Three-moment equation for spans (1) and (2)
M1L1+2M2(L1+L2)+M3L2+6A1ˉa1L1+6A2ˉb2L2=0
Where,
M1=0
L1=L2=4 m
6A1ˉa1L1=−ML(3a2−L2)=−84[3(22)−42]=8 kN⋅m2
6A2ˉb2L2=14woL3=14(2)(43)=32 kN⋅m2
Thus,
0+2M2(4+4)+M3(4)+8+32=0
16M2+4M3=−40 ← equation (1)
Apply three-moment equation to spans (2) and (3)
M2L2+2M3(L2+L3)+M4L3+6A2ˉa2L2+6A3ˉb3L3=0
Where,
L2=4 m
L3=3 ft
M4=0
6A2ˉa2L214woL3=14(2)(43)=32 kN⋅m2
6A3ˉb3L3=PbL(L2−b2)=6(2)b3(32−22)=20 kN⋅m2
Thus,
M2(4)+2M3(4+3)+0+32+20=0
4M2+14M3=−52 ← equation (2)
From equations (1) and (2)
M2=−1.6923 kN⋅m=−1692.3 N⋅m answer
M3=−3.2307 kN⋅m=−3230.7 N⋅m answer