For the beam loaded as shown in Fig. P-632, compute the value of (AreaAB) barred(X)A. From this result, is the tangent drawn to the elastic curve at B directed up or down to the right? (Hint: Refer to the deviation equations and rules of sign.)

 For the beam loaded as shown in Fig. P-632, compute the value of (AreaAB) barred(X)A. From this result, is the tangent drawn to the elastic curve at B directed up or down to the right? (

Solution

ΣMB=0$\mathrm{\Sigma }{M}_B=0$

3R1+200(1)=800(2)(2)$3R_1+200(1)=800(2)(2)$

R1=1000N$R_1=1000\text{N}$
 

ΣMA=0$\mathrm{\Sigma }{M}_A=0$

3R2=200(4)+800(2)(1)$3R_2=200(4)+800(2)(1)$

R1=800N$R_1=800\text{N}$
 

 

(AreaAB)X¯A=12(2)(2000)(43)+12(1)(800)(73)−13(2)(1600)(32)−12(1)(400)(73)−12(1)(200)(83)$(Are{a}_{AB}){\overline{X}}_A=\frac{1}{2}(2)(2000)(\frac{4}{3})+\frac{1}{2}(1)(800)(\frac{7}{3})-\frac{1}{3}(2)(1600)(\frac{3}{2})-\frac{1}{2}(1)(400)(\frac{7}{3})-\frac{1}{2}(1)(200)(\frac{8}{3})$

(AreaAB)X¯A=1266.67N⋅m3$(Are{a}_{AB}){\overline{X}}_A=1266.67\text{N}\cdot {\text{m}}^3$           answer
 

The value of (AreaAB) barred(X)A is positive, therefore point A is above the tangent through B, thus the tangent through B is upward to the right. See the approximate elastic curve shown above and refer to the rules of sign for more information.