Determine the midspan value of EIδ for the beam shown in Fig.

Determine the midspan value of EIδ for the beam shown in Fig. 
 

solution

From the load diagram
ΣMR1=0$\mathrm{\Sigma }{M}_{R1}=0$

6R2=12(4)(600)[13(4)]$6R_2=\frac{1}{2}(4)(600)[\frac{1}{3}(4)]$

R2=266.67 N$R_2=266.67\text{ N}$
 

y1=6004${\displaystyle \frac{y}{1}}={\displaystyle \frac{600}{4}}$

y=150 N/m$y=150\text{ N/m}$
 

From the moment diagram
a=3R2=3(266.67)$a=3R_2=3(266.67)$

a=800 N⋅m$a=800\text{ N}\cdot \text{m}$
 

b=−12(1)(y)[13(1)]$b=-\frac{1}{2}(1)(y)[\frac{1}{3}(1)]$

b=−16y=−16(150)$b=-\frac{1}{6}y=-\frac{1}{6}(150)$

b=−25 N⋅m$b=-25\text{ N}\cdot \text{m}$
 

From the conjugate beam
ΣMF1=0$\mathrm{\Sigma }{M}_{F1}=0$

6F2+14(4)(1600)[15(4)]=12(6)(1600)[13(6)]$6F_2+\frac{1}{4}(4)(1600)[\frac{1}{5}(4)]=\frac{1}{2}(6)(1600)[\frac{1}{3}(6)]$

F2=1386.67 N⋅m2$F_2=1386.67\text{ N}\cdot {\text{m}}^2$
 

 

Mmidspan=12(3a)[13(3)]−14(1b)[15(1)]−3F2$M_{midspan}=\frac{1}{2}(3a)[\frac{1}{3}(3)]-\frac{1}{4}(1b)[\frac{1}{5}(1)]-3F_2$

Mmidspan=12(3)(800)[13(3)]−14(1)(25)[15(1)]−3(1386.67)$M_{midspan}=\frac{1}{2}(3)(800)[\frac{1}{3}(3)]-\frac{1}{4}(1)(25)[\frac{1}{5}(1)]-3(1386.67)$

Mmidspan=−2961.25 N⋅m3$M_{midspan}=-2961.25\text{ N}\cdot {\text{m}}^3$
 

Thus, the deflection at the midspan is
EIδm=Mmidspan$EI{\delta }_m=M_{midspan}$

EIδm=−2961.25 N⋅m3$EI{\delta }_m=-2961.25\text{ N}\cdot {\text{m}}^3$

EIδm=2961.25 N⋅m3$EI{\delta }_m=2961.25\text{ N}\cdot {\text{m}}^3$   below the neutral axis           answer