Built-in Beams 149 and,dividing areas A2 and A4 into the convenient triangles shown, 67.5×103×15)+(生×28.8×103×1.82×318+（生×28.8×103×121.8+青×1.2 +(2MA×3×号×3)+（全MB×3×3×3)=(位×14.4×103×1.2）号×1.2 +×144x0x1(2+》 (101.25+31.1+38.0)103+1.5M4+3MB=(6.92+23.3)103 1.5M4+3MB=-140×103 MA+2Mg=-93.4×103 (2) (2)-(1, MB=-34×103Nm=-34kNm and from (1), M4=-25.4×103Nm=-25.4kNm The fixing moments are therefore negative and not positive as assumed in Fig.6.7.The total B.M.diagram is then found by combining all the separate loading diagrams and the fixing moment diagram to produce the result shown in Fig.6.7.It will be seen that the maximum B.M.occurs at the built-in end B and has a value of 34kNm.This will therefore be the position of the maximum bending stress also,the value being determined from the simple bending theory 0n=M_34×103×100×10-3 42×10-6 =81×106=81MN/m2 Example 6.2 A built-in beam,4 m long,carries combined uniformly distributed and concentrated loads as shown in Fig.6.8.Determine the end reactions,the fixing moments at the built-in supports and the magnitude of the deflection under the 40kN load.Take EI =14 MNm2. 30 kN/m 40kN Fig.6.8. Solution Using Macaulay's method (see page 106) EI d2y M最-M+Rx-0x-1-0[-6]Built-in Beams 149 and, dividing areas A, and A, into the convenient triangles shown, 2 x 1.8 3 (67.5 x IO3 x 1.5)+ (3 x 28.8 x lo3 x 1.8)- + (5 x 28.8 x lo3 x 1.2)(1.8+ $ X 1.2) + (3 MA x 3 x $ x 3) + (f MB x 3 x 5 x 3) = (4 x 14.4 x lo3 x 12)3 x 1.2 + (f x 14.4 x lo3 x 1.8) 1.2 + - ( Y) (101.25 + 31.1 + 38.0)103 + 1.5MA+ 3MB = (6.92 + 23.3)103 1.5 MA+ 3MB = - 140 X lo3 MA + 2MB = - 93.4 x lo3 (2) M B = - 3 4 x 1 0 3 N m = -34kNm and from (l), The fixing moments are therefore negative and not positive as assumed in Fig. 6.7. The total B.M. diagram is then found by combining all the separate loading diagrams and the fixing moment diagram to produce the result shown in Fig. 6.7. It will be seen that the maximum B.M. occurs at the built-in end B and has a value of 34kNm. This will therefore be the position of the maximum bending stress also, the value being determined from the simple bending theory MA= -25.4~ 103Nm = -25.4kNm MY 34 x 103 x io0 x 10-3 omax= - - I 42 x = 81 x lo6 = 81 MN/mZ Example 6.2 A built-in beam, 4 m long, carries combined uniformly distributed and concentrated loads as shown in Fig. 6.8. Determine the end reactions, the fixing moments at the built-in supports and the magnitude of the deflection under the 40 kN load. Take El = 14 MN m2. 30 kN/m 40 kN I X Fig. 6.8. Solution Using Macaulay's method (see page 106)