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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+// exa 5.8 Pg 156
+clc;clear;close;
+
+// Given Data
+P=20;// kN
+e=80;//mm
+tau=150;// MPa
+
+
+Pd=P/4;// kN
+C=P*e;// kN.mm (Couple)
+// As C.G. lies at 45mm from top rivet
+l1=45;l4=45;//mm
+l2=15;l3=15;//mm
+//(F1/l1)*(2*l1*l4+2*l2*l3) = C
+F1= C*1000/(2*l1*l4+2*l2*l3)*l1;//N
+R1=sqrt(Pd**2+F1**2);// N
+//R1=%pi/4*d0**2*tau
+d0=sqrt(R1/(%pi/4*tau));//mm
+printf('Diameter of rivets - \n d0 = %.3f mm',d0)
+printf('\n Use d0 = 13.5 mm & d = 12 mm')