## Info

If the N.A. is drawn as shown in Fig. 1.14 at an angle of 47°41' to the XX axis through the centroid of the section, then this is the axis about which bending takes place. The points of maximum stress are then obtained by inspection as the points which are the maximum perpendicular distance from the N.A.

Thus B is the point of maximum tensile stress and C the point of maximum compressive stress.

Now from eqn (1.19) the stress at any point is given by cr = Px + Qy stress at A = -4897 x 106(57 x 10"3) + 4460 x 106(31 x 10~3)

= —141 MN/m2 (compressive) stress at B = -4897 x 106(—19 x 10-3) + 4460 x 106(44 x 10"3)

-- 289 MN/m2 (tensile) stress at C = -4897 x 106(-6 x 10-3) + 4460 x 106{-83 x 10~3) — —341 MN/m (compressive)

### Example 13

(a) A horizontal cantilever 2 m long is constructed from the Z-section shown in Fig. 1.15. A load of 10 kN is applied to the end of the cantilever at an angle of 60° to the horizontal as shown. Assuming that no twisting moment is applied to the section, determine the stresses at points ^ and B. (7XV x 48.3 x 10"6 m4, 7VV = 4.4 x 1(T6 m4.)

(b) Determine the principal second moments of area of the section and hence, by applying the simple bending theory about each principal axis, check the answers obtained in part (a).

(c) What will be the deflection of the end of the cantilever? E = 200 GN/m2. Fig. 1.15.

Solution

(a) For this section Ixy for the web is zero since its centroid lies on both axes and hence h and k are both zero. The contributions to Ixy of the other two portions will be negative since in both cases either h or k is negative.

Now, at the built-in end,

M, = +10 000 sin 60° x 2 = +17 320 Nm My = —10000cos60° x 2 = -10000 Nm Substituting in eqns. (1.20) and (1.21),

17 320 = PIxy + Qlxx = (-9.91 P + 48.30) 10-6 -10000 = -PIyy - QIxy = (~4.4P + 9.91Q) 10^6

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