Question 9.1: A bending moment M1 is applied at the free end of a cantilev......
A bending moment M_1 is applied at the free end of a cantilever of length L and constant flexural rigidity E I. Find an expression for w(x).
Given: Load applied to beam.
Find: Deflection, or “equation of elastic curve.”
Assume: Hooke’s law applies; long, slender beam.
We start with an FBD and the external reaction forces and/or moments. In this case, this procedure is straightforward: the fixed support exerts a reaction moment equal and opposite to M_1 on the beam. If we made an imaginary cut at any x and used the method of sections to find the local internal bending moment, we would similarly find that at each x, the internal bending moment was M_1. M(x) = M_1 = constant, as shown in the bending moment diagram.
At the fixed end (x = 0), we know that deflection and slope are both zero; at the free end (x = L), we know the moment is M_1 and the shear is zero. We can thus begin integrating the second-order equation for deflection w(x):
E I \frac{\mathrm{d}^2 w(x)}{\mathrm{d} x^2}=-M(x)=-M_1 ,
Integrate: E I \frac{\mathrm{d} w(x)}{\mathrm{d} x}=-M_1 x+C_3 ,
Apply BC: \left.\frac{\mathrm{d} w}{\mathrm{~d} x}\right|_{x=0}=0 \rightarrow C_3=0 \text { (fixed end) } ,
Integrate: E I w(x)=-\frac{1}{2} M_1 x^2+C_4,
Apply BC: w(0)=0 \rightarrow C_4=0 \text { (fixed end). }
So,
w(x)=-\frac{M_1 x^2}{2 E I}.
This deflection is negative, which means that the deflection due to M_1 is downward. The maximum deflection is at x = L, and the neutral axis has the general shape sketched here.
