I am trying to find the internal forces in the member at point G in the following illustration:

My answer is different from the solution in the textbook and I think it has to do with how I'm looking at the 1500 lb external force acting on point A.

I've included the 1500 lb force in my free body diagrams for both AB and AE. I'm getting the feeling that it should only be acting on one of these members; however, I am unsure.

Does the load at the joint act on only one member, or on all members in the joint?

  • $\begingroup$ Please edit your question showing us what you've actually done. $\endgroup$
    – Wasabi
    Commented May 19, 2016 at 21:31
  • $\begingroup$ If it acted on two members then it would be 3000lbs total? $\endgroup$
    – Jodes
    Commented May 19, 2016 at 21:51

2 Answers 2


An externally applied load is only ever applied once; it is then transferred through the connection (via internal forces) to all other members of the joint. If there are multiple members connected at the point of application, you can make an arbitrary choice of where to apply that load, as long as you stick with that choice for the entire analysis. (This is correct for static analysis of trusses and frames, at least.)

I find it sometimes helps to "zoom in" conceptually on that linkage and go through the transfer of force step-by-step, with as many free-body diagrams as necessary, even including the pin itself if you're at all uncertain about how force transfers from member to member.

I'll start with the wood member, with the pin hole exaggerated for convenience and cutting at G to save me some time sketching:

I've applied the external force here. My force balance for this FBD will include the equation:

$$\Sigma F_y = F_{A,y} + F_{G,y} - 1,500\ \mathrm{lb_f} = 0 \tag 1$$

From this point on, the value of the externally applied force is reflected in the values of $F_{A,y}$ and $F_{G,y}$. Now I'll draw the FBD of the pin:

A note on FBD conventions: I like to draw every force, the first time I define it on a diagram, in the positive direction. It doesn't really matter which way you draw it the first time, as long as you always respect the initial definition in any later step, but in this case it works out nicely for analysis by inspection. Here's the equation in the y-direction anyway:

$$\Sigma F_y = F_{rod,y} - F_{A,y} = 0 \tag 2$$

Now we can look at the other member in the joint, which is the metal rod. You can consider this to show a crimped end rather than the forked clevis end shown in the original illustration, if you like; it makes no difference to the analysis but saves time sketching the FBD.

So why didn't I draw the external 1,500 lbf load on this third diagram? The answer is simple:

  • We have defined $F_{rod,y}$ originally in terms of all forces present in the previous FBD of the connection;
  • We have defined those forces present in the FBD of the connection all in terms of the forces present in the previous FBD of the first member;
  • The external force was applied in the FBD of the first member.

Our equations are all linked by shared variables, so the effect of the external force when applied just once propagates through to all of the other members and connectors at that joint—and potentially to the entire frame.

Rearrange equation $(1)$:

$$F_{A,y} = 1,500\ \mathrm{lb_f} - F_{G,y}$$

Substitute into equation $(2)$:

$$F_{rod,y} - (1,500\ \mathrm{lb_f} - F_{G,y}) = 0$$

Since $F_{rod,y}$ is applied directly to the metal rod in the third FBD and its definition takes into account the external load we already applied in the first FBD, you should now be able to see that there's no need to apply the external load to more than one member in a joint. If you did apply it again, you'd be double-counting and your analysis would be incorrect.

Incidentally, as long as we're not concerned the pin will fail, we can skip drawing its FBD and just apply the third law action/reaction pairs directly between the members of the joint.


Using the method of virtual work, assume that $AE$ is given a small deflection $d\theta$ downwards and that the unknown force is $Y$.

$$ 1500 (6+2+2) d\theta + Y (6+2) d\theta + \left(\int_0^6 \frac{300 l}{6}l \, dl\right)d\theta =0$$

$$ \Rightarrow 15000 + 8 Y+ 3600 =0$$

$$\Rightarrow Y = -2325 $$


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