An aluminum I-beam has a flange width of 150 mm, web depth of 200 mm, and [7M flange thickness of 10 mm. If the beam is subjected to a bending moment of 60kN-m about the x-axis and 30kN-m about the y-axis, determine the principal axes, the moments of inertia about the principal axes, and the maximum normal stress in the section
To determine the principal axes, moments of inertia about the principal axes, and maximum normal stress in the section, we first need to calculate the moment of inertia of the section about the x and y axes.
The moment of inertia about the x-axis (I_x) can be calculated using the parallel axis theorem:
I_x = I_xx + A_x * d_x^2
where I_xx is the moment of inertia of the beam about its centroidal x-axis, A_x is the area of the beam's cross-section, and d_x is the distance from the centroid of the cross-section to the x-axis.
Given that the beam is an I-beam with a flange width of 150 mm and flange thickness of 10 mm, the centroid of the cross-section lies at the center of the web depth of 200 mm. Therefore, the distance from the centroid to the x-axis (d_x) is 100 mm.
Then, the moment of inertia about the x-axis can be calculated as follows:
I_xx = (1/12) * b * h^3 + A_f * d_x^2 + A_w * (h/2)^2
I_xx = (1/12) * 150 * 200^3 + 150 * 10 * 100^2 + 200 * 10 * (100)^2
I_xx = 500 * 10^6 mm^4
A_x = 150 * 10 + 200 * 10 = 3500 mm^2
I_x = 500 * 10^6 + 3500 * 100^2 = 8500 * 10^6 mm^4
Similarly, the moment of inertia about the y-axis (I_y) can be calculated using the parallel axis theorem by exchanging x and y in the calculations above:
I_yy = (1/12) * h * b^3 + A_f * (b/2)^2 + A_w * (b/2 + d_x)^2
I_yy = (1/12) * 200 * 150^3 + 150 * 10 * (150/2)^2 + 200 * 10 * (150/2 + 100)^2
I_yy = 140.56 * 10^6 mm^4
A_y = 200 * 10 + 150 * 10 = 3500 mm^2
I_y = 140.56 * 10^6 + 3500 * (150/2 + 100)^2 = 10056.25 * 10^6 mm^4
The principal axes can be found by rotating the x and y axes by an angle theta that satisfies the equation:
tan(2 * theta) = (2 * I_xy) / (I_x - I_y)
where I_xy is the product of inertia. For a symmetric section like the I-beam, I_xy = 0.
Therefore, tan(2 * theta) = 0 / (8500 * 10^6 - 10056.25 * 10^6) = 0
This indicates that the principal axes coincide with the x and y axes.
The maximum normal stress (σ_max) occurs at the extreme fiber of the section and can be calculated using the flexure formula:
σ = M * c / I
where M is the bending moment, c is the distance from the neutral axis to the extreme fiber, and I is the moment of inertia about the axis of bending.
For the x-axis bending moment of 60 kN-m:
c_x = h / 2 = 200 / 2 = 100 mm
σ_x = 60 * 10^6 * 100 / 8500 * 10^6 = 705.88 N/mm^2
For the y-axis bending moment of 30 kN-m:
c_y = b / 2 = 150 / 2 = 75 mm
σ_y = 30 * 10^6 * 75 / 10056.25 * 10^6 = 223.53 N/mm^2
Therefore, the maximum normal stress in the section is 705.88 N/mm^2.