MT-1620 Fall 2002 du dv zk+zk=0 xy dy dx = cross-section does not change shape(as assumed!) dv dw kX W dz dy (10-4) N dw (10-5) Now the Stress-Strain equations let's first do isotropic VOn+σ ZZ V+6 yEI yy Vo、+0 0 Paul A. Lagace @2001 Unt10-p.10MIT - 16.20 Fall, 2002 ∂u ∂v εxy = + = − zk + zk = 0 ∂y ∂x ⇒ cross - section does not change shape (as assumed!) ∂v ∂w ∂w εyz = ∂z + ∂y = kx + ∂y (10 - 4) ∂w ∂u ∂w εzx = ∂x + ∂z = − ky + ∂x (10 - 5) Now the Stress-Strain equations: let’s first do isotropic 1 εxx = E [σxx − ν(σyy + σzz )] = 0 εyy = 1 [σyy − ν(σxx + σzz )] = 0 E 1 εzz = E [σzz − ν(σxx + σyy )] = 0 ⇒ σxx, σyy, σzz = 0 Paul A. Lagace © 2001 Unit 10 - p. 10