Chapter 4, Problem 73P

The cantilevered handle in Prob. 3–84 is made from mild steel. Let F_y = 250 lbf and F_x = F_z = 0. Using Castigliano’s theorem, determine the vertical deflection (along the y axis) at the tip. Repeat the problem with shaft OC simplified to a uniform diameter of 1 in for its entire length. What is the percent error from this simplification?

To determine

The vertical deflection at the tip along the y-axis.

The percentage error of the deflection in case of simplified shaft.

Answer to Problem 73P

The vertical deflection at the tip along the y-axis is $0.848\text{in}$

The percentage error of the deflection in case of simplified shaft is $20.11\%$.

Explanation of Solution

The given force is the y-direction is $F_y=250\text{lbf}$, force in the x-direction is $F_x=300\text{lbf}$, the force in z-direction is $F_z=0$, the diameter of bar OA is $1\frac{1}{2}\text{in}$, the diameter of rod AB is $1\text{in}$, length of rod AB is $9\text{in}$, the length of rod CD is $12\text{in}$, the diameter of the rod CD is $\frac{3}{4}\text{in}$, the length of rod BC is $2\text{in}$, and the diameter of rod BC is $1\frac{1}{2}\text{in}$.

Write the moment of inertia of the rod $OA$.

$I_{OA}=\frac{\pi d_{OA}^4}{64}$ (I)

Here, the diameter of the rod $OA$ is $d_{OA}$.

The moment of inertia of the rod $BC$ is equal to the moment of inertia of the rod $OA$.

Write the polar moment of inertia of the rod $OA$.

$J_{OA}=\frac{\pi d_{OA}^4}{32}$ (II)

The polar moment of inertia of the rod $BC$ is equal to that of $OA$.

Write the moment of inertia of the rod $AB$.

$I_{AB}=\frac{\pi d_{AB}^4}{64}$ (III)

Here, the diameter of the rod $AB$ is $d_{AB}$.

Write the polar moment of inertia of the rod $AB$.

$J_{AB}=\frac{\pi d_{AB}^4}{32}$ (IV)

Write the moment of inertia of the rod $CD$.

$I_{CD}=\frac{\pi d_{CD}^4}{64}$ (V)

Here, the diameter of the rod $CD$ is $d_{CD}$.

Write the deflection at the tip by Castigliano’s theorem.

${\left({\delta }_D\right)}_y=\frac{\partial U}{\partial F}={\displaystyle \sum \left(\frac{Tl}{JG}\right)}\frac{\partial T}{\partial F}+{\displaystyle \sum \left(\frac{1}{EI}\right){\displaystyle \int M\frac{\partial M}{\partial F}}d\overline{x}}$ (VI)

Here, the strain energy is $U$, the torque is $T$, the moment is $M$, the polar moment of inertia is $J$, the moment of inertia is $I$, the force is $F$, the axial force is $F_a$, the modulus of elasticity is $E$, the modulus of rigidity is $G$, the length of the member is $l$, and the dummy variable of position is $\overline{x}$.

The dummy variable $\overline{x}$ is used to originate at the ends where the loads are applied on each segment of the structure.

Write the moment function for $OC$.

${\left(M\right)}_{OC}=F\overline{x}$ (VII)

Differentiate the Equation (VII) with respect to $F$.

$\begin{array}{c}{\left(\frac{\partial M}{\partial F}\right)}_{OC}=\frac{\partial }{\partial F}\left\{F\cdot \overline{x}\right\}\\ =\overline{x}\end{array}$

Write the torque on rod $OC$.

${\left(T\right)}_{OC}=F\cdot l_{CD}$ (VIII)

Differentiate the Equation (VIII) with respect to $F$.

$\begin{array}{c}{\left(\frac{\partial T}{\partial F}\right)}_{OC}=\frac{\partial }{\partial F}\left[F\cdot l_{CD}\right]\\ =l_{CD}\end{array}$

Write the moment function of the link $CD$.

${\left(M\right)}_{CD}=F\cdot \overline{x}$ (IX)

Differentiate the Equation (IX) with respect to $F$.

