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A plane element in the body is subjected to the stresses as shown in fig.

A plane element in the body is subjected to the stresses as shown in fig. Determine 1) The magnitude of principle stresses and respective principal planes. 2) Maximum shear stresses and the plane on which they occur. Sketch the stresses on properly oriented planes. Solve analytically or graphically.

fig.

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Solving by Analytical Method:-

1 2

Sketch 3

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enter image description here

Data:

$P_x = 80 \ N/mm^2$ (tensile)

$P_y = -20 \ N/mm^2$ (Compressive)

$\zeta = 25 \ N/mm^2$

A] Magnitude of principal stresses:

Let $P_1$ and $P_2$ be the principal stresses.

$P_1, \ P_2 = \frac{Px + Py}{2} \pm \sqrt{ (\frac{Px – Py}{2})^2 + \zeta^2}$

$= \frac{80 – 20}{2} \pm \sqrt{ (\frac{80 – (-20)}{2})^2 + 25^2}$

$P_1 , P_2 = 30 \pm \sqrt{50^2 + 25^2}$

$= 30 \pm 25\sqrt{5}$

$P_1 = 85.90 \ Nmm^2$ (tensile) $\rightarrow \ 30 + 25 \sqrt{5}$

$P_2 = -25.90 \ N/mm^2$ (Compressive) $\rightarrow \ 30 – 25 \sqrt{5}$

Now, the inclination of the principal planes is

$tan \ 2 \ \theta_1 = \frac{2 \zeta}{Px – Py}$

$tan \ 2 \ \theta_1 = \frac{2 \times 25}{80 – (-20)} = 0.5$

$\therefore \ \theta_1 = 13°16’$

$\theta_2 = \theta_1 + 90 = 103°16’$

B] The maximum shear stress is,

$\zeta_{max} = \frac{P_1 – P_2}{2} = \frac{85.90 – (- 25.90)}{2}$

$\zeta_{max} = 111.8 \ N/mm^2$

And its inclination is $\theta = \theta_1 + 45° = 13°16’ + 45°$

$\theta = 58°16°$

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Sketch

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