Chapter 5, Problem 42P

To determine

The factor of safety for yielding from distortion-energy theory.

The factor of safety for yielding from maximum-shear-stress theory.

Answer to Problem 42P

The factor of safety for yielding from distortion-energy theory is $4.57$.

The factor of safety for yielding from maximum-shear-stress theory is $4.43$.

Explanation of Solution

The given assumption is that the belt tension on the loose side at $B$ is $15\%$ of the tension on the tight side.

Write the relationship between tension on the loose side with respect to tension on the tight side.

    $T_2=0.15T_1$                                                                                         (I)

Here, the tension on the tight side is $T_1$ and the tension on the loose side is $T_2$.

Write the equation to balance the tension on the counter shaft.

    $\begin{array}{c}{\displaystyle \sum T}=0\\ \left(T_{A1}-T_{A2}\right)\frac{d_A}{2}-\left(T_2-T_1\right)\frac{d_B}{2}=0\end{array}$                                                           (II)

Substitute $0.15T_1$ for $T_2$ in Equation (II).

Calculate the tension on the loose side.

    $\begin{array}{l}\left(T_{A1}-T_{A2}\right)\frac{d_A}{2}+\left(0.15T_1-T_1\right)\frac{d_B}{2}=0\\ T_1=\frac{1}{0.85}\left(\left(T_{A1}-T_{A2}\right)\frac{d_A}{d_B}\right)\end{array}$                                                          (III)

Here, the tension on the tight side of pulley $A$ is $T_{A1}$, the tension on the loose side of pulley $A$ is $T_{A2}$, diameter of shaft $A$ is $d_A$ and the diameter of shaft $B$ is $d_B$.

Write the magnitude of bearing reaction force at $C$ in $z\text{-}$ direction.

    $\begin{array}{l}{\displaystyle \sum M_{Oy}}=0\\ \left[\begin{array}{l}\left(T_{A1}+T_{A2}\right)\sin \theta \left(l_{OA}\right)-\left(T_1+T_2\right)\left(l_{OA}+l_{AB}\right)\\ -R_{Cz}\left(l_{OA}+l_{AB}+l_{BC}\right)\end{array}\right]=0\\ R_{Cz}=\frac{\left[\begin{array}{l}\left(T_{A1}+T_{A2}\right)\sin \theta \left(l_{OA}\right)-\\ \left(T_1+T_2\right)\left(l_{OA}+l_{AB}\right)\end{array}\right]}{\left(l_{OA}+l_{AB}+l_{BC}\right)}\end{array}$                                          (IV)

Here, the tension on tight side of pulley $A$ is $T_{A1}$, the tension on the loose side of pulley $A$ is $T_{A2}$, the magnitude of the bearing force at $C$ in $z\text{-}$ direction is $R_{Cz}$, distance between $O$ and $A$ is $l_{OA}$, distance between $A$ and $B$ is $l_{AB}$, distance between $B$ and $C$ is $l_{BC}$ and the angle between the tension on the pulley $A$ and $B$ is $\theta$.

Write the magnitude of bearing reaction force at $O$ in $z$ direction.

    $\begin{array}{l}{\displaystyle \sum F_z}=0\\ \left[\begin{array}{l}{R}_{Oz}-\left(T_{A1}+T_{A2}\right)\cos \theta \\ +\left(T_1+T_2\right)+R_{Cz}\end{array}\right]=0\\ R_{Oz}=\left(T_{A1}+T_{A2}\right)\cos \theta -\left(T_1+T_2\right)-R_{Cz}\end{array}$                                                         (V)

Write the magnitude of bearing reaction force at $C$ in $y$ direction.

    $\begin{array}{l}{\displaystyle \sum M_{Oz}}=0\\ R_{Cy}\left(l_{OA}+l_{AB}+l_{BC}\right)+\left(T_{A1}+T_{A2}\right)\sin \theta \left(l_{OA}\right)=0\\ R_{Cy}=-\frac{\left(T_{A1}+T_{A2}\right)\sin \theta \left(l_{OA}\right)}{\left(l_{OA}+l_{AB}+l_{BC}\right)}\end{array}$                                            (VI)

Here, the magnitude of bearing force at $C$ in $y$ direction is $R_{Cy}$.

Write the magnitude of bearing force at $O$ in $y$ direction.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{Oy}+\left(T_{A1}+T_{A2}\right)\cos \theta +R_{Cy}=0\\ R_{Oy}=-\left(T_{A1}+T_{A2}\right)\cos \theta -R_{Cy}\end{array}$                                                                 (VII)

Here, the magnitude of bearing reaction force at $O$ in $z$ direction is $R_{Oz}$.

Calculate the bearing reaction force at $B$.

