How to design an RCC beam with BM in both direction of the lateral axis acting simultaneously?
609585
Use the biaxial bending provisions for columns.  If you want to make the math easy, give it a small unit axial load.  You used to have to do this in steel, too.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
800938
WELL YOU CAN MAKE A PRISM OUT OF A CARDBORD AND \nADD 2 LEGS BELOW IT AND IT WILL WORK
387124
http://jef.raskincenter.org/pictures/paper_beams.html\n\nhttp://www.serve.com/chunter/index/info/beams.html
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Did this way back in Boy Scouts....Relatively easy...just need to build 2 electromagnets and a series circuit..... I'll see if instructions are online.\nhttp://www.terrax.org/comm/morsecode/buzzer2.aspx\nhttp://chss3.montclair.edu/~pererat/perbuild.htm\nthe second one is similar to what I built
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You can Use Dean and Stark Apparatus To Separate them \n\nAlso one Source i m providing you which involve the separation using PTC\n\nRefer it \n\nhttp://www.ptcorganics.com/Top10PTC.pdf
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A is on the x-axis (horizontally), while B is on y-axis (vertically).
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The deflection of a simply supported beam is given by (5/384)*(wL^4)/EI\n\nWhere w is the distributed loading, l is the beam length, E is the youngs modulous a material property and I is the moment of inertia also known as the second moment of Area.  (Slope has a similar equation but is dependent on L^3)\n\nIf both cross sections are made of the same material then the only variance in the two deflections is the moment of inertia I.\n\nBecause the I beam is designed to resist moment  it has most of its mass in the flanges  where it is furthest away from the centre of mass of the section and will therefore contribute most effectivly.  Engineered materials often take advantage of this property of I beams to save materials, you can reduce the amount of material needed to satisfy strength and deflection requirements by using an I beam over a solid rectangle.\n\n\nSo for the same area an I beam will deflect less, resist more moment and be effecient where as a solid section will deflect more have a greater slope, resist less moment and be very inefficient
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Here are two (see refs.). Note that the second one is a rotating 3-D model.
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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'
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I doubt that a "planning board" would likely follow or be much swayed by your calculations of barycenter, especially since they use the generalization "most" which is anything over 50%.  Perhaps you can concentrate each geometric building at its vertical centerline (center of square(s) and rectangle(s), circle(s), etc.  Then assign a "mass" to each "vertical point" on a ground plan.  Does "most" refer to building volume or mass?  Likely volume which you can scale from 3-D?\n\nDraw two lines (north-south and east-west) through approx. center oof plan.  Fix one end of the first line (north-south?) and assume the buildings rotate about the fixed point (with the plan held in a vertical plane and the N-S line horizontal).  Multiply each building "mass" times the distance to the fixed point (in feet) to obtain all the "moments" about the point.  Find the balancing moment in the opposite direction (a single force for the combined mass of buildings times a distance "x").  Repeat the process with the E-W line horizontal to find "y".  The barycenter should be at the intersection of the balancing moments ("x" and "y").\n\nRepeat the process with the new buildings included.  Compare the pre- and post-building project barycenters.  Good luck.
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put an arm perpindicular on the shaft, put another arm pinned to the first, and slide the second through a sleeve.\n\njust like an old fasioned steam engine turned linear movement (piston) into rotary(drive wheel
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Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
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On paper, it is simple, u will use a y-axis for elev,, x-axis for distances --- like a Profile layout.\nBut i reckon u want a display 'model', well same logic, and u can use diff color.. only use very fine plumb lines or plastic threads as interconnections. Better encase the result in a glass/Plexiglas cabinet.
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NO way to do it cheap and safe.\nAlso, due to the size, moving is difficult, unless you can build it in parts, and keep it together while using it.
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Add all sides together
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I certainly hope you have visited the link below for visual and mathematical insights into this problem.\n\nHowever, the only materials that come to mind which lend themselves to a physical model of this transformation are:\n\n o thin-but-rigid springs or "lead" wire(s) as supports/frames\n o plastic wrap as a surface substrate\n o flexible, rubberized, quick-drying paint as a protective layer\n\nI regret that my imagination is limited with respect to this problem, but good luck with your project!