Question

The 50-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed and concentrated torsional loadings shown where T = 140 N-m. 2 kNm 600 Nm B 0.8 m 0.6 m A Part A Determine the angle of twist at the free end A of the shaft due to these loadings. Use Gst = 75.0 GPa Express your answer to three significant figures and include appropriate units. μΑ ? 22 % Å O 0.303 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining

Answer 1

SOLUTION:

A solid shaft of Diameter D, is subjected with Distributed and concentrated torsion as shown in the Figure.

Given:

Diameter(D) = 50mm

TA = 140N.m
(cw) ;

  

G = 75 GPa

BY SUPERPOSITION METHOD:

Let, Twisting Angle , due to T(A) , T(C) and t, is $\theta _{a},\theta _{c}and\theta _{t}$

so,
ва TALAB IPG 140 * 1.4 (0.0504) * (75 * 109) 0.00425(CW)) 32
$\theta _{a}= \frac{T_{A}L_{AB}}{I_{P}G}=\frac{140*1.4}{\frac{\pi }{32}(0.050^4)*(75*10^9)}=0.00425(CW)$

SIMILARLY,

$\theta _{c}= \frac{T_{c}L_{AC}}{I_{P}G}=\frac{600*0.8}{\frac{\pi }{32}(0.050^4)*(75*10^9)}=0.01043(ACW)$

Now for, twist angle due to uniform distributed torsion,

$d\theta _{t}=\frac{t.x}{I_{P}.G}.dx$

$\theta _{t}=\int_{0}^{L_{AC}}\frac{t.x}{I_{P}.G}.dx$

$\theta _{t}=\int_{0}^{0.8}\frac{2000.x}{46019.42}.dx=0.01390 (CW)$

So, twist angle at free end,

$\theta=\theta _{a}-\theta _{c}+\theta _{t}$

$\theta=0.00425-0.01390-0.01043$

$\theta=0.00772 rad...$