Chapter 3, Problem 73P

(a)

To determine

The force $F_B$, assuming the shaft is running at constant speed.

Answer to Problem 73P

The force $F_B$, assuming the shaft is running at constant speed is $22810\text{N}$.

Explanation of Solution

The figure below shows the free body diagram of pulley A.

Figure-(1)

The figure below shows the free body diagram of pulley B.

Figure-(2)

Convert the forces from $\text{kN}$ to $\text{N}$.

$\begin{array}{c}{F}_A=11\text{kN}\\ =\left(11\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\\ =11000\text{N}\end{array}$

Calculate the force $F_B$, using the net torque equation.

$\begin{array}{l}{\displaystyle \sum T}=0\\ -\left(F_A\times \frac{d_A}{2}\times \cos {\theta }_1\right)+\left(F_B\times \frac{d_B}{2}\times \cos {\theta }_2\right)=0\\ F_B=F_A\left(\frac{d_A\cos {\theta }_1}{d_B\cos {\theta }_2}\right)\end{array}$ (I)

Here, the force acting on pulley $A$ is $F_A$, the diameter of pulley $A$ is $d_A$, the angle at which force acts on pulley $A$ is ${\theta }_1$, the force acting on pulley $B$ is $F_B$, the diameter of pulley $B$ is $d_B$ and the angle at which force acts on pulley $B$ is ${\theta }_2$.

Conclusion:

Substitute $11000\text{N}$ for $F_A$, $600\text{mm}$ for $d_A$, $20°$ for ${\theta }_1$, $300\text{mm}$ for $d_B$ and $25°$ for ${\theta }_2$ in Equation (I).

$\begin{array}{c}{F}_B=\left(11000\text{N}\right)\times \left(\frac{600\text{mm}\left(\cos 20°\right)}{300\text{mm}\left(\cos 25°\right)}\right)\\ =\left(11000\text{N}\right)\left(2.0736\right)\\ =22810.39\text{N}\\ \approx 22810\text{N}\end{array}$

Thus, the force $F_B$, assuming the shaft is running at constant speed is $22810\text{N}$.

(b)

To determine

The bearing reaction forces, assuming the bearing act as simple supports.

Answer to Problem 73P

The bearing reaction forces, assuming the bearing act as simple supports at $O$ is $7926\text{N}$ and at $C$ is $\text{13657}\text{N}$.

Explanation of Solution

Write the expression for moment about bearing $O$ in $z\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum M_{Oz}}=0\\ \left[\begin{array}{l}-\left(F_A\sin {\theta }_1\times l_{OA}\right)\\ -\left(F_B\sin {\theta }_2\times \left(l_{OA}+l_{AB}\right)\right)\\ +R_{Cy}\times \left(l_{OA}+l_{AB}+l_{BC}\right)\end{array}\right]=0\\ R_{Cy}=\frac{\left[\begin{array}{l}\left(F_B\sin {\theta }_2\times \left(l_{OA}+l_{AB}\right)\right)\\ +\left(F_A\sin {\theta }_1\times l_{OA}\right)\end{array}\right]}{\left(l_{OA}+l_{AB}+l_{BC}\right)}\end{array}$ (II)

Here, the reaction force at bearing $C$ in $y\text{-}$ direction is $R_{Cy}$, the distance between $O$ and $A$ is $l_{OA}$, the distance between $A$ and $B$ is $l_{AB}$ and the distance between $B$ and $C$ is $l_{BC}$.

Write expression for the equation to balance the forces in $y\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{Oy}-F_A\sin {\theta }_1+R_{Cy}-F_B\sin {\theta }_2=0\\ R_{Oy}=F_B\sin {\theta }_2-R_{Cy}+F_A\sin {\theta }_1\end{array}$ (III)

Here, the reaction force at bearing $O$ in $y\text{-}$ direction is $R_{Oy}$.

