Reaction Forces on a Beam

Question

I'm new to beams and I'm struggling to find the correct vertical reactions forces at support C and E.

The correct answer is 8.58 kN for $V_c$ and -3.58 kN for $V_e$

However, when I calculate the sum of the moments and set equal to 0 I get

$$\begin{gather} 1.5 + 2V_c - (3\cdot3) + (2\cdot(2+\frac{2}{3}))= 0 \\ V_c = 2.4\text{ kN} \therefore V_e= 2.6\text{ kN} \end{gather}$$

What is it that I'm doing wrong in my calculation?

Answer 1

You have your signs wrong and the CG of distributed load is at 3.33m from E.

Assuming counter clockwise moment positive, and the reaction direction up positive:

$ M_E=0 \rightarrow \ 1.5 +2*3.33 + 3*3 -2 V_c =0 $

$ V_c=\frac{1.5+6.66+9}{2}= \frac{17.16}{2}=8.58kN $

I leave the rest to you.

Answer 2

Step one:

Choose a right handed coordinate system. As you can see in my schematic:

The $X$ axis lies along the beam:

So, in $ZX$ plane, the counterclockwise moment is positive.

Step 2:

Horizontal and vertical equilibrium:

The horizontal equilibrium is evident so i skip it.

The vertical equilibrium:

$$R_E +R_C + -2 [\frac{kN}{m}]\times 2 \times \frac{1}{2}-3 [kN] = 0 $$

$R_C$ and $R_E$ are the reaction forces, i demonstrate those as they point upward but you can choose those pointing down. The third term represents the resultant of the distributed load, (the area under curve), notice its point of application.

Moment equilibrium:

I choose point $C$ as the reference.

$$1.5 [kN.m]+3[kN] \times 1[m]+2[kN] \times \frac{4}{3}[kN.m]+V_E \times 2 [m]=0$$

Now we can eventually find the value of $R_E$ and substituting it in the first equation gives us the value of $R_C$.

Answer 3

Here are the mistakes:

1. The triangle load acts on the point 2 + 4/3 = 10/3 or 3 + 1/3. That's the load center, the baricenter of the triangle.

2. The sign of this term should be equal to the sign of 3*3 because both loads tend to bend the beam counter-clowisely.