150 Mechanics of Materials Note that the unit of load of kilonewton is conveniently accounted for by dividing EI by 103.It can then be assumed in further calculation that R is in kN and M in kNm. Integrating, x240。 =Mx+R72Tc-1.6]-30-16J+h 103dx and E x2 x340 30 0y=M,2+R6-6[x-1.6]-24[x-1.6]+A+B Now,when x=0,y=0..B=0 and when x=0, 盘=0 ,A=0 When x=4,y=0 0=M,x+Rx石-24-双4 42 4340, 0=8MA+10.67R4-92.16-41.47 133.6=8M4+10.67R Nin' (1) When x=4, dy =0 dx 0=4M,+R-924-9e4 42 0=4Ma+8R4-115.2-69.12 184.32=4MA+8RA (2) Multiply(2)×2, 368.64=8MA+16R (3) (3)-(1), 235.04=5.33R 235.04 RA=5.33 =44.1kN Now RA+Ra=40+(2.4×30)=112kN Rg=112-44.1=67.9kN Substituting in(2), 4M4+352.8=184.32 MA=(184.32-352.8)=-42.12kNm i.e.M is in the opposite direction to that assumed in Fig.6.8.150 Mechanics of Materials Note that the unit of load of kilonewton is conveniently accounted for by dividing EZ by lo3. It can then be assumed in further calculation that RA is in kN and MA in kNm. Integrating, _- EI dy = M,x+RA---[(~-1.6)~]--[(~-1.6)~] x2 40 30 +A lo3 dx 22 6 and EI x2 x3 40 30 2 66 24 my = MA- + RA- - -[ (x- 1.6y] - -[(x- 1.6)4] +AX +B Now, when x =0, y=O ... B=O dY and whenx=O, - =O .’. A =O dx When x =4, y =O 42 43 40 30 2 66 24 0 = MA X - + RA x - - - (2.4)3 - - (2.4)4 0 = 8MA+ 10.67 RA-92.16-41.47 133.6 = 8MA+ 10.67 RA N lq ’ dY When x = 4, - =O dx 42 40 30 2 2 6 0 = MA+ -RA- -(2.4)’- -(2.4)3 Multiply (2) x 2, (3) - (I), Now .. 368.64 = 8MA + 16RA = 44.1 kN 235.04 5.33 RA=- RA+RB = 40-t (2.4 X 30) = 112kN RB = 112 -44.1 = 67.9 kN Substituting in (2), 4MA+ 352.8 = 184.32 .. MA = (184.32 - 352.8) = - 42.12 kN m i.e. MA is in the opposite direction to that assumed in Fig. 6.8