Download homework assignments and more Study notes Stress Analysis in PDF only on Docsity!
Problem 6. 20
A bar of steel has the minimum properties S e = 50 kpsi , S y = 60 kpsi , S ut = 80 kpsi. The bar is
subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 kpsi. Find the factor of
safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use: ( a ) Goodman criterion. ( b ) Gerber criterion. ( c ) Morrow criterion. Solution: Obtain von Mises stresses for the alternating, mean, and maximum stresses. (a) Goodman, Equation (6-41) (b) Gerber, Equation (6-48) (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44): Eq. (6-46): Problem 6. 22
Repeat Problem 6–20 but with a steady torsional stress of 15 kpsi , an alternating torsional stress of 10 kpsi ,
and an alternating bending stress of 12 kpsi.
Solution: Obtain von Mises stresses for the alternating, mean, and maximum stresses.
(a) Goodman, Equation (6-41) (b) Gerber, Equation (6-48) (c) Morrow Estimate the fatigue strength coefficient. Eq. (6-44): Eq. (6-46): Problem 6. 25 The cold-drawn AISI 1040 steel bar shown in the figure is subjected to a completely
reversed axial load fluctuating between 28 kN in compression to 28 kN in
tension. Estimate the fatigue factor of safety based on achieving infinite life and the yielding factor of safety. If infinite life is not predicted, estimate the number of cycles to failure. Solution: Given:. From Table A-20, for AISI 1040 CD, Check for yielding Determine the fatigue factor of safety based on infinite life Eq. (6-10): Eq. (6-18): Eq. (6-20):
Determine the fatigue factor of safety based on infinite life. From Prob. 6-25: Goodman criteria, Equation (6-41): Gerber criteria, Equation (6-48): Morrow criteria: Estimate the fatigue strength coefficient. Eq. (6-44): Eq. (6-46): The results are consistent with Fig. 6-36, where for a mean stress that is about half of the yield strength, the Goodman line should predict failure significantly before the other two. Problem 6. 28 The figure shows a formed round-wire cantilever spring subjected to a varying
force. The hardness tests made on 50 springs gave a minimum hardness of
400 Brinell. It is apparent from the mounting details that there is no
stress concentration. A visual inspection of the springs indicates that the surface finish corresponds closely to a hot-rolled finish. Ignore curvature effects on the bending stress. What number of applications is likely to cause failure? Solve using: ( a ) Goodman criterion. ( b ) Gerber criterion. Solution 6-
Eq. (2-36): Sut = 0.5(400) = 200 kpsi Eq. (6-10): Eq. (6-18): Eq. (6-24): Eq. (6-19): Since we have used the equivalent diameter method to get the size factor, and in doing so introduced greater uncertainties, we will choose not to use a size factor greater than one. Let kb = 1. Eq. (6-17): (a) Goodman criterion, Eq. (6-41): Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. Using the Goodman criterion, Eq. (6-58): Fig. 6-23: f = 0. Eq. (6-13): Eq. (6-14): Eq. (6-15): (b) Gerber criterion, Eq. (6-48):
Since the mean stress is negative, use Eq. (6-42). Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of safety of Problem 6. 31 The figure shows the free-body diagram of a connecting-link portion having stress concentration at three sections. The dimensions are
r =0.25∈ , d =0.40∈ , h =0.50∈ ,
w 1 =3.50∈ ,and w 2 =3.00∈¿ The forces F fluctuate
between a tension of 5 kip and a compression of 16 kip.
Neglect column action and find the least factor of safety if the material is cold-drawn AISI 1018 steel. Solution: Eq. (6-10): Eq. (6-18): Eq. (6-19): Eq. (6-25): Eq. (6-17): Fillet: Fig. A-15-5: Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph. q = 0.
Using Goodman criteria, Eq. (6-41), Hole: Fig. A-15-1: Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the graph. q = 0. Using Goodman criteria, Eq. (6-41), Thus, the design is controlled by the threat of fatigue at the fillet with a minimum factor of safety of Problem 6. 50 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 Goodman criterion. If the life is not infinite, conservatively
Using Goodman, Eq. (6-41): Ans. Problem 6. 56 In the figure shown, shaft A, made of AISI 1020 hot-rolled steel, is welded to a fixed support and is subjected to loading by equal and
opposite forces F via shaft B. A theoretical stress-concentration factor
Kts of 1.6 is induced in the shaft by the 1 / 8 ∈¿ weld fillet. The length
of shaft A from the fixed support to the connection at shaft B is 2 ft.
The load F cycles from 150 to 500 lbf.
( a ) For shaft A, find the factor of safety for infinite life using the Goodman fatigue failure criterion. ( b ) Repeat part ( a ) using the Gerber fatigue failure criterion. Solution: Eqs. (6-33) and (6-36), or Fig. 6-27: qs = 0. Eq. (6-32): Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. Find the modifiers and endurance limit. Eq. (6-10): Eq. (6-18): Eq. (6-23): Eq. (6-19):
Eq. (6-25): Eq. (6-17): Since the stress is entirely shear, we will use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-16. From Eq. (6-58), Ssu = 0.67 Su = 0.67 (55) = 36.9 kpsi. (a) Goodman, Eq. (6-41): Ans. (b) Gerber Eq. (6-48): Problem 6 - 59
A flat leaf spring has fluctuating stress of σ max = 360 MPa and σ min = 160 MPa applied for 8 × 104
cycles. If the load changes to^ σ^ max =^320 MPa^ and^ σ^ min =−^200 MPa^ how many cycles should the
spring survive, using the Goodman criterion? The material is AISI 1020 CD and has a fully corrected endurance
strength of S e = 175 MPa.
( a ) Use Miner’s method. ( b ) Use Manson’s method. Solution: For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175 MPa. First Loading: Goodman, Eq. (6-58): Fig. 6-23: Off the graph, so let f = 0.9.
Firstly, number of cycles to each regime of loading is calculated: Using Miner’s method, the total damage due to N number of cycles to failure for the three regimes of loading must be equal to one, Thus,
