[static equilibrium] pin and link support type

AI Thread Summary
The discussion clarifies the differences between pin and link supports in static equilibrium. A pin support typically provides two reaction force components, fx and fy, while a link support, like at point G, may only provide one component due to being a 2-force member that can only handle axial loads. The confusion arises because, although point G is classified as a pin support, it only exhibits a horizontal reaction force due to equilibrium conditions. Both support types direct their reactions along the member's longitudinal axis, resulting in tension or compression without bending or shear stresses. Understanding these distinctions is crucial for analyzing structures in static equilibrium.
kougou
Messages
80
Reaction score
0
Hello every one

I am confused between two support type:
the pin, and the link

On the textbook, it says, the pin prevents the object from translating vertically and horzontally, therefore, it provides fx and fy reaction forces.

but another type of support is much identical to the pin, which is the link support, the photo is here: http://imageshack.us/f/105/pinconnectiontw4.jpg/

at point A, the beam AB is subjected to the pin support, so it has two components of reaction forces, namely fx and fy

but what about at point G, how come it only provides one force components, namely fx, but not fy as well? is the point G also classified as pin support?
i am very confused; instructor said that whenever u see a pin support, there's two force components; how come this is not a case here?
 
Engineering news on Phys.org


is the link support the beam which connects from GB, and pin support the node at G and A?
 


G is a pinned support as well as A. The reason there is reaction in x only at G is because member GB can support axial loads only.
 
kougou said:
Hello every one

I am confused between two support type:
the pin, and the link

On the textbook, it says, the pin prevents the object from translating vertically and horzontally, therefore, it provides fx and fy reaction forces.

but another type of support is much identical to the pin, which is the link support, the photo is here: http://imageshack.us/f/105/pinconnectiontw4.jpg/

at point A, the beam AB is subjected to the pin support, so it has two components of reaction forces, namely fx and fy

but what about at point G, how come it only provides one force components, namely fx, but not fy as well? is the point G also classified as pin support?
i am very confused; instructor said that whenever u see a pin support, there's two force components; how come this is not a case here?
Yes, point G is a pin support. In general, there are 2 force components at a pin, in the x and y direction, but in some cases, such as joint G, there is only one. This is due to equilibrium considerations for so called '2-force' members. A 2-force member is a member subjected to a force (applied or a reaction force) at each end with no forces applied in between. All members of a truss are 2 force members when loads are applied at the joints only. Member BG is a 2-force member. As such, it can only take axial loading in tension or compression along its longitudinal axis. Thus , the reaction at G must be entirely horizontal; otherwise, the resultant reaction force could not be directed along the BG axis.

EDIT: I notice you are posting the same question in the ME sub-forum. You should stick to that forum and communicate further with the responder if you have additional questions.
 
Last edited:
PhanthomJay said:
Yes, point G is a pin support. In general, there are 2 force components at a pin, in the x and y direction, but in some cases, such as joint G, there is only one. This is due to equilibrium considerations for so called '2-force' members. A 2-force member is a member subjected to a force (applied or a reaction force) at each end with no forces applied in between. All members of a truss are 2 force members when loads are applied at the joints only. Member BG is a 2-force member. As such, it can only take axial loading in tension or compression along its longitudinal axis. Thus , the reaction at G must be entirely horizontal; otherwise, the resultant reaction force could not be directed along the BG axis.

EDIT: I notice you are posting the same question in the ME sub-forum. You should stick to that forum and communicate further with the responder if you have additional questions.

apologize, this is my first time to use this forum... . I will never do that again.

and what about this http://www.ecourses.ou.edu/cgi-bin/view_anime.cgi?file=512LIN.swf&course=st&chap_sec=05.1, the link support, which is very similar to the pin support? Same reasoning as well?
 
kougou said:
apologize, this is my first time to use this forum... . I will never do that again.

and what about this http://www.ecourses.ou.edu/cgi-bin/view_anime.cgi?file=512LIN.swf&course=st&chap_sec=05.1, the link support, which is very similar to the pin support? Same reasoning as well?
Welcome to Physics forums, kougou! Yes, same reasoning for a link support; the support reactions are always directed along the longitudinal axis of the member, creating only tension or compression axial forces in the member, regardless of the angle of the link, provided that there are no forces applied in between the 2 ends. NOTE that when the link is at an angle, there are still x and y components of the link reaction forces, but their resultant sum is a single force directed axially along the link, no bending or shear stresses.
 
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Back
Top