Bar forces (1-6) should be calculated?

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The discussion revolves around calculating bar forces in a symmetrical framework, with particular focus on the effects of a movable bearing. Participants emphasize the importance of applying equilibrium equations for forces and moments to solve the problem. There is uncertainty about the direction of forces, with advice to initially assume a direction and adjust if calculations yield negative values. A user shares their attempt at a solution but struggles to arrive at the correct answers for the bar forces. The conversation highlights the need for clarity in problem setup and the application of relevant equations to reach a solution.
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Homework Statement


The pictue shows a symmetrical framework. I should calculate the bar forces. and make a "Clear Cut".I*m not shure because of the left movable bearing (on the left side)!?

1.jpg


Homework Equations


total x=0
total y=0
total M=0

The Attempt at a Solution


Here is my "Clear Cut" is it correctly?

2.jpg
 

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You are right that the force at lower left acts at 45 degrees, but do you think it acts up or down on the frame?
 
haruspex said:
You are right that the force at lower left acts at 45 degrees, but do you think it acts up or down on the frame?

I'm not shure, but our professor said, we should just set it in one direction, and after we calculate it we can see which the right direction was (if it is negative, we have to switch the direction).

But do you know, how I can solve this ...? I'm not shure
 
Dan123 said:

Homework Statement


The pictue shows a symmetrical framework. I should calculate the bar forces. and make a "Clear Cut".I*m not shure because of the left movable bearing (on the left side)!?

View attachment 218785

Homework Equations


total x=0
total y=0
total M=0

The Attempt at a Solution


Here is my "Clear Cut" is it correctly?

View attachment 218786
i forgot...there is given

F, h, F1=2F and F2=3F
 
Dan123 said:
our professor said, we should just set it in one direction, and after we calculate it we can see which the right direction was (if it is negative, we have to switch the direction).
That's correct, but it seemed you were asking for comments on your diagram.
Dan123 said:
how I can solve this
That's not what you asked originally. As per forum rules you need to show an attempt at a solution. Apply the relevant equations you quoted.
 
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haruspex said:
That's correct, but it seemed you were asking for comments on your diagram.

That's not what you asked originally. As per forum rules you need to show an attempt at a solution. Apply the relevant equations you quoted.
ok, I will send my solution...or my try :)
 
haruspex said:
That's correct, but it seemed you were asking for comments on your diagram.

That's not what you asked originally. As per forum rules you need to show an attempt at a solution. Apply the relevant equations you quoted.
So here is my try...I can't finish it...I don't get the right answer (the right answer is: S1=6F ; S2=(√5)/2 F ; S3=1,5F ; S4=7F ; S5=(√5)/2 F ; S6=1,5F )

IMG_20180123_203046_376.jpg


IMG_20180123_203116_656.jpg


I thought of dividing it in two parts (special calculation method...?!

Could you pplease help me :)
 

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Dan123 said:
(the right answer is: S1=6F ; S2=(√5)/2 F ; S3=1,5F )
Doesn't look right to me.
I was suspicious that you may have specified the wrong values for the applied forces, so I made them arbitrary.
Consider the leftmost triangle and take moments about its lower right corner. Let N be the normal force from the lower left support, Gt the downward load at the top of the triangle, and Gr the load at the bottom right.
Take members 1 and 2 to be under compression, 3 under tension.
Y: N/√2-2S2/√5=Gr+Gt
X: N/√2=S2/√5-S3+S1
M: N/√2=Gt/2+S1
If I substitute the answers in the X and M equations I get N=5F√2, Gt=-F.
 
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