Unishivaji Structural Mechanics - II May 2016 Solved Question Paper Q1 B)

In a two - dimensional problem the stresses at a Point are σ_x = 100 MPa tensile, σ_y =60 MPa both tensile. lf the principle stresses is limited to 150 MPa. Find the value of shear stress q. Also find the inclination of the principle plane and magnitude of the maximum shear.

12 - Marks

>>Given:
σ_x = 100 MPa = 100 N/mm² (Tensile);
σ_y = 60 MPa = 60 N/mm² (Tensile);
Principal stress limited to 150 MPa = σ_p1 = 150 N/mm²;
Value of q =? ; θ₁ =? ; θ₂ =? ; q_tmax =?

Solution:

$\sigma _{n1}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+q^{2}}$
$150=\left( \frac{100 + 60}{2} \right) + \sqrt{\left ( \frac{100-60}{2} \right )^{2}+q^{2}}$

$150=80 + \sqrt{400+q^{2}}$

$70= \sqrt{400+q^{2}}$

$4900= 400+q^{2}$

$q^{2}= 4500$

∴ $q= 67.082\:N/mm^{2}$

$tan2\theta_{1}=\frac{2q}{\left (\sigma _{x} - \sigma _{y}\right )}$
$tan2\theta_{1}=\frac{2 \times 67.082}{\left (100 - 60\right )}=3.3541$
$2\theta_{1}=tan^{-1}3.3541=73.39^{0}$

∴ $\theta_{1}=36.69^{0}$