How to get shear force using the "Truss Model Theory" from Hsu and Mo?
456459
Hi contact me at blade_trinity_mr@yahoo.com and if you can send me the papers you have for this subject and i'll try to help you.
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
811039
I will try. \nNot because it is difficult to understand but difficult to describe. Let us take a three member truss.\n\nA truss must be in equilibrium otherwise it will fall into its three component pieces. Lets assume it is for a roof\n\nFrom its apex to the two walls its load will come from the tiles, and for these the angles and weights are calculated to establish what portion of it goes to which joint.\nThe two portions at the top are examined together to make sure their loads will be stable against each other.\nThe portions at the bottom are being supported by two walls but they have two components each. One wants to rest on the wall while the other is trying to move away from the wall.\nTo keep this from happening we have to introduce our third truss member and make sure it can take the tension required of it.\nNow we know all the points we can show the summation of all the loads and ascertain if they are in compression or tension.\nWhen we have selected or designed each of our truss members we can fix them to keep the whole assembly stable.\nNow the truss it in equilibrium.\nTry it with large match sticks and think of what you are reading.\nIt might help.
254925
I'm assuming that the pivot point will land between the 15 and 30 Kg loads because it seems more logical. If you try putting the X distance between the 20 and 15 loads, you'd probably get a negative distance. \n\nActually, the weights have to be multiplied by the pull of gravity (9.81m/ss) to make them actual forces but then again, they would all just cancel out.\n\ni hope you understand the crude diagram\n\n|-------10------|       - distance where weight of beam is applied\n|---------x-------|    - pivot point dist from left\n ____________________  - beam (20 m)\n|_________|__________|\n20_______15_________30 - applied forces (weights)\n\n20kg(X) + 15kg(X-10m)  = 30kg(20m-X)\n20kgX + 15kgX - 150kgm = 600kgm - 30kgX\n20kgX + 15kgX + 30kgX  = 600kgm + 150kgm\n                65kgX  = 750kgm\n                    X  = 750kgm/65kg\n                    X  = 11.53846 m from left end
1055779
1 ksi = 1000 psi\n\nWhat do your points represent and how did you get that answer?\n\nLater...\nE= stress * length/deflection, or force*length/(deflection*area)\nIf the tester was zeroed, your result is non-linear - .004 deflection for the first 150# and .006 for the next 150#. Could indicate yielding - or slipping. If you're sure of your work, check the data and the setup.
1171597
It sounds like a Civil Engineering problem, maybe for steel structures? Well, regardless, I suggest going to the ASCE (Am. Society of Civil Engineers) web site. They should have some good examples including some mesh - free finite element method solutions. Should give strengths, buckling info, etc.
472579
Depends on the structural design.  Shear will take place near to supports.  Load pressures will take place all the way down to the foundations (getting larger all the way.\n\nSome structures will tend to push out (like a pitched roof) and some will push down (like a flat roof). \n\nIf you need anything more specific drop me a note.
418175
it will shear at 45 deg, which is, as previously mentioned, the direction of greatest shear stress.  This occurs despite the random structure because it is caused by stress, not structure.  A similar behavior occurs in metals pulled in slow tension, with "shear lips" forming 45 deg from the direction of the tension.
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
402913
well, I don't know about the testing, but for design, you use 0.4Fy as your shear strength. If you have 36 ksi steel, then you need 14.5 ksi to create a shear failure. Don't know if that works forpunching holes; it's probably less. Good question.
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
432088
*copied from previous post\n\nYour question is unclear about what set of assumptions you want to use. You seem to imply a perfectly constructed stool, with a perfectly centered force, using incompressible material. You want the stool to have no mass, and you imply that you want to use statics to solve this problem. Strictly applying statics, this system is indeterminate, NOT the ridiculous answer of 0 [N] in two legs and W/2 [N] in the others. There are only 3 equations of equilibrium here, 1 for vertical forces and 2 for moments. Since you have four legs, this system cannot be solved by statics alone. Treating the stool as a deformable structure, you can use geometric compatibility to get a fourth equation. I challenge you to find a competently built stool that has 0 [N] in any leg (a physical stool built of real materials), or an incompetently built stool with 0 [N] in any two legs (unstable system that is obviously not static). The answer indicated as "correct" requires an incorrect application of statics. A large part of studying physics is learning when to use a particular tool, and the physical limits of that tool.
