Question

name: quiz #03 ce 23100 - statics prof. saldarriaga problem: find the reactions at pin support a and at roller support c for the simply supported beam under the loads shown below. the weight of the beam is negligible.

Explanation:

Step1: Sum of vertical forces

Let the vertical reaction at $A$ be $A_y$ and at $C$ be $C_y$. The sum of vertical forces $\sum F_y=0$. So, $A_y + C_y-10 - 4-2=0$, which simplifies to $A_y + C_y=16$ kips.

Step2: Sum of moments about point $A$

The sum of moments about $A$, $\sum M_A = 0$. Taking counter - clockwise moments as positive, we have $C_y\times(4 + 3)-10\times4-4\times(4 + 3/2)-2\times(4 + 3+3)-624/12=0$.
First, calculate the right - hand side terms:
$10\times4 = 40$, $4\times(4 + 3/2)=4\times5.5 = 22$, $2\times(4 + 3+3)=2\times10 = 20$, and $624/12 = 52$.
The equation becomes $7C_y-40 - 22-20 - 52=0$, or $7C_y-134 = 0$.
Solve for $C_y$: $7C_y=134$, so $C_y=\frac{134}{7}\approx19.14$ kips.

Step3: Find $A_y$

Substitute $C_y$ into the $\sum F_y = 0$ equation. $A_y=16 - C_y$. $A_y=16-\frac{134}{7}=\frac{112 - 134}{7}=-\frac{22}{7}\approx - 3.14$ kips.

Answer:

The vertical reaction at $A$, $A_y\approx - 3.14$ kips and the vertical reaction at $C$, $C_y\approx19.14$ kips.