Question 4.7: Finding Reactions and Deflection of Statically Indeterminate...
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 l = 10 in, a = 4 in, and b = 7 in. The beam’s I = 0.08 in4 and E = 30 Mpsi. The magnitude of the distributed force is w = 500 in/lb.
Assumptions: Ignore the beam weight as negligible compared to the applied load.

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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.
⟨x−a⟩2 (3.17a)
⟨x−a⟩1 (3.17b)
⟨x−a⟩0 (3.17c)
⟨x−a⟩−1 (3.17d)
⟨x−a⟩−2 (3.17e)
∫−∞x⟨λ−a⟩2dλ=3⟨x−a⟩3 (3.18a)
∫−∞x⟨λ−a⟩1dλ=2⟨x−a⟩2 (3.18b)
∫−∞x⟨λ−a⟩0dλ=⟨x−a⟩1 (3.18c)
∫−∞x⟨λ−a⟩−1dλ=⟨x−a⟩0 (3.18d)
∫−∞x⟨λ−a⟩−2dλ=⟨x−a⟩−1 (3.18e)
q=R1⟨x−0⟩−1−w⟨x−a⟩0+R2⟨x−b⟩−1+R3⟨x−l⟩−1 (a)
V=∫qdx=R1⟨x−0⟩0−w⟨x−a⟩1+R2⟨x−b⟩0+R3⟨x−l⟩0+C1 (b)
M=∫Vdx=R1⟨x−0⟩1−2w⟨x−a⟩2+R2⟨x−b⟩1+R3⟨x−l⟩1+C1x+C2 (c)
θ=∫EIMdx=EI1⎩⎪⎪⎪⎧2R1⟨x−0⟩2−6w⟨x−a⟩3+2R2⟨x−b⟩2+2R3⟨x−l⟩2+2C1x2+C2x+C3⎭⎪⎪⎪⎫ (d)
y=∫θdx=EI1⎩⎪⎪⎪⎧6R1⟨x−0⟩3−24w⟨x−a⟩4+6R2⟨x−b⟩3+6R3⟨x−l⟩3+6C1x3+2C2x2+C3x+C4⎭⎪⎪⎪⎫ (e)
2 There are 3 reaction forces and 4 constants of integration to be found. The constants C1 and C2 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 x=0 (denoted as x=0–), the shear and moment will both be zero. The same conditions obtain at a point infinitesimally to the right of x=l (denoted as x=l+). Also, the deflection y 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 x=0–,V=0,M=0; when x=0,y=0; when x=b,y=0; when x=l,y=0; when x=l+,V=0,M=0.
4 Substitute the boundary conditions x=0,y=0,x=b,y=0, and x=l,y=0 into (e).
for x = 0 :
y(0)C4=0=EI1⎩⎪⎪⎧6R1⟨0−0⟩3−24w⟨0−a⟩4+6R2⟨0−b⟩3+6R3⟨0−l⟩3+C3(0)+C4⎭⎪⎪⎫=0 (f)
for x=b :
y(b)=0=EI1⎩⎪⎧6R1⟨b−0⟩3−24w⟨b−a⟩4+6R2⟨b−b⟩3+6R3⟨b−l⟩3+C3b+C4⎭⎪⎫C3=b1⎩⎪⎧(−6R1b3+24w⟨b−a⟩4⎭⎪⎫=71⎩⎪⎧−6R173+100(7−4)4⎭⎪⎫=385.7−8.17R1 (g)
for x=l :
y(l)C3C3=0=EI1⎩⎪⎪⎧6R1⟨l−0⟩3−24w⟨l−a⟩4+6R2⟨l−b⟩3+6R3⟨l−l⟩3+C3l+C4⎭⎪⎪⎫=l1⎩⎪⎪⎧−6R1l3+24w⟨l−a⟩4−6R2⟨l−b⟩3⎭⎪⎪⎫=101⎩⎪⎪⎧−6R1103+24100⟨10−4⟩4−6R2⟨10−7⟩3⎭⎪⎪⎫=540−16.67R1−4.5R2 (h)
5 Two more equations can be written using equations (c) and (b) and noting that at a point l+, infinitesimally beyond the right end of the beam, both V and M are zero. We can substitute l for l+ since their difference is vanishingly small.
M(l)0R1R1=0=R1⟨l−0⟩1−2w⟨l−a⟩2+R2⟨l−b⟩1+R3⟨l−l⟩1=R1l−2w(l−a)2+R2(l−b)=l1[2w(l−a)2−R2(l−b)]=101[2100(10−4)2−R2(10−7)]=180−0.3R2 (i)
V(l)0R2=0=R1⟨l⟩0−w⟨l−a⟩1+R2⟨l−b⟩0+R3⟨l−l⟩0=0=R1−w(l−a)+R2+R3=w(l−a)−R1−R3=600−R1−R3 (j)
6 Equations (f) through (j) provide 5 equations in the 5 unknowns, R1,R2,R3,C3,C4 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.
y=EI1⎩⎪⎪⎪⎪⎪⎧6R1x3+b1⎩⎪⎧24w⟨b−a⟩4−6R1b3⎭⎪⎫x−24w⟨x−a⟩4+6R2⟨x−b⟩3+6R3⟨x−l⟩3⎭⎪⎪⎪⎪⎪⎫ (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 |
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Variable | Value | Unit |
R1 | 158.4 | lb |
R2 | 2471.9 | lb |
R3 | 369.6 | lb |
C1 | 0.0 | lb |
C2 | 0.0 | lb-in |
C3 | -1052.7 | rad |
C4 | 0.0 | in |
Vmin | -1291.6 | lb |
Vmax | 1130.4 | lb |
Mmin | -1141.1 | lb-in |
Mmax | 658.7 | lb-in |
θmin | -0.025 | deg |
θmax | 0.027 | deg |
ymin | -0.0011 | in |
ymin | 0.0001 | in |
