Calculate forces on beam with hook

  • Thread starter kaffekjele
  • Start date
  • Tags Tags
    Beam Forces
5 replies · 2K views · last reply
kaffekjele
Messages
20
Reaction score
0
Calculate forces on beam with "hook"

I have a beam which looks roughly like the attached file. The aim is to calculate forces in A and B. Would I have to look at AB and CDE independently, or could I proceed as "usual" by calculating the moment in A and proceed with forces calculation in x and y direction?
 
Attachments
  • Diagram of a horizontal beam from A to B with a bent hook: a vertical segment rises from point C on the beam to D, then a short horizontal segment runs from D to E.
    Bjelke.jpg
    2.2 KB · Views: 456


You could proceed "as usual". The entire system must be in equilibrium, subsystems may not and typically will not be in equilibrium per se.
 


I've done the forces and moment calculations on the beam, but I'm a bit unsure if it's done correctly. Every force is working on the center lines according to the figure, so I might be missing something here. I'd appreciate it if someone could take a look at my calculations and perhaps give some input.
A better figure showing all the measurements is attached.ƩMA= 0 → B*6,3+F*sin 44,4*3,6 =0 (direction: counter clockwise.)
B*6,3+19,5*sin 44,4*3,6 =0
B= -7,796 kN

ƩFx=0 → Ax-F*cos44,4=0
Ax-19,5*cos 44,4=0
Ax-13,932=0
Ax= 13,932kN
ƩFy=0 → Ay+B-F*sin44,4=0
Ay+7,796-19,5*sin 44,4=0
Ay-5,847=0
Ay= 5,847
 
Attachments
  • Handwritten sketch of a beam with a hook: horizontal beam supported at A (pin) and B (roller), with dimensions 2,2 and 4,1 labelled below, a vertical hook section with a 4,0 height and 1,4 width, and a force F = 19,5 kN at 44,4° applied at point E.
    20130217_201938 (2).jpg
    27.6 KB · Views: 475
Last edited:


How did you get "3.6" in ƩMA= 0 → B*6,3+F*sin 44,4*3,6 =0? The distance from A to E is not 3.6.
 


That's one of the things where I'm not sure if it's done correctly, but what I did was add 2,2 and 1,4. I suppose the height has to come into play somewhere, but I'm not sure how. Up until now I've only been doing calculations on regular beams without any sort of "protrusion".(Sorry, I don't know the correct English term for a beam like this.)
 


The moment of a force is the product of the "lever arm", the force's magnitude, and the sine of the angle between the lever arm and the force. The lever arm is the distance from the point around which the moment is taken to the point of the force's application. In this case, the lever arm is AE. You need to find its length and the angle between AE and the force.

Alternatively, the moment is defined as a vectorial product of the lever arm and the force, which allows you to get around without computing angles and lengths.

You can use either method.