problem 7.6 - determine all forces acting on member fcb. the 2 - kip force is applied to pin f.

Explanation:

Step1: Take moments about point E

Let the vertical reaction at B be $V_B$ and the horizontal and vertical reactions at E be $H_E$ and $V_E$ respectively. The sum of moments about E, $\sum M_E=0$. The 4 - kip - ft couple at A, the 2 - kip force at F and the reaction at B contribute to the moments. The moment due to the 4 - kip - ft couple is 4 kip - ft (counter - clockwise). The moment due to the 2 - kip force at F about E is $2\times(6 + 3)$ kip - ft (counter - clockwise) and the moment due to $V_B$ about E is $V_B\times(8 + 4+3)$ ft. $$4+2\times(6 + 3)-V_B\times(8 + 4+3)=0$$ $$4 + 18-15V_B=0$$ $$15V_B=22$$ $$V_B=\frac{22}{15}\text{ kip}$$

Step2: Consider the equilibrium of forces in the x - direction for the whole structure

$\sum F_x = 0$. The only horizontal force acting on the whole structure is the 2 - kip force at F. So, $H_E = 2$ kip.

Step3: Consider the equilibrium of forces in the y - direction for the whole structure

$\sum F_y=0$. Let's consider the vertical forces. $V_B+V_E = 0$. Since $V_B=\frac{22}{15}$ kip, then $V_E=-\frac{22}{15}$ kip.

Step4: Analyze member FCB

For member FCB, at point B, there is a vertical reaction $V_B=\frac{22}{15}\text{ kip}$ (upward). At point C, there are internal forces. To find the forces at C, we can consider the equilibrium of a section just to the right of C. Let the shear force at C be $V_C$ and the bending moment at C be $M_C$. Taking moments about C for the part of the structure to the right of C (including the 2 - kip force at F). The moment due to the 2 - kip force at F about C is $2\times6$ kip - ft (counter - clockwise) and the moment due to $V_B$ about C is $V_B\times8$ ft (clockwise). $M_C=2\times6 - V_B\times8$ Substitute $V_B=\frac{22}{15}$ into the above formula: $M_C = 12-\frac{22}{15}\times8=12-\frac{176}{15}=\frac{180 - 176}{15}=\frac{4}{15}\text{ kip - ft}$ The shear force $V_C$ can be found by considering the vertical forces on the part of the structure to the right of C. $V_C=2 - V_B=2-\frac{22}{15}=\frac{30 - 22}{15}=\frac{8}{15}\text{ kip}$

The forces acting on member FCB are: a vertical reaction of $\frac{22}{15}\text{ kip}$ at B (upward), a shear force of $\frac{8}{15}\text{ kip}$ and a bending moment of $\frac{4}{15}\text{ kip - ft}$ at C.

Answer:

Vertical reaction at B: $\frac{22}{15}\text{ kip}$ (upward), Shear force at C: $\frac{8}{15}\text{ kip}$, Bending moment at C: $\frac{4}{15}\text{ kip - ft}$