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

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

Calculate the bearing reaction force at $O$.

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

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

The calculations for shear force and bending moment diagram in $y$ direction.

Calculate the shear force at $O$ in $y$ direction.

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

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

Calculate the shear force at $A$ in $y$ direction.

    $S{F}_{Ay}=S{F}_{Oy}+(T_{A1}+T_{A2})\cos \theta$                                                              (XI)

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

Calculate the shear force at $C$ in $y$ direction.

    $S{F}_{Cy}=S{F}_{Ay}+R_{Cy}$                                                                                (XII)

Here, the shear force at $C$ in $y$ direction is $S{F}_{Cy}$.

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

    $M_O=M_C=0$                                                                                 (XIII)

The moment at the supports of the simply supported beam is zero.

Calculate the moment at $A$ in $y$ direction.

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

Here, the moment at $A$ is $M_A$ in $y$ direction.

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

Calculate the shear force at $O$ in $z$ direction.

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

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

Calculate the shear force at $A$ in $z$ direction.

    $S{F}_{Az}=S{F}_{Oz}+\left(T_{A1}+T_{A2}\right)\sin 45°$                                                            (XVI)

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

Calculate the shear force at $B$ in $z$ direction.

    $S{F}_{Bz}=S{F}_{Az}-\left(T_1+T_2\right)$                                                                     (XVII)

Calculate the shear force at $C$ in $z$ direction.

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

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

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

    $M_O=M_C=0$                                                                                    (XIX)

The moment at the supports of the simply supported beam is zero.

Calculate the moment at $A$ in $z$ direction.

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

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

Calculate the moment at $B$ in $z$ direction.

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

Here, the moment at $A$ is $M_{By}$ in $z$ direction.

It is clear from the bending moment diagrams, that the critical location is at $A$. At this point, there will be maximum bending moment.

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

    $T=\left(T_{A1}-T_{A2}\right)\times \frac{d_A}{2}$                                                             (XXIII)

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

Calculate the bending stress.

    $\sigma =\frac{32M_A}{\pi d^3}$                                                                            (XXIV)

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

Calculate the shear stress.

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

Here, the shear stress is $\tau$.

Calculate the maximum principal stress.

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

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

Calculate the minimum principal stress.

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

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

Calculate the maximum shear stress.

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

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

Calculate the factor of safety from maximum-shear-stress theory.

    $n=\frac{S_y}{{\sigma }_1-{\sigma }_2}$                                                                                    (XXIX)

Here, the maximum yield stress for $1018$ CD steel is $S_y$.

Calculate the factor of safety from distortion-energy theory.

    $n=\frac{S_y}{\sigma \text{'}}$                                                                                          (XXX)

Here, the Von Mises stress is $\sigma \text{'}$.

Write the expression for von Mises stress.

    $\sigma \text{'}={\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}$

Substitute $\left[{\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}\right]$ for $\sigma \text{'}$ in the Equation (XXX).

    $n=\frac{S_y}{\left[{\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}\right]}$                                                               (XXXI)

Conclusion:

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$, $250\text{mm}$ for $d_A$ and $300\text{mm}$ for $d_B$  in Equation (III).

    $\begin{array}{c}{T}_1=\frac{1}{0.85}\left(\left(300\text{N}-45\text{N}\right)\times \frac{250\text{mm}}{300\text{mm}}\right)\\ =\frac{1}{0.85}\left(\left(255\text{N}\right)\times \frac{5\text{mm}}{6\text{mm}}\right)\\ =250\text{N}\end{array}$

Substitute $250\text{N}$ for $T_1$ in Equation (I)

    $\begin{array}{c}{T}_2=0.15\times 250\text{N}\\ =37.5\text{N}\end{array}$

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$, $250\text{N}$ for $T_1$, $37.5\text{N}$ for $T_2$, $300\text{mm}$ for $l_{OA}$, $400\text{mm}$ for $l_{AB}$, $150\text{mm}$ for $l_{BC}$ and $45°$ for $\theta$ Equation (IV).

    $\begin{array}{c}{R}_{Cz}=\frac{\left(\left(300\text{N}+45\text{N}\right)\sin 45°\times 300\text{mm}\right)-\left(\left(250\text{N}+37.5\text{N}\right)\left(300\text{mm}+400\text{mm}\right)\right)}{\left(300\text{mm}+400\text{mm}+150\text{mm}\right)}\\ =\frac{\left(345\text{N}\times \sin 45°\times 300\text{mm}\right)-\left(\left(287.5\text{N}\right)\left(700\text{mm}\right)\right)}{\left(850\text{mm}\right)}\\ =-150.7\text{N}\end{array}$

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$, $250\text{N}$ for $T_1$, $37.5\text{N}$ for $T_2$, $45°$ for $\theta$ and $-150.7\text{N}$ for $R_{Cz}$ in Equation (V).

