Stucco Aluminium Sheet-AA3XXX

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1 Specifications of PE Coated Aluminum Coil/Sheet

Alloy

AA1050,AA1060, AA1070, AA1100, AA3003, AA3004, AA3005, AA3105, AA5005, AA5052, AA5754, AA5083, AA8011

Temper:

H12, H14, H16, H18, H22, H24, H26, H32,HO, F

Thickness:

0.10-4.0mm

Width:

10mm- 2000mm

Coating

PE(Polyester)

Painting Thickness

Standard 16-25 microns, max 40 microns

Color

Acording to Ral colors or customer’s samples

Standard:

GB/T17748-1999, ASTM, ISO, EU standard

Special Specification is available on customer’s requirement

PE(polyester) Coating

PE(polyester) coating:high molecular polymer as monomer and addition of alkyd, is an UV-resistant coating. It can be classified matt and glossy according to coating gloss. The compact molecule structure, makes paint surface luster and smooth,which assure good printing on the panel surface. With an warranty of 8-10 years for weather resistance, it is specially applied for internal decoration and sign board.

2 Usage/Applications of PE Coated Aluminum Coil/Sheet

Our company's PE 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 pipe, sanwich panels and drainage, etc.

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Q:how to connect copper pipe and aluminum sheet?
welding!
Q:How Strong is Aluminum?
How strong aluminum is, depends on the specific alloy of aluminum, whether it is the aluminum 1000 series, or the 3000 series, or the 6000 series, or other variants. The 6000 series is most common for structural purposes. Do a search on its yield strength and its ultimate tensile strength, to get an idea of how strong your particular variant is. The alloy Aluminum 6061-T6 has a yield strength of 35000 lb/in^2, and an ultimate tensile strength of 42000 lb/in^2. This is for the case of pure tension with a uniform load. The square inches refers to its cross sectional area. The ultimate strength occurs at the point where it will completely rupture. The yield strength occurs at the point, where it still supports the load, but ceases to be reversible after the load is removed. Pure aluminum metal is much weaker than its alloys. It has a yield strength of about 1500 psi, and an ultimate strength of 10000 psi. For structural purposes, you almost always want to stay below the yield strength. How stiff aluminum is, is common among all alloys, as 10000000 psi. That is to say 10 million pounds of tension on an aluminum member with a cross sectional area of 1 inch, would hypothetically cause an increase equal to 1 length, if it were still in-tact and linear-elastic at that amount of stress. It isn't. A more realistic case, would be that 10000 pounds of tension on this member, would cause an 0.1% increase in length. As for how light aluminum is per square inch, that depends upon how long your rod is. A 3 ft rod weighs 3 times as much per square inch as does a 1 ft rod. The weight of a material is measured PER UNIT VOLUME rather than per unit area. Unless you are talking about a standard thickness. Aluminum 6061 alloys have a density of 0.0975 pounds/cubic inch. Pure aluminum has a this density as well, as do most of its alloys.
Q:How many square meters does the aluminum plate engrave?
The price of aluminum sheet is calculated according to the price of aluminum ingot + processing fee (aluminum plate state), and the specific price is also different according to the size and size of aluminum sheet. If you have any questions, please ask, thank you.
Q:Analysis of a Magnesium-Aluminum Alloy?
P_H2 = 0.950 atm (Dalton's Law of partial pressures) n=Pv/RT = (0.950 atm)(0.311 L) / (0.08206 (L*atm)/(mol*K))(302.15 K) n_H2 = 0.011915983 mol Balanced equations: Al + 3HCl -- 3/2H2 + AlCl3 Mg + 2HCl -- H2 + MgCl2 By these equations, we know that every mole of Al will give us 1.5 moles of H2, and every mole of Mg will give 1 mole of H2. We can therefore set up an equation for the mass of Al like this: *Let a = the mass of MAGNESIUM* Al = 0.250 g - a With this equation in mind, we can setup two equations solving for 'n' of each element by dividing by its molar mass and multiplying by the molar ratio: n_Mg = a / 24.30 (1:1 ratio, so we don't have to multiply) -- number of moles of H2 produced by the reaction of Mg (now written as n_H') = a / 24.30 n_Al = (0.250 g - a) / (26.98 g/mol) Because of the molar ratio shown above, we must multiply n_Al by 1.5 in order to get n_H2 produced by the reaction of aluminum, hereafter known as n_H2 Since we know the number of moles produced by the sum of the reactions, we can add these equations together and solve for n_H2. (**note that your value will be different because you have a different volume**) Set up the equation like this: n_H2' + n_H2 = n_H2 = 0.011915983 mol Sub in your individual equations for n_H2' and n_H2: (a/24.3) + 1.5[(0.250-a)/26.98] = 0.011915983 mol Rearrange and solve for a (mass of MAGNESIUM): (26.98a + 9.1125 - 36.45a) / (24.3)(26.98) = 0.011915983 0.011915983 = 9.47a a = 0.137298281 g Once you have your 'a' value, divide it by the total mass (0.250 g) and multiply by 100%. This gives you the percentage of Mg. (0.137298281 g / 0.250 g) * 100% = 54.9193 % Since you want ALUMINUM, you must subtract the percentage of Mg from 100. 100 - 54.9193 = 45.08% So, the mass percentage of aluminum is 45.08%. I hope this is helpful!
Q:Chevy S10 Aluminum head?
Could be 1 of 5 things Blown intake gasket Blown head gasket Cracked head Cracked block Warped head. Most of the times you will not see a crack with your eyes. The head needs to be magnafluced or pressure tested to see if theres a crack. Check your phone book for machine shops. An engine machine shop will check a head for cracks for around $20-40 per head To check if its Warper you can do that with a true flat, straight edge like a long steel rule, and a feeler gauge. Lay rule on bottom of head where gasket goes (all traces of old gasket cleaned off 1st), Try to slide a .001 feeler gauge under rule. If warped a machine shop can mill head flat for around $35-50 per head
Q:An aluminum clock pendulum?
A for a swinging pendulum period T =2pi*√(L/g), where L is length of the pendulum, g=9.8m/s^2; metals shrink when cooled, so period will decrease, number of ticks per hour will increase, the clock will hurry up; B♣ period at t1°=20C° is T1=2pi*√(L1/g); ♣ period at t2°=-5C° is T2=2pi*√(L2/g); ♣ the law of linear extension says L2/L1= 1+s*Δt°, where s=23.1·10-6 (1/К°) is specific linear extension of aluminum, Δt°=t2°-t1°; ♦ thus T2/T1 =√(L2/L1) = √(1+s*Δt°) =f1/f2, where frequency f1= 1 Hz at 20C° or 3600 ticks per hour, hence frequency at -5C° is f2= f1/√(1+s*Δt°); therefore instead of 3600 ticks per hour the clock will do 3600/√(1+s*Δt°) ticks per hour; thus it will gain 3600*(1/√(1+s*Δt°) –1) = = 3600*(1/√(1 -23.1·10-6 *25°) –1) = 1.04 s/hour;
Q:why does the hardness of 2024 aluminum sheet not increase but decrease after thermal treatment?
The hardness of 2024 aluminum sheet will decrease after thermal treatment(solution treatment), and it will increase after aging treatment.
Q:which kind of enterprises are aluminium sheet circles used in?
user use cold squezz method to transform the aluminum sheet circle into various standard capacitor shell, aluminium collapsible tube shell, which are broadly used in electron industry, daily chemical industry, medicine, education and automobile products,electrical appliance, heat preservation, machine manufacturing, automobile,spaceflight,military industry,mould, construction, printing and other industries.
Q:calculating ions in Aluminum Nitrate?
? number of NO3 = 3.99g AlNO3 x(1mol AlNO3/212.996 g AlNO3)x(1mol NO3/1mol AlNO3)x(6.02*10^23 number of NO3/1mol NO3)=11*10^21 number of NO3
Q:4.0mm five bars, aluminum plate, 4 aluminum plate, one ton, how many square?
The price is calculated according to the price of aluminum ingot + processing fees (aluminum state), the specific price according to the size of the aluminum plate, zero shear and the whole board price also has the difference, Jiangsu Yi Heng Liu Guangxi Henan Mingtai aluminum has the.

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