Question about Statics problems involving frames

  • Thread starter Thread starter Yossi33
  • Start date Start date
  • Tags Tags
    Frames Statics
AI Thread Summary
In mechanics, when analyzing frames with applied moments, it is crucial to determine how the moment affects each member. A couple moment applied at a specific point influences the entire member it acts upon, and this effect may or may not transfer to adjacent members based on their connection. The discussion emphasizes that while moments are conceptual, they manifest as forces at the ends of members, which must be considered in calculations. Understanding the geometry and connections of the frame is essential for accurate analysis. Ultimately, the interaction between members and the applied moments dictates the approach to solving statics problems in frame mechanics.
Yossi33
Messages
22
Reaction score
3
Homework Statement
solving frames problems
Relevant Equations
couple of forces
hello, i have a question in this topic of frames in mechanic engineering, if i have a problem with 3 or 4 members that connected and a moment of couple is applied at specific frame,a moment of couple is general and is applied at any given point of the body. so when i dismember the frame and draw a free-body diagram of each member do i need take the couple in each member or just the member it works on ? thanks
 
Physics news on Phys.org
What type of frame are we discussing?
Is the moment co-planar with the frame?
How are the members connected to each other and to the ground?
 
question 8.png
for example in this problem a 3knm moment is applied at point G of CDE member and now if i dismember CBA and try to calculate moment in Z axis do i need to take this moment?
 
Once a punctual moment is applied onto a member, any point of that member is under the influence of that moment.
Whether or not that influence is transferred to a neighbor member, depends on the capability of the point at which they meet to transfer torque; hence, the reason behind my third question.

A moment is just a concept; in the real world, you only have forces and levers.
Therefore, at the end of a member loaded with a "moment", you always have a force, which also implies a direction of application.

Please, see:
https://www.engineersedge.com/beam_bending/beam_bending35.htm

https://www.engineersedge.com/beam_bending/beam_bending51.htm

https://www.engineersedge.com/beam_bending/beam_bending51.htm

https://www.engineersedge.com/beam_bending/beam_bending55.htm

https://www.engineersedge.com/calculators/frame_deflections_concentrated_moment_14346.htm

https://www.engineersedge.com/calculators/frame_deflections_concentrated_moment_14346.htm

:cool:
 
Lnewqban said:
in the real world, you only have forces and levers.
Sure, but in the diagram it is a twist around a vertical axis, so there is no shown lever to apply the force to.
I assume ABC and CDE are rigid elbows freely hinged to each other and to rod BD.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top