Question

Q 4. A beam is shown in the figure given below where A is hinged, and B and C are roller supports. Use Three-Moment-Theorem to determine the end moments and draw the BMD for the beam. w kN/m P2 kN P1 kN B A D 2EI 2EI EI L4 L3 L3 L2 L1. in Given values: L1=4m, L2=3m, L3=3m, L4=2m, P1=6KN, P2=8KN, W=8kn/m

Answer 1

Solution:- the figure(diagram) given in the question are as follows:

P1=6 kN P2=8 kN w=8 kN/m A 4 m 3m 3 m 3m 2EI TITIE EI 2E 7 m 6 m 2 m

considering each span simply supported-

Calculating bending moment under load:-

P2LI4=8 6/4=12 kN-m (P1*a*b)/L=(6*4~3/7=10.285 kN-m A1 A2 2m N A B D 7 m 6 m 16 kN-m

bending moment at C(Mc)=16 kN-m

A1=(1/2)*10.285*7=35.9975

A1=35.9975

X_L1=7+4/3

X_L1=3.667 m

X_R1=7+3/3

X_R1=3.333 m

A2=(1/2)*12*6=36

A2=36

X_L2=X_R2=6/2=3 m

Apply three moment equation-

for A-B-C

M_A(L1/I1)+M_B(L1/I1+L2/I2)+Mc(L2/I2)+(6*A1*X_L1)/(L1*I1)+(6*A2*X_R2)/(L2*I2)=0 , [Eq-1]

where, M_A=0

L1=7 m , L2=6 m

I1=1 , I2=2

A1=35.9975 , A2=36

values put in above equation-(1)

0+M_B(7+6/2)+Mc(6/2)+(6*35.9975*3.667)/(7*1)+(6*36*3)/(6*2)=0

10*M_B+3*MC+113.145+54=0

10*M_B+3M_C+167.145=0 , [Eq-2]

where, M_C=16 kN-m

10*M_B+3*(-16)+167.145=0

10*M_B+119.145=0

M_B=-119.145/10

M_B=-11.9145 kN-m ,(negative sign represent only anticlockwise moment)

bending moment at support B(M_B)=11.914 kN-m (anticlockwise)

bending moment at A(M_A)=0

bending moment at C(Mc)=16 kN-m , (anticlockwise)

16 kN-m 10.285 kN-m 11.914. in 12 kN-m fixed end moment moment dure to load 2 m D B с 7 m 6 m bending moment diagram

[Ans]