written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 12 Marks
Year: Dec 2014
written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 12 Marks
Year: Dec 2014
written 7.8 years ago by |
Fixed End Moment :
FE MAB = + wL² / 12 = 24*4² / 12 = 32 KN-m
FE MBA = - wL² / 12 = - 24*4² / 12 = - 32 KN-m
FE MBC = + wL² / 12 = 20*4² / 12 = 26.67 KN-m
FE MCB = - wL² / 12 = - 20*4² / 12 = - 26.67 KN-m
FE MDB = + wL² / 8 = 20*4² / 8 = + 15 KN-m
FE MBD = - wL / 8 = -30*4 / 8 = - 15 KN-m
Application of Slope deflection Method Equation
Application of condition of Equilibrium ,
ΣM @B = 0
MBA + M BC + MBD = 0
-32 + ΘB EI + 26.67 + ΘB EI + 0.5 ΘC EI - 15 + ΘB EI = 0
3 ΘB EI + 0.5 ΘC EI = 20.33
ΣM @ C = 0 , MCB = 0
0.5 ΘB EI + ΘC EI = 26.67
ΘB = 2.54 / EI
ΘC = 25.4 / EI
Final end moments
MAB = 32 + 0.5 * 2.54 = 33.27 KN-m
MBA = -32 + 2.5 = - 29.46 KN-m
MBC = 26.67 + 2.54 + 0.5 * 25.4 = 41.92 KN-m
MCB = -26.67 + 0.5 * 2.54 + 25.4 = 0
MDB = 15 + 0.5 * 2.54 = 16.27 KN-m
MBD = -15 + 2.54 = -12.46 KN-m