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

Question

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 [latex]l[/latex] = 10 in, and load location [latex]a[/latex] = 4 in. The beam’s [latex]I[/latex] = 0.5 in[latex]^{4}[/latex] and [latex]E[/latex] = 30 Mpsi. The magnitude of the applied force is [latex]F[/latex] = 400 lb.

Assumptions: Ignore the beam weight as negligible compared to the applied load.

Accepted Answer

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,

[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_1\langle x-0 angle^{-2}+R_1\langle x-0 angle^{-1}-F\langle x-a angle^{-1} [/latex] (a)

[latex] V=\int q d x=-M_1\langle x-0 angle^{-1}+R_1\langle x-0 angle^0-F\langle x-a angle^0+C_1 [/latex] (b)

[latex] M=\int V d x=-M_1\langle x-0 angle^0+R_1\langle x-0 angle^1-F\langle x-a angle^1+C_1 x+C_2 [/latex] (c)

[latex] heta=\int \frac{M}{E I} d x=\frac{1}{E I}\left\lgroup \begin{array}{c} -M_1\langle x-0 angle^1+\frac{R_1}{2}\langle x-0 angle^2-\frac{F}{2}\langle x-a angle^2 \ +\frac{C_1 x^2}{2}+C_2 x+C_3 \end{array} ight group [/latex] (d)

[latex] y=\int heta d x=\frac{1}{E I}\left\lgroup \begin{array}{c} -\frac{M_1}{2}\langle x-0 angle^2+\frac{R_1}{6}\langle x-0 angle^3-\frac{F}{6}\langle x-a angle^3 \ +\frac{C_1 x^3}{6}+\frac{C_2 x^2}{2}+C_3 x+C_4 \end{array} ight group [/latex] (e)

The reaction moment [latex]M_{1}[/latex] at the wall is in the [latex]z[/latex] direction and the forces [latex]R_{1}[/latex] and [latex]F[/latex] are in the [latex]y[/latex] direction in equation ([latex]b[/latex]) . All moments in equation ([latex]c[/latex]) are in the [latex]z[/latex] 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 [latex]C_{1} = C_{2} = 0[/latex].

3 The reaction force [latex]R_{1}[/latex] and reaction moment [latex]M_{1}[/latex] are calculated from equations ([latex]b[/latex]) and ([latex]c[/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} V\left(l^{+} ight) &=0=R_1\langle l-0 angle^0-F\langle l-a angle^0 \ 0 &=R_1-F \ R_1 &=F=400 lb \end{aligned} [/latex] (f)

[latex] \begin{aligned} M\left(l^{+} ight) &=0=-M_1\langle l-0 angle^0+R_1\langle l-0 angle^1-F\langle l-a angle^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 - ext { in } c w \end{aligned} [/latex] (g)

Since [latex]w, l[/latex], and [latex]a[/latex] are known from the given data, equation ([latex]f[/latex]) can be solved for [latex]R_{1}[/latex], and this result substituted in equation ([latex]g[/latex]) to find [latex]M_{1}[/latex]. Note that equation ([latex]f[/latex]) is just [latex]\Sigma F_{y}[/latex] = 0, and equation ([latex]g[/latex]) is [latex]\Sigma M_{z}[/latex] = 0. [latex]M_{1}[/latex] does not appear in equation ([latex]f[/latex]) because it is in a different vector direction than the [latex]y[/latex] forces.

