6063 Alloy H22 Aluminum Rolled Sheet Aluminium Coil

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Loading Port:
Shanghai
Payment Terms:
TT OR LC
Min Order Qty:
5 m.t.
Supply Capability:
500 m.t./month

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Item specifice:

Grade: 1000 Series,3000 Series,5000 Series,6000 Series,7000 Series,2000 Series Surface Treatment: Coated,Embossed,Mill Finish Shape: Square,Round,Flat
Temper: O-H112,Soft,Half Hard,Hard Application: Decorations,Door & Window,Transportation Tools,Kitchen Use Technique: Extruded
Thickness: 0.2mm--7mm Width: 280MM -1500MM Outer Diameter: 700mm
Net Weight (kg): 1500 Packaging: Wooden pallet

Product Description:

6063 Alloy H22 Aluminum Rolled Sheet Aluminium Coil


aluminum coil specifications:

1) Alloy :1050, 1060,1100,  3003  3004 3105 3A21 5005 5052 etc 

2) Temper: O/H12/H14/H1/H18/H32/H34/H36/H38//H111/H112/H116/H321/T6/T651/T3/T351 etc

3) Thickness: 0.1mm to 6mm

4) Width:20mm to 3300mm

5)Coil weight: 100kgs to 6 tons depends on actual requirement

6)Core material: Aluminum or paper

7)Coil Inner diameter: 75mm, 150mm, 200mm, 300mm, 405mm, 505mm or as required 

8) Protective film can be added


item

 3003 Aluminum coil

Standard

GB/T3190-2008,GB/T3880-2006,ASTM B209,JIS H4000-2006,etc

Material

1060,1050,1100

3003,3103,3004,3005,3105

5052, 5454,5754

Size

Thickness

0.5mm-3.5mm

Width

800-1500mm

Weight/Roll

About 1.5MT/3MT

Quality control

Mill Test Certificate is supplied with shipment, Third Part Inspection is acceptable.

Surface

Bright, polished, hair line, brush, checkered, embossed, etc

Trade terms

Price term

,FOB, CNF, CIF, etc

Payment Term

TT,L/C

MOQ

2MT

20 GP Capacity

About 20-25MT

Delivery time

1.The products will delivery immediately after receiving the payment.

2.According to the order quantity, prompt delivery.

Export to

Ireland,Singapore,Indonesia,Ukraine,Spain,Canada,USA,Brazil,Thailand,Korea,Iran,India,Egypt,Kuwait,

Oman,Viet Nam, South Africa, Dubai, Russia, etc

Package

Stick blue film→plastic film→waterproof paper→1~2 tons on a export standard pallet(corner protection)

Application

1)Further making utensil.2)Solar reflective film3)The appearance of the building4)Interior decorating:ceilings,walls,etc.5)Furniture cabinets6)Elevator decoraction7)Signs,nameplate,bags making.8)Decoration inside and outside the car9)Household appliances:refrigerators,microwave ovens,audio equipment,etc.10)The consumer electronics:mobile phones,digital cameras,MP3,etc.


