diff options
Diffstat (limited to '3532')
89 files changed, 2138 insertions, 0 deletions
diff --git a/3532/CH1/EX1.10/Ex1_10.sce b/3532/CH1/EX1.10/Ex1_10.sce new file mode 100644 index 000000000..3c869fd4c --- /dev/null +++ b/3532/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,93 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.8.2\n')
+//given data
+T=0.1//time period of periodic motion in sec
+W=2*%pi/T
+k=12/2//number of elements in half cycle
+mprintf('\tNo of elements in one cycle 2k=12,t(j) in degrees\n')
+mprintf('t(j) f(j) cos(t(j)) f(j)*cos(t(j)) sin(t(j)) f(j)*sin(t(j) cos(2*t(j)) f(j)*cos(2*t(j)) sin(2*t(j)) f(j)*sin(2*t(j)) cos(3*t(j)) f(j)*cos(3*t(j) sin(3*t(j)) f(j)*sin(3*t(j)\n')
+f(1)=10/6
+for j=1:6
+ t(j)=j*(%pi/k)
+ m(j)=cos(t(j))
+ n(j)=f(j)*m(j)
+ o(j)=sin(t(j))
+ p(j)=f(j)*o(j)
+ q(j)=cos(2*t(j))
+ r(j)=f(j)*q(j)
+ s(j)=sin(2*t(j))
+ u(j)=f(j)*s(j)
+ v(j)=cos(3*t(j))
+ x(j)=f(j)*v(j)
+ y(j)=sin(3*t(j))
+ z(j)=f(j)*y(j)
+ f(j+1)=f(j)+f(1)
+mprintf('%3.0f\t',t(j)*(180/%pi))
+mprintf('%3.4f\t\t',f(j))
+mprintf('%3.4f\t\t',m(j))
+mprintf('%3.4f\t\t',n(j))
+mprintf('%3.4f\t\t',o(j))
+mprintf('%3.4f\t\t',p(j))
+mprintf('%3.4f\t\t',q(j))
+mprintf('%3.4f\t\t',r(j))
+mprintf('%3.4f\t\t',s(j))
+mprintf('%3.4f\t\t',u(j))
+mprintf('%3.4f\t\t',v(j))
+mprintf('%3.4f\t\t',x(j))
+mprintf('%3.4f\t\t',y(j))
+mprintf('%3.4f\n',z(j))
+end
+f(7)=f(j)-f(1)
+for j=7:12
+ t(j)=j*(%pi/k)
+ m(j)=cos(t(j))
+ n(j)=f(j)*m(j)
+ o(j)=sin(t(j))
+ p(j)=f(j)*o(j)
+ q(j)=cos(2*t(j))
+ r(j)=f(j)*q(j)
+ s(j)=sin(2*t(j))
+ u(j)=f(j)*s(j)
+ v(j)=cos(3*t(j))
+ x(j)=f(j)*v(j)
+ y(j)=sin(3*t(j))
+ z(j)=f(j)*y(j)
+ f(j+1)=f(j)-f(1)
+ mprintf('%3.0f\t',t(j)*(180/%pi))
+mprintf('%3.4f\t\t',f(j))
+mprintf('%3.4f\t\t',m(j))
+mprintf('%3.4f\t\t',n(j))
+mprintf('%3.4f\t\t',o(j))
+mprintf('%3.4f\t\t',p(j))
+mprintf('%3.4f\t\t',q(j))
+mprintf('%3.4f\t\t',r(j))
+mprintf('%3.4f\t\t',s(j))
+mprintf('%3.4f\t\t',u(j))
+mprintf('%3.4f\t\t',v(j))
+mprintf('%3.4f\t\t',x(j))
+mprintf('%3.4f\t\t',y(j))
+mprintf('%3.4f\n',z(j))
+end
+sumf(j)=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)
+sumcos(t(j))=m(1)+m(2)+m(3)+m(4)+m(5)+m(6)+m(7)+m(8)+m(9)+m(10)+m(11)+m(12)
+sumfjcos(t(j))=n(1)+n(2)+n(3)+n(4)+n(5)+n(6)+n(7)+n(8)+n(9)+n(10)+n(11)+n(12)
