Truss Analysis: Find Reaction Forces with Explanation

[Paladashe]

Hi, I'm new on this website and it seems great :p So, I'm struggling with something really simple... trusses analysis. I'm struggling with moments etc , I've just drawn an example linked in this post , could you try to find the reaction forces (roller pin etc) with some explanation ? Thanks

 

[tiny-tim]

Hi Paladashe! Welcome to PF!

(please use the homework forum in future)

If I'm understanding the question correctly, no truss analysis is involved …

you have a rigid body with one applied force (plus its own weight), and you have to find the two reaction forces, so just use linear components of force, and moments …

show us what you get. ​

 

[Paladashe]

Thanks Tiny-tim for your response. In fact, I now for vertical external forces and horizontal ones so I divided the 100N vector in two vectors (one with a value of 50N orientated downwards and one with a value of 50•(square root of 3). I found that each pin has a vertical vector orientated upwards with a value of 25 N and for the fixed one, also an horizontal vector with a value of 50•(square roots of 3 orientated to the left). Am I right? I'm sure that I've made some mistake...what value did you find?

 

[tiny-tim]

Hi Paladashe!

Thanks Tiny-tim for your response. In fact, I now for vertical external forces and horizontal ones so I divided the 100N vector in two vectors (one with a value of 50N orientated downwards and one with a value of 50•(square root of 3).

Correct.

I found that each pin has a vertical vector orientated upwards with a value of 25 N…

Nooo …

you can't assume they're equal, can you?

all you know is that they add to 50N, so you need one more equation, which is … ?

 

[Paladashe]

The moment equation, right?
... EMa=-50•3•12,1244 + 50(square root of 3)•7 + 4•12,1244•Rb ?

 

[Paladashe]

Where a is the pin fixed and b the roller pin

 

[tiny-tim]

The moment equation, right?

Yes.

(either about A or about B)

... EMa=-50•3•12,1244 + 50(square root of 3)•7 + 4•12,1244•Rb ?

(where 12.1244 = 14*√3/2)

Yes …

or you could avoid splitting the 100 N by doing 100 times the lever arm (the perpendicular distance from the line of the 100 N to the point A). ​

 

[Paladashe]

Yes but don't know how to do this because its not perpendicular to the line of action:)

 

[Paladashe]

Yes, but by doing my equation , I found that Rb=50N so R(vertical) is equal to 0??

 

[tiny-tim]

use trig

 

[Paladashe]

yes but it's not working for me :p could you please try to explain why my equation is wrong ? Please ! I'm getting a headache by trying and trying :p

 

[tiny-tim]

Yes, but by doing my equation , I found that Rb=50N …

I don't get 50

 

[Paladashe]

-50?

 

[tiny-tim]

i mean, using your equation (which is a little confusing ) i don't get 50

 

[Paladashe]

Can you give me your result ? Is it 37,5?

 

[tiny-tim]

from your equation (am i misreading it?) i got 25

 

[Paladashe]

No as you said before, we can't assume that both pin have the same vertical forces (25-25). Or, if Rby is equal to 25 and that EFy=0 <=> -50+25+Ray=0 :) so Ray is equal to 0 which is impossible

 

[tiny-tim]

(just got up :zzz:)

No as you said before, we can't assume that both pin have the same vertical forces (25-25).

yes, we can't assume it …

but that doesn't mean that we can't prove it (if it's true) ​

Or, if Rby is equal to 25 and that EFy=0 <=> -50+25+Ray=0 :) so Ray is equal to 0 which is impossible

not following you …

doesn't -50+25+Ray=0 mean Ray = 25 ? ​