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

Question

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 beam as shown. Length [latex]l[/latex] = 10 in, [latex]a[/latex] = 4 in, and [latex]b[/latex] = 7 in. The beam’s [latex]I[/latex] = 0.08 in[latex]^{4}[/latex] and [latex]E[/latex] = 30 Mpsi. The magnitude of the distributed force is [latex]w[/latex] = 500 in/lb.

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

Accepted Answer

See Figures 4-22d (repeated here) and 4-27 (p. 173).

1 Write an equation for the load function in terms of equations 3.17 (pp. 113–114) and integrate the resulting function four times using equations 3.18 (pp. 114–115) to obtain the shear, moment, slope, and deflection functions.

[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=R_1\langle x-0 angle^{-1}-w\langle x-a angle^0+R_2\langle x-b angle^{-1}+R_3\langle x-l angle^{-1} [/latex] (a)

[latex] V=\int q d x=R_1\langle x-0 angle^0-w\langle x-a angle^1+R_2\langle x-b angle^0+R_3\langle x-l angle^0+C_1 [/latex] (b)

[latex] M=\int V d x=R_1\langle x-0 angle^1-\frac{w}{2}\langle x-a angle^2+R_2\langle x-b angle^1+R_3\langle x-l 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} \frac{R_1}{2}\langle x-0 angle^2-\frac{w}{6}\langle x-a angle^3+\frac{R_2}{2}\langle x-b angle^2+\frac{R_3}{2}\langle x-l 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{R_1}{6}\langle x-0 angle^3-\frac{w}{24}\langle x-a angle^4+\frac{R_2}{6}\langle x-b angle^3+\frac{R_3}{6}\langle x-l 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)

2    There are 3 reaction forces and 4 constants of integration to be found. The constants [latex]C_{1}[/latex] and [latex]C_{2}[/latex] are zero because the reaction forces and moments acting on the beam are included in the loading function. This leaves 5 unknowns to be found.
3    If we consider the conditions at a point infinitesimally to the left of [latex]x = 0[/latex] (denoted as [latex]x = 0^{–}[/latex]), the shear and moment will both be zero. The same conditions obtain at a point infinitesimally to the right of [latex]x = l[/latex] (denoted as [latex]x = l^{+}[/latex]). Also, the deflection [latex]y[/latex] must be zero at all three supports. These observations provide the 5 boundary conditions needed to evaluate the 3 reaction forces and 2 remaining integration constants: i.e., when [latex]x = 0^{–}, V = 0, M = 0[/latex]; when [latex]x = 0, y = 0[/latex]; when [latex]x=b, y = 0[/latex]; when [latex]x=l, y = 0[/latex]; when [latex]x = l+, V = 0, M = 0[/latex].
4    Substitute the boundary conditions [latex]x = 0, y = 0, x = b, y = 0[/latex], and [latex]x = l, y = 0[/latex] into ([latex]e[/latex]).
for [latex] x [/latex] = 0 :

[latex] \begin{aligned} y(0) &=0=\frac{1}{E I}\left\lgroup \frac{R_1}{6}\langle 0-0 angle^3-\frac{w}{24}\langle 0-a angle^4+\frac{R_2}{6}\langle 0-b angle^3+\frac{R_3}{6}\langle 0-l angle^3+C_3(0)+C_4 ight group \ C_4 &=0 \end{aligned} [/latex] (f)

for [latex] x = b [/latex] :

[latex] \begin{array}{c} y(b)=0=\frac{1}{E I}\left\lgroup\frac{R_1}{6}\langle b-0 angle^3-\frac{w}{24}\langle b-a angle^4+\frac{R_2}{6}\langle b-b angle^3+\frac{R_3}{6}\langle b-l angle^3+C_3 b+C_4 ight group \ C_3=\frac{1}{b}\left\lgroup(-\frac{R_1}{6} b^3+\frac{w}{24}\langle b-a angle^4 ight group=\frac{1}{7}\left\lgroup-\frac{R_1}{6} 7^3+100(7-4)^4 ight group=385.7-8.17 R_1 \end{array} [/latex] (g)

for [latex] x = l [/latex] :

