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 ll = 10 in, aa = 4 in, and bb = 7 in. The beam’s II = 0.08 in4^{4} and EE = 30 Mpsi. The magnitude of the distributed force is ww = 500 in/lb.

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

F4-22d
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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.

xa2 \langle x-a\rangle^2       (3.17a)

xa1 \langle x-a\rangle^1        (3.17b)

xa0 \langle x-a\rangle^0        (3.17c)

xa1 \langle x-a\rangle^{-1}         (3.17d)

xa2 \langle x-a\rangle^{-2}      (3.17e)

xλa2dλ=xa33 \int_{-\infty}^x\langle\lambda-a\rangle^2 d \lambda=\frac{\langle x-a\rangle^3}{3}   (3.18a)

xλa1dλ=xa22 \int_{-\infty}^x\langle\lambda-a\rangle^1 d \lambda=\frac{\langle x-a\rangle^2}{2}   (3.18b)

xλa0dλ=xa1 \int_{-\infty}^x\langle\lambda-a\rangle^0 d \lambda=\langle x-a\rangle^1     (3.18c)

xλa1dλ=xa0 \int_{-\infty}^x\langle\lambda-a\rangle^{-1} d \lambda=\langle x-a\rangle^0     (3.18d)

xλa2dλ=xa1 \int_{-\infty}^x\langle\lambda-a\rangle^{-2} d \lambda=\langle x-a\rangle^{-1}     (3.18e)

q=R1x01wxa0+R2xb1+R3xl1 q=R_1\langle x-0\rangle^{-1}-w\langle x-a\rangle^0+R_2\langle x-b\rangle^{-1}+R_3\langle x-l\rangle^{-1}     (a)

V=qdx=R1x00wxa1+R2xb0+R3xl0+C1 V=\int q d x=R_1\langle x-0\rangle^0-w\langle x-a\rangle^1+R_2\langle x-b\rangle^0+R_3\langle x-l\rangle^0+C_1     (b)

M=Vdx=R1x01w2xa2+R2xb1+R3xl1+C1x+C2 M=\int V d x=R_1\langle x-0\rangle^1-\frac{w}{2}\langle x-a\rangle^2+R_2\langle x-b\rangle^1+R_3\langle x-l\rangle^1+C_1 x+C_2     (c)

θ=MEIdx=1EIR12x02w6xa3+R22xb2+R32xl2+C1x22+C2x+C3 \theta=\int \frac{M}{E I} d x=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{2}\langle x-0\rangle^2-\frac{w}{6}\langle x-a\rangle^3+\frac{R_2}{2}\langle x-b\rangle^2+\frac{R_3}{2}\langle x-l\rangle^2 \\ +\frac{C_1 x^2}{2}+C_2 x+C_3 \end{array} \right\rgroup       (d)

y=θdx=1EIR16x03w24xa4+R26xb3+R36xl3+C1x36+C2x22+C3x+C4 y=\int \theta d x=\frac{1}{E I}\left\lgroup \begin{array}{c} \frac{R_1}{6}\langle x-0\rangle^3-\frac{w}{24}\langle x-a\rangle^4+\frac{R_2}{6}\langle x-b\rangle^3+\frac{R_3}{6}\langle x-l\rangle^3 \\ +\frac{C_1 x^3}{6}+\frac{C_2 x^2}{2}+C_3 x+C_4 \end{array} \right\rgroup      (e)

2    There are 3 reaction forces and 4 constants of integration to be found. The constants C1C_{1} and C2C_{2} 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=0x = 0 (denoted as x=0x = 0^{–}), the shear and moment will both be zero. The same conditions obtain at a point infinitesimally to the right of x=lx = l (denoted as x=l+x = l^{+}). Also, the deflection yy 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=0x = 0^{–}, V = 0, M = 0; when x=0,y=0x = 0, y = 0; when x=b,y=0x=b, y = 0; when x=l,y=0x=l, y = 0; when x=l+,V=0,M=0x = l+, V = 0, M = 0.
4    Substitute the boundary conditions x=0,y=0,x=b,y=0x = 0, y = 0, x = b, y = 0, and x=l,y=0x = l, y = 0 into (ee).
for x x = 0 :

y(0)=0=1EIR16003w240a4+R260b3+R360l3+C3(0)+C4C4=0 \begin{aligned} y(0) &=0=\frac{1}{E I}\left\lgroup \frac{R_1}{6}\langle 0-0\rangle^3-\frac{w}{24}\langle 0-a\rangle^4+\frac{R_2}{6}\langle 0-b\rangle^3+\frac{R_3}{6}\langle 0-l\rangle^3+C_3(0)+C_4 \right\rgroup \\ C_4 &=0 \end{aligned}      (f)

