Chapter 2, Problem 2.1P

To determine

The shear and normal stress.

Answer to Problem 2.1P

So, the shear and normal stress are:

${\text{σ}}_{shear}=-433.01+250=-183.01$ lbf/ft² ${\text{σ}}_{normal}=2500\pm 683.01$ lbf/ft²

Explanation of Solution

Given information:

σ xx = 3000 lbf/ft²

σ yy = 2000lbf/ft²

σ xy = 500 lbf/ft²

angle Θ = 30⁰

We know that:

${\text{σ}}_{normal}=({\text{σ}}_{xx}+{\text{σ}}_{yy})/2\pm \left(({\text{σ}}_{xx}-{\text{σ}}_{yy}\right)/2cos2\theta +{\text{σ}}_{xy}sin2\theta )$

${\text{σ}}_{normal}=\frac{3000+2000}{2}\pm \left(\frac{3000-2000}{2}\times cos\left(2\times 30\right)+500\times sin\left(2\times 30\right)\right)$

${\text{σ}}_{normal}=2500\pm \left(250+433.01\right)$

${\text{σ}}_{normal}=2500\pm 683.01$ lbf/ft² ${\text{σ}}_{shear}=-({\text{σ}}_{xx}-{\text{σ}}_{yy})/2sin2\theta +{\text{σ}}_{xy}cos2\theta$

${\text{σ}}_{shear}=-\frac{3000-2000}{2}\times sin\left(2\times 30\right)+500cos\left(2\times 30\right)$

${\text{σ}}_{shear}=-433.01+250=-183.01$ lbf/ft²