Write expression for the moment about bearing $O$ in $y\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum M_{Oy}}=0\\ \left[\begin{array}{l}\left(F_A\cos {\theta }_1\times l_{OA}\right)\\ -\left(F_B\cos {\theta }_2\times \left(l_{OA}+l_{AB}\right)\right)\\ -R_{Cz}\times \left(l_{OA}+l_{AB}+l_{BC}\right)\end{array}\right]=0\\ R_{Cz}=\frac{\left[\begin{array}{l}\left(F_A\cos {\theta }_1\times l_{OA}\right)\\ -\left(F_B\cos {\theta }_2\times \left(l_{OA}+l_{AB}\right)\right)\end{array}\right]}{\left(l_{OA}+l_{AB}+l_{BC}\right)}\end{array}$ (IV)

Here, the reaction force at bearing $C$ in $z\text{-}$ direction is $R_{Cz}$.

Write expression for the equation to balance the forces in $z\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum F_z}=0\\ R_{Oz}-F_A\cos {\theta }_1+R_{Cz}+F_B\cos {\theta }_2=0\\ R_{Oz}=F_A\sin {\theta }_1-R_{Cz}-F_B\cos {\theta }_2\end{array}$ (V)

Here, the reaction force at bearing $O$ in $z\text{-}$ direction is $R_{Oz}$.

Calculate the reaction forces at bearing $O$.

$R_O=\sqrt{R_{Oy}^2+R_{Oz}^2}$ (VI)

Here, the reaction force at bearing $O$ is $R_O$.

Calculate the reaction forces at bearing $C$.

$R_C=\sqrt{R_{Cy}^2+R_{Cz}^2}$ (VII)

Here, the reaction force at bearing $C$ is $R_C$.

Conclusion:

Substitute $22810\text{N}$ for $F_B$, $20°$ for ${\theta }_1$, $11000\text{N}$ for $F_A$, $25°$ for ${\theta }_2$, $400\text{mm}$ for $l_{OA}$, $350\text{mm}$ for $l_{AB}$ and $300\text{mm}$ for $l_{BC}$ in Equation (II).

$\begin{array}{c}{R}_{Cy}=\frac{\left[\begin{array}{l}\left(\left(22810\text{N}\right)\left(\sin 25°\right)\left(400\text{mm}+350\text{mm}\right)\right)\\ +\left(\left(11000\text{N}\right)\left(\sin 20°\right)\left(400\text{mm}\right)\right)\end{array}\right]}{\left(400\text{mm}+350\text{mm}+300\text{mm}\right)}\\ =\frac{8734830.543\text{N}\cdot \text{mm}}{1050\text{mm}}\\ =\left(8318.8862\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \approx 8.319\text{kN}\end{array}$

Convert $R_{Cy}$ from $\text{kN}$ to $\text{N}$.

$\begin{array}{c}{R}_{Cy}=8.319\text{kN}\\ =\left(8.319\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\\ =8319\text{N}\end{array}$

Substitute $22810\text{N}$ for $F_B$, $25°$ for ${\theta }_2$, $8319\text{N}$ for $R_{Cy}$, $11000\text{N}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (III).

$\begin{array}{c}{R}_{Oy}=\left[\left(22810\text{N}\right)\left(\sin 25°\right)-\left(8319\text{N}\right)+\left(11000\text{N}\right)\left(\sin 20°\right)\right]\\ =9639.9225\text{N}-8319\text{N}+3762.2215\text{N}\\ =\left(5083.144\text{N}\right)\\ \approx 5083\text{N}\end{array}$

Substitute $22810\text{N}$ for $F_B$, $20°$ for ${\theta }_1$, $11000\text{N}$ for $F_A$, $25°$ for ${\theta }_2$, $400\text{mm}$ for $l_{OA}$, $350\text{mm}$ for $l_{AB}$ and $300\text{mm}$ for $l_{BC}$ in Equation (IV).

$\begin{array}{c}{R}_{Cz}=\frac{\left[\begin{array}{l}\left(\left(11000\text{N}\right)\left(\cos 20°\right)\left(400\text{mm}\right)\right)\\ -\left(\left(22810\text{N}\right)\left(\cos 25°\right)\left(400\text{mm}+350\text{mm}\right)\right)\end{array}\right]}{\left(400\text{mm}+350\text{mm}+300\text{mm}\right)}\\ =\frac{-11370012.94\text{N}\cdot \text{mm}}{1050\text{mm}}\\ \approx -10830\text{N}\end{array}$

Substitute $11000\text{N}$ for $F_A$, $20°$ for ${\theta }_1$, $-10830\text{N}$ for $R_{Cz}$, $22810\text{N}$ for $F_B$ and $25°$ for ${\theta }_2$ in Equation (V).

