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author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
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committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3862/CH3/EX3.10 | |
parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3862/CH3/EX3.10')
-rw-r--r-- | 3862/CH3/EX3.10/Ex3_10.sce | 36 |
1 files changed, 36 insertions, 0 deletions
diff --git a/3862/CH3/EX3.10/Ex3_10.sce b/3862/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..da5656eff --- /dev/null +++ b/3862/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,36 @@ +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) |