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 7, Problem 29P
To determine

The demonstration of how rapidly Rayleigh’s method converges for the uniform-diameter solid shaft.

Expert Solution & Answer
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Answer to Problem 29P

The Rayleigh method for uniform diameter shaft is converging rapidly by using a static deflection beam equation.

Explanation of Solution

Write the expression for moment of inertia.

    I=π64d4                                                                        (I)

Here, the diameter of the shaft is d.

Write the expression for area of the shaft.

    A=π4d2                                                                            (II)

Write the expression for weight of the shaft.

    w=Aγl                                                                          (III)

Here, the specific weight is γ

Write the expression for influence coefficient.

    δij=bjxi6EIl(l2bj2xi2)                                                        (IV)

Here, the length of the shaft is l, the elastic constant is E.

Write the expression for deflection at point 1.

    y1=w1δ11                                                                         (V)

Write the expression for Rayleigh method.

    wy=w1y1                                                                         (VI)

Write the expression for Rayleigh method.

    wy2=w1y21                                                                   (VII)

Write the expression for first critical speed.

    ω1=gwywy2                                                                 (VIII)

Draw the diagram for the two elements system.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 7, Problem 29P , additional homework tip  1

Figure-(1)

The figure-(1) shows the required dimension.

Write the expression for the deflection at point 1 for two element.

    y1=w1δ11+w2δ12                                                                 (IX)

Write the expression for the deflection at point 2 for two element.

    y2=w1δ11+w2δ12                                                                      (X)

Write the expression for Rayleigh method for two element.

    wy=w1y1+w2y2                                                                  (XI)

Write the expression for Rayleigh method for two element.

    wy2=w1y21+w2y22                                                              (XII)

Write the expression for first critical speed for two element.

    ω2=gwywy2                                                                         (XIII)

Draw the diagram for the three element system.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 7, Problem 29P , additional homework tip  2

Figure-(2)

The Figure-(2) shows all the dimensions for the three elements.

Write the expression for the deflection at point 1 for three elements

    y1=w1(δ11+δ12+δ13)                                                             (XIV)

Write the expression for the deflection at point 2 for three element.

    y2=w1(δ12+δ22+δ32)                                                                 (XV)

Write the expression for the deflection at point 2 for three element.

    y3=w1(δ13+δ32+δ33)                                                                  (XVI)

Write the expression for Rayleigh method for three elements.

    wy=w1(y1+y2+y3)                                                               (XVII)

Write the expression for Rayleigh method for three elements.

    wy2=w1(y12+y22+y32)                                                          (XVIII)

Write the expression for first critical speed for three elements.

    ω3=gwywy2                                                                              (XIX)

Conclusion:

Substitute 25mm for d in Equation (I).

    I=π(25mm)464=(19174.75985mm4)(1m41012mm4)=1.917×108m4

Substitute 25mm for d in Equation (II).

    A=π4(25mm)2=(490.8738mm2)(1m2106mm2)=4.909×104m2

Substitute 4.909×104m2 for A, 76.5kN/m2 for γ and 600mm for l in Equation (III).

    w=(4.909×104m2)(76.5kN/m2)(600mm)=(4.909×104m2)(76.5kN/m2)(1000N1kN)(600mm)(1m1000mm)=22.53N

Substitute 0.3m for b1, 0.3m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 1 for j in Equation (VI).

    δ11=(0.3m)(0.3m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.3m)2(0.3m)2)=(0.09m2)(0.18m2)(1242×109N/m2)(1.1502×108m5)=1.134×106m/N

Substitute 1.134×106m/N for δ11,and 22.53N for w1 in Equation (VII).

    y1=(22.53N)(1.134×106m/N)=2.555×105m

Calculate the square of the deflection at point 1 of element 1.

    y12=(2.555×105m)2y12=6.528×1010m2

Substitute 2.555×105m for y1 and 22.53N for w1 in Equation (VIII).

    wy=(22.53N)(2.555×105m)=5.756×104Nm

Substitute 6.528×1010m2 for y12 and 22.53N for w1 in Equation (IX).

    wy2=(22.53N)(6.528×1010m2)=1.471×108Nm2

Substitute 5.756×104Nm for wy, 9.81m/s2 for g and 1.471×108Nm2 for wy2 in Equation (X).

    ω1=9.81m/s2(5.756×104Nm1.471×108Nm2)=383863.766rad/s620rad/s

Substitute 0.45m for b1, 0.15m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 1 for j in Equation (VI).

    δ11=(0.45m)(0.15m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.45m)2(0.15m)2)=(0.0675m2)(0.135m2)(1242×109N/m2)(1.1502×108m5)=6.37×107m/N

Substitute 0.45m for b2, 0.15m for x2, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 2 for i and 2 for j in Equation (VI).

    δ22=(0.45m)(0.15m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.45m)2(0.15m)2)=(0.0675m2)(0.135m2)(1242×109N/m2)(1.1502×108m5)=6.37×107m/N

Substitute 0.15m for b2, 0.15m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 2 for j in Equation (VI).

    δ12=(0.15m)(0.15m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.15m)2(0.15m)2)=(0.0225m2)(0.315m2)(1242×109N/m2)(1.1502×108m5)=4.961×107m/N

Substitute 0.15m for b1, 0.15m for x2, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 2 for i and 1 for j in Equation (VI).

