how do I construct a model that illustrates Shear stress in a beam?
361455
This question is both complex and easy; easy because the way to illustrate or model the shear stresses in a beam requires a simple diagram that cannot be drawn on here, and easy because once you see the procedure you will realize it is actually quite straightfoward. \n\nThe best thing for you to do is grab a Materials' Engineering book -- most will have, at some point, an accurate answer to this question.  Try your mechanics and dynamics books too, as this bordelines on many subjects.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
789492
An engineer, would require the weight "per foot" of the beams, and the number and location of support locations under each beam.\n \nFirst is to calculate the "dead load" and "deflection" of beams.\n\nSecond, afterwards one may calculate the live load and prescribe its bearing upon beams....\n\nIt's all in the details, the details, the details...\n\nThis is the best I can do , with what information you presented...
317313
If you need an exact balance, you will need too weights (combined).  You will need 3/4 the weight of the beam at the end of the short end to balance the beam (unless you can neglect that).  You then need to balance the weight (w) at the end of the long end with a counter weight CW) at the end of the short end.  A formula that woorks should look like this:\n\nW x 3ft = CW x 1ft\n\nCW = (W x 3ft)/1ft\n\nCW = 3W (plus 3/4 beam weight)
1022160
I hope this will help. \nFor dynamics see reference 2
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.
1087933
FEM employs the power of the computer to rapidly apply the principles of structural deformation and 3 dimensional stress / strain analysis to every point on models of any geometry in order to predict the affects of forces on the model. These computations would take forever but for the computer. Even with the computer FEM will take hours sometimes to compile, depending upon the density of nodes or points in the model to be analyzed. The virtue is that design can evolve with the tool of FEM. Changes in structure and shape can be finely tuned in a relatively short time. In addition aerodynamic (fluid) analysis is likewise performed when desired such that the effects of wind and speed etc is incorporated into the final design.
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.
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
196712
http://builder.com.com/5100-6373_14-5109284.html has a\nmini tutorial on it for you
921019
First,\ncompute the components of force perpendicular to the boom due to the weight of the boom and the object:\nBp=cos(48)*588\nOp=cos(48)*1770\n\nthe tension in the support cable can be computed by summing the torques around the pivot point:\n\nTC+TBp+TOp=0\nTC=tension*.66*24\nTBp=-Bp*.5*24\nTOp=-Op*24\nTension=(Bp*.5+Op)/.66\n=2093N\n\nThe vertical component of the tension is\ncos(48)*2093\nThe horizontal component is \nsin(48)*2093\n=1555N\nwhich is equal to the horizontal reaction force in the floor\n\nSince we know the components of force on the floor parallel to the beam as\nBh=sin(48)*588\nOh=sin(48)*1770\n\nthen v^2+h^2=parallel^2\nv=sqrt((sin(48)*588+sin(48)*1770)^2-1555^2)\n=807N\n\n\nj
284837
Answering that question depends on many things:\n\n1- What material do you want to construct with (concrete, steel, timber....)\n\n2- The type of stresses acting on the member under study\n\nthese two keywords shall guide you to the design criteria, however, I can tell you a general criteria for designing anything.... you have one equation, one of its sides is the actual stress acting which can be calculated as follows\n\nstress= force / area of cross section\n\nand the other side is to be the allowable stress stated in the material's code under the corresponding case of loading (if the member is designed for moment and sheer, the flexure and the sheering allowable stresses should be taken into consideration)\n\nthe actual stress should'nt exceed the allowable but even be less that the allowable with a reasonable factor of safety.
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
417256
If you have a shear wall, you need to design it according to chapter 11 of ACI 318. Your wall will obviously have some axial load, but shear due to wind or seismic will control. It's really not that hard to design it if you follow the examples in a good RC book. \nSince you have a  shear wall, then the rest of your structure just carries gravity loads, so effectively, it is common practice (and academically accepted) to design all the gravity load resisting system. And then, design the shear wall. You lateral loads do not affect the size or reinforcement of you beams and columns. Do make sure, however, that the shear walls (you better have at least one in each direction) are big enough that their stiffeness is a lot bigger than that of the columns. So, if you have for example 25 18" x 18" columns, but you wall is only 8" thick and 36" long then obviously, that's not a shear wall since the sum of your column stiffnesses is a lot bigger than the shear wall. PS I need some "best answer" points
1053010
On a bisic model there are some basic questions:\n\n1: What is the velosity the item is to achieve to produce the results your are after. \n\n2: What mass are you allowed in you deisgn, as shaft design will need to be considered. \n\n3. What materials can withstand the velosity, and yet maintain you mass variable. \n\n4. Diameter is dependent on velocity, material,\n\nas far as buliding it... you must consider the G factor...dependent upon weather it is earth bound or space bound.
1041136
Read in Detail at this site\nhttp://books.google.com/books?vid=ISBN0768917425&id=Ca58uytAJoAC&printsec=toc&dq=mechanical+engineering%3F+CIVIL+%26+ELECTRICAL%3F