A 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
To calculate the maximum normal stress, we first need to calculate the moment of inertia of the section about the x-axis and y-axis.
For calculating the moment of inertia about the x-axis (Ix), we use the formula:
Ix = (b*h^3)/12 + 2*A*(h/2)^2
Where:
b = flange width = 75 mm
h = depth - 2*(flange thickness) = 200 - 2*10 = 180 mm
A = thickness * flange width = 10 * 75 = 750 mm^2
Ix = (75*180^3)/12 + 2*750*(180/2)^2
Ix = 6075000 mm^4
Similarly, for the moment of inertia about the y-axis (Iy), we have:
Iy = (h*b^3)/12 + 2*A*(b/2)^2
Iy = (180*75^3)/12 + 2*750*(75/2)^2
Iy = 3360937.5 mm^4
Now, the maximum normal stress (σmax) can be calculated using the bending moment (Mx and My), the distance from the neutral axis (c) to the extreme fiber, and the section modulus (Sx and Sy) for x and y-axis respectively.
σmax = M*c/I
For x-axis:
σmax(x) = Mx*c/Ix
c for x-axis = h/2 = 180/2 = 90 mm
σmax(x) = 50*90/6075000
σmax(x) = 0.000741 MPa
For y-axis:
σmax(y) = My*c/Iy
c for y-axis = b/2 = 75/2 = 37.5 mm
σmax(y) = 20*37.5/3360937.5
σmax(y) = 0.000022 MPa
The location of the neutral axis can be calculated by finding the centroidal neutral axis distance (c) from the bottom of the channel section.
c = (A1*y1 + A2*y2)/(A1 + A2)
Where:
A1 = Area of top flange = b*t
y1 = Distance of top flange centroid from the bottom = h - t/2
A2 = Area of web = (h - 2*t) * t
y2 = Distance of web centroid from the bottom = (h - 2*t)/2
c = [75*10*(180-10/2) + (200-2*10)*10*((180-2*10)/2)] / (75*10 + (200-2*10)*10)
c = [67500 + 34000]/(750 + 1800)
c = 37400/2550
c = 14.67 mm
Therefore, the maximum normal stress is 0.000741 MPa for the x-axis and 0.000022 MPa for the y-axis. The location of the neutral axis is 14.67 mm from the bottom of the channel section.