Question

A uniform steel bar of 2 cm × 2 cm area of cross-section is subjected to an axial pull of 40000 kg. Calculate the intensity of “normal stress, shear stress and resultant stress on a plane normal to which is inclined at 30° to the axis of the bar. Solve the problem graphically by drawing Mohr Circle.

Answer 1

Given that : 

Load applied ‘P’= 4000 × 9.8 = 39.2 kN 

σ_x = 39.2/(2 × 10 ^{– 2})² = 98MN/m²

Steps to draw Mohr’s circle. 

Step1: Take origin ‘O’ and draw a horizontal line OX. 

Step2: Cut off OA equal to σ_x by taking scale 1 mm = 1MN/m²

Step 3: Bisect OA at C 

Step 4: With C as center and radius CA draw a circle. 

Step 5: At C draw a line CP at an angle 2θ with OX meeting the circle. At P (θ is angle made by oblique plane with minor principle stress, here zero). 

Step 6: Through P draw perpendicular to OX, it intersect OX at Q, join OP. Measure OQ, PQ and OQ as σ, τ and σ_r. respectively. Therefore; 

Normal stress on the plane σ = OQ = 73.5 × 1 = 73.5MN/m² 

Tangential or shear stress on the plane τ = PQ = 43 × 1 = 43 MN/m² 

And Resultant stress σ_r = OP = 85 × 1 = 85 MN/m².