Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Chapter 4, Problem 21P

Consider the uniformly loaded simply supported steel beam with an overhang as shown. The second-area moment of the beam is I = 0.05 in4. Use superposition (with Table A–9 and the results of Prob. 4–20) to determine the reactions and the deflection equations of the beam. Plot the deflections.

Problem 4–21

Chapter 4, Problem 21P, Consider the uniformly loaded simply supported steel beam with an overhang as shown. The second-area

Expert Solution & Answer
Check Mark
To determine

The net reaction at A.

The net reaction at B.

The expression for the deflection in the beam of portion AB using superposition.

The expression for the deflection in the beam of portion BC using superposition.

The plot of deflection verses length of the beam.

Answer to Problem 21P

The net reaction at A is 420lbf.

The net reaction at B is 980lbf.

The expression for the deflection in the beam of portion AB using superposition is

yAB=[2.777×106x((20)x2x31000)+((8.888×106)x(100x2))].

The expression for the deflection in the beam of portion BC using superposition is

yBC={(2.777×103)(x10)(2.777×106)[(14+x)4+896x9216]}.

The plot of deflection verses length of the beam is

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 4, Problem 21P , additional homework tip  1

Explanation of Solution

Write the expression for the reaction at A (one end) due to uniform load between the supports.

    R1=wl2 (I)

Here, the uniform load on the beam is w, the length of the beam is l and the reaction at the A is R1.

Write the expression for the reaction at A (other end) due to uniform load between the supports.

    R2=wl2                                                                                          (II)

Here, the reaction at the A is R2.

Write the expression for the deflection of the beam between A and B.

    yAB=wx24EI(2lx2x3l3)                                                           (III)

Here, the moment of inertia of the beam is I and the modulus of the elasticity is E

Write the expression for the slope of the deflection in the beam.

    θAB=dyABdx                                                                                  (IV)

Substitute wx24EI(2lx2x3l3) for yAB in Equation (IV).

    θAB=dyABdx=d(wx24EI(2lx2x3l3))dx=w24EI(6lx24x3l3)                                                     (V)

Substitute l for x in Equation (V).

    θAB=w24EI(6l(l2)4l3l3)=wl324EI                                                       (VI)

Write the expression for the deflection of the beam for BC portion.

    yBC=θAB|x=l(xl)                                                                     (VII)

Substitute wl324EI for θAB in Equation (VII).

    yBC=θAB|x=l(xl)=wl324EI(xl)                                                                  (VIII)

Write the expression for the reaction at one end of the overhang section.

    RA=wa22l                                                                                    (IX)

Here, the distance between B and C is a.

Write the expression for the reaction at other end of the overhang section.

    RB=wa2l(2la)                                                                            (X)

Write the expression for the deflection in the beam section AB due to overhung load.

    yAB=wa2x12EIl(l2x2)                                                                   (XI)

Write the expression for the deflection in the beam section BC due to overhung load.

    yBC=w24EI[(l+ax)44a2(lx)(l+a)a4]               (XII)

Write the expression for the net reaction at A.

    RnA=R1RA                                                                          (XIII)

Write the expression for the net reaction at B.

    RnB=R2+RB                                                                          (XIV)

Add equation (III) and (XI).

    yAB=wx24EI(2lx2x3l3)+wa2x12EIl(l2x2)=wx12EI[12(2lx2x3l3)+a2l(l2x2)]                              (XV)

Add equation (VII) and (XII).

    yBC=wl324EI(xl)w24EI[(l+ax)44a2(lx)(l+a)a4]=w24EI[l3(xl)[(l+ax)44a2(lx)(l+a)a4]] (XVI)

Conclusion:

Substitute 100lbf/in for w and 10in for l in Equation (I).

    R1=(100lbf/in)(10in)2=500lbf

Substitute 100lbf/in for w and 10in for l in Equation (II).

    R2=(100lbf/in)(10in)2=500lbf

Substitute 100lbf/in for w4in for a and 10in for l in Equation (IX).

    RA=(100lbf/in)(4in)22(10in)=1600lbf/in(in2)20in=80lbf

Substitute 100lbf/in for w, 4in for a and 10in for l in Equation (X).

    RB=(100lbf/in)(4in)2(10in)(2(10in)+4in)=9600lbf/in(in2)20in=480lbf

Substitute 80lbf for RA and 500lbf for R1 in Equation (XIII).

    RnA=500lbf80lbf=420lbf

Thus, the net reaction at A is 420lbf.

Substitute 480lbf for RB and 500lbf for R2 in Equation (XIV).

    RnB=500lbf+480lbf=980lbf

Thus, the net reaction at B is 980lbf.

Substitute 100lbf/in for w, 30×106psi for E, 0.05in4 for I, 4in for a and 10in for l in Equation (XV).

    yAB=(100lbf/in)x12(30×106psi)(0.05in4)[12(2(10in)x2x3(10in)3)(4in)2(10in)((10in)2x2)]=5.555×106x[(0.5(20in)x2x31000in3)(1.6in(100in2x2))]=[2.777×106x((20)x2x31000)+((8.888×106)x(100x2))] (XVII)

Thus, the expression for the deflection in the beam of portion AB using superposition is yAB=[2.777×106x((20)x2x31000)+((8.888×106)x(100x2))].

Substitute 100lbf/in for w, 30×106psi for E, 0.05in4 for I, 4in for a and 10in for l in Equation (XVI).

    yBC=(100lbf/in)24(30×106psi)(0.05in4)[((10in)3(x10in))[(10in+4inx)44(4in)2(10inx)(10in+4in)(4in)4]]=2.777×106[((1000)(x10))[(14+x)4896(10x)256]]={(2.777×103)(x10)(2.777×106)[(14+x)4+896x9216]}. (XVIII)

Thus, the expression for the deflection in the beam of portion BC using superposition is yBC={(2.777×103)(x10)(2.777×106)[(14+x)4+896x9216]}.

Substitute different values of x to calculate the deflection at different points of the beam.

Substitute 0 for x in Equation (XVII).

    yAB=[2.777×106(0)((20)(0)2(0)31000)+((8.888×106)(0)(100(0)2))]=0

Substitute 1 for x in Equation (XVII).

    yAB=[2.777×106(1)((20)(1)2(1)31000)+((8.888×106)(1)(100(1)2))]=(2.7242×103)+(8.7991×104)=0.001844

Similarly,

Use excel spread sheet to calculate the y values for different values of x.

The Table-(1) shows the different values of the deflection at different point of the beam.

Table-(1)

S. No.length (x) indeflection (y) in
100.0000
20.50.00093
31.00.00184
41.50.00269
520.00344
62.50.00410
730.00463
83.50.00502
940.00528
104.50.00538
1150.00534
125.50.00516
1360.00485
146.50.00442
1570.00388
167.50.00326
1780.00259
188.50.00189
1990.00120
209.50.00055
21100.00000
2210.50.00043
23110.00077
2411.50.00103
25120.00124
2612.50.00141
27130.00157
2813.50.00172
29140.00186

Draw the plot of length of the beam verses deflection of the beam.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 4, Problem 21P , additional homework tip  2

Figure-(1)

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Chapter 4 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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