a prestress beam were lifted at 2 points what will hapen to the stress at top and bottom at this 2 points?
547007
Hello Hassan,\n\nDepending upon the materials used in the construction of said beam, it is possible that the beam has been lifted at the points of a harmonic. When the slightest breeze passes by it will induce a vibration that will continue to ascend, impelled by the very little help of the breeze. Soon, all ears nearby will be held and crying screams will be heard while the beam rings ever increasingly loud.\n\nAfter some point in time, the stresses in the beam caused by the vibration will overcome the tensile strength of the entire beam and cause it to shatter like glass.\nYour question is moot unless the construction workers can place the beam in position before the crescendo of vibration has time to escalate.\n\n-Gabe
701346
the smaller one would be stronger, as it would not suffer as much tensile stress (or strain) per volume area..... all materials have a certain amount of pressure and stress they can withstand before buckling or deforming in someway, and is known as 'youngs modulus' (this is a ratio between the stress and strain limits of a material). the forces on a bridge that enginneers worry about are more often that caused by the bridges own structure than of vehicles and other things across it, such as the weight of the steel and concreate been pulled down between two supporting pillars by gravity, or the weight carried on suspension cables.\n\nso even if the scale was increased 5 times, the limits of the material (steel for example) would not increase, the thicker the steel supports, the more spread the pressure (force over area) is, tru..., however, this is not sufficient enough to support the new weight load; for example... imagine a cylindrical steel support.... even if the scale increased 5 times, the forces would exceed the limits, the structure would be 5 times greater... but unless the limits of the material became 5 times greater... it would not hold....\n....... i think.\n\nhope that helps!\n\nthat psychics explanation!!!
665747
Working Stress Method\n• Material is assumed to behave elastically.\n• The allowable stresses are obtained by dividing the limiting stresses of material by factor of safety.\n• Ignores uncertainties in different types of load.\n• Does not use the full range of strains in the material.\n• Therefore, disregards the nonlinear part of the material\ncurve.\n• Considers failure as a function of stress while it is a\nfunction of strain.\n• Stress as a measure of safety does not give true margin of\nsafety against failure. A stress factor of safety 3 for\nconcrete does not mean that the member will fail at a load\nthree times the working load.\n• The structure must carry loads safely. It is logical to use\nthe method based on load causing failure.\n• The additional load carrying capacity of the structure due\nto redistribution of moment can not be accounted for.\n• Produces conservative designs.\n\n\n\nUltimate Load Method\n• The loads are enhanced by load factor-\n• Design load = LF * service load\n• The method uses total stress-strain curves of the\nmaterial.\n• Strain based failure.\n• LF=3 means the structure has 3 times more\ncapacity\n• Since the method considers the plastic region of\nthe stress-strain curve also, it utilizes the reserve\ncapacity of the member.\n• Uncertainty in load only is considered. Uncertainty\nin material is ignored.
376669
They are the stresses normal to a rectangular element of the item being analyzed for stress.  A good explanation can be found here.......\n\n\nhttp://acm.uva.es/p/v104/10429.html
652894
the amount of pressure exerted on an object
959600
need to know the mechanics of the lift you are proposing. Otherwise the answer is just 826 N.\n\nAre you worrying about compresive load? Tensile Load?  Which Joint?
856591
You left out a LOT of info, like\n1) Loaded as column or beam?\n2) If beam, concentrated or distributed load?\n3) What kind of wood?\n4) Answer based on engineering equations or building code?
712075
Calculate the perimeter of the pile, multiply by the thickness of the raft (assuming it is made of concrete or equivalent solid material) to give the area that resists shear.  Divide the design tension of the pile (or the compression in case of low tide, if applicable) by the above area that resists shear to give the theoretical shear stress.\n\nThis is the basic theory behind it.  \n\nYour local building code should give more details.  For example, the circumference may be increased according to the thickness of the raft, and so on.\n\nIf you are not experienced with these calculations, you may want to have someone experienced and competent to check it, because any failure may involve a lot of money, and/or lives.
507550
Wow... a lot of complicated answers...simply put... the majority of your force (or pressure) will be on the arch sideways with the most force on it at the initial take off point. It will gradually decrease as it goes round and up as gravity will take hold and it looses it momentum due to the friction against the arch(not allowing it to go straight. Remember that a body in motion wants to stay in motion. If the arch is not tight, it will carry most of its (the projectile) energy to the top platform, so make sure it is prepared for it.
606209
The technical answer to your question is that the moment of inertia of the beam's cross section is greater when speaking of the I-beam, and this means less bending.\n\n\nDumbing it down a little. The I beam will be able to support more weight.  Look at the cross-section of the materials.  The I-Beam has a height of 6 inches.  the square tube has a height of 3 inches.  The bending of a 6" piece of steel vs 2 3" pieces along their thickest portions is harder when the material is farther from the midpoint, ie, 6" is farther than 3".\n\nAs far as your max deflection and max load are concerned, you need to specify the types of steel, and the type of I-beam would be useful as well (S type, W type), as would the type of square tube (squared corners or rounded).  These won't make a huge difference, but it will make a little.
83815
The I-beam puts its mass where it will do the most good, at the top and bottom (assuming a vertically applied load) where the compressive and tensile stresses are greatest.
461030
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.\n\nFor each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.\n\n16N/(9.8 m/s^2)=1.63 kg\n1.63 kg*30 m/s^2=(force of feet)-16N\nForce of feet=64.9N\n\nAnswers for the second and third scenario:\nb) 2600N\nc) 260N
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The stress in the cable is m*g/A.  The Young's modulus of steel is about 200*10^9 N/m^2 (pascal).  Therefore the strain will be\n\nm*g*/(A*200*10^9).\n\nThe elongation is strain*length.\n\nFor the second part, replace g above with (g+a), where a =2.7m/s^2\n\nFor the last, compute the mass from S = m*(g+a)/A, The elastic limit is Sm, so the max mass is m = Sm*A/(g+a)\n\nNote: keep the units straight.  Use kg, m, sec, N in all calcs then convert to the asked-for units last.
572487
Distribution of force over area.
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The tension is the weight + the force Fb required to accelerate the bundle along with the rest of the elevator. \nNeed to calculate the elevator's acceleration a\ny = (1/2)*a*t^2 where y = 1m and t = 1.6s. Solve for a.\n\nNow calculate Fb\nFb = m*a. Solve for Fb.\n\nT = W + Fb
183249
2 times the distance from one end to the center.
826246
smaller tension = 246 N\nlarger tension = 654 N\n\narrived at by summing the moments around one end of the scaffold (equation #1) and then doing a force balance on the scaffold (equation #2)