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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
commit7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch)
tree449d555969bfd7befe906877abab098c6e63a0e8 /3862/CH9/EX9.3/Ex9_3.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+clear
+//
+
+//variable declaration
+
+//summation of all horizontal forces is zero & vertical forces is zero.
+P1=(20) //vertical down Load at 2m from A,KN
+P2=(30) //uniform distributed load from 2m to 6m from A,KN/m(in 4m of span)
+P3=(60) //Inclined down to right Load at angle 45° at 7m from A,KN
+
+theta3=45
+//horizontal,vertical component at B is Hb,Vb respectively.
+
+Ra=(P1*7+P2*4*5+P3*2*sin(theta3*%pi/180))/(9) //reaction at B point,KN
+
+printf("\n RA= %0.4f KN",Ra)
+
+Hb=P3*cos(theta3*%pi/180)
+printf("\n HB= %0.4f KN",Hb)
+//now vertical component
+Vb=P1+P2*4+P3*sin(theta3*%pi/180)-Ra
+printf("\n VB= %0.4f KN",Vb)
+
+Rb=sqrt((Hb**2)+(Vb**2))
+
+printf("\n RB= %0.4f KN",Rb)
+
+alpha=(atan(Vb/Hb))*180/%pi
+
+printf("\n alpha= %0.2f °",alpha)