148 Mechanics of Materials 40KN 30 kN/m cocoococccccxxeoccooccocxecco 出B 2m+06m+-2m 20 kN Bending moment diagrams 晋3a75w A 8 (o)u.d.. (b)40kN load ab.40x8x1.2:288kNm A2 A3wO2.延：44kNm (c)20 kN load Me--34 kN m Ma=-25.4 kN m Aa (d)Fixing-moment diagram 51.6 28.84 Total bending-moment 25.4 272 3056 diagram on bose of fixing moment line 2104 -1.64 Total bending-moment diagram re-drawn on conventional horizontal -25.4 base Fig.6.7.Illustration of the application of the"principle of superposition"to Mohr's area-moment method of solution. Now from eqn.(6.1) A1+A2+A4=A3 (号×33.75×103×3)+(3×28.8×103×3)+[(M4+M)3]=(生×14.4×103×3) 67.5×103+43.2×103+1.5(M4+M)=21.6×103 MA+MB=-59.4×103 (1) Also,from egn.(6.2),taking moments of area about A, A1X1+A2X2+A44=A3x3148 Mechanics of Materials 40 kN A 0 20 kN I I 1 I M,= -34 kN rn M,=-25.4 kN rn I ---_ -- -- --- -- Bending moment diagrams (a) u.d.1. (b) 40kN load (c) 20 kN load (d) Fixing-moment diagram Total bending- moment diagram on base of fixing moment line Total bending-moment diagram re-drawn on conventiona I horizonta I base -34 Fig. 6.7. Illustration of the application of the “principle of superposition” to Mohr’s area-moment method of solution. Now from eqn. (6.1) A1 + A2 + A4 = A3 (~X33.75X103x3)+(~ ~28.8~10~X3)+[$(MA+M~)3]=(~~14.4~10~~3) 67.5 x lo3 +43.2 x lo3 + 1.5(MA + MB) = 21.6 x lo3 MA + MB = - 59.4 x 103 (1) Also, from eqn. (6.2), taking moments of area about A, A121 + A222 + A424 = A323