√ME 5463-001 Fracture Mechanics, Final Exam #3

— David Wagner 2009/12/08 16:52

A thin-walled cylindrical pressure container (70 mm inner diameter and 90 mm outer diameter) has a semicircular surface flaw (a = 1.5 mm), which is oriented normal to the longitudinal direction of the cylinder. If the maximum pressure is 55 MPa and the minimum pressure is 20 MPa, estimate the number of pressure cycles the cylinder could withstand before failure. Assume that $da/dN = 5*10^{-39} (Delta K)^4$ , the fracture toughness of the material is 20.9 MPa√m.

Note this is not the worst case. A crack oriented parallel to the longitudinal direction would have twice the opening stress applied.

Assume the plate is approximately flat with infinite x- and y-dimensions around the crack.

Given

r = 0.090 m;
t = 0.020 m;
;
m;
Pa√m;
${da}/{dN} = 5*10^{-39} (Delta K)^4 = A(Delta K)^m$ .

Found

Hoop Stress: $sigma_x = {pr}/t$ (does not introduce singular stresses at the crack tip).
Longitudinal Stress: $sigma_y = {pr}/{2t}$ ;
There are no shear stresses.
. Edge condition making half of a penny crack.

Determined

$A = 5*10^{-39}$ ;
m = 4;
;
$p_mean = {55+20}/2 = 37.5*10^6 Pa$ ;
.
$sigma_max = {55*10^6 * 0.090}/{2*0.020}$ = Pa

The crack length at brittle failure will be:

$a_f = {1/pi} ({K_c}/{Y sigma_max})^2$ = $a_f = {1/pi} ({20.9*10^6}/{0.7168 * 123.75*10^6})^2$ = 0.043636 m.

Delta sigma = $Delta sigma_y = {Delta p r}/{2t}$ = ${35*10^6 * 0.090}/{2 * 0.020}$ = 78.75*10^6 Pa

$N_f = 2 { {a_0}^{{2-m}/2} - {a_f}^{{2-m}/2} } / {(m-2)A(Y sqrt{pi} Delta sigma)^m}$ = $2 { {a_0}^{-1} - {a_f}^{-1} } / {2 A(Y sqrt{pi} Delta sigma)^4}$ = ${ {0.0015}^{-1} - 0.043636^{-1} } / {5*10^{-39} * 0.7168^4 * pi^2 * {78.75*10^6}^4}$ = 1.2849*10^9 cycles.

1.3 Billion Cycles