Two approaches in statics not adding up

[davidwinth]

Homework Statement

This is not homework.

Relevant Equations

Sum of the forces equals zero, sum of the moments equals zero.

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.

Method of Joints on joint C.

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

Solving these leads to

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

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

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

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. What have I missed?

Thanks.

 

[Lnewqban]

I believe that the calculated values of Ay and Ey are incorrect, since Cy=200 N only.

 

[davidwinth]

I believe that the calculated values of Ay and Ey are incorrect, since Cy=200 N only.

A force balance on the entire structure shows that these compoments are correct. The Cy is only the reaction of pin C on member CDE. There is another member that has a Cy too (member CBA).

 

[Master1022]

I'm a bit rusty on this type of thing (will need to do some searching before I can pin-point the exact location of the error), but a few points:

Homework Statement:: This is not homework.
Relevant Equations:: Sum of the forces equals zero, sum of the moments equals zero.

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

View attachment 296018

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.

I agree with this part.

Method of Joints on joint C.

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

Solving these leads to

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

I don't see anything immediately wrong with this.

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

View attachment 296020

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

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.

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

I think the error is somewhere here, but need to do a closer examination to see what it is. Just a few things that come to mind:
- Why did you resolve components of $F_c$ back onto $F_{cd}$ in the end? I think we ought to do $F_{cd} = \sqrt{F_{cx}^2 + F_{cy} ^2}$ or have I missed something? Is it because we aren't assuming $F_c$ lies along CD? Nonetheless that doesn't change the answer by very much.
- I think there may be something with the forces at the top. That is, I think something hasn't been accounted for there. Are we sure the horizontal force from pin/reaction has been fully accounted for?

Apologies if its of no help

 

[davidwinth]

Thanks, Master1022.

As for the part at the end, I was just doing a force balance on the cut section. I may have made a mistake there, but I don't see it yet. I appreciate your checking this out!

 

[T1m0]

The reason for the discrepancy is that the Method of Joints is only valid for Truss structures. A truss structure consists of members that carry axial loads only (no bending). These are bar structures where each bar is pin-connected to another bar AT THE END. Your structure has two bars that have a connection in their middle which will result in an internal bending moment.