142 Mechanics of Materials $6.2 W Free'moment diagram Fixing moment diagrom M、 ⊙ Total B.M.diagram WL Γ Fig.6.1. By symmetry the fixing moments are equal at both ends.Now from eqn.(6.1) Aa+Ap=0 WL 支×L×年=-ML WL M= 8 (6.3) The B.M.diagram is therefore as shown in Fig.6.1,the maximum B.M.occurring at both the ends and the centre. Applying Mohr's second theorem for the deflection at mid-span, first moment of area of B.M.diagram between centre and 1 one end about the centre E =[经✉竖xL)*)+(竖x)] 贤+]紧既] WL =-192E1 (i.e.downward deflection) (6.4) 6.2.Built-in beam carrying uniformly distributed load across the span Consider now the uniformly loaded beam of Fig.6.2.142 Mechanics of Materials 46.2 iA- Free' ment diagram Fixing moment diagmm %I1 M=-% 8 Fig. 6.1. By symmetry the fixing moments are equal at both ends. Now from eqn. (6.1) A,+& = 0 .. WL 3 x LX -= -ML 4 The B.M. diagram is therefore as shown in Fig. 6.1, the maximum B.M. occurring at both the ends and the centre. Applying Mohr's second theorem for the deflection at mid-span, first moment of area of B.M. diagram between centre and one end about the centre 6=[ 1L ML L 1 [ WL~ ML~ 1 W L ~ W L ~ EZ 96 8 El 96 (i.e. downward deflection) WLJ 192EZ - -- 6.2. Built-in beam carrying uniformly distributed load across the span Consider now the uniformly loaded beam of Fig. 6.2