Chapter_6_Solutions

# Chapter_6_Solutions - Chapter 6 Eq(2-21 Eq(6-8 Table 6-2...

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Chapter 6 6-1 Eq. (2-21): 3.4 3.4(300) 1020 MPa ut B SH  Eq. (6-8): 0.5 0.5(1020) 510 MPa eu t SS Table 6-2: 1.58, 0.085 ab Eq. (6-19): 0.085 1.58(1020) 0.877 b au t ka S Eq. (6-20): 0.107 0.107 1.24 1.24(10) 0.969 b kd  Eq. (6-18): (0.877)(0.969)(510) 433 MPa . ea b e Sk k S A n s ______________________________________________________________________________ 6-2 (a) Table A-20: S ut = 80 kpsi Eq. (6-8): 0.5(80) 40 kpsi . e SA n s n s n s (b) Table A-20: S ut = 90 kpsi Eq. (6-8): 0.5(90) 45 kpsi . e (c) Aluminum has no endurance limit. Ans. (d) Eq. (6-8): S ut > 200 kpsi, 100 kpsi . e   ______________________________________________________________________________ 6-3 rev 120 kpsi, 70 kpsi ut S Fig. 6-18: 0.82 f Eq. (6-8): 0.5(120) 60 kpsi ee Eq. (6-14):   2 2 0.82(120) () 161.4 kpsi 60 ut e fS a S Eq. (6-15): 1 1 0.82(120) log log 0.0716 33 6 0 ut e b S     Eq. (6-16): 1 1/ 0.0716 rev 70 116 700 cycles . 161.4 b NA a n s ______________________________________________________________________________ 6-4 rev 1600 MPa, 900 MPa ut S Fig. 6-18: S ut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77. Eq. (6-8): S ut > 1400 MPa, so S e = 700 MPa Eq. (6-14):   2 2 0.77(1600) 2168.3 MPa 700 ut e a S Chapter 6 - Rev. A, Page 1/66

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Eq. (6-15): 1 1 0.77(1600) log log 0.081838 3 3 700 ut e fS b S     Eq. (6-16): 1 1/ 0.081838 rev 900 46 400 cycles . 2168.3 b NA a  n s ______________________________________________________________________________ 6-5 230 kpsi, 150 000 cycles ut SN Fig. 6-18, point is off the graph, so estimate: f = 0.77 Eq. (6-8): S ut > 200 kpsi, so 100 kpsi ee SS   Eq. (6-14):   2 2 0.77(230) () 313.6 kpsi 100 ut e a S Eq. (6-15): 1 1 0.77(230) log log 0.08274 3 3 100 ut e b S Eq. (6-13): 0.08274 313.6(150 000) 117.0 kpsi . b f Sa N A n s ______________________________________________________________________________ 6-6 = 160 kpsi 1100 MPa ut S Fig. 6-18: f = 0.79 Eq. (6-8): 0.5(1100) 550 MPa Eq. (6-14):   2 2 0.79(1100) 1373 MPa 550 ut e a S Eq. (6-15): 1 1 0.79(1100) log log 0.06622 3 3 550 ut e b S Eq. (6-13): 0.06622 1373(150 000) 624 MPa . b f N A n s ______________________________________________________________________________ 6-7 150 kpsi, 135 kpsi, 500 cycles ut yt SSN Fig. 6-18: f = 0.798 From Fig. 6-10, we note that below 10 3 cycles on the S-N diagram constitutes the low- cycle region, in which Eq. (6-17) is applicable. Chapter 6 - Rev. A, Page 2/66
Eq. (6-17):  log 0.798 /3 log /3 150 500 122 kpsi . f fu t SS N A n s    The testing should be done at a completely reversed stress of 122 kpsi, which is below the yield strength, so it is possible. Ans. ______________________________________________________________________________ 6-8 The general equation for a line on a log S f - log N scale is S f = aN b , which is Eq. (6-13). By taking the log of both sides, we can get the equation of the line in slope-intercept form. log log log f SbN  a a Substitute the two known points to solve for unknowns a and b . Substituting point (1, S ut ), log log(1) log ut Sb From which . Substituting point ut aS 3 (10 , ) and ut ut f Sa S 3 log log10 log ut ut f S From which 1/3 log bf (log )/3 3 1 10 f t N N  N N ______________________________________________________________________________ 6-9 Read from graph: From 36 10 ,90 and (10 ,50). b N 11 22 log log log log log log b b From which 12 2 21 log log log log log log / SN a NN 1 63 log90log10 log50log10 log10 /10 2.2095 log 2.2095 0.0851 3 6 10 10 162.0 kpsi log50 / 90 0.0851 3 ( ) 162 10 10 in kpsi . a fa x a b N A n s Chapter 6 - Rev. A, Page 3/66

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Check: 3 6 3 0.0851 10 6 0.0851 10 ( ) 162(10 ) 90 kpsi ( ) 162(10 ) 50 kpsi fa x x S S    The end points agree. ______________________________________________________________________________ 6-10 d = 1.5 in, S ut = 110 kpsi Eq. (6-8): 0.5(110) 55 kpsi e S Table 6-2: a = 2.70, b = 0.265 Eq. (6-19): 0.265 2.70(110) 0.777 b au t ka S Since the loading situation is not specified, we’ll assume rotating bending or torsion so Eq. (6-20) is applicable. This would be the worst case.
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## This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_6_Solutions - Chapter 6 Eq(2-21 Eq(6-8 Table 6-2...

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