# Why does the method of joints not match a force balance result?

I am trying to figure out why the two approaches do not match in solving this statics problem. This involves the image below.

The first thing to do is to find the reactions at A and E. This is easy enough, with the results being that E only has a vertical component equal to 400 N and A has Ax = -346.4 N, Ay = -200 N. With this out of the way, we can find the force in CD and CB by using the method of joints. Assume the right hand member is in compression and the left hand member is in tension.

Method of Joints

$$\sum F_x = 0 = 400cos(30)-F_{cd}cos(60)-F_{bc}cos(60)$$ $$\sum F_y = 0 = -400sin(30)+F_{cd}cos(60)-F_{bc}cos(60)$$

$$F_{cd} = 461.88 N$$ $$F_{bc} = 230.94 N$$

So far so good. But now what if instead we look at individual members and solve the forces on them? Take member CDE for example. Here we only look at the reactions acting on the member. This includes the reactions at the pin at C, D, and E. A force balance on member CDE gives,

Method of Force Balance on Member CDE

$$R_D = 305.4N$$ $$R_{Cx} = 305.4N$$ $$R_{Cy} = -400 N$$

And we already found the reaction force at E. But here's where the trouble comes in. Say we take a cut between C and D to find the internal force in this member in order to see if it matches what we found by the Method of Joints. The internal force will have to balance with the reactions at C for equilibrium. Thus, we sum the projections of the components of the force at C along the member.

$$F_{cd} = R_{cx}cos(60) + R_{cy}cos(30) = 499 N$$

Thus the method of joints does not agree with the force balance approach on just what the internal force in member CDE really is. What have I missed?

• The model is unstable because the roller can't resist the lateral load.
– r13
Commented Jan 25, 2022 at 21:23
• It's not unstable because it's pinned at A. Commented Jan 25, 2022 at 21:31
• The member DE will need to resist an axial force that has a horizontal component that causes the instability.
– r13
Commented Jan 25, 2022 at 21:40

The method of joints is usually be applied to trusses.

The structure you have here is a frame and its members behave differently to trusses rods. More specifically a rod (in a truss) develops/transfers only axial loads. This is done because forces are only applied at the end of the rod ( only two forces are applied and that requires that they are equal and opposite).

On the example above there are more that two forces applied on the member so the member behaves as a beam that transfers axial shear and bending moments.

So at point B the moment does not have to be zero. And the reactions from ABC affect the BD beam.

• That makes sense. So the reason why one method does not match the other is that one method is inapplicable to the problem. Commented Jan 25, 2022 at 22:00
• @Karlton yes that is exaclty right Commented Jan 26, 2022 at 7:35
• +1. Just to make it a bit more explicit: the moment of beam AC at point B does not have to be zero. Beam BD however, will have zero moment at B. Commented Feb 27, 2022 at 23:53

Your structure would be a truss and you could apply any conventional truss method to solve it if the support at E would be pin support. You would then get the BD as a zero force member.

As it is both AC and CE are acting as a beam with both moment and axial load. It is not a truss and as you noticed you will get wrong answers if you try to solve it like a truss.