Section Modulus calculation and optimization
Case 1 : Section modulus, S1(min.) = I(min.)/y(max.)
= (2.606877 x 10e5 mm4)/(440.4773 mm),
where y(max.)= l/2 = (880.9546 mm / 2) = 440.4773 mm
= 591.83004 mm3
Case 2 : Section modulus of a modified structure , S2(min.) = I(min.)/y(max.)
= (2.608076323 x 10e5 mm4) / (440.4773 mm),
where y(max.)= l/2 = (880.9546 mm / 2) = 440.4773 mm
= 592.1023224 mm3
Percentage increase in Section Modulus, % S = ((S2 (min.) - S1 (min.))/ S1(min.)) x 100 %
= (( 592.1023224 mm3 - 591.83004 mm3 ) / 591.83004 mm3) x 100%
= 0.046 %
Where,
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object
Case study : From the above two cases, it can be inferred that, due to a slight change in the structural form of the section, the section modulus is getting modified as the moment of inertia changes. In this case, a percentage increase of 0.046 % is observed. Thus due to slight increase in thickness of some of the portions of the section, it lead to increase in the magnitude of the moment of inertia, and hence more value of section modulus and hence the result is improved strength.