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//The given composite section can be divided into two rectangles
//
//variable declaration
A1=150.0*10.0 //Area of 1,mm^2
A2=140.0*10.0 //Area of 2,mm^2
A=A1+A2 //Total area,mm^2
//Due to symmetry, centroid lies on the symmetric axis y-y. The distance of the centroid from the top most fibre is given by:
Y1=5.0
Y2=10.0+70.0
yc=(A1*Y1+A2*Y2)/A
//Referring to the centroidal axis x-x and y-y, the centroid of A1 is g1 (0.0, yc-5) and that of A2 is g2 (0.0, 80-yc)
//Moment of inertia of the section about x-x axis Ixx = moment of inertia of A1 about x-x axis + moment of inertia of A2 about x-x axis.
Ixx=(150*(10**3)/12)+(A1*((yc-5)**2))+(10*(140**3)/12)+(A2*((80-yc)**2))
printf("\n Ixx= %0.1f mm^4",Ixx)
Iyy=(10*(150**3)/12)+(140*(10**3)/12)
printf("\n Iyy= %0.1f mm^4",Iyy)
//Hence, the moment of inertia of the section about an axis passing through the centroid and parallel to the top most fibre is Ixxmm^4 and moment of inertia of the section about the axis of symmetry is Iyy mm^4.
//The radius of gyration is given by
kxx=sqrt(Ixx/A)
printf("\n kxx= %0.2f mm",kxx)
kyy=sqrt(Iyy/A)
printf("\n kyy= %0.2f mm",kyy)
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