Question 3.2B: Shear and Moment Diagrams of a Simply Supported Beam Using S...
Shear and Moment Diagrams of a Simply Supported Beam Using Singularity Functions
Problem: Determine and plot the shear and moment functions for the simply supported beam with uniformly distributed load shown in Figure 3-22a.
Given: Beam length l = 10 in, and load location a = 4 in. The magnitude of the uniform force distribution is w = 10 lb/in.
Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored.
See Figures 3-22a and 3-24.
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. For the beam in Figure 3-22a,
\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 =R_1\langle x-0\rangle^{-1}-w\langle x-a\rangle^0+R_2\langle x-l\rangle^{-1} (a)
V =\int q d x=R_1\langle x-0\rangle^0-w\langle x-a\rangle^1+R_2\langle x-l\rangle^0+C_1 (b)
M =\int V d x=R_1\langle x-0\rangle^1-\frac{w}{2}\langle x-a\rangle^2+R_2\langle x-l\rangle^1+C_1 x+C_2 (c)
There are two reaction forces and two constants of integration to be found. We are integrating along a hypothetical infinite beam from –∞ to x. The variable x can take on values both before and beyond the end of the beam. If we consider the conditions at a point infinitesimally to the left of x = 0 (denoted as x = 0^{–}), the shear and moment will both be zero there. The same conditions apply at a point infinitesimally to the right of x = l (denoted as x = l^{+}). These observations provide the four boundary conditions needed to evaluate the four constants C_{1}, C_{2}, R_{1}, R_{2}: when x = 0^{–}, V = 0, M = 0; when x = l+, V = 0, M = 0.
2 The constants C_{1} and C_{2} are found by substituting the boundary conditions x = 0^{–}, V = 0, and x = 0^{–}, M = 0 in equations (b) and (c), respectively:
\begin{aligned} V\left(0^{-}\right) &=0=R_1\left\langle 0^{-}-0\right\rangle^0-w\left\langle 0^{-}-a\right\rangle^1+R_2\left\langle 0^{-}-l\right\rangle^0+C_1 \\ C_1 &=0 \\ M\left(0^{-}\right) &=0=R_1\left\langle 0^{-}-0\right\rangle^1-\frac{w}{2}\left\langle 0^{-}-a\right\rangle^2+R_2\left\langle 0^{-}-l\right\rangle^1+C_1\left(0^{-}\right)+C_2 \\ C_2 &=0 \end{aligned} (d)
Note that in general, the constants C_{1} and C_{2} will always be zero if the reaction forces and moments acting on the beam are included in the loading function, because the shear and moment diagrams must close to zero at each end of the beam.
3 The reaction forces R_{1} and R_{2} can be calculated from equations (c) and (b) respectively by substituting the boundary conditions x = l^{+}, V = 0, M = 0. Note that we can substitute l for l^{+} to evaluate it since their difference is vanishingly small.
\begin{aligned} M\left(l^{+}\right) &=R_1\left\langle l^{+}-0\right\rangle^1-w \frac{\left\langle l^{+}-a\right\rangle^2}{2}+R_2\left\langle l^{+}-l\right\rangle^1=0 \\ 0 &=R_1 l^{+}-\frac{w\left(l^{+}-a\right)^2}{2} \\ R_1 &=\frac{w\left(l^{+}-a\right)^2}{2 l^{+}}=\frac{w(l-a)^2}{2 l}=\frac{10(10-4)^2}{2(10)}=18 \end{aligned} (e)
\begin{aligned} V\left(l^{+}\right) &=R_1\left\langle l^{+}-0\right\rangle^0-w\left\langle l^{+}-a\right\rangle^1+R_2\left\langle l^{+}-l\right\rangle^0=0 \\ 0 &=R_1-w(l-a)+R_2 \\ R_2 &=w(l-a)-R_1=10(10-4)-18=42 \end{aligned} (f)
Since w, l, and a are known from the given data, equation (e) can be solved for R_{1}, and this result substituted in equation (f) to find R_{2}. Note that equation (f) is just \Sigma F = 0, and equation (e) is the sum of moments taken about point l and set to 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 R_{2} in them. The independent variable x was run from 0 to l = 10 at 0.1 increments. The reactions, loading function, shear-force function, and moment function were calculated from equations (a) through (f) above and are plotted in Figure 3-24. The files EX03-02 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 is largest at x = l and the moment has a maximum M_{max} near the center. The value of x at M_{max} can be found by setting V to 0 in equation (b) and solving for x. (The shear function is the derivative of the moment function and so must be zero at each of its minima and maxima.) This gives x = 5.8 at M_{max}. The function values at these points of maxima or minima can then be calculated from equations b and c respectively by substituting the appropriate values of x and evaluating the singularity functions. For the maximum absolute value of shear force at x = l,
\begin{aligned} V_{\max }=V_{@ x=l^{-}} &=R_1\left\langle l^{-}-0\right\rangle^0-w\left\langle l^{-}-a\right\rangle^1+R_2\left\langle l^{-}-l\right\rangle^0 \\ &=R_1-w\left(l^{-}-a\right)+0 \\ &=18-10(10-4)+0=-42 \end{aligned} (g)
Note that the first singularity term evaluates to 1 since l^{– } > 0 (see Eq. 3.17c, p. 114), the second singularity term evaluates to (l – a) because l^{–} > a in this problem (see Eq. 3.17b), and the third singularity term evaluates to 0 as defined in equation 3.17c. The maximum moment is found in similar fashion:
\begin{aligned} M_{\max }=M_{@ x=5.8} &=R_1\langle 5.8-0\rangle^1-w \frac{\langle 5.8-a\rangle^2}{2}+R_2\langle 5.8-l\rangle^1 \\ &=R_1\langle 5.8\rangle^1-w \frac{\langle 5.8-4\rangle^2}{2}+R_2\langle 5.8-10\rangle^1 \\ &=18(5.8)-10 \frac{(5.8-4)^2}{2}+0=88.2 \end{aligned} (h)
The third singularity term evaluates to 0 because 5.8 < l (see Eq. 3.17b).
6 The results are
R_1=18 \quad R_2=42 \quad V_{\max }=-42 \quad M_{\max }=88.2 (i)
