This set of recommended problems is designed to help you reinforce and apply the concepts covered in Module 2. The problem numbers correspond to those in Shigley’s Mechanical Engineering Design, 11th Edition.

For a review of static failure theories for ductile materials:

• 5-51: This problem builds on 3-80 by applying static failure theories of ductile materials to the critical element(s).
• 5-60: This problem builds on 3-91 by applying static failure theories of ductile materials to the critical element(s).

For practice with the stress-life method and S-N diagrams:

• 6-3: Solve for the expected life of a specimen given a stress amplitude.
• 6-5: Solve for the fatigue strength given a number of cycles to failure.

For practice with Marin factors:

• 6-10 (a) & (b): Determine the fatigue characteristics of a simply supported rotating shaft (see Table A-20 for material properties).

For practice predicting life of a part under completely reversing simple loading:

• 6-14 & 6-17: Follow the road map provided in your equation sheet for completely reversing simple loading.

For practice with Goodman criteria:

• 6-38: Build on the solution from problem 3-80 to perform a fatigue failure analysis.

For practice with combined loading modes:

• 6-47 & 6-54: Build on solutions from previous homework problems to perform fatigue failure analysis.
• 6-57: Analyze a clutch-testing machine (commonly used in automotive research) to determine if the machine has an adequate factor of safety as designed.

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Problem 5-51 Solution

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This question asks us to build on the Problem 3-80 Solution to determine the minimum factor of safety for yielding. The material is given as 1018 CD steel. (See Table A-20 of your textbook to get the yield strength for this material, which is given as 370 MPa.)

MSS: n = 1.28

DE: n = 1.32

You can also find a YouTube video below to guide you through the solution.

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Problem 5-60 Answer

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Note: To get full credit for questions like this on a quiz or test, you'll need to show all of your work. Here, I may only providing the final answer for you to check. If you have questions on how to get here, please feel free to ask me during office hours, or send me an email.

This question asks us to build on the Problem 3-91 Solution to determine the minimum factor of safety for yielding. The material is given as 1018 CD steel. (See Table A-20 of your textbook to get the yield strength for this material, which is given as 370 MPa.)

MSS: n = 2.79

DE: n = 2.89

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Problem 6-3 Answer

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N = 117,000 cycles

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Problem 6-5 Answer

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S_f = 806.69 MPa

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Problem 6-10 Answer

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n = 0.98

S_e = 192 MPa

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Problem 6-14 Answer

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Note: To get full credit for questions like this on a quiz or test, you'll need to show all of your work. Here, I may only providing the final answer for you to check. If you have questions on how to get here, please feel free to ask me during office hours, or send me an email.

F_a = 17,000 N

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Problem 6-17 Solution

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Solve for the reactions at A and B

$R_A=8896.443 \mathrm{N}$

$R_B=6672.333 \mathrm{N}$

Find the critical cross-section

Drawing our shear and moment diagrams, we see that the maximum bending moment is at the location where F1 is applied (we'll call this location A). Just to the right, a stress raiser exists at the shoulder (we'll call this location B). The bending moment at A is 1807.757 Nm. The bending moment at B is 1694.772 Nm. There is also a stress concentration at the other shoulder pictured, to the right of F2. For a complete and thorough analysis, you should check stresses at all three of these locations. Only the solution at location B (which turns out to be the most critical) is shown below.

Calculate the endurance limit

For AISI 1040 CD steel, we find the ultimate tensile strength in Table A-20 to be $S_{ut}=590$ MPa.

