Chapter 16, Problem 4P

The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The face width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of 1000 kPa.

Problem 16–4

The dimensions in millimeters are a = 150, c = 165, R = 200, and d = 50.

(a)    Determine the maximum actuating force.

(b)    Estimate the brake capacity.

(c)    Noting that rotation may be in either direction, estimate the hinge-pin reactions.

To determine

The maximum actuating force.

Answer to Problem 4P

The maximum actuating force is $5.7\text{kN}$.

Explanation of Solution

Write the expression for moment of frictional forces.

    $M_f=\frac{f{p}_{a}br}{\sin {\theta }_a}{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta }\left(r-a\cos \theta \right)d\theta$                                       (I)

Here, coefficient of friction is $f$, face width of shoe is $b$, maximum pressure on the shoe is $p_a$, outer radius is $a$, angle at which frictional material is located from hinge is $\theta$ and inner radius is $r$.and moment of frictional forces is $M_f$.

Write the expression for moment of normal forces.

    $M_N=\frac{p_{a}bra}{\sin {\theta }_a}{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }$                                                      (II)

Here, moment of normal forces is $M_N$.

Write the expression for actuating force.

    $F=\frac{M_N-M_f}{c}$                                                                  (III)

Here, actuating force is $F$.

Conclusion:

Calculate ${\theta }_2$.

    $\begin{array}{c}{\theta }_2=90°-\left(15°+25°\right)\\ =90°-40°\\ =50°\end{array}$

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $1000\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (I).

    $\begin{array}{c}{M}_f=\frac{\left(0.24\right)\left(1000\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)}{\sin \left(75°\right)}{\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta }\left(200\text{mm}-150\text{mm}\cos \theta \right)d\theta \\ =\left[\frac{\begin{array}{l}\left(0.24\right)\left(1000\text{kPa}\right)\\ \left(\left(15000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2\right)\end{array}}{\sin \left(75°\right)}\right]\left[{\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta }\left(\begin{array}{l}\left(200\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)-\\ \left(150\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\cos \theta \end{array}\right)d\theta \right]\\ =3727\left[{\left(-0.2\cos \theta \right)}_{10}^{75}+0.15{\left(\frac{1}{4}\cos 2\theta \right)}_{10}^{75}\right]\\ =288.9\text{N}\cdot \text{m}\end{array}$

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $1000\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (II).

    $\begin{array}{c}{M}_N=\frac{\left(1000\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)\left(150\text{mm}\right)}{\sin \left(70°\right)}{\displaystyle \underset{10°}{\overset{75°}{\int }}{\sin }^2\theta d\theta }\\ =\left(1000\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(2250000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3{\left[\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right]}_{10}^{75}\\ =2329.4\left[\frac{13\pi }{72}-\frac{1}{4}\left(0.158\right)\right]\\ =1229.3\text{N}\cdot \text{m}\end{array}$

Substitute $1229.3\text{N}\cdot \text{m}$ for $M_N$, $165\text{mm}$ for $c$ and $288.9\text{N}\cdot \text{m}$ for $M_f$ in Equation (III).

    $\begin{array}{c}F=\frac{1229.3\text{N}\cdot \text{m}-288.9\text{N}\cdot \text{m}}{165\text{mm}}\\ =\frac{940.4\text{N}\cdot \text{m}}{\left(165\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\left(5699.39\text{N}\cdot \text{m}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \approx 5.7\text{kN}\end{array}$

Thus, the maximum actuating force is $5.7\text{kN}$.

To determine

The total braking capacity.

Answer to Problem 4P

The total braking capacity is $1752.4\text{N}\cdot \text{m}$.

Explanation of Solution

Write the expression for torque applied by primary shoe.

    $T_P=\frac{f{p}_{a}b{r}^2\left(\cos {\theta }_1-\cos {\theta }_2\right)}{\sin {\theta }_a}$                     (IV)

Here, braking torque applied by primary shoe is $T_P$.

Write the expression for moment of frictional forces.