    δ21=(0.15m)(0.15m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.15m)2(0.15m)2)=(0.0225m2)(0.315m2)(1242×109N/m2)(1.1502×108m5)=4.961×107m/N

Substitute 6.37×107m/N for δ11,11.625N for w1, 11.625N for w2 and 4.961×107m/N for δ12 in Equation (XI).

    y1=(11.625N)(6.37×107m/N)+(11.625N)(4.961×107m/N)=1.035×106m+5.7671×106m=1.277×105m

Substitute 6.37×107m/N for δ11, 11.625N for w1, 11.625N for w2 and 5.67×107m/N for δ12 in Equation (XII).

    y2=(11.625N)(6.37×107m/N)+(11.625N)(4.961×107m/N)=1.035×106m+5.7671×106m=1.277×105m

Calculate the square of the deflection at point 1 of element 2.

    y12=(1.277×105m)2=1.632×1010m2

Calculate the square of the deflection at point 2 of element 2.

    y22=(1.277×105m)2=1.632×1010m2

Substitute 1.277×105m for y1, 11.265N for w1, 1.277×105m for y2 and 11.265N for w2 in Equation (XIII).

    wy=(11.265N)(1.277×105m)+(11.265N)(1.277×105m)=1.438×104Nm+1.438×104Nm=2.877×104Nm

Substitute 1.632×1010m2 for y12, 11.265N for w2, 1.632×1010m2 for y22 and 11.265N for w1 in Equation (XIV).

    wy2=(11.265N)(1.632×1010m2)+(11.265N)(1.632×1010m2)=1.838×109Nm2+1.838×109Nm2=3.677×109Nm2

Substitute 2.877×104Nm for wy, 9.81m/s2 for g and 3.677×109Nm2 for wy2 in Equation (XV).

    ω2=9.81m/s2(2.877×104Nm3.677×109Nm2)=767565.1346rad/s876rad/s

Substitute 0.5m for b1, 0.1m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 1 for j in Equation (VI).

    δ11=(0.5m)(0.1m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.5m)2(0.1m)2)=(0.05m2)(0.1m2)(1242×109N/m2)(1.1502×108m5)=3.500×107m/N

Substitute 0.5m for b3, 0.1m for x3, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 3 for i and 3 for j in Equation (VI).

δ33=(0.5m)(0.1m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.5m)2(0.1m)2)=(0.05m2)(0.1m2)(1242×109N/m2)(1.1502×108m5)=3.500×107m/N

Substitute 0.3m for b2, 0.3m for x2, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 2 for i and 2 for j in Equation (VI).

δ22=(0.3m)(0.3m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.3m)2(0.3m)2)=(0.09m2)(0.18m2)(1242×109N/m2)(1.1502×108m5)=1.13×106m/N

Substitute 0.3m for b2, 0.1m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 2 for j in Equation (VI).

    δ12=(0.3m)(0.1m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.3m)2(0.1m)2)=(0.03m2)(0.26m2)(1242×109N/m2)(1.1502×108m5)=5.46×107m/N

Substitute 0.3m for b2, 0.1m for x3, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 3 for i and 2 for j in Equation (VI).

    δ32=(0.3m)(0.1m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.3m)2(0.1m)2)=(0.03m2)(0.26m2)(1242×109N/m2)(1.1502×108m5)=5.46×107m/N

Substitute 0.1m for b3, 0.1m for x1, 207×109N/m2 for E, 1.917×108m4 for I, 0.6m for l, 1 for i and 3 for j in Equation (VI).

    δ13=(0.1m)(0.1m)6(207×109N/m2)(1.917×108m4)(0.6m)((0.6m)2(0.1m)2(0.1m)2)=(0.01m2)(0.34m2)(1242×109N/m2)(1.1502×108m5)=2.38×107m/N

Substitute 3.500×107m/N for δ11, 7.51N for w1, 2.38×107m/N for δ13 and 5.46×107m/N for δ12 in Equation (XVI).

    y1=(7.51N)[3.500×107m/N+5.46×107m/N+2.38×107m/N]=(7.51N)(11.34×107m/N)=8.51×106m

Substitute 1.13×106m/N for δ22, 7.51N for w1, 5.46×107m/N for δ32 and 5.46×107m/N for δ12 in Equation (XVII).

    y2=(7.51N)[5.46×107m/N+1.13×106m/N+5.46×107m/N]=(7.51N)(2.22×106m/N)=1.672×105m

Substitute 2.38×107m/N for δ13, 7.51N for w1, 5.46×107m/N for δ32 and 3.500×107m/N for δ33 in Equation (XVIII).

    y3=(7.51N)[2.38×107m/N+5.46×107m/N+3.500×107m/N]=(7.51N)(11.34×107m/N)=8.51×106m

Calculate the square of the deflection at point 1 of element 3.

    y12=(8.51×106m)2=7.25×1011m2

Calculate the square of the deflection at point 2 of element 3.

    y22=(1.672×105m)2=2.79×1010m2

Calculate the square of the deflection at point 3 of element 3.

    y32=(8.51×106m)2=7.24×1011m2

Substitute 8.51×106m for y1, 7.51N for w1, 1.672×105m for y2 and 8.51×106m for y3 in Equation (XIX).

    wy=(7.51N)(8.51×106m+1.672×105m+8.51×106m)=(7.51N)(3.374×105m)=2.535×104Nm

Substitute 7.25×1011m2 for y12, 7.51N for w1, 2.79×1010m2 for y22 and 7.24×1011m2 for y32 in Equation (XX).

    wy2=(7.51N)(7.25×1011m2+2.79×1010m2+7.24×1011m2)=(7.51N)(4.238×1010m2)=3.189×109Nm2

Substitute 2.535×104Nm for wy, 9.81m/s2 for g and 3.189×109Nm2 for wy2 in Equation (XXI).

    ω3=9.81m/s2(2.535×104Nm3.189×109Nm2)=779816.5569rad/s=883rad/s

Since the static bending equation is available, and satisfied the moment-free and deflection-free ends, so the convergence is rapid using a static deflection beam equation.

Thus, the Rayleigh method for uniform diameter shaft is converging rapidly by using a static deflection beam equation.

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Chapter 7 Solutions

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

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