- #1

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- 15

- Homework Statement
- For the given loads, determine the reactions at the beam supports.

- Relevant Equations
- ##\sum F=0, \sum M=0##

Figure:

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:

**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

$$\sum F_y=0:\,\, -720\, \textrm{lb}+B_y-1800\, \textrm{lb}+2360\, \textrm{lb}-1200\, \textrm{lb}=0$$

$$B_y=1360\, \textrm{lb}\qquad \mathbf{B}=1360\, \textrm{lb}\uparrow$$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

$$\sum F_y=0:\,\, -720\, \textrm{lb}+B_y-1800\, \textrm{lb}+2360\, \textrm{lb}-1200\, \textrm{lb}=0$$

$$B_y=1360\, \textrm{lb}\qquad \mathbf{B}=1360\, \textrm{lb}\uparrow$$