Chapter 4, Problem 4P

An engineer is forced by geometric considerations to apply the torque on the spring of Prob. 4–3 at the location x = 0.4l. For a uniform-diameter spring, this would cause one leg of the span to be underutilized when both legs have the same diameter. For optimal design the diameter of each leg should be designed such that the maximum shear stress in each leg is the same. This problem is to redesign the spring of part (b) of Prob. 4–3. Using x = 0.4l, l = 10 in, T = 1500 lbf · in, and G = 11.5 Mpsi, design the spring such that the maximum shear stresses in each leg are equal and the spring has the same spring rate (angle of twist) as part (b) of Prob. 4–3. Specify d₁, d₂, the spring rate k, and the torque and the maximum shear stress in each leg.

To determine

The diameter $d_1$.

The diameter $d_2$.

The spring rate $k$.

The individual torque in each leg.

The maximum stress in each leg.

Answer to Problem 4P

The diameter $d_1$ is $0.27494\text{in}$.

The diameter $d_2$ is $0.41241\text{in}$.

The spring rate is $7056.167\text{lbf}\cdot \text{in}$.

The torque in the left leg is $342.849\text{lbf}\cdot \text{in}$.

The torque in the right leg is $1157.1156\text{lbf}\cdot \text{in}$.

The maximum stress in the left leg is $83.987\text{kpsi}$.

The maximum stress in the right leg is $84.015\text{kpsi}$.

Explanation of Solution

Write the expression for the angular displacement of the torsional bar for the distance $x$ from the fixed end.

    ${\theta }_1=\frac{T_{1}x}{G{J}_1}$                                                                                               (I)

Here, the torque in the left leg is $T_1$, the angular deformation at portion $x$ is ${\theta }_1$, the modulus of rigidity is $G$ and the polar moment of inertia is $J_1$.

Write the expression for the polar moment of inertia of the torsional bar for the distance $x$ from the fixed end.

    $J_1=\frac{\pi }{32}{d}_1^4$

Here, diameter of the spring in the portion $x$ is $d_1$.

Substitute $\frac{\pi }{32}{d}_1^4$ for $J_1$ in Equation (I).

    ${\theta }_1=\frac{T_{1}x}{G\left(\frac{\pi }{32}{d}_1^4\right)}$

Write the expression for the angular displacement of the torsional bar for the distance $\left(l-x\right)$ from the fixed end.

    ${\theta }_2=\frac{T_2\left(l-x\right)}{G{J}_2}$                                                                                  (II)

Here, the torque in the right leg is $T_2$, the length is $l$, the angular deformation at portion $l-x$ is ${\theta }_2$, and the polar moment of area is $J_2$.

Write the expression for the polar moment of inertia of the torsional bar for the distance $l-x$ from the fixed end.

    $J_2=\frac{\pi }{32}{d}_2^4$

Here, diameter of the spring in the portion $l-x$ is $d_2$.

Substitute $\frac{\pi }{32}{d}_2^4$ for $J_2$ in Equation (II).

    ${\theta }_2=\frac{T_2\left(l-x\right)}{G\left(\frac{\pi }{32}{d}_2^4\right)}$

Write the expression for maximum shear stress at section of length $x$.

    ${\tau }_1=\frac{16T_1}{\pi d_1^3}$                                                                                               (III)

Here, the shear stress is ${\tau }_1$.

Write the expression for maximum shear stress at section of length $l-x$.

    ${\tau }_2=\frac{16T_2}{\pi d_2^3}$                                                                                           (IV)

Here, the shear stress is ${\tau }_2$.

Since the spring is under uniform stress, so the shear stress in both the legs are considered as equal, that is

    ${\tau }_1={\tau }_2$                                                                                                   (V)

Substitute $\frac{16T_1}{\pi d_1^3}$ for ${\tau }_1$ and $\frac{16T_2}{\pi d_2^3}$ for ${\tau }_2$ in Equation (V).

    $\begin{array}{l}\frac{16T_1}{\pi d_1^3}=\frac{16T_2}{\pi d_2^3}\\ \frac{T_1}{d_1^3}=\frac{T_2}{d_2^3}\end{array}$                                                                                         (VI)

Write the expression of the net torque on the system.

    $T=T_1+T_2$                                                                                           (VII)

Here, the net torque is $T$.

Write the expression for net angular deflection in the bar.

    $\theta =\frac{Tl}{\frac{\pi }{32}{d}^{4}G}$                                                                                       (VIII)

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

Since the spring has uniform deflection throughout, so the deflection in the left leg is equal to the overall deflection.

