At a point in an elastic material under strain, there are normal stresses of 50 N/mm² and 30 N/mm² respectively at right angles to each other with a shearing stress of 25 N/mm².
Find the principle stresses the position of principal planes if
(a) 50 N/mm² and 30 N/mm² are both tensile.’
(b) 50 N/mm² is tensile and 30 N/mm² is compressive.
Find also the maximum shear stress and its plane in both the cases. [KU, June 2010]

Question

Accepted Answer

(a) [latex]\sigma_x=50 \mathrm{~N} / \mathrm{mm}^2, \sigma_y=30 \mathrm{~N} / \mathrm{mm}^2, au_{x y}=25 \mathrm{~N} / \mathrm{mm}^2[/latex]

[latex]\begin{aligned}\sigma_{1,2} &=\frac{1}{2}\left(\sigma_x+\sigma_y ight) \pm \frac{1}{2} \sqrt{\left(\sigma_x-\sigma_y ight)^2+4 au_{x y}^2} \&=\frac{1}{2}(50+30) \pm \frac{1}{2} \sqrt{(50-30)^2+4 imes 25^2} \&=40 \pm 26.926 \\sigma_1 &=66.926 \mathrm{~N} / \mathrm{mm}^2, \sigma_2=13.074 \mathrm{~N} / \mathrm{mm}^2 \ au_{\max } &=26.926 \mathrm{~N} / \mathrm{mm}^2\end{aligned}[/latex]

Principle planes are given by,

[latex]\begin{aligned} an 2 heta &=\frac{2 au_{x y}}{\sigma_x-\sigma_y}=\frac{2 imes 25}{50-30}=\frac{50}{20}=2.5 \2 heta &=68.2^{\circ} \ heta_1 &=34.1^{\circ}, heta_2=34.1^{\circ}+90^{\circ}=124.1^{\circ}\end{aligned}[/latex]

(b) [latex]\sigma_x=50 \mathrm{~N} / \mathrm{mm}^2, \sigma_y=-30 \mathrm{~N} / \mathrm{mm}^2, au_{x y}=25 \mathrm{~N} / \mathrm{mm}^2[/latex]

[latex]\begin{aligned}\sigma_{1,2} &=\frac{1}{2}(50-30) \pm \frac{1}{2} \sqrt{(50+30)^2+4 imes 25^2} \&=10 \pm 47.17 \\sigma_1 &=57.17 \mathrm{~N} / \mathrm{mm}^2, \sigma_2=-37.17 \mathrm{~N} / \mathrm{mm}^2 \ au_{\max } &=47.17 \mathrm{~N} / \mathrm{mm}^2 \ an 2 heta &=\frac{2 imes 25}{50+30}=0.625 \2 heta &=32^{\circ}, heta_1=16^{\circ}, heta_2=106^{\circ}\end{aligned}[/latex]

Question A2.14: At a point in an elastic material under strain, there are no...

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