Chapter 6, Problem 39P

6–37* to

6–46* 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.

Problem Number	Original Problem, Page Number
6–37*	3–68, 151
6–38*	3–69, 151
6–39*	3–70, 151
6–40*	3–71, 151
6–41*	3–72, 152
6–42*	3–73, 152
6–43*	3–74, 152
6–44*	3–76, 153
6–45*	3–77, 153
6–46*	3–79, 153

Problem 3–70*

Dimensions in inches.

To determine

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

Answer to Problem 39P

The minimum factor of safety for fatigue on infinite life is $1.01$.

Explanation of Solution

The Free body diagram of pulley $A$ is shown below.

Figure-(1)

The free body diagram of pulley $B$ is shown below.

Figure-(2)

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 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 equation to balance the tension on the counter shaft.

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

Calculate the tension on the loose side.

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

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

    $\begin{array}{l}{\displaystyle \sum M_{Oy}}=0\\ -\left(T_1+T_2\right)\left(l_{OA}+l_{AB}\right)-R_{Cz}\left(l_{OA}+l_{AB}+l_{BC}\right)=0\\ R_{Cz}=\frac{-\left(T_1+T_2\right)\left(l_{OA}+l_{AB}\right)}{\left(l_{OA}+l_{AB}+l_{BC}\right)}\end{array}$           (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 $l_{AB}$ and distance between $B$ and $C$ is $l_{BC}$.

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

    $\begin{array}{l}{\displaystyle \sum F_z}=0\\ R_{Oz}+\left(T_1+T_2\right)+R_{Cz}=0\\ R_{Oz}=-\left(T_2+T_1\right)-R_{Cz}\end{array}$         (V)

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

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

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

Write the magnitude of bearing force at $O$ in $y\text{-}$ direction.

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

Here, the magnitude of bearing reaction force at $O$ in $z\text{-}$ 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 $R_O$.

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

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

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

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

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

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

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

    $M_{By}=R_{Cy}\times l_{BC}$                                                                (XV)

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

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

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

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

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

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

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

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

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

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

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

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 $B$ is $M_B$.

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

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

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

Here, the shear stress is $\tau$.

Write the expression for von Mises stress for alternating

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

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

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

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

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.

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

Write the expression for yielding by using distortion energy theory.

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

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

Write the expression for endurance limit of rotary test specimen.

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

Write the expression for surface condition modification factor.

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

Write the expression for size modification factor.

    $K_b=0.879d^{-0.107}$                                                 (XXXIII)

Write the expression for modified endurance limit.

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

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

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

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

Conclusion:

Substitute $300\text{lbf}$ for $T_{A1}$, $50\text{lbf}$ for $T_{A2}$, $8\text{in}$ for $d_A$ and $6\text{in}$ for $d_B$  in Equation (III).

    $\begin{array}{c}{T}_1=\frac{1}{0.85}\left(\left(300\text{lbf}-50\text{lbf}\right)\times \frac{8\text{in}}{6\text{in}}\right)\\ =392.156\text{lbf}\\ \approx 392.16\text{lbf}\end{array}$

Substitute $392.16\text{lbf}$ for $T_1$ in Equation (I)

    $\begin{array}{c}{T}_2=0.15\times \left(392.16\text{lbf}\right)\\ =58.824\text{lbf}\\ \approx 58.82\text{lbf}\end{array}$

Substitute $392.16\text{lbf}$ for $T_1$, $58.82\text{lbf}$ for $T_2$, $8\text{in}$ for $l_{OA}$, $8\text{in}$ for $l_{AB}$ and $6\text{in}$ for $l_{BC}$ Equation (IV).

    $\begin{array}{c}{R}_{Cz}=\frac{\left[\left(-(392.16\text{lbf}+58.82\text{lbf})\left(8\text{in}+8\text{in}\right)\right)\right]}{\left(8\text{in}+8\text{in}+6\text{in}\right)}\\ =-327.985\text{lbf}\\ \approx -327.99\text{lbf}\end{array}$

Substitute $392.16\text{lbf}$ for $T_1$, $58.82\text{lbf}$ for $T_2$ and $-327.99\text{lbf}$ for $R_{Cz}$ in Equation (V).

    $\begin{array}{c}{R}_{Oz}=-\left(58.82\text{lbf}+392.16\text{lbf}\right)+327.99\text{lbf}\\ =-\left(450.98\text{lbf}\right)+\left(327.99\text{lbf}\right)\\ =-122.99\text{lbf}\end{array}$

Substitute $300\text{lbf}$ for $T_{A1}$, $50\text{lbf}$ for $T_{A2}$, $8\text{in}$ for $l_{OA}$, $8\text{in}$ for $l_{AB}$ and $6\text{in}$ for $l_{BC}$ in Equation (VI.)

