Question 4.12: Impact Loading on a Beam A weight W is dropped from a height...
Impact Loading on a Beam
A weight W is dropped from a height h, striking at midspan a simply supported steel beam of length L. The beam is of rectangular cross section of width b and depth d (Figure 4.18). Calculate the maximum deflection and maximum stress for these two cases:
a. The beam is rigidly supported at each end.
b. The beam is supported at each end by springs.
Given: W = 100 N, h = 150 mm, L = 2 m, b = 30 mm, and d = 60 mm
Assumptions: Modulus of elasticity E = 200 GPa and spring rate k = 200 kN/m.
We have M_{max} = WL /4 at point C and I = bd^{3} /12. The maximum deflection, due to a static load, is (from case 5 of Table A.9)
\delta_{s t}=\frac{W L^3}{48 E I}=\frac{100(2)^3(12)}{48\left(200 \times 10^9\right)(0.03)(0.06)^3}=0.154 mm
The maximum static stress equals
\sigma_{s t}=\frac{M_{\max } C}{I}=\frac{100(2)(0.03)(12)}{4(0.03)(0.06)^3}=2.778 MPa
a. The impact factor, using Equation 4.32, is
K=1+\sqrt{1+\frac{2 h}{\delta_{s t}}} (4.32)
K=1+\sqrt{1+\frac{2(0.15)}{0.154\left(10^{-3}\right)}}=45.15
Therefore
\delta _{max} = 45.15(0.154) = 6.95 mm
\sigma _{max} = 45.15(2.778) = 125 MPa
b. The static deflection of the beam due to its own bending and the deformation of the springs is
\delta_{s t}=0.154+\frac{50}{200}=0.404 mm
The impact factor is then
K=1+\sqrt{1+\frac{2(0.15)}{0.404\left(10^{-3}\right)}}=28.27
Hence,
\delta _{max} = 28.27(0.404) = 11.42 mm
\sigma _{max} = 28.27(2.778) = 78.53 MPa
Comments: Comparing the results, we observe that dynamic loading considerably increases deflection and stress in a beam. Also noted is a reduction in stress with increased flexibility, owing to the spring added to the supports. However, the values calculated are probably somewhat higher than the actual quantities, because of our simplifying assumptions 3 and 4.