## Ei

Therefore component of deflection perpendicular to the V axis

_ WVL3 _ 10 000 cos 47°5T x 23 " ~ 3EIV ~ 3 x 200 x 109 x 2.27 x 10"6

= 39.4 x 10~3 = 39.4 mm and component of deflection perpendicular to the U axis

_ WUL} _ 10000sin47°51' x 23 " ~ 3EIU ~ 3 x 200 x 109 x 50.43 x 10~6 = 1.96 x 10"3 = 1.96 mm The total deflection is then given by

= yj{82u + S2V) = 10~3a/(39.42 + 1.962) = 39.45 x 10"3 = 39.45 mm

Alternatively, since bending actually occurs about the N.A., the deflection can be found from

its direction being normal to the N.A.

From Mohr's circle of Fig. 1.16, /N A. = 2.39 x 10~6 m4

Example 1.4

Check the answer obtained for the stress at point B on the angle section of Example 1.2 using the momental ellipse procedure.

Solution

The semi-axes of the momental ellipse are given by ku = \ — and k,, = \ — V A V A

The ellipse can then be constructed by setting off the above dimensions on the principal axes as shown in Fig. 1.17 (The inclination of the N.A. can be determined as in Example 1.2 or from eqn. (1.24).) The second moment of area of the section about the N.A. is then obtained from the momental ellipse as

Thus for the angle section of Fig. 1.14

/vv = 1.186 x 10"6 m4, /„ = 4 x 10~6 m4, Iyy = 1.08 x 10~6 m4

The principal second moments of area are then given by Mohr's circle of Fig. 1.18 or from the equation lu, lr = + /.vv) ± (/„ - / vv) sec 29]

where

Tangent ellipse parallel to NA

Tangent ellipse parallel to NA  Fig. 1.18.

28 = —39°5', 9 = —19°33' and sec 29 = -1.2883

0 0