4 Substitute [latex]x = 0, heta = 0[/latex] and [latex]x = 0, y[/latex] = 0 in ([latex]d[/latex]) and ([latex]e[/latex]) and solve for [latex]C_{3 }[/latex] and [latex]C_{4}[/latex]:

[latex] \begin{aligned} heta(0) &=0=\frac{1}{E I}\left\lgroup -M_1\langle 0-0 angle^1+\frac{R_1}{2}\langle 0-0 angle^2-\frac{F}{2}\langle 0-a angle^2+C_3 ight group \ C_3 &=M_1\langle 0-0 angle^1-\frac{R_1}{2}\langle 0-0 angle^2+\frac{F}{2}\langle 0-4 angle^2=0 \end{aligned} [/latex] (h)

[latex] \begin{aligned} y(0) &=0=\frac{1}{E I}\left\lgroup -\frac{M_1}{2}\langle 0-0 angle^2+\frac{R_1}{6}\langle 0-0 angle^3-\frac{F}{6}\langle 0-a angle^3+C_3(0)+C_4 ight group \ C_4 &=\frac{M_1}{2}\langle 0-0 angle^2-\frac{R_1}{6}\langle 0-0 angle^3+\frac{F}{6}\langle 0-4 angle^3=0 \end{aligned} [/latex] (i)

5 Substitution of the expressions for [latex]C_{3}, C_{4}, R_{1}[/latex], and [latex]M_{1}[/latex] from ([latex]f[/latex]), ([latex]g[/latex]), ([latex]h[/latex]), and ([latex]i[/latex]) into equation ([latex]e[/latex]) gives the deflection equation for the cantilever beam in Figure 4-22b (p. 165):

[latex] y=\frac{F}{6 E I}\left[x^3-3 a x^2-\langle x-a angle^3 ight] [/latex] (j)

6 The maximum deflection of a cantilever beam is at its free end. Substitute [latex]x = l[/latex] in equation ([latex]j[/latex]) to find [latex]y_{max}[/latex].

[latex] \begin{array}{l} y_{\max }=\frac{F}{6 E I}\left[l^3-3 a l^2-(l-a)^3 ight]=\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} [/latex] (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.

Question 4.5: Finding Beam Slope and Deflection of a Cantilever Beam Using...

Related Answered Questions

Crimping-Tool Stress and Deflection Analysis Problem Determine the stresses and deflections at critical points in the crimping tool shown in Figures 3-3 (repeated here) and 4-49. Given The geometry and loading are known from Case Study 2A on p. 84. The thickness of link 1 is 0.313 in, of links 2 ...

Verified Answer:

Verified Answer:

Automobile Scissors-Jack Stress and Deflection Analysis Problem Determine the stresses and deflections at critical points in the scissors-jack assembly shown in Figures 3-5 (repeated here) and 4-52. Given The geometry and loading are known from Case Study 3A (p. 88). The design load is 2 000 lb ...

Verified Answer:

Verified Answer:

Column Design for Concentric Loading Problem: A beachfront house is to be jacked up 10 ft above grade and placed on a set of steel columns. The weight to be supported by each column is estimated to be 200 000 lb. Two designs are to be considered, one using square steel tubes and the other using ...

Verified Answer:

Combined Bending and Torsional Stresses Problem Find the most highly stressed locations on the bracket shown in Figure 4-30 and determine the applied and principal stresses at those locations. Given The rod length l = 6 in and arm a = 8 in. The rod outside diameter d = 1.5 in. Load F = 1 000 lb. ...

Verified Answer:

Design of a Torsion Bar Problem: Determine the best cross-sectional shape for a hollow torsion bar to be made from a sheet of steel of known dimensions in order to withstand a pure torsional load with minimum angular deflection. Also find the maximum shear stress. Given: The applied torque is 10 ...

Verified Answer:

Finding Reactions of a Statically Indeterminate Beam Using Castigliano’s Method Problem: Find the reaction forces for the indeterminate beam in Figure 4-22e. Given: The load is uniformly distributed over the beam as shown. Length is l. The magnitude of the distributed force is w. Assumptions: ...

Verified Answer:

Finding Reactions and Deflection of Statically Indeterminate Beams Using Singularity Functions Problem: Determine and plot the loading, shear, moment, slope, and deflection functions for the beam in Figure 4-22d. Find maximum deflection. Given: The load is uniformly distributed over part of the ...

Verified Answer:

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). ...

Verified Answer: