1 files changed, 62 insertions, 0 deletions
diff --git a/3750/CH1/EX1.10/Ex1_10.sce b/3750/CH1/EX1.10/Ex1_10.sce
new file mode 100644
index 000000000..f3cf4252a
--- /dev/null
+++ b/3750/CH1/EX1.10/Ex1_10.sce
@@ -0,0 +1,62 @@
+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 10
+//(a) To Calculate exiting stress in steel
+//(b)To Calculate final stress in steel in steel if additional end trust is applied
+
+clc();
+
+//Initialization of variables
+Ds=18;  //Diameter of steel rod, Unit in mm
+
+Dc=39;  //Outside diameter of copper sleeve, Unit i mm
+dc=24;//Inside diameter of copper sleeve, Unit in mm
+EsbyEc=2;//Ratio of Young's modulus of steel to young's of copper
+d=1.5//depth of copper removed, Unit in mm
+
+
+//Computations
+SigmaS1=10;//Tension stress set up in steel, Unit in N/mm^2
+
+As=(%pi/4)*Ds^2;  //Cross section Area of steel rod, Unit in mm^2
+Ac=(%pi/4)*(Dc^2-dc^2);  //Cross section Area of copper sleeve, Unit in mm^2
+Acr=(%pi/4)*((Dc-2*d)^2-dc^2);  //Area of reduce sectoin of copper, Unit in mm^2
+SigmaC1=(As/Ac)*SigmaS1;  //Stress set up in copper tube, Unit in N/mm^2
+
+//(a)When tube reduced in area for half it's length
+//Let SigmaC2 be stress in reduced secton in copper  & SigmaCdash in the reminder
+//Let SigmaS2 be stress in rod
+//Equilibrium equation:  Load on tube=Load on Rod
+         //SigmaC2*Acr=SigmaC2dash*Ac=SigmaS2*As
+         //SigmaC2=(As/Acr)*SigmaS2......(i)
+         //SigmaC2dash=(As/Ac)*SigmaC2.....(ii)
+         
+         
+//Compatibility Equation: Reduction in lenght of rod=Reduction in length of tube
+        //(SigmaS1-SigmaS2)*l/Es=(SigmaC2-SigmaC1)*l/(2*Ec) + (SigmaC3dash-SigmaC1)*l/(2*Ec)
+        
+//Solving Equilibrium Equations and compatibility equation  
+SigmaS2=(SigmaS1+EsbyEc*SigmaC1)/(1+As*EsbyEc/(2*Acr)+As*EsbyEc/(2*Ac));   //Unit in N/mm^2       The answer vary due to round off error
+
+//Result (a)
+printf("The exiting stress in steel, SigmaS2= %.1fN/mm^2\n",SigmaS2)
+
+
+
+//(b)An additonal end thrust of 5000N is applied
+
+P=5000;//Additonal end thrust, Unit in N
+//Let SigmaS3 And SigmaC3 be stresses in reduce section of steel and copper respectively
+//Let SigmaC3dash be stress in remainder section of copper
+   
+   
+   //Equlibrium Equation:
+         //P=SigmaC3*Acr-SigmaS3*As
+         //SigmaC3=P/Acr+(As/Acr)*SigmaS3............(iii)
+         //SigmaC3dash=P/Ac+(As/Ac)*SigmaS3............(iv)
+
+
+SigmaS3=(SigmaS1+EsbyEc*SigmaC1-(EsbyEc/2)*(P/Acr+P/Ac))/(1+EsbyEc*As/(2*Acr)+EsbyEc*As/(2*Ac));      //Unit in N/mm^2,      The answer vary due to round off error   
+
+//Result (b)
+printf("Final Stress in Steel,SigmaS3=%.1f N/mm^2",SigmaS3)