Homework_05_Sol_V2

pdf

School

Purdue University *

*We aren’t endorsed by this school

Course

35401

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by ColonelHeat13546

Homework #5 HW#5.1 Problem 6.55 (Shigley 11 th Edition) The shaft shown in the figure is machined from AISI 1040 CD steel. The shaft rotates at 1600 rpm and is supported in rolling bearings at A and B . The applied forces are ? 1 = 1200 ??𝑓 and ? 2 = 2400 ??𝑓 . A steady torque ? = 2200 ??𝑓 ∗ 𝑖𝑛 is being transmitted through the shaft between the points of application of the forces. Determine the minimum fatigue factor of safety based on achieving infinite life. If infinite life is not predicted, estimate the number of cycles to failure. Also check for yielding. Use the Goodman criterion. Solution: (1) Draw a free body diagram: (2) Sum forces and moments to solve for reactionary forces at bearings A and B : ∑ ? ? = ? ? + ? ? − ? 1 − ? 2 = 0  ? ? = ? 1 + ? 2 − ? ? = 1600 ??𝑓 ∑ ? ? = −8*F 1 − 16*? 2 + 24*? ? = 0  ? ? = 1 24 (8*? 1 + 16*? 2 ) = 2000 ??𝑓 (3) Draw the shear ( V(x) ), bending moment ( M(x) ), and torque ( T(x) ) diagrams to aid in selecting point(s) on the beam to analyze:

As seen in our moment diagram, the point most likely to be critical on our beam occurs at the shoulder fillet between the 1 5 8 𝑖𝑛. diameter and the 1 7 8 𝑖𝑛. diameter ( ? = 10.5 𝑖𝑛. ) due to the stress concentration and exaggerated magnitude of the bending moment as compared to the other points. (4) Calculate the stresses at the selected point of 𝒙 = ??. 𝟓 𝒊?. (note that we’re using the smaller diameter of 𝒅 = ? 𝟓 𝟖 𝒊?. at this location): ? = 8*? ? + (10.5 − 8)*(? ? − ? 1 ) = 8*1600 + 2.5*400 = 13800 𝑖𝑛.-??𝑓 ? = ? 2 = 1.625 2 = 0.8125 𝑖𝑛. ? = 𝜋 64 ? 4 = 𝜋 64 (1.625) 4 = 0.342281 𝑖𝑛. 4 𝜎 ? = ?? ? = (13800 𝑖?.-???)(0.8125 𝑖?.) 0.342281 𝑖?. 4 = 32.758130 ???𝑖 Fully reversed loading cycle  𝜎 ? = 𝜎 ? = 32.758130 ???𝑖  𝜎 ? = 0 ???𝑖