+sumsin(t(j))=o(1)+o(2)+o(3)+o(4)+o(5)+o(6)+o(7)+o(8)+o(9)+o(10)+o(11)+o(12)
+sumfjsin(t(j))=p(1)+p(2)+p(3)+p(4)+p(5)+p(6)+p(7)+p(8)+p(9)+p(10)+p(11)+p(12)
+sumcos2(t(j))=q(1)+q(2)+q(3)+q(4)+q(5)+q(6)+q(7)+q(8)+q(9)+q(10)+q(11)+q(12)
+sumfjcos2(t(j))=r(1)+r(2)+r(3)+r(4)+r(5)+r(6)+r(7)+r(8)+r(9)+r(10)+r(11)+r(12)
+sumsin2(t(j))=s(1)+s(2)+s(3)+s(4)+s(5)+s(6)+s(7)+s(8)+s(9)+s(10)+s(11)+s(12)
+sumfjsin2(t(j))=u(1)+u(2)+u(3)+u(4)+u(5)+u(6)+u(7)+u(8)+u(9)+u(10)+u(11)+u(12)
+sumcos3(t(j))=v(1)+v(2)+v(3)+v(4)+v(5)+v(6)+v(7)+v(8)+v(9)+v(10)+v(11)+v(12)
+sumfjcos3(t(j))=x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)+x(12)
+sumsin3(t(j))=y(1)+y(2)+y(3)+y(4)+y(5)+y(6)+y(7)+y(8)+y(9)+y(10)+y(11)+y(12)
+sumfjsin3(t(j))=z(1)+z(2)+z(3)+z(4)+z(5)+z(6)+z(7)+z(8)+z(9)+z(10)+z(11)+z(12)
+a0=sumf(j)/(2*k)
+a1=sumfjcos(t(j))/k
+b1=sumfjsin(t(j))/k
+a2=sumfjcos2(t(j))/k
+b2=sumfjsin2(t(j))/k
+a3=sumfjcos3(t(j))/k
+b3=sumfjsin3(t(j))/k
+disp('The fourier components of periodic motion shown in example 1.8.1 are as follows')
+mprintf('\nao=%f\na1=%f\nb1=%f\na2=%f\nb2=%f\na3=%f\nb3=%f\n',a0,a1,b1,a2,b2,a3,b3)
diff --git a/3532/CH1/EX1.3/Ex1_3.sce b/3532/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..3d7d67f0e --- /dev/null +++ b/3532/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.4.2\n')
+//given data
+//case 1
+//x1=(1/2)*cos((%pi/2)*t)...x1=a*cos(W1*t)
+//x2=sin(%pi*t)...x2=b*sin(W2*t)
+//calculations
+W1=(%pi/2)
+W2=%pi
+t1=2*%pi/(W1)
+t2=2*%pi/(W2)
+p1=[t1 t2]
+T1=lcm(p1)
+//case 2
+//x1=2*cos((%pi*t)...x1=a*cos(W3*t)
+//x2=2*cos(2*t)...x2=a*cos(W4*t)
+W3=%pi
+W4=2
+t3=2*%pi/(W3)
+t4=2*%pi/(W4)
+p2=[t3 t4]
+T2=lcm(p2)
+//output
+mprintf('Case(i)\nTime period of first wave is %f sec\nTime period of first wave is %f sec\nThe time period of combined wave is %f sec\nCase(ii)\nTime period of first wave is %f sec\nTime period of first wave is %f sec\nThe time period of combined wave is %f sec',t1,t2,T1,t3,t4,T2)
+mprintf('\nNOTE: The time period of combined motion in case (ii) cannot be calculated\n since pi is a non-terminating and non recurring number.\n But SCILAB takes the value of pi to be 3.141593 and therefore\n calculates the LCM of pi and the time period of first wave in case (ii.')
