Looking for an I-Beam size. Clear span is 35 ft, caring a total of 70,000lbs evenly distributed over span.?
923963
Traffic timing for this much weight is also considerable. A\n   building code shows that you might need 15 or more for\n   this area so titanium is probably out.  The core value of\n   bronzed steel is ok for the average busy building.  Size again\n   must be measurables ranging from 2.7 to 3.7 made in 20 foot\n   can be made in 35 precise.
759565
Okay, letting h be length of support beam, and x be length of ground, then h=3x-1, and 6^2+x^2=h^2.\nSo, 36+x^2=(3x-1)^2 =9x^2-6x+1.\nSo, 8x^2-6x-35=0.\nAnd solving I get x=-7/4 and 5/2.\nSo, of course -7/4 wouldn't make sence, so x=5/2=2.5.\nSo the support beam 3x-1 = (3*2.5)-1=6.5 foot.
461466
What did your M3 weigh before? Maybe 3500lbs? Then it's easy enough:\n\n(3500 + 100) / 3500 x 240 = 246.9\n\nYou need another 7bhp. But pay attention to the other answer that mentioned the effect on handling.
36563
the number ranges from 350 to 700 depending on size and weight.
963852
The original deck has a total area of 120ft².  The new deck has an area of 192 ft².  That's 72 ft² bigger.\n\nWe need to look at what percent of 120 is 72... Division is the key here...\n\n72/120 = .60\n\nThat means the new deck is 60% larger, hence the cost would be increased by 60% also.
1116334
easy, just do 70-53 = 17 X 10^-6/c = answer\n70-83 times 10^-6/c = answer\n\nthen just 350 ft - answer\n\nthis is how u get it
692400
I would go to Google and type in the search window 70.75 inches in centimeters\n\nIt tells the answer.  Then do the other 2.
201698
The volume in cu.ft will be 70'x1'x1/3' = 23 1/3 cubic feet \n27cu.ft = 1 cubic yard\nYou'll need 0.86 cubic yard
1187485
HOW MANY RIVET'S OR BOLTS WOULD BE  NEEDED FOR A 789 FOOT BRIDGE WITH 2 DECKS AND 2 LANES FOR VECHILE ON EACH DECK CROSSING A RIVER WITH NO STEEL CABLES USED IN THE CONSTRUCTION OF THE BRIDGE     \n\nANSWER 1,234,345  GIVE OR TAKE 2 OR 3 THOUSAND
753895
PV=nRT\nn=PV/RT\n\nR=10.7316 ft³ · psi · °R-1 · lbmol-1\nP=14.7 psi\nT=530 R, 540 R\nV=100ft^3\n@70F=530R\nn=14.7*100/(10.7316*530)=.2585\nmol wt=28.8\nm=28.8*.2584=7.44 lb\n@ 80f=540R\nn=14.7*100/(10.7316*540)=.2537\nm=.2537*28.8=7.31 lb
456275
9.81 m/s² or 32.2 ft/s²
473534
71.5972 or 178993/2500
545032
I work with a builder in the DFW area that does that.  Shoot me a message with more specific questions.\n\nThanks, and good luck.
150631
300 ft, with a repeater more.
242811
www.constructionbook.com/xq/ASP/ ProductID.3422/id.341/subID.752/qx/default2.htm \n\ntry here....i find it in google
296473
Check with some local contractors or place your add with the local newspaper.  Good Luck!
254925
I'm assuming that the pivot point will land between the 15 and 30 Kg loads because it seems more logical. If you try putting the X distance between the 20 and 15 loads, you'd probably get a negative distance. \n\nActually, the weights have to be multiplied by the pull of gravity (9.81m/ss) to make them actual forces but then again, they would all just cancel out.\n\ni hope you understand the crude diagram\n\n|-------10------|       - distance where weight of beam is applied\n|---------x-------|    - pivot point dist from left\n ____________________  - beam (20 m)\n|_________|__________|\n20_______15_________30 - applied forces (weights)\n\n20kg(X) + 15kg(X-10m)  = 30kg(20m-X)\n20kgX + 15kgX - 150kgm = 600kgm - 30kgX\n20kgX + 15kgX + 30kgX  = 600kgm + 150kgm\n                65kgX  = 750kgm\n                    X  = 750kgm/65kg\n                    X  = 11.53846 m from left end