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Diffstat (limited to '3862/CH2/EX2.18')
| -rw-r--r-- | 3862/CH2/EX2.18/Ex2_18.sce | 48 |
1 files changed, 48 insertions, 0 deletions
diff --git a/3862/CH2/EX2.18/Ex2_18.sce b/3862/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..9f2b7260d --- /dev/null +++ b/3862/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,48 @@ +clear +// + +//variable declaration +//The two 40 kN forces acting on the smooth pulley may be replaced by a pair of 40 kN forces acting at centre of pulley C and parallel to the given forces, since the sum of moments of the two given forces about C is zero + +PA=20.0 //inclined at 45° loading at A,KN +PB=30.0 //inclined at 60° loading at B,KN + +PC1=40.0 //inclined at 30° loading at C,KN +PC2=40.0 //inclined at 20° loading at C,KN +PD=50.0 //inclined at 30.0 at distance 2m form A,KN +PE=20.0 //inclined at alpha at distance xm form A,KN +P=20.0 //vertical loading at distance 4m,KN + + + +thetaA=45.0*%pi/180.0 +thetaB=60.0*%pi/180.0 +thetaC1=30.0*%pi/180.0 +thetaC2=20.0*%pi/180.0 +thetaD=30.0*%pi/180.0 +AD=2.0 +AC=3.0 +AB=6.0 + +//sum of vertical Fy & sum of horizontal forces Fx is zero +//Assume direction of Fx is right +//Assume direction of Fy is up +Fx=PA*cos(thetaA)-PB*cos(thetaB)-PD*cos(thetaD)-PC1*sin(thetaC1)+PC2*cos(thetaC2) + +Fy=-PA*sin(thetaA)-P+P-PB*sin(thetaB)-PD*sin(thetaD)-PC2*sin(thetaC2)-PC1*cos(thetaC1) + + +R=sqrt((Fx**2)+(Fy**2)) +printf("\n R= %0.2f KN",R) + +alpha=atan(Fy/Fx)*180/%pi +printf("\n alpha= %0.2f °",alpha) + +//Let the resultant intersect AB at a distance x from A. Then, + + +X=(-P*4+P*4+PB*sin(thetaB)*AB+PD*sin(thetaD)*AD-PD*cos(thetaD)*AD+PC2*AC*cos(thetaC2)-PC1*AC*sin(thetaC1))/R + +printf("\n x= %0.2f m",X) + +printf("\n The equilibriant is equal and opposite to the resultant in which E = 116.515 kN, alpha= 76.82° and x= %0.2f m",X) |