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For the truss shown above, x$1=3$m$,$ x$2=5$m$,$ and y$=6$m$.($Figure is NOT to scale$).$ The bold members $($AB and DE$)$ have a weight per unit length of W$=9$kN$/$m$,$ and all other members have negligible weight. The magnitude of the applied forces are P$1=12$kN and P$2=6$kN$.$

Solve for the force in members BD$,$ BE$,$ and CD $($FBD$,$ FBE, and FCD$).$

Using the convention that tension forces are positive and compression forces are negative. solve the equation Z $=-$FBD$+$FBE$-$FCD$.$

Write your final answer for Z in kN units in the answer box below. Round your answer to two decimal places.

Hint: Here is a method to solve, whilst checking if your answer is correct as you go:

Take moments about the left reaction support to find the magnitude of the right reaction support. Then take moments about the right reaction support to find the magnitude of the left reaction support. Then check that vertical equilibrium correctly sums to zero.

Evaluate the left sectioned shape and then evaluate the right sectioned shape. Confirm that you obtain the same answers using both sectioned shapes. Perhaps even try solving the problem using Method of Joints.

When you have confirmed that you have the correct answer, complete the algebraic equation.