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| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH4/EX4.8/Ex4_8.sce | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3886/CH4/EX4.8/Ex4_8.sce')
| -rw-r--r-- | 3886/CH4/EX4.8/Ex4_8.sce | 15 |
1 files changed, 15 insertions, 0 deletions
diff --git a/3886/CH4/EX4.8/Ex4_8.sce b/3886/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..47b6ff003 --- /dev/null +++ b/3886/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,15 @@ +//Finding forces in members
+//Refer fig. 4.15 (a)
+RA=7*20/2 //kN
+RB=RA //kN (symmetry)
+//CE is perpendicular on AB
+CE=5.196 //m
+DE=3 //m
+theta=atand(5.196/3) //degree
+//The fact that 20 kN loads are equidistant can be used to find out horizontal distances of loads from A
+//Consider equilibrium of left hand side portion of section (1)-(1)
+//taking moment about A
+F2=(20*2.25+20*4.5+20*6.75)/(6*sind(60)) //kN (Tension)
+F1=(70-20-20-20+51.96*sind(60))/sind(30) //kN (compression)
+F3=-51.96*cosd(60)+110*cosd(30) //kN (Tension)
+printf("The required forces are:-\nF1=%.2d kN (Compression)\nF2=%.2d kN (Tension)\nF3=%.2d kN (Tension)",F1,F2,F3)
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