Question 3.3B: Shear and Moment Diagrams of a Cantilever Beam Using Singula...
Shear and Moment Diagrams of a Cantilever Beam Using Singularity Functions
Problem: Determine and plot the shear and moment functions for the cantilever beam with a concentrated load as shown in Figure 3-22b.
Given: Beam length l = 10 in, and load location a = 4 in. The magnitude of the applied force is F = 40 lb.
Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored.
See Figures 3-22b and 3-25.
1 Write equations for the load function in terms of equations 3.17 (pp. 113–114) and integrate the resulting function twice using equations 3.18 (pp. 114–115) to obtain the shear and moment functions. Note the use of the unit doublet function to represent the moment at the wall. For the beam in Figure 3-22b,
\langle x-a\rangle^2 (3.17a)
\langle x-a\rangle^1 (3.17b)
\langle x-a\rangle^0 (3.17c)
\langle x-a\rangle^{-1} (3.17d)
\langle x-a\rangle^{-2} (3.17e)
\int_{-\infty}^x\langle\lambda-a\rangle^2 d \lambda=\frac{\langle x-a\rangle^3}{3} (3.18a)
\int_{-\infty}^x\langle\lambda-a\rangle^1 d \lambda=\frac{\langle x-a\rangle^2}{2} (3.18b)
\int_{-\infty}^x\langle\lambda-a\rangle^0 d \lambda=\langle x-a\rangle^1 (3.18c)
\int_{-\infty}^x\langle\lambda-a\rangle^{-1} d \lambda=\langle x-a\rangle^0 (3.18d)
\int_{-\infty}^x\langle\lambda-a\rangle^{-2} d \lambda=\langle x-a\rangle^{-1} (3.18e)
q =-M_1\langle x-0\rangle^{-2}+R_1\langle x-0\rangle^{-1}-F\langle x-a\rangle^{-1} (a)
V =\int q d x=-M_1\langle x-0\rangle^{-1}+R_1\langle x-0\rangle^0-F\langle x-a\rangle^0+C_1 (b)
M =\int V d x=-M_1\langle x-0\rangle^0+R_1\langle x-0\rangle^1-F\langle x-a\rangle^1+C_1 x+C_2 (c)
The reaction moment M_{1} at the wall is in the z direction and the forces R_{1} and F are in the y direction in equation (b). All moments in equation (c) are in the z direction.
2 Because the reactions have been included in the loading function, the shear and moment diagrams both close to zero at each end of the beam, making C_{1} = C_{2} = 0.
3 The reaction force R_{1} and reaction moment M_{1} are calculated from equations (b) and (c) respectively by substituting the boundary conditions x = l^{+}, V = 0, M = 0. Note that we can substitute l for l^{+} since their difference is vanishingly small. M_{1} does not appear in equation (d) because its singularity function is not defined at l = l^{+}.
\begin{aligned} V &=M\langle l\rangle^{-1}+R_1\langle l-0\rangle^0-F\langle l-a\rangle^0=0 \\ 0 &=M(0)+R_1-F \\ R_1 &=F=40 lb \end{aligned} (d)
\begin{aligned} M &=-M_1\langle l-0\rangle^0+R_1\langle l-0\rangle^1-F\langle l-a\rangle^1=0 \\ 0 &=-M_1+R_1(l)-F(l-a) \\ M_1 &=R_1(l)-F(l-a)=40(10)-40(10-4)=160 lb -\text { in cW } \end{aligned} (e)
Since F, l, and a are known from the given data, equation (d) can be solved for R_{1}, and this result substituted in equation (e) to find M_{1}. Note that equation (d) is just \Sigma F_{y} = 0, and equation (e) is \Sigma M_{z} = 0.
4 To generate the shear and moment functions over the length of the beam, equations (b) and (c) must be evaluated for a range of values of x from 0 to l, after substituting the above values of C_{1}, C_{2}, R_{1}, and M_{1} in them. The independent variable x was varied from 0 to l = 10 at 0.1 increments. The reactions, loading function, shearforce function and moment function were calculated from equations (a) through (e) above and are plotted in Figure 3-25. The files EX03-03 that generate these plots are on the CD-ROM.
5 The largest absolute values of the shear and moment functions are of interest for the calculation of stresses in the beam. The plots show that the shear force and the moment are both largest at x = 0. The function values at these points can be calculated from equations (b) and (c) respectively by substituting x = 0 and evaluating the singularity functions:
R_1=40 \quad V_{\max }=40 \quad\left|M_{\max }\right|=160 (f)
