Shear and Moment Diagrams of an Overhung Beam Using Singularity Functions Problem: Determine and plot the shear and moment functions for the overhung beam with an applied moment and ramp load as shown in Figure 3-22c. Given: Beam length l = 10 in, and load location a = 4 in. The magnitude of the

Question

Shear and Moment Diagrams of an Overhung Beam Using Singularity Functions

Problem: Determine and plot the shear and moment functions for the overhung beam with an applied moment and ramp load as shown in Figure 3-22c.

Given: Beam length [latex]l[/latex] = 10 in, and load location [latex]a[/latex] = 4 in. The magnitude of the applied moment [latex]M[/latex] = 20 lb-in, and the slope of the force distribution is [latex]w[/latex] = 10 lb/in/in.

Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored.

Accepted Answer

See Figures 3-22c and 3-26.

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 3-22c,

[latex] \langle x-a angle^2 [/latex] (3.17a)

[latex] \langle x-a angle^1 [/latex] (3.17b)

[latex] \langle x-a angle^0 [/latex] (3.17c)

[latex] \langle x-a angle^{-1} [/latex] (3.17d)

[latex] \langle x-a angle^{-2} [/latex] (3.17e)

[latex] \int_{-\infty}^x\langle\lambda-a angle^2 d \lambda=\frac{\langle x-a angle^3}{3} [/latex] (3.18a)

[latex] \int_{-\infty}^x\langle\lambda-a angle^1 d \lambda=\frac{\langle x-a angle^2}{2} [/latex] (3.18b)

[latex] \int_{-\infty}^x\langle\lambda-a angle^0 d \lambda=\langle x-a angle^1 [/latex] (3.18c)

[latex] \int_{-\infty}^x\langle\lambda-a angle^{-1} d \lambda=\langle x-a angle^0 [/latex] (3.18d)

[latex] \int_{-\infty}^x\langle\lambda-a angle^{-2} d \lambda=\langle x-a angle^{-1} [/latex] (3.18e)

[latex] q =M\langle x-0 angle^{-2}+R_1\langle x-a angle^{-1}-w\langle x-a angle^1+R_2\langle x-l angle^{-1} [/latex] (a)

[latex] V =\int q d x=M\langle x-0 angle^{-1}+R_1\langle x-a angle^0-\frac{w}{2}\langle x-a angle^2+R_2\langle x-l angle^0+C_1[/latex] (b)

[latex] M =\int V d x=M\langle x-0 angle^0+R_1\langle x-a angle^1-\frac{w}{6}\langle x-a angle^3+R_2\langle x-l angle^1+C_1 x+C_2 [/latex] (c)

2 As demonstrated in the previous two examples, the constants of integration [latex]C_{1}[/latex] and [latex]C_{2}[/latex] will always be zero if the reaction forces are included in the equations for shear and moment. So we will set them to zero.

3 The reaction forces [latex]R_{1}[/latex] and [latex]R_{2}[/latex] can be calculated from equations ([latex]c[/latex]) and ([latex]b[/latex]) respectively by substituting the boundary conditions [latex]x = l^{+}, V = 0, M = 0[/latex]. Note that we can substitute [latex]l[/latex] for [latex]l^{+}[/latex] since their difference is vanishingly small.

[latex] \begin{aligned} M &=M_1\langle l angle^0+R_1\langle l-a angle^1-\frac{w}{6}\langle l-a angle^3+R_2\langle l-l angle^1=0 \ 0 &=M_1+R_1(l-a)-\frac{w}{6}(l-a)^3 \ R_1 &=\frac{w}{6}(l-a)^2-\frac{M_1}{(l-a)} \ &=\frac{10}{6}(10-4)^2-\frac{20}{(10-4)}=56.67 lb \end{aligned} [/latex] (d)

[latex] \begin{aligned} V &=M\langle l angle^{-1}+R_1\langle l-a angle^0-\frac{w}{2}\langle l-a angle^2+R_2\langle l-l angle^0=0 \ 0 &=M(0)+R_1-\frac{w}{2}(l-a)^2+R_2 \ R_2 &=\frac{w}{2}(l-a)^2-R_1=\frac{10}{2}(10-4)^2-56.67=123.33 lb \end{aligned} [/latex] (e)

Note that equation ([latex]d[/latex]) is just [latex]\Sigma M_{z}[/latex] = 0, and equation ([latex]e[/latex]) is [latex]\Sigma F_{y}[/latex] = 0.

4 To generate the shear and moment functions over the length of the beam, equations ([latex]b[/latex]) and ([latex]c[/latex]) must be evaluated for a range of values of [latex]x[/latex] from 0 to [latex]l[/latex], after substituting the values of [latex]C_{1} = 0, C_{2} = 0, R_{1}[/latex], and [latex]R_{2}[/latex] in them. The independent variable [latex]x[/latex] was varied from 0 to [latex]l[/latex] = 10 at 0.1 increments. The reactions, loading function, shearforce function, and moment function were calculated from equations ([latex]a[/latex]) through ([latex]f[/latex]) above and are plotted in Figure 3-26.[latex]^{*}[/latex]

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 [latex]x = l[/latex] and the moment has a maximum to the right of the beam center. The value of [latex]x[/latex] at [latex]M_{max}[/latex] can be found by setting [latex]V[/latex] to 0 in equation ([latex]b[/latex]) and solving for [latex]x[/latex]. The shear function is the derivative of the moment function and so must be zero at each of its minima and maxima. This gives [latex]x[/latex] = 7.4 at [latex]M_{max}[/latex]. The function values at these points of maxima or minima can be calculated from equations ([latex]b[/latex]) and ([latex]c[/latex]) respectively by substituting the appropriate values of [latex]x[/latex] and evaluating the singularity function:

[latex] R_1=56.7 \quad R_2=123.3 \quad V_{\max }=-120 \quad M_{\max }=147.2 [/latex] (f)

* The files EX03-04 that generate these plots are on the CD-ROM.

Question 3.4: Shear and Moment Diagrams of an Overhung Beam Using Singular...

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