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

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Answer 1

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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

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(l2−bj2−xi2) (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=g∑wy∑wy2 (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=g∑wy∑wy2 (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=g∑wy∑wy2 (XIX)

Conclusion:

Substitute 25 mm for d in Equation (I).

I=π(25 mm)464=(19174.75985 mm4)(1 m41012 mm4)=1.917×10−8 m4

Substitute 25 mm for d in Equation (II).

A=π4(25 mm)2=(490.8738 mm2)(1 m2106 mm2)=4.909×10−4 m2

Substitute 4.909×10−4 m2 for A, 76.5 kN/m2 for γ and 600 mm for l in Equation (III).

w=(4.909×10−4 m2)(76.5 kN/m2)(600 mm)=(4.909×10−4 m2)(76.5 kN/m2)(1000 N1 kN)(600 mm)(1 m1000 mm)=22.53 N

Substitute 0.3 m for b1, 0.3 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 1 for j in Equation (VI).

δ11=(0.3 m)(0.3 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.3 m)2−(0.3 m)2)=(0.09 m2)(0.18 m2)(1242×109 N/m2)(1.1502×10−8 m5)=1.134×10−6 m/N

Substitute 1.134×10−6 m/N for δ11,and 22.53 N for w1 in Equation (VII).

y1=(22.53 N)(1.134×10−6 m/N)=2.555×10−5 m

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

y12=(2.555×10−5 m)2y12=6.528×10−10 m2

Substitute 2.555×10−5 m for y1 and 22.53 N for w1 in Equation (VIII).

∑wy=(22.53 N)(2.555×10−5 m)=5.756×10−4 N⋅m

Substitute 6.528×10−10 m2 for y12 and 22.53 N for w1 in Equation (IX).

∑wy2=(22.53 N)(6.528×10−10 m2)=1.471×10−8 N⋅m2

Substitute 5.756×10−4 N⋅m for ∑wy, 9.81 m/s2 for g and 1.471×10−8 N⋅m2 for ∑wy2 in Equation (X).

ω1=9.81 m/s2(5.756×10−4 N⋅m1.471×10−8 N⋅m2)=383863.766 rad/s≈620 rad/s

Substitute 0.45 m for b1, 0.15 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 1 for j in Equation (VI).

δ11=(0.45 m)(0.15 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.45 m)2−(0.15 m)2)=(0.0675 m2)(0.135 m2)(1242×109 N/m2)(1.1502×10−8 m5)=6.37×10−7 m/N

Substitute 0.45 m for b2, 0.15 m for x2, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 2 for i and 2 for j in Equation (VI).

δ22=(0.45 m)(0.15 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.45 m)2−(0.15 m)2)=(0.0675 m2)(0.135 m2)(1242×109 N/m2)(1.1502×10−8 m5)=6.37×10−7 m/N

Substitute 0.15 m for b2, 0.15 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 2 for j in Equation (VI).

δ12=(0.15 m)(0.15 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.15 m)2−(0.15 m)2)=(0.0225 m2)(0.315 m2)(1242×109 N/m2)(1.1502×10−8 m5)=4.961×10−7 m/N

Substitute 0.15 m for b1, 0.15 m for x2, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 2 for i and 1 for j in Equation (VI).

δ21=(0.15 m)(0.15 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.15 m)2−(0.15 m)2)=(0.0225 m2)(0.315 m2)(1242×109 N/m2)(1.1502×10−8 m5)=4.961×10−7 m/N

Substitute 6.37×10−7 m/N for δ11,11.625 N for w1, 11.625 N for w2 and 4.961×10−7 m/N for δ12 in Equation (XI).

y1=(11.625 N)(6.37×10−7 m/N)+(11.625 N)(4.961×10−7 m/N)=1.035×10−6 m+5.7671×10−6 m=1.277×10−5 m

Substitute 6.37×10−7 m/N for δ11, 11.625 N for w1, 11.625 N for w2 and 5.67×10−7 m/N for δ12 in Equation (XII).

y2=(11.625 N)(6.37×10−7 m/N)+(11.625 N)(4.961×10−7 m/N)=1.035×10−6 m+5.7671×10−6 m=1.277×10−5 m

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

y12=(1.277×10−5 m)2=1.632×10−10 m2

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

y22=(1.277×10−5 m)2=1.632×10−10 m2

Substitute 1.277×10−5 m for y1, 11.265 N for w1, 1.277×10−5 m for y2 and 11.265 N for w2 in Equation (XIII).

∑wy=(11.265 N)(1.277×10−5 m)+(11.265 N)(1.277×10−5 m)=1.438×10−4 N⋅m+1.438×10−4 N⋅m=2.877×10−4 N⋅m

Substitute 1.632×10−10 m2 for y12, 11.265 N for w2, 1.632×10−10 m2 for y22 and 11.265 N for w1 in Equation (XIV).