    $\begin{array}{c}{R}_{Oz}=\left(300\text{N}+45\text{N}\right)\cos 45°-\left(250\text{N}+37.5\text{N}\right)+150.7\text{N}\\ =107.2\text{N}\end{array}$

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$, $300\text{mm}$ for $l_{OA}$, $400\text{mm}$ for $l_{AB}$, $150\text{mm}$ for $l_{BC}$ and $45°$ for $\theta$ in Equation (VI.)

    $\begin{array}{c}{R}_{Cy}=-\frac{\left(300\text{N}+45\text{N}\right)\times \sin 45°\times \left(300\text{mm}\right)}{\left(300\text{mm}+400\text{mm}+150\text{mm}\right)}\\ =-\frac{\left(345\text{N}\right)\times \sin 45°\times \left(300\text{mm}\right)}{\left(850\text{mm}\right)}\\ =-86.10\text{N}\end{array}$

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$, $-86.10\text{N}$ for $R_{Cy}$ and $45°$ for $\theta$  in Equation (VII)

    $\begin{array}{c}{R}_{Oy}=-\left(300\text{N}+45\text{N}\right)\cos 45°+86.10\text{N}\\ =-\left(345\text{N}\right)\cos 45°+86.10\text{N}\\ =-157.9\text{N}\end{array}$

Substitute $-150.7\text{N}$ for $R_{Cz}$ and $-86.1\text{N}$ for $R_{Cy}$ in Equation (VIII)

    $\begin{array}{c}{R}_C=\sqrt{{\left(-150.7\text{N}\right)}^2+{\left(-86.1\text{N}\right)}^2}\\ =173.56\text{N}\end{array}$

Substitute $-157.9\text{N}$ for $R_{Oy}$ and $107.2\text{N}$ for $R_{Oz}$ in Equation (IX).

    $\begin{array}{c}{R}_O=\sqrt{{\left(-157.9\text{N}\right)}^2+{\left(107.2\text{N}\right)}^2}\\ =190.85\text{N}\end{array}$

Substitute $-157.9\text{N}$ for $R_{Oy}$ in Equation (X).

    $S{F}_{Oy}=-157.9\text{N}$

Substitute $-157.9\text{N}$ for $S{F}_{Oy}$, $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$ and $45°$ for $\theta$ in Equation (XI).

    $\begin{array}{c}S{F}_{Ay}=-157.9\text{N}+(300\text{N}+45\text{N})\cos 45°\\ =86.1\text{N}\end{array}$

Substitute $-86.1\text{N}$ for $R_{Cy}$ and $86.1\text{N}$ for $S{F}_A$ in Equation (XII).

    $\begin{array}{c}S{F}_{Cy}=86.1\text{N}-86.1\text{N}\\ =0\text{N}\end{array}$

Substitute $-157.9\text{N}$ for $S{F}_{Oy}$ and $300\text{mm}$ for $l_{OA}$ in Equation (XIV).

    $\begin{array}{c}{M}_{Ay}=-157.9\text{N}\times \left(300\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =-47.37\text{N}\cdot \text{m}\end{array}$

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

Figure (1)

Substitute $107.2\text{N}$ for $R_{Oz}$ in Equation (XV).

    $S{F}_{Oz}=-107.2\text{N}$

Substitute $-107.2\text{N}$ for $S{F}_{Oz}$, $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$ and $45°$ for $\theta$ in Equation (XVI).

    $\begin{array}{c}S{F}_{Az}=-107.2\text{N}+\left(300\text{N}+45\text{N}\right)\cos 45°\\ =136.8\text{N}\end{array}$

Substitute $136.8\text{N}$ for $S{F}_{Az}$, $250\text{N}$ for $T_1$ and $37.5\text{N}$ for $T_2$ in Equation (XVII).

    $\begin{array}{c}S{F}_{Bz}=136.8\text{N}-\left(250\text{N}+37.5\text{N}\right)\\ =-150.7\text{N}\end{array}$

Substitute $-150.7$ for $S{F}_{Bz}$ and $-150.7$ for $R_{Cz}$ in Equation (XVIII).

    $\begin{array}{c}S{F}_{Cz}=-150.7\text{N}+150.7\text{N}\\ =0\text{N}\end{array}$

Substitute $-107.2\text{N}$ for $S{F}_{Oz}$ and $300\text{mm}$ for $l_{OA}$ in Equation (XX).

    $\begin{array}{c}{M}_{Az}=-107.2\text{N}\times \left(300\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =-32.16\text{N}\cdot \text{m}\end{array}$

Substitute $-150.7\text{N}$ for $R_{Cz}$ and $150\text{mm}$ for $l_{BC}$ in Equation (XXI).

