Chapter 7, Problem 1P

A shaft is loaded in bending and torsion such that M_a = 70 N · m, T_a = 45 N · m, M_m = 55 N · m, and T_m = 35 N · m. For the shaft, S_u = 700 MPa and S_y = 560 MPa, and a fully corrected endurance limit of S_e = 210 MPa is assumed. Let K_f = 2.2 and K_fs = 1.8. With a design factor of 2.0 determine the minimum acceptable diameter of the shaft using the

(a) DE-Gerber criterion.

(b) DE-ASME Elliptic criterion.

(c) DE-Soderberg criterion.

(d) DE-Goodman criterion.

(a) Discuss and compare the results.

To determine

The diameter of shaft of shaft using DE-Gerber criterion.

Answer to Problem 1P

The diameter of shaft is $25.85\text{mm}$.

Explanation of Solution

Write expression for parameter.

    $A=\sqrt{4{\left(K_f{M}_a\right)}^2+3{\left(k_{fs}{T}_a\right)}^2}$ (I)

Here, fatigue stress concentration factor for bending is $K_f$, fatigue stress concentration factor for torsion is $K_{fs}$, alternative bending moment is $M_a$ and alternative torque is $T_a$.

Write expression for parameter.

    $B=\sqrt{4{\left(K_f{M}_m\right)}^2+3{\left(k_{fs}{T}_m\right)}^2}$ (II)

Here, mid range bending moment is $M_m$ and mid-range torque is $T_m$.

Write expression for diameter by applying DE-Gerber criterion.

    $d={\left[\frac{8nA}{\pi S_e}\left(1+{\left\{1+{\left(\frac{2B{S}_e}{A{S}_{ut}}\right)}^2\right\}}^{\frac{1}{2}}\right)\right]}^{\frac{1}{3}}$ (III)

Here, design factor is $n$, endurance limit is $S_e$ and ultimate tensile stress is $S_{ut}$.

Conclusion:

Substitute $2.2$ for $K_f$, $1.8$ for $K_{fs}$, $70\text{N}\cdot \text{m}$ for $M_a$ and $45\text{N}\cdot \text{m}$ for $T_a$ in Equation (I).

    $\begin{array}{c}A=\sqrt{4{\left(2.2\times 70\text{N}\cdot \text{m}\right)}^2+3{\left(1.8\times 45\text{N}\cdot \text{m}\right)}^2}\\ =\sqrt{4{\left(154\text{N}\cdot \text{m}\right)}^2+3{\left(81\text{N}\cdot \text{m}\right)}^2}\\ =\sqrt{114547}\text{N}\cdot \text{m}\\ =338.4\text{N}\cdot \text{m}\end{array}$

Substitute $2.2$ for $K_f$, $1.8$ for $K_{fs}$, $55\text{N}\cdot \text{m}$ for $M_m$ and $35\text{N}\cdot \text{m}$ for $T_m$ in Equation (II).

    $\begin{array}{c}A=\sqrt{4{\left(2.2\times 55\text{N}\cdot \text{m}\right)}^2+3{\left(1.8\times 35\text{N}\cdot \text{m}\right)}^2}\\ =\sqrt{4{\left(121\text{N}\cdot \text{m}\right)}^2+3{\left(63\text{N}\cdot \text{m}\right)}^2}\\ =\sqrt{70471}\text{N}\cdot \text{m}\\ =265.5\text{N}\cdot \text{m}\end{array}$

Substitute $338.4\text{N}\cdot \text{m}$ for $A$, $265.5\text{N}\cdot \text{m}$ for $B$, $210\text{MPa}$ for $S_e$, $700\text{MPa}$ for $S_{ut}$, $2$ for $n$ in Equation (III).

    $\begin{array}{c}d={\left[\frac{8\left(2\right)\left(338.4\text{N}\cdot \text{m}\right)}{\pi \left(210\text{MPa}\right)}\left(1+{\left\{1+{\left(\frac{2\left(265.5\text{N}\cdot \text{m}\right)\left(210\text{MPa}\right)}{\left(338.4\text{N}\cdot \text{m}\right)\left(700\text{MPa}\right)}\right)}^2\right\}}^{\frac{1}{2}}\right)\right]}^{\frac{1}{3}}\\ ={\left[\frac{8\left(2\right)\left(338.4\text{N}\cdot \text{m}\right)}{\pi \left(210\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\left(1+{\left\{1+{\left(\frac{2\left(265.5\text{N}\cdot \text{m}\right)\left(210\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}{\left(338.4\text{N}\cdot \text{m}\right)\left(700\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\right)}^2\right\}}^{\frac{1}{2}}\right)\right]}^{\frac{1}{3}}\\ ={\left[\left(8.2069\times {10}^{-6}{\text{m}}^3\right)\left(1.28164\right)\right]}^{\frac{1}{3}}\left|\frac{1000\text{mm}}{1\text{m}}\right|\\ =25.85\text{mm}\end{array}$

Thus, the diameter of shaft is $25.85\text{mm}$

To determine

The diameter of shaft using DE-ASME Elliptic criterion.

Answer to Problem 1P

The diameter of shaft is $25.77\text{mm}$.

Explanation of Solution

Write expression for diameter by applying DE-Elliptical criterion.

