A circular hole of diameter d = 10cm is to be punched out of a...

Shear Area,& ${A}_{s} = 2\pir \times t$

where,~

$2\pir \to$the~perimeter of the hole and

$t \to$the~thickness of the hole.

$⇒ A = 2\pi\left(\dfrac{0 . 1}{2}\right)t = 0 . 1\pit$.

Force = shearstress \times Area$</p><p>$⇒ F = {σ}_{s} \times {A}_{s}$</p><p>$⇒ F = 2 \times {10}^{8} \times 0 . 1\pit$</p><p>$⇒ F = 2 \times {10}^{7}\pit$.</p></blockquote><blockquote><p>$compressivestress = \dfrac{Force}{Area}$</p><p>$⇒ {σ}_{c} = \dfrac{F}{\pi\left(\dfrac{d}{2}\right)2}$</p><p>$⇒ 4 \times {10}^{8} = \dfrac{2 \times {10}^{7}\pit}{\pi\left(\dfrac{0 . 1}{2}\right)2}$</p><p>$⇒ t = \dfrac{4 \times {10}^{8}}{8 \times {10}^{9}} = \dfrac{1}{20} = 0 . 05m$</p><p>$∴ t = 5cm.