12 files changed, 249 insertions, 0 deletions
diff --git a/3750/CH1/EX1.1/Ex1_1.sce b/3750/CH1/EX1.1/Ex1_1.sce
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+// Strength of Materials By G.H. Ryder
+// Chapter 1
+//Example 1
+
+// To Calculate Young's Modulus, Stress at limit of proportion, the yield stress, ultimate stress, % elongation and %contraction.
+clc();
+
+//Given (from question) i.e. Initialization of variables
+P=80000;      //load at limit of proportionality , unit in newton(N)
+D= 2;       //original diameter, unit in cm
+l=4;        // gauge length , unit in cm 
+x=0.048;     //extension at limit of proportionality , unit in mm
+Py=85000;    //Load at yield point , unit in newton (N)
+Pmax=150000;   //maximum or ultimate load, units in newton(N)
+l1=5.56;      //elongation, unit in cm
+D1=1.58;      //contracted diameter at neck , unit in cm
+
+//Calculations
+A=%pi*100*(D^2)/4;  //Cross section area , unit in mm^2
+E=(P*l*10)/(A*x);  //Youngs Modulus , unit in N/(mm^2)
+stress1=P/A;      //Stress at limit of proportionality,unit in N/(mm^2) 
+stressY=Py/A;       //yield stress N/(mm^2) ,unit in N/(mm^2) 
+stressuts=Pmax/A;      //ultimate tensile stress,unit in N/(mm^2)  
+el=(l1-l)*100/l;   //percentage elongation
+co=(D^2-D1^2)*100/D^2;      //percentage contraction 
+
+//Outputs
+printf("Young Modulus = %.2fN/mm^2\n",E)    //The answers vary due to round off error
+printf("stress at limit of proportionality = %.0fN/mm^2\n",stress1)    
+printf("yield stress = %.0fN/mm^2\n",stressY)   
+
+printf("ultimate tensile stress = %.2fN/mm^2\n",stressuts)    //The answers vary due to round off error
+printf("percentage elongation = %.0f percent\n",el)
+printf("percentage contraction = %.0f percent\n",co)      
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
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+//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)
diff --git a/3750/CH1/EX1.11/Ex1_11.sce b/3750/CH1/EX1.11/Ex1_11.sce
new file mode 100644
index 000000000..f2d0a5dc6
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+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 11
+// To Calculate thermal Stress in rod & tube
+SteelOD=2.4;  //External diameter of steel tube, Unit in cm
+SteelID=1.8;   //Internal diameter of steel tube, Unit in cm
+CopperDia=1.5;  //Diameter of copper rod, unit in mm
+Es=210,000;  //Young's Modulus for steel , Unit in N/mm^2
+Ec=100,00;  //Young's Modulus for copper , Unit in N/mm^2
+alphaS=11e-6;  //co-efficient of linear expansion for steel, Unit in perdegreeC
+alphaC=18e-6;  //co-efficient of linear expansion for copper, Unit in perdegreeC
+AreaSteel=%pi*(SteelOD^2-SteelID^2)/4;   //cross section Area  of steel tube, Unit in cm^2
+AreaCopper=%pi*CopperDia^2/4;   //cross section Area  of copper bar, Unit in cm^2
+
+//Equillibrium Equation : SigmaC*AreaCopper=SigmaS*AreaSteel
+Ti=10; //Initial Temperature, Unit in perdegreeC
+Tf=200;//Final Temperature, Unit in perdegreeC
+//Compatibility Equation:
+               //alphaC*(Tf-Ti)-SigmaC/Ec=alphaS*(Tf-Ti)-SigmaS/Es
+SigmaS=(alphaS+alphaC)*(Tf-Ti)/((1/Es)*AreaSteel/(AreaCopper*Ec));
+//Stress in steel, Expression from compatability & Equillibrium Equation, Unit in N/mm^2
+
+SigmaC=AreaSteel*SigmaS/AreaCopper;
+printf("Stress in Copper rod= %f N/mm^2\n",SigmaC)
+printf("Stress in Steel Tube= %f N/mm^2\n",SigmaS)
+
+
+
diff --git a/3750/CH1/EX1.12/Ex1_12.sce b/3750/CH1/EX1.12/Ex1_12.sce
new file mode 100644
index 000000000..a0d8d225c
