For the beam loaded as shown in Fig. P-624, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction.

 For the beam loaded as shown in Fig. P-624, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction.

 

 

Solution 
ΣMR2=0$\mathrm{\Sigma }{M}_{R2}=0$

6R1=400+1000(2)$6R_1=400+1000(2)$

R1=400N$R_1=400\text{N}$
 

ΣMR1=0$\mathrm{\Sigma }{M}_{R1}=0$

6R2+400=1000(2)$6R_2+400=1000(2)$

R2=600N