# Torsion of Shafts

 Question 1
A cylindrical transmission shaft of length 1.5 m and diameter 100 mm is made of a linear elastic material with a shear modulus of 80 GPa. While operating at 500 rpm, the angle of twist across its length is found to be 0.5 degrees.
The power transmitted by the shaft at this speed is _______kW. (Rounded off to two decimal places)
Take $\pi= 3.14$.
 A 238.64 B 254.35 C 632.25 D 456.35
GATE ME 2023   Strength of Materials
Question 1 Explanation: Given: Length, $L=1.5 \mathrm{~m}$, Dia, $d=100 \mathrm{~mm}=0.1 \mathrm{~mm}$
Shear modulus, $\mathrm{G}=80 \mathrm{GPa}$
$N=500 \mathrm{rpm} \text {, }$
angle of twist,
$\theta=0.5^{\circ}$
$\theta=0.5 \times \frac{\pi}{180^{\circ}} \mathrm{rad}$

$\because$ As, We know,
$\frac{\mathrm{T}}{\mathrm{J}}=\frac{\mathrm{G} \theta}{\mathrm{L}}$
where $\mathrm{J}$ for cylinder solid shaft $=\frac{\pi \mathrm{d}^{4}}{32}$

Torque,
$T=\frac{G \theta \cdot J}{L}=\frac{\left(80 \times 10^{9}\right) \times\left(0.5 \times \frac{\pi}{180}\right)}{1.5} \times \frac{\pi}{32} \times(0.1)^{4} =4.56 \times 10^{3} \mathrm{~N}-\mathrm{m}$

Now, power,
$\quad P=T \times \omega=T \times \frac{2 \pi N}{60} =4.56 \times 10^{3} \times \frac{2 \pi \times 500}{60}$
$P =238.64 \mathrm{~kW}$
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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

There are 5 questions to complete.

### 10 thoughts on “Torsion of Shafts”

1. in question 24 torque is 1.0 Nm not 10 Nm. Please correct it

• In this topic only total 23 questions are there. So can you please verify your question number

• 2. Question 11, in explanation factors should be ((r4⁴-r3⁴)(r4⁴ + r3⁴)) in both numerator and denominator.

3. Question 21, torque is 1 Nm

4. Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=0 at x=-L or assuming Mz=M at x=0. But we actually get different values of ‘c’ in those different assumptions. Hence, it is proved that the question is wrong.

• Correction*

Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=M at x=0 since the expression is given to be valid at 0≤x≤L . But we actually get different value of ‘c’ in this assumption. Hence, it is proved that the question is wrong.

• I am really sorry.. the question is absolutely correct.. but correct your solution where it is written that c=2L but it would be c=-2L..

• 5. 