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+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 12
+// To Find New Tension in bolts
+SideB=20;  //side of square liquid , Unit in cm
+Pc=5;   //central load, Unit in KN
+n=4;   //Numbers of bolts
+SideF=16;  //side of square holding foundation
+Pb=0.5;   //tension in bolt, Unit in KN
+shift=2;   //shift in line of  action of load, Unit i cm
+InitialLoad=Pc+n*Pb;   //Initial load on ??????? , Unit in KN
+//This load is distributed over 20cm causing compression of x
+RateOfLoading=(5*8)*(xbye/2)*(7/20); Unit in KN/cm;
+F=RateOfLoading*SideB/2;   //Total force in ?????? in oneside, Unit in KN
+C=2/3*SideB/2;    //Distance of centroid from centre line , Unit in cm
+//Moment Equation:   2*xbye*SSideB/2+2*F*C=Pc*2
+xbye=Pc*2/(2*SideF/2+2*F*C)    //Unit in KN
diff --git a/3750/CH1/EX1.2/Ex1_2.sce b/3750/CH1/EX1.2/Ex1_2.sce
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+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 2
+// To Calculate Strain energy
+P=10,000;    //Tension 100d, Unit in KN
+E=205,000;   //Young's Modulus, Unit in N/mm^2
+RootDia=16.6; //Root diameter of thread, Unit in mm
+AreaOfCore=%pi*(RootDia^2)/4   //Unit in mm^2
+ShankDia=20; // Diameter at Shank, Unit in mm
+AreaAtShank=%pi*(ShankDia^2)/4; //Unit in mm^2
+ThreadLength=25;   //Unit in mm
+ShankLength=50;   // Unit in mm
+StressInScrew=P/AreaOfCore;   //Unit in N/mm^2
+StressInShank=P/AreaAtShank;   //Unit in N/mm^2
+TotalSE=(StressInScrew^2)*AreaOfCore*ThreadLength+(StressInShank^2)*(AreaAtShank*ShankLength)/(2*E);
+// Total Strain Energy, Unit in N/mm^2
+//If Shank is turned down to root diameter(16.6mm) for same maximum stress 
+StressInBolt=P/AreaOfCore;  //Unit in N/mm^2
+NewSE=((StressInBolt^2)*(AreaOfCore)*(ThreadLength+ShankLength))/(2*E)
+//Strain Energy after shank is turned down to root diameter, Unit in Nmm
+printf("Total Strain Energy=%f Nmm\n", TotalSE)
+printf("Strain Energy after Shank is turned down to root diameter=%f Nmm\n", NewSE)
+
+
diff --git a/3750/CH1/EX1.3/Ex1_3.sce b/3750/CH1/EX1.3/Ex1_3.sce
new file mode 100644
index 000000000..9b8f654d7
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+++ b/3750/CH1/EX1.3/Ex1_3.sce
@@ -0,0 +1,16 @@
+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 3
+// To Calulate Stress
+g=9.8;   //Acceleration due to Gravity, Unit in m/sec^2
+m=100;  //Falling Mass , Unit in Kg
+W=m*g;   //Falling weight , Unit in N
+D=2;  // Diameter of steel bar, Unit in cm
+A=%pi*(D^2)*100/4;   //Cross section Area of steel bar, Unit in mm^2
+h=4;  //height from which W is falling, Unit in cm
+l=3;   //lenght of steel bar, Unit in cm
+E=205,000;  //Young's Modulus of steel, Unit in m
+P=W*(1+(1+2*h*10*E*A/(W*l*1000))^(1/2));
+//Formula for Equivalent load from Energy Equation, Unit in N
+Stress=P/4;  //Stress set up in bar,unit in N/mm^2
+printf("The Stress set up in steel bar is %f N.mm^2",Stress)
diff --git a/3750/CH1/EX1.4/Ex1_4.sce b/3750/CH1/EX1.4/Ex1_4.sce
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+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 4
+// To Calulate Stress & Extension
+g=9.8;   //Acceleration due to Gravity, Unit in m/sec^2
+m=100;  //Falling Mass , Unit in Kg
+W=m*g;   //Falling weight , Unit in N
+D1=1; // diameter of first part of bar, Unit in cm
+l1=1.5;  //Lenght fo first part of bar, Unit in m
+D2=2;  // diameter of second part of bar, Unit in cm
+l2=1.5;   //Lenght fo second part of bar, Unit in m
+A1=%pi*(D1^2)/4*100;  //Area of first part of bar, Unit in mm^2
