how to calculate a balcony structure?
371498
About the same as any raised floor.  Posts, beams, joists, decking.  There are guides for this at lumber places.  Also check local codes, they ALL vary.
790576
You have to find basic wind speed, K1,K2,K3 factors, Design wind pressure, Gust factors, exposed area & solidity ratio, drag co-efficient.\nWind load on panel = ExpArea*Drag*Gust*Design Wind Pr.\nPanel load is distributed to nodes & you can analyse and find the axial forces (Comp & ten) treating these are truss members.\nThen design of these members is as explained in IS802.
679755
I would take half of the area between adjacent rafters and multiply by 220 to get the load in pounds for end rafters.\n\nI would take double that for interior rafters.\n\nThis assumes that you have the same spacing between all rafters.
959992
depends on which city you are building that that model for ,as different cities have different rules regarding the building heights,but if its just for fun then divide the total height of your model with the number of floors you want ,to get the exact height of each floor.
359093
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
765616
Since the volume is 1000, \npi*(r^2)*h = 1000, and so you can express h in terms of r.\n\nCall the cost per sq cm of the wall 1 unit.  Then the cost of the ends is 2 unit per sq cm.  Apply the areas, and get\ncost of wall = 2*pi*r*h  =  (sub in the expression you got above for h in terms of r)\nCost of ends = 2*pi*(r^2) * 2\n\nWrite f(r) for total cost, and as usual differentiate etc to find minimum value of r.  Obviously r and h must both be positive, and that's the only restriction on the domains.  Hope you can do it yourself from here.  You need the exercise and I don't.
10513
Area = base * height\nBase = Area / height\n     = 50.15/ 5.9\n     = 8.5
831079
Then you measure the length of the shadow of the building (call it L)\n\nThe height of the building (H) is to 3 feet as the length of building's shadow is to 4 feet or\n\nH/3 = L/4 or H = 3/4 * L
703001
To find the force P, its equal in size, but opposite in direction to the weight of the crate,resolved in the direction of the slope:\n\nP=Wcos60 or P=Wsin30\nP=(80*9.81)cos60=392.4N\n\nR is equal in size, but opposite in direction to the weight of the crate, resolved perpindicular to the slope\n\nR=Wcos30 or R=Wsin60\nR=(80*9.81)cos30=679.66N\n\nPlease note you need to use weight, not mass ( W=m*g)\n\nGo do the math!
902316
Let y be required distance from the far end.\n\nThe weight is acting at the C.G i.e., at 2 m from the far end.\n\n2/3 of the weight w is acting upward at y and \n\n1/3 of the weight w is acting upward at 4m from the far end.\n\nTaking moments about the far end, \n\n2/3 W y + 1/3 W 4 = W x 2. \n\n Dividing both sides by 2W and multiplying by 3 on both sides\n\ny + 2 = 3 \n\ny = 1m.
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
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'
168445
first find a vicinity plan. \nget a straw(in sodas) and a triangle\nstand on a specific location with the distance of X from the building.\nmake sure to know your height.\ncreate a transit by putting the straw along the hypoteneous of your 45x45 triangle.\nlook through the straw (try to see the zenith of the building) and measure the angle of depression.\nthen, simply apply the law of sines and cosines etc etc.\nthis will not give you exact building height but an instant approximation. this can also be used in finding approximate height of towers, canyons, mountains, and volcanoes.
284409
Ex:\nA man stands 60 feet away from a building and sees the top of the building at a 60 degree angle. How tall is the bulding?\n \n1)Choosing an equation: (sin, cosine or tan)\nGoing to those three equations which one do we use?\nWell, we want to find the hight of the bulding (opposite b/c it is opposite the angle), so lets use tan.\n \n2)Setting the equation:\n \ntan 60 (the measure of the angle is θ) = (x) (opposite) / 60 (side adjacent to angle is adjacent)\n \n3)Press on Calculator: tan 60\n \nU should get: 1.732050...\n \n4) Multiply by 60. (b/c ur solving for x , see original equation),  \n5)Answer:  about 103.9230485\n \n6)Using 30-60-90 Triangle tricks we check our answer and find the height is 60 X (sqrt 3) which is aproximately 103.9230485 so we are right.
821443
A.\nwidth=one side of the existing shed\n        =20m\n\nlet \nlength=x\nwidth=y\ny=20\nfence=160\n160=2(length)+(width)\n160=2(x)+y\n160=2x+20\n2x+20=160\n2x=140\nx=70\nlength=70m\nmeasurement=70x20\narea=1400m^2\nB.\nperimeter=180\n180=2x+2y\n2x+2y=180\n(/20):x+y=90\ny=90-x\n\nArea=x\nsubs y=90-x into eqt.A=xy\nA=x(90-x)\nA=90x-x^2\n-differentiate dA/dx\nda/dx=90-2x\nda/dx=0\n90-2x=0\n-2x=-90\nx=45\nsubs x=45 into eqt. y=90-x\ny=90-45\ny=45\n\nlength=45,width=45
565159
20 vertical and 20 horizontal  then fix them on each other using tacks with add separating material .. then u will have 10 wood left left .. put two of them over each other in each side .. so u will have two left .. to keep u move easily in the sea .. :D .. good luck .. nice Q
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area