# Bending moment when there are no constraints

I faced this problem in an old book (Hall Allen S., Theory and Problems of Machine Design; 1961)

The answer according to the book is 592 [in.lb], but I think the bending moment should be 0 because there are no constraints.

Could anybody help me to understand this?

• Have you forgotten inertia? – Transistor Jul 4 '20 at 21:28
• The key word in this question is “suddenly”... – Jonathan R Swift Jul 4 '20 at 21:30
• The self weight of the beam (about 5.5lb) is negligible compared with the 200 lb force, so I would ignore it in a first attempt at a solution. There is a constraint on the motion. If the beam was not resting on the surface, the beam would rotate when the impulse was applied and the left hand end would move down. – alephzero Jul 4 '20 at 22:33
• The question doesn't say which direction the force was applied, but I assumed from the picture it was vertically upwards. If you think it is horizontal, the table does not apply any constraint to the motion. – alephzero Jul 4 '20 at 22:46
• For a horizontal force, if you assume the beam is rigid you can find its linear and angular accelerations. Then find the internal force distribution in the beam which is consistent with that motion. You should find the mass of the beam cancels out in the final answer. – alephzero Jul 4 '20 at 22:50

The impulse causes lateral acceleration of the rod and rotation of the rod.

$$P=m\frac{dv}{dt} \quad \text{for lateral acceleration}$$

$$P=I\frac{d\omega}{dt} \quad \text{for rotational acceleration}$$

We calculate the net acceleration for each point along the length of the rod and after multiplying that by the density of the rod we get the forces' acting on the rod and calculate the bending moment.

The net acceleration is the difference between linear acceleration and rotational acceleration for each small dx length of the rod. The net acceleration varies linearly from just linear acceleration at the middle of the rod to maximum at the ends of the rod. we calculated the max at the end and build the triangle of net acceleration and hence the load.

$$\alpha_{net \ x=L/2} =(1/(\rho d_{x_{L/2}}) - \frac{\ L/2}{\rho d_{x_{L/2}} I})P=1/\rho (1-6/L ) )P$$ So we can consider the beam loaded with two triangles of $$\rho(1-6/L)(L/2)(2L/3)P$$

I leave the rest to you. Check my arithmetic please.

Note: many have questions on pure linear acceleration and combined rotation and linear acceleration. The reason the impulse P works the same for both cases is that it gets to accomplish more work in case of combined accelerations.