Question 4.6: Finding Beam Slope and Deflection of an Overhung Beam Using ...
Finding Beam Slope and Deflection of an Overhung Beam Using Singularity Functions
Problem: Determine and plot the slope and deflection functions for an overhung beam with a uniformly distributed load over part of its length and a concentrated force at its end as shown in Figure 4-22c (p. 165).
Given: Beam length l = 10 in, and load locations a = 4 in and b = 7 in. The beam’s I = 0.2 in^{4} and E = 30 Mpsi. The magnitude of the concentrated force is F = 200 lb and the distributed force is w = 100 lb/in.
Assumptions: The weight of the beam is negligible compared to the applied loads and so can be ignored.
See Figures 4-22c (repeated here), 4-25, 4-26.
1 The distributed load does not extend over the entire length of this beam. All singularity functions extend from their initial point to the end of the beam. So, to terminate the uniform load’s step function at some point short of the end of the beam it is necessary to apply another step function of equal amplitude and opposite sign in order to cancel it for all points beyond length a as shown in Figure 4-25. The sum of the two step functions of opposite sign is then zero to the right of distance a.
q=R_1\langle x-0\rangle^{-1}-w\langle x-0\rangle^0+w\langle x-a\rangle^0+R_2\langle x-b\rangle^{-1}-F\langle x-l\rangle^{-1} (a)
V=\int q d x=R_1\langle x-0\rangle^0-w\langle x-0\rangle^1+w\langle x-a\rangle^1+R_2\langle x-b\rangle^0-F\langle x-l\rangle^0+C_1 (b)
M=\int V d x=\left\lgroup \begin{array}{c} R_1\langle x-0\rangle^1-\frac{w}{2}\langle x-0\rangle^2+\frac{w}{2}\langle x-a\rangle^2+R_2\langle x-b\rangle^1 \\ -F\langle x-l\rangle^1+C_1 x+C_2 \end{array} \right\rgroup (c)
\theta=\int \frac{M}{E I} d x=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{2}\langle x-0\rangle^2-\frac{w}{6}\langle x-0\rangle^3+\frac{w}{6}\langle x-a\rangle^3+\frac{R_2}{2}\langle x-b\rangle^2 \\ -\frac{F}{2}\langle x-l\rangle^2+\frac{C_1}{2} x^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{R_1}{6}\langle x-0\rangle^3-\frac{w}{24}\langle x-0\rangle^4+\frac{w}{24}\langle x-a\rangle^4+\frac{R_2}{6}\langle x-b\rangle^3 \\ -\frac{F}{6}\langle x-l\rangle^3+\frac{C_1}{6} x^3+\frac{C_2}{2} x^2+C_3 x+C_4 \end{array} \right\rgroup (e)
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 Since both the shear and moment are zero at x = l^{+}, the reactions R_{1} and R_{2} can be calculated simultaneously from (b) and (c) with x = l+ = l:
\begin{aligned} V(l) &=0=R_1\langle l-0\rangle^0-w\langle l-0\rangle^1+w\langle l-a\rangle^1+R_2\langle l-b\rangle^0-F\langle l-l\rangle^0 \\ 0 &=R_1-w l+w(l-a)+R_2-F \\ R_2 &=-R_1+w l-w(l-a)+F=400 lb \end{aligned} (f)
\begin{aligned} M(l) &=0=\left\lgroup R_1\langle l\rangle^1-\frac{w}{2}\langle l\rangle^2+\frac{w}{2}\langle l-a\rangle^2+R_2\langle l-b\rangle^1-F\langle l-l\rangle^1 \right\rgroup \\ 0 &=R_1 l-\frac{w l^2}{2}+\frac{w(l-a)^2}{2}+R_2(l-b) \\ R_1 &=\frac{1}{l}\left[\frac{w l^2}{2}-\frac{w(l-a)^2}{2}-R_2(l-b)\right]=200 lb \end{aligned} (g)
Note that the equations (f) are just the sum of forces = 0, and the sum of moments taken about point l and set to 0.
4 Substitute x = 0, y = 0, and x = b, y = 0 in equation (e) and solve for C_{3} and C_{4}:
\begin{array}{l} y(0)=0=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{6}\langle 0-0\rangle^3-\frac{w}{24}\langle 0-0\rangle^4+\frac{w}{24}\langle 0-a\rangle^4+\frac{R_2}{6}\langle 0-b\rangle^3 \\ -\frac{F}{6}\langle 0-l\rangle^3+\frac{C_1}{6}(0)^3+\frac{C_2}{2}(0)^2+C_3(0)+C_4 \end{array} \right\rgroup \\ C_4=0 \end{array} (h)
\begin{array}{l} y(b)=0=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{6}\langle b-0\rangle^3-\frac{w}{24}\langle b-0\rangle^4+\frac{w}{24}\langle b-a\rangle^4+\frac{R_2}{6}\langle b-b\rangle^3 \\ -\frac{F}{6}\langle b-l\rangle^3+\frac{C_1}{6}(b)^3+\frac{C_2}{2}(b)^2+C_3(b)+C_4 \end{array} \right\rgroup\\ C_3=\frac{1}{b}\left[-\frac{R_1}{6} b^3+\frac{w}{24} b^4-\frac{w}{24}\langle b-a\rangle^4\right]\\ =\frac{1}{7}\left[-\frac{200}{6}(7)^3+\frac{100}{24}(7)^4-\frac{100}{24}(7-4)^4\right]=-252.4 \end{array} (i)
5 Substitution of the expressions for C_{1}, C_{2}, C_{3}, C_{4}, R_{1}, and R_{2} from equations (f), (g), (h), and (i) into equation (e) gives the resulting deflection equation
y=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{6} x^3-\frac{w}{24} x^4+\frac{w}{24}\langle x-a\rangle^4+\frac{R_2}{6}\langle x-b\rangle^3-\frac{F}{6}\langle x-l\rangle^3 \\ +\frac{1}{b}\left[-\frac{R_1}{6}(b)^3+\frac{w}{24}(b)^4-\frac{w}{24}\langle b-a\rangle^4\right] x \end{array} \right\rgroup (j)
6 Since an overhung beam is a form of cantilever beam, the maximum deflection is most likely at the free end. Substitute x = l in equation (f) to find y_{max}.
\begin{array}{l} y_{\max }=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{6} l^3-\frac{w}{24} l^4+\frac{w}{24}\langle l-a\rangle^4+\frac{R_2}{6}\langle l-b\rangle^3-\frac{F}{6}\langle l-l\rangle^3 \\ +\frac{1}{b}\left[-\frac{R_1}{6}(b)^3+\frac{w}{24}(b)^4-\frac{w}{24}\langle b-a\rangle^4\right] l \end{array} \right\rgroup\\ =\frac{1}{3 E 7(0.2)}\left\lgroup \begin{array}{l} \frac{200}{6} 10^3-\frac{100}{24} 10^4+\frac{100}{24}\langle 10-4\rangle^4+\frac{400}{6}(10-7)^3 \\ -\frac{200}{6} 0^3+\frac{1}{7}\left[-\frac{200}{6}(7)^3+\frac{100}{24}(7)^4-\frac{100}{24}(7-4)^4\right] 10 \end{array} \right\rgroup\\ y_{\max }=-0.0006 \text { in } \end{array} (k)
7 Plots of the loading, shear, moment, slope, and deflection functions for part (c) are shown in Figure 4-26. The files EX04-06 can be opened in the program of your choice to examine the model and see larger-scale plots of the functions in Figure 4-26.
