Forces on a truss (to calculate stress)

[ana111790]

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.

 

[p21bass]

Use the method of joints to find the forces within each member.

If you don't remember method of joints (which you should have had in your statics class), here's a refresher.

http://en.wikibooks.org/wiki/Statics/Method_of_Joints

 

[ana111790]

Thank you for you reply. I never took a statics class so I'm having trouble with this stress analysis course now.

I read over the method of joints, but don't I need the angles though to figure out the internal forces? This is not a right triangle so I'm not sure how to solve for them...

 

[p21bass]

Wow, I would never imagine that your school wouldn't have statics as a prerequisite for this type of course. I suggest you talk to your instructor about getting some help with statics, as you'll need it throughout the course.

In any case, you have dimensions, so you can construct your own right triangles. And you could either solve for the angles, or use trig identities to solve for the forces. I'd suggest starting with joint B, as you already have the vertical external force, and there's only one beam at joint B that has a vertical component to it.

One other thing I noticed - your F_ax should actually point the other way. It doesn't make sense that, since sum(F_x = 0), and F_cx points to the right. Which, I figure, you either know, or didn't realize because you haven't taken a statics course. Also, your sum of moments about A is way off, as you didn't include the 2kip at point C, and it should have been 5kip*168in (6ft + 8ft) for F_cy.

 

[ana111790]

I fixed the moments, I don't know how I ended up with that earlier but here's what I have:

Sum of moments = (72*Fby)-(168*5)-(84*2)=0
Fby = 14 kips

Then with sum of forces:
Fax = -2 kips (compression)
Fay = -9 kips (compression)

So I went to joint B like you said:
FBD:

and Theta = arctan(7/8) = 41.2 degrees.

So Fby + F_CB*sin(theta) = 0
according to Fby above, F_CB*sin(theta) = 14 kips
so F_CB=14 kips/sin41.2 = 21.3kips

So this will be the normal force for member CB of the truss.
I calculated the cross sectional area of the pipe:
(Pi*Diameter^2)/4 - (Pi*(Diameter-thickness)^2)/4
so Area=1.38 in sq.

So finally:
Normal stress=Normal force /Area
21.3 kips/1.38in^2=16.0 klbs

but the answer at the end of the book is different. I don't understand what I'm doing wrong... :(

 

[p21bass]

Well, you did it correctly, but 21.3kip/1.38(in^2) = 15.4 ksi (kilopounds per square inch), not 16.0 ksi. When I did it, I didn't round - I just got all of the variables into one equation - and I came up with 15.1 ksi. 15.1 vs. 15.4 should be fine, as it's just a matter of rounding. Is one of those values what the book has? Because, like I said - you actually did it correctly.