Gnl

Now for the resultant force across the section to be zero, F\ + F2 = Fi + F 4

a-7 < 1 « 1.35(113 — y)2 135(238 - 2y) 67.5 + 1.35y ---1--

61.5y + 1.35y2 = 17.24 x 103 - 305y + 1.35/ + 32.13 x 103 - 210y 642.5 y = 49370 y — 76.8 mm

Substituting back,

F\ = 67.5 kN F2 = 103.7 kN Ft, — 23 kN F4 = 148.1 kN

The moment of resistance of the beam can now be obtained by taking the moments of these forces about the N.A. Here, for ease of calculation, it is assumed that F4 acts at the mid-point of the web. This, in most cases, is sufficiently accurate for practical purposes.

3(113-30

Moment of resistance = |

= (6030 + 5312 + 554 + 6243)10"3 kNm = 18.14 kNm

Now the maximum B.M. present on a cantilever carrying a u.d.l. is wL2/2 at the support wL2 ,

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