$\begin{array}{c}{\left(\frac{\partial M}{\partial F}\right)}_{CD}=\frac{\partial }{\partial F}\left[F\cdot \overline{x}\right]\\ =\overline{x}\end{array}$

Substitute $\overline{x}$ for ${\left(\frac{\partial M}{\partial F}\right)}_{CD}$, $l_{CD}$ for ${\left(\frac{\partial T}{\partial F}\right)}_{OC}$, $\left(F\cdot l_{CD}\right)$ for ${\left(T\right)}_{OC}$, $\overline{x}$ for ${\left(\frac{\partial M}{\partial F}\right)}_{OC}$, $\left(F\overline{x}\right)$ for ${\left(M\right)}_{OC}$, and $\left(F\cdot \overline{x}\right)$ for ${\left(M\right)}_{CD}$ in Equation (VI).

$\begin{array}{l}{\delta }_D=\left(\frac{T_{OC}\cdot \left(l_{OA}+l_{BC}\right)}{J_{OA}\cdot G}\right){\left(\frac{\partial T}{\partial F}\right)}_{OC}+\left(\frac{T_{OC}\left(l_{OA}+l_{BC}\right)}{J_{AB}\cdot G}\right){\left(\frac{\partial T}{\partial F}\right)}_{OC}\\ +\frac{1}{E\cdot I_{OA}}{\displaystyle \underset{0}{\overset{l_{OA}}{\int }}{M}_{OC}\cdot {\left(\frac{\partial M}{\partial F}\right)}_{OC}d\overline{x}}+\frac{1}{E\cdot I_{AB}}{\displaystyle \underset{l_{OA}}{\overset{l_{OB}}{\int }}{M}_{OC}\cdot {\left(\frac{\partial M}{\partial F}\right)}_{OC}}\\ +\frac{1}{E.I_{OA}}{\displaystyle \underset{l_{OB}}{\overset{l_{BC}}{\int }}{M}_{OC}\cdot {\left(\frac{\partial M}{\partial F}\right)}_{OC}}+\frac{1}{E.I_{CD}}{\displaystyle \underset{0}{\overset{l_{CD}}{\int }}{M}_{CD}\cdot {\left(\frac{\partial M}{\partial F}\right)}_{CD}}\\ =\left(\frac{\left[\left(F\right)\cdot \left(l_{CD}\right)\right]\left(l_{OA}+l_{BC}\right)}{\left(J_{OA}\right)\cdot G}\right)\left(l_{CD}\right)+\left(\frac{F\cdot l_{CD}}{J_{AB}\cdot G}\right)\left(l_{CD}\right)\\ +\frac{1}{E\cdot I_{OA}}{\displaystyle \underset{0}{\overset{l_{OA}}{\int }}\left(F\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}+\frac{1}{E\cdot I_{AB}}{\displaystyle \underset{l_{OA}}{\overset{l_{OB}}{\int }}\left(F\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\\ +\frac{1}{E.I_{OA}}{\displaystyle \underset{l_{OB}}{\overset{l_{OC}}{\int }}\left(F\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}+\frac{1}{E.I_{CD}}{\displaystyle \underset{0}{\overset{\left(l_{CD}\right)}{\int }}\left(F\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\end{array}$ (X)

For the case of simplified  shaft $OC$.

Substitute $\overline{x}$ for ${\left(\frac{\partial M}{\partial F}\right)}_{CD}$, $l_{CD}$ for ${\left(\frac{\partial T}{\partial F}\right)}_{OC}$, $\left(F\cdot l_{CD}\right)$ for ${\left(T\right)}_{OC}$, $\overline{x}$ for ${\left(\frac{\partial M}{\partial F}\right)}_{OC}$, $\left(F\overline{x}\right)$ for ${\left(M\right)}_{OC}$, and $\left(F\cdot \overline{x}\right)$ for ${\left(M\right)}_{CD}$ in Equation (VI).

