Question 6.SP.1: Using the method of joints, determine the force in each memb......

MODELING and ANALYSIS: You can combine these steps for each joint of the truss in turn. Draw a free-body diagram; draw a force polygon or write the equilibrium equations; and solve for the unknown forces.

Entire Truss. Draw a free-body diagram of the entire truss (Fig. 1); external forces acting on this free body are the applied loads and the reactions at C and E. Write the equilibrium equations, taking moments about C.

+↺ \Sigma M_C=0: \quad(2000 ~\mathrm{lb})(24 ~\mathrm{ft})+(1000 ~\mathrm{lb})(12 ~\mathrm{ft})-E(6 ~\mathrm{ft})=0

E=+10,000 ~\mathrm{lb} \quad \mathbf{E}=10,000 ~\mathrm{lb} \uparrow

\stackrel{ +}{\rightarrow} \Sigma F_x=0: \quad \mathbf{C}_x=0

+\uparrow \Sigma F_y=0: \quad-2000 ~\mathrm{lb}-1000 ~\mathrm{lb}+10,000 ~\mathrm{lb}+ C_y=0

C_y=-7000 ~\mathrm{lb} \quad \mathbf{C}_y=7000 ~\mathrm{lb} \downarrow

Joint A. This joint is subject to only two unknown forces: the forces exerted by AB and those by AD. Use a force triangle to determine \mathbf{F}_{A B} and \mathbf{F}_{A D} (Fig. 2). Note that member AB pulls on the joint so AB is in tension, and member AD pushes on the joint so AD is in compression. Obtain the magnitudes of the two forces from the proportion

\frac{2000 ~\mathrm{lb}}{4}=\frac{F_{A B}}{3}=\frac{F_{A D}}{5}

F_{A B}=1500 ~\mathrm{lb} ~T

F_{A D}=2500 ~\mathrm{lb} ~C

Joint D. Since you have already determined the force exerted by member AD, only two unknown forces are now involved at this joint. Again, use a force triangle to determine the unknown forces in members DB and DE (Fig. 3).

F_{D B}=F_{D A} \quad F_{D B}=2500 ~\mathrm{lb} ~T

F_{D E}=2\left(\frac{3}{5}\right) F_{D A} \quad F_{D E}=3000 ~\mathrm{lb} ~C

Joint B. Since more than three forces act at this joint (Fig. 4), determine the two unknown forces \mathbf{F}_{B C} \text { and } \mathbf{F}_{B E} by solving the equilibrium equations \Sigma F_x=0 \text { and } \Sigma F_y=0. Suppose you arbitrarily assume that both unknown forces act away from the joint, i.e., that the members are in tension. The positive value obtained for F_{B C} indicates that this assumption is correct; member BC is in tension. The negative value of F_{B E} indicates that the second assumption is wrong; member BE is in compression.

+\uparrow \Sigma F_y=0: \quad-1000-\frac{4}{5}(2500)-\frac{4}{5} F_{B E}=0

F_{B E}=-3750 ~\mathrm{lb} \quad F_{B E}=3750 ~\mathrm{lb} ~C

\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad F_{B C}-1500-\frac{3}{5}(2500)-\frac{3}{5}(3750)=0

F_{B C}=+5250 ~\mathrm{lb} \quad F_{B C}=5250 ~\mathrm{lb} ~T

Joint E. Assume the unknown force \mathbf{F}_{E C} acts away from the joint (Fig. 5). Summing x components, you obtain

\stackrel{ \pm}{\rightarrow} \Sigma F_x=0: \quad \frac{3}{5} F_{E C}+3000+\frac{3}{5}(3750)=0

F_{E C}=-8750 ~\mathrm{lb} \quad F_{E C}=8750 ~\mathrm{lb} ~C

Summing y components, you obtain a check of your computations:

\begin{aligned}+\uparrow \Sigma F_y & =10,000-\frac{4}{5}(3750)-\frac{4}{5}(8750) \\& =10,000-3000-7000=0\quad \text{(checks)}\end{aligned}

REFLECT and THINK: Using the computed values of \mathbf{F}_{C B} \text { and } \mathbf{F}_{C E}, you can determine the reactions \mathbf{C}_x \text { and } \mathbf{C}_y by considering the equilibrium of Joint C (Fig. 6). Since these reactions have already been determined from the equilibrium of the entire truss, this provides two checks of your com putations. You can also simply use the computed values of all forces acting on the joint (forces in members and reactions) and check that the joint is in equilibrium:

\stackrel{+}{\rightarrow} \Sigma F_x=-5250+\frac{3}{5}(8750)=-5250+5250=0 (checks)

+\uparrow \Sigma F_y=-7000+\frac{4}{5}(8750)=-7000+7000=0 (checks)