What forces do flanged bushes on a door hinge experience?

[ghobbs]

Hi. I have a door mounted on 2 hinges, aligned on a vertical axis. The hinge shafts run in flanged plain bushes.

I'm trying to work out what forces the flanged bushes will see. The door weighs 100Kg, with the centre of mass at the centre of the door. The door is 2m wide and 2m vertically between hinges.

I think that the flanges will see the mass of the door, i.e. each flange sees 500N.

The bushes will also see radial forces, counteracting the moment of the door about the hinges. Based on the moment and geometry, will the top bush see 1000N and the lower bush -1000N?

 

[CWatters]

my experience is limited to hanging a field gate. In that case its impossible to get both hinges to share the load because they are never the same distance apart. One or the other "bottoms out" first and carries the vertical load.

 

[sophiecentaur]

As CW wrote, only one of the hinges will have a shear / vertical force on it but the upper hinge will have a force outwards and the lower will have a force inwards. This simple setup gives the horizontal forces as + and - 500N, assuming the cm of the door is half way between the levels of the hinges the total moment being 1000Nm and each hinge provides half the necessary torque against this( 500+500, one way = 1000 the other way). Unless you have some resilience in the hinge mountings plus some way of fine adjustment of the vertical height of the hinges, you can't be sure which hinge will take all the additional vertical load of 1000N.

 

[ghobbs]

Ah yes, I understand now. Thanks for the clarification.

 

[pmacias]

I would approach this question by first considering the principles of mechanics and forces. Flanged bushes on a door hinge experience several forces, including the weight of the door, the radial forces from the moment of the door, and any external forces applied to the door.

The weight of the door, which is 100kg in this case, will be evenly distributed between the two hinges, resulting in 500N of force on each flanged bush. This is due to the principle of equilibrium, where the downward force of the door is balanced by an equal and opposite force from the hinges.

In addition, the flanged bushes will also experience radial forces from the moment of the door about the hinges. This is due to the door's weight being offset from the hinges, creating a moment or rotational force. The top bush will experience a 1000N force in the direction of the moment, while the lower bush will experience a -1000N force in the opposite direction. This is in accordance with the principles of torque and moments.

It is important to note that these forces may vary depending on the specific design and materials of the hinges and bushes. For example, if the hinges are not aligned perfectly on the vertical axis, the forces may be slightly different. Additionally, any external forces applied to the door, such as a person pushing or pulling on it, will also affect the forces on the flanged bushes.

In conclusion, the flanged bushes on a door hinge experience a combination of forces, including the weight of the door, radial forces from the moment of the door, and any external forces applied to the door. By understanding the principles of mechanics and forces, we can accurately predict and analyze the forces on the flanged bushes in this scenario.