? = 2200 𝑖𝑛.-??𝑓 ? = ? 2 = 1.625 2 = 0.8125 𝑖𝑛. ? = 𝜋 32 ? 4 = 𝜋 32 (1.625) 4 = 0.684563 𝑖𝑛. 4 𝜏 ?? = ?? ? = (2200 𝑖?.-???)(0.8125 𝑖?.) 0.684563 𝑖?. 4 = 2.611155 ???𝑖 Constant applied torque  𝜏 ? = 0 ???𝑖  𝜏 ? = 𝜏 ?? = 2.611155 ???𝑖 (5) Calculate bending and torsion stress concentration factors: Required parameters for line-fit plot: 𝐷 ? = 1.875 1.625 = 1.1538 ; ? ? = 0.0625 1.625 = 0.03846 ? ? = 1.95 (fig. A-15-9) (bending) ? ?? = 1.6 (fig. A-15-8) (torsion) Given material is AISI 1040 CD Steel: ? ?? = 85 ???𝑖 ; ? ? = 71 ???𝑖 (table A-20) ? = 0.75 (fig. 6-26) (bending) ? ? = 0.8 (fig. 6-27) (torsion) ? ? = 1 + ?(? ? − 1) = 1 + 0.75(1.95 − 1) = 1.7125 (eq. 6-32) (bending) ? ?? = 1 + ? ? (? ?? − 1) = 1 + 0.8(1.6 − 1) = 1.48 (eq. 6-32) (torsion) (6) Calculate Von-Mises Stresses: 𝜎 ? ′ = [(? ? 𝜎 ? ) 2 + 3(? ?? 𝜏 ? ) 2 ] 1 2 = [(1.7125*32.758130) 2 ] 1 2 = 56.098297 ???𝑖 (eq. 7-4) 𝜎 ? ′ = [(? ? 𝜎 ? ) 2 + 3(? ?? 𝜏 ? ) 2 ] 1 2 = [3*(1.48*2.611155) 2 ] 1 2 = 6.693527 ???𝑖 (eq. 7-5) 𝜎 ??? ′ = [(? ? 𝜎 ? + ? ? 𝜎 ? ) 2 + 3(? ?? 𝜏 ? + ? ?? 𝜏 ? ) 2 ] 1 2 = [(1.7125*32.758130) 2 + 3*(1.48*2.611155) 2 ] 1 2 = 56.496214 ???𝑖 (eq. 7-15) (7) Check for yielding: 𝑛 ? = ? ? 𝜎′ ??? = 71 ?𝑝?𝑖 56.496213 ?𝑝?𝑖 = 1.3 (eq. 7-16) 𝑛 ? > 1  specimen not expected to yield. (8) Calculate endurance limit: ? = 2.00 ; ? = −0.217 (table 6-2) ? ? = ?? ?? ? = 2.00(85 ???𝑖) −0.217 = 0.762687 (eq. 6-18) ? ? = 0.879? −0.107 = 0.879(1.625 𝑖𝑛. ) −0.107 = 0.834502 (eq. 6-19) ? ? = 1 - (torsion + bending) (eq. 6-25) ? ? = ? ? = 1 ?′ ? = 0.5? ?? = 0.5(85 ???𝑖) = 42.5 ???𝑖 (eq. 6-10) ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ′ = (0.762687)(0.834502)(1)(1)(1)(42.5 ???𝑖) = 27.049708 ???𝑖 (eq. 6-17) (9) Check for infinite life prediction (Goodman): 𝑛 ? = ( 𝜎 ? ′ ? ? + 𝜎 ? ′ ? ?? ) −1 = ( 56.098297 ?𝑝?𝑖 27.049708 ?𝑝?𝑖 + 6.693527 ?𝑝?𝑖 85 ?𝑝?𝑖 ) −1 = 0.5 (eq. 6-41) 𝑛 ? < 1  specimen predicted to have finite life.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

(10) Find cycles to failure (Goodman): 𝜎 ?? = 𝜎 ? ′ 1− 𝜎 ? ′ 𝑆 ?? = 56.098297 ?𝑝?𝑖 1− 6.693527 𝑘𝑝?𝑖 85 𝑘𝑝?𝑖 = 60.893500 ???𝑖 (eq. 6-59) 𝑓 = 0.87 (fig. 6-23) ? = (?? ?? ) 2 ? ? = (0.87*85 ?𝑝?𝑖) 2 27.049708 ?𝑝?𝑖 = 202.168632 ???𝑖 (eq. 6-13) ? = − 1 3 log ( ?? ?? ? ? ) = − 1 3 log ( 0.87*85 ?𝑝?𝑖 27.049708 ?𝑝?𝑖 ) = −0.145592 (eq. 6-14) ? = ( 𝜎 ?? ? ) 1 ? = ( 60.893500 ?𝑝?𝑖 202.168632 ?𝑝?𝑖 ) − 1 0.145592 = 3536 cycles (eq. 6-15)

HW#5.2 Perform the rainflow cycle counting technique on a given load time history. It is assumed that the loading history is periodic. Solution (WATERFALL METHOD): (1) Adjust plot such that points A-B-C-D follow point I (original label is in parentheses):

(2) T urn the adjusted plot sideways. Points have been re-lettered so that they are in order. (3) Start rainfall from valleys: a. Starting at A, rain falls past C since it is a lower valley (-5 < -3), and falls off the roof at D. b. Starting at C, rain meets with flow from A-D and stops at B. c. Starting at E, rain falls past G since it is a lower valley (-4 < -1), and falls off the roof at H. d. Starting at G, rain meets with flow from E-H and stops at F. (4) Start rainfall from peaks: a. Starting from B, rain stops at D since it is a lower peak (1 < 4). b. Starting from D, rain falls past F since it is a higher peak (4 > 2), and falls past H since it is a higher peak (4 > 3), and falls off the roof at I/A. c. Starting from F, rain stops at H since it is a lower peak (2 < 3). d. Starting from H, rain meets with flow from D-I/A and stops at E. (5) Count the number of cycles based on the waterfall chart: Num. Cycles From To From To Range Mean 1 A D -5 4 9 -0.5 1 B C 1 -3 4 -1 1 E H -4 3 7 -0.5 1 F G 2 -1 3 0.5