$\begin{array}{c}{R}_{Oz}=\left(11000\text{N}\right)\left(\sin 20°\right)+10830\text{N}-\left(22810\text{N}\right)\left(\cos 25°\right)\\ =14592.22158\text{N}-20672.88062\text{N}\\ =-6080.66\text{N}\\ \approx -\text{6081}\text{N}\end{array}$

Substitute $5083\text{N}$ for $R_{Oy}$ and $-\text{6081}\text{N}$ for $R_{Oz}$ in Equation (VI).

$\begin{array}{c}{R}_O=\sqrt{{\left(5083\text{N}\right)}^2+{\left(-\text{6081}\text{N}\right)}^2}\\ =\sqrt{25836889{\text{N}}^2+36978561{\text{N}}^2}\\ =\sqrt{62815450{\text{N}}^2}\\ \approx 7926\text{N}\end{array}$

Substitute $8319\text{N}$ for $R_{Cy}$ and $-10830\text{N}$ for $R_{Cz}$ in Equation (VII).

$\begin{array}{c}{R}_C=\sqrt{{\left(8319\text{N}\right)}^2+{\left(-10830\text{N}\right)}^2}\\ =\sqrt{186494661{\text{N}}^2}\\ =13656.30481\text{N}\\ \approx \text{13657}\text{N}\end{array}$

Thus, the bearing reaction forces, assuming the bearing act as simple supports at $O$ is $7926\text{N}$ and at $C$ is $\text{13657}\text{N}$.

(c)

To determine

The shear force and bending moment diagram for the shaft.

Answer to Problem 73P

The shear force and bending moment diagram for the shaft in $y\text{-}$ direction is as follows.

Figure-(2)

The shear force and bending moment diagram for the shaft in $z\text{-}$ direction is as follows.

Figure-(3)

Explanation of Solution

The calculations for shear force and bending moment diagram in $y\text{-}$ direction.

Calculate the shear force at $O$ in $y\text{-}$ direction.

$S{F}_{Oy}=R_{Oy}$ (VIII)

Here, the shear force at $O$ in $y\text{-}$ direction is $S{F}_{Oy}$.

Calculate the shear force at $A$ in $y\text{-}$ direction.

$S{F}_{Ay}=S{F}_{Oy}-F_A\sin {\theta }_1$ (IX)

Here, the shear force at $A$ in $y\text{-}$ direction is $S{F}_{Ay}$.

Calculate the shear force at $B$ in $y\text{-}$ direction.

$S{F}_{By}=S{F}_{Ay}-\left(F_B\sin {\theta }_2\right)$ (X)

Here, the shear force at $B$ in $y\text{-}$ direction is $S{F}_{By}$.

Calculate the shear force at $C$ in $y\text{-}$ direction.

$S{F}_{Cy}=S{F}_{By}+R_{Cy}$ (XI)

Here, the shear force at $B$ in $y\text{-}$ direction is $S{F}_{By}$.

Calculate the moment at $O$ and $C$.

$M_O=M_C=0$ (XII)

Here, the moment at $O$ is $M_O$ and the moment at $C$ is $M_C$.

Calculate the moment at $A$ in $y\text{-}$ direction.

$M_{Ay}=S{F}_{Oy}\times l_{OA}$ (XIII)

Here, the moment at $A$ in $y\text{-}$ direction is $M_{Ay}$.

Calculate the moment at $B$ in $y\text{-}$ direction.

$M_{By}=F_B\sin {\theta }_2\times l_{CB}$ (XIV)

Here, the moment at $B$ in $y\text{-}$ direction is $M_{By}$.

The calculations for shear force and bending moment diagram in $z\text{-}$ direction.

Calculate the shear force at $O$ in $y\text{-}$ direction.

$S{F}_{Oz}=-R_{Oz}$ (XV)

Here, the shear force at $O$ in $z\text{-}$ direction is $S{F}_{Oz}$.

Calculate the shear force at $A$ in $z\text{-}$ direction.

