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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) 336 Beam analysis P=o -t-— tlength MAt f L (a) H MAf a1 b1 C (b) (c) Fig. 10.9 Plastic failure of a fixed-end beam MB the simply-supported sagging parabolic moment diagram ACB (peak value of wL 2 /8) on the uniform moment diagram Aa 1b 1B due to the end hogging moments M A and M B. From the moment–curvature relation applied over the full length of the beam, it is readily deduced, by any one of a number of standard methods, that the rotations of the end sections at A (q A clockwise) and at B (q B anti-clockwise) and the central deflection D C are given by q A q B D C 3 wL MAL MBL = - - 24EI 3EI 6EI 3 wL MAL MBL = - - 24EI 6EI 3EI 5 wL M L M L = - - 384 EI 16EI 16EI 4 2 2 A B P=o (10.4) (10.5) (10.6) The compatibility conditions that now have to be satisfied are the directional restraint of the end sections of the beam, i.e. qA = q B = 0, giving the well-known
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) result M A = M B = wL 2 /12. Equilibrium considerations now lead to the derivation of the central sagging moment, M C. It follows from Fig. 10.9(b) that (10.7) whence MC = wL2 /24, i.e. the end moments are twice the central moment. Finally, application of Equation (10.3) gives the central deflection as wL4 /384EI. Suppose the load intensity w is increased until the fibres yield at some point in the beam. It is assumed that the cross section has an idealized moment–curvature relationship, i.e. the shape factor is unity (see Section 9.3.5). Up to this stage the ratio of the end to central moments remains at 2, as represented by the dashed line in Fig. 10.9(c). Due to the symmetry of the structure and loading, plastic hinges will develop simultaneously at the fixed ends, i.e. at the points of maximum moment. At the moment when the hinges form at the ends, the fixing moments have become equal to the Mp of the beam and the loading intensity w has reached a value of 12Mp /L2 2 wL Ê MA MBˆ MC = - + 8 Ë 2 2 ¯ . A slight load increase then causes the plastic hinges to rotate while sustaining this constant moment Mp. This means that thereafter the beam behaves as a simply-supported beam with constant end moments of Mp. The structure is now statically determinate, and therefore the two degrees of redundancy in the original problem no longer exist. In other words, two plastic hinges have formed, eliminating a corresponding number of redundancies. At the same time two compatibility requirements have been eliminated; the condition that end slopes are zero is no longer correct because the ends of the member are now rotating as plastic hinges. The central deflection at this stage, derived from Equation (10.6), has the value D C and, from Equation (10.7), the central sagging moment M C becomes [(wL 2 /8) - M p]. Further increases in the load intensity cause a third plastic hinge to form at midspan: see the full line in Fig. 10.9(c). This means that the moment MC attains a value of M p and therefore at C M wL2 = -M 8 hence 5 Ê 12M L = Á 384 Ë L ¯ EI p p ML ML 8EI 32EI 4 2 2 pˆp p ˜ - - 2 M = wL 16 or w= 16M L 2 2 p p Plastic failure of single members 337 The ratio of the end and central moments is now unity; as a result of the formation of the plastic hinges there has been a redistribution of moments. Substituting w = 16M p/L 2 and M A = M B = M p in Equation (10.6) gives a central deflection of D C = M pL 2 /12EI. When the final hinge has formed the deflection increases rapidly without any further increase in the load. The beam is said to have failed as a hinged mechanism.
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