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Solution (4-POINT): 1) Adjust plot such that points A-B-C-D follow point I (original label is in parentheses): 2) Identify sets of four consecutive peak/valley points. a. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ contained by range of ? and ? , cycle [? − ?] counted and removed from chart. b. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ extends outside range of ? and ? , no cycle counted. c. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ extends outside range of ? and ? , no cycle counted. d. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ contained by range of ? and ? , cycle [? − ?] counted and removed from chart. e. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ extends outside range of ? and ? , no cycle counted. f. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ extends outside range of ? and ? , no cycle counted. g. [? > ? > ? > ?]; segment ?? ̅̅̅̅ extends outside range of ? and ? , no cycle counted. h. [? > ? > ? > ?] ; segment ?? ̅̅̅̅ contained by range of ? and ? , cycle [? − ?] counted and removed from chart. i. [? > ? > ?]; final cycle remaining, cycle [? − ?] counted. 3) Count the number of cycles based on the 4-point counting technique: Num. Cycles From To From To Range Mean 1 A D -5 4 9 -0.5 1 B C 1 -3 4 -1 1 E H -4 3 7 -0.5 1 F G 2 -1 3 0.5

HW#5.3 A connecting link is made of Grade 30 gray cast iron. All surfaces are machined. The dimensions are r =0.25 in, d =0.40 in, h =0.50 in, w 1 =3.50 in, and w 2 =3.0 in.: 1) The part is subject to constant tension forces F at 20 kips. Will the part fail? If no, determine the factor of safety. 2) The part is subject to F fluctuating between a tension of 5 kips and a compression of 16 kips. Will the part saft for infinite life? If yes, determine the factor of safety. Solution:

(1) Determine maximum stress value for the center hole: From (fig. A-15-1) description and plot: ? ℎ = (? − ?)? = (? 1 − ?)ℎ = (3.5 − 0.40 𝑖𝑛. )(0.50 𝑖𝑛. ) = 1.55 𝑖𝑛. 2 Required plot parameter: ? ? = ? ? 1 = 0.40 3.5 = 0.114286 ? ?,ℎ = 2.65 𝜎 ℎ = 𝐹 ? ℎ = 20 ?𝑖𝑝? 1.55 𝑖?. 2 = 12.903226 ???𝑖 (2) Determine maximum stress value for the shoulder fillets: From (fig. A-15-5) description and plot: ? ? = ?? = ? 2 ℎ = (3 𝑖𝑛. )(0.5 𝑖𝑛. ) = 1.5 𝑖𝑛. 2 Required plot parameters: ? ? = ? ? 2 = 0.25 3 = 0.083333 ; 𝐷 ? = ? 1 ? 2 = 3.5 3 = 1.166667 ? ?,? = 1.8 𝜎 ? = 𝐹 ? ? = 20 ?𝑖𝑝? 1.5 𝑖?. 2 = 13.333333 ???𝑖 (3) Check both points for failure: Given material is Grade 30 gray cast iron: ? ?? = 31 ???𝑖 ? ? ? ? ? ? ′ = 14 ???𝑖 ? ? = 1.10 (table A-24) 𝑛 ℎ = ? ?? 𝜎 ℎ = 31 ?𝑝?𝑖 12.903226 ?𝑝?𝑖 = 2.4 𝑛 ? = ? ?? 𝜎 ? = 31 ?𝑝?𝑖 13.333333 ?𝑝?𝑖 = 2.3 𝑛 ℎ , 𝑛 ? > 1  part will not fail. (4) Calculate alternating and mid-range values loads: ? ? = | 𝐹 ??? −𝐹 ?𝑖? 2 | = | 5−(−16) ?𝑖𝑝? 2 | = 10.5 ?𝑖?? (eq. 6-8) ? ? = 𝐹 ??? +𝐹 ?𝑖? 2 = 5+(−16) ?𝑖𝑝? 2 = −5.5 ?𝑖?? (eq. 6-9) (5) Calculate alternating and mid-range stresses for the center hole: ? = 0.20 (section 6-10, pg. 323 of 11 th edition) ? ?,ℎ = 1 + ?(? ?,ℎ − 1) = 1 + 0.20(2.65 − 1) = 1.33 (eq. 6-32) 𝜎 ?,ℎ = ? ?,ℎ 𝐹 ? ? ℎ = (1.33)(10.5 ?𝑖𝑝?) 1.55 𝑖?. 2 = 9.009677 ???𝑖 𝜎 ?,ℎ = ? ?,ℎ 𝐹 ? ? ℎ = (1.33)(−5.5 ?𝑖𝑝?) 1.55 𝑖?. 2 = −4.719355 ???𝑖 (6) Calculate alternating and mid-range stresses for the shoulder fillets: ? ?,? = 1 + ?(? ?,? − 1) = 1 + 0.20(1.8 − 1) = 1.16 (eq. 6-32) 𝜎 ?,? = ? ?,ℎ 𝐹 ? ? ? = (1.18)(10.5 ?𝑖𝑝?) 1.5 𝑖?. 2 = 8.120 ???𝑖 𝜎 ?,? = ? ?,ℎ 𝐹 ? ? ? = (1.18)(−5.5 ?𝑖𝑝?) 1.5 𝑖?. 2 = −4.253333 ???𝑖 (7) Calculate endurance limit: ? ? = 0.9 (eq. 6-25) ? ? = ? ? = 1 ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ′ = (0.9)(14 ???𝑖) = 12.6 ???𝑖 (eq. 6-17) (8) Calculate slope of fluctuating-stress plot ( 𝝈 ? vs 𝝈 𝒂 ) in the second quadrant (brittle materials):

