Design for Shock Load - Deflection and Stress Generated

Theory

Comparison Between Slowly and Suddenly Applied Load


Worked Example

Assume a mass of 2kg travelling at 0.5 m/s hits the free end of a 3m long steel cantilever beam which has a cross section 50mm deep (in the direction that the load is travelling) by 60mm wide. Assume E for steel is 200GPa.

Determine the maximum dynamic deflection and stress and what the equivalent drop height is.

I = b d3/12 = 0.06(0.05)3/12 = 625.0E-9m4 The static (due to gravity) deflection is given by: y = P(L)3/3 E I = 2 x 9.81(3)3/(3 x 200E9 x 625.0E-9) = 1.41E-3m or 1.41mm The maximum stress = Mmax x half depth of beam/I = 2 x 9.81 x 3 x 0.025/625.0E-9 = 2.35MPa For the impact load:
The maximum deflection = v (m L3 / 3 E I)0.5 = 0.5(2 x (3)3/3 x 200E9 x 625.0E-9)0.5 = 6.0E-3m or 6mm Maximum stress is proportional to maximum deflection and in the dynamic case = 6 x 2.35/1.41 = 10MPa

To determine the equivalent drop height - use the equation of constant acceleration:

v2 = u2 + 2 a s

(0.5)2 = 2 x 9.81 x s

s = (0.5)2/2 x 9.81 = 12.7mm

Dropping a load from a height of just 12.7mm generates a stress exceeding four times the stress generated by the same load applied slowly.

David J Grieve, 1st March 2005.