    $\theta ={\theta }_1$                                                                                                    (IX)

Substitute $\frac{Tl}{\frac{\pi }{32}{d}^{4}G}$ for $\theta$ and $\frac{T_{1}x}{G\left(\frac{\pi }{32}{d}_1^4\right)}$ for ${\theta }_1$ in Equation

    $\begin{array}{l}\frac{T_{1}x}{G\left(\frac{\pi }{32}{d}_1^4\right)}=\frac{Tl}{\frac{\pi }{32}{d}^{4}G}\\ T_1=T\frac{l}{x}{\left(\frac{d_1}{d}\right)}^4\end{array}$                                                                      (X)

Since the spring has uniform deflection throughout, so the deflection in the right leg is equal to the overall deflection.

    $\theta ={\theta }_2$                                                                                             (XI)

Substitute $\frac{Tl}{\frac{\pi }{32}{d}^{4}G}$ for $\theta$ and $\frac{T_2\left(l-x\right)}{G\left(\frac{\pi }{32}{d}_2^4\right)}$ for ${\theta }_2$ in Equation

    $\begin{array}{l}\frac{T_2\left(l-x\right)}{G\left(\frac{\pi }{32}{d}_2^4\right)}=\frac{Tl}{\frac{\pi }{32}{d}^{4}G}\\ T_2=T\frac{l}{l-x}{\left(\frac{d_2}{d}\right)}^4\end{array}$                                                                    (XII)

Since the spring has uniform deflection throughout, so the deflection in the right leg is equal to the deflection in the left leg. That is,

    ${\theta }_1={\theta }_2$                                                                                         (XIII)

Substitute $\frac{T_{1}x}{G\left(\frac{\pi }{32}{d}_1^4\right)}$ for ${\theta }_1$ and $\frac{T_2\left(l-x\right)}{G\left(\frac{\pi }{32}{d}_2^4\right)}$ for ${\theta }_2$ in Equation

    $\begin{array}{l}\frac{T_{1}x}{G\left(\frac{\pi }{32}{d}_1^4\right)}=\frac{T_2\left(l-x\right)}{G\left(\frac{\pi }{32}{d}_2^4\right)}\\ \frac{T_{1}x}{\left(d_1^4\right)}=\frac{T_2\left(l-x\right)}{\left(d_2^4\right)}\end{array}$                                                                     (XIV)

Write the stiffness of the combine system.

    $k=\frac{T}{\theta }$                                                                                                  (XV)

Here, $k$ is the stiffness of the combined system.

Conclusion:

Substitute $1500\text{lbf}\cdot \text{in}$ for $T$, $10\text{in}$ for $l$, $4\text{in}$ for $x$ and $0.5\text{in}$ for $d$ in Equation (X).

    $\begin{array}{c}{T}_1=1500\text{lbf}\cdot \text{in}\times \left(\frac{10\text{in}}{4\text{in}}\right){\left(\frac{d_1}{0.5\text{in}}\right)}^4\\ =\left(3750\text{lbf}\cdot \text{in}\right){\left(\frac{d_1}{0.5\text{in}}\right)}^4\\ =\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(d_1\right)}^4\end{array}$                                                    (XVI)

Substitute $1500\text{lbf}\cdot \text{in}$ for $T$, $10\text{in}$ for $l$, $4\text{in}$ for $x$ and $0.5\text{in}$ for $d$ in Equation (XII).

    $\begin{array}{c}{T}_2=1500\text{lbf}\cdot \text{in}\times \left(\frac{10\text{in}}{6\text{in}}\right){\left(\frac{d_2}{0.5\text{in}}\right)}^4\\ =\left(2500\text{lbf}\cdot \text{in}\right){\left(\frac{d_2}{0.5\text{in}}\right)}^4\\ =\left(40000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(d_2\right)}^4\end{array}$                                               (XVII)

Substitute $10\text{in}$ for $l$, $4\text{in}$ for $x$ in Equation (XIV).

    $\frac{4\text{in}\times T_1}{\left(d_1^4\right)}=\frac{6\text{in}\times T_2}{\left(d_2^4\right)}$                                                                          (XVIII)

Divide Equation (VI) and Equation (XVIII).

    $\begin{array}{l}\frac{\frac{T_1}{d_1^3}}{\frac{4\text{in}\times T_1}{\left(d_1^4\right)}}=\frac{\frac{T_2}{d_2^3}}{\frac{6\text{in}\times T_2}{\left(d_2^4\right)}}\\ \frac{d_1}{4\text{in}}=\frac{d_2}{6\text{in}}\\ d_2=1.5d_1\end{array}$                                                                       (XIX)

Substitute $1.5d_1$ for $d_2$ in Equation (XVII).

    $\begin{array}{c}{T}_2=\left(40000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(1.5d_1\right)}^4\\ =\left(40000\text{lbf}\cdot \text{in}/{\text{in}}^4\right)\times 5.0625\times d_1^4\\ =\left(202500\text{lbf}\cdot \text{in}/{\text{in}}^4\right)d_1{}^4\end{array}$                                           (XX)

Substitute $\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(d_1\right)}^4$ for $T_1$ and $\left(202500\text{lbf}\cdot \text{in}/{\text{in}}^4\right)d_1{}^4$ for $T_2$ in Equation (VII).

