Torsion of Shafts and Pressure Vessels

Question 1
A solid circular torsional member OPQ is subjected to torsional moments as shown in the figure (not to scale). The yield shear strength of the constituent material is 160 MPa.

The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is ________
A
15.3
B
12.6
C
18.2
D
11.8
GATE CE 2021 SET-2   Solid Mechanics
Question 1 Explanation: 


\begin{aligned} \tau_{\max _{O P}}=\frac{16 T_{O P}}{\pi d_{O P}^{3}}=\frac{16 \times 3 \times 10^{3}}{\pi \times 100^{3}}=15.27 \mathrm{~N} / \mathrm{mm}^{2} \\ \tau_{\max \mathrm{PQ}}=\frac{16 T_{P Q}}{\pi d_{P Q}^{3}}=\frac{16 \times 1 \times 10^{6}}{\pi \times 80^{3}}=9.94 \mathrm{~N} / \mathrm{mm}^{2}\\ \tau_{\max }=9.94 \mathrm{~N} / \mathrm{mm}^{2} \end{aligned}
Absolute max shear stress = 15.27 \mathrm{~N} / \mathrm{mm}^{2}
Question 2
A plane frame shown in the figure has linear elastic springs at node H. The spring constants are k_x=k_y=5 \times 10^5kN/m and k_\theta =3 \times 10^5kNm/rad.

For the externally applied moment of 30 kNm at node F, the rotation (in degrees, round off to 3 decimals) observed in the rotational spring at node H is_____
A
0.001
B
0.003
C
0.005
D
0.009
GATE CE 2019 SET-2   Solid Mechanics
Question 2 Explanation: 
No moment can be taken by segment FE.
\therefore \;\;M_{FE}=0

\begin{aligned} M_\theta &=k_\theta \times \theta \\ 30kNm&=3 \times 10^5 \times \theta \\ \theta &=1 \times 10^{-4} \; \text{radians} \\ &=0.0057^{\circ} \end{aligned}
Question 3
A closed thin-walled tube has thickness, t, mean enclosed area within the boundary of the centerline of tube's thickness, A_m, and shear stress, \tau. Torsional moment of resistance, T, of the section would be
A
0.5 \tau A_m t
B
\tau A_m t
C
2 \tau A_m t
D
4 \tau A_m t
GATE CE 2019 SET-2   Solid Mechanics
Question 3 Explanation: 


Shear Stress,
\begin{aligned} \tau &=\frac{T}{J}R=\frac{T}{2 \pi R^3 t}R \\ \tau &=\frac{T}{2 \pi R^2 t}=\frac{T}{2 A_m t} \\ T &=2\tau A_mt \end{aligned}
Question 4
A solid circular beam with radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The shear stress component \tau_{xy} at Point 'M' in the cross-section of the beam at a distance of 1 m from the fixed end is
A
0.0 MPa
B
0.51 MPa
C
0.815 MPa
D
2.0 MPa
GATE CE 2018 SET-1   Solid Mechanics
Question 4 Explanation: 


