# Reactions at the beam supports

• Guillem_dlc
In summary, the problem involves finding the point of the centre of gravity for a system with three different loads applied at different points. By using the equations for summing forces and moments, we can calculate the values for B and C, which represent the forces at the supports of the beam. However, the calculated location for the centroid of area 2 is incorrect and should be adjusted to account for the weight distribution of the triangle load.

#### Guillem_dlc

Homework Statement
For the given loads, determine the reactions at the beam supports.
Relevant Equations
##\sum F=0, \sum M=0##
Figure:

My attempt at a solution:

We know that ##Q=A_T##
We calculate ##Q##:
$$Q=\dfrac{3\cdot 480}{2}+\dfrac{600\cdot 6}{2}+600\cdot 2=3720\, \textrm{lb}$$
Then we look for the point ##\overline{x}## of the centre of gravity:
$$\overline{x_1}=1\, \textrm{ft},\quad \overline{x_2}=3+\dfrac63=5\, \textrm{ft},\quad \overline{x_3}=3+6+\dfrac22=10\, \textrm{ft}$$
$$\overline{x}=\dfrac{\sum x_iQ_i}{Q}=5,84\, \textrm{ft}$$
$$\sum Fx=\boxed{Bx=0}$$
$$\sum Fy=0=By+C-Q=0\rightarrow \boxed{By=1959,2\, \textrm{lb}}$$
$$\sum M_B=2,84Q-6C=0\rightarrow \boxed{C=1760,8\, \textrm{lb}}$$

Would this not be the case in this one? It's just that the solution tells me the following and I don't get that centre of gravity in area 2:
Official solution:

We have
$$R_I=\dfrac12 (3\, \textrm{ft})(480\, \textrm{lb}/\textrm{ft})=720\, \textrm{lb}$$
$$R_{II}=\dfrac12 (6\, \textrm{ft})(600\, \textrm{lb}/\textrm{ft})=1800\, \textrm{lb}$$
$$R_{III}=(2\, \textrm{ft})(600\, \textrm{lb}/\textrm{ft})=1200\, \textrm{lb}$$
Then
$$\xrightarrow{+}\sum F_x=0:\,\, B_x=0$$
$$\sum M_B=0:\,\, (2\, \textrm{ft})(720\, \textrm{lb})-(4\, \textrm{ft})(1800\, \textrm{lb})+(6\, \textrm{ft})C_y-(7\, \textrm{ft})(1200\, \textrm{lb})=0$$
$$C_y=2360\, \textrm{lb}\qquad \mathbf{C}=2360\, \textrm{lb} \uparrow$$
or
$$\sum F_y=0:\,\, -720\, \textrm{lb}+B_y-1800\, \textrm{lb}+2360\, \textrm{lb}-1200\, \textrm{lb}=0$$
$$B_y=1360\, \textrm{lb}\qquad \mathbf{B}=1360\, \textrm{lb}\uparrow$$

The effective loads are applied at the centroids of the areas. Check ##x_2##

Last edited:
Lnewqban
The location that you have calculated for the centroid of A2 is not correct.
Much of the weight is located towards the right side of the 11-foot beam, therefore the calculated concentrated total weight should be located far from 5.84 feet from the left end.

Of course, I have taken it as if the triangle is rotated. It should be ##2/3##

Lnewqban