The uncorrected endurance limit is:

${S^{\text{'}}}_e=0.5S_{ut}=$ 295 MPa

The Marin factors are:

$k_a=3.04{\left(S_{ut}\right)}^{-0.217}$=0.76

$k_b={\left(d/7.62\right)}^{-0.107}$=0.82

$k_c=1$

$k_d=1$

$k_e=1$

So, the corrected endurance limit is $S_e=\left(0.76\right)\left(0.83\right)\left(1\right)\left(1\right)\left(1\right){S^{\text{'}}}_e$ = 186.1 MPa

Determine the fatigue stress-concentration factor(s)

$K_f=1+q\left(K_t-1\right)$

The notch sensitivity q can be found in Figure 6-26 using the given fillet radius r = 1.588 mm: q=0.75

$K_t$ can be found in Figure A-15-9, using r/d = 1.588/41.275 = 0.038, D/d = 47.625/41.275 = 1.15: $K_t$ = 2.0

$K_f$ = 1 + 0.75*(2.0-1) = 1.75

Apply Kf to the completely reversed stress

${\sigma }_{a0}=\frac{Mc}{I}=\frac{(1694.772 \mathrm{N}·\mathrm{m})(0.041275 \mathrm{m}/2)}{\pi {\left(0.041275 \mathrm{m}\right)}^4/64}=245.499 \mathrm{MPa}$

${\sigma }_a=K_f{\sigma }_{a0}=$ (1.75)(245.499 MPa) = 425.623 MPa

${\sigma }_m=0$

Find factor of safety for infinite life using Goodman criteria

$n_f={\left[\frac{{\sigma }_a}{S_e}+\frac{{\sigma }_m}{S_{ut}}\right]}^{-1}$

Since ${\sigma }_m=0$, $n_f=\frac{S_e}{{\sigma }_a}$

$n_f$=0.44

So, infinite life is not predicted

Find the predicted number of cycles to failure

$f=1.06-\left(4.1*{10}^{-4}\right)S_{ut} + \left(1.5 * {10}^{-7}\right){S_{ut}}^2 =$0.87

$a = \frac{{\left(f{S}_{ut}\right)}^2}{S_e}$=1415.8

$b=-\frac{1}{3}\left[\log \left(\frac{f*S_{ut}}{S_e}\right)\right]$= -0.147

${\sigma }_{ar}={\sigma }_a$ (because ${\sigma }_m$ = 0)

$N = {\left({\sigma }_{ar}/a\right)}^{1/b}$= 3555 cycles

(Note that these answers differ slightly from those in the book, due to error in interpreting the graphs and rounding errors.)

Check for 1st cycle yielding

$n_y=\frac{S_y}{{\sigma }_{max}}=\frac{S_y}{{\sigma }_a+{\sigma }_m}=\frac{490 \mathrm{MPa}}{425.623 \mathrm{MPa}}=1.15$

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Problem 6-38 Answer

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Note: To get full credit for questions like this on a quiz or test, you'll need to show all of your work. Here, I may only providing the final answer for you to check. If you have questions on how to get here, please feel free to ask me during office hours, or send me an email.

n_f = 0.51

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Problem 6-47 Answer

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Note: To get full credit for questions like this on a quiz or test, you'll need to show all of your work. Here, I may only providing the final answer for you to check. If you have questions on how to get here, please feel free to ask me during office hours, or send me an email.

n_f = 1.3

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Problem 6-54 Solution

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This problem is an extension of problem 6-17, where we now have an additional steady torque applied

$T_m=282.5 N·m$ 

$T_a=0$

The steady applied torque is constant across the shaft, so our selected critical cross-section doesn't change.

Calculate the endurance limit

The only part of this that might change from 6-17 is $k_c$. We already had $k_c=1$, which is the value used for combined loading. So, the endurance limit doesn't change from problem 6-17.

$S_e=\left(0.76\right)\left(0.83\right){S^{\text{'}}}_e$ = 186.1 MPa

Calculate von Mises stresses for combined loading

$$\begin{gathered} {{\sigma }^{\text{'}}}_a={\left\{{\left[{\left(K_f\right)}_{bending}{\left({\sigma }_{a0}\right)}_{bending}+{\left(K_f\right)}_{axial}{\left({\sigma }_{a0}\right)}_{axial}\right]}^2+3*{\left[{\left(K_{fs}\right)}_{torsion}{\left({\tau }_{a0}\right)}_{torsion}\right]}^2\right\}}^{1/2} \\ {{\sigma }^{\text{'}}}_m={\left\{{\left[{\left(K_f\right)}_{bending}{\left({\sigma }_{m0}\right)}_{bending}+{\left(K_f\right)}_{axial}{\left({\sigma }_{m0}\right)}_{axial}\right]}^2+3*{\left[{\left(K_{fs}\right)}_{torsion}{\left({\tau }_{m0}\right)}_{torsion}\right]}^2\right\}}^{1/2} \end{gathered}$$

