A short 6061-T6 aluminum cylindrical block, having an original diameter of 20mm and a length of 75mm, is placed in a compression
machine and tightened until the applied axial load is 5kN. Determine (a) the decrease in its length and (b) its new diameter.
Un bloque cilíndrico corto de aluminio 6061-T6, que tiene un diámetro original de 20 mm y una longitud de 75 mm, se coloca en una máquina de compresión y se aprieta hasta que la carga axial aplicada sea de 5 kN. Determine (a) la disminución de su longitud y (b) su nuevo diámetro.
Un bloque cilíndrico corto de aluminio 6061-T6, que tiene un diámetro original de 20 mm y una longitud de 75 mm, se coloca en una máquina de compresión y se aprieta hasta que la carga axial aplicada sea de 5 kN. Determine (a) la disminución de su longitud y (b) su nuevo diámetro.
1 answer:
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In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
Solution :
a). Using Gauss's law :
,
.........(1)
Let in equation (1)
Therefore, .............(2)
....................(3)
Therefore,
.............(4)
Now differentiating the equation (4) w.r.t. 'b', we get
Thus the radius for the inner cylinder conductor is
b). For the energy storage, substitute the radius in (4), we get
This is the amount of energy stored in the conductor.