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

STRATEGY: To use the method of joints, you start with an analysis of the free-body diagram of the entire truss. Then, look for a joint connecting only two members as a starting point for the calculations. In this example, we start at joint A and proceed through joints D, B, E, and C, but you could also start at joint C and proceed through joints E, B, D, and A.

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.

+↺ΣM_{C} = 0:        ( 2000 lb )( 24 ft) + ( 1000 lb )( 12 ft) − E( 6 ft) = 0

E = +10,000 lb                           E = 10,000 lb ↑

\underrightarrow{+}\Sigma F_{x}=0\colon                                 C_{x} = 0

+↑ΣF_{y} = 0:      − 2000 lb − 1000 lb + 10,000 lb + C_{y} = 0

C_{y} = − 7000 lb                                 C_{y} = 7000 lb ↓

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 F_{AB}~and~F_{AD} (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 lb T ◂

F_{A D} = 2500 lb C ◂

As an alternative to the force triangle approach, remember that a more general analytic solution can also be used. This alternate method is especially conducive to joint equilibrium problems that involve more than three forces, and is illustrated later in this sample problem for the analysis of joints B, E, and C.

Joint D. Because 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}\qquad\qquad\qquad\qquad\qquad\qquad F_{D B}=2500\,{\mathrm{lb}}~T   ◂

F_{D E}=2({\frac{3}{5}})F_{D A}\qquad\qquad\qquad\qquad\qquad F_{D E}=3000\;\mathrm{lb}\;C   ◂

Joint B. Because more than three forces act at this joint (Fig. 4), determine the two unknown forces \mathbf{F}_{B C}~\mathrm{and\:}\mathbf{F}_{B E} by solving the equilibrium equations ΣF_{x} = 0~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 \mathbf{F}_{B C} indicates that this assumption is correct; member BC is in tension. The negative value of \mathbf{F}_{B E} indicates that the second assumption is wrong; member BE is in compression.

+↑ΣF_{y} = 0: \qquad -1000-{\frac{4}{5}}(2500)-{\frac{4}{5}}F_{B E}=0

F_{B E}=-3750\,\mathrm{lb}\qquad\qquad\qquad F_{B E}=3750\,\mathrm{lb}\ C   ◂

\underrightarrow{+}\Sigma F_{x}=0: \qquad F_{B C}-1500-{\frac{3}{5}}(2500)-{\frac{3}{5}}(3750)=0

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

Joint E. Assume the unknown force F_{EC} acts away from the joint (Fig. 5).Summing x components, you obtain

\underrightarrow{+}\Sigma F_{x}=0: \qquad {\frac{3}{5}}F_{E C}+3000+{\frac{3}{5}}(3750)=0

F_{E C}=-8750\,\mathrm{lb}\qquad\qquad\qquad F_{E C}=8750\,\mathrm{lb}\ C   ◂

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

+↑ΣF_{y} = 10,000-{\frac{4}{5}}(3750)-{\frac{4}{5}}(8750)

                                                 = 10,000 − 3000 − 7000 = 0                        ( checks )

REFLECT and THINK: Using the computed values of F_{CB}~and~F_{CE}, you can determine the reactions C_{x}~and~C_{y} by considering the equilibrium of joint C (Fig. 6). Because these reactions have already been determined from the equilibrium of the entire truss, this provides two checks of your computations. 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:

\underrightarrow{+}\Sigma F_{x} =-~5250~+~\frac {3}{5}(8750)~=-~5250 + 5250 = 0                        ( checks )

+↑\Sigma F_{y} =-~7000~+~\frac {4}{5}(8750)~=-~7000 + 7000 = 0                        ( checks )