Chapter 7, Problem 29P

To determine

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

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.

    $I=\frac{\pi }{64}{d}^4$                                                                        (I)

Here, the diameter of the shaft is $d$.

Write the expression for area of the shaft.

    $A=\frac{\pi }{4}{d}^2$                                                                            (II)

Write the expression for weight of the shaft.

    $w=A\gamma l$                                                                          (III)

Here, the specific weight is $\gamma$

Write the expression for influence coefficient.

    ${\delta }_{ij}=\frac{b_j{x}_i}{6EIl}\left(l^2-b_j^2-x_{{}_i}^2\right)$                                                        (IV)

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

Write the expression for deflection at point 1.

    $y_1=w_1{\delta }_{11}$                                                                         (V)

Write the expression for Rayleigh method.

    $\sum wy=w_1{y}_1$                                                                         (VI)

Write the expression for Rayleigh method.

    $\sum w{y}^2=w_1{y}^2{}_1$                                                                   (VII)

Write the expression for first critical speed.

    ${\omega }_1=\sqrt{g\frac{\sum wy}{\sum w{y}^2}}$                                                                 (VIII)

Draw the diagram for the two elements system.

Figure-(1)

The figure-(1) shows the required dimension.

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

    $y_1=w_1{\delta }_{11}+w_2{\delta }_{12}$                                                                 (IX)

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

    $y_2=w_1{\delta }_{11}+w_2{\delta }_{12}$                                                                      (X)

Write the expression for Rayleigh method for two element.

    $\sum wy=w_1{y}_1+w_2{y}_2$                                                                  (XI)

Write the expression for Rayleigh method for two element.

    $\sum w{y}^2=w_1{y}^2{}_1+w_2{y}^2{}_2$                                                              (XII)

Write the expression for first critical speed for two element.

    ${\omega }_2=\sqrt{g\frac{\sum wy}{\sum w{y}^2}}$                                                                         (XIII)

Draw the diagram for the three element system.

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

    $y_1=w_1\left({\delta }_{11}+{\delta }_{12}+{\delta }_{13}\right)$                                                             (XIV)

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

    $y_2=w_1\left({\delta }_{12}+{\delta }_{22}+{\delta }_{32}\right)$                                                                 (XV)

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

    $y_3=w_1\left({\delta }_{13}+{\delta }_{32}+{\delta }_{33}\right)$                                                                  (XVI)

Write the expression for Rayleigh method for three elements.

    $\sum wy=w_1\left(y_1+y_2+y_3\right)$                                                               (XVII)

Write the expression for Rayleigh method for three elements.

    $\sum w{y}^2=w_1\left(y_1^2+y_2^2+y_3^2\right)$                                                          (XVIII)

Write the expression for first critical speed for three elements.

    ${\omega }_3=\sqrt{g\frac{\sum wy}{\sum w{y}^2}}$                                                                              (XIX)

Conclusion:

Substitute $25\text{mm}$ for $d$ in Equation (I).

    $\begin{array}{c}I=\frac{\pi {\left(25\text{mm}\right)}^4}{64}\\ =\left(19174.75985{\text{mm}}^4\right)\left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)\\ =1.917\times {10}^{-8}{\text{m}}^4\end{array}$

Substitute $25\text{mm}$ for $d$ in Equation (II).

    $\begin{array}{c}A=\frac{\pi }{4}{\left(25\text{mm}\right)}^2\\ =\left(490.8738{\text{mm}}^2\right)\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\\ =4.909\times {10}^{-4}{\text{m}}^2\end{array}$

Substitute $4.909\times {10}^{-4}{\text{m}}^2$ for $A$, $76.5\text{kN}/{\text{m}}^2$ for $\gamma$ and $600\text{mm}$ for $l$ in Equation (III).

