Chapter 6, Problem 6.17P

Interpretation Introduction

Interpretation:

The thickness of the aluminum plate is 0.5cm and it can withstand the force of 50,000N without any plastic deformation. Yield strength of aluminum plate is 125MPa. Minimum width of the plate is to be determined.

Concept introduction:

Minimum width of the plate is determined by using the following formulas-

  $Stress(\sigma )=\frac{F}{A}$

  $A=\left(\frac{\pi }{4}{d}^2\right)$

Or

  $A=width\times thickness$