$S{F}_{Az}=S{F}_{Oz}+F_A\cos {\theta }_1$ (XVI)

Here, the shear force at $A$ in $z\text{-}$ direction is $S{F}_{Az}$.

Calculate the shear force at $B$ in $z\text{-}$ direction.

$S{F}_{Bz}=S{F}_{Az}-F_B\cos {\theta }_2$ (XVII)

Here, the shear force at $B$ in $z\text{-}$ direction is $S{F}_{Bz}$.

Calculate the shear force at $C$ in $z\text{-}$ direction.

$S{F}_{Cz}=S{F}_{Bz}-R_{Cz}$ (XVIII)

Here, the shear force at $C$ in $z\text{-}$ direction is $S{F}_{Cz}$.

Calculate the moment at $O$ and $C$.

$M_O=M_C=0$ (XIX)

Here, the moment at $O$ is $M_O$ and the moment at $C$ is $M_C$.

Calculate the moment at $A$ in $z\text{-}$ direction.

$M_{Az}=S{F}_{Az}\times l_{OA}$ (XX)

Here, the moment at $A$ in $z\text{-}$ direction is $M_{Az}$.

Calculate the moment at $B$ in $z\text{-}$ direction.

$M_{Bz}=R_{Cz}\times l_{BC}$ (XXI)

Here, the moment at $B$ in $z\text{-}$ direction is $M_{Bz}$.

Conclusion:

Substitute $5083\text{N}$ for $R_{Oy}$ in Equation (VIII).

$S{F}_{Oy}=5083\text{N}$

Substitute $5083\text{N}$ for $S{F}_{Oy}$, $11000\text{N}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (IX.)

$\begin{array}{c}S{F}_{Ay}=5083\text{N}-\left(11000\text{N}\right)\left(\sin 20\right)\\ =1320.778\text{N}\\ =1321\text{N}\end{array}$

Substitute $1321\text{N}$ for $S{F}_{Ay}$, $22810\text{N}$ for $F_B$ and $25°$ for ${\theta }_2$ in Equation (X).

$\begin{array}{c}S{F}_{By}=\left(1321\text{N}\right)-\left(\left(22810\text{N}\right)\left(\sin 25°\right)\right)\\ =-8318.92255\text{N}\\ \approx -8319\text{N}\end{array}$

Substitute $-8319\text{N}$ for $S{F}_{By}$ and $8319\text{N}$ for $R_{Cy}$ in Equation (XI).

$\begin{array}{c}S{F}_{Cy}=-8319\text{N}+8319\text{N}\\ =0\text{N}\end{array}$

Substitute $5083\text{N}$ for $S{F}_{Oy}$ and $400\text{mm}$ for $l_{OA}$ in Equation (XIII).

$\begin{array}{c}{M}_{Ay}=\left(5083\text{N}\right)\left(400\text{mm}\right)\\ =\left(5083\text{N}\right)\left(400\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =2033.2\text{N}\cdot \text{m}\\ \approx 2033\text{N}\cdot \text{m}\end{array}$

Substitute $22810\text{N}$ for $F_B$, $25°$ for ${\theta }_2$ and $300\text{mm}$ for $l_{BC}$ in Equation (XIV).

$\begin{array}{c}{M}_{By}=\left(22810\text{N}\right)\left(\sin 25°\right)\left(300\text{mm}\right)\\ =\left(22810\text{N}\right)\left(\sin 25°\right)\left(300\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ \approx 2892\text{N}\cdot \text{m}\end{array}$

Thus, the shear force and bending moment diagram in $y\text{-}$ direction is as follow.

Figure-(4)

Substitute $-\text{6081}\text{N}$ for $R_{Oz}$ in Equation (XV).

$S{F}_{Oz}=-\text{6081}\text{N}$

Substitute $-\text{6081}\text{N}$ for $S{F}_{Oz}$, $11000\text{N}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (XVI).

$\begin{array}{c}S{F}_{Az}=-\text{6081}\text{N}+\left(11000\text{N}\right)\left(\cos 20°\right)\\ =4255.618\\ \approx 4256\text{N}\end{array}$

Substitute $4256\text{N}$ for $S{F}_{Az}$, $22810\text{N}$ for $F_B$ and $25°$ for ${\theta }_2$ in Equation (XVII).