893876
Simply keep in mind that summation of forces equal to zero.\n\nSum (Fx)=0\nSum (Fy)=0\nlet th\nF1- force o string supporting W1 (weight on the left)\nF2- force o string supporting W2 (weight on the right)\nF3- force o string between the weights\n\nFx=0=F1cos(54) - F2cos(68) +F3cos (13)\nFy=0= W1+W2-F1sin(54) - F2sin(68) -F3sin (13)\n\nwe have 2 equations and three unknowns.\nSince the dimensions or length of the ropes ar enot given we can't employ moments.\n\nlet's use superposition and solve one side at a time\n\nW1=F1sin(54) then F1=W1/sin(54)=37.73/sin(54)=46.6N\nsimilarly\nF2=W2/sin(68)=90.944/sin(68)=98.08\n\nsince \nF1cos(54) - F2cos(68) +F3cos (13)\n\nthen  F3=(F1cos(54) - F2cos(68))/cos (13)\n\nF3=(46.6cos (54)-98.08cos(68))/cos(13)= - 9.6 N\n\n(Negative sign means that the originally assumed 'positive' direction to the left was wrong. F3 is pointing to the right.)
102434
Part (a): \n   Given three values (Mrock = 1.2 kg, Xrock = 5 cm, Xstand = 14 cm), and we need to solve for one unknown (Mstick).  To clarify, Mrock means mass of rock, and Xrock means position of the rock on the meter stick.\n   Since you're being tested on this concept, I'll assume you know the equation for torque: T = Force x Lever arm.  We will also use the concept of Static Equilibrium, which is a fancy term that means that nothing is moving.  Since this system is in static equilibrium, we can use the equation T(clockwise) = T(counterclockwise), using the stand as our pivot point (14 cm).\n   To find clockwise torque, we need to find the Lever arm from the center of mass to the pivot point.  The center of mass is located at 50 cm, and the pivot point is at 14 cm, so our lever arm is 50 cm - 14 cm = 36 cm.  T(clockwise) = Mstick * g * 36 cm  (It's okay to use units of cm in this problem because it cancels out in the end.)\n   If the rock is hung from a massless string, we can assume that the force on the meter stick from the string, which is the tension in the string, is equal to Mrock * g.\n   Counterclockwise torque: Lever arm = 14 cm - 5 cm = 9 cm.  Torque(counterclockwise) = Mrock * g * 9 cm (plug in numbers later to simplify calculation)\n   Torque(counterclockwise) = Torque(clockwise) ==>  Mrock * g * 9 cm = Mstick * g * 36 cm  ==> g's cancel, manipulate equation to get: Mstick = (9 cm / 36 cm) * 1.2 kg = 0.3 kg\n\nPart (b):\n   Since the stand is not causing the meter stick to accelerate upwards, it must be exerting a force upward equal to the force the meter stick exerts downward.  The total force from the meter stick is equal to (Mstick * g) + (Mrock * g) ===>  (Mstick + Mrock) * g ===>  1.5 kg * 9.8 m/s^2  ===> 14.7 N\n   I hope this helps (I'm a physics major...this should be a reliable answer)
1115091
The best answer to this question would very long. The quick and dirty response is:\n\nCutting tools in metalwork apply a compressive force to the work piece. The compressive force overcomes the forces holding the atoms together and and the material begins to slip along planes of irregular crystal stucture. The effected zone is reffered to as the shear zone. \n\nAgain this is short answer. Probably raised more questions than it answered.
848001
Look up Vector Mechanics for Engineers and Scientists.  This deals with basic Newtonian physics as applied to static structures.\n\nBone up on chemistry.  It is easy to forget your chemistry, and as far as I know you will be required to take a few semesters of basic Chem, even if you are not planning on being a Chem Eng.\n\nDo not lose your Maths either!  Grab a pre-calc or a calculus book and keep practicing.
913014
F(net) = m x a\n\nF (net) = F (forward) - F(air friction)\n\nso  F(air friction)  = F (net) - F(forward)\n\nMass is given 20 kg\n \nnet acceleration given 2.0  m/s^2\n\n\nSolve for the Net Force, subtract the forward force, the difference is the force due to air friction.