    $d={\left(\frac{16n}{\pi }\left[\sqrt{\left({\left(\frac{A}{S_e}\right)}^2+{\left(\frac{B}{S_y}\right)}^2\right)}\right]\right)}^{\frac{1}{3}}$ (III)

Here, design factor is $n$, endurance limit is $S_e$ and yield stress is $S_y$.

Conclusion:

Substitute $338.4\text{N}\cdot \text{m}$ for $A$, $265.5\text{N}\cdot \text{m}$ for $B$, $210\text{MPa}$ for $S_e$, $560\text{MPa}$ for $S_y$, $2$ for $n$ in Equation (III).

    $\begin{array}{c}d={\left(\frac{16\left(2\right)}{\pi }\left[\sqrt{\left({\left\{\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\text{MPa}\right)}\right\}}^2+{\left\{\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(560\text{MPa}\right)}\right\}}^2\right)}\right]\right)}^{\frac{1}{3}}\\ ={\left(\frac{32}{\pi }\left[\sqrt{\left({\left\{\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\right\}}^2+{\left\{\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(560\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\right\}}^2\right)}\right]\right)}^{\frac{1}{3}}\\ =\left(2.16770\right){\left[\sqrt{\left(2.5967\times {10}^{-12}{\text{m}}^3+0.2247\times {10}^{-12}{\text{m}}^3\right)}\right]}^{\frac{1}{3}}\\ =25.77\text{mm}\end{array}$

Thus, the diameter of shaft is $25.77\text{mm}$.

To determine

The diameter of shaft using DE-Soderberg criterion.

Answer to Problem 1P

The diameter of shaft is $27.7\text{m}$.

Explanation of Solution

Write expression for diameter by applying DE-Soderberg criterion.

    $d={\left(\frac{16n}{\pi }\left[\left(\frac{A}{S_e}+\frac{B}{S_y}\right)\right]\right)}^{\frac{1}{3}}$ (III)

Here, design factor is $n$, endurance limit is $S_e$ and yield stress is $S_y$.

Conclusion:

Substitute $338.4\text{N}\cdot \text{m}$ for $A$, $265.5\text{N}\cdot \text{m}$ for $B$, $210\text{MPa}$ for $S_e$, $560\text{MPa}$ for $S_y$ and $2$ for $n$ in Equation (III).

    $\begin{array}{c}d={\left(\frac{16\left(2\right)}{\pi }\left[\left(\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\text{MPa}\right)}+\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(560\text{MPa}\right)}\right)\right]\right)}^{\frac{1}{3}}\\ ={\left(\frac{32}{\pi }\left[\left(\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}+\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(560\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\right)\right]\right)}^{\frac{1}{3}}\\ ={\left(\frac{32}{\pi }\left[\left(1.6114\times {10}^{-6}\text{m}+0.4741\times {10}^{-6}\text{m}\right)\right]\right)}^{\frac{1}{3}}\\ =27.7\text{mm}\end{array}$

Thus, the diameter of shaft is $27.7\text{m}$.

To determine

The diameter of shaft using DE-Goodman criterion.

Answer to Problem 1P

The diameter of shaft is $20.27\text{mm}$.

Explanation of Solution

Write expression for diameter by applying DE-Goodman criterion.

    $d={\left(\frac{16n}{\pi }\left[\left(\frac{A}{S_e}+\frac{B}{S_{ut}}\right)\right]\right)}^{\frac{1}{3}}$ (III)

Conclusion:

Substitute $338.4\text{N}\cdot \text{m}$ for $A$, $265.5\text{N}\cdot \text{m}$ for $B$, $210\text{MPa}$ for $S_e$, $700\text{MPa}$ for $S_{ut}$, $2$ for $n$ in Equation (III).

    $\begin{array}{c}d={\left(\frac{16\left(2\right)}{\pi }\left[\left(\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\text{MPa}\right)}+\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(700\text{MPa}\right)}\right)\right]\right)}^{\frac{1}{3}}\\ ={\left(\frac{32}{\pi }\left[\left(\frac{\left(338.4\text{N}\cdot \text{m}\right)}{\left(210\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}+\frac{\left(265.5\text{N}\cdot \text{m}\right)}{\left(700\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}\right)\right]\right)}^{\frac{1}{3}}\\ ={\left(\frac{32}{\pi }\left[\left(1.6114\times {10}^{-6}\text{m}+0.37928\times {10}^{-6}\text{m}\right)\right]\right)}^{\frac{1}{3}}\\ =27.27\text{mm}\end{array}$

Thus, the diameter of shaft is $20.27\text{mm}$.

Comparison with diameter obtained from DE-Geber Criteria.

The diameters obtain from DE-Geber Criteria is $25.85\text{mm}$.

The following table shows the comparison of the diameters obtain from different criterion with respect to DE-Gerber criteria.

Criteria	Diameter	Percentage (Relative to DE-Geber Criteria)
DE-Elliptical criterion	$25.77\text{mm}$	$0.31\%$ (Lower)
DE-Soderberg criterion	$27.70\text{mm}$	$7.2\%$(Higher)
DE-Goodman criterion	$27.27\text{mm}$	$5.5\%$(Higher)