summaryrefslogtreecommitdiff
path: root/3774/CH3
diff options
context:
space:
mode:
Diffstat (limited to '3774/CH3')
-rw-r--r--3774/CH3/EX3.1/Ex3_1.sce23
-rw-r--r--3774/CH3/EX3.10/Ex3_10.sce27
-rw-r--r--3774/CH3/EX3.11/Ex3_11.sce26
-rw-r--r--3774/CH3/EX3.12/Ex3_12.sce25
-rw-r--r--3774/CH3/EX3.13/Ex3_13.sce48
-rw-r--r--3774/CH3/EX3.2/Ex3_2.sce25
-rw-r--r--3774/CH3/EX3.3/Ex3_3.sce23
-rw-r--r--3774/CH3/EX3.4/Ex3_4.sce45
-rw-r--r--3774/CH3/EX3.5/Ex3_5.sce28
-rw-r--r--3774/CH3/EX3.6/Ex3_6.sce25
-rw-r--r--3774/CH3/EX3.8/Ex3_8.sce22
11 files changed, 317 insertions, 0 deletions
diff --git a/3774/CH3/EX3.1/Ex3_1.sce b/3774/CH3/EX3.1/Ex3_1.sce
new file mode 100644
index 000000000..357bcaf6b
--- /dev/null
+++ b/3774/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,23 @@
+// exa 3.1 Pg 62
+
+clc;clear;close;
+
+// Given Data
+P=30;// kN
+Sut=350;// MPa
+n=2.5;// factor of safety
+
+sigma_w=Sut/n;// MPa (Working stress for the link)
+
+t=poly(0,'t');// thickness of link
+A=2.5*t**2;// mm.sq.
+I=t*(2.5*t)**3/12;// mm^4 (Moment of Inertia about N-A)
+sigma_d=P/A;// N/mm.sq.
+e=10+1.25*t;//mm
+M=P*10**3*e;// N.mm
+sigma_t=M*1.25*t/I;// N/mm.sq.
+//maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w ...eqn(1)
+expr=sigma_d+sigma_t-sigma_w ;// expression of polynomial from above eqn.
+t=roots(numer(expr));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+t=t(1);// mm // discarding -ve roots
+printf('dimension of cross section of link, t=%.f mm. Adopt t=21 mm. ',t)
diff --git a/3774/CH3/EX3.10/Ex3_10.sce b/3774/CH3/EX3.10/Ex3_10.sce
new file mode 100644
index 000000000..a9d5db7fa
--- /dev/null
+++ b/3774/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,27 @@
+// exa 3.10 Pg 71
+
+clc;clear;close;
+
+// Given Data
+d=4;// cm
+M=15000;// N.cm
+Syt=20000;// N/cm.sq.
+
+printf('\n (i) Maximum Principal Stress Theory-')
+z=%pi*d**3/32;// cm.cube.
+sigma_b=M/z;// N/cm.sq.
+T=poly(0,'T')
+tau=16*T/(%pi*d**3);// N/cm.sq.
+//sigma1=(1/2)*(sigma_b+sqrt(sigma_b**2+4*tau**2)) // Maximum principal stress
+//sigma1=(sigma_b/2+sqrt(sigma_b**2/4+tau**2)) // on solving
+//tau=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)
+sigma1=Syt;// N/cm.sq.
+T=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)*(%pi*d**3)/16;// N.cm.
+printf('\n Maximum value of torque, T = %.f N.cm.',T)
+
+printf('\n (ii) Maximum Shear Stress Theory')
+tau_d=0.5*Syt;// N.cm.
+//Te=sqrt(M**2+T**2)=(%pi/16)*d**3*tau_d
+T=sqrt(((%pi/16)*d**3*tau_d)**2-M**2);// N.cm.
+printf('\n Maximum value of torque, T = %.f N.cm.',T)
+// Answer in the textbook is not accurate.
diff --git a/3774/CH3/EX3.11/Ex3_11.sce b/3774/CH3/EX3.11/Ex3_11.sce
new file mode 100644
index 000000000..529217bd8
--- /dev/null
+++ b/3774/CH3/EX3.11/Ex3_11.sce
@@ -0,0 +1,26 @@
+// exa 3.11 Pg 72
+
+clc;clear;close;
+
+// Given Data
+N=200;// rpm
+P=25;// kW
+tau_d=42;// MPa
+W=900;// N
+L=3;// m
+Syt=56;// MPa
+Syc=56;// MPa
+sigma_d=56;// MPa
+
+T=P*60*10**3/(2*%pi*N);// N.m
+M=W*L/4;// N.m
+Te=sqrt(M**2+T**2);// N.m
+// Te=(%pi/16)*d**3*tau_d
+d=(Te*10**3/((%pi/16)*tau_d))**(1/3);// mm
+printf('\n shaft diameter(using equivalent torque)-\n d=%.f mm.',d)
+
+Me=(1/2)*(M+sqrt(M**2+T**2));//N.m
+// Me=(%pi/32)*d**3*sigma_d
+d=(Me*10**3/((%pi/32)*sigma_d))**(1/3);// mm
+printf('\n shaft diameter(using equivalent bending moment)-\n d=%.f mm.',d)
+printf('\n adopt d=57 mm.')
