Question

State of stress at a point in a material is as shown in the Fig. (a). Determine 

(i) principal stresses 

(ii) maximum shear stress 

(iii) plane of maximum shear stress and 

(iv) the resultant stress on the plane of maximum shear stress.

Answer 1

Selecting x and y-axis as shown in figure, 

p_x = – 50 N/mm² , 

p_y = 100 N/mm² , 

and q = 75 N/mm² .

= 25 + 106.07 

= 131.07 N/mm²

The principal plane makes an angle θ to y-axis in anticlockwise direction. Then

Plane of maximum shear makes 45° to it 

θ = – 31.72 + 45.00 = 13.28°. 

Normal stress on this plane is given by

p_x = \(\frac{p_x+p_y}2\) + \(\frac{p_x-p_y}2\) cos 2θ + q sin 2θ

= \(\frac{-50-100}2\) cos 2(13.28) + 75 sin (2 x 13.28)

= 25 – 67.08 + 33.54 

= – 8.54 N/mm² 

p_t = q_max = 106.07 N/mm²

∴ Resultant stress p = \(\sqrt{(-8.51)^2+106.07^2}\)

= 106.41 N/mm²

Let ‘p’ make angle φ to tangential stress (maximum shear

stress plane). Then referring to Fig. (b)

tan φ = \(\frac{p_n}{p_t}\) = \(\frac{8.54}{106.07}\)

∴ φ = 4.6° as shown in Fig. (b).