ub mae 311

MAE 311: Machines and Mechanisms I Spring 2014

Name: Solution Key Section: NA Exam #2

Copyright 2014 Phillip M. Cormier 1

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 17th, 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

infinite life) when performing fatigue analysis for component design.

Modifying factors are used to account for effects noticed during testing (e.g., size) and differences

between the experimental test specimens and production components (e.g., surface finish).

Only need to identify one of the reasons.

2. For static loading, stress concentration factors themselves are a function of what two parameters?

Geometry and load type

3. For 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 theories were 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 stress

component for ductile and brittle materials? Briefly describe how a negative midrange stress

component is handled when evaluating safety against fatigue failure (for both ductile and brittle

materials). Hint: see Figure 1 and 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 component. For ductile materials, you replace the negative midrange

stress with zero when calculating a factor of safety to keep a conservative estimate of the factor of

safety. For brittle materials you leverage the negative midrange stress when calculating the factor

of safety.


Figure 1: Fatigue Failure Envelope for Brittle Material

Figure 2: Fatigue Failure Data for Ductile Material




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 (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


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)

(

)

(

)

( (

) )

(

)

( )

( )




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.


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 Fy 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)

[( ) ( )]( )

( )




[( ) ( )]( )

( )

[

]

Principal stresses:

[( )

( ) ( )

]

[( ) ( ) ( )

]

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).

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