6063 Alloy H22 Aluminum Rolled Sheet Aluminium Coil


6063 Alloy H22 Aluminum Rolled Sheet Aluminium Coil


6063 Alloy H22 Aluminum Rolled Sheet Aluminium Coil


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Q:Calculate the resistance of aluminium cable of length 1o km and diameter 2mm?
Let ρ be the resistivity of the aluminum cable, A the cross sectional area of the cable and l the cable's length. The resistance R of the aluminum cable is now given by: R = ρ·l / A = 2.7·10^-8·(10·10^3) / (π·r²) = 2.7·10^-8·(10·10^3) / (π·(1·10^-3)²) = 85.9 Ω I guess the tricky part is remembering: R = ρ·l / A but when you realize how intuitive this equation is you don't have to memorize it. The resistance R is proportional to the resistivity ρ, so we must have the following in our equation: R = ρ·k , where k is some value governed by the dimensions of the cable! It is logical that the resistance is also proportional to the wires length l. This is logical when you realize that the longer the wire the more obstacles the electrons will have to pass by! Now we have that: R = ρ·l·c , where c is another value governed now by the cross sectional area (only dimensional property left). If we think of the wire as a highway with cars instead of electrons we realize that there is less resistance (higher speed) when the highway has many lanes and is very broad. From this we realize that the greater the cross sectional area the smaller the resistance. R must therefore be inversely proportional to A. We now have: R = ρ·l / A I hope there are no mistakes here :P
Q:Is there anyway to strengthen aluminum?
Aluminum bronze is used commercially to make non-sparking knives. Why try to reinvent the wheel? I'm trying to think of a way I can make an aluminum knife that will hold an edge. You can strengthen aluminum by either cold working or heat treating a suitable alloy, but it will never be anything close to a good steel knife. Would heat treatment plus 10 tons of pressure repeated have a similar affect? No. Assuming that you are talking about the article linked to below: - you would have to duplicate the alloy process, even then you could only make a small thin sheet. They used a heat treatable alloy. The process involves torsional strain under 60 metric tons per square centimeter compression. How big a press do you think you would need for a knife size blank? And remember, you have to apply torsional cold work while compressing it. -Their results ( 1 gPA YS 5% elongation in a very thin sheet) don't look that great compared to a good knife steel. Or a good titanium knife.
Q:Which companies will use aluminum sheet and coil?
Plants to make auto parts such as exhaust pipe thermal baffle, muffler thermal baffle and front thermal baffle.
Q:Powdered aluminum is used in the thermite reaction for welding metals.?
Aluminium okorder.com/
Q:Is it a good idea putting Aluminum foil on my windows?
looks really tacky from the outside
Q:Engines: Aluminum block vs. Iron block?
The latter points were right. Iron blocks are heavier, although much cheaper to make. Aluminum ones are significantly lighter and handle heat better, but it is a bit more expensive to make them.
Q:Thermodynamics - Steel Ring and Aluminum disk?
Steel ring: - Inner diameter: 2.00 cm at 430-deg C - thermal expansion coefficient: 17.3e-6 = s Aluminum ring: - Outer diameter: 2.01 cm at 430-deg C - thermal expansion coefficient: 23.1e-6 = a a) The aluminum ring is too large. If they are both cooled, the aluminum ring will shrink at a faster rate, so at some cooler temperature, there will be a match. What we need: R_new = 2.01*(1+ a*dT) = 2.00*(1 + s*dT) 2.01 + (2.01*a)dT = 2.00 + (2.00*s)dT 0.01 = dT*(2*s - 2.01*a) dT = 0.01*e6/(2*17.3 - 2.01*23.1) = -8.5e2 (deg-C) = -850 (deg-C) Hmm, so we need a reduction in temperature by 850 deg. But we started at T-celsius = 430 = T-kelvin = 803. A reduction by 850 deg gives us -47 deg-K. But that's impossible! Either the coefficients of thermal expansion are extremely different at 430-C than at 20-C (quite possible) or this degree of accommodation is not possible. b) This cannot be answered, based on the impasse in a).
Q:how to weld aluminum with fluxcore?
You can't. you need a TIG welder to weld aluminum. It's an especially hot welder for aluminum specifically.
Q:Sodium Aluminum Phosphate, is in food products?
make your own pancakes and you have nothing to worry about.
Q:Using principles of chemical bonding, explain why solid aluminum is a good conductor of electricity but solid?
Pure aluminium is a good conductor because it's atoms are metalically bonded. This means it's electrons are delocalised and can therefore produce a current. NaCl has a high boiling point because in order to evaporate it first needs to melt. And to melt it you need to break it's lattice structure. Na and Cl atoms are held by strong ionic bonds. BrCl on the other hand is covalently bonded. It's molecular structure does not need to be broken in order for it to melt or boil. All that has to be done is weak van der waal bonds need to be broken.

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