Question

At a point in a bracket, the stress on two mutually perpendicular planes are 100 N/mm2 (tensile) and 50 N/mm2 (tensile). The shear stress across the planes is 80 N/mm2. Find using Mohr stress circle or otherwise calculate:

a. Magnitude and direction of the resultant stress on plane making an angle of 200 with the plane of the first stress.

b. Maximum shear stress and location of its plane.

c. Principal stresses and the location of principal planes.
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Answer 1

The following three points calculation can be defined as follows:

Explanation:

$\sigma_x= 100 \frac {N}{mm^2}\\\\\sigma_y= 50 \frac {N}{mm^2}\\\\T_{xy}=80 \frac {N}{mm^2}\\\\\theta =20^{\circ}$

The resultant stress, principle stress, and maximum shear stress status are determined by the analytic method which can be defined as follows:

$\to \sigma_\theta=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2} \cos 2 \theta + T_{xy} \sin 2\theta$

         $=\frac{\100+50}{2}+\frac{100-50}{2} \cos 40+80\sin 40$

         $=145.57 \ Mpa$

$\to T_\theta=\frac{\sigma_x-\sigma_y}{2}\sin 2\theta-T_{xy}\cos 2 \theta$

        $=\frac{\100-50}{2}\sin 40-80\cos 40\\\\=-45.21 \ Mpa$

$\to \sigma_R=\sqrt{\sigma_B^2+T_\theta^2}$              

     $=\sqrt{145.57^2-45.21^2}\\\\=152.43 \ Mpa$

$\to \tan \theta= \frac{T_\theta}{\sigma_\theta}\\\\ \theta= \tan^{-1} \frac{T_\theta}{\sigma_\theta}$

  $= \tan^{-1} \frac{ -45.21}{145.57}\\\\=17.25^{\circ}$

Calculating  Maximum shear stress:

$T_{max}=\sqrt{(\frac{\sigma_x-\sigma_y}{2})^2+T_{xy}}$

        $=\sqrt{25^2 +80^2}\\\\=83. 81 \ Mpa$

$\theta_s=\frac{1}{2} \tan^{-1}\frac{\sigma_x-\sigma_y}{2T_{xy}}$

   $=8.677^{\circ}$

$\sigma_{avg}=75\\\\\sigma_{p1}=158.81 \ Mpa\\\\\sigma_{p2}= -8.81 \ Mpa\\\\\sigma_{\theta\ p1} =36.62^{\circ}$

Scale 1 MPa = cm in the mohr circle  

Its location with maximum shear stress container and main aircraft are interconnected