$\begin{array}{c}S{F}_{Bz}=4256\text{N}-\left(22810\text{N}\right)\left(\cos 25°\right)\\ =-16416.88\text{N}\\ \approx -16417\text{N}\end{array}$

Substitute $-16417\text{N}$ for $S{F}_{Bz}$ and $-10830\text{N}$ for $R_{Cz}$ in Equation (XVIII).

$\begin{array}{c}S{F}_{Cz}=\left(-16417\text{N}\right)-\left(-10830\text{N}\right)\\ =-16417\text{N}+10830\text{N}\\ =-5587\text{N}\end{array}$

Substitute $4256\text{N}$ for $S{F}_{Az}$ and $400\text{mm}$ for $l_{OA}$ in Equation (XX).

$\begin{array}{c}{M}_{Az}=\left(4256\text{N}\right)\left(400\text{mm}\right)\\ =\left(4256\text{N}\right)\left(400\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =1702.4\text{N}\cdot \text{m}\\ \approx 1702\text{N}\cdot \text{m}\end{array}$

Substitute $10830\text{N}$ for $F_B$ and $300\text{mm}$ for $l_{BC}$ in Equation (XXI).

$\begin{array}{c}{M}_{Bz}=\left(10830\text{N}\right)\left(300\text{mm}\right)\\ =\left(10830\text{N}\right)\left(300\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =3249\text{N}\cdot \text{m}\end{array}$

Thus, the shear force and bending moment diagram in $z\text{-}$ direction is as follow.

Figure-(5)

(d)

To determine

The bending stress at point of maximum bending moment.

The shear stress at point of maximum bending moment.

Answer to Problem 73P

The bending stress at point of maximum bending moment is $355\text{MPa}$.

The shear stress at point of maximum bending moment is $127\text{MPa}$.

Explanation of Solution

Write the net moment at $A$.

$M_A=\sqrt{M_{Ay}^2+M_{Az}^2}$ (XXII)

Here, the net moment at $A$ is $M_A$.

Write the net moment at $B$.

$M_B=\sqrt{M_{By}^2+M_{Bz}^2}$ (XXIII)

Here, the net moment at $C$ is $M_C$.

Write the torque transmitted by shaft from $A$ to $B$.

$T=F_A\cos {\theta }_1\times \frac{d_A}{2}$ . (XXIV)

Here, the torque transmitted by shaft from $A$ to $B$ is $T$.

Calculate the bending stress.

$\sigma =\frac{32M_B}{\pi d^3}$ (XXV)

Here, the bending stress is $\sigma$ and diameter of shaft is $d$.

Calculate the shear stress.

$\tau =\frac{16T}{\pi d^3}$ (XVI)

Here, the shear stress is $\tau$.

Conclusion:

Substitute $2033\text{N}\cdot \text{m}$ for $M_{Ay}$ and $1702\text{N}\cdot \text{m}$ for $M_{Az}$ in Equation (XXII).

$\begin{array}{c}{M}_A=\sqrt{{\left(2033\text{N}\cdot \text{m}\right)}^2+{\left(1702\text{N}\cdot \text{m}\right)}^2}\\ =\left(\sqrt{7029893}\right)\text{N}\cdot \text{m}\\ \approx 2652\text{N}\cdot \text{m}\end{array}$

Substitute $2892\text{N}\cdot \text{m}$ for $M_{By}$ and $3249\text{N}\cdot \text{m}$ for $M_{Bz}$ in Equation (XXIII).

$\begin{array}{c}{M}_B=\sqrt{{\left(2892\text{N}\cdot \text{m}\right)}^2+{\left(3249\text{N}\cdot \text{m}\right)}^2}\\ =\sqrt{18919665{\text{N}}^2\cdot {\text{m}}^2}\\ =4349.674\text{N}\cdot \text{m}\\ \approx \text{4350}\text{N}\cdot \text{m}\end{array}$

Since, $M_B>M_A$, the critical location is at $B$.

Substitute $11000\text{N}$ for $F_A$, $20°$ for ${\theta }_1$ and $600\text{mm}$ for $d_A$ in Equation (XXIV).