+A2=%pi*(D2^2)/4*100;  //A;rea of Second part of bar, Unit in mm^2
+E=205,000;  //Young's Modulus of the bar, Unit in N/mm^2
+h=4;  //height from which weight is falling, Unit in cm
+P=W*(1+(1+2*h*10*E/((l1*1000/A1)+(l2*1000/A2)))^(1/2)); //Formula for Equivalent load, from energy equation, Unit in N
+x=P*l1/A1*E+P*l2/(A2*E);  //Extension in rod, unit in mm
+//The maximum stress will occur in smallest section. so, 
+maxstress=P/A1;
+printf("maximum stress=%f N/mm^2",maxstress)
+printf("Extension =%f mm",x)
diff --git a/3750/CH1/EX1.5/Ex1_5.sce b/3750/CH1/EX1.5/Ex1_5.sce
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+++ b/3750/CH1/EX1.5/Ex1_5.sce
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+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 5
+// To Calulate  the maximum laod which can be carried
+ m1=100; //mass of cage , Unit in Kg
+ m2=0.9;  //mass of rope, Unit in Kg/m
+ l2=25;  //lenght of ropewire, Unit in m
+ n=49; //No. of wires
+ D2=1.6; //Diameter of each wire, Unit in mm
+ StressAll=90;// Max allowable stress, Unit in N/mm^2
+ E=70,000;// Young's Modulus for rope, unit in N/mm^2
+ h=10;  //height from which lift is dropped , Unit in cm
+ AreaOfRope=%pi*n*(D2^2)/4;  //Unit in mm^2
+ g=9.8;   //Acceleration due to Gravity, Unit in m/sec^2
+ StressInitial=(m1+m2*l2)/AreaOfRope;  //Initial stress unit in N/mm^2
+ StressImpact=StressAll-StressInitial; // Increase stress due to Impact, Unit in N/mm^2
+ P=StressImpact*AreaOfRope/g;  //Equivalent static load,Unit in Kg
+ x=StressImpact*l2*100/E  //Entension, Unit in cm
+ W=P*x/(2*(h+x))   //max load that can be dropped, Unit in Kg
+ printf("The maximum load which can be carried is %f Kg",W)
+ 
diff --git a/3750/CH1/EX1.6/Ex1_6.sce b/3750/CH1/EX1.6/Ex1_6.sce
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index 000000000..85630bbab
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+++ b/3750/CH1/EX1.6/Ex1_6.sce
@@ -0,0 +1,4 @@
+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 6
+//No Numerical Compution - Therotical Derivation to find extension
diff --git a/3750/CH1/EX1.7/Ex1_7.sce b/3750/CH1/EX1.7/Ex1_7.sce
new file mode 100644
index 000000000..43d95025c
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+++ b/3750/CH1/EX1.7/Ex1_7.sce
@@ -0,0 +1,4 @@
+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 7
+//No Numerical Compution - Therotical Derivation to find condition for a column to have uniform strength
diff --git a/3750/CH1/EX1.8/Ex1_8.sce b/3750/CH1/EX1.8/Ex1_8.sce
new file mode 100644
index 000000000..a16e227b9
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+++ b/3750/CH1/EX1.8/Ex1_8.sce
@@ -0,0 +1,14 @@
+//Strength Of Material By G.H.Ryder
+//Chapter 1
+//Example 8
+// To Find the Maximum Stress
+l=1;  //lenght of steel rod, Unit in m
+N=1000;   //rpm of rod, Unit in rmp
+rho=7.8;   //density of the material, Unit in g/cm^3
+Omega=%pi*2*N/60;   //Angular Velocity, Unit in rad/sec
+//sigma a=-rhox^2*Omega*2/2+c, formula obtain from integration
+//At x=l, sigma=0, c=rho*l^2*Omega*2/2
+x=0; //x is distance from axis of rod
+//Maximum Stress occur at axis, so
+sigma=((rho*(Omega^2))/2)*((l^2)-(x^2));   //Stress in bar, Unit in N/mm^2
+printf("The maximum Stress %f N/mm^2",sigma)
diff --git a/3750/CH1/EX1.9/Ex1_9.sce b/3750/CH1/EX1.9/Ex1_9.sce
new file mode 100644
index 000000000..2c0bacb0a
--- /dev/null
+++ b/3750/CH1/EX1.9/Ex1_9.sce
@@ -0,0 +1,4 @@
+//Strength Of Material By G.Hyder
+//Chapter 1
+//Example 9
+//No Numerical Compution - Therotical Derivation to find how the load is shared