Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II

Marks: 12 Marks

Year: Dec 2014

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

1. MAB = FE MAB + 2 EI / L [ 2 ΘA + Θ B ] = 32 + 2 EI * [ ΘB ] / 4
   = 32 + 2EI / 4 [ ΘB ] ………………………... A is fixed , ΘA = 0

2. MBA = FE MBA + 2 EI / L [ ΘA + 2 ΘB ] = -32 + 2EI * [ 2 ΘB ] / 4 = -32 + ΘB EI ………………….……… A is fixed , ΘA = 0

3. MBC = FE MBC + 2 EI / L [ 2 ΘB + ΘC ] = 26.67 + 2 EI * [ 2 ΘB + ΘC ] / 4 = 26.67 + ΘB EI + 0.5 [ ΘC ] EI

4. MCB = FE MCB + 2 EI / L [ ΘB + 2 ΘC ] = - 26.67 + 2 EI [ ΘB + 2 ΘC ] = -26.67 + 0.5 ΘB + [ ΘC ] EI

5. MDB = FE MDB + 2 EI / L [ 2 ΘD + Θ B ] = 15 + 2 EI [ ΘB ] / 4
   = 15 + 0.5 EI [ ΘB ]
   ………………………….. ΘD = 0 as D is Fixed .

6. MBD = FE MBD + 2 EI [ ΘA +2 Θ B ] / L = -15 + 2 EI [2 ΘB ] / 4 = -15 + ΘB EI …………………………… D is Fixed, ΘD = 0.

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