Answer:
a=42.86 mm
b=160.7 mm
Explanation:
Using the attached figure
Shearing strain=[tex]\frac {\sigma}{a}=\frac {\tau}{G}[/tex]
[tex]a=\frac {G\sigma}{\tau}=\frac {12 Mpa\times 0.005 m}{1.4 Mpa}=0.042857143 m\approx 42.86 mm[/tex]
The shearing stress will be given by
[tex]\tau=\frac {0.5P}{A}=\frac {P}{2bc}[/tex]
[tex]b=\frac {P}{2c\tau}=\frac {45000 N}{2\times 0.1\times 1.4\times 10^{6}}=0.160714286 m\approx 160.7 mm[/tex]
