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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/CH2/EX2.11/Ex2_11.sce | |
parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3862/CH2/EX2.11/Ex2_11.sce')
-rw-r--r-- | 3862/CH2/EX2.11/Ex2_11.sce | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/3862/CH2/EX2.11/Ex2_11.sce b/3862/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..b8fca588f --- /dev/null +++ b/3862/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,31 @@ +clear +// + +//variable declaration + +PB=2.0 //loading at B,KN +PC=sqrt(3.0) //loading at C,KN +PD=5.0 //loading at D,KN +PE=PC //loading at E,KN +PF=PB //loading at F,KN + +//Let O be the centre of the encircling circle A, B, C, D, E and F. In regular hexagon each side is equal to the radius AO. Hence OAB is equilateral triangle. + +angleoab=60.0*%pi/180 +anglecab=angleoab/2.0 +theta1=anglecab +theta2=(angleoab-theta1) +theta3=theta1 +theta4=theta1 + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=PB*cos(theta1+theta2)+PC*cos(theta2)+PD+PE*cos(theta3)+PF*cos(theta3+theta4) + +Fy=-PB*sin(theta1+theta2)-PC*sin(theta2)+0+PE*sin(theta3)+PF*sin(theta3+theta4) + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f KN",R) + +theta=atan(Fy/Fx)*180/%pi |