diff --git a/3532/CH1/EX1.4.2/Ex1_3.sce b/3532/CH1/EX1.4.2/Ex1_3.sce new file mode 100644 index 000000000..3d7d67f0e --- /dev/null +++ b/3532/CH1/EX1.4.2/Ex1_3.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.4.2\n')
+//given data
+//case 1
+//x1=(1/2)*cos((%pi/2)*t)...x1=a*cos(W1*t)
+//x2=sin(%pi*t)...x2=b*sin(W2*t)
+//calculations
+W1=(%pi/2)
+W2=%pi
+t1=2*%pi/(W1)
+t2=2*%pi/(W2)
+p1=[t1 t2]
+T1=lcm(p1)
+//case 2
+//x1=2*cos((%pi*t)...x1=a*cos(W3*t)
+//x2=2*cos(2*t)...x2=a*cos(W4*t)
+W3=%pi
+W4=2
+t3=2*%pi/(W3)
+t4=2*%pi/(W4)
+p2=[t3 t4]
+T2=lcm(p2)
+//output
+mprintf('Case(i)\nTime period of first wave is %f sec\nTime period of first wave is %f sec\nThe time period of combined wave is %f sec\nCase(ii)\nTime period of first wave is %f sec\nTime period of first wave is %f sec\nThe time period of combined wave is %f sec',t1,t2,T1,t3,t4,T2)
+mprintf('\nNOTE: The time period of combined motion in case (ii) cannot be calculated\n since pi is a non-terminating and non recurring number.\n But SCILAB takes the value of pi to be 3.141593 and therefore\n calculates the LCM of pi and the time period of first wave in case (ii.')
diff --git a/3532/CH1/EX1.4/Ex1_4.sce b/3532/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..a9383af80 --- /dev/null +++ b/3532/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.5.1\n')
+//given data
+//x1=a*sin(W1*t)
+//x2=b*sin(W2*t)
+//calculations
+a=1.90//amplitude of first wave in cm
+b=2.00//amplitude of second wave in cm
+W1=9.5//frequency of first wave in rad/sec
+W2=10.0//frequency of second wave in rad/sec
+xmax=b+a//maximum amplitude of motion in cms
+xmin=abs(a-b)//minimum amplitude of motion in cms
+f=abs(W1-W2)/(2*%pi)//beat frequency in Hz
+t=1/f//time period of beat in sec
+//output
+mprintf('The maximum amplitude of motion is %4.4f cms\nThe minimum amplitude of motion is %4.4f cms\n The beat frequency is %4.4f Hz\n the time period is %4.4f sec',xmax,xmin,f,t)
diff --git a/3532/CH1/EX1.5.1/Ex1_4.sce b/3532/CH1/EX1.5.1/Ex1_4.sce new file mode 100644 index 000000000..a9383af80 --- /dev/null +++ b/3532/CH1/EX1.5.1/Ex1_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.5.1\n')
+//given data
+//x1=a*sin(W1*t)
+//x2=b*sin(W2*t)
+//calculations
+a=1.90//amplitude of first wave in cm
+b=2.00//amplitude of second wave in cm
+W1=9.5//frequency of first wave in rad/sec
+W2=10.0//frequency of second wave in rad/sec
+xmax=b+a//maximum amplitude of motion in cms
+xmin=abs(a-b)//minimum amplitude of motion in cms
+f=abs(W1-W2)/(2*%pi)//beat frequency in Hz
+t=1/f//time period of beat in sec
+//output
+mprintf('The maximum amplitude of motion is %4.4f cms\nThe minimum amplitude of motion is %4.4f cms\n The beat frequency is %4.4f Hz\n the time period is %4.4f sec',xmax,xmin,f,t)