    $M_{fs}=\frac{M_f}{{10}^6}{p}_a$                          (V)

Here, moment of frictional forces for secondary shoe is $M_{fs}$.

Write the expression for normal forces for secondary shoes.

    $M_{Ns}=\frac{M_N}{{10}^6}{p}_a$                        (VI)

Here, normal forces for secondary shoes is $M_{Ns}$.

Write the expression for actuating forces.

    $F=\frac{M_{Ns}+M_{Fs}}{c}$                       (VII)

Here actuating force is $F$.

Write the expression\n for braking torque on left hand side shoe.

    $T_L=\frac{f{p}_{a}b{r}^2\left(\cos {\theta }_1-\cos {\theta }_2\right)}{\sin {\theta }_a}$                           (VIII)

Write the expression for total braking torque.

    $T_{Total}=T_p+T_s$                   (IX)

Here, total braking torque is $T_{Total}$.

Conclusion:

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $1000\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (IV).

    $\begin{array}{c}{T}_P=\frac{0.24\left(1000\text{kPa}\right)\left(75\text{mm}\right){\left(200\text{mm}\right)}^2\left(\cos 10°-\cos 75°\right)}{\sin {75}^{\circ }}\\ =\frac{0.24\left(1000\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(3000000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3\left(0.9848-\left(0.2588\right)\right)}{0.9659}\\ =\frac{522.72}{0.9659}\\ =541.17\text{N}\cdot \text{m}\end{array}$

Substitute $288.9\text{N}\cdot \text{m}$ for $M_f$ in Equation (V).

    $\begin{array}{c}{M}_{fs}=\frac{288.9\text{N}\cdot \text{m}}{{10}^6}{p}_a\\ =288.9\times {10}^{-6}{p}_a\end{array}$

Substitute $1229.3\text{N}\cdot \text{m}$ for $M_n$ in Equation (VI).

    $\begin{array}{c}{M}_{Ns}=\frac{1229.3\text{N}\cdot \text{m}}{{10}^6}{p}_a\\ =1229.3\times {10}^{-6}{p}_a\end{array}$

Substitute $5.7\text{kN}$ for $F$, $288.9\times {10}^{-6}{p}_a$ for $M_{fs}$ and $1229.3\times {10}^{-6}{p}_a$ for $M_{Ns}$ in Equation (VII).

    $\begin{array}{l}5.7\text{kN}=\frac{1229.3\times {10}^{-6}{p}_a+288.9\times {10}^{-6}{p}_a}{165\text{mm}}\\ \left(5.7\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\text{=}\frac{1518.2\times {10}^{-6}{p}_a}{\left(165\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ p_a=\frac{940.5}{1518.2\times {10}^{-6}}\\ p_a=619.5\text{kPa}\end{array}$

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $619.5\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (VIII).

    $\begin{array}{c}{T}_s=\frac{0.24\left(619.5\text{kPa}\right)\left(75\text{mm}\right){\left(200\text{mm}\right)}^2\left(\cos 10°-\cos 75°\right)}{\sin {75}^{\circ }}\\ =\frac{0.24\left(619.5\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(3000000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3\left(0.9848-\left(0.2588\right)\right)}{0.9659}\\ =\frac{323.82}{0.9659}\\ =335.32\text{N}\cdot \text{m}\end{array}$

Substitute $335.32\text{N}\cdot \text{m}$ for $T_s$ and $541.17\text{N}\cdot \text{m}$ for $T_p$ in Equation (IX).

    $\begin{array}{c}{T}_{Total}=2\left(541.17\text{N}\cdot \text{m}\right)+2\left(335.32\text{N}\cdot \text{m}\right)\\ =1752.4\text{N}\cdot \text{m}\end{array}$

Thus, the total braking capacity is $1752.4\text{N}\cdot \text{m}$.

To determine

The horizontal reaction on primary shoes.

The vertical reaction on primary shoe.

The horizontal reaction on secondary shoes.

The vertical reaction on secondary shoe.

The resultant reaction.

Answer to Problem 4P

The horizontal reaction on primary shoes is $-\text{0}\text{.66}\text{kN}$.

