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## Q. 3.7

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

## 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