∑wy2=(11.265 N)(1.632×10−10 m2)+(11.265 N)(1.632×10−10 m2)=1.838×10−9 N⋅m2+1.838×10−9 N⋅m2=3.677×10−9 N⋅m2

Substitute 2.877×10−4 N⋅m for ∑wy, 9.81 m/s2 for g and 3.677×10−9 N⋅m2 for ∑wy2 in Equation (XV).

ω2=9.81 m/s2(2.877×10−4 N⋅m3.677×10−9 N⋅m2)=767565.1346 rad/s≈876 rad/s

Substitute 0.5 m for b1, 0.1 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 1 for j in Equation (VI).

δ11=(0.5 m)(0.1 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.5 m)2−(0.1 m)2)=(0.05 m2)(0.1 m2)(1242×109 N/m2)(1.1502×10−8 m5)=3.500×10−7 m/N

Substitute 0.5 m for b3, 0.1 m for x3, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 3 for i and 3 for j in Equation (VI).

δ33=(0.5 m)(0.1 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.5 m)2−(0.1 m)2)=(0.05 m2)(0.1 m2)(1242×109 N/m2)(1.1502×10−8 m5)=3.500×10−7 m/N

Substitute 0.3 m for b2, 0.3 m for x2, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 2 for i and 2 for j in Equation (VI).

δ22=(0.3 m)(0.3 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.3 m)2−(0.3 m)2)=(0.09 m2)(0.18 m2)(1242×109 N/m2)(1.1502×10−8 m5)=1.13×10−6 m/N

Substitute 0.3 m for b2, 0.1 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 2 for j in Equation (VI).

δ12=(0.3 m)(0.1 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.3 m)2−(0.1 m)2)=(0.03 m2)(0.26 m2)(1242×109 N/m2)(1.1502×10−8 m5)=5.46×10−7 m/N

Substitute 0.3 m for b2, 0.1 m for x3, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 3 for i and 2 for j in Equation (VI).

δ32=(0.3 m)(0.1 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.3 m)2−(0.1 m)2)=(0.03 m2)(0.26 m2)(1242×109 N/m2)(1.1502×10−8 m5)=5.46×10−7 m/N

Substitute 0.1 m for b3, 0.1 m for x1, 207×109 N/m2 for E, 1.917×10−8 m4 for I, 0.6 m for l, 1 for i and 3 for j in Equation (VI).

δ13=(0.1 m)(0.1 m)6(207×109 N/m2)(1.917×10−8 m4)(0.6 m)((0.6 m)2−(0.1 m)2−(0.1 m)2)=(0.01 m2)(0.34 m2)(1242×109 N/m2)(1.1502×10−8 m5)=2.38×10−7 m/N

Substitute 3.500×10−7 m/N for δ11, 7.51 N for w1, 2.38×10−7 m/N for δ13 and 5.46×10−7 m/N for δ12 in Equation (XVI).

y1=(7.51 N)[3.500×10−7 m/N+5.46×10−7 m/N+2.38×10−7 m/N]=(7.51 N)(11.34×10−7 m/N)=8.51×10−6 m

Substitute 1.13×10−6 m/N for δ22, 7.51 N for w1, 5.46×10−7 m/N for δ32 and 5.46×10−7 m/N for δ12 in Equation (XVII).

y2=(7.51 N)[5.46×10−7 m/N+1.13×10−6 m/N+5.46×10−7 m/N]=(7.51 N)(2.22×10−6 m/N)=1.672×10−5 m

Substitute 2.38×10−7 m/N for δ13, 7.51 N for w1, 5.46×10−7 m/N for δ32 and 3.500×10−7 m/N for δ33 in Equation (XVIII).

y3=(7.51 N)[2.38×10−7 m/N+5.46×10−7 m/N+3.500×10−7 m/N]=(7.51 N)(11.34×10−7 m/N)=8.51×10−6 m

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

y12=(8.51×10−6 m)2=7.25×10−11 m2

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

y22=(1.672×10−5 m)2=2.79×10−10 m2

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

y32=(8.51×10−6 m)2=7.24×10−11 m2

Substitute 8.51×10−6 m for y1, 7.51 N for w1, 1.672×10−5 m for y2 and 8.51×10−6 m for y3 in Equation (XIX).

∑wy=(7.51 N)(8.51×10−6 m+1.672×10−5 m+8.51×10−6 m)=(7.51 N)(3.374×10−5 m)=2.535×10−4 N⋅m

Substitute 7.25×10−11 m2 for y12, 7.51 N for w1, 2.79×10−10 m2 for y22 and 7.24×10−11 m2 for y32 in Equation (XX).

∑wy2=(7.51 N)(7.25×10−11 m2+2.79×10−10 m2+7.24×10−11 m2)=(7.51 N)(4.238×10−10 m2)=3.189×10−9 N⋅m2

Substitute 2.535×10−4 N⋅m for ∑wy, 9.81 m/s2 for g and 3.189×10−9 N⋅m2 for ∑wy2 in Equation (XXI).

ω3=9.81 m/s2(2.535×10−4 N⋅m3.189×10−9 N⋅m2)=779816.5569 rad/s=883 rad/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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