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

? = 𝜎 ? 𝜎 ? = 𝐹 ? 𝐹 ? = 10.5 −5.5 = −1.909091 (fig. 6-39) (9) Calculate the allowable stress (brittle materials): ? ? = ? ? ? (section 6-15, pg. 345 of 11 th edition) ? ? = ? ? + ( ? ? ? ?? − 1) ? ? (eq. 6-65) ? ? = ? ? + ( ? ? ? ?? − 1) ? ? ?  ? ? − ( ? ? ? ?? − 1) ? ? ? = ? ?  ? ? = ? ? 1−( 𝑆 ? 𝑆 ?? −1) 1 ? = 12.6 ?𝑝?𝑖 1−( 12.6 𝑘𝑝?𝑖 31 𝑘𝑝?𝑖 −1) 1 −1.909091 = 18.284886 ???𝑖 (10) Check for infinite life prediction (brittle materials): At the center hole: 𝑛 ℎ = ? ? 𝜎 ?,ℎ = 18.284886 ?𝑝?𝑖 9.009677 ?𝑝?𝑖 = 2.0 At the shoulder fillets: 𝑛 ? = ? ? 𝜎 ?,? = 18.284886 ?𝑝?𝑖 9.009677 ?𝑝?𝑖 = 2.3 𝑛 ℎ , 𝑛 ? > 1  specimen predicted to have infinite life. HW#5.4 Problem 6.61 (Shigley 11 th Edition) A machine part will be cycled at ±350 ?𝑃? for 5 ∗ 10 3 cycles. Then the loading will be changed to ±260 ?𝑃? for 5 ∗ 10 4 cycles. Finally, the load will be changed to ±225 ?𝑃? . How many cycles of operation can be expected at this stress level? For the part, ? ?? = 530 ?𝑃? , 𝑓 = 0.9 , and has a fully corrected endurance strength of ? ? = 210 ?𝑃? . (a) Use Miner’s method. (b) Use Manson’ method. Solution: (1) (Miner’s method) calculate ? and ? parameters of the S-N curve-fit equation: ? = (?? ?? ) 2 ? ? = (0.9*530 ?𝑃?) 2 210 ?𝑃? = 1083.471429 ?𝑃? (eq. 6-13) ? = − 1 3 log ( ?? ?? ? ? ) = − 1 3 log ( 0.9*530 ?𝑃? 210 ?𝑃? ) = −0.118766 (eq. 6-14) Assume given material is steel: 𝜎 ? ′ = ? ?? + 345 ?𝑃? = 530 + 345 ?𝑃? = 875 ?𝑃? (eq. 6-44) (2) Check for infinite life prediction after cycle 1: 𝑛 ? = ( 𝜎 ? ? ? + 𝜎 ? 𝜎 ? ′ ) −1 = ( 350 ?𝑃? 210 ?𝑃? ) −1 = 0.6 (eq. 6-46)

