- #1

ledphones

- 17

- 0

## Homework Statement

A 3/4-in.-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.