A bar of cross-section 20 mm X 20 mm is axially pulled by a force ‘P’ KN. If the maximum stress induced in the bar is 50 MPa, what is “P”?

 

A bar of cross-section 20 mm X 20 mm is axially pulled by a force ‘P’ KN. If the maximum stress induced in the bar is 50 MPa, what is “P”?

Solution

Area resisting the load= here, CS Area= 400 mm2.

Maximum (tensile, as load is axial pull) stress in bar under the given load= 50 MPa= 50 N/mm2.

So, using formula for stress, = Resistance÷ Area resisting the load.

as resistance = Load here P (given in kN)

Therefore, 50 = (Px 1000) ÷ 400

=> P = 20