Question

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor principal stresses.

Answer 1

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

• $\sigma_x$ is the normal stress which is 1750 psf
• $\sigma_y$ is 0
• $\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$