    $\begin{array}{c}{R}_{Cy}=-\frac{\left(300\text{lbf}+50\text{lbf}\right)8\text{in}}{\left(8\text{in}+8\text{in}+6\text{in}\right)}\\ =\frac{2800\text{lbf}\cdot \text{in}}{22\text{in}}\\ =-127.27\text{lbf}\end{array}$

Substitute $300\text{lbf}$ for $T_{A1}$, $50\text{lbf}$ for $T_{A2}$ and $-127.27\text{lbf}$ for $R_{Cy}$ in Equation (VII)

    $\begin{array}{c}{R}_{Oy}=-\left(T_{A1}+T_{A2}\right)-R_{Cy}\\ =-\left(300\text{lbf}+50\text{lbf}\right)+127.27\text{lbf}\\ =-222.73\text{lbf}\end{array}$

Substitute $-327.99\text{N}$ for $R_{Cz}$ and $-127.27\text{lbf}$ for $R_{Cy}$ in Equation (VIII)

    $\begin{array}{c}{R}_C=\sqrt{{\left(327.99\text{lbf}\right)}^2+{\left(127.27\text{lbf}\right)}^2}\\ =\sqrt{123775.093{\text{lbf}}^2}\\ =351.8\text{lbf}\end{array}$

Substitute $-222.73\text{lbf}$ for $R_{Oy}$ and $-122.99\text{lbf}$ for $R_{Oz}$ in Equation (IX).

    $\begin{array}{c}{R}_O=\sqrt{{\left(222.73\text{lbf}\right)}^2+{\left(122.99\text{lbf}\right)}^2}\\ =\sqrt{64735.193{\text{lbf}}^2}\\ =254.43\text{lbf}\end{array}$

Substitute $-222.73\text{lbf}$ for $R_{Oy}$ in Equation (X).

    $S{F}_{Oy}=-222.73\text{lbf}$

Substitute $-222.73\text{lbf}$ for $S{F}_{Oy}$ and $8\text{in}$ for $l_{OA}$ in Equation (XIV).

    $\begin{array}{c}{M}_{Ay}=\left(-222.73\text{lbf}\right)\left(8\text{in}\right)\\ =-1781.84\text{lbf}\cdot \text{in}\end{array}$

Substitute $-127.27\text{lbf}$ for $R_O$ and $6\text{in}$ for $l_{BC}$ in Equation (XV).

    $\begin{array}{c}{M}_{By}=\left(-127.27\text{lbf}\right)\left(6\text{in}\right)\\ =-763.65\text{lbf}\cdot \text{in}\end{array}$

Substitute $-122.99\text{lbf}$ for $R_{Oz}$ in Equation (XVI).

    $S{F}_{Oz}=122.99\text{lbf}$

Substitute $122.99\text{lbf}$ for $S{F}_{Oy}$ and $8\text{in}$ for $l_{OA}$ in Equation (XX).

    $\begin{array}{c}{M}_{Az}=\left(122.99\text{lbf}\right)\left(8\text{in}\right)\\ =983.92\text{lbf}\cdot \text{in}\end{array}$

Substitute $-327.99\text{lbf}$ for $R_O$ and $6\text{in}$ for $l_{BC}$ in Equation (XXI).

    $\begin{array}{c}{M}_{Bz}=-\left(-327.99\text{lbf}\right)\times 6\text{in}\\ =1967.84\text{lbf}\cdot \text{in}\end{array}$

Substitute $983.92\text{lbf}\cdot \text{in}$ for $M_{Az}$ and $-1781.84\text{lbf}\cdot \text{in}$ for $M_{Ay}$ in Equation (XXII).

    $\begin{array}{c}{M}_A=\sqrt{{\left(983.92\text{lbf}\cdot \text{in}\right)}^2+{\left(-1781.84\text{lbf}\cdot \text{in}\right)}^2}\\ =2035.4\text{lbf}\cdot \text{in}\\ \approx 2035\text{lbf}\cdot \text{in}\end{array}$

Substitute $1967.84\text{lbf}\cdot \text{in}$ for $M_{Bz}$ and $-763.65\text{lbf}\cdot \text{in}$ for $M_{By}$ in Equation (XXIII)

    $\begin{array}{c}{M}_B=\sqrt{{\left(1967.84\text{lbf}\cdot \text{in}\right)}^2+{\left(-763.65\text{lbf}\cdot \text{in}\right)}^2}\\ =2111.08\text{lbf}\cdot \text{in}\\ \approx 2111\text{lbf}\cdot \text{in}\end{array}$

Since $M_B>M_A$, so the point of maximum bending stress is critically located at $B$.

