A statically indeterminate structure is one in which there are more unknown reaction forces (or internal forces) than there are the static equilibrium equations. In this problem, there are 3 static equilibrium equations: Sum of forces in X direction =0, Sum of forces in Y direction =0, and sum of torques (moments) about any point =0. But there are 4 unknown reaction forces, namely, Fax, Fay, Fbx, and Fby. Using the static equilibrium equations, you can solve for Fax and Fbx, but try as you might, you can't solve for Fby or Fay without having to consider compatability and deformations. All that the static equilibrium equation will tell you in the Y direction is that Fay + Fby = F (the applied force); it cannot determine how that force F splits amongst each of those terms. That is why you need another equation beyond those 3 equilibrium equations.Dafe said:
Might seem that way, but not true. Put in some simple numbers like F =1, and L1 = L2 =L3 =1, and try it. You won't be able to get Fay and Fby no matter how hard you try.moemoney said:
Ahh yes very true. Fay and Fby in this case can't be represented in any moment equations other then having Fay = Fby (moment about point F1) just because they are in-line with each other right?PhanthomJay said:
Be careful, if you take moments about F1, you have to consider the moments due to support reactions Ax and Bx as well. Ay does not generally = By. You have to use the compatability/deformation approach to find them, in conjunction with Ay + By = F. Statics alone and taking moments won't cut it.moemoney said:
Yes, excellent. You might want to rearrange terms a bit, to get [tex]F_{ay} = FL2/(L2 + L3)[/tex], but it's just an algebraic manipulation to put it in a somewhat better form. You'll also find that [tex]F_{by} = FL3/(L2 + L3)[/tex].Dafe said:
Fax and Fbx are equal, but they point in opposite directions, as you show on your diagram. Their values are functions of F, L1, L2 and L3. If you are comfortable with your result, fine. I notice, however, that you have not shown the proper direction for the support reactions Fay and Fby. Be sure to catch it.Dafe said:
your call. Latex is nice, but you've done well so far without it.
My pleasure.
I think I may have unfortunately left my imagination in Disney World, but regardless, I was envisioning a free body diagram of member AB, where at that joint you have a downward internal force F and an internal counterclockwise couple of (F)(L1). The actual applied force F is where it is. Nice job!
PhanthomJay said:
That is correct. Fay points upward (toward the member, compression forces in segment L3), and Fby also points upward (away from the member, tension forces in segment L2). It takes a little imagination to determine this.Dafe said:
You are confusing the axial forces in AB, which causes no twisting, with the moment couple, (F)*(L1), which does the twisting. It is this couple which produces the horizontal reactions Fax and Fbx in the directions you have shown, which, for some reason, you have been very secretive of the values you have determined for same. Care to disclose them? I'd very much like to see what you got.
Thanks to you and John Lennon, I'm a dreamer again. Imagine!Dafe said:
Yup, you got it! I was just testing you, because sometimes as we focus on the difficult, we mess up the simpler stuff. Also, way back in your original post you noted that Fa^2 = Fax^2 + Fay^2, which is quite correct. But it is perfectly acceptable (and preferred) to leave the support reactions in their x and y component forms. Peace.
Classical mechanics is a branch of physics that deals with the motion and behavior of objects under the influence of forces, such as gravity or friction. It is based on the principles of Newton's laws of motion and has been the foundation of many other areas of physics.
Some common examples of classical mechanics problems include calculating the motion of a pendulum, determining the trajectory of a projectile, and analyzing the forces acting on a car moving on a curved road.
To solve a classical mechanics problem, you will need to understand and apply the principles of Newton's laws of motion, as well as other relevant concepts such as energy, momentum, and rotational motion. It is important to carefully identify and draw diagrams of all the forces acting on the object in question and use mathematical equations to calculate the desired quantities.
One of the main challenges in solving classical mechanics problems is correctly identifying and considering all the relevant forces acting on an object. This may involve taking into account friction, air resistance, and other factors that can affect an object's motion. Another challenge is applying the correct equations and solving the problem using appropriate mathematical techniques.
Classical mechanics is the foundation of many other branches of physics, including thermodynamics, electromagnetism, and quantum mechanics. It provides a framework for understanding the motion and behavior of macroscopic objects, while other branches of physics deal with the behavior of subatomic particles and systems on a smaller scale.