    $\begin{array}{c}w=\left(4.909\times {10}^{-4}{\text{m}}^2\right)\left(76.5\text{kN}/{\text{m}}^2\right)\left(600\text{mm}\right)\\ =\left(4.909\times {10}^{-4}{\text{m}}^2\right)\left(76.5\text{kN}/{\text{m}}^2\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\left(600\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =22.53\text{N}\end{array}$

Substitute $0.3\text{m}$ for $b_1$, $0.3\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $1$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{11}=\frac{\left(0.3\text{m}\right)\left(0.3\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.3\text{m}\right)}^2-{\left(0.3\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.09{\text{m}}^2\right)\left(0.18{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =1.134\times {10}^{-6}\text{m}/\text{N}\end{array}$

Substitute $1.134\times {10}^{-6}\text{m}/\text{N}$ for ${\delta }_{11}$,and $22.53\text{N}$ for $w_1$ in Equation (VII).

    $\begin{array}{c}{y}_1=\left(22.53\text{N}\right)\left(1.134\times {10}^{-6}\text{m}/\text{N}\right)\\ =2.555\times {10}^{-5}\text{m}\end{array}$

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

    $\begin{array}{c}{y}_1^2={\left(2.555\times {10}^{-5}\text{m}\right)}^2\\ y_1^2=6.528\times {10}^{-10}{\text{m}}^2\end{array}$

Substitute $2.555\times {10}^{-5}\text{m}$ for $y_1$ and $22.53\text{N}$ for $w_1$ in Equation (VIII).

    $\begin{array}{c}\sum wy=\left(22.53\text{N}\right)\left(2.555\times {10}^{-5}\text{m}\right)\\ =5.756\times {10}^{-4}\text{N}\cdot \text{m}\end{array}$

Substitute $6.528\times {10}^{-10}{\text{m}}^2$ for $y_1^2$ and $22.53\text{N}$ for $w_1$ in Equation (IX).

    $\begin{array}{c}\sum w{y}^2=\left(22.53\text{N}\right)\left(6.528\times {10}^{-10}{\text{m}}^2\right)\\ =1.471\times {10}^{-8}\text{N}\cdot {\text{m}}^2\end{array}$

Substitute $5.756\times {10}^{-4}\text{N}\cdot \text{m}$ for $\sum wy$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $1.471\times {10}^{-8}\text{N}\cdot {\text{m}}^2$ for $\sum w{y}^2$ in Equation (X).

    $\begin{array}{c}{\omega }_1=\sqrt{9.81\text{m}/{\text{s}}^2\left(\frac{5.756\times {10}^{-4}\text{N}\cdot \text{m}}{1.471\times {10}^{-8}\text{N}\cdot {\text{m}}^2}\right)}\\ =\sqrt{383863.766}\text{rad}/\text{s}\\ \approx 620\text{rad}/\text{s}\end{array}$

Substitute $0.45\text{m}$ for $b_1$, $0.15\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $1$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{11}=\frac{\left(0.45\text{m}\right)\left(0.15\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.45\text{m}\right)}^2-{\left(0.15\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.0675{\text{m}}^2\right)\left(0.135{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =6.37\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.45\text{m}$ for $b_2$, $0.15\text{m}$ for $x_2$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $2$ for $i$ and $2$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{22}=\frac{\left(0.45\text{m}\right)\left(0.15\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.45\text{m}\right)}^2-{\left(0.15\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.0675{\text{m}}^2\right)\left(0.135{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =6.37\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.15\text{m}$ for $b_2$, $0.15\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $2$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{12}=\frac{\left(0.15\text{m}\right)\left(0.15\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.15\text{m}\right)}^2-{\left(0.15\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.0225{\text{m}}^2\right)\left(0.315{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =4.961\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.15\text{m}$ for $b_1$, $0.15\text{m}$ for $x_2$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $2$ for $i$ and $1$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{21}=\frac{\left(0.15\text{m}\right)\left(0.15\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.15\text{m}\right)}^2-{\left(0.15\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.0225{\text{m}}^2\right)\left(0.315{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =4.961\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $6.37\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{11}$,$11.625\text{N}$ for $w_1$, $11.625\text{N}$ for $w_2$ and $4.961\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{12}$ in Equation (XI).

