MAE 311: Machines Machines and and Mechanisms Mechanisms I
Spring 2014
Name: Solution Key
Section: NA
Exam #2
Print your name at the top each page. You will turn in both this exam sheet and your equation sheet. Point distribution is shown for each problem (this may be helpful in allocating time for each question. Any instance of academic dishonesty will result in a zero for the exam, and has the potential for further disciplinary action (including expulsion from the university).
By printing your name you
acknowledge you will not discuss this exam with anyone prior 5:00 PM on April 17 th, 2014.
Theory Section (35 Points Total, 5 Points Each ) 1. Briefly describe why modifying factors are used to adjust fatigue strength values (either for finite or or infinite life) when performing fatigue analysis for component design. Modifying factors are used to account for effects noticed during testing (e.g., (e.g., size) size) and differences between the experimental test test specimens and production components (e.g., (e.g., surface surface finish). Only need to identify one of the reasons. 2. For static loading, stress concentration factors themselves are are a function of what two parameters? Geometry and load type 3. For fatigue fatigue failure, the notch sensitivity (q) adjusts the stress concentration to account for the impact of what? Material properties or material choice 4. Fatigue failure failure theories were developed developed largely with experimental data. Identify one impact this has when designing a component against fatigue. Typically use higher factors of safety when designing against fatigue and should do component testing once the part is manufactured. Only need to identify one of the impacts. 5. What is the impact of a negative midrange stress component on the allowable alternating alternating stress component for ductile and brittle materials? Briefly describe how a negative midrange midrange stress component is handled when evaluating safety against fatigue failure (for both ductile and brittle materials). Hint: see Figure Figure 1 and Figure Figure 2. A negative midrange stress has little impact on the allowable alternating component for ductile materials; for brittle materials a negative midrange stress component has the effect of increasing the allowable alternating stress stress component. For ductile materials, materials, you replace the negative midrange midrange stress with zero when calculating a factor of safety to keep a conservative estimate of the factor of safety. For brittle materials materials you leverage the negative midrange midrange stress when calculating the the factor of safety.
Copyright 2014 –Phillip M. Cormier
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Figure 1: Fatigue Failure Envelope for Brittle Material
Figure 2: Fatigue Failure Data for Ductile Material
Copyright 2014 –Phillip M. Cormier
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MAE 311: Machines and Mechanisms I
Spring 2014
Name: Solution Key
Section: NA
Exam #2
6. Consider the ASME Elliptic failure envelope shown in Figure 3. a) Identify the relative magnitudes of the factor of safety for points A, B, C, and D (<, ≤,=, etc.) b) What is the yield strength of the material being used? a) A>B=D>C (3 Points) b) 84 kpsi (2 Points) 7. Sketch the following on Figure 3: (2.5 Points Each) a) The new ASME Elliptic failure envelope if instead of using the endurance limit of 33.9 kpsi a fatigue strength of 40 kpsi is used. b) The rough shape of the Gerber Fatigue line if the ultimate tensile strength is 100 kpsi.
Figure 3: Failure Envelope for Questions 5 and 6
Copyright 2014 –Phillip M. Cormier
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Mini Problems (32 Points Total ) 8. Calculate the factor of safety for infinite life if the steel used has an adjusted endurance limit of 40 kpsi, an ultimate tensile strength of 100 kpsi, and a yield strength of 80 kpsi, provided the midrange stress and alternating stress components are both found to be 20 kpsi. (10 Points) Using ASME Elliptic (any ductile failure theory acceptable):
( ) () () ( ) ( ) 9. Calculate the factor of safety for infinite life if the cast iron used has an adjusted endurance limit of 20 kpsi and an ultimate tensile strength of 40 kpsi, provided the midrange stress and alternating stress components are found to be -20 kpsi and 40 kpsi respectively. (10 Points)
( ) ( ) ( ( ))
Copyright 2014 –Phillip M. Cormier
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MAE 311: Machines and Mechanisms I
Spring 2014
Name: Solution Key
Section: NA
Exam #2
10. Calculate the midrange and alternating stress components for each case shown below. Identify which case would be the critical case to consider and explain why (assuming all other things are equal). (12 Points , 3 Points Each) a)
b)
c)
a)
b)
c)
Case b would be the critical case because it has a tensile midrange stress; this is the case regardless of whether or not the material is brittle or ductile.
Copyright 2014 –Phillip M. Cormier
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Design Problem (33 Points Total ) 11. Consider the component and its loading shown in Figure 4. The steel being used has an adjusted endurance limit of 40 kpsi, an ultimate tensile strength of 130 kpsi, and a yield strength of 100 kpsi. The applied load F y does not fluctuate. a) Determine the current factor of safety at Point A. Show all work. (21 Points) b) Plot the critical stress state on the set of principal axes provided. (3 Points) c) If you used a different failure theory, would your lever arm (currently 6 inches) have to increase, decrease, or stay the same to maintain the same factor of safety? (3 Points) d) If the desired factor of safety was 3, would you have to increase, decrease, or keep the current magnitude of the applied load? Identify this new point on the principal axes. (6 Points) Bonus: If you used cast iron, what factor of safety should be used if you want to maintain a similar risk level (not change your risk attitude)? (1 Point) Bonus: Determine the required change in length of the lever arm that results in a factor of safety of 3? (4 Points)
Figure 4: Component and Loading for Design Problem (Question 11) a)
Copyright 2014 –Phillip M. Cormier
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MAE 311: Machines and Mechanisms I
Spring 2014
Name: Solution Key
Principal stresses:
Section: NA
Exam #2
[ ]
b)
b) See Graph c) DE -> MSS: Deccrease; MSS -> DE: Increase d) The failure theory would have to decrease
Bonus 1: Double Bonus 2: Cannot obtain this factor of safety (cannot shrink the lever arm enough to reduce the stresses to the appropriate level).
Copyright 2014 –Phillip M. Cormier
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Copyright 2014 –Phillip M. Cormier
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