what affects the deflection of a cantilever?
101115
It is depended on the moment and the surface moments if inertia of cross section of cantilever.\n\nMoment is force time distance : M=r*f\n\nand I is used to show surface moments if inertia.\n\nthe more the moment more is the deflection.\nthe more the I less is the deflection.
123992
I answered your Treb question, and the same holds true here.  Size, construction materials, etc. all can affect the throwing of an onager.\n\nThe http://www.algobeautytreb.com/ site has a section about half-way down called 'Onager Physic' with a link to a pdf.  You might find that helpful.
552283
First of all, a maximum deflection of L/720 (I think) is a bit of an "overkill" if you are designing a floor for residential use. I believe the code allows us to use a minimum of L/240 deflection for our floors joists. A stronger design deflection criteria of L/360 for our joists is more than sufficient to carry a live load of 40 pounds per square foot and a dead load of about 15 pounds per square foot.\n\nResidential codes require a minimum of 1-1/8" thickness for both the deck and subfloor. We normally use 1/2" decking under a 5/8" thk. sub-floor for joists spaced at 12" and 16"  o.c.\n\nI understand that spacing your floor joists (or your floor trusses...a truss is a structural system made up of horizontal, vertical and diagonal members joined together to act as a single member....are you really using trusses on your floor?) at 24" o.c. is more economical than spacing them at 12", 16" or 19.2" o.c. Assuming you are using 12" square tiles (granite), the spacing between the joists would only give you 2 tiles on that spacing. I don't think that you would need a subfloor thicker than 5/8" if you have a decking of 1/2" underneath. The sheathing's (plywood or OSB) main use is to distribute the loads unto the joists. I don't think two tiles will deflect that much in a spacing of 24" o.c. Besides, the deck and the subfloor are laid out in a certain pattern (which I assume you already know) that will make it stronger and thus have some fire rating qualities. \n\nThis means that you need not worry too much regarding the thickness of your sub-floor as long as you install the minimum thickness with the minimum decking thickness required. What you should worry about is if your joist (or your "trusses" as you term it) will be able to carry the load superimposed on it and be able to transfer these loads into the supporting walls (or beams, which is another question) and unto the foundation wall or footings below.\n\nMy suggestion, since your question is a bit unclear (what is the use of the floor, is it for residential or commercial, is it located on the first floor or on the upper floors, is it to be used as a cantilevered balcony or a supported deck,  is it to be used for the exterior or interior since exterior use may mean snow load, etc.)is for you to call the tile manufacturer and give you some answers giving them the description of how you plan to lay out your joists (trusses).\n\nI am an architect and one thing I always remember regarding structural questions....if I can't do it myself, let the manufacturer do it for you (be sure to put their answers in writing) or better yet, ask a structural engineer.
1153005
It depends on the material of the bow, and how far it has been stretched. Every material has a stress/strain curve that determines (1) how much the material bends under a given load; and (2) how much load is needed to cause permanent deformation. If the bow and the force on it are well below the limit of deformation, the bow will keep its energy permanently until disturbed. \n\nIf the bow is at its deformation point, it will bend permanently and the kinetic energy will be lost in changing the shape of the bow. For a well designed bow, this should never happen even at the maximum strain point.
1180056
Consider a ball that is spinning about an axis perpendicular to the flow of air across it (see left). The air travels faster relative to the center of the ball where the periphery of the ball is moving in the same direction as the airflow. This reduces the pressure, according to Bernouilli's principle. The opposite effect happens on the other side of the ball, where the air travels slower relative to the center of the ball. There is therefore an imbalance in the forces and the ball deflects - or, as Sir J J Thomson put it in 1910, "the ball follows its nose". This lateral deflection of a ball in flight is generally known as the "Magnus effect".
1108087
Check the shafting and bearings for deflection.
489536
By strength I suppose you mean elastic calculation when the body is subjected to deforming forces. Actually, the centroid of a shape having an unifrom mass distirbution coincides with its center of mass. This point is central to the analysis carried out in rigid body statics and dynamics.
207863
see:\nhttp://en.wikipedia.org/wiki/EngineerIng
863216
It has to do more with the material of the ground rather than with the surface. It all depends on it's elastic coeficient. The least elastic, the more it will bounce because of the transmission of the kinetic energy.
572487
Distribution of force over area.
667120
At the spring's original configuration, there is a force stretching it in the form of the mass.  The force and the amount of deflection are related through\n\nF = -kx\n\nSo the spring is stretched -F/k, or -mg/k, initially.\n\nWork to lift the weight is given by\n\nW = Fd\n\nWe have a force and a distance, so the work is then\n\nW = (mg)(-mg/k)\n= -(mg)^2/k
774877
The equation doesn't differ for different springs, only the value of spring stiffness k does. A different value of k means the deflection x resulting from a given force Fm will differ also since Fm = k*x.
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.
280266
Force exerted on a material can cause stress and strain.  If you apply hand "pressure" to a glob of J-ello it will cause "stress" and deformation "strain."  If you set a weight on a surface, the force of the weight will cause "compressive stress" as it compresses the molecules of the surface that must strain slightly to bear the added force.  If you suspend a heavy magnet from a steel surface it causes "tensile stress" from tension that causes slight strain in the surface to support it.\n\nIf you take say a one square inch rod and press it down against a surface with a force of 100 pounds, the stress in the rod will be 100 PSI (pounds per square inch) compression and the stress on the surface will also be 100 PSI compression.  However, the force within the surface will become better distributed away from the contact point and the local stress at any such point will be less than 100 PSI.  Therefore, the cause and effect are both measured in PSI but depend for value on the distribution of the stress.  Although stress is calculated or measured in PSI the area in question may be much larger or much smaller than a square inch (like driving a car 60 MPH regardless of how long the trip is?).  Hope that helps a little.
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.
858223
weight (second order effects), roughness, types of materials, speed (second order effects), moisture content (humidity. affects materials such as wood), temperature (ie hot rubber sticks). if the object is moving fast, then aerodynamics can actually effect it by forcing air into the area of contact for example (eg hydroplaning on wet road where water gets between rubber and the road).
303242
depend on the allawable deflection \nfor 10 mm deflection \nif the 12 is vertical 8 m span \nif 6 is vertical 6 m span