The, the vertical reaction on primary shoe is $9.88\text{kN}$.

The horizontal reaction on secondary shoes is $-\text{0}\text{.137}\text{kN}$.

The, the vertical reaction on secondary shoe is $3.9\text{kN}$.

The resultant reaction is $6.03\text{kN}$.

Explanation of Solution

Write the expression for horizontal reaction on hinge pin for primary shoe.

    $R_{x_1}=\frac{p_{a}br}{\mathrm{Sin}{\theta }_a}\left({\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta -f{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }}\right)-F_x$                  (X)

Here, horizontal reaction is $R_{x_1}$.

Write the expression for vertical reaction on hinge pin for primary shoe.

    $R_{y_1}=\frac{p_{a}br}{\mathrm{Sin}{\theta }_a}\left({\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }+f{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta }\right)-F_y$              (XI)

Here, vertical reaction is $R_{y_1}$.

Write the expression for horizontal reaction on hinge pin for secondary shoe.

    $R_{x2}=\frac{p_{a}br}{\mathrm{Sin}{\theta }_a}\left({\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta +f{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }}\right)-F_x$        (XII)

Here, horizontal reaction on hinge pin for secondary shoe is $R_{x2}$.

Write the expression for vertical reaction on hinge pin for secondary shoe.

    $R_{y2}=\frac{p_{a}br}{\mathrm{Sin}{\theta }_a}\left({\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }+f{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta }\right)-F_y$          (XIII)

Here, vertical reaction on hinge pin for secondary shoe is $R_{y2}$.

Write the expression for net horizontal reaction.

    $R_H=R_{x1}+R_{x2}$                                                             (XIV)

Here, net horizontal reaction is $R_H$.

Write the expression for net vertical reaction.

    $R_V=R_{y1}-R_{y2}$                                                           (XV)

Here, net vertical reaction is $R_V$.

    $R=\sqrt{R_H^2+R_V^2}$                                                          (XVI)

Here, resultant reaction is $R$.

Conclusion

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $5.7\text{kN}$ for $F_x$, $1000\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (X).

    $\begin{array}{c}{R}_{x_1}=\frac{\left(1000\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)}{\mathrm{Sin}75°}\left(\begin{array}{l}{\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta \cos \theta d\theta }\\ -0.24{\displaystyle \underset{10}{\overset{75}{\int }}{\sin }^2\theta d\theta }\end{array}\right)-\left(5.7\text{kN}\right)\\ \text{=}\frac{\left(1000\text{kPa}\right)\left(15000{\text{mm}}^2\right)}{\mathrm{Sin}75°}\left[\begin{array}{l}{\left(\frac{-1}{4}\cos 2\theta \right)}_{10}^{75}\\ -0.24{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right)}_0^{75}\end{array}\right]-\left(5.7\text{kN}\right)\\ \text{=}\frac{\left(1000\text{kPa}\right)\left(15000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{\mathrm{Sin}75°}\left[0.4515-0.24\left(0.528\right)\right]-5.7\text{kN}\\ =-\text{0}\text{.66}\text{kN}\end{array}$

Thus horizontal reaction on primary shoes is $-\text{0}\text{.66}\text{kN}$.

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $0$ for $F_y$, $1000\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (XI).

    $\begin{array}{c}{R}_{y_1}=\frac{\left(1000\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)}{\mathrm{Sin}75°}\left({\displaystyle \underset{10}{\overset{75}{\int }}{\sin }^2\theta d\theta }+0.24{\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta \cos \theta d\theta }\right)-0\\ =\frac{\left(1000\text{kPa}\right)\left(15000{\text{mm}}^2\right)}{\mathrm{Sin}75°}\left[{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right)}_0^{75}+0.24{\left(\frac{-1}{4}\cos 2\theta \right)}_{10}^{75}\right]\\ =\frac{\left(1000\text{kPa}\right)\left(15000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{\mathrm{Sin}75°}\left[0.528+0.24\left(0.4515\right)\right]\\ =9.88\text{kN}\end{array}$

Thus, the vertical reaction on primary shoe is $9.88\text{kN}$.