$\begin{array}{c}{\left({\delta }_B\right)}_y=\frac{T_{OC}\cdot l_{OC}}{J_{AB}\cdot G}{\left(\frac{\partial T}{\partial F}\right)}_{OC}+\frac{1}{E\cdot I_{AB}}{\displaystyle \underset{0}{\overset{l_{OC}}{\int }}{\left(M\right)}_{OC}{\left(\frac{\partial M}{\partial F}\right)}_{OC}d\overline{x}}\\ +\frac{1}{E\cdot I_{CD}}{\displaystyle \underset{0}{\overset{l_{CD}}{\int }}{\left(M\right)}_{CD}{\left(\frac{\partial M}{\partial F}\right)}_{CD}d\overline{x}}\\ =\left[\begin{array}{l}\frac{\left(F\cdot l_{CD}\right)\cdot l_{OC}}{J_{AB}\cdot G}\left(l_{CD}\right)+\frac{1}{E\cdot I_{AB}}{\displaystyle \underset{0}{\overset{l_{OC}}{\int }}\left(F\cdot \overline{x}\right)\left(\overline{x}\right)d\overline{x}}\\ +\frac{1}{E\cdot I_{CD}}{\displaystyle \underset{0}{\overset{l_{CD}}{\int }}\left(F\cdot \overline{x}\right)\left(\overline{x}\right)d\overline{x}}\end{array}\right]\end{array}$ . (XI)

Calculate  the percentage error of the simplified shaft deflection.

$k=\frac{{\left({\delta }_B\right)}_y-{\left({\delta }_D\right)}_y}{{\left({\delta }_D\right)}_y}\times 100$ (XII)

Conclusion:

Refer to the Table A-5 “Physical Constants of Materials” and obtain the value of $E=30\text{Mpsi}$ and $G=11.5\text{Mpsi}$ for steel.

Substitute $1\frac{1}{2}\text{in}$ for $d_{OA}$ in Equation (I).

$\begin{array}{c}{I}_{OA}=\frac{\pi {\left(1\frac{1}{2}\text{in}\right)}^4}{64}\\ =0.2485{\text{in}}^4\end{array}$

Substitute $1\frac{1}{2}\text{in}$ for $d_{OA}$ in Equation (II).

$\begin{array}{c}{J}_{OA}=\frac{\pi {\left(1\frac{1}{2}\text{in}\right)}^4}{32}\\ =0.4970{\text{in}}^4\end{array}$

Substitute $1\text{in}$ for $d_{AB}$ in Equation (III).

$\begin{array}{c}{I}_{AB}=\frac{\pi {\left(1\text{in}\right)}^4}{64}\\ =0.04909{\text{in}}^4\end{array}$

Substitute $1\text{in}$ for $d_{AB}$ in Equation (IV).

$\begin{array}{c}{J}_{AB}=\frac{\pi {\left(1\text{in}\right)}^4}{32}\\ =0.9819{\text{in}}^4\end{array}$

Substitute $\frac{3}{4}\text{in}$ for $d_{CD}$ in Equation (V).

$\begin{array}{c}{I}_{CD}=\frac{\pi {\left(\frac{3}{4}\text{in}\right)}^4}{64}\\ =0.01553{\text{in}}^4\end{array}$

Substitute $12\text{in}$ for $l_{CD}$, $\left(250\text{lbf}\right)$ for $F$, $\left(0.4970{\text{in}}^4\right)$ for $J_{OA}$, $2\text{in}$ for $l_{OA}$, $0.04909{\text{in}}^4$ for $I_{AB}$, $0.01553{\text{in}}^4$ for $I_{CD}$, $11\text{in}$ $l_{OB}$, $13\text{in}$ for $l_{OC}$, $0.09819{\text{in}}^4$ for $J_{AB}$, $0.2485{\text{in}}^4$ for $I_{OA}$, $30\text{Mpsi}$ for $E$, and $11.5\text{Mpsi}$ for $G$ in Equation (X).