We have no axial components of stress, so we can remove those terms. We also know that our mean bending stress is zero, and our alternating torsional stress is zero. Plugging that in, our von Mises stress equations reduce to

$$\begin{gathered} {{\sigma }^{\text{'}}}_a={\left(K_f\right)}_{bending}{\left({\sigma }_{a0}\right)}_{bending} \\ {{\sigma }^{\text{'}}}_m=\sqrt{3}{\left(K_{fs}\right)}_{torsion}{\left({\tau }_{m0}\right)}_{torsion} \end{gathered}$$

From problem 6-17, we already know that ${{\sigma }^{\text{'}}}_a=$ 425.623 MPa

Before we can find ${{\sigma }^{\text{'}}}_m$, we need to find the fatigue stress-concentration factor for torsion.

Determine the fatigue stress-concentration factor(s)

We can look at Figure A-15-8 to find $K_{ts}$. As in problem 6-17, r/d = 1.588/41.275 = 0.038, D/d = 47.625/41.275 = 1.15, so

$K_{ts}=$  1.6

To find $q_s$, we can look in Figure 6-27. The given fillet radius is still r = 1.588 mm. So, $q_s$ = 0.8

$K_{fs}=1+q_s\left(K_{ts}-1\right)$ = 1+0.8(1.6-1) = 1.48

Apply Kfs to the torsional stress

${\tau }_{m0}=\frac{T_{m}c}{J}=\frac{(282.5 \mathrm{N}·\mathrm{m})*(0.041275/2 \mathrm{m})}{\pi *{\left(0.041275 \mathrm{m}\right)}^4/32}=20.461 \mathrm{MPa}$

${{\sigma }^{\text{'}}}_m=\sqrt{3}\left(1.48\right)\left(20.461 \mathrm{MPa}\right) =52.451 \mathrm{MPa}$

Find factor of safety for infinite life using Goodman criteria

$n_f={\left[\frac{{{\sigma }^{\text{'}}}_a}{S_e}+\frac{{{\sigma }^{\text{'}}}_m}{S_{ut}}\right]}^{-1} = {\left[\frac{425.623}{186.1}+\frac{52.451}{590}\right]}^{-1}$=0.42

So we do not predict infinite life. This makes sense, as we have made the stress state more severe at our critical cross-section, so if we did not have infinite life in problem 6-17, we should not have infinite life in this problem either.

Find the predicted number of cycles to failure

f, a, and b will still be the same as calculated in problem 6-17

$f=1.06-\left(4.1*{10}^{-4}\right)S_{ut} + \left(1.5 * {10}^{-7}\right){S_{ut}}^2 =$0.87

$a = \frac{{\left(f{S}_{ut}\right)}^2}{S_e}$=1415.8

$b=-\frac{1}{3}\left[\log \left(\frac{f*S_{ut}}{S_e}\right)\right]$= -0.147

${\sigma }_{ar}$ will be different, because we are now concerned with a mean stress

${\sigma }_{ar}=\frac{{{\sigma }^{\text{'}}}_a}{1-{{\sigma }^{\text{'}}}_m/S_{ut}}=\frac{425.623}{1-52.451/590}=467.153 \mathrm{MPa}$

$N = {\left({\sigma }_{ar}/a\right)}^{1/b}$= 1887 cycles

Again, it makes sense that the number of cycles to failure is less in this case as compared to problem 6-17, because we have a more severe stress state. We have added a torsional load to the existing bending load we had in the previous problem.

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Problem 6-57 Answer

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P = 18.24 N

Yielding is not predicted.

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