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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
commit7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch)
tree449d555969bfd7befe906877abab098c6e63a0e8 /3862/CH2/EX2.16/Ex2_16.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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Added new codeHEADmaster
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+clear
+//
+
+//variable declaration
+P1=1120.0 //vertical down Loading at 2m distance from O,KN
+P2=120.0 //vertical up loading at 4m distance from O,KN
+P3=420.0 //vertical downloading at 5m distance from O,KN
+H=500.0 //Horizontal loading at 4m ditance from O,KN
+ah=4.0
+a1=2.0
+a2=4.0
+a3=5.0
+a=7.0
+//assume Resulat R at distance x from O,
+//sum of vertical Fy & sum of horizontal forces Fx is zero
+//Assume direction of Fx is right
+//Assume direction of Fy is up
+Rx=H
+Ry=P1-P2+P3
+
+printf("\n Ry= %0.2f KN downward",Ry)
+
+//Let x be the distance from O where the resultant cuts the base.
+//moment at O
+x=(H*ah+P1*a1-P2*a2+P3*a3)/(Ry)
+
+printf("\n x= %0.3f m",x)
+
+printf("\n The resultant passes through the middle third of the base i.e., between 7/3m, and 2*7/3 m.Hence, the dam is safe.")
+