$\begin{array}{c}T=\left(11000\text{N}\right)\left(\cos 20°\right)\frac{\left(600\text{mm}\right)}{2}\\ =\left(11000\text{N}\right)\left(\cos 20°\right)\frac{\left(600\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\\ =3100.9856\text{N}\cdot \text{m}\\ \approx 3101\text{N}\cdot \text{m}\end{array}$

Substitute $\text{4350}\text{N}\cdot \text{m}$ for $M_B$ and $50\text{mm}$ for $d$ in Equation (XXV).

$\begin{array}{c}\sigma =\frac{\left(32\right)\left(\text{4350}\text{N}\cdot \text{m}\right)}{\pi {\left(50\text{mm}\right)}^3}\\ =\frac{\left(32\right)\left(\text{4350}\text{N}\cdot \text{m}\right)}{\pi {\left(\left(50\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^3}\\ =\frac{139200\text{N}\cdot \text{m}}{\left(3.92699\times {10}^{-4}\right){\text{m}}^3}\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^{\text{2}}}\right)\\ \approx 355\text{MPa}\end{array}$

Thus, the bending stress at point of maximum bending moment is $355\text{MPa}$.

Substitute $3101\text{N}\cdot \text{m}$ for $T$ and $50\text{mm}$ for $d$ in Equation (XXVI).

$\begin{array}{c}\tau =\frac{16\times 3101\text{N}\cdot \text{m}}{\pi {\left(50\text{mm}\right)}^3}\\ =\frac{\left(16\right)\left(3101\text{N}\cdot \text{m}\right)}{\pi {\left(\left(50\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^3}\\ =\frac{49616\text{N}\cdot \text{m}}{\left(3.92699\times {10}^{-4}\right){\text{m}}^3}\left(\frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^{\text{2}}}\right)\\ \approx 127\text{MPa}\end{array}$

Thus, the shear stress at point of maximum bending moment is $127\text{MPa}$.

(e)

To determine

The principal stresses at point of maximum bending moment.

The maximum shear stress at point of maximum bending moment.

Answer to Problem 73P

The principal stresses at point of maximum bending moment are $396\text{MPa}$ and $-41\text{MPa}$.

The maximum shear stress at point of maximum bending moment is $218\text{MPa}$.

Explanation of Solution

Calculate the maximum principal stress.

${\sigma }_1=\frac{\sigma }{2}+\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXVII)

Here, the maximum principal stress is ${\sigma }_1$.

Calculate the minimum principal stress.

${\sigma }_2=\frac{\sigma }{2}-\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXVIII)

Here, the minimum principal stress is ${\sigma }_2$.

Calculate the maximum shear stress.

${\tau }_{\max }=\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXIX)

Here, maximum shear stress is ${\tau }_{\max }$.

Conclusion:

Substitute $355\text{MPa}$ for $\sigma$ and $127\text{MPa}$ for $\tau$ in Equation (XXVII).

$\begin{array}{c}{\sigma }_1=\frac{355\text{MPa}}{2}+\sqrt{{\left(\frac{355\text{MPa}}{2}\right)}^2+{\left(127\text{MPa}\right)}^2}\\ =177.5\text{MPa}+218.255\text{MPa}\\ \approx 396\text{MPa}\end{array}$

Substitute $355\text{MPa}$ for $\sigma$ and $127\text{MPa}$ for $\tau$ n Equation (XXVIII).

$\begin{array}{c}{\sigma }_2=\frac{355\text{MPa}}{2}-\sqrt{{\left(\frac{355\text{MPa}}{2}\right)}^2+{\left(127\text{MPa}\right)}^2}\\ =177.5\text{MPa}-218.255\text{MPa}\\ \approx -41\text{MPa}\end{array}$

Thus, the principal stresses at point of maximum bending moment are $396\text{MPa}$ and $-41\text{MPa}$.

Substitute $355\text{MPa}$ for $\sigma$ and $127\text{MPa}$ for $\tau$ in Equation (XXIX).

$\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{355\text{MPa}}{2}\right)}^2+{\left(127\text{MPa}\right)}^2}\\ =\sqrt{47635.25{\text{MPa}}^2}\\ =218.255\text{MPa}\\ \approx 218\text{MPa}\end{array}$

Thus, the maximum shear stress at point of maximum bending moment is $218\text{MPa}$.