A rectangular beam, with cross-sectional width b, height h, Young's modulus Erect, has the same flexural rigidity as a circular beam with the same cross-sectional area.
What is the Young's modulus, Ecirc, of the circular beam in terms of Erect,b, and h?
What aspect ratio of the rectangular beam, h/b, will result in the same Young's modulus for the rectangular and circular beam Ecirc=Erect?
4*pi*E_rect*h/(12*b)
To find the Young's modulus, Ecirc, of the circular beam in terms of Erect, b, and h, we can use the formula for the flexural rigidity of a beam:
I = (1/12) * b * h^3 * Erect
where I is the moment of inertia. For a circular beam with the same cross-sectional area as the rectangular beam, the moment of inertia can be expressed as:
Icirc = (π/64) * d^4 * Ecirc
where d is the diameter of the circular beam.
Since the cross-sectional areas for both beams are equal, we have:
b * h = π * (d/2)^2
b * h = (π/4) * d^2
Solving for d in terms of b and h:
d^2 = (4 * b * h) / π
d = sqrt((4 * b * h) / π)
Substituting this expression for d into the equation for Icirc, we get:
Icirc = (π/64) * [(sqrt((4 * b * h) / π))^4] * Ecirc
Icirc = (π/64) * (16 * b^2 * h^2 / π^2) * Ecirc
Icirc = (1/4) * b^2 * h^2 * Ecirc / π
Now, equating the expressions for I and Icirc:
(1/12) * b * h^3 * Erect = (1/4) * b^2 * h^2 * Ecirc / π
Simplifying the equation:
3 * Erect = (b * h * Ecirc) / (4 * π)
Ecirc = (12 * Erect * π) / (b * h)
Therefore, the Young's modulus, Ecirc, of the circular beam in terms of Erect, b, and h is given by:
Ecirc = (12 * Erect * π) / (b * h)
To find the aspect ratio of the rectangular beam (h/b) that will result in the same Young's modulus for both the rectangular and circular beams (Ecirc = Erect), we can substitute Ecirc = Erect into the equation:
Erect = (12 * Erect * π) / (b * h)
Simplifying:
1 = (12 * π) / (b * h)
b * h = 12 * π
Therefore, the aspect ratio of the rectangular beam, h/b, that will result in the same Young's modulus for both beams is:
h/b = 12 * π / b * h
To find the Young's modulus, Ecirc, of the circular beam in terms of Erect, b, and h, we need to use the concept of flexural rigidity. Flexural rigidity is a property that describes the resistance of a beam to bending.
The flexural rigidity of a rectangular beam is given by the equation:
ERect = (b * h^3)/12
For a circular beam, the flexural rigidity can be calculated using the moment of inertia. The moment of inertia for a circular cross-section is given by:
Icirc = (π * d^4)/64, where d is the diameter.
Since the circular beam has the same cross-sectional area as the rectangular beam, we can equate the moment of inertia of the circular beam to the flexural rigidity of the rectangular beam:
(b * h) = (π * d^2)/4
Now, we can express the diameter of the circular beam, d, in terms of b and h:
d = √((4 * b * h)/π)
Next, we can substitute this value for d into the equation for the moment of inertia, Icirc:
Icirc = (π * (√((4 * b * h)/π))^4)/64
Simplifying this expression, we get:
Icirc = (π * (4 * b * h)/π^2)/64
Reducing further, we have:
Icirc = (4 * b * h)/64π
The flexural rigidity, which is equal to the Young's modulus of the circular beam, Ecirc, can be calculated by multiplying the moment of inertia, Icirc, by the Young's modulus formula:
Ecirc = (Icirc * c^2)/2, where c is the distance from the neutral axis to the outermost fiber.
Since the circular beam has a radius of d/2, we can express c as:
c = d/2 = √((4 * b * h)/π)/2
Now, we can substitute this value for c and the expression for Icirc into the equation for Ecirc:
Ecirc = ((4 * b * h)/64π) * (√((4 * b * h)/π)/2)^2 / 2
Simplifying this expression gives us the final result for the Young's modulus of the circular beam, Ecirc, in terms of Erect, b, and h:
Ecirc = (2 * b * h * (4 * b * h)^2) / (64 * π^2 * 8)
To find the aspect ratio of the rectangular beam, h/b, that results in the same Young's modulus for the rectangular and circular beam, Ecirc = Erect, we can set the two expressions for Ecirc and Erect equal to each other and solve for h/b.