- #1

ana111790

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## Homework Statement

A simple pin-connected truss is loaded and supported as shown in the

figure. All members of the truss are aluminum pipes that have an

outside diameter of 4 in and a wall thickness of .226. Determine the

normal stress in each truss member.

## Homework Equations

Normal stress = Normal force/ Cross sectional area.

Sum of moments = 0

Sum of forces in x direction = 0

Sum of forces in y direction = 0

## The Attempt at a Solution

In order to use this formula [Normal stress = Normal force/ Cross sectional area] I calculated the cross sectional area as diameter* thickness= .904 in^2

I also need the normal forces of each member of the truss. So I drew the FBD of the structure showing the reaction forces:

Solving for reaction forces:

1) Sum of moments around A = (72in * Fby) - (36 in*Fcy)=0

Therefore: Fby= 2.5 kips

2) Sum of forces in the x direction = Fax+Fcx=0

Therefore: Fax=2 kips

3) Sum of forces in they direction = Fay+Fby-Fcy=0

Therefore Fay=2.5 kips

Now I'm not sure how to figure out the normal force on each member given the reaction forces (I calculated the reaction forces because of an example my prof. did in class...)

Can anyone help me? Thank you lots.