for x=b x = b :

y(b)=0=1EIR16b03w24ba4+R26bb3+R36bl3+C3b+C4C3=1b(R16b3+w24ba4=17R1673+100(74)4=385.78.17R1 \begin{array}{c} y(b)=0=\frac{1}{E I}\left\lgroup\frac{R_1}{6}\langle b-0\rangle^3-\frac{w}{24}\langle b-a\rangle^4+\frac{R_2}{6}\langle b-b\rangle^3+\frac{R_3}{6}\langle b-l\rangle^3+C_3 b+C_4\right\rgroup \\ C_3=\frac{1}{b}\left\lgroup(-\frac{R_1}{6} b^3+\frac{w}{24}\langle b-a\rangle^4\right\rgroup=\frac{1}{7}\left\lgroup-\frac{R_1}{6} 7^3+100(7-4)^4\right\rgroup=385.7-8.17 R_1 \end{array}    (g)

for x=l x = l :

y(l)=0=1EIR16l03w24la4+R26lb3+R36ll3+C3l+C4C3=1lR16l3+w24la4R26lb3C3=110R16103+100241044R261073=54016.67R14.5R2 \begin{aligned} y(l) &=0=\frac{1}{E I}\left\lgroup\frac{R_1}{6}\langle l-0\rangle^3-\frac{w}{24}\langle l-a\rangle^4+\frac{R_2}{6}\langle l-b\rangle^3+\frac{R_3}{6}\langle l-l\rangle^3+C_3 l+C_4\right\rgroup \\ C_3 &=\frac{1}{l}\left\lgroup-\frac{R_1}{6} l^3+\frac{w}{24}\langle l-a\rangle^4-\frac{R_2}{6}\langle l-b\rangle^3\right\rgroup \\ C_3 &=\frac{1}{10}\left\lgroup-\frac{R_1}{6} 10^3+\frac{100}{24}\langle 10-4\rangle^4-\frac{R_2}{6}\langle 10-7\rangle^3\right\rgroup=540-16.67 R_1-4.5 R_2 \end{aligned}       (h)

5    Two more equations can be written using equations (cc) and (bb) and noting that at a point l+l^{+}, infinitesimally beyond the right end of the beam, both VV and MM are zero. We can substitute ll for l+l^{+} since their difference is vanishingly small.

M(l)=0=R1l01w2la2+R2lb1+R3ll10=R1lw2(la)2+R2(lb)R1=1l[w2(la)2R2(lb)]R1=110[1002(104)2R2(107)]=1800.3R2 \begin{aligned} M(l) &=0=R_1\langle l-0\rangle^1-\frac{w}{2}\langle l-a\rangle^2+R_2\langle l-b\rangle^1+R_3\langle l-l\rangle^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)\right] \\ R_1 &=\frac{1}{10}\left[\frac{100}{2}(10-4)^2-R_2(10-7)\right]=180-0.3 R_2 \end{aligned}     (i)

V(l)=0=R1l0wla1+R2lb0+R3ll0=00=R1w(la)+R2+R3R2=w(la)R1R3=600R1R3 \begin{aligned} V(l) &=0=R_1\langle l\rangle^0-w\langle l-a\rangle^1+R_2\langle l-b\rangle^0+R_3\langle l-l\rangle^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}      (j)

6    Equations (ff) through (jj) provide 5 equations in the 5 unknowns, R1,R2,R3,C3,C4R_1, R_2, R_3, C_3, C_4 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=1EIR16x3+1bw24ba4R16b3xw24xa4+R26xb3+R36xl3 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\rangle^4-\frac{R_1}{6} b^3\right\rgroup x-\frac{w}{24}\langle x-a\rangle^4 \\ +\frac{R_2}{6}\langle x-b\rangle^3+\frac{R_3}{6}\langle x-l\rangle^3 \end{array}\right\rgroup     (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
R1  R_1  158.4 lb
R2  R_2  2471.9 lb
R3  R_3  369.6 lb
C1  C_1  0.0 lb
C2  C_2  0.0 lb-in
C3  C_3  -1052.7 rad
C4  C_4  0.0 in
Vmin  V_{min }  -1291.6 lb
Vmax  V_{max }  1130.4 lb
Mmin  M_{min }  -1141.1 lb-in
Mmax  M_{max }  658.7 lb-in
θmin  \theta_{min }  -0.025 deg
θmax  \theta_{max }  0.027 deg
ymin  y_{min }  -0.0011 in
ymin y_{min } 0.0001 in
F4-27

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