**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 d^{3}/12 = 0.06(0.05)^{3}/12 = 625.0E-9m^{4}
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 = M_{max} 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 L^{3} / 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:

v^{2} = u^{2} + 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.