## Mechanics of Materials, 7th Edition

$$D_{max} = 43.4 \space mm$$
Solution: 1)Area of Shear Force Distribution: Equal to the surface area on the inside of the punched hole: $A = \pi * D * t$ , where D is hole punch Diameter and t is thickness of polysterene Sheet 2)Shear Stress required for material failure: $\tau_{FAIL} = \frac {F_{MAX}}{A} = 55 \space MPa$ $\tau_{FAIL} = \frac{45 * 10^{3}}{ \pi * D_{MAX} * 6*10^{-3}} \geq 55 *10^6$ $\Rightarrow \space\space D_{MAX}\leq \frac{45*10^{3}}{\pi * 6*10^{-3} * 55 * 10^{6}} = 0.0434 \space m$ $\boxed{D_{MAX} = 43.4 \space mm } \space\space\leftarrow\space ANS$