diff --git a/3774/CH3/EX3.12/Ex3_12.sce b/3774/CH3/EX3.12/Ex3_12.sce
new file mode 100644
index 000000000..7b94e8c01
--- /dev/null
+++ b/3774/CH3/EX3.12/Ex3_12.sce
@@ -0,0 +1,25 @@
+// exa 3.12 Pg 72
+
+clc;clear;close;
+
+// Given Data
+sigma_w=60;// MPa
+F=10;// kN
+alfa=30;// degree
+
+FH=F*sind(alfa);// kN
+FV=F*cosd(alfa);// kN
+t=poly(0,'t');// mm
+A=t*t;// mm.sq.
+sigma_d=FV*10**3/A
+M=FV*10**3*120+FH*10**3*150;// N.mm
+I=t*(2*t)**3/12;// mm^4
+sigma_t=M*t/I;// N/mm.sq.
+// Tensile stress at A=sigma_d+sigma_t=sigma_w ...eqn(1)
+expr = sigma_d+sigma_t-sigma_w;// polynomial from above eqn.
+t=roots(numer(expr));// roots of the polynomial
+t=t(1);// mm // discarding -ve roots
+printf('\n value of t = %.1f mm',t)
+A=2*t**2;// mm.sq.
+printf('\n Area of cross-section of Hanger, A = %.f mm.sq.',A)
+// Note-Answer in the textbook is slighly wrong and cross section not calculated.
diff --git a/3774/CH3/EX3.13/Ex3_13.sce b/3774/CH3/EX3.13/Ex3_13.sce
new file mode 100644
index 000000000..bbbecc0e8
--- /dev/null
+++ b/3774/CH3/EX3.13/Ex3_13.sce
@@ -0,0 +1,48 @@
+// exa 3.13 Pg 74
+
+clc;clear;close;
+
+// Given Data
+P=15;// kW
+n1=200;// rpm
+l=600;// mm
+z2=18;// no. of teeth
+m2=5;// mm
+alfa2=14.5;// degree
+l2=120;// mm
+z1=72;// no. of teeth
+m1=5;// mm
+alfa1=14.5;// degree
+l1=150;// mm
+sigma_d=80;// MPa
+
+d1=m1*z1;// mm
+v1=%pi*d1*n1/(60*10**3);// m/s
+Ft1=10**3*P/v1;// N (outwards)
+Fr1=Ft1*tand(alfa1);// N (Downwards)
+d2=m2*z2;// mm
+v2=%pi*d2*n1/(60*10**3);// m/s
+Ft2=10**3*P/v2;// N (outwards)
+Fr2=Ft2*tand(alfa2);// N (Upwards)
+
+// RAV*600=Fr1*450+Fr2*120 (Taking moments about bearing B)
+RAV=(Fr1*450+Fr2*120)/600;// N (Downwards)
+RBV=(Fr1-Fr2-RAV);// N (upwards)
+MCV=RAV*l1;// N.mm
+MBV=Fr2*l2;// N.mm
+
+// RAH*600=-Ft1*450+Ft2*120 (Taking moments about bearing B)
+RAH=(-Ft1*450+Ft2*120)/600;// N (Outwards)
+RBH=Ft1+Ft2+RAH;// N (inwards)
+MCH=RAH*l1;// N.mm
+MBH=Ft2*l2;// N.mm
+
+// Resultant Bending Moments
+MC=sqrt(MCV**2+MCH**2);// N.mm
+MB=sqrt(MBV**2+MBH**2);// N.mm
+Mmax=max(MC,MB);// N.mm
+T=10**3*P/(2*%pi*n1);// N.m
+Me=(1/2)*(Mmax+sqrt(Mmax**2+T**2));// N.mm
+// Me=(%pi/32)*d**3*sigma_d
+d=(Me/((%pi/32)*sigma_d))**(1/3)
+printf('\n shaft diameter is : %.f mm',d)
diff --git a/3774/CH3/EX3.2/Ex3_2.sce b/3774/CH3/EX3.2/Ex3_2.sce
new file mode 100644
index 000000000..5414dc4be
--- /dev/null
+++ b/3774/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,25 @@
+// exa 3.2 Pg 63
+
+clc;clear;close;
+
+// Given Data
+P=6;// kN
+alfa=30;// degree
+Sut=250;// MPa
+n=2.5;// factor of safety
+
+sigma_w=Sut/n;// MPa (Working stress for the link)
+PH=P*10**3*cosd(alfa);// kN
+PV=P*10**3*sind(alfa);// kN
+
+t=poly(0,'t');// thickness of link
+A=2*t*t;// mm.sq.
