Page 68 - Civil Engineering Formulas

CASE 11. Beam Overhanging Both Supported Unsymmetrically Placed—Continuous Load, Uniformly Distributed

Rx 1
–
x 1) 2 x 2 ) 2
x) 2 + M M1
a – –
– (a (a (b b
R 2w w w w R 1
1 2 1 2 1 2 x 2
a = = = L
– wa 2 wb 2 M M M W
R w (max) = R 1 = 2 1 R xx 1
x 1 : when x 1 = M M1 M1 = x: when x < a 2 x 1: when x 1 < L x 2: when x 2 < b a M 1

At R: At R1: At At (a + x 1) At At
= load per unit of length [(a + L) 2 + b 2 ] 2L [(b + L) 2 + a 2 ] 2L

W R = w R 1 = w (max) = wa or R – wa x: when x < a V = w (a – x) x 1 : when x 1 < L V = R – w x 2 : when x 2 < b V = w(b – x 2 )
=
L + b
W V
+ At At At
a

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