$\begin{array}{c}{\delta }_D=\left[\begin{array}{c}\left(\frac{\left[\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right]\cdot \left(2\text{in}+2\text{in}\right)}{\left(0.4970{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\right)\left(12\text{in}\right)\\ +\left(\frac{\left[\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right]\left(2\text{in}+2\text{in}\right)}{\left(0.09819{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\right)\left(12\text{in}\right)\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.2485{\text{in}}^4\right)}{\displaystyle \underset{0}{\overset{2\text{in}}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.04909{\text{in}}^4\right)}{\displaystyle \underset{2\text{in}}{\overset{11\text{in}}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\\ +\frac{1}{\left(30\text{Mpsi}\right).\left(0.2485{\text{in}}^4\right)}{\displaystyle \underset{11\text{in}}{\overset{13\text{in}}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\\ +\frac{1}{\left(30\text{Mpsi}\right).\left(0.01553{\text{in}}^4\right)}{\displaystyle \underset{0}{\overset{\left(12\text{in}\right)}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\cdot \overline{x}d\overline{x}}\end{array}\right]\\ =\left[\begin{array}{c}\left(\frac{\left[\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right]\cdot \left(2\text{in}+2\text{in}\right)}{\left(0.4970{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\right)\left(12\text{in}\right)\\ +\left(\frac{\left[\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right]\left(2\text{in}+2\text{in}\right)}{\left(0.09819{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\right)\left(12\text{in}\right)\\ +\frac{\left(250\text{lbf}\right)}{\left(30\text{Mpsi}\right)\cdot \left(0.2485{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_0^{2\text{in}}\\ +\frac{\left(250\text{lbf}\right)}{\left(30\text{Mpsi}\right)\cdot \left(0.04909{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_2^{11\text{in}}\\ +\frac{\left(250\text{lbf}\right)}{\left(30\text{Mpsi}\right).\left(0.2485{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_{11}^{13\text{in}}\\ +\frac{\left(250\text{lbf}\right)}{\left(30\text{Mpsi}\right).\left(0.01553{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_0^{12\text{in}}\end{array}\right]\\ =\left[\begin{array}{c}1.008\times {10}^{-4}\times 250+1.148\times {10}^{-3}\times 250+3.58\times {10}^{-7}\times 250+2.994\times {10}^{-4}\times 250\\ +3.872\times {10}^{-5}\times 250+1.2363\times {10}^{-3}\times 250\end{array}\right]\\ \approx \text{0}\text{.706}\text{in}\end{array}$

Substitute $\left(250\text{lbf}\right)$ for $F$, $12\text{in}$ for $l_{CD}$, $13\text{in}$ for $l_{OC}$, $0.09819{\text{in}}^4$ for $J_{AB}$, $0.04909{\text{in}}^4$ for $I_{AB}$, $0.01553{\text{in}}^4$ for $I_{CD}$, $30\text{Mpsi}$ for $E$, and $11.5\text{Mpsi}$ for $G$ in Equation (XI).

$\begin{array}{c}{\left({\delta }_B\right)}_y=\left[\begin{array}{c}\frac{\left(\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right)\cdot \left(13\text{in}\right)}{\left(0.09819{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\left(12\text{in}\right)\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.04909{\text{in}}^4\right)}{\displaystyle \underset{0}{\overset{13}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\left(\overline{x}\right)d\overline{x}}\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.01553{\text{in}}^4\right)}{\displaystyle \underset{0}{\overset{12}{\int }}\left(\left(250\text{lbf}\right)\cdot \overline{x}\right)\left(\overline{x}\right)d\overline{x}}\end{array}\right]\\ =\left[\begin{array}{c}\frac{\left(\left(250\text{lbf}\right)\cdot \left(12\text{in}\right)\right)\cdot \left(13\text{in}\right)}{\left(0.09819{\text{in}}^4\right)\cdot \left(11.5\text{Mpsi}\right)}\left(12\text{in}\right)\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.04909{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_0^{13}\\ +\frac{1}{\left(30\text{Mpsi}\right)\cdot \left(0.01553{\text{in}}^4\right)}\left(250\text{lbf}\right){\left[\frac{{\overline{x}}^3}{3}\right]}_0^{12}\end{array}\right]\\ =0.848\text{in}\end{array}$

Thus, the vertical deflection at the tip along the y-axis is $0.848\text{in}$.

Substitute $0.848\text{in}$ for ${\left({\delta }_B\right)}_y$, and $\text{0}\text{.706}\text{in}$ for ${\left({\delta }_D\right)}_y$ in Equation (XII).

$\begin{array}{c}k=\frac{\left(0.848\text{in}\right)-\left(\text{0}\text{.706}\text{in}\right)}{\left(\text{0}\text{.706}\text{in}\right)}\times 100\\ =20.11\%\end{array}$

Thus, the percentage error of the deflection in case of simplified shaft is $20.11\%$.