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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
commit7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch)
tree449d555969bfd7befe906877abab098c6e63a0e8 /3862/CH4/EX4.12/Ex4_12.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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Added new codeHEADmaster
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+clear
+//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)