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Homework Statement
A 3/4in.diameter rod made of the same material as rods AC and AD in the truss shown was tested to failure and an ultimate load of 29 kips was recorded. Using a factor of safety of 3.0, determine the required diameter (a) of rod AC, (b) of rod AD.
Homework Equations
τ(allowable)=τ(ultimate)/f.s.
A(req) = P/σ(allowable)
ƩM=0
ƩF(y)=0
ƩF(x)=0
Area=π/4*d^2
The Attempt at a Solution
First I found the allowable load. 29/3= 9.66667
Next I took the moment about B finding the reaction of A in the x direction to be 60 kips.
By summing the forces in the x and y direction i found the reaction at B to be 60 and the reaction of A in the y to be 20.
This is where I go wrong. I take the force in rod AC to be √(20^2+60^2) = 63.25 kips + 10 kips from C. So the total force is 73.25.
Then the next step would be dividing this force by the allowable force to find the area. 73.25/9.66667=7.578
From here I the area is equal to pi/4*d^2 ... d=√(A*4/pi) which would be 3.106 in for bar AC. The correct answer is 1.141 in in AC and 1.549 in in AD.
Thanks for the help in advance.
Homework Statement
Homework Equations
The Attempt at a Solution
Attachments

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