Question 4.5: Finding Beam Slope and Deflection of a Cantilever Beam Using...
Finding Beam Slope and Deflection of a Cantilever Beam Using Singularity Functions
Problem: Determine and plot the slope and deflection functions for the beam shown in Figure 4-22b (repeated here).
Given: The load is the concentrated force shown. Let beam length l = 10 in, and load location a = 4 in. The beam’s I = 0.5 in^{4} and E = 30 Mpsi. The magnitude of the applied force is F = 400 lb.
Assumptions: Ignore the beam weight as negligible compared to the applied load.
See Figures 4-22b and 4-24.
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 4-22b,
\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)
\theta=\int \frac{M}{E I} d x=\frac{1}{E I}\left\lgroup \begin{array}{c} -M_1\langle x-0\rangle^1+\frac{R_1}{2}\langle x-0\rangle^2-\frac{F}{2}\langle x-a\rangle^2 \\ +\frac{C_1 x^2}{2}+C_2 x+C_3 \end{array} \right\rgroup (d)
y=\int \theta d x=\frac{1}{E I}\left\lgroup \begin{array}{c} -\frac{M_1}{2}\langle x-0\rangle^2+\frac{R_1}{6}\langle x-0\rangle^3-\frac{F}{6}\langle x-a\rangle^3 \\ +\frac{C_1 x^3}{6}+\frac{C_2 x^2}{2}+C_3 x+C_4 \end{array} \right\rgroup (e)
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.
\begin{aligned} V\left(l^{+}\right) &=0=R_1\langle l-0\rangle^0-F\langle l-a\rangle^0 \\ 0 &=R_1-F \\ R_1 &=F=400 lb \end{aligned} (f)
\begin{aligned} M\left(l^{+}\right) &=0=-M_1\langle l-0\rangle^0+R_1\langle l-0\rangle^1-F\langle l-a\rangle^1 \\ 0 &=-M_1+l R_1(l)-F(l-a) \\ M_1 &=R_1(l)-F(l-a)=400(10)-400(10-4)=1600 lb -\text { in } c w \end{aligned} (g)
Since w, l, and a are known from the given data, equation (f) can be solved for R_{1}, and this result substituted in equation (g) to find M_{1}. Note that equation (f) is just \Sigma F_{y} = 0, and equation (g) is \Sigma M_{z} = 0. M_{1} does not appear in equation (f) because it is in a different vector direction than the y forces.
4 Substitute x = 0, \theta = 0 and x = 0, y = 0 in (d) and (e) and solve for C_{3 } and C_{4}:
\begin{aligned} \theta(0) &=0=\frac{1}{E I}\left\lgroup -M_1\langle 0-0\rangle^1+\frac{R_1}{2}\langle 0-0\rangle^2-\frac{F}{2}\langle 0-a\rangle^2+C_3 \right\rgroup \\ C_3 &=M_1\langle 0-0\rangle^1-\frac{R_1}{2}\langle 0-0\rangle^2+\frac{F}{2}\langle 0-4\rangle^2=0 \end{aligned} (h)
\begin{aligned} y(0) &=0=\frac{1}{E I}\left\lgroup -\frac{M_1}{2}\langle 0-0\rangle^2+\frac{R_1}{6}\langle 0-0\rangle^3-\frac{F}{6}\langle 0-a\rangle^3+C_3(0)+C_4 \right\rgroup \\ C_4 &=\frac{M_1}{2}\langle 0-0\rangle^2-\frac{R_1}{6}\langle 0-0\rangle^3+\frac{F}{6}\langle 0-4\rangle^3=0 \end{aligned} (i)
5 Substitution of the expressions for C_{3}, C_{4}, R_{1}, and M_{1} from (f), (g), (h), and (i) into equation (e) gives the deflection equation for the cantilever beam in Figure 4-22b (p. 165):
y=\frac{F}{6 E I}\left[x^3-3 a x^2-\langle x-a\rangle^3\right] (j)
6 The maximum deflection of a cantilever beam is at its free end. Substitute x = l in equation (j) to find y_{max}.
\begin{array}{l} y_{\max }=\frac{F}{6 E I}\left[l^3-3 a l^2-(l-a)^3\right]=\frac{F a^2}{6 E I}(a-3 l) \\ y_{\max }=\frac{400(4)^2}{6(3 E 7)(0.5)}[4-3(10)]=-0.00185 in \end{array} (k)
7 Plots of the loading, shear, moment, slope, and deflection functions are shown in Figure 4-24. Note that the beam slope becomes increasingly negative for the portion of beam between the support and the load and then becomes constant to the right of the load. While not very apparent at the small scale of the figure, the beam deflection becomes a straight line to the right of the point of application of the load.
The files EX04-05 can be opened in the program of your choice to examine the model and see larger-scale plots of the functions in Figure 4-24.
