i think 15
Anal
249$
well , reinforcement steel required for slab is nearly 0.7 to 0.8% of the volume of concrete so if concrete is 1 cum then steel will be 1*0.7/100 =0.007 but density of steel is 7850 kg/ cum so 0.007*7850=54.95 say 55 kg per cum so steel required to reinforce 1 cum concrete for slab is approximately 55 kgs.
i dn k=12*45
The steel sheet in question has a volume of 1 x 1 x 0.002 m3. Mild steel has a density of 7800 kg/m3.[A]As such the mass = 7800 x 0.002Mass = 15.6 kg[A] Cobb, F. (2009) Structural Engineer's Pocket Book, 2nd Edition. Amsterdam, Elsevier.
Concrete is composed of cement and other cementitious materials such as fly ash and slag cement, coarse aggregate made of crushed stone, fine aggregate such as sand, water, and chemical admixtures. In reinforced concrete, steel is introduced in to the concrete. In plain concrete, no steel reinforcement is introduced. Generally tensile and compressive strength is taken by reinforced concrete and only compressive strength is taken by plain concrete
Let us consider a tor steel of diameter 'd' mm and length 1 m d mm=d*10^-6 m Area of cross section,A=(3.14*d^2*10^-6)/4 Volume of steel bar of length 1 m=((3.14*d^2*10^-6)/4)*1 m Sp.Weight of steel=7850 kg/m3 so,Weight of tor steel=((3.14*d^2*10^-6)/4)*7850 Kg Hence 162 is nothing but a constant used for calculating weight of all type of tor steel and it is found out from; (4*10^6)/(3.14*7850)=162
7600 kg
Cost is dependant on the grade and form and where it is produced - labour costs.
well , reinforcement steel required for slab is nearly 0.7 to 0.8% of the volume of concrete so if concrete is 1 cum then steel will be 1*0.7/100 =0.007 but density of steel is 7850 kg/ cum so 0.007*7850=54.95 say 55 kg per cum so steel required to reinforce 1 cum concrete for slab is approximately 55 kgs.
If 1 kg of tomatoes costs 5 CHF, 0.1 kg will cost 1/10th of 5 CHF, so the cost will be 0.5 kg.
The cost would be 1.5 p.
i dn k=12*45
That depends on whether you are talking about 1 °C or 1 °F. It also depends somewhat on the type of steel. Warming carbon steel by 1 °C requires about 0.49 J/kg of steel. Warming 316 stainless steel 1 °C requires about 0.45 J/kg of steel. To find the answer for 1 °F, multiply the values by 5/9.
983363
potato
1 kg
Steel is not a standard alloy: its composition is varied to meet different requirements. As a result its density varies. A cubic metre of steel will have a mass of between 7750 kg and 8050 kg.
The steel sheet in question has a volume of 1 x 1 x 0.002 m3. Mild steel has a density of 7800 kg/m3.[A]As such the mass = 7800 x 0.002Mass = 15.6 kg[A] Cobb, F. (2009) Structural Engineer's Pocket Book, 2nd Edition. Amsterdam, Elsevier.