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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
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parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+clear
+//Given
+//we will divide this into two equal parts and other part
+l = 10.0 // in - The height
+t = 0.1 // in - The width
+b = 5.0 //mm- The width of the above part
+A = t* b //in2 - area of part
+y_net = l/2 // The com of the system
+y_1 = l // The position of teh com of part_2
+I_1 = t*(l**3)/12 //in^4 The moment of inertia of part 1
+I_2 = 2*A*((y_1-y_net)**2) //in^4 The moment of inertia of part 2
+I = I_1 + I_2 //in^4 The moment of inertia
+e = (b**2)*(l**2)*t/(4*I) //in the formula of channels
+l_sc = e - t/2 //in- The shear centre
+printf("\n The shear centre from outside vertical face is %0.3f in",l_sc )