— David Wagner 2009/10/18 12:39

√ME 5463-001 Fracture Mechanics, Homework #12

If εx'x' is measured as 0.01 using the single strain gage approach with the strain gage placed 50mm away with θ=30° from the crack tip in a loaded body (E=200GPa and nu=0.3), determine the values of α as well as the K-factor.

  • epsilon_{x prime x prime}=0.01.
  • r = 0.050 m.
  • θ=30° ≡ 30° * pi/180 = 0.52360 radian.
  • E = 200E9 Pa.
  • mu = E/{2(1+nu)} = 200E9/{2+0.6} = 76.923e9.
  • k={1-nu}/{1+nu} approx {0.7}/{1.3} ≈ 0.53846.

K is a function of three unknowns: A1, B0, and α.

Since εy'y' is also unknown, there is only one equation available to solve for K, and it can, at best, result in an equation of the form

K_I = 1/{B_0 a_0(alpha) + A_1 a_1(alpha) + a_3(alpha) },

where a_1, a_2, and a_3 can be calculated from the information given.

Desired α

If cos 2 alpha = -k = -{{1 - nu}/{1 + nu}} ⇒ alpha = {1/2}acos{-{{1 - nu}/{1 + nu}}} = {1/2}acos{-{{1 - 0.3}/{1 + 0.3}}} = {1/2}acos{-{{0.7}/{1.3}}} ≈ 1.0697 ≈
α ≈ 1.07 radian, then B0 is not needed.

If tan{theta/2} =-cot 2 alpha ⇒ alpha = {1/2}acot(-tan{theta/2}) ≈ {1/2}acot(-tan{0.26180}) =
α ≈ -0.65450 radian (or 0.91630 rad), then A1 is not needed.

However, the problem does not state that the strain gage is oriented at either of these values for α. And, even it had been, K would still depend on one of the unknowns B_0 or A_1.

Determine K

If the strain gage had been placed at a theta that eliminates A_1, and then oriented at an alpha to eliminate B_0, K could be determined directly from the strain reading as K_I = E sqrt{2 pi r} epsilon_{x prime x prime}, when

  • alpha = 1.0697 radian, about 61°.
  • theta = 2 atan(-cot 2 alpha) = 1.1372 radian, about 65°.

Assuming the remote stress B_0 term is negligible and the strain gage is oriented at α ≈ -0.65450 radian (or 0.91630 rad), thus making the A_1 term zero, K can be calculated.

K_I = {2 mu epsilon_{x prime x prime} sqrt{2 pi r}}/ { k cos{theta/2} - {1/2}sin theta sin{3 theta/2} cos 2 alpha +{1/2}sin theta cos{3 theta/2} sin 2 alpha }. 

epsilon_xpxp = 0.01;
           r = 0.050;
       theta ≡ 30*pi/180;
           E = 200e9;
          nu = 0.3;
          mu = E/(2*(1+nu))
           k = (1-nu)/(1+nu)
       alpha = 1.0697;
 
K_I = 2*mu*epsilon_xpxp*sqrt(2*pi*r)/( k*cos(theta/2)...
 - (1/2)*sin(theta) *sin(3*theta/2) *cos(2*alpha)...
 + (1/2)*sin(theta) *cos(3*theta/2) *sin(2*alpha) )

This make K_I = 1.1283e9, which seems ridiculous, but that is the answer by this formula.

Perhaps Pa is the wrong unit for E?


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