Given:
Let flange be b1t1, and web be hw
flange width, b1 = 350
flange thickness, t1 = 30
web depth, h = 250
web width, w = 25
Solution:
let Y be the distance of the Neutral Axis(NA) from the top of the section
Section area = A = b1t1 + hw = 350.030.0 + 250.025.0 = 16750.0 mm^2
From the area properties, Ay = ∑(AiYi), where Ai is area of any segment from the total section, and Yi is its centroidal distance from the top hence, 16750.0Y = 350.030.015.0 + 250.025.0155.0
hence, neutral axis is at Y = 67.239 mm from the top face
Using the parallel axis theorem,
moment of inertia about the neutral axis is given as
Ixx = ∑(Ic.g. + Aidy^2) = (b1t13/12 + b1t1(Y-t1/2)2) + (wh3/12 + wh*(t1+h/2-Y)2)
= 110130628.11 mm^4
Section Modulus, Zxx = Ixx/Ymax = 110130628.11/212.761 = 517626.013 mm^3
Let x1x1 be the equal area axis at a distance y1 from top.
b1x1 = A/2
then x1 = A/(2b1) = 23.929 mm
Plastic Section Modulus, Zp = A(y1+y2)/2
Where y1 is the centroid of the area above the equal area axis.
distance y1 = 2(b1t1(x1-t1/2) + (x1-t1)w*(x1-t1)/2)/A
putting all values, y1 = 11.249577673134326 mm
similarly, y2 = (h+t1-x1)/2 = 128.0355 mm
hence, Zp = A(y1+y2)/2 = 16750.0*(11.249577673134326 + 128.0355)/2 = 1166512.5255125002 mm^3
Shape Factor = Zp/Zxx = 1166512.5255125002/517626.013 = 2.254