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $5.7\text{kN}$ for $F_x$, $619.5\text{kPa}$ for $p_a$ and $0.24$ for $f$ in Equation (XII).

    $\begin{array}{c}{R}_{x2}=\frac{\left(619.5\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)}{\mathrm{Sin}75°}\left({\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta \cos \theta d\theta +0.24{\displaystyle \underset{10}{\overset{75}{\int }}{\sin }^2\theta d\theta }}\right)-5.7\text{kN}\\ \text{=}\frac{\left(619.5\text{kPa}\right)\left(15000{\text{mm}}^2\right)}{\mathrm{Sin}75°}\left[{\left(\frac{-1}{4}\cos 2\theta \right)}_{10}^{75}+0.24{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right)}_0^{75}\right]-5.7\text{kN}\\ \text{=}\frac{\left(619.5\text{kPa}\right)\left(15000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{\mathrm{Sin}75°}\left[0.4515+0.24\left(0.528\right)\right]-5.7\text{kN}\\ =-\text{0}\text{.137}\text{kN}\end{array}$

Thus horizontal reaction on secondary shoes is $-\text{0}\text{.137}\text{kN}$.

Substitute $10°$ for ${\theta }_1$, $50°$ for ${\theta }_2$, $75°$ for ${\theta }_a$, $150\text{mm}$ for $a$, $200\text{mm}$ for $r$, $0$ for $F_y$, $619.5\text{kPa}$  for $p_a$ and $0.24$ for $f$ in Equation (XIII).

    $\begin{array}{c}{R}_{y_2}=\frac{\left(619.5\text{kPa}\right)\left(75\text{mm}\right)\left(200\text{mm}\right)}{\mathrm{Sin}75°}\left({\displaystyle \underset{10}{\overset{75}{\int }}{\sin }^2\theta d\theta }+0.24{\displaystyle \underset{10}{\overset{75}{\int }}\sin \theta \cos \theta d\theta }\right)-0\\ =\frac{\left(619.5\text{kPa}\right)\left(15000{\text{mm}}^2\right)}{\mathrm{Sin}75°}\left[{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right)}_0^{75}+0.24{\left(\frac{-1}{4}\cos 2\theta \right)}_{10}^{75}\right]\\ =\frac{\left(619.5\text{kPa}\right)\left(15000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{\mathrm{Sin}75°}\left[0.528-0.24\left(0.4515\right)\right]\\ =3.9\text{kN}\end{array}$

Thus, the vertical reaction on secondary shoe is $3.9\text{kN}$.

Substitute $-\text{0}\text{.66}\text{kN}$ for $R_{x1}$  and $-\text{0}\text{.137}\text{kN}$ fro $R_{x2}$ in Equation (XIV).

    $\begin{array}{c}{R}_H=-\text{0}\text{.66}\text{kN+}\left(-\text{0}\text{.137}\text{kN}\right)\\ =-0.797\text{kN}\\ \approx \text{0}\text{.8}\text{kN}\end{array}$

Substitute $9.88\text{kN}$ for $R_{y1}$ and $3.9\text{kN}$ for $R_{y2}$ in Equation (XV).

    $\begin{array}{c}{R}_V=9.88\text{kN}-3.9\text{kN}\\ =\text{5}\text{.98}\text{kN}\end{array}$

Substitute $\text{5}\text{.98}\text{kN}$ for $R_v$ and $\text{0}\text{.8}\text{kN}$ for $R_H$ in Equation (XVI).

    $\begin{array}{c}R=\sqrt{{\left(\text{0}\text{.8}\text{kN}\right)}^2+{\left(\text{5}\text{.98}\text{kN}\right)}^2}\\ =\sqrt{0.64{\text{kN}}^2+35.7604{\text{kN}}^2}\\ =\sqrt{36.4004{\text{kN}}^2}\\ =6.03\text{kN}\end{array}$

Thus, the resultant reaction is $6.03\text{kN}$.