Pre-painted Aluminium Coil 1060 3003 3105

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Loading Port:
Shanghai
Payment Terms:
TT OR LC
Min Order Qty:
8 m.t.
Supply Capability:
2000 m.t./month

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Product Description:

Pre-painted Aluminium Coil 1060 3003 3105 Description:

Pre-painted Aluminium Coil 1060 3003 3105 are of a wide range of colors, which gives wonderful appearance no matter in residential and commercial constructions of great exhibition centers.The coated aluminum coil/sheet have been widely used in the fields of construction and decoration( garage doors, ceiling etc.), electronic appliances, lighting decoration, air-condition air pipes, sandwich panels and drainages etc.

 

Main Features of Pre-painted Aluminium Coil 1060 3003 3105:

1) High flexibility 
2) Impact resistance 
3) Excellent weather-proof durability 
4) Anti-ultraviolet 
5) High erosion resist

Images of  the Pre-painted Aluminium Coil 1060 3003 3105:

Pre-painted Aluminium Coil 1060 3003 3105

Pre-painted Aluminium Coil 1060 3003 3105

Pre-painted Aluminium Coil 1060 3003 3105

Pre-painted Aluminium Coil 1060 3003 3105 Specification:

Alloy

A1100,A3003,A1050,A8011   etc

Temper

H16,H18,H24

Thickness

From   0.024mm to 1.2mm

Width

Standard   width:1240mm

Special   width:1300mm,1520mm,1570mm,1595mm

Diameter

Standard   dia:1200mm

Interior   dia:150mm,405mm,505mm

Weight

2.5   T/coil,3.0 T/coil

Coating

PE, PVDF,   AC

Surface

Embossed,   mill finish, coated

Color

AS to   code RAL

Gloss

10-90%(EN   ISO-2813:1994)

Coating   Thickness

PE: more   than 18 micron

PVDF: more   than 25 micron

Coating   Hardness

(pencil   resistance)

More   than 2h

Coating   adhesion

5J(EN   ISO-2409:1994)

Impact   Resistance

No   peeling or cracking(50 kg/cm,ASTMD-2794:1993)

Flexibility

(T-bend)

2T

MEK   resistance

More   than 100

 

FAQ:

a.What is monthly capacity

---CNBM is one stated own company and our monthly capacity is about  2000tons.

b. Now which countries do you export your goods?

---Now we export to  South East Asia,Africa, North America,South America  ect.


