Help calculating force required to bend

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To calculate the force required to bend a 2mm thick piece of 6063 T5 aluminum, the width of the material is a crucial factor. A U-shaped configuration with one arm bolted down and a 30" lever applied at the end can be bent, but the exact force needed depends on the width of the aluminum piece. Based on experience, bending a wider piece (like 4 inches) with a long lever can be done relatively easily, though 6063 aluminum is prone to cracking if bent too tightly. Therefore, it's essential to consider both the dimensions and the bending technique to avoid damage. Proper calculations and care are necessary for successful bending.
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Can anyone help me figure out how much force would be required to bend a 2mm thick piece of 6063 T5 aluminum?

The end product is a little difficult to explain so I think the following is the best way to visualize the situation:

Imagine a U shaped piece made out of 2mm thick 6063 T5 aluminum. One of the arms of the U is bolted down to an immovable object and the other arm of the U is bolted to an aluminum tube that is perpendicular (for the purposes of the exercise assume the tube will never bend) to the arm of the U. The tube is 30" in length. Assuming the force is applied to the end of the 30" tube for max leverage, how much force would it take to bend the U shaped object?

Is it something that would be able to be bent by hand?

Thanks
 
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Give me a lever long enough and a place to put it and I can move the world... Short answer, yes, you can bend anything by hand...

One critical dimension you're leaving out is how wide the 2mm piece is. Just going by the experience I have with 5252 T8 aluminum in 3/16 (~4mm), I can quite easily bend a 4" wide piece with a 30" lever in a vice... I've done some work with 6063, and used to bend 4" wide strips of 3/8" routinely with about a 48" lever. 6063 isn't terribly friendly to bending though, it'll crack if you bend it too tight.. take care about that
 
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