Chapter 6, Problem 40P

For the problem specified in the table, build upon the results of the original problem to determine the minimum factor of safety for fatigue based on infinite life, using the modified Goodman criterion. The shaft rotates at a constant speed, has a constant diameter, and is made from cold-drawn AISI 1018 steel.

To determine

The minimum factor of safety for fatigue on infinite life, using the modified Goodman criterion.

Answer to Problem 40P

The minimum factor of safety for fatigue on infinite life is $2.04$.

Explanation of Solution

Write the relationship between tensions 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 expression to balance the tension on the counter shaft.

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

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 $d_B$.

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

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

Write the expression for 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 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 expression for the magnitude of bearing reaction force at $O$ in $z\text{-}$ 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)

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

Write the expression for 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 expression for 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 $z\text{-}$ direction is $R_{Oz}$.

Write the expression for the bearing reaction force at $B$.

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

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

Write the expression for the bearing reaction force at $O$.

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

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

Write the expression for the moment at $A$ in $y\text{-}$ direction.

    $M_{Ay}=R_{Oy}\times l_{OA}$                                                                                   (X)

Here, the moment at $A$ is $M_A$ in $y\text{-}$ direction, $l_{oA}$ is distance of maximum bending moment in $y-$ direction.

Write the expression for the moment at $A$ in $z\text{-}$ direction.

    $M_{Az}=-R_{Oz}\times l_{OA}$                                                                               (XI)

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

It is clear from the bending moment diagram, that the critical location is at $A$. The bending moment will be maximum at the critical point.

Write the expression for the net moment at point $A$.

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

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

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

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

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

Write the expression for the bending stress.

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

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

Write the expression for the shear stress.

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

Here, the shear stress is $\tau$.

Write the expression for von Misses stress for alternating

    ${{\sigma }^{\prime }}_a=\sqrt{{\sigma }_a{}^2+3{\tau }_a{}^2}$                                                                            (XVI)

Here, alternating stress due to completely reversed is ${\sigma }_a$ and ${\tau }_s$ for alternating torsional stress.

Write the expression for von Misses stress for mid-range.

    ${{\sigma }^{\prime }}_m=\sqrt{{\sigma }_m{}^2+3{\tau }_m{}^2}$                                                                        (XVII)

Here, mean stress due to completely reversed is ${\sigma }_m$ and ${\tau }_m$ for mean torsional stress.

Write the expression for von Mises for maximum stress.

    ${{\sigma }^{\prime }}_{\max }=\sqrt{{\sigma }^2+3{\tau }^2}$                                                                         (XVIII)

Write the expression for yielding by using distortion energy theory.

$n_y=\frac{S_y}{{{\sigma }^{\prime }}_{\max }}$                                                                                   (XIX)

Here the $n_y$ is factor of safety according to distortion energy theory, and $S_y$ stands for endurance limit

Write the expression for endurance limit of rotary test specimen.

${S^{\prime }}_e=0.5\left(S_{ut}\right)$                                                                                   (XX)

Write the expression for surface condition modification factor.

    $K_a=a{S}_{ut}{}^b$                                                                             (XXI)

Write the expression for size modification factor.

    $K_b=1.24d^{-0.107}$                                                                     (XXII)

Write the expression for modified endurance limit.

    $S_e=K_a{K}_b{S}_e{}^{\prime }$                                                                        (XXIII)

Write the expression to find out factor of safety by using modified Goodman.

    $\frac{1}{n_f}=\frac{{{\sigma }^{\prime }}_a}{S_e}+\frac{{{\sigma }^{\prime }}_m}{S_{ut}}$                                                                      (XXIV)

Here modified endurance limit is $S_e$ and the ultimate strength of material is $S_{ut}$.

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).

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

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 $\theta$ in Equation (IV).

    $\begin{array}{c}{R}_{Cz}=\frac{\left(\left(300\text{N}+45\text{N}\right)\sin 45°\times 300\text{mm}\right)-\left(\left(250\text{N}+37.5\text{N}\right)\left(\begin{array}{l}300\text{mm}\\ +400\text{mm}\end{array}\right)\right)}{\left(300\text{mm}+400\text{mm}+150\text{mm}\right)}\\ =\frac{\left(\left(300\text{N}+45\text{N}\right)\sin 45°\times 300\text{mm}\right)-\left(\left(250\text{N}+37.5\text{N}\right)\left(700\text{mm}\right)\right)}{\left(850\text{mm}\right)}\\ \approx -228\text{N}\end{array}$

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 $-228\text{N}$ for $R_{Cz}$ in Equation (V).

    $\begin{array}{c}{R}_{Oz}=\left(300\text{N}+45\text{N}\right)\cos 45°-\left(250\text{N}+37.5\text{N}\right)+228\text{N}\\ \text{=}\left(345\text{N}\right)\cos 45°-\left(287.5\text{N}\right)+228\text{N}\\ \approx 184.5\text{N}\end{array}$

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.)

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

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)

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

Substitute $-228\text{N}$ for $R_{Cz}$ and $-86.1\text{N}$ for $R_{Cy}$ in Equation (VIII).

    $\begin{array}{c}{R}_C=\sqrt{{\left(-228\text{N}\right)}^2+{\left(-86.1\text{N}\right)}^2}\\ =243.7\text{N}\end{array}$

Substitute $-157.9\text{N}$ for $R_{Oy}$ and $184.5\text{N}$ for $R_{Oz}$ in Equation (IX).

