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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
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tree449d555969bfd7befe906877abab098c6e63a0e8 /3862/CH3/EX3.10
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+clear
+//
+
+//Each load is 20 kN.
+
+//variable declaration
+
+P=20.0
+AB=18.0
+A=3.0
+
+RA=P*7/2
+RB=RA
+
+theta1=30.0*%pi/180
+a=(3*A)/(4*cos(theta1))
+//Take Section (A)–(A) and consider the equilibrium of left hand side part of the French Truss
+//Drop perpendicular CE on AB.
+
+CE=3*A*tan(theta1)
+DE=A
+
+theta=atan(CE/DE)*180/%pi
+printf("\n theta= %0.0f °",theta)
+
+//moment at point A
+
+F2=(P*a*cos(theta1)*6)/(A*2*sin(theta*%pi/180))
+printf("\n F2= %0.4f KN (Tension)", F2)
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+F1=(F2*sin(theta*%pi/180)+RA-P*3)/(sin(theta1))
+printf("\n F1= %0.4f KN (Comp)",F1)
+
+F3=F1*cos(theta1)-F2*cos(theta*%pi/180)
+printf("\n F3= %0.4f KN (Tension)",F3)