    $\begin{array}{c}{y}_1=\left(11.625\text{N}\right)\left(6.37\times {10}^{-7}\text{m}/\text{N}\right)+\left(11.625\text{N}\right)\left(4.961\times {10}^{-7}\text{m}/\text{N}\right)\\ =1.035\times {10}^{-6}\text{m}+5.7671\times {10}^{-6}\text{m}\\ =1.277\times {10}^{-5}\text{m}\end{array}$

Substitute $6.37\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{11}$, $11.625\text{N}$ for $w_1$, $11.625\text{N}$ for $w_2$ and $5.67\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{12}$ in Equation (XII).

    $\begin{array}{c}{y}_2=\left(11.625\text{N}\right)\left(6.37\times {10}^{-7}\text{m}/\text{N}\right)+\left(11.625\text{N}\right)\left(4.961\times {10}^{-7}\text{m}/\text{N}\right)\\ =1.035\times {10}^{-6}\text{m}+5.7671\times {10}^{-6}\text{m}\\ =1.277\times {10}^{-5}\text{m}\end{array}$

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

    $\begin{array}{c}{y}_1^2={\left(1.277\times {10}^{-5}\text{m}\right)}^2\\ =1.632\times {10}^{-10}{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{y}_2^2={\left(1.277\times {10}^{-5}\text{m}\right)}^2\\ =1.632\times {10}^{-10}{\text{m}}^2\end{array}$

Substitute $1.277\times {10}^{-5}\text{m}$ for $y_1$, $11.265\text{N}$ for $w_1$, $1.277\times {10}^{-5}\text{m}$ for $y_2$ and $11.265\text{N}$ for $w_2$ in Equation (XIII).

    $\begin{array}{c}\sum wy=\left(11.265\text{N}\right)\left(1.277\times {10}^{-5}\text{m}\right)+\left(11.265\text{N}\right)\left(1.277\times {10}^{-5}\text{m}\right)\\ =1.438\times {10}^{-4}\text{N}\cdot \text{m}+1.438\times {10}^{-4}\text{N}\cdot \text{m}\\ =2.877\times {10}^{-4}\text{N}\cdot \text{m}\end{array}$

Substitute $1.632\times {10}^{-10}{\text{m}}^2$ for $y_1^2$, $11.265\text{N}$ for $w_2$, $1.632\times {10}^{-10}{\text{m}}^2$ for $y_2^2$ and $11.265\text{N}$ for $w_1$ in Equation (XIV).

    $\begin{array}{c}\sum w{y}^2=\left(11.265\text{N}\right)\left(1.632\times {10}^{-10}{\text{m}}^2\right)+\left(11.265\text{N}\right)\left(1.632\times {10}^{-10}{\text{m}}^2\right)\\ =1.838\times {10}^{-9}\text{N}\cdot {\text{m}}^2+1.838\times {10}^{-9}\text{N}\cdot {\text{m}}^2\\ =3.677\times {10}^{-9}\text{N}\cdot {\text{m}}^2\end{array}$

Substitute $2.877\times {10}^{-4}\text{N}\cdot \text{m}$ for $\sum wy$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $3.677\times {10}^{-9}\text{N}\cdot {\text{m}}^2$ for $\sum w{y}^2$ in Equation (XV).

    $\begin{array}{c}{\omega }_2=\sqrt{9.81\text{m}/{\text{s}}^2\left(\frac{2.877\times {10}^{-4}\text{N}\cdot \text{m}}{3.677\times {10}^{-9}\text{N}\cdot {\text{m}}^2}\right)}\\ =\sqrt{767565.1346}\text{rad}/\text{s}\\ \approx 876\text{rad}/\text{s}\end{array}$

Substitute $0.5\text{m}$ for $b_1$, $0.1\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $1$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{11}=\frac{\left(0.5\text{m}\right)\left(0.1\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left({\left(0.6\text{m}\right)}^2-{\left(0.5\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2\right)\\ =\frac{\left(0.05{\text{m}}^2\right)\left(0.1{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =3.500\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.5\text{m}$ for $b_3$, $0.1\text{m}$ for $x_3$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $3$ for $i$ and $3$ for $j$ in Equation (VI).

