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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
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parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+//Analysing the truss
+//Refer fig. 4.10(a)
+//inclined members make an angle theta with the horizontal
+theta=atand(4/3) //Degree
+//Joint E
+//Refer fig. 4.10 (c)
+//applying equilibrium conditions
+FED=20/sind(theta) //kN (Tension)
+FEF=25*cosd(theta) //kN (Compression)
+//Refer fig 4.10 (b)
+//Taking moment about A
+RC=20*6/8 //kN
+VA=20 //kN
+HA=RC //kN
+//Joint A
+//Refer fig.4.10 (d)
+//applying equilibrium conditions
+FAB=VA //kN (Compression)
+FAF=HA //kN (Compression)
+//Joint C
+//Refer fig. 4.10 (E)
+FCB=RC/cosd(theta) //kN (Compression)
+FCD=FCB*sind(theta) //kN (Tension)
+//Joint B
+//Refer fig. 4.10 (f)
+FBF=(FBC*sind(theta)-FAB)/sind(theta) //kN
+FBD=0+25*cosd(theta) //kN (Tension)
+//Joint F
+//Refer Fig. 4.10(g)
+FFD=0
+FBF=0
+printf("FED=%.2f kN (Tension)\nFEF=%.2f kN (Compression)\nRC=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN (Compression)\nFAF=%.2f kN (Compression)\nFCB=%.2f kN (Compression)\nFCD=%.2f kN (Tension)\nFBF=%.2f kN\nFBD=%.2f kN (Tension)\nFFD=%.2f kN\nFBF=%.2f kN",FED,FEF,RC,VA,HA,FAB,FAF,FCB,FCD,FBF,FBD,FFD,FBF)