Chapter 5, Problem 1P

A link in a mechanism is made from a round bar having a diameter of 10.0 mm. It is subjected to a tensile force that varies from 3500 to 500 N in a cyclical fashion as the mechanism runs.

To determine

The minimum stress, maximum stress, mean stress, stress ratio, and alternating stress.

Answer to Problem 1P

Maximum stress is 44.6 MPa

Minimum stress is 6.37 MPa

Mean stress is 25.5 MPa

Alternating stress = 19.11 MPa

Stress ratio is 0.143

Explanation of Solution

Given Information:

Diameter of the bar, D = 10 mm

Maximum tensile force = $P_{\max }=3500N$

Minimum tensile force = $P_{\min }=500N$

Area of the bar, A

  $\begin{array}{c}A=\frac{\pi }{4}D\\ =\frac{\pi }{4}\left({10}^2\right)\\ =78.54{\text{ mm}}^2\end{array}$

Stress in the bar is defined as,

  $Stress=\frac{Load}{Area}$

For maximum stress,

  $\begin{array}{c}{\sigma }_{\max }=\frac{P_{\max }}{A}\\ =\frac{3500}{78.54}\\ =44.6{\text{ N/mm}}^2\\ =44.6\text{ MPa}\end{array}$

For minimum stress,

  $\begin{array}{c}{\sigma }_{\min }=\frac{P_{\min }}{A}\\ =\frac{500}{78.54}\\ =6.37{\text{ N/mm}}^2\\ =6.37\text{ MPa}\end{array}$

Mean stress, ${\sigma }_m$

  $\begin{array}{c}{\sigma }_m=\frac{{\sigma }_{\max }+{\sigma }_{\min }}{2}\\ =\frac{44.6+6.37}{2}\\ =25.5\text{ MPa}\end{array}$

Alternating stress, ${\sigma }_a$

  $\begin{array}{c}{\sigma }_a=\frac{{\sigma }_{\max }-{\sigma }_{\min }}{2}\\ =\frac{44.6-6.37}{2}\\ =19.11\text{ MPa}\end{array}$

Stress Ratio, R

  $\begin{array}{c}R=\frac{{\sigma }_{\min }}{{\sigma }_{\max }}\\ =\frac{6.37}{44.6}\\ =0.143\end{array}$

Stress vs time plot