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This material is copyright - all rights reserved. Reproduced under licence from The Steel Construction Institute on 12/2/2007 To buy a hardcopy version of this document call 01344 872775 or go to http://shop.steelbiz.org/ Steel Designers' Manual - 6th Edition (2003) 332 Beam analysis 160 kNm 1 93.3 kN LA 4.Om 20m I- I 6.Om 360 kN load ]1320 kNm 12667 Fig. 10.6 Bending moment and shear force diagrams for fixed-end beam bending moment It will be seen that for symmetrical loads, where MA = MB, the reactions will be the same as for simply-supported beams. An example of bending moment and shear force diagrams for a built-in beam carrying a point load is given in Fig. 10.6. 10.4 Continuous beams The solution of this type of beam consists, in the first instance, of the evaluation of the fixing or negative moments at the supports. The most general method is the use of Clapeyron’s Theorem of Three Moments. The theorem applies only to any two adjacent spans in a continuous beam and in its simplest form deals with a beam which has all the supports at the same level, and has a constant section throughout its length. The proof of the theorem results in the following expression: Ê A1 ¥ x1 A2 ¥ x2ˆ MA ¥ L1+ 2MB( L1 + L2)+ MC ¥ L2 = 6 + Ë L L ¯ 1 2 shear
This material is copyright - all rights reserved. Reproduced under licence from The Steel Construction Institute on 12/2/2007 To buy a hardcopy version of this document call 01344 872775 or go to http://shop.steelbiz.org/ Steel Designers' Manual - 6th Edition (2003) where MA, MB and MC are the numerical values of the hogging moments at the supports A, B and C respectively, and the remaining terms are illustrated in Fig. 10.7. In a continuous beam the conditions at the end supports are usually known, and these conditions provide starting points for the solution. The types of end conditions are three in number: (1) simply supported (2) partially fixed, e.g. a cantilever (3) completely fixed, i.e. the end of the beam is horizontal as in the case of a fixed beam. The SF at the end of any span is calculated after the support moments have been evaluated, in the same manner as for a fixed beam, each span being treated separately. It is essential to note the difference between SF and reaction at any support, e.g. with reference to Fig. 10.7 the SF at support B due to span AB added to the SF at B due to span BC is equal to the total reaction at the support. If the section of the beam is not constant over its whole length, but remains constant for each span, the expression for the moments is rewritten as follows: M L M I L L M I I L 1 Ê 1 2 ˆ 2 Ê A1 ¥ x1 A2 ¥ x2ˆ ¥ + 2 + + ¥ = 6 + Ë ¯ I Ë L ¥ I L ¥ I ¯ A B C 1 1 2 any load system fA L1 L2 I area A1 area A2 x1 - Fig. 10.7 Clapeyron’s Theorem of Three Moments 2 1 1 Continuous beams 333 2 2 c_f
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