+sigma_d=PH/A;// N/mm.sq.
+M=PH*100+PV*250;// N.mm
+I=t*(2*t)**3/12;// mm^4 (Moment of Inertia)
+sigma_t=M*t/I;// N/mm.sq.
+//maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w ...eqn(1)
+expr=sigma_d+sigma_t-sigma_w ;// expression of polynomial from above eqn.
+t=roots(numer(expr));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+t=t(1);// mm // discarding -ve roots
+printf('dimension of cross section of link, t=%.f mm.',t)
diff --git a/3774/CH3/EX3.3/Ex3_3.sce b/3774/CH3/EX3.3/Ex3_3.sce
new file mode 100644
index 000000000..6a1fcf9f1
--- /dev/null
+++ b/3774/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,23 @@
+// exa 3.3 Pg 64
+
+clc;clear;close;
+
+// Given Data
+P=20;// kN
+Sut=300;// MPa
+n=3;// factor of safety
+
+sigma_w=Sut/n;// MPa (Working stress for the link)
+
+t=poly(0,'t');// thickness of link
+A=4*t*t;// mm.sq.
+sigma_d=P*10**3/A;// N/mm.sq.
+e=6*t;//mm
+M=P*10**3*e;// N.mm
+z=t*(4*t)**2/6;// mm^3 (section modulus at x1-x2)
+sigma_b=M/z;// N/mm.sq.
+//maximum tensile stress at x1 = sigma_d+sigma_b=sigma_w ...eqn(1)
+expr=sigma_d+sigma_b-sigma_w ;// expression of polynomial from above eqn.
+t=roots(numer(expr));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+t=t(2);// mm // discarding -ve roots
+printf('dimension of cross section of link, t=%.2f mm. Use 23 mm.',t)
diff --git a/3774/CH3/EX3.4/Ex3_4.sce b/3774/CH3/EX3.4/Ex3_4.sce
new file mode 100644
index 000000000..b8f3b6496
--- /dev/null
+++ b/3774/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,45 @@
+// exa 3.4 Pg 65
+
+clc;clear;close;
+
+// Given Data
+P=15;// kN
+sigma_t=20;// MPa
+sigma_c=60;// MPa
+n=3;// factor of safety
+
+a=poly(0,'a');// from the diagram.
+// Area of cross section
+A1=2*a*a;// mm.sq.
+A2=2*a*a/2;// mm.sq.
+A=A1+A2;// mm.sq.
+
+// Location of neutral axis
+//3*a**2*y_bar=2*a**2*a/2+a**2*(a+a/2)
+y_bar=(2*a**2*a/2+a**2*(a+a/2))/(3*a**2);// mm
+
+// Moment of Inertia about neutral axis N-A
+I=2*a*a**3/12+2*a**2*(y_bar-0.5*a)**2+2*((a/2)*(a**3/12)+(a**2/2)*(1.5*a-y_bar)**2);// mm^4
+yt=y_bar;//mm
+yc=2*a-y_bar;// mm
+e=y_bar-0.5*a;//mm
+M=P*10**3*e;// N.mm
+sigma_d=P*10**3/A;// N/mm.sq.
+sigma_t1=M*yt/I;//N/mm.sq.
+sigma_c1=M*yc/I;//N/mm.sq.
+sigma_r_t=sigma_d+sigma_t1;// N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)
+sigma_r_c=sigma_c1-sigma_d;// N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)
+
+//equating resulting tensile stress with given value sigma_t-sigma_r_t=0...eqn(1)
+expr1=sigma_t-sigma_r_t;// expression of polynomial from above eqn.
+a1=roots(numer(expr1));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+a1=a1(2);// mm // discasrding -ve roots
+printf('Equating resultant tensile stress gives, a = %.2f mm',a1)
+
+//equating resulting compressive stress with given value sigma_c-sigma_c_t=0...eqn(1)
+expr2=sigma_c-sigma_r_c;// expression of polynomial from above eqn.