Substitute $300\text{lbf}$ for $T_{A1}$, $50\text{lbf}$ for $T_{A2}$ and $8\text{in}$ for $d_A$ in Equation (XXIV).

    $\begin{array}{c}T=\left(300\text{lbf}-50\text{lbf}\right)\frac{8\text{in}}{2}\\ =\left(250\text{lbf}\right)\left(4\text{in}\right)\\ =1000\text{lbf}\cdot \text{in}\end{array}$

Substitute $2111\text{lbf}\cdot \text{in}$ for $M_B$ and $1\text{in}$ for $d$ in Equation (XXV).

    $\begin{array}{c}\sigma =\frac{32\times 2111\text{lbf}\cdot \text{in}}{\pi {\left(1\text{in}\right)}^3}\\ =\left(21502.47\text{psi}\right)\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ =21.502\text{kpsi}\\ \approx 21.5\text{kpsi}\end{array}$

Substitute $1000\text{lbf}\cdot \text{in}$ for $T$ and $1\text{in}$ for $d$ in Equation (XXVI).

    $\begin{array}{c}\tau =\frac{16\times 1000\text{lbf}}{\pi {\left(1\text{in}\right)}^3}\\ =\left(5092.96\text{psi}\right)\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ \approx 5.09\text{kpsi}\end{array}$

Substitute $21.5\text{kpsi}$ for ${\sigma }_a$ and $0\text{kpsi}$ for ${\tau }_a$ in Equation (XXVII).

    $\begin{array}{c}{\sigma }_a=\sqrt{{\left(21.5\text{kpsi}\right)}^2+3{(0\text{kpsi})}^2}\\ =\sqrt{462.5{\text{kpsi}}^2}\\ =21.5\text{kpsi}\end{array}$

Substitute $0\text{kpsi}$ for ${\sigma }_m$ and $5.09\text{kpsi}$ for ${\tau }_m$ in Equation (XXVIII).

    $\begin{array}{c}{{\sigma }^{\prime }}_m=\sqrt{{\left(0\text{kpsi}\right)}^2+3{\left(5.09\text{kpsi}\right)}^2}\\ =\sqrt{77.7243{\text{kpsi}}^2}\\ =8.82\text{kpsi}\end{array}$

Substitute $21.5\text{kpsi}$ for $\sigma$ and $5.09\text{kpsi}$ for $\tau$.in equation (XXIX).

    $\begin{array}{c}{{\sigma }^{\prime }}_{\max }=\sqrt{{\left(21.5\text{kpsi}\right)}^2+3{(5.09\text{kpsi})}^2}\\ =\sqrt{539.9743{\text{kpsi}}^2}\\ =23.24\text{kpsi}\end{array}$

Substitute $54\text{kpsi}$ for $S_y$ and $23.24\text{kpsi}$ for ${{\sigma }^{\prime }}_{\max }$ in equation (XXX).

    $\begin{array}{c}{n}_y=\frac{54\text{kpsi}}{23.24\text{kpsi}}\\ =2.32\end{array}$

Substitute $S_{ut}$ from Table A-20.in equation (XXXI)

    $\begin{array}{c}{S^{\prime }}_e=0.5\left(64\text{kpsi}\right)\\ =32\text{kpsi}\end{array}$

Substitute $2.70$ for $a$ and $-0.265$ for $b$ in equation (XXXII)

    $\begin{array}{c}{K}_a=2.70{\left(64\right)}^{-0.265}\\ =0.90\end{array}$

Substitute $1$ for $d$ in equation (XXXIII)

    $\begin{array}{c}{K}_b=0.879{(1)}^{-0.107}\\ =0.88\end{array}$

Substitute $0.90$ for $K_a$,$0.88$ for $K_b$ and $32$ for ${S^{\prime }}_e$ in equation (XXXIV)

    $\begin{array}{c}{S}_e=0.90\left(0.88\right)\left(32\right)\text{kpsi}\\ =25.3\text{kpsi}\end{array}$

Substitute $21.5$ for ${{\sigma }^{\prime }}_a$,$8.82$ for ${{\sigma }^{\prime }}_m$,$25.3$ for $S_e$, $32$ for $S_{ut}$ in equation (XXXV)

    $\begin{array}{l}\frac{1}{n_f}=\frac{21.5}{25.3}+\frac{8.82}{64}\\ \frac{1}{n_f}=0.9876\\ n_f\approx 1.01\end{array}$

Thus the minimum factor of safety by using modified Goodman criterion is $1.01$.