Answer
$$D_{max} = 43.4 \space mm $$
Work Step by Step
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:
$\tau_{FAIL} = \frac {F_{MAX}}{A} = 55 \space MPa$
$\tau_{FAIL} = \frac{45 * 10^{3}}{ \pi * D_{MAX} * 6*10^{-3}} \geq 55 *10^6$
$\Rightarrow \space\space D_{MAX}\leq \frac{45*10^{3}}{\pi * 6*10^{-3} * 55 * 10^{6}} = 0.0434 \space m$
$\boxed{D_{MAX} = 43.4 \space mm } \space\space\leftarrow\space ANS$