example: 2.the compound beam shown in fig is pin connected at b.determine the components of reaction at its…

Explanation:

Step1: Analyze vertical - force equilibrium for whole beam

$\sum F_y = 0$, let the vertical reaction at $A$ be $A_y$, at $C$ be $C_y$. The vertical component of the $10 - kN$ force is $10\times\frac{3}{5}=6\ kN$ (down - ward), and the distributed load is $4\ kN/m\times2\ m = 8\ kN$ (down - ward). So $A_y + C_y-6 - 8=0$, or $A_y + C_y=14\ kN$.

Step2: Analyze moment equilibrium about point $A$

$\sum M_A = 0$. The moment of the vertical component of the $10 - kN$ force about $A$ is $6\times(2)\ kN\cdot m$ (clock - wise), the moment of the distributed load about $A$ is $8\times(2 + 1)\ kN\cdot m$ (clock - wise), and the moment of the reaction at $C$ about $A$ is $C_y\times(2 + 2)\ kN\cdot m$ (counter - clockwise). So $6\times2+8\times3 - 4C_y=0$. $12 + 24-4C_y = 0$, $36-4C_y = 0$, then $C_y = 9\ kN$.

Step3: Find the vertical reaction at $A$

Substitute $C_y = 9\ kN$ into $A_y + C_y=14\ kN$. We get $A_y=14 - 9=5\ kN$.

Step4: Analyze horizontal - force equilibrium for whole beam

$\sum F_x = 0$. The horizontal component of the $10 - kN$ force is $10\times\frac{4}{5}=8\ kN$ (to the right). Let the horizontal reaction at $A$ be $A_x$. So $A_x-8 = 0$, then $A_x = 8\ kN$.

Answer:

$A_x = 8\ kN$, $A_y = 5\ kN$, $C_y = 9\ kN$