An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
1 answer:
Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm
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Answer:
For the point inside the cylinder:
For the point outside the cylinder:
where x0 is the position of the point on the x-axis and σ is the surface charge density.
Explanation:
Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.
<u>x = -x0:</u>
The cylinder is consist of the sum of the rings with the same radius.
First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.
We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.
The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.
Now, the electric field created by the small portion is 'dE'.
The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.
We have to integrate it over the ring, which is an angular integration.
This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:
The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.
<u>x = +x0: </u>
We will only change the boundaries of the last integration.