For AISI 3140 steel (http://www.efunda.com/materials/alloys/alloy_steels/show_alloy.cfm?ID=AISI_3140&prop=all&Page_Title=AISI%203140 this link gives the properties of AISA 3140 steel )

The only thing which is going to decide that whether the failure would occur first at the fillet and not at the hole is the stress concentration factor.

Kt = A(r/D) b

D/d = 2, which gives

A= .90879

b= -.28598

Kt = 1.9

Now for the hole we will consider following tables

In the case of hole

Now for Kt = stress concentration factor (Pg 997 Appendix D Machine Design by Robert Norton )

First we have d/D = .5

Kta = 3.9702- 9.292(d/D)+27.159(d/D)2 + 30.231(d/D)3 – 393.19(d/D)4 + 650.39(d/D)5 +15.451(d/D)6 = 5.8846

Ktb = 3.92150 – 24.435(d/D) + 234.06(d/D)2 – 1200.5(d/D)3 + 3059.5(d/D)4 – 3042.4(d/D)5 = -3.604

Kt = Kta+Ktb = 2.22

So the stress concentration factor is higher in the case of hole so we will consider it for the failure calculation

Sut = 689.5 MPa

For steel bar under torsion

Se’ = .29 Sut ≈ 200 MPa

For 50% reliability, reliability factor

Kr = 1

For ground surface finish, surface finish factor

Ks = 1.58x(689.5)-.085 (slide 238 MEL 311.pdf)

Temperature factor = Ktemp = 1

Size factor = Ksize = 1.24(20)-.107 (slide 241 MEL 311.pdf)

Se = Ks*Kr* Ktemp *KsizeSe’/Kt = 74.45 MPa

Yield stress = Syt = .75 Sut

= .75 * 689.5 = 517.125 MPa

Now stress σ= 16T/πd3

Table 1

S. No. Tmax Tmin Tm Ta σm σa Sti

1 180 0 90 90 57.32484076 57.32484076 -------2 500 150 325 175 207.0063694 111.4649682 185.86863 300 -25 137.5 162.5 87.57961783 103.5031847 124.60644 200 -200 0 200 0 127.388535 127.3885

We can see that case 1 lies in region of safety (and hence imparts infinite life) but case 2,3 don’t lie in region of safety and case 4 is a case of reversed loading . So we will extrapolate the goodman graph and find St which we will find through similar triangles. We will use St in SN line equation to find the life cycle due to individual stresses. We will use the following formula

Sti = OQ = AO*MD/AM 2.791, 1.873

The results are given in table 1

For the SN curve we will solve the following two equations

log(.9Sut) = k1*log103 + k2

log(Se) = k1*log106 + k2

here Sut and Se are in Mpa

Solving the 2 equations for k1 and k2 we get

k1 = -.306

k2 = 3.709

so now we will solve the following equation for cases 2,3 and 4

log(Sti) = -.306logNi + 3.709

this gives following table 2

Ni no of cycles in 10 seconds period (ni)

αi

infinite 5 0.227273

50723.48 8 0.363636

173780.08 3 0.136364

186208.714

6 0.272727

SUM 22

αi = ni/sum

now we will use miner’s formula

Σ αi/Ni = 1/N

Where N = total life

Solving we get N = 106209.41 cycles

Q2.

a) Generally a preload value is determined experimentally but a 90% preload is generally safe for all dynamic loading applications For M12 SAE grade 5.8 bolt At = 84.27 mm2

Sp = proof strength = 380 So Fi = .9SpAt = .9*380xE6*84.27xE-6 = 28.82 kN

This preload can be achieved by torquing the bolt with a torque wrench b) Force applied on the cap = PA = P*π*d2/4 = .071P

Now there are 10 bolts so 10*preload = .071P So P = 10*28820/.071 = 4059154.92 Pa = 4.06MPa

c) Bolt properties for SA grade 5.8 bolt are international standards and hence can be found out in every book Sut = 520 MPa Yield strength = Sy = 420 Mpa For axial loading Se’ = .45Sut =234 For the below values of different factors this link has been referred http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Machine%20design1/pdf/Module-3_lesson-3.pdfSize factor C1 = .85 Load factor C2 = .85 Surface finish factor C3 = .76 (for machined surfaces)Temperature factor C4 = 1 Reliability factor C5 = .897 (90% reliability)Kf = 2.2 (from slide 103 MEL 311 part 2)Se = C1C2C3C4C5Se’/kf = 52.38 Mpa Now we will use modified goodman line and factor of safety 1 we have (refer to slides 104 and 105 of MEL 311 part 2 )σmSut

+ σaSe

=1 --------------1

σmSy

+ σaSy

=1 ------------------2

Solving equations 1 and 2 we have σm= 408.798

σa = 11.2 tanθ = σa/ σm = .027 here F = load on the bolt assembly Fbmax = Fi + Fmax Kb/(kb + kp) = Fi + Fmax [Kb/(kb + .25kb)] = Fi + .8Fmax Fi = preload Kb = bolt stiffness Kp = part stifnessFbmin = Fi + Fmin Kb/(kb + kp) = Fi σa = (Fbmax –Fbmin)/(2*A) = .4Fmax/A σm = Fi+.4Fmax/Awhere A = 84.27 mm2

nowσm/520xE6 + σa/52.38xE6 ≤ 1solving we get Fmax ≤ 3432.694 N Now tan θ = σa/ σm = .045 which is greater than .027 hence this line will be applicable Similarly solving σm/420xE6 + σa/420xE6 ≤ 1solving we get Fmax ≤ 8216.75 N From the two values of Fmax we chose the smaller one Now there are 10 bolts so this Fmax will be multiplied with 10 PA = .071P = 10*3432.694 P = .483 Mpa

d) Maximum stress in a thin walled vessel (whether cylindrical or spherical)is= pr/2tWhere r = radius t = thickness p = pressure assuming the vessel to be of steel and taking the UTS of this steel to be same as that AISI 3140 properties of this steel on this link http://www.efunda.com/materials/alloys/alloy_steels/show_alloy.cfm?ID=AISI_3140&prop=all&Page_Title=AISI%203140 we have UTS = 689.5 Mpa and yield strength = 422 MPa so 422xE6 = pr/2t t = (483000 * .15)/(2*422xE6) = .00009 m = .09 mm

this is the minimum thickness the wall of the vessel must have

References :Robert L. Norton --Machine Design: An integrated Approach Second Edition

MEL 311.pdf MEL 311 part 2.pdf http://books.google.co.in/books?id=d-eNe-VRc1oC&pg=PA176&lpg=PA176&dq=life+of+a+component+in+fluctuating+stresses&source=bl&ots=by9GIcFn3Z&sig=1ggvzVgIcw-B1s5p5kFs92L6Pmg&hl=en&ei=KPVvTJqbJY7qvQPrqvlB&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBoQ6AEwAQ#v=onepage&q=life%20of%20a%20component%20in%20fluctuating%20stresses&f=false pg 178