Solutions_2

Substituting for mx and my from eqs ii and iii in eq i

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (l − z) My = 0 In BA Mx = −W (l − z) M y = −2 W l −z 2 (ii) (iii) (i) Now referring to Fig. P.16.4 the centroid of area, C, of the beam cross-section is at the centre of antisymmetry. Then Ixx = 2 td Iyy = 2 td Ixy = td d 4 d 2 d 4 2 + 2 td 3 12 = 2td 3 3 d 4 2 + d 2 td 3 + td 12 d 4 = − d 2 5td 3 12 td 3 4 + td − = Substituting for Ixx , Iyy and Ixy in Eq. (16.18) gives σz = 1 [(3.10My − 1.16Mx )x + (1.94Mx − 1.16My )y] td 3 (iv) Solutions to Chapter 16 Problems 189 Along the edge l, x = 3d /4, y = d /2. Equation (iv) then becomes σz,1 = 1 (1.75My + 0...
View Full Document

Ask a homework question - tutors are online