Pin Jointed Frame

[qaisjc]

Hey Guys, I've got a pretty simple (I think) pin joint frame question. Im pretty weak in this area of mechanics so if someone could advise me on how to go about doing this question i would really appreciate it.

The Figure Q1 (The Attatchment) shows a pin jointed frame.

The frame is supported

• At A by a roller joint and
• At D by a pin joint.
• AD = CD = 1m and BC = 2m.

a) Evaluate the reactions at A and D, i.e. the forces exerted on the frame by the supports.

b) Calculate the forces carried by the following members, indicating if they are in tension or compression.

i) AB
ii) BC
iii) AC

[PhanthomJay]

Hello gaisjc and welcome to PF.

You must first calculate the reaction forces using the standard 3 equilibrium equations. Note that the roller support by definition must just have a vertical reaction. Then you can use the method of joints to solve for the member forces. You must show some attempt at a solution before we can be of further assistance, however weak that attempt may be..

[qaisjc]

Oops, my appologies. I havent attempted a solution because I honastly dont know what to do. I realise I have to fine the moment at D using \SigmaMD=0 and use that to find Dx and Dy, then use the method of joints to solve for the member forces, but as to what numbers I use to find the moment at D, thats what confuses me.

[PhanthomJay]

Oops, my appologies. I havent attempted a solution because I honastly dont know what to do. I realise I have to fine the moment at D using \SigmaMD=0 and use that to find Dx and Dy, then use the method of joints to solve for the member forces, but as to what numbers I use to find the moment at D, thats what confuses me.Summing moments about D is a good idea, but first you should identify all unknown support reactions. You've already noted Dx and Dy. What about Ay?(As I noted earlier, there can be no Ax, just an Ay at support A). Now sum moments due to the applied force, and due to Ay, about D ,to solve for Ay. The member lengths are given. Note that the moment of a force about a point is the product of the force times the perpendicular distance from the line of action of that force to that point.

[qaisjc]

I really appreciate your help PhanthomJay. Ok, so this is what i've got so far;

\SigmaMDz=0, therefore (FAyx1)-(5x2)=0
therefore FAy=10

\SigmaFy=0, therefore FDy+10-5=0
therefore FDy=5

\SigmaFx=0 therefore FDx=0

Right? Or have I missed something?

[PhanthomJay]

I really appreciate your help PhanthomJay. Ok, so this is what i've got so far;

\SigmaMDz=0, therefore (FAyx1)-(5x2)=0
therefore FAy=10Yes, but you must specify in which direction (up or down) that force acts on the frame.

\SigmaFy=0, therefore FDy+10-5=0
therefore FDy=5 well, actually, that works out to F_Dy = -5. What does that minus sign mean in regard to the direction of F_Dy?

\SigmaFx=0 therefore FDx=0 yes, correct.

Right? Or have I missed something?Looks Ok, as long as you have your directions correct. Please watch your plus and minus signs. Once you're clear on the directions (up or down) that the vertical reaction forces exert on the frame, you can move onto the member force calculations.