The only non-zero stresses are \tau_{\theta z}=\tau_{z \theta}=\tau
if \theta is 90^{\circ} then \theta=y
\begin{aligned} \text{Hence}\quad\tau_{z y} &=\tau_{y z}=\tau_{\max } \\ &=16 T / \pi \mathrm{d}^{3}=0.815 \mathrm{MPa} \end{aligned}
But in rest of the planes shear stresses are zero, hence, \tau_{x y}=\tau_{y x}=0
Question 5
A hollow circular shaft has an outer diameter of 100 mm and inner diameter of 50 mm. If the allowable shear stress is 125 MPa, the maximum torque (in kN-m) that the shaft can resist is____
A
2
B
231
C
230.9
D
23
GATE CE 2017 SET-2   Solid Mechanics
Question 5 Explanation: 
\begin{aligned} \therefore \quad \frac{T}{I_{p}}&=\frac{\tau}{r} \\ \Rightarrow \quad \tau&=\frac{T}{Z_{p}}=\frac{T}{\frac{\pi\left(D^{4}-d^{4}\right)}{16 \times D}}=\frac{16 T}{\frac{\pi}{D}\left(D^{4}-d^{4}\right)} \\ \Rightarrow \quad \tau&=\frac{16 T}{\pi D^{3}\left[1-\left(\frac{d}{D}\right)^{4}\right]} \\ \Rightarrow \quad 125&=\frac{16 T}{\pi(100)^{3}\left[1-\left(\frac{50}{100}\right)^{4}\right]} \\ \Rightarrow \quad T&=23 \mathrm{kNm} \end{aligned}
Question 6
A solid circular shaft of diameter d and length L is fixed at one end free at the other end. A torque T is applied at the free end. The shear modulus of the material is G. The angle of twist at the free end is
A
\frac{16TL}{\pi d^{4}G}
B
\frac{32TL}{\pi d^{4}G}
C
\frac{64TL}{\pi d^{4}G}
D
\frac{128TL}{\pi d^{4}G}
GATE CE 2010   Solid Mechanics
Question 6 Explanation: 
Angle of twist is given by,
\begin{aligned} \theta &=\frac{T L}{G I_{P}} \text { but } I_{P}=\frac{\pi d^{4}}{32} \\ \Rightarrow \quad \theta&=\frac{T L}{G \times \frac{\pi d^{4}}{32}}=\frac{32 T L}{\pi d^{4} G} \end{aligned}
Question 7
A hollow circular shaft has an outer diameter of 100 mm and a wall thickness of 25 mm. The allowable shear stress in the shaft is 125 MPa. The maximum torque the shaft can transmit is
A
46 kNm
B
24.5 kNm
C
23 kNm
D
11.5 kNm
GATE CE 2009   Solid Mechanics
Question 7 Explanation: 
\begin{aligned} \therefore\quad \frac{T}{I_{P}} &=\frac{\tau}{R} \\ \Rightarrow \quad T &=\frac{\tau}{R} \times I_{P} \\ &=125 \times \frac{\pi}{32} \times\left(100^{4}-50^{4}\right) \times \frac{2}{100} \times 10^{-6} \\ &=23.00 \mathrm{kN}-\mathrm{m} \end{aligned}
Question 8
A thin walled cylindrical pressure vessel having a radius of 0.5 m and wall thickness of 25 mm is subjected to an internal pressure of 700 kPa. The hoop stress developed is
A
14MPa
B
1.4MPa
C
0.14MPa
D
0.014MPa
GATE CE 2009   Solid Mechanics
Question 8 Explanation: 
\begin{aligned} \text { Hoop stress } &=\frac{p d}{2 t}=\frac{700 \times 10^{3} \times 2 \times 0.5}{2 \times 25 \times 10^{-3}} \\ &=14 \times 10^{6} \mathrm{Pa}=14 \mathrm{MPa} \end{aligned}
Question 9
The maximum shear stress in a solid shaft of circular cross-section having diameter d subjected to a subject to a torque T is \tau. If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be
A
2\tau
B
\tau
C
\tau/2
D
\tau/4
GATE CE 2008   Solid Mechanics
Question 9 Explanation: 
\begin{aligned} \therefore \quad \frac{\tau}{R}&=\frac{T}{I_{P}} \\ \Rightarrow \quad \frac{\tau_{1}}{\tau_{2}}&=\frac{R_{1}}{R_{2}} \times \frac{T_{1}}{T_{2}} \times \frac{I_{P_{2}}}{I_{P_{1}}} \\ \Rightarrow \quad \frac{\tau_{1}}{\tau_{2}}&=\frac{R_{1}}{2 R_{1}} \times \frac{T_{1}}{4 T_{1}} \times\left(\frac{2 R_{1}}{R_{1}}\right)^{4} \\ \Rightarrow \quad \frac{\tau_{1}}{\tau_{2}}&=\frac{1}{2} \times \frac{1}{4} \times 16 \\ \Rightarrow\quad\tau_{2}&=\frac{\tau_{1}}{2} \\ \Rightarrow\quad\tau_{2}&=\frac{\tau}{2} \end{aligned}
Question 10
The maximum and minimum shear stresses in a hollow circular shaft of outer diameter 20 mm and thickness 2 mm, subjected to a torque of 92.7 Nm will be
A
59 MPa and 47.2 MPa
B
100 MPa and 80 MPa
C
118 MPa and 160 MPa
D
200 MPa and 160 MPa
GATE CE 2007   Solid Mechanics
Question 10 Explanation: 
\begin{aligned} \because \quad \frac{f_{s}}{R} &=\frac{T}{J} \\ \text { Here, } \quad J &=\frac{\pi}{32}\left(20^{4}-16^{4}\right) \mathrm{mm}^{4} \\ R_{1} &=\frac{20}{2}=10 \mathrm{mm} \\ T &=92.7 \mathrm{N}-\mathrm{m} ; \\ R_{2} &=\frac{16}{2}=8 \mathrm{mm} \\ f_{s_{1}} &=\frac{T R_{1}}{J}=\frac{92.7 \times 10^{3} \times 10}{(\pi / 32) \times\left(20^{4}-16^{4}\right)} \\ &=99.96 \mathrm{MPa} \approx 100 \mathrm{MPa} \\ \text { and } \quad f_{{9}_{2}} &=\frac{T R_{2}}{J}=\frac{92.7 \times 10^{3} \times 8}{(\pi / 32) \times\left(20^{4}-16^{4}\right)} \\ &=79.96 \mathrm{MPa} \simeq 80 \mathrm{MPa} \end{aligned}
There are 10 questions to complete.

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