This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: the size of the flaw that initiated fracture. Assume thatl l.l. Sofution: Kr = fa,,lE or a = (t t n)lK6  falz a = (l I x)[2S,ONpsi6 [email protected],OWpsi)]2 = 0.093in. c. 73 ,t' el r{u 7:2il The highstrength steel in Figure 721 is subjected to a stress alternating at20O revolutions per minute between 600 MPa and 200 MPa (both tension). Calculate the growth rate of a surface crack when it reaches a length of 0.2 mm in both m/cycle and m/s. Assume mxf = lX Sotutlon: For the steel, C = L .62 x l0t2 and n = 3.2.'l\e change in the stess intensity factor AK is b,DZ AK  ttt^,I *= qiFfooor"rPa  200MPa\fifroooz tnl =t!llMPa.6il The crack growth rate is dntdN=r.62x 10r'\W;; z,yq N6tq dnldN = !.62x 10r2(ffi)3'2  ffi m/cycle The Sclence and Englneerlng of Materlals 'lnstructor,s Soluflons Manual z.fl fuldt= (€ x lOa m/cycle[20O cycleVmin/ 60 Vmin datdt=E&b ?"b3XtE,? n lS ? q t .  a . q . I...
View
Full Document
 Spring '08
 snively
 Tensile strength, Fracture toughness, The Flaws, strain fracture toughness, srain fracture toughness, initiatedfracture. ssumethatl l.l

Click to edit the document details