[latex] \begin{aligned} y(l) &=0=\frac{1}{E I}\left\lgroup\frac{R_1}{6}\langle l-0 angle^3-\frac{w}{24}\langle l-a angle^4+\frac{R_2}{6}\langle l-b angle^3+\frac{R_3}{6}\langle l-l angle^3+C_3 l+C_4 ight group \ C_3 &=\frac{1}{l}\left\lgroup-\frac{R_1}{6} l^3+\frac{w}{24}\langle l-a angle^4-\frac{R_2}{6}\langle l-b angle^3 ight group \ C_3 &=\frac{1}{10}\left\lgroup-\frac{R_1}{6} 10^3+\frac{100}{24}\langle 10-4 angle^4-\frac{R_2}{6}\langle 10-7 angle^3 ight group=540-16.67 R_1-4.5 R_2 \end{aligned} [/latex] (h)

5 Two more equations can be written using equations ([latex]c[/latex]) and ([latex]b[/latex]) and noting that at a point [latex]l^{+}[/latex], infinitesimally beyond the right end of the beam, both [latex]V[/latex] and [latex]M[/latex] are zero. We can substitute [latex]l[/latex] for [latex]l^{+}[/latex] since their difference is vanishingly small.

[latex] \begin{aligned} M(l) &=0=R_1\langle l-0 angle^1-\frac{w}{2}\langle l-a angle^2+R_2\langle l-b angle^1+R_3\langle l-l angle^1 \ 0 &=R_1 l-\frac{w}{2}(l-a)^2+R_2(l-b) \ R_1 &=\frac{1}{l}\left[\frac{w}{2}(l-a)^2-R_2(l-b) ight] \ R_1 &=\frac{1}{10}\left[\frac{100}{2}(10-4)^2-R_2(10-7) ight]=180-0.3 R_2 \end{aligned} [/latex] (i)

[latex] \begin{aligned} V(l) &=0=R_1\langle l angle^0-w\langle l-a angle^1+R_2\langle l-b angle^0+R_3\langle l-l angle^0=0 \ 0 &=R_1-w(l-a)+R_2+R_3 \ R_2 &=w(l-a)-R_1-R_3=600-R_1-R_3 \end{aligned} [/latex] (j)

6 Equations ([latex]f[/latex]) through ([latex]j[/latex]) provide 5 equations in the 5 unknowns, [latex]R_1, R_2, R_3, C_3, C_4[/latex] and can be solved simultaneously. The deflection function can be expressed in terms of the geometry plus the loading and reaction forces, but, in this case, a simultaneous solution is necessary.

[latex] y=\frac{1}{E I}\left\lgroup\begin{array}{c} \frac{R_1}{6} x^3+\frac{1}{b}\left\lgroup\frac{w}{24}\langle b-a angle^4-\frac{R_1}{6} b^3 ight group x-\frac{w}{24}\langle x-a angle^4 \ +\frac{R_2}{6}\langle x-b angle^3+\frac{R_3}{6}\langle x-l angle^3 \end{array} ight group [/latex] (k)

7 Plots of the loading, shear, moment, slope, and deflection functions are shown in Figure 4-27 and their extreme values in Table 4-2. The files EX04-07 can be opened in the program of your choice to examine the model and see larger-scale plots of the functions shown in Figure 4-27.

Table 4-2 Example 4-7 Calculated Data
Variable	Value	Unit
[latex] R_1 [/latex]	158.4	lb
[latex] R_2 [/latex]	2471.9	lb
[latex] R_3 [/latex]	369.6	lb
[latex] C_1 [/latex]	0.0	lb
[latex] C_2 [/latex]	0.0	lb-in
[latex] C_3 [/latex]	-1052.7	rad
[latex] C_4 [/latex]	0.0	in
[latex] V_{min } [/latex]	-1291.6	lb
[latex] V_{max } [/latex]	1130.4	lb
[latex] M_{min } [/latex]	-1141.1	lb-in
[latex] M_{max } [/latex]	658.7	lb-in
[latex] heta_{min } [/latex]	-0.025	deg
[latex] heta_{max } [/latex]	0.027	deg
[latex] y_{min } [/latex]	-0.0011	in
[latex] y_{min } [/latex]	0.0001	in

Question 4.7: Finding Reactions and Deflection of Statically Indeterminate...

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:

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: