Chapter 16, Problem 1P

The figure shows an internal rim-type brake having an inside rim diameter of 300 mm and a dimension R = 125 mm. The shoes have a face width of 40 mm and are both actuated by a force of 2.2 kN. The drum rotates clockwise. The mean coefficient of friction is 0.28.

1. (a)   Find the maximum pressure and indicate the shoe on which it occurs.
2. (b)   Estimate the braking torque effected by each shoe, and find the total braking torque.
3. (c)   Estimate the resulting hinge-pin reactions.

Problem 16-1

To determine

The maximum pressure.

The shoe on which maximum pressure occurs.

Answer to Problem 1P

The maximum pressure is $734\text{kPa}$.

The shoe on which maximum pressure occurs is right side.

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.

$F=\frac{M_N-M_f}{2a\cos \theta }$ . (III)

Here, actuating force is $F$.

Write the expression for actuating force in reversed rotation.

$F=\frac{M_N+M_f}{2a\cos \theta }$ . (IV).

Conclusion:

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $125\text{mm}$ for $a$ in Equation (I).

$\begin{array}{c}{M}_f=\left[\begin{array}{l}\frac{\left(0.28\right)p_a\left(40\text{mm}\right)150\text{mm}}{\sin \left(90°\right)}\\ {\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta }\left(150\text{mm}-125\text{mm}\cos \theta \right)d\theta \end{array}\right]\\ =\left[\begin{array}{l}\frac{\left(0.28\right)p_a\left(6000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{\sin \left(90°\right)}\\ {\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta }\left(150\text{mm}-125\text{mm}\cos \theta \right)d\theta \end{array}\right]\\ =1.68\times {10}^{-3}{p}_a\left[{\left(-150\cos \theta \right)}_0{}^{120}+125{\left(\frac{1}{4}\cos 2\theta \right)}_0^{120}\right]\\ =1.68\times {10}^{-3}{p}_a\left[150\left(-0.5-1\right)+31.25\left(-0.5-1\right)\right]\end{array}$

$\begin{array}{c}{M}_f=\left[1.68\times {10}^{-3}\left(p_a\right)\left[\begin{array}{l}-150\text{mm}\left(-0.5-1\right)\\ +31.25\text{mm}\left(-0.5-1\right)\end{array}\right]\right]\\ =1.68\times {10}^{-3}\left(p_a\left[225\text{mm}-46.875\text{mm}\right]\right)\\ =1.68\times {10}^{-3}\left(p_a\right)\left[178.125\text{mm}\right]\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =\left(2.99\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a\end{array}$

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $125\text{mm}$ for $a$ in Equation (II).

$\begin{array}{c}{M}_N=\frac{p_a\left(40\text{mm}\right)\left(150\text{mm}\right)\left(125\text{mm}\right)}{\sin \left(90°\right)}{\displaystyle \underset{0°}{\overset{120°}{\int }}{\sin }^2\theta d\theta }\\ =p_a\left(750000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3{\left[\frac{\theta }{2}-\frac{1}{4}\sin 2\theta \right]}_0^{120}\\ =7.5\times {10}^{-4}{p}_a\left[\frac{2\pi }{6}-\frac{1}{4}\sin \left(240°\right)\right]\\ =9.47\times {10}^{-4}{p}_a\text{N}\cdot \text{m}\end{array}$

Substitute  $9.47\times {10}^{-4}{p}_a\text{N}\text{.m}$ for $M_N$, $\left(2.99\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a\text{N}\cdot \text{m}$ for $M_f$, $125\text{mm}$ for $a$, $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in equation (III).

$\begin{array}{l}2.2\text{kN}=\frac{\left(9.47\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a-\left(2.99\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a\text{N}\cdot \text{m}}{2\left(125\text{mm}\right)\cos 30°}\\ 2.2\text{kN}=\frac{\left(6.48\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a}{150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(0.866\right)}\\ p_a=\left(733.9\text{kN}/{\text{m}}^{\text{2}}\right)\left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\\ p_a\approx 734\text{kPa}\end{array}$

Substitute $9.47\times {10}^{-4}{p}_a\text{N}\cdot \text{m}$ for $M_N$, $2.99\times {10}^{-4}{p}_a\text{N}\cdot \text{m}$ for $M_f$, $125\text{mm}$ for $a$, $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in equation (IV).

$\begin{array}{l}2.2\text{kN}=\frac{\left(9.47\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a+\left(2.99\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a\text{N}\cdot \text{m}}{2\left(125\text{mm}\right)\cos 30°}\\ 2.2\text{kN}=\frac{\left(12.46\times {10}^{-4}{\text{m}}^{\text{3}}\right)p_a}{150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(0.866\right)}\\ p_a=\left(386\text{kN}/{\text{m}}^{\text{2}}\right)\left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\\ p_a=386\text{kPa}\end{array}$

Since the maximum pressure occur on the right side of the shoe. So the maximum pressure occurs on the right shoe.

Thus, the maximum pressure is $734\text{kPa}$.

To determine

The braking torque effected by right shoe.

The braking torque effected by left shoe.

The total braking torque.

Answer to Problem 1P

The braking torque effected by right shoe is $277.45\text{N}\cdot \text{m}$.

The braking torque effected by left shoe is $145.9\text{N}\cdot \text{m}$.

The total braking torque is $\text{423}\text{.35}\text{N}\cdot \text{m}$.

Explanation of Solution

Write the expression for torque applied by the right hand shoe.

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

Here, braking torque on right hand side shoe is $T_R$.

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

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

Here, braking torque on left hand side shoe is $T_L$.

Write the expression for total braking torque.

$T_{Total}=T_R+T_L$ (VII)

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

Conclusion:

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $734\text{kPa}$ for $p_a$$150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation (V).

$\begin{array}{c}{T}_R=\left[\frac{0.28\left(734\text{kPa}\right)\left(40\text{mm}\right){\left(150\text{mm}\right)}^2\left(\cos \left(0°\right)-\cos \theta \right)}{\sin \left(90°\right)}\right]\\ =\left[0.28\left(734\text{kPa}\right)\left(900000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3\left(1-\left(-0.5\right)\right)\right]\\ =0.28\left(734\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(9\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(1.5\right)\\ =277.45\text{N}\cdot \text{m}\end{array}$

Thus, braking torque on right hand side shoe is $277.45\text{N}\cdot \text{m}$.

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $386\text{kPa}$ for $p_a$ $150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation (VI).

$\begin{array}{c}{T}_R=\frac{0.28\left(386\text{kPa}\right)\left(40\text{mm}\right){\left(150\text{mm}\right)}^2\left(\cos 0°-\cos \theta \right)}{\sin 90°}\\ =\frac{0.28\left(386\text{kPa}\right)\left(900000{\text{mm}}^3\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3\left(1-\left(-0.5\right)\right)}{1}\\ =0.28\left(386\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(9\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(1.5\right)\\ =145.9\text{N}\cdot \text{m}\end{array}$

Thus, braking torque on left hand side shoe is $145.9\text{N}\cdot \text{m}$.

Substitute $145.9\text{N}\cdot \text{m}$ for $T_L$ and $277.45\text{N}\cdot \text{m}$ for $T_R$ in Equation (VII).

$\begin{array}{c}{T}_{Total}=277.45\text{N}\cdot \text{m}+145.9\text{N}\cdot \text{m}\\ =\text{423}\text{.35}\text{N}\cdot \text{m}\end{array}$

Thus, the total torque is $\text{423}\text{.35}\text{N}\cdot \text{m}$.

To determine

The resulting hinge pin reaction at right hand shoe.

The resulting hinge pin reaction at left hand shoe.

Answer to Problem 1P

The resulting hinge pin reaction at right hand shoe is $4.24\text{kN}$.

The resulting hinge pin reaction at left hand shoe is $\text{0}\text{.98}\text{kN}$.

Explanation of Solution

Write the expression for force in horizontal direction for right hand shoe.

$F_x=F\sin \theta$ (VIII)

Here, horizontal force is $F_x$.

Write the expression for vertical direction force on right hand side shoe.

$F_y=F\cos \theta$ (IX)

Here, vertical direction force is $F_y$.

Write the expression for horizontal reaction for right hand side shoe.

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

Here, reaction in horizontal direction is $R_x$.

Write the expression for reaction in vertical direction for right hand side shoe.

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

Here, reaction in vertical direction for left hand shoe is $R_y$.

Write the expression for resultant reaction.

$R=\sqrt{R_x{}^2+R_y{}^2}$ (XII)

Write the expression for force in horizontal direction for left hand shoe.

$F_x=F\sin \theta$ (XIII)

Here, horizontal force is $F_x$.

Write the expression for vertical direction force on left hand side shoe.

$F_y=F\cos \theta$ (XIV)

Here, vertical direction force is $F_y$.

Write the expression for horizontal reaction for left hand side shoe.

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

Here, reaction in horizontal direction is $R_x$.

Write the expression for reaction in vertical direction for left hand side shoe.

$R_y=\frac{p_{a}br}{\sin {\theta }_a}\left({\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\sin }^2\theta d\theta }-f{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta \cos \theta d\theta }\right)-F_y$ (XVI)

Here, reaction in vertical direction for right hand shoe is $R_y$.

Conclusion:

Substitute $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in equation (VIII).

$\begin{array}{c}{F}_x=2.2\text{kN}\left(\sin 30°\right)\\ =2.2\text{kN}\left(0.5\right)\\ =1.1\text{kN}\end{array}$

Substitute $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in equation (IX).

$\begin{array}{c}{F}_y=2.2\text{kN}\left(\cos 30°\right)\\ =2.2\text{kN}\left(0.866\right)\\ =1.9052\text{kN}\end{array}$

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $734\text{kPa}$ for $p_a$ $150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation(X).

$\begin{array}{c}{R}_x=\left[\frac{\left(734\text{kPa}\right)\left(40\text{mm}\right)\left(150\text{mm}\right)}{\sin {90}^{\circ }}\left({\displaystyle \underset{0}{\overset{120}{\int }}\sin \theta \cos \theta d\theta }-0.28{\displaystyle \underset{0}{\overset{120}{\int }}{\sin }^2\theta d\theta }\right)-1.1\text{kN}\right]\\ \text{=}\left[\begin{array}{l}\frac{\left(734\text{kPa}\right)\left(6000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{1}\\ \left[{\left(-\frac{1}{2}\cos \theta \right)}_0^{120}-0.28{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta d\theta \right)}_0^{120}\right]-1.1\text{kN}\end{array}\right]\\ =\left(734\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(6\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left[0.375-0.28\left(1.264\right)\right]-1.1\text{kN}\\ =-1.007\text{kN}\end{array}$

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $734\text{kPa}$ for $p_a$ $150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation(XI).

$\begin{array}{c}{R}_y=\left[\begin{array}{l}\frac{\left(734\text{kPa}\right)\left(40\text{mm}\right)\left(150\text{mm}\right)}{\sin {90}^{\circ }}\left({\displaystyle \underset{0}{\overset{120}{\int }}{\sin }^2\theta d\theta }-0.28{\displaystyle \underset{0}{\overset{120}{\int }}\sin \theta \cos \theta d\theta }\right)\\ -1.9052\text{kN}\end{array}\right]\\ \text{=}\left[\begin{array}{l}\frac{\left(734\text{kPa}\right)\left(6000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{1}\\ \left[{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta d\theta \right)}_0^{120}-0.28{\left(-\frac{1}{2}\cos \theta \right)}_0^{120}\right]-1.9052\text{kN}\end{array}\right]\\ =\left(734\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(6\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left[\left(1.264\right)-0.28\left(0.375\right)\right]-1.9052\text{kN}\\ =4.12\text{kN}\end{array}$

Substitute $-1.007\text{kN}$ for $R_x$ and $4.12\text{kN}$ for $R_y$ in Equation (XII).

$\begin{array}{c}R=\sqrt{{\left(-1.007\text{kN}\right)}^2+{\left(4.12\text{kN}\right)}^2}\\ =\sqrt{17.988449{\text{kN}}^2}\\ =4.24\text{kN}\end{array}$

The resulting hinge pin reaction at right hand shoe is $4.24\text{kN}$.

Substitute $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in Equation (XIII).

$\begin{array}{c}{F}_x=2.2\text{kN}\left(\sin 30°\right)\\ =2.2\text{kN}\left(0.5\right)\\ =1.1\text{kN}\end{array}$

Substitute $2.2\text{kN}$ for $F$ and $30°$ for $\theta$ in Equation (XIV).

$\begin{array}{c}{F}_y=2.2\text{kN}\left(\cos 30°\right)\\ =2.2\text{kN}\left(0.866\right)\\ =1.9052\text{kN}\end{array}$

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $386\text{kPa}$ for $p_a$ $150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation(XV).

$\begin{array}{c}{R}_x=\left[\begin{array}{l}\frac{386\text{kPa}\left(40\text{mm}\right)\left(150\text{mm}\right)}{\sin {90}^{\circ }}\left({\displaystyle \underset{0}{\overset{120}{\int }}\sin \theta \cos \theta d\theta }+0.28{\displaystyle \underset{0}{\overset{120}{\int }}{\sin }^2\theta d\theta }\right)\\ -1.1\text{kN}\end{array}\right]\\ \text{=}\left[\begin{array}{l}\frac{\left(386\text{kPa}\right)\left(6000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{1}\\ \left[{\left(-\frac{1}{2}\cos \theta \right)}_0^{120}+0.28{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta d\theta \right)}_0^{120}\right]-1.1\text{kN}\end{array}\right]\\ =\left(386\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(6\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left[0.375+0.28\left(1.264\right)\right]-1.1\text{kN}\\ =0.588\text{kN}\end{array}$

Substitute $0°$ for ${\theta }_1$, $120°$ for ${\theta }_2$, $90°$ for ${\theta }_a$, $40\text{mm}$ for $b$, $386\text{kPa}$ for $p_a$ $150\text{mm}$ for $r$ and $0.25$ for $f$ in Equation(XVI).

$\begin{array}{c}{R}_y=\left[\begin{array}{l}\frac{\left(386\text{kPa}\right)\left(40\text{mm}\right)\left(150\text{mm}\right)}{\sin {90}^{\circ }}\\ \left({\displaystyle \underset{0}{\overset{120}{\int }}{\sin }^2\theta d\theta }-0.28{\displaystyle \underset{0}{\overset{120}{\int }}\sin \theta \cos \theta d\theta }\right)-1.9052\text{kN}\end{array}\right]\\ \text{=}\left[\begin{array}{l}\frac{\left(386\text{kPa}\right)\left(6000{\text{mm}}^2\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^2}{1}\\ \left[{\left(\frac{\theta }{2}-\frac{1}{4}\sin 2\theta d\theta \right)}_0^{120}-0.28{\left(-\frac{1}{2}\cos \theta \right)}_0^{120}\right]-1.9052\text{kN}\end{array}\right]\\ =\left(386\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(6\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left[\left(1.264\right)-0.28\left(0.375\right)\right]-1.9052\text{kN}\\ =0.779\text{kN}\end{array}$

Substitute $0.588\text{kN}$ for $R_x$ and $0.779\text{kN}$ for $R_y$ in equation (XIII).

$\begin{array}{c}R=\sqrt{{\left(0.588\text{kN}\right)}^2+{\left(0.779\text{kN}\right)}^2}\\ =\sqrt{0.952585{\text{kN}}^2}\\ =0.976\text{kN}\\ \approx \text{0}\text{.98}\text{kN}\end{array}$

Thus the resultant reaction on left hand side shoe is $\text{0}\text{.98}\text{kN}$.