$\begin{array}{c}{\delta }_{33}=\frac{\left(0.5\text{m}\right)\left(0.1\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left({\left(0.6\text{m}\right)}^2-{\left(0.5\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2\right)\\ =\frac{\left(0.05{\text{m}}^2\right)\left(0.1{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =3.500\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.3\text{m}$ for $b_2$, $0.3\text{m}$ for $x_2$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $2$ for $i$ and $2$ for $j$ in Equation (VI).

$\begin{array}{c}{\delta }_{22}=\frac{\left(0.3\text{m}\right)\left(0.3\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left({\left(0.6\text{m}\right)}^2-{\left(0.3\text{m}\right)}^2-{\left(0.3\text{m}\right)}^2\right)\\ =\frac{\left(0.09{\text{m}}^2\right)\left(0.18{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =1.13\times {10}^{-6}\text{m}/\text{N}\end{array}$

Substitute $0.3\text{m}$ for $b_2$, $0.1\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $2$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{12}=\frac{\left(0.3\text{m}\right)\left(0.1\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.3\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.03{\text{m}}^2\right)\left(0.26{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =5.46\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.3\text{m}$ for $b_2$, $0.1\text{m}$ for $x_3$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $3$ for $i$ and $2$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{32}=\frac{\left(0.3\text{m}\right)\left(0.1\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left(\begin{array}{l}{\left(0.6\text{m}\right)}^2\\ -{\left(0.3\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2\end{array}\right)\\ =\frac{\left(0.03{\text{m}}^2\right)\left(0.26{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =5.46\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $0.1\text{m}$ for $b_3$, $0.1\text{m}$ for $x_1$, $207\times {10}^9\text{N}/{\text{m}}^2$ for $E$, $1.917\times {10}^{-8}{\text{m}}^4$ for $I$, $0.6\text{m}$ for $l$, $1$ for $i$ and $3$ for $j$ in Equation (VI).

    $\begin{array}{c}{\delta }_{13}=\frac{\left(0.1\text{m}\right)\left(0.1\text{m}\right)}{6\left(207\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.917\times {10}^{-8}{\text{m}}^4\right)\left(0.6\text{m}\right)}\left({\left(0.6\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2-{\left(0.1\text{m}\right)}^2\right)\\ =\frac{\left(0.01{\text{m}}^2\right)\left(0.34{\text{m}}^2\right)}{\left(1242\times {10}^9\text{N}/{\text{m}}^2\right)\left(1.1502\times {10}^{-8}{\text{m}}^5\right)}\\ =2.38\times {10}^{-7}\text{m}/\text{N}\end{array}$

Substitute $3.500\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{11}$, $7.51\text{N}$ for $w_1$, $2.38\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{13}$ and $5.46\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{12}$ in Equation (XVI).

    $\begin{array}{c}{y}_1=\left(7.51\text{N}\right)\left[3.500\times {10}^{-7}\text{m}/\text{N}+5.46\times {10}^{-7}\text{m}/\text{N}+2.38\times {10}^{-7}\text{m}/\text{N}\right]\\ =\left(7.51\text{N}\right)\left(11.34\times {10}^{-7}\text{m}/\text{N}\right)\\ =8.51\times {10}^{-6}\text{m}\end{array}$

Substitute $1.13\times {10}^{-6}\text{m}/\text{N}$ for ${\delta }_{22}$, $7.51\text{N}$ for $w_1$, $5.46\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{32}$ and $5.46\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{12}$ in Equation (XVII).

