Chapter 4, Problem 1P

For the sets of given stresses on an element given in Table4−2, draw a complete Mohr’s circle, find the principal stresses and the maximum shear stress, and draw the principal stress element and the maximum shear stress element. Any stress components not shown are assumed to be zero

To determine

To draw: Mohr’s circle to determine the principal stresses and maximum shear stress. Also, draw the maximum shear stress element and principal stress element.

Answer to Problem 1P

A complete Mohr’s circle is given as below,

  

The maximum principal stress is $24.14\text{ ksi}$ and the minimum principal stress is $-4.14\text{ ksi}$ . The maximum shear stress is $14.14\text{ ksi}$ .

The principal stress element is given as,

  

The maximum shear stress element is given as,

  

Explanation of Solution

Given Information:

Write the stresses in the x-direction, y-direction, and shear stress in the x-y plane.

  $\begin{array}{c}{\sigma }_x=2\text{0 ksi}\\ {\sigma }_y=\text{0 ksi}\\ {\tau }_{xy}=1\text{0 ksi}\end{array}$

Calculate the center of Mohr’s circle.

  $\begin{array}{c}\left(a,0\right)=\left(\frac{{\sigma }_x+{\sigma }_y}{2},0\right)\\ =\left(\frac{20\text{ ksi}+0\text{ ksi}}{2},0\right)\\ =\left(10\text{ ksi},0\right)\end{array}$

Calculate the radius of Mohr’s circle.

  $\begin{array}{c}R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ =\sqrt{{\left(\frac{20\text{ ksi}-0\text{ ksi}}{2}\right)}^2+{\left(10\text{ ksi}\right)}^2}\\ =14.14\text{ ksi}\end{array}$

Draw the complete Mohr’s circle diagram.

  

Write the expression of principal stresses.

  ${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Substitute $20\text{ ksi}$ for ${\sigma }_x$ , $0\text{ ksi}$ for ${\sigma }_y$ , and $10\text{ ksi}$ for ${\tau }_{xy}$ to calculate the principal stresses.

  $\begin{array}{c}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ =\frac{20\text{ ksi}+0\text{ ksi}}{2}\pm \sqrt{{\left(\frac{20\text{ ksi}-0\text{ ksi}}{2}\right)}^2+{\left(10\text{ ksi}\right)}^2}\\ =10\pm 14.14\text{ ksi}\\ {\sigma }_1=24.14\text{ ksi}\\ {\sigma }_2=-4.14\text{ ksi}\end{array}$

Write the expression of maximum shear stress.

  ${\tau }_{\max }=\left|\frac{{\sigma }_1-{\sigma }_2}{2}\right|$

Substitute $24.14\text{ ksi}$ for ${\sigma }_1$ , and $-\text{4}\text{.14 ksi}$ for ${\sigma }_2$ to calculate the maximum shear stress.

  $\begin{array}{c}{\tau }_{\max }=\left|\frac{{\sigma }_1-{\sigma }_2}{2}\right|\\ =\left|\frac{24.14\text{ ksi}-\left(-4.14\text{ ksi}\right)}{2}\right|\\ =14.14\text{ ksi}\end{array}$

Draw the principal stress elements.

  

Draw the maximum shear stress element.