pi(500)} \\ &=763943.7268 \mathrm{N}-\mathrm{mm} \\ \tau_{\max } &=\frac{16 T}{\pi d^{3}}=\frac{16 \times 763943.7268}{\pi(40)^{3}} \\ &=60.792 \mathrm{MPa} \end{aligned}
Question 6
Two circular shafts made of same material, one solid (S) and one hollow (H), have the same length and polar moment of inertia. Both are subjected to same torque. Here, \theta_{s} is the twist and \tau_{s} is the maximum shear stress in the solid shaft, whereas \theta_{H} is the twist and \tau_{H} is the maximum shear stress in the hollow shaft. Which one of the following is TRUE?
A
\theta_{s}=\theta_{h} and \tau_{s}=\tau_{H}
B
\theta_{s} \gt \theta_{h} and \tau_{s} \gt \tau_{H}
C
\theta_{s}\lt\theta_{h} and \tau_{s}\lt\tau_{H}
D
\theta_{s}=\theta_{h} and \tau_{s}\lt\tau_{H}
GATE ME 2016 SET-3   Strength of Materials
Question 6 Explanation: 
\begin{aligned} L_s=L_H, & T_s=T_H=T,\\ I_{ps} &=I_{pH}\\ \therefore \; \frac{\pi}{32}d^4&=\frac{\pi}{32}(d_o^4-d_i^4)\\ \therefore \; d_o & \gt d\\ \therefore \;r_o &\ gt r\\ \text{Angle of twist}\\ \theta &=\frac{TL}{GI_p}\\ \therefore \; \theta _s&=\theta _H\\ \tau _{max_s}&=\frac{I}{I_p}r\\ \tau _{max_H}&=\frac{I}{I_p}r_o\\ \therefore \; \tau _{max_H}& \gt \tau _{max_s} \end{aligned}
Question 7
The cross-sections of two solid bars made of the same material are shown in the figure. The square cross-section has flexural (bending) rigidity I1, while the circular cross-section has flexural rigidity I2. Both sections have the same cross-sectional area. The ratio I1/I2 is

A
1/\pi
B
2/\pi
C
\pi/3
D
\pi/6
GATE ME 2016 SET-3   Strength of Materials
Question 7 Explanation: 


\begin{aligned} \therefore \quad b^{2}&=\frac{\pi}{4} d^{2} \\ \Rightarrow \quad \frac{b^{2}}{d^{2}}&=\frac{\pi}{4} \\ I_{1} &=\frac{b^{4}}{12} \\ I_{2} &=\frac{\pi d^{4}}{64} \\ \therefore \qquad \frac{I_{1}}{I_{2}} &=\frac{b^{4}}{12} \times \frac{64}{\pi d^{4}}=\frac{16}{3 \pi} \frac{b^{4}}{d^{4}} \\ &=\frac{16}{3 \pi} \cdot \frac{\pi^{2}}{16}=\frac{\pi}{3} \end{aligned}
Question 8
A simply supported beam of length 2L is subjected to a moment M at the mid-point x = 0 as shown in the figure. The deflection in the domain o\leq x\leq L is given by
w=\frac{-Mx}{12EIL}(L-x)(x+c)
where E is the Young's modulus, I is the area moment of inertia and c is a constant (to be determined)

The slope at the center x = 0 is
A
ML/(2EI)
B
ML/(3EI)
C
ML/(6EI)
D
ML/(12EI)
GATE ME 2016 SET-2   Strength of Materials
Question 8 Explanation: 


Equation of deflection
\begin{aligned} W &=\frac{-M}{12 E I L}\left[L x^{2}-x^{3}+L c x-C x^{2}\right] \\ \frac{d W}{d x} &=\frac{-M}{12 E I L}\left[2 L x-3 x^{2}+L C-2 C x\right] \\ \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 E I L}[2 L-6 x-2 C] \\ \therefore \quad E I \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 L}[2 L-6 x-2 C] \\ \text { now } \quad E I \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 L}[2 L-6 x-2 C]\\ \text{at }\quad x&=L, M_{x}=0 \\ \therefore \quad C&=2 L\\ Slope\\ \frac{d W}{d x} &=\frac{-M}{12 E I L}\left[2 L x-3 x^{2}+2 L^{2}-4 L x\right] \\ \therefore \quad \left|\frac{d W}{d x}\right|_{x=0} &=\frac{-M L}{6 E I} \end{aligned}
Question 9
A rigid horizontal rod of length 2L is fixed to a circular cylinder of radius R as shown in the figure. Vertical forces of magnitude P are applied at the two ends as shown in the figure. The shear modulus for the cylinder is G and the Young's modulus is E.

The vertical deflection at point A is
A
PL^{3}/(\pi R^{4}G)
B
PL^{3}/(\pi R^{4}E)
C
2PL^{3}/(\pi R^{4}E)
D
4PL^{3}/(\pi R^{4}G)
GATE ME 2016 SET-2   Strength of Materials
Question 9 Explanation: 


Angle of twist \theta
\theta =\frac{(2PL)L}{GI_p}=\frac{2PL^2}{G\frac{\pi}{2}R^4}=\frac{4PL^2}{\pi GR^4}
\therefore \; Vertical deflection of A
\Delta _A=\theta L=\frac{4PL^3}{\pi G R^4}
Question 10
A machine element XY, fixed at end X, is subjected to an axial load P, transverse load F, and a twisting moment T at its free end Y. The most critical point from the strength point of view is
A
a point on the circumference at location Y
B
a point at the center at location Y
C
a point on the circumference at location X
D
a point at the center at location X
GATE ME 2016 SET-2   Strength of Materials
Question 10 Explanation: 
Both bending stress and shear stress (due to torque) will be max at the circumference. The bending stress will be maximum at x. Hence, critical section will be at a point on the circumference at location x.
There are 10 questions to complete.

5 thoughts on “Torsion of Shafts”

Leave a Comment

Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.