updated the code

1 files changed, 19 insertions, 20 deletions

diff --git a/278/CH4/EX4.3/ex_4_3.sce b/278/CH4/EX4.3/ex_4_3.sce
index 1467c605c..8f502e032 100755
--- a/278/CH4/EX4.3/ex_4_3.sce
+++ b/278/CH4/EX4.3/ex_4_3.sce
@@ -1,21 +1,20 @@
-//to find 1.)diameter of the rods,2.)extension in each rod in the length of 2.5m

-clc

-//solution

-//given

-P=3.5*10^6//N//load applied

-f1=85//(N/mm^2)// safe stress

-E=210*10^3//(N/mm^2)//young's modulus

-l=2.5*10^3//mm

-pi=3.14

-//1)diameter of rod(d)

-//let d be diameter of rods in mm

-//since both rods carries equal load ,therefore load on single rod is

-P1=P/2//N

-d=sqrt(4*P1/(f1*pi))//using f1=P/A//mm

-printf("the diameter of rods is,%f mm\n",d)

-//2)extension in rod

-//let x be extension in rod

-//E=(P1*l)/(A*x)

- //P1/A=f1

-x=(f1*l)/E

+clc
+//solution
+//given
+P=3.5*10^6//N//load applied
+f1=85//(N/mm^2)// safe stress
+E=210*10^3//(N/mm^2)//young's modulus
+l=2.5*10^3//mm
+pi=3.14
+//1)diameter of rod(d)
+//let d be diameter of rods in mm
+//since both rods carries equal load ,therefore load on single rod is
+P1=P/2//N
+d=sqrt(4*P1/(f1*pi))//using f1=P/A//mm
+printf("the diameter of rods is,%f mm\n",d)
+//2)extension in rod
+//let x be extension in rod
+//E=(P1*l)/(A*x)
+ //P1/A=f1
+x=(f1*l)/E
 printf("the extension of rod is,%f mm",x)
\ No newline at end of file