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Chapter 3

Q. 3.7

Determine the reactions at the supports for the beam shown in Fig. 3.21(a).

Step-by-Step

Verified Solution

Free-Body Diagram See Fig. 3.21(b).

Static Determinacy The beam is internally unstable. It is composed of three rigid members, AB , BE, and EF, connected by two internal hinges at B and E. The structure has r = 5 and e_c = 2; because r = 3 + e_c, the structure is statically determinate.

Support Reactions

+\longrightarrow \sum{F_x}=0

 

A_x=0                   Ans.

Next, we apply the equation of condition, \sum{M^{AB}_B}=0 , which involves the summation of moments about B of all the forces acting on the portion AB.

+\circlearrowleft \sum{M^{AB}_B}=0

 

-A_y(20) + [5(20)](10) = 0

 

A_y = 50 kN

 

A_y = 50 kN \uparrow                                    Ans.

Similarly, by applying the equation of condition \sum{M^{EF}_E}=0 , we determine the reaction F_y as follows:

+\circlearrowleft \sum{M^{EF}_E}=0

 

-[3(20)](10) + F_y(20) = 0

 

F_y = 30 kN

 

F_y = 30 kN \uparrow                                 Ans.

The remaining two equilibrium equations can now be applied to determine the remaining two unknowns, C_y and D_y:

+\circlearrowleft \sum{M_D}=0

 

-50(90) + [5(40)](70) – C_y(50) + [3(90)](5) + 30(40) = 0

 

C_y = 241 kN

 

C_y = 241 kN \uparrow                         Ans.

It is important to realize that the moment equilibrium equations involve the moments of all the forces acting on the entire structure, whereas, the moment equations of condition involve only the moments of those forces that act on the portion of the structure on one side of the internal hinge.

Finally, we compute D_y by using the equilibrium equation,

+\uparrow \sum{F_y}=0

 

50 – 5(40) + 241 – 3(90) + D_y + 30 = 0

 

D_y = 149 kN

 

D_y = 149 kN \uparrow                                   Ans.

Checking Computations

+\circlearrowleft \sum{M_F}=-50(130) + [5(40)](110) – 241(90) + [3(90)](45) – 149(40)

 

= 0                                    Checks