Some simple questions regarding engineering like whats yielding, strength to weight ratio and plastic deformat
955958
Hi!\n\nYield point (or yield stress) is the stress at which a material starts being plastically deformed, which means that, once you reach the yield point, the deformation becomes permanent. \n\nPoint load is a force applied to a specific point and distributed load is a force distributed throughout the analyzed structure.
518951
The yield stress is actually more valuable for design. You never want to exceed the yield stress of a material (unless you need it to fail for some reason). Comparing the yield stress with the ultimate stress can give a quick idea of how brittle the material is. \n\nThe working stress is usually set to some fraction of the yield stress. For example customers will often require that the maximum working stress not exceed 2/3rds of the minimum allowed yield stress for the material.
876819
Currently websites are missing for basic engineering. If u can built website related engineering basics it will be of great helpful to growing buddies.
282242
that's a real vague question, try to be more specific or at least say what king of structure do u want? (steel, conc, ..?)\nanyways this is an exremely simple example.\n\nlets' say u want to design a tension member -carrying tension force = 100 tons, -between 2 supports and this member is a steel member of yield stress= 1.4\nso to get the area u need :\n100/1.4= 71.43 cm square let's say again u want it to be I-shaped steel then u'll need  I number 400 i.e IPE 400
75327
TO FIND STRESS FACTORS FOR MATERIALS, IE METALS, PLASTICS, ETC. GO TO YOUR LOCAL BOOKSTORE AND ORDER A PIPEFITTERS MANUAL, WELDERS MANUAL, OR MILLWRIGHTING MANUAL..THEY WILL HAVE ALL THE GRAPHS AND CHARTS YOU NEED. THE COST WILL RANGE FROM $12.00 ON UP BUT ANY OF THEM WILL BE WELL WORTH THE COST...\nTHERE ARE SEVERAL FACTORS THAT INDUCE STRESS UPON ANY MATERIAL IN INDUSTRIES, WEIGHT, WIND, CHEMICAL/EROSION/RUST/HEAT/COLD/ ETC., IF YOU HAVE A PARTICULAR QUESTION, I WOULD BE HAPPY TO TRY TO ANSWER IT IF I CAN..
796030
Depends on the supports, whether they are simple (pivot possible), or completely fixed.\n\nAlso depends on whether it is a point load, or a distributed load.\n\nThere is a technique that is taught in engineering such that you can work it out for almost any configuration, but it is a bit too involved to detail. Do a web search on 'Bending Moment Diagrams' and you will find lots on the techniques.\n\nFor specific calculation, I found a neat online calculator, go to http://www.mechengcalculations.com/index.html, and look for your beam under section 'Beam Flexure'
95760
The answer you're looking for is here...\nhttp://www.geocities.com/yielding_insanity/answer.html
606521
Try this:\n\nhttp://education.yahoo.com/reference/weights_and_measures/
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.
852384
I can't help you with the first question, but to figure out the second one you just need to use Hooke's law (I'm assuming the spring to be almost perfectly elastic.) In equation form the law looks like: F=k*x\nWhere F is the force applied, k is the spring constant, and x is the deformation.\nThe force applied will be the objects weight, so 9.8*.5=4.9N, and the deformation is .04m.\nSo 4.9N=k*.04m\nk=4.9N/.04m\nk=122.5N/m
404227
I think you will have much better success if you narrow the subject area. The tighter the niche the better. For example,if you want to know about fracture mechanics of aluminum substrates, you will not be interested in someone dealing with die bond failure of copper interconnects in semiconductors. Failure analysis is too segmented for a generalist to have much of an input. (I'm a generalist fwiw)
444257
The ultimate strength of a beam is found by US = max load/CSA.  There's not a lot of math to it, but you do need to measure pretty carefully.\n\nThere comes a problem, however, where some materials "stretch" before actually failing, thus making the measurement of the max load difficult.  As an example, if you've ever pulled plastic wrap apart, it gets easier at a point right before it breaks into two.  The point where after you pull on it and it gets easier is called the yield strength.  After you pass the yield strength, you don't need as much force to cause failure in the material.\n\nNow in steel, the yield and ultimate strengths are pretty close to each other that the above discussion really doesn't affect it if you are trying to break a beam on your own.  But the ultimate strength of steel is probably something you can look up in a table somewhere.  Then you can use the above formula to see how much your beam can take based on the US and the cross sectional area.
960760
1 Yes xeroxed material will serve the purpose. Try & get from diff sources like IMS , Career Launcher etc\n\n2. Career Launcher is better if you are non Engineering & need more help in Maths
238500
need to know the type of metal, then find the shear force & plug it into your equation.  don't forget to factor in each thread  (Basic solid mechanics)
600829
We need a little more info.  How is the bar supported?  (Cantilever, simple at both ends, etc)\n\nAlso, how is it tapered?  (Height, width, diameter...)\n\nHow much is it tapered?  (Percent of taper parameter at small end)\n\nAll the above will make a huge difference not only in the actual displacement, but the form of the equation as well.
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
660156
http://www.physlink.com/education/askexperts/ae25.cfm