    $\begin{array}{c}{M}_{Bz}=-\left(-150.7\text{N}\right)\times \left(150\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =22.56\text{N}\cdot \text{m}\end{array}$

Thus, the shear force diagram and bending moment diagram for the shaft in $y$ direction is as follows.

Figure (2)

Substitute $-47.37\text{N}\cdot \text{m}$ for $M_{Ay}$ and $-32.16\text{N}\cdot \text{m}$ for $M_{Az}$ in Equation (XXII).

    $\begin{array}{c}{M}_A=\sqrt{{\left(-47.37\text{N}\cdot \text{m}\right)}^2+{\left(-32.16\text{N}\cdot \text{m}\right)}^2}\\ =57.26\text{N}\cdot \text{m}\end{array}$

Substitute $300\text{N}$ for $T_{A1}$, $45\text{N}$ for $T_{A2}$ and $250\text{mm}$ for $d_A$ in Equation (XXIII).

    $\begin{array}{c}T=\left(300\text{N}-45\text{N}\right)\times \frac{\left(250\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{2}\\ =31.88\text{N}\cdot \text{m}\end{array}$

Convert diameter of shaft into $\text{m}$.

    $\begin{array}{c}d=20\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\\ =0.02\text{m}\end{array}$

Substitute $57.26\text{N}\cdot \text{m}$ for $M_A$ and $0.02\text{m}$ for $d$ in Equation (XXV).

    $\begin{array}{c}\sigma =\frac{32\times 57.26\text{N}\cdot \text{m}}{\pi {\left(0.02\text{m}\right)}^3}\\ =72.9\text{MPa}\end{array}$

Substitute $31.88\text{N}\cdot \text{m}$ for $T$ and $0.02\text{m}$ for $d$ in Equation (XXVI).

    $\begin{array}{c}\tau =\frac{16\times 31.88\text{N}}{\pi {\left(0.02\text{m}\right)}^3}\\ =20.3\text{MPa}\end{array}$

Substitute $72.9\text{MPa}$ for $\sigma$ and $20.3\text{MPa}$ for $\tau$ in Equation (XXVI).

    $\begin{array}{c}{\sigma }_1=\frac{72.9\text{MPa}}{2}+\sqrt{{\left(\frac{72.9\text{MPa}}{2}\right)}^2+{\left(20.3\text{MPa}\right)}^2}\\ =36.45\text{MPa}+41.7\text{MPa}\\ =78.15\text{MPa}\\ \approx 78.2\text{MPa}\end{array}$

Substitute $72.9\text{MPa}$ for $\sigma$ and $20.3\text{MPa}$ for $\tau$ in Equation (XXVII).

    $\begin{array}{c}{\sigma }_2=\frac{72.9\text{MPa}}{2}-\sqrt{{\left(\frac{72.9\text{MPa}}{2}\right)}^2+{\left(20.3\text{MPa}\right)}^2}\\ =36.45\text{MPa}-41.7\text{MPa}\\ =-5.25\text{MPa}\end{array}$

Substitute $72.9\text{MPa}$ for $\sigma$ and $20.3\text{MPa}$ for $\tau$ in Equation (XXVIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{72.9}{2}\right)}^2+{20.3}^2}\\ =41.7\text{MPa}\end{array}$

Refer to the Table A-20 “Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels” and obtain $S_y=370\text{MPa}$ for $1018$ CD steel.

Substitute $78.2\text{MPa}$ for ${\sigma }_1$, $370\text{MPa}$ for $S_y$ and $\left(-5.27\text{MPa}\right)$ for ${\sigma }_2$ in Equation (XXIX).

    $\begin{array}{c}n=\frac{370\text{MPa}}{78.2\text{MPa}-\left(-5.25\text{MPa}\right)}\\ \approx 4.43\end{array}$

Thus, the factor of safety for yielding from maximum-shear-stress theory is $4.43$.

Substitute $78.2\text{MPa}$ for ${\sigma }_1$, $370\text{MPa}$ for $S_y$ and $\left(-5.27\text{MPa}\right)$ for ${\sigma }_2$ in Equation (XXXI).

    $\begin{array}{c}n=\frac{370\text{MPa}}{{\left[{\left(78.2\text{MPa}\right)}^2-\left(78.2\text{MPa}\right)\cdot \left(-5.27\text{MPa}\right)+{\left(-5.27\text{MPa}\right)}^2\right]}^{\frac{1}{2}}}\\ =4.569\\ \approx 4.57\end{array}$

Thus, the factor of safety for yielding from distortion-energy theory is $4.57$.