    $\begin{array}{c}{y}_2=\left(7.51\text{N}\right)\left[5.46\times {10}^{-7}\text{m}/\text{N}+1.13\times {10}^{-6}\text{m}/\text{N}+5.46\times {10}^{-7}\text{m}/\text{N}\right]\\ =\left(7.51\text{N}\right)\left(2.22\times {10}^{-6}\text{m}/\text{N}\right)\\ =1.672\times {10}^{-5}\text{m}\end{array}$

Substitute $2.38\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{13}$, $7.51\text{N}$ for $w_1$, $5.46\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{32}$ and $3.500\times {10}^{-7}\text{m}/\text{N}$ for ${\delta }_{33}$ in Equation (XVIII).

    $\begin{array}{c}{y}_3=\left(7.51\text{N}\right)\left[2.38\times {10}^{-7}\text{m}/\text{N}+5.46\times {10}^{-7}\text{m}/\text{N}+3.500\times {10}^{-7}\text{m}/\text{N}\right]\\ =\left(7.51\text{N}\right)\left(11.34\times {10}^{-7}\text{m}/\text{N}\right)\\ =8.51\times {10}^{-6}\text{m}\end{array}$

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

    $\begin{array}{c}{y}_1^2={\left(8.51\times {10}^{-6}\text{m}\right)}^2\\ =7.25\times {10}^{-11}{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{y}_2^2={\left(1.672\times {10}^{-5}\text{m}\right)}^2\\ =2.79\times {10}^{-10}{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{y}_3^2={\left(8.51\times {10}^{-6}\text{m}\right)}^2\\ =7.24\times {10}^{-11}{\text{m}}^2\end{array}$

Substitute $8.51\times {10}^{-6}\text{m}$ for $y_1$, $7.51\text{N}$ for $w_1$, $1.672\times {10}^{-5}\text{m}$ for $y_2$ and $8.51\times {10}^{-6}\text{m}$ for $y_3$ in Equation (XIX).

    $\begin{array}{c}\sum wy=\left(7.51\text{N}\right)\left(8.51\times {10}^{-6}\text{m}+1.672\times {10}^{-5}\text{m}+8.51\times {10}^{-6}\text{m}\right)\\ =\left(7.51\text{N}\right)\left(3.374\times {10}^{-5}\text{m}\right)\\ =2.535\times {10}^{-4}\text{N}\cdot \text{m}\end{array}$

Substitute $7.25\times {10}^{-11}{\text{m}}^2$ for $y_{{}_1}^2$, $7.51\text{N}$ for $w_1$, $2.79\times {10}^{-10}{\text{m}}^2$ for $y_2^2$ and $7.24\times {10}^{-11}{\text{m}}^2$ for $y_3^2$ in Equation (XX).

    $\begin{array}{c}\sum w{y}^2=\left(7.51\text{N}\right)\left(7.25\times {10}^{-11}{\text{m}}^2+2.79\times {10}^{-10}{\text{m}}^2+7.24\times {10}^{-11}{\text{m}}^2\right)\\ =\left(7.51\text{N}\right)\left(4.238\times {10}^{-10}{\text{m}}^2\right)\\ =3.189\times {10}^{-9}\text{N}\cdot {\text{m}}^2\end{array}$

Substitute $2.535\times {10}^{-4}\text{N}\cdot \text{m}$ for $\sum wy$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $3.189\times {10}^{-9}\text{N}\cdot {\text{m}}^2$ for $\sum w{y}^2$ in Equation (XXI).

    $\begin{array}{c}{\omega }_3=\sqrt{9.81\text{m}/{\text{s}}^2\left(\frac{2.535\times {10}^{-4}\text{N}\cdot \text{m}}{3.189\times {10}^{-9}\text{N}\cdot {\text{m}}^2}\right)}\\ =\sqrt{779816.5569}\text{rad}/\text{s}\\ =883\text{rad}/\text{s}\end{array}$

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.