Machine Design Problems

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 fil let 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 fac tor (Pg 997 Appendix D Machine Design by Robert Norton ) First we have d/D = .5 2

3

4

5

6

Kta = 3.9702- 9.292(d/D)+27.159(d/D) + 30.231(d/D)   393.19(d/D) + 650.39(d/D) +15.451(d/D) = 5.8846 2

3

4

5

Ktb = 3.92150  24.435(d/D) + 234.06(d/D)  1200.5(d/D) + 3059.5(d/D)  3042.4(d/D) = -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 -.107

Size factor = Ksize = 1.24(20)

(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 = 16T/d

3

Table 1

S. No.

Tmax

Tmin

Tm

Ta

m

a

Sti

1

180

0

90

90

57.32484076

57.32484076

-------

2

500

150

325

175

207.0063694

111.4649682

185.8686

3

300

-25

137.5

162.5

87.57961783

103.5031847

124.6064

4

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 3

log(.9Sut) = k1*log10 + k2 6

log(Se) = k1*log10 + 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)

infinite

i 5

0.227273

50723.48

8

0.363636

173780.08

3

0.136364

186208.714

6

0.272727

SUM  22

i = ni/sum now we will use miners formula  i/Ni = 1/N Where N = total life Solving we get N = 106209.41 cycles

Q2. a) Generally a preload value is determined e xperimentally but a 90% preload is generally safe for all dynamic loading applications For M12 SAE grade 5.8 bolt At = 84.27 mm

2

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)

2

Force applied on the cap = PA = P**d /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/Webcoursecontents/IIT%20Kharagpur/Machine%20design1/pdf/Module-3_lesson-3.pdf  Size 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 )    





   



--------------1

  ------------------2

Solving equations 1 and 2 we have m= 408.798 a = 11.2 tan = a/ m = .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 stifness Fbmin = Fi + Fmin Kb/(kb + kp) = Fi a = (Fbmax Fbmin)/(2*A) = .4Fmax/A m = Fi+.4Fmax/A 2

where A = 84.27 mm now

m/520xE6 + a/52.38xE6  1 solving we get Fmax  3432.694 N Now tan  = a/ m = .045 which is greater than .027 hence this line will be applicable Similarly solving m/420xE6 + a/420xE6  1 solving 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/2t Where 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-eNeVRc1oC&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&resnu m=2&ved=0CBoQ6AEwAQ#v=onepage&q=life%20of%20a%20component%20in%20fluctuating% 20stresses&f=false

pg 178