ME 5463-001 Fracture Mechanics Exam #1 — David Wagner 2009/10/11 14:16

1. b) what are the boundary conditions (e.g. external loads) over the region bounded by lines of y = 0, y = d, and x = 0?

From part a),
$sigma_x = {6P}/{d^3} (2 x^2 y -d x^2)$ ; $sigma_y = {2P}/{d^3} (2 y^3 -3 d y^2)$ ; $tau_{xy} = -{12P}/{d^3} (xy^2 -d xy)$ .

At y=0

$sigma_x = {6P}/{d^3} (2 x^2 y -d x^2)$ = ${6P}/{d^3} (0 -d x^2)$ = $-{6Px^2}/{d^2}$ ;

$sigma_y = {2P}/{d^3} (2 y^3 -3 d y^2)$ = ${2P}/{d^3} (0)$ = 0;

$tau_{xy} = -{12P}/{d^3} (xy^2 -d xy)$ = $-{12P}/{d^3} (0)$ = 0.

At y=d

$sigma_x = {6P}/{d^3} (2 x^2 y -d x^2)$ = ${6P}/{d^3} (2 x^2 d -d x^2)$ = ${6P}/{d^2} (x^2)$ = ${6Px^2}/{d^2}$ ;

$sigma_y = {2P}/{d^3} (2 y^3 -3 d y^2)$ = ${2P}/{d^3} (2 d^3 -3 d^3)$ = {2P} (2 -3) = -2P ;

$tau_{xy} = -{12P}/{d^3} (xy^2 -d xy)$ = $-{12P}/{d^3} (xd^2 -xd^2)$ = 0.

$sigma_x = {6P}/{d^3} (2 x^2 y -d x^2)$ = ${6P}/{d^3} (0)$ = 0;

$sigma_y = {2P}/{d^3} (2 y^3 -3 d y^2)$ = ${2P}/{d^3} (2 y^3 -3 d y^2)$ ;

$tau_{xy} = -{12P}/{d^3} (xy^2 -d xy)$ = $-{12P}/{d^3} (0)$ = 0.