Question

for this case of beam bending, what is the internal bending moment at x = 0.2 m if =69.2 n and =1.1 m in units of nm

Answer 1

The internal bending moment at x = 0.2 m on a beam with a force of 69.2 N applied at 1.1 m from the support is 62.28 Nm.

Step-by-step explanation:

The internal bending moment at a specific point x = 0.2 m on a beam can be found by calculating the torque produced by the forces on the beam at that point. Since the force (F = 69.2 N) is applied at the end of the beam (1.1 m from the support), the bending moment at x = 0.2 m can be calculated using the equation for torque (M = F × d), where M is the moment, F is the force, and d is the distance from the point of rotation. In this case, the distance d from the point of rotation to where the force is applied is 1.1 m - 0.2 m = 0.9 m. Therefore, the bending moment at x = 0.2 m is M = 69.2 N × 0.9 m = 62.28 Nm.