1 Derive an expression for the J integral for an axisymmetri
1.
Derive an expression for the J integral for an axisymmetrically notched bar in
tension shown below, where the notch depth is sufficient to confine plastic
deformation to the ligament.
Solution
GIVEN:-
For three zones 1,2 & 3 shown in the sketch in the problem, the three different diameters are D, d and again D.
SOLUTION:-
For Zone-1, Stress intensity factor K1 = 2P / (D)^1.5
For Zone-2 K2 = 4P / (d)^1.5
For Zone-3 K3 = 2P / (D)^1.5
So by Formula J-integral = K^2/E which gives the total J-integral Jtotal = J1 + J2 + J3
That is Jtotal = K1^2/E + K2^2/E + K3^2/E
Therefore, Jtotal = [2P / (D)^1.5]^2/E + [4P / (d)^1.5]^2/E + [2P / (D)^1.5]^2/E (ANSWER)
