For the beam loaded as shown in Fig. P-625, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left.)

 For the beam loaded as shown in Fig. P-625, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left.)

 

Solution

ΣMR2=0$\mathrm{\Sigma }{M}_{R2}=0$

4R1=400(3)(2.5)+500(2)$4R_1=400(3)(2.5)+500(2)$

R1=1000N$R_1=1000\text{N}$
 

ΣMR1=0$\mathrm{\Sigma }{M}_{R1}=0$

4R2=400(3)(1.5)+500(2)$4R_2=400(3)(1.5)+500(2)$

R2=700N$R_2=700\text{N}$
 

 

(AreaAB)X¯A=12(4)(2800)(43)−12(2)(1000)(23)−13(3)(1800)(34)$(Are{a}_{AB}){\overline{X}}_A=\frac{1}{2}(4)(2800)(\frac{4}{3})-\frac{1}{2}(2)(1000)(\frac{2}{3})-\frac{1}{3}(3)(1800)(\frac{3}{4})$

(AreaAB)X¯A=5450N⋅m3$(Are{a}_{AB}){\overline{X}}_A=5450\text{N}\cdot {\text{m}}^3$           answer
 

(AreaAB)X¯B=12(4)(2800)(83)−12(2)(1000)(103)−13(3)(1800)(134)$(Are{a}_{AB}){\overline{X}}_B=\frac{1}{2}(4)(2800)(\frac{8}{3})-\frac{1}{2}(2)(1000)(\frac{10}{3})-\frac{1}{3}(3)(1800)(\frac{13}{4})$

(AreaAB)X¯B=5750N⋅m3$(Are{a}_{AB}){\overline{X}}_B=5750\text{N}\cdot {\text{m}}^3$           answer