I'm doing a research on the prediction of the behavior of shear wall. I need to use the Truss Model Theory from Hsu and Mo journal titled "Softening of Concrete in Low-Rise Shearwall". I have problems understanding the equation and substituting the values into the equations. I hope anyone out there can help me with a relevant example. Thanks!
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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.
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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.
644924
That is a very tricky question, I am doing my undergraduate thesis on this topic in comparing the 1995 NBC with the more current 2005 edition.  The loading for which you design must be specified by your countries or provinces building code.  Because wind speed varies all the time and earthquakes come in all different magnitudes, a reasonable value for each must be used in the desing of the lateral supports for the building.  Generally a statistical anaylsis will determine the design wind force or quake strength.  The find speed can be converted to a pressure using the density of air and must be considered on the windward and leeward sides of a building.  It increases with height.  Earthquake loading is much trickier since the ground is shaking but generally triangular loading is used with the most force at the top and the least at the bottom.  the 95 code also specifies a concentrated force at the roof
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Normal fault: hanging wall moves down relative to the footwall.\n\nReverse fault: hanging wall moves up relative to the footwall.\n\nSorry don't have time to search for fault diagrams...\n\nwriting lesson plans right now\n\nyou really should do your own homework. :P
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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.
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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)
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Looking at your very specific question, I'm assuming you have a reasonable handle on the engineering basics of what happens when you apply a force or load to something. \n\nLet's start by defining 'Strain'. If you take a metal rod, for example, and pull it by applying a large load at a prescribed rate, then the 'Strain', at any given point of time in the rod's elongation, is the ratio of the original length of the rod to the rod's current length. So 'Strain' is simply a measure of how much something has changed from its original shape, 'length' in this case. This is also called 'Engineering Strain'. \n\nWhen you keep pulling the rod, at a certain load it starts to yield at a point along its length where it forms a 'neck' before failure (assuming a 'ductile' or reasonably soft material). If you measure the cross-sectional area at the neck you can see that it reduces as the rod elongates. The 'Engineering Stress' is the ratio of the applied Load to the ORIGINAL cross-sectional area of the rod.\n\nThe point at which the metal rod starts 'yielding' is considered to be beyond its 'elastic' limit - its capability to 'recover' or get back to the original length. The strain beyond this 'yield' point is called 'Plastic Strain'.\n\nIn contrast with 'Engineering Strain', 'True Strain' is the natural log of the ratio of the original cross-sectional area to the current cross-sectional area at the neck. Correspondingly, the 'True Stress' is the ratio of the Load to the current cross-sectional Area. \n\n'Effective Plastic Strain' or 'EPS' is simply the Total True strain minus the Recoverable strain. So, it is the Residual True Strains after unloading elastically, or, in other words, its the unrecoverable portion of the true strain beyond the yield limit. \n\nUsually EPS is just called Plastic Strain, but when you are dealing with True Strains input in specific engineering applications or 'solver' software they like to define it as 'Effective' Plastic Strain. In that case, where you are defining a stress vs. effective plastic strain curve, the first (plastic) strain value would be zero and the first stress value would be the initial yield stress.\n\nSee below link for more details.
54845
Go to http://rutchem.rutgers.edu/Courses_f04/Chem161/Lecture19k.pdf#search='lewis%20structure%20for%20nitrous%20oxide' and scroll down until you see it
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So basically we're resolving all forces (no friction forces) for their horizontal components, summing them and then calling this the resultant force Fww that the wall exerts on the wedge.\n\nFirst step: resolve the forces acting on the block (of mass M) on the plane at angle theta to the horizontal\n\nF1 = normal force to the plane = Mgcos(theta)\nF2 = force in direction of (tangent to) the plane = Mgsin(theta)\n\nSecond step: resolve both forces found above into their horizontal components (i.e. normal to the wall)\n\nF1x = Mg sin(theta) cos(theta)\nF2x =Mg cos (theta) sin (theta)\n\nFww = F1x + F2x\nFww = 2Mg cos(theta) sin (theta)
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a) The stress in the wire is given by force per unit area = m*g/A\nIf Y is the Young's modulus of the material of the wire, \nY = stress / strain. Therefore \nstrain = stress / Y.\nElongation = strain*original length.\nb) When the object is accelerated upward with an acceleration of a, the force acting in the wire = m*(g+a)\nThe rest of the equations above are valid.\nc) If the elastic limit of the wire is E, the greatest mass that can be accelerated up with an acceleration of a is got from \nE = Mmax*(g+a). Hence\nMmax = E/(g+a)
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This is your garden variety related rates problem.\n\nAt any instant when the monkey is on the wall, you have yourself a right triangle, with the taut rope being the hypotenuse, the distance from fastening point to the wall being the horizontal leg, and the distance the monkey climbs vertically being the vertical leg. The one equation which relates all three of those quantities together is the Pythagorean Theorem: x^2 + y^2 = z^2 (x and y being the legs and z being the hypotenuse).\n\nSince this is a related rates problem, you need to differentiate this equation with respect to time (t). The result is 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt). All of the d*/dt terms represent the rate * is changing over time.\n\nSo, now it's a matter of plugging in the appropriate values given in the word problem. According to the word problem, x=8 ft, dy/dt=4 ft/s, y=6 ft (at that instant). From this info, one can derive that z=10 ft (via Pythagorean Theorem) and dx/dt=0 ft/s (because the distance between the fasten point and the wall is fixed). The problem is asking for dz/dt. Now, plug in:\n2(8)(0) + 2(6)(4) = 2(10)(dz/dt)\nSolving for dz/dt, your answer should be 2.4 ft/s (if the math is correct).
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This site should help\n\nhttp://www.technologystudent.com/struct1/model1.htm
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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.
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HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
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click on the below link, and choose any one of the results. I would advise reading at least 3 of them to get a feel for theory before choosing one of the walkthroughs.\n\nBest of luck to you!
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This link for a PDF File in Rapid Share website, it a book called "Bridges Design Manaual"\n\nI hope it will help you.\n\nhttp://rapidshare.de/files/37074217/bridgedesignmanual.rar
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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.