𝑛 ? < 1  specimen predicted to have finite life (damage) (3) Calculate equivalent completely reversed stress after cycle 1: 𝜎 ?? = 𝜎 ? 1− 𝜎 ? 𝜎 ? ′ = 𝜎 ? = 350 ?𝑃? (eq. 6-60) (4) Calculate the maximum number of cycles at the cycle 1 load: ? 1 = ( 𝜎 ?? ? ) 1 ? = ( 350 ?𝑃? 1083.471429 ?𝑃? ) − 1 0.118766 = 13554 cycles (eq. 6-15) (5) Repeat (2)-(4) for cycle 2: 𝑛 ? = ( 𝜎 ? ? ? + 𝜎 ? 𝜎 ? ′ ) −1 = ( 260 ?𝑃? 210 ?𝑃? ) −1 = 0.8 (eq. 6-46) 𝑛 ? < 1  specimen predicted to have finite life (damage) 𝜎 ?? = 𝜎 ? 1− 𝜎 ? 𝜎 ? ′ = 𝜎 ? = 260 ?𝑃? (eq. 6-60) ? 2 = ( 𝜎 ?? ? ) 1 ? = ( 260 ?𝑃? 1083.471429 ?𝑃? ) − 1 0.118766 = 165585 cycles (eq. 6-15) (6) Repeat (2)-(4) for cycle 3: 𝑛 ? = ( 𝜎 ? ? ? + 𝜎 ? 𝜎 ? ′ ) −1 = ( 225 ?𝑃? 210 ?𝑃? ) −1 = 0.9 (eq. 6-46) 𝑛 ? < 1  specimen predicted to have finite life (damage) 𝜎 ?? = 𝜎 ? 1− 𝜎 ? 𝜎 ? ′ = 𝜎 ? = 225 ?𝑃? (eq. 6-60) ? 3 = ( 𝜎 ?? ? ) 1 ? = ( 225 ?𝑃? 1083.471429 ?𝑃? ) − 1 0.118766 = 559388 cycles (eq. 6-15) (7) Calculate the number of cycles at the final loading condition: ∑ ? 𝑖 ? 𝑖 = ? ; ? = 1 (eq. 6-68) ? 1 ? 1 + ? 2 ? 2 + ? 3 ? 3 = 1  𝑛 3 = ? 3 (1 − ? 1 ? 1 − ? 2 ? 2 ) = (559388)(1 − 5∗10 3 13554 − 5∗10 4 165585 ) = 184115 cycles (8) (Manson’s Method) Calculate life remaining after cycle 1 (use solutions from (1)-(4) above): ? 1 ′ = ? 1 − 5*10 3 = 13554 − 5*10 3 = 8554 cycles; ? 1 = 350 ?𝑃? (9) Calculate updated S-N curve coefficients after cycle 1: ? 0 = 10 3 ; ? 0 = 𝑓? ?? = 0.9*530 ?𝑃? = 477 ?𝑃? (We will use this point on the S-N plot which will hold for each updated curve.) ? ? = ?? ? (eq. 6-12) ? 1 = ? 2 ? 1 ′ ? 2 ? 0 = ? 2 ? 0 ? 2  ? 1 ? 0 = ? 1 ′ ? 2 ? 0 ? 2  ? 1 ? 0 = ( ? 1 ′ ? 0 ) ? 2  ? 2 = log(? 1 /? 0 ) log(? 1 ′ /? 0 ) = log(350/477) log(8554/10 3 ) = −0.144236 ? 2 = ? 0 ? 0 ? 2 = 477 ?𝑃? 1000 −0.144236 = 1291.895210 ?𝑃?

(10) Calculate life remaining after cycle 2: 𝜎 ?? = 𝜎 ? 1−𝜎 ? /𝜎 ? ′ = 𝜎 ? = 260 ?𝑃? (eq. 6-60) ? 2 = ( 𝜎 ?? ? 2 ) 1 ? 2 = ( 260 ?𝑃? 1291.895210 ?𝑃? ) − 1 0.144236 = 67170 cycles (eq. 6-15) ? 2 ′ = ? 2 − 5*10 4 = 67170 − 5*10 4 = 17170 cycles (11) Calculate updated S-N curve coefficients after cycle 2: ? ? = ?? ? (eq. 6-12) ? 2 = ? 3 ? 2 ′ ? 3 ? 0 = ? 3 ? 0 ? 3  ? 2 ? 0 = ? 2 ′ ? 3 ? 0 ? 3  ? 2 ? 0 = ( ? 2 ′ ? 0 ) ? 3  ? 3 = log(? 2 /? 0 ) log(? 2 ′ /? 0 ) = log(260/477) log(17170/10 3 ) = −0.213436 ? 3 = ? 0 ? 0 ? 3 = 477 ?𝑃? 1000 −0.213436 = 2083.663749 ?𝑃? (12) Calculate the number of cycles at the final loading condition: 𝜎 ?? = 𝜎 ? 1−𝜎 ? /𝜎 ? ′ = 𝜎 ? = 260 ?𝑃? (eq. 6-60) ? 3 = ( 𝜎 ?? ? 3 ) 1 ? 3 = ( 225 ?𝑃? 2083.663749 ?𝑃? ) − 1 0.213436 = 33803 cycles (eq. 6-15)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version