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Q:Why does a ring of aluminum interact with a magnetic field?
Aluminum is not a magnetic material and will not be affected by stationary magnets. However, if a changing magnetic field is present, an electrical voltage will be induced in the aluminum ring and an electric current will flow as a result. This current will produce a magnetic field which will interact with the externally produced magnetic field. The key here is a changing magnetic field - because the voltage induced in the aluminum ring will be proportional to the rate of change of the field. If an aluminum disk is passed between the poles of a powerful magnet, eddy currents will be induced in the aluminum producing a magnetic field that will tend to retard the motion of the aluminum relative to the magnet. The changing field in this case is due to the motion of the aluminum. Iron is attracted to a magnetic field because it is a ferro-magnetic material. Wood is not a magnetic material, and it is not an electrical conductor, so it will not behave as aluminum does in a magnetic field.
Q:What are the production processes of aluminum coil and what are the functions of various industrial furnaces in the production of aluminum? The more detailed, the better. Thank you! Points!
General steps are as follows: melting - hot rolling - cold rolling - annealing – shearing – straightening – punching
Q:bmx racing aluminum vs chromoly frame?
I'm not a racer, so don't quote me on anything. The metal doesn't make you faster. The difference between aluminum and chromoly is strength and weight. Aluminum is much weaker than chromoly, but also much lighter. I believe aluminum parts are generally used for racers under 110 lbs, but, again, I'm not a racer so I'm not exactly sure.
Q:Heat, vaporization, and melting points of aluminum.?
Calculate the quantity of energy required to heat 1.58 mol of aluminum from 33°C to its normal melting point in KJ- 1.58 mol x 26.98 g/mol x 0.902 J/g°C x (658 - 33)°C x 1 kJ/1000 J = 24.03 kJ Calculate the quantity of energy required to melt 1.02 mol of aluminum at 658°C In KJ- 1.02 mol x 26.98 g/mol x 3.95 kJ/g = 108.7 kJ Calculate the amount of energy required to vaporize 1.02 mol of aluminum at 2467°C In KJ- 1.02 mol x 26.98 g/mol x 10.52 kJ/g = 289.5 kJ
Q:ladders can be made using aluminium alloy,fiber glass and wood.?
get fiber glass,they will last a long time
Q:How does aluminum ingot be processed to aluminum coil (aluminum sheet)?
This depends on what aluminum ingot is. Small ingot needs to be casted into big one by putting into the melting furnace, and then be rolled to aluminum coil through rolling mill.
Q:Why does galvanic reaction occur so intensely in aluminum vs. copper?
This problem arises between various metals such as iron aluminium where some form of plastic barrier is needed to stop electrochemical reaction taking place where these metals are in long term contact with each other especially in a wet environment. As far as aluminium copper are concerned, the difference in their chemical reactivity is particularly large; aluminium is a highly reactive metal that quickly becomes dull by coating itself with a very thin layer of aluminium oxide whilst copper is one of the least reactive of all metals. However, once covered with a fine layer of aluminium oxide, aluminium will tend not to corrode further. The effect of this is that any copper that does go into solution as a result of contact with weak atmospheric acids will very readily precipitate again as metallic copper onto the aluminium. Reactions similar to those below will readily take place: H2O (rain) + CO2 --- H2CO3 (carbonic acid - carbon dioxide dissolved in rain water) Cu + H2CO3 --- CuCO3 + H2 In an aqueous environment, 3 CuCO3 + 2 Al --- Al2(CO3)3 + 3 Cu Copper aluminium are regularly used because they are relatively low cost when compared with potential alternatives, even with recent commodity price increases. Toughened plastic has replaced copper in certain plumbing applications which have obviously eliminated this galvanic reaction problem.
Q:Copper and Aluminum Initial Temperature?
Example Problem Statement: An aluminum wing on a passenger jet is 29 m long when its temperature is 29°C. At what temperature would the wing be 7 cm (0.07 m) shorter? Step 1: Write down what you know in symbolic form: The wing is made of aluminum, therefore: α = 25 x 10-6 /oC Length: l = 29 m Temperature: T1 = 29 oC Change in length: Δl = -7 cm (-0.07 m) Step 2: Write down what you don't know in symbolic form: Temperature: T2 = ? oC Step 3: Find an equation that contains what you know and what you don't know: Δl = αlΔT = αl(T2 - T1) Step 4: Solve the equation for what you don't know: Δl = αl(T2 - T1); divide both sides by αl Δl/ αl = (αl/ αl )(T2 - T1) = T2 - T1; add T1 to both sides Δl/αl + T1 = T2 - T1 + T1 = T2 Δl/ αl + T1 = T2 Step 5: Plug in (substitute) what you know and you will find out what you don't know: Δl/ αl + T1 = T2 = (-0.07 m)/( 25 x 10-6 /oC x 29 m) + 29 oC = -96.551 oC + 29 oC = -67.551 oC = -67.6 oC rounded off.
Q:Why does fillet appear when rolling aluminum coil?
How thick and wide is it when there is such problem? Which is the rolling pass? And what is the type of the rolling mill? The simple 4 rollers or CVC and HC rolling mill with roller control. Does the filler appear in rolling or out of roller? Different thickness will lead to different analysis.
Q:What is a good slogan for Aluminum?
Aluminum: It's Shiny in paper form, yo!

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