Answer:
Explanation:
Given
Ultimate shear strength [tex]\tau =3\times 10^9\ Pa[/tex]
thickness of steel Plate [tex]t=1.60\ cm[/tex]
Cross-sectional [tex]A=1.20\times 10^2\ cm^2[/tex]
suppose d is the diameter of cross-section
[tex]\frac{\pi }{4}d^2=1.20\times 10^2[/tex]
[tex]d=12.359\ cm[/tex]
Shear area of circular cross-section
[tex]A_s=\pi dt[/tex]
[tex]A_s=\pi \times 12.359\times 1.6[/tex]
[tex]A_s=62.13\ cm^2[/tex]
Shear Force [tex]F_s=A_s\times \tau[/tex]
[tex]F_s=62.13\times 10^2\times 3000[/tex]
[tex]F_s=18.639\ MN[/tex]