Question 4.SP.4: The frame shown supports part of the roof of a small buildin......

STRATEGY: Draw a free-body diagram of the frame and of the cable BDF. The support at E is fixed, so the reactions here include a moment. To determine its value, sum moments about point E.

MODELING:

Free-Body Diagram. Represent the reaction at the fixed end E by the force components E_{x}~and~E_{y} and the couple M_{E} (Fig. 1). The other forces acting on the free body are the four 20-kN loads and the 150-kN force exerted at end F of the cable.

ANALYSIS:

Equilibrium Equations. First note that

DF = {\sqrt{(4.5~\mathrm{m})^{2}+(6~\mathrm{m})^{2}}}=7.5~\mathrm{m}

Then, you can write the three equilibrium equations and solve for the reactions at E.

\underrightarrow{+}\Sigma F_{x} = 0\colon       E_{x}+{\frac{4.5}{7.5}}(150\,\mathrm{kN})=0

             E_{x} = − 90.0 kN                   E_{x} = 90.0 kN ← ◂

+ ↑ ΣF_{y} = 0\colon         E_{y}-4(20\,\mathrm{kN})-{\frac{6}{7.5}}(150\,\mathrm{kN})=0

                      E_{y} = +200 kN                   E_{y} = 200 kN ↑ ◂

+↺ΣM_{E} = 0\colon                   (20 kN)(7.2 m) + (20 kN)(5.4 m) + (20 kN)(3.6 m) +(20 kN)(1.8 m) -{\frac{6}{7.5}}(150\,\mathrm{kN})(4.5\,\mathrm{m})+M_{E}=0

                 M_{E} = +180.0 kN·m                   M_{E} = 180.0 kN·m ↺ ◂

REFLECT and THINK: The cable provides a fourth constraint, making this situation statically indeterminate. This problem therefore gave us the value of the cable tension, which would have been determined by means other than statics. We could then use the three available independent static equilibrium equations to solve for the remaining three reactions.