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At a point in a strained material, a direct tensile stress of 60 N/mm^2 along x-direction and compressive stress of 40 N/mm^2 along y-direction are applied on a plane right angle to each other.

At a point in a strained material, a direct tensile stress of 60 N/mm^2 along x-direction and compressive stress of 40 N/mm^2 along Y direction are applied on a plane right angle to each other. If maximum principal stress is limited to 75 N/mm^2. Find the shear stress that may be allowed on the plane. Also, find minimum principal stress and maximum shear stress.

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Data:

Direct Stress:

$P_x = 60 \ N/mm^2$ (tensile)

$P_y = -40 \ N/mm^2$ (compressive)

Maximum principal stress $P_1 = 75 \ N/mm^2$

To Find:

[1] Shear stress ($\zeta$)

[2] Minimum principal stress $(P_2)$

[3] Maximum shear stress $(\zeta _{max})$

A] The sum of the normal stresses on two mutually perpendicular plane is constant.

$\therefore \ P_1 + P_2 = P_x + P_y$

$75 + P_2 = 60 + (-40)$

$P_2 = -55 \ N/mm^2$ (Compressive)

Now, B] Maximum principal stress $P_1$ is given by,

$P_1 = \frac{px + py}{2} + \sqrt{(\frac{px – py}{2})^2 + \zeta ^2}$

$\therefore \ 75 = \frac{60 + (-40)}{2} + \sqrt{ [ \frac{60 – (-40)}{2}]^2 + \zeta ^2}$

$65 = \sqrt{ 2500 + \zeta ^2}$

$\therefore$ Squaring on both sides,

$2500 + \zeta ^2 = 65^2$

$\therefore \ \zeta = 41.53 \ N/mm^2$

C] Maximum shear stress.

$\zeta_{max} = \frac{P_1 – P_2}{2} = \frac{75 – (-55)}{2} = 65 \ N/mm^2$

$\therefore \zeta_{max} = 65 \ N/mm^2$