In a bi-axial stress system the stresses along the two directions are $\sigma_{x}= 60 N/mm^{2}$ (tensile) and $\sigma_{y}=40 N/mm^{2}$ (Compressive).Find the max strain.Take $E = 200 kN/mm^{2}$ , m=4.

648views

written 5.6 years ago by

teamques10 ★ 69k

$\begin{aligned} \mathrm{e}_{\mathrm{x}} &=\left(\frac{\sigma_{\mathrm{x}}}{\mathrm{E}}\right)-\left(\mu \frac{\sigma_{\mathrm{y}}}{\mathrm{E}}\right) \ &=\frac{1}{\mathrm{E}}\left(\sigma_{\mathrm{x}}-\mu \times \sigma_{\mathrm{y}}\right) \end{aligned}$

$\begin{aligned} &=\frac{1}{200 \times 10^{3}}(60+0.25 \times 40) \ e_{x} &=3.5 \times 10^{4} \end{aligned}$

$\begin{aligned} \mathrm{e}_{\mathrm{y}} &=\left(\frac{\sigma_{\mathrm{y}}}{\mathrm{E}}\right)-\left(\mu \frac{\sigma_{\mathrm{x}}}{\mathrm{E}}\right) \ &=\frac{1}{\mathrm{E}}\left(\sigma_{\mathrm{y}}-\mu \times \sigma_{\mathrm{x}}\right) \end{aligned}$

$\begin{aligned} &=\frac{1}{200 \times 10^{3}}(-40-0.25 \times 60) \ e_{y} &=-2.75 \times 10^{4} \end{aligned}$

Maximum strain is $e_{x} =3.5 \times 10^{4}$

Create a free account to keep reading this post.

and 5 others joined a min ago.

ADD COMMENT EDIT