diff --git a/3532/CH1/EX1.6.1/Ex1_6.sce b/3532/CH1/EX1.6.1/Ex1_6.sce new file mode 100644 index 000000000..6d4966f45 --- /dev/null +++ b/3532/CH1/EX1.6.1/Ex1_6.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.6.1\n')
+//given data
+//case 1
+//a complex number is represented as Z=X+j*Y where j is imaginary
+//V=3 +j*7
+x1=3
+y1=7
+//calculations
+r1=sqrt(x1^2+y1^2)
+ if (y1/x1)>0 then theta1=atan(y1/x1)
+ else theta1=%pi-atan(abs(y1/x1))
+ end
+theta1=atan(y1/x1)
+//case 2
+//V=-5 +j*4
+x2=-5
+y2=4
+//calculations
+r2=sqrt(x2^2+y2^2)
+ if (y2/x2)>0 then theta1=atan(y2/x2)
+ else theta2=%pi-atan(abs(y2/x2))
+ end
+//output
+mprintf('case(i) V=3+j*7 is represented as V=%3.3f*e^(j*(%3.3f))\ncase(ii) V=-5+j*4 is represented as V=%3.3f*e^(j*(%3.3f))',r1,theta1,r2,theta2)
diff --git a/3532/CH1/EX1.6.2/Ex1_7.sce b/3532/CH1/EX1.6.2/Ex1_7.sce new file mode 100644 index 000000000..fe16f356e --- /dev/null +++ b/3532/CH1/EX1.6.2/Ex1_7.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.6.2\n')
+//given data
+//Z=r*e^(i*theta) is represented as Z=r*cos(theta) + i*r*sin(theta)= x +i*y
+//where r*cos(theta)=x and r*sin(theta)=y
+//case 1
+//V=5*e^(j*0.10)
+r1=5
+theta1=0.1
+x1=r1*cos(theta1)
+y1=r1*sin(theta1)
+v1=complex(x1,y1)
+//case 2
+//V=17*e^(-j*3.74)
+r2=17
+theta2=-3.74
+x2=r2*cos(theta2)
+y2=r2*sin(theta2)
+v2=complex(x2,y2)
+//output
+mprintf('case(i):V=5*e^(j*0.10) is represented as')
+disp(v1)
+mprintf('\ncase(ii):V=17*e^(-j*3.74) is represented as')
+disp(v2)
+mprintf('\nNOTE:complex number is represented as x+y*i in SCILAB')
diff --git a/3532/CH1/EX1.6/Ex1_6.sce b/3532/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..6d4966f45 --- /dev/null +++ b/3532/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.6.1\n')
+//given data
+//case 1
+//a complex number is represented as Z=X+j*Y where j is imaginary
+//V=3 +j*7
+x1=3
+y1=7
+//calculations
+r1=sqrt(x1^2+y1^2)
+ if (y1/x1)>0 then theta1=atan(y1/x1)
+ else theta1=%pi-atan(abs(y1/x1))
+ end
+theta1=atan(y1/x1)
+//case 2
+//V=-5 +j*4
+x2=-5
+y2=4
+//calculations
+r2=sqrt(x2^2+y2^2)
+ if (y2/x2)>0 then theta1=atan(y2/x2)
+ else theta2=%pi-atan(abs(y2/x2))
+ end
+//output
+mprintf('case(i) V=3+j*7 is represented as V=%3.3f*e^(j*(%3.3f))\ncase(ii) V=-5+j*4 is represented as V=%3.3f*e^(j*(%3.3f))',r1,theta1,r2,theta2)
diff --git a/3532/CH1/EX1.7.1/Ex1_8.sce b/3532/CH1/EX1.7.1/Ex1_8.sce new file mode 100644 index 000000000..5d4c7f118 --- /dev/null +++ b/3532/CH1/EX1.7.1/Ex1_8.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.7.1\n')
+//given data
+Po=25//amplitude of force in N
+Xo=0.05//ampliyude of displacement in m
+W=20*%pi
+//calculations
+//case 1
+t0=0