    $\begin{array}{c}{R}_O=\sqrt{{\left(-157.9\text{N}\right)}^2+{\left(184.5\text{N}\right)}^2}\\ \approx 243\text{N}\end{array}$

Substitute $-157.9\text{N}$ for $R_{Oy}$ and $300\text{mm}$ for $l_{OA}$ in Equation (X).

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

Substitute $-107.2\text{N}$ for $R_{Oz}$ and $300\text{mm}$ for $l_{OA}$ in Equation (XI).

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

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

    $\begin{array}{c}{M}_A=\sqrt{{\left(-47.37\text{N}\right)}^2+{\left(-32.16\text{N}\right)}^2}\\ =57.255\text{N}\cdot \text{m}\\ \approx 57.26\text{N}\cdot \text{m}\end{array}$

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

    $\begin{array}{c}T=\left(300\text{N}-45\text{N}\right)\times \frac{\left(250\text{mm}\right)}{2}\\ =31875\text{N}\cdot \text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ \approx 31.88\text{N}\cdot \text{m}\end{array}$

Convert diameter of shaft from millimeter to meter.

    $\begin{array}{c}d=20\text{mm}\\ =\left(20\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =0.02\text{m}\end{array}$

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

    $\begin{array}{c}{\sigma }_a=\frac{32\times 57.26}{\pi {\left(0.02\right)}^3}\\ =\left(72905696.3\text{Pa}\right)\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =72.9056963\text{MPa}\\ \approx 72.9\text{MPa}\end{array}$

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

    $\begin{array}{c}{\tau }_a=\frac{16\times 31.88}{\pi {\left(0.02\right)}^3}\\ ==\left(20295438.34\text{Pa}\right)\times \left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =20.29543834\text{MPa}\\ \approx 20.3\text{MPa}\end{array}$

Substitute $72.9\text{MPa}$ for ${\sigma }_a$ and $0\text{MPa}$ for ${\tau }_a$ in Equation (XVI).

    $\begin{array}{c}{{\sigma }^{\prime }}_a=\sqrt{{\left(72.9\text{MPa}\right)}^2+3{(0\text{MPa})}^2}\\ =72.9\text{MPa}\end{array}$

Substitute $0\text{MPa}$ for ${\sigma }_m$ and $20.3\text{MPa}$ for ${\tau }_m$ in Equation (XVII).

    $\begin{array}{c}{{\sigma }^{\prime }}_m=\sqrt{{\left(0\text{MPa}\right)}^2+3{\left(20.3\text{MPa}\right)}^2}\\ =35.2\text{MPa}\end{array}$

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

    $\begin{array}{c}{\sigma }_{\max }{}^{\prime }=\sqrt{{\left(72.9\text{MPa}\right)}^2+3{(20.3\text{MPa})}^2}\\ =80.9\text{MPa}\end{array}$

Refer to Table A-20 “Deterministic ASTM tensile and yield strengths for some hot-rolled (HR) and cold-drawn (CD) steels” to obtain the yield strength for AISI 1018 CD steel as $370\text{MPa}$.

Substitute $370\text{MPa}$ for $S_y$ and $80.9\text{MPa}$ for ${{\sigma }^{\prime }}_{\max }$ in Equation (XIX).

    $\begin{array}{c}{n}_y=\frac{370\text{MPa}}{80.9\text{MPa}}\\ =4.57\end{array}$

Refer to Table A-20 “Deterministic ASTM tensile and yield strengths for some hot-rolled (HR) and cold-drawn (CD) steels” to obtain the ultimate strength for AISI 1018 CD steel as $440\text{MPa}$.

Substitute $440\text{MPa}$ for $S_{ut}$ in Equation (XX).

    $\begin{array}{c}{S^{\prime }}_e=0.5\left(440\text{MPa}\right)\\ =220\text{MPa}\end{array}$

Substitute $4.51$ for $a$ and $-0.265$ for $b$ in Equation (XXI).

    $\begin{array}{c}{K}_a=4.51{\left(440\right)}^{-0.265}\\ =0.90\end{array}$

Substitute $20\text{mm}$ for $d$ in Equation (XXII).

    $\begin{array}{c}{K}_b=1.24{(20\text{mm})}^{-0.107}\\ =0.90\end{array}$

Substitute $0.90$ for $K_a$, $0.90$ for $K_b$ and $220$ for ${S^{\prime }}_e$ in Equation (XXIII).

    $\begin{array}{c}{S}_e=0.90\left(0.90\right)\left(220\text{MPa}\right)\\ =178.2\text{MPa}\end{array}$

Substitute $72.9$ for ${{\sigma }^{\prime }}_a$, $35.2$ for ${{\sigma }^{\prime }}_m$, $178.2$ for $S_e$, $440$ for $S_{ut}$ in Equation (XXIV).

    $\begin{array}{l}\frac{1}{n_f}=\left(\frac{72.9\text{MPa}}{178.2\text{MPa}}\right)+\left(\frac{35.2\text{MPa}}{440\text{MPa}}\right)\\ \frac{1}{n_f}=0.4090+0.08\\ n_f=\frac{1}{0.489}\\ n_f=2.04\end{array}$

Thus minimum factor of safety by using modified Goodman criterion is $2.04$.