I am looking for the weight of Dense Select Structure 2x10 lumber, in pounds per linear foot?
447021
From the ref. (see table 22), Douglas fir at 15% humidity has a weight per foot of 0.233 lb * cross-sectional area in square inches. (Douglas fir happens to be the only wood the ref. lists with a Dense Select Structure rating.)
651326
approx:200 per sqr.inch.
312260
It's about 10 lbs per square foot. (Multiply the density of steel, about 490 pounds per cubic foot, times the thickness of the plate in feet, 1/4 in. = .02 ft.) I'm assuming you mean 1/4 in. plate.
430952
Well, what quality to you want to measure -  size?  The weight?  The density?
296138
Each mix design for concrete will vary the weight and density of the mix. Air entrainment, water to cement ratio, coarse stone% and fine aggregate%, cement content, admixture content, fly ash substitution all contribute to the variation in weight. I do agree that 150 lbs per cubic foot is a recognized average when calculating concrete weight. since there is 27 cubic feet in a cubic yard, (150 lbs x 27 cubic feet) one cubic yard equals 4050 lbs or 2.025 english tons.
27505
Conventional concrete has a density of about 2300 kg/m3.\nThere are about 10.76 cubic ft in a cubic meter and 2.2 lbs in a kilogram. So that means that a cubic foot of concrete would weigh 2300 x 2.2 / 10.76 = 470 lbs.\nA half a cubic foot would be 235 lbs.\nSpread over 144 square inches that would be 235/144 = 1.63 psi.\n\nThat is a 6" slab of concrete would exert 1.64 psi on the ground underneath it. Was that our question?
282242
that's a real vague question, try to be more specific or at least say what king of structure do u want? (steel, conc, ..?)\nanyways this is an exremely simple example.\n\nlets' say u want to design a tension member -carrying tension force = 100 tons, -between 2 supports and this member is a steel member of yield stress= 1.4\nso to get the area u need :\n100/1.4= 71.43 cm square let's say again u want it to be I-shaped steel then u'll need  I number 400 i.e IPE 400
1055994
Try www.warehousingonline.com
332070
Partition this into parts of known sizes ,then calculate weights  of each part and its cent of gravity. Then take a point outside the assy. The moment of each part about this point equals the total weight times  the dist of the CG from this point. Repeat this process to find the vert and horiz position of the CG.Consider a letter L shape.,height l1,cross secton a1,b1 inches.weight=w1=l1*a1*b1*0.284 (steel) lbs.Bottom of L  , length l2, cross sectio a2,b2. weight=w2=l2*a2*b2*0.284 lbs.Then dist of CG from corner of L is w2*l2/(2*(w1+w2)), and dist up from corner is \nw1*l1/(2*(w1+w2)).- I hope ?
629615
The easiest way to find out would be to find a pack of 24 pencils (#2 I'm assuming) and look at the packaging.  Divide the weight measurement by 24 to get the weight.\n\nI'll look around and see if I have some...\n\nEdit:  I searched and searched and searched, but couldn't find any here, or even pictures on the web.  The resolution of the pictures was too low.  And even looking through the different sites of pencil manufacturers, online office supply stores, etc. I couldn't find the answer.\n\nFinally I did a search for "pencil gram" and found a cached page from a National Geophric site that indicates that a standard wooden pencil (new) weighs about 7 grams.\n\nSo my answer is about 7 grams or 1/4 oz.
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?
569185
First off, just take the cube root (or use that x^y button on your calculator and raise the volume in question to the 0.3333333333 power). 1.8x10^6 cu ft would then be about 121 feet on a side.  Now, yourreal question is how big is 27,000 tons (I assume english units, not metric tons since you went to cubic feet).  That will depend on the density of the substance in question.  27,000 tons of water will take up more room than 27,000 tons of iron. The volume you gave gives a density of about 30lbs/cubic foot, which is about 1/2 the density of water.\n\nThis does sound like something hollow, like a boat, or maybe it is wood. That is a big block of wood. You gonna carve yourself an office building or something? A ten story cubic building is about that volume.
1133778
http://online.unitconverterpro.com/metric-conversion-tables/convert-alpha/density.html
242811
www.constructionbook.com/xq/ASP/ ProductID.3422/id.341/subID.752/qx/default2.htm \n\ntry here....i find it in google
881707
Determining the mass is trivial: just weigh the block.  For the density, you will need to know the volume, which will require some measuring with a ruler.  Then divide the mass by the volume for your answer.
681402
assuming 1.39/lb.\nit would take 7.5 lbs.\nso it would cost $10.43
859844
The previous answerer has the right idea but the wrong details.\n\nThe four parts should weigh 1, 3, 9 and 27 pounds.  The logic is that the desired weight is written in "balanced ternary" notation, where the values of the digit positions are 1, 3, 9, 27, 81, . . . (that's "ternary"), but the allowed digit values instead of being 0, 1, 2 (that's regular ternary) are -1, 0, 1 (that's balanced ternary).  Then the digit values for any desired total weight written down in balanced ternary tell you whether the weight for that column is to go in one pan, go in the other pan, or stay on the floor.