+a2=roots(numer(expr2));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+a2=a2(2);// mm // discarding -ve roots
+printf('\n Equating resultant compressive stress gives, a = %.2f mm',a2)
+a=ceil(a1);//mm
+printf('\n dimension of cross section of link, a=%.2f mm. adopt a=%.f mm.',a1,a)
diff --git a/3774/CH3/EX3.5/Ex3_5.sce b/3774/CH3/EX3.5/Ex3_5.sce
new file mode 100644
index 000000000..a09bdbdd0
--- /dev/null
+++ b/3774/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,28 @@
+// exa 3.5 Pg 67
+
+clc;clear;close;
+
+// Given Data
+Syt=760;// MPa
+M=15;// kN.m
+T=25;//kN.m
+n=2.5;// factor of safety
+E=200;// GPa
+v=0.25;// Poisson's ratio
+
+sigma_d=Syt/n;// MPa
+// let d is diameter of the shaft
+sigma_b_into_d_cube=32*M*10**6/%pi;// N/mm.sq. (where sigma_b_into_d_cube = sigma_d*d**3)
+tau_into_d_cube=16*T*10**6/%pi//d**3;// N/mm.sq. (where tau_into_d_cube = tau*d**3)
+sigma1_into_d_cube=sigma_b_into_d_cube/2+1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2) ; // (where sigma1_into_d_cube=sigma1*d**3)
+sigma2_into_d_cube=sigma_b_into_d_cube/2-1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2); // (where sigma2_into_d_cube=sigma2*d**3)
+printf('\n (i) Maximum shear stress theory')
+tau_max_into_d_cube=(sigma1_into_d_cube-sigma2_into_d_cube)/2; //(where tau_max_into_d_cube = tau_max*d**3)
+d=(tau_max_into_d_cube/(sigma_d/2))**(1/3);//mm
+printf('diameter of shaft, d=%.1f mm or %.f mm',d,ceil(d))
+
+printf('\n (ii) Maximum strain energy theory')
+//sigma1**2+sigma2**2-2*v*sigma1*sigma2=sigma_d**2
+d=((sigma1_into_d_cube**2+sigma2_into_d_cube**2-2*v*sigma1_into_d_cube*sigma2_into_d_cube)/sigma_d**2)**(1/6)
+printf('diameter of shaft, d=%.1f mm',d)
+printf('\n Adopt d=100mm')
diff --git a/3774/CH3/EX3.6/Ex3_6.sce b/3774/CH3/EX3.6/Ex3_6.sce
new file mode 100644
index 000000000..fc7afb40c
--- /dev/null
+++ b/3774/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,25 @@
+// exa 3.6 Pg 69
+
+clc;clear;close;
+
+// Given Data
+N=200;// rpm
+P=200;// kW
+tau_d=42;// Mpa
+W=900;// N
+L=3;// m
+sigma_t=56;// MPa
+sigma_c=56;// MPa
+
+T=P*60*10**3/(2*%pi*N);// N.m
+M=W*L/4;// N.m
+Te=sqrt(M**2+T**2);// N.m
+//Te=(%pi/16)*d**3*tau_d
+d=(Te/((%pi/16)*tau_d)*1000)**(1/3);// mm
+printf('\n Using equivalent torque equation,\n shaft diameter d = %.f mm',d)
+
+Me=(1/2)*(M+sqrt(M**2+T**2));// N.m
+//Me=(%pi/32)*d**3*sigma_d
+d=(Me/((%pi/32)*sigma_c)*10**3)**(1/3);//mm
+printf('\n Using equivalent bending moment equation,\n shaft diameter d = %.2f mm or %.f mm',d, ceil(d))
+printf('\n Adopt d=105 mm.')
diff --git a/3774/CH3/EX3.8/Ex3_8.sce b/3774/CH3/EX3.8/Ex3_8.sce
new file mode 100644
index 000000000..1003c0c77
--- /dev/null
+++ b/3774/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,22 @@
+// exa 3.8 Pg 70
+
+clc;clear;close;
+
+// Given Data
+M=15;// N.m
+P=5;// kW
+N=500;// rpm
+tau_d=40;// Mpa
+sigma_d=58;// MPa
+
+T=P*60*10**3/(2*%pi*N);// N.m
+Te=sqrt(M**2+T**2);// N.m
+//Te=(%pi/16)*d**3*tau_d
+d=(Te/((%pi/16)*tau_d)*1000)**(1/3);// mm
+printf('\n Using equivalent torque equation,\n shaft diameter d = %.f mm',d)
+
+Me=(1/2)*(M+sqrt(M**2+T**2));// N.m
+//Me=(%pi/32)*d**3*sigma_d
+d=(Me/((%pi/32)*sigma_d)*10**3)**(1/3);//mm
+printf('\n Using equivalent bending moment equation,\n shaft diameter d = %.2f mm or %.f mm',d, ceil(d))
+printf('\n Adopt d=23 mm.')