    $T=\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(d_1\right)}^4+\left(202500\text{lbf}\cdot \text{in}/{\text{in}}^4\right)d_1{}^4$             (XXI)

Substitute $1500\text{lbf}\cdot \text{in}$ for $T$ in Equation (XXI).

    $\begin{array}{l}\left(1500\text{lbf}\cdot \text{in}\right)=\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(d_1\right)}^4+\left(202500\text{lbf}\cdot \text{in}/{\text{in}}^4\right)d_1{}^4\\ \left(1500\text{lbf}\cdot \text{in}\right)=\left(262500\text{lbf}\cdot \text{in}/{\text{in}}^4\right)d_1^4\\ d_1={\left(\frac{1500\text{lbf}\cdot \text{in}}{262500\text{lbf}\cdot \text{in}/{\text{in}}^4}\right)}^{\frac{1}{4}}\\ d_1=0.27494\text{in}\end{array}$

Thus, the diameter $d_1$ is $0.27494\text{in}$.

Substitute $0.27494\text{in}$ for $d_1$ in Equation (XIX).

    $\begin{array}{c}{d}_2=1.5\times \left(0.27494\text{in}\right)\\ =0.41241\text{in}\end{array}$

Thus, the diameter $d_2$ is $0.41241\text{in}$.

Substitute $0.27494\text{in}$ for $d_1$ in Equation (XVI).

  $\begin{array}{c}{T}_1=\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(0.27494\text{in}\right)}^4\\ =\left(60000\text{lbf}\cdot \text{in}/{\text{in}}^4\right)\left(5.71\times {10}^{-3}\right){\text{in}}^4\\ =342.849\text{lbf}\cdot \text{in}\end{array}$

Thus, the torque in the left leg is $342.849\text{lbf}\cdot \text{in}$.

Substitute $0.41241\text{in}$ for $d_2$ in Equation (XVII).

    $\begin{array}{c}{T}_2=\left(40000\text{lbf}\cdot \text{in}/{\text{in}}^4\right){\left(0.41241\text{in}\right)}^4\\ =\left(40000\text{lbf}\cdot \text{in}/{\text{in}}^4\right)\left(0.0289{\text{in}}^4\right)\\ =1157.1156\text{lbf}\cdot \text{in}\end{array}$

Thus, the torque in the right leg is $1157.1156\text{lbf}\cdot \text{in}$.

Substitute the value $1500\text{lbf}\cdot \text{in}$ for $T$, $10\text{in}$ for $l$, $0.5\text{in}$ for $d$, $11.5\times {10}^6\text{psi}$ for $G$ in Equation (XI).

    $\begin{array}{c}\theta =\frac{\left(1500\text{lbf}\cdot \text{in}\right)\times \left(10\text{in}\right)}{\frac{\pi }{32}{\left(0.5\text{in}\right)}^4\times \left(11.5\times {10}^6\text{psi}\right)}\\ =\frac{15000\text{lbf}\cdot {\text{in}}^2}{70563.12\text{lbf}\cdot {\text{in}}^2}\\ =0.21258\text{rad}\end{array}$

Substitute the value $1500\text{lbf}\cdot \text{in}$ for $T$,$0.21258\text{rad}$ for $\theta$ in Equation (XV).

    $\begin{array}{c}k=\frac{\left(1500\text{lbf}\cdot \text{in}\right)}{\left(0.21258\text{rad}\right)}\\ =7056.167\text{lbf}\cdot \text{in}\end{array}$

Thus, the spring rate is $7056.167\text{lbf}\cdot \text{in}$.

Substitute the value $342.849\text{lbf}\cdot \text{in}$ for $T_1$, $0.27494\text{in}$ for $d_1$ in Equation (V)

    $\begin{array}{c}{\tau }_1=\frac{16\left(342.849\text{lbf}\cdot \text{in}\right)}{\pi {\left(0.27494\text{in}\right)}^3}\\ =83987.96\text{psi}\\ =83.987\text{kpsi}\end{array}$

Thus, the maximum stress in the left leg is $83.987\text{kpsi}$.

Substitute the value $1157.1156\text{lbf}\cdot \text{in}$ for $T_2$, $0.41241\text{in}$ for $d_2$ in Equation (V)

    $\begin{array}{c}{\tau }_2=\frac{16\left(1157.1156\text{lbf}\cdot \text{in}\right)}{\pi {\left(0.41241\text{in}\right)}^3}\\ =84015.4768\text{psi}\\ =84.015\text{kpsi}\end{array}$

Thus, the maximum stress in the right leg is $84.015\text{kpsi}$.