Answer to Question #213045 in Mechanical Engineering for Evans

Question

A solid circular steel shaft is designed as shown in Figure P1. It consists of two

segments: AB has a diameter of 35 mm and BC has a diameter of 55 mm. If the

lengths of the sections AB and BC are 25 cm and 30 cm respectively and the

modulus of rigidity of steel is 77.2 GPa,

A. Draw the free body diagrams of the rod segments AB and BC.

B. Find the torques TAB and TBC of the sections AB and BC respectively.

C. Find the shearing stress in section AB.

D. Find the angle of twist of section AB.

E. Find the angle of twist of section BC.

F. Find the angle of twist of the point A relative to point C, leaving your answer

Answer 1

Part a

Part b

$\frac{T_{AB}}{J}=\frac{G \theta}{l}$

$\frac{T_{AB}}{\frac{\pi}{32}*35^2}=\frac{72.2*1.3*\frac{\pi}{180}}{250} \implies T_{AB}=0.78805 Nmm$

$\frac{T_{BC}}{\frac{\pi}{32}*55^2}=\frac{72.2*1.3*\frac{\pi}{180}}{300} \implies T_{BC}=1.6217 Nmm$

Part c

$τ_{AB} = \frac{F}{A}=\frac{0.78805}{\pi*17.5^2}=0.000819 N/mm^2$

Part d

$\frac{T_{AB}}{J}=\frac{G \theta}{l}$

$\frac{0.78805}{\frac{\pi}{32}*35^2}=\frac{72.2*\theta}{250} \implies \theta = 1.3^0$

Part e

$\frac{T_{BC}}{J}=\frac{G \theta}{l}$

$\frac{0.78805}{\frac{\pi}{32}*55^2}=\frac{72.2*\theta}{300} \implies \theta = 1.3^0$

Part f

$\frac{{AB}}{BC}= \frac{1.3}{1.3}=1$