steel channel section has a depth of 200 mm, flange width of 75 mm, and thickness of 10 mm. If the section is subjected to a bending moment of 50 kN-m about the x-axis and 20 kN-m about the y-axis, calculate the maximum normal stress and the location of the neutral axis.

Answer 1

To calculate the maximum normal stress and the location of the neutral axis, we first need to determine the section properties of the steel channel.

Given:
Depth (d) = 200 mm
Flange width (b) = 75 mm
Thickness (t) = 10 mm
Bending moment about x-axis (Mx) = 50 kN-m
Bending moment about y-axis (My) = 20 kN-m

The centroid of the section can be calculated as follows:
Yc = (d - t) / 2 = (200 - 10) / 2 = 95 mm

The moment of inertia about the x-axis (Ix) can be calculated as follows:
Ix = (1/12) * b * d^3 + t * d * (Yc - t/2)^2 + b * t^3 = (1/12) * 75 * 200^3 + 10 * 200 * (95 - 5)^2 + 75 * 10^3
Ix ≈ 8875000 mm^4

The moment of inertia about the y-axis (Iy) can be calculated as follows:
Iy = (1/12) * d * b^3 + b * t * (b/2 - t/2)^2 + t * d^3 = (1/12) * 200 * 75^3 + 75 * 10 * (75/2 - 5/2)^2 + 10 * 200^3
Iy ≈ 5343750 mm^4

The maximum normal stress can be calculated using the bending stress formula:
σ = M * y / I

For bending about the x-axis:
σx = (Mx * (d/2 - Yc)) / Ix
σx = (50 * (200/2 - 95)) / 8875000
σx = 0.02831 N/mm^2

For bending about the y-axis:
σy = (My * (b/2)) / Iy
σy = (20 * (75/2)) / 5343750
σy = 0.01410 N/mm^2

The maximum normal stress is the higher of the two:
Maximum normal stress = σx = 0.02831 N/mm^2

The neutral axis is located along the centroid of the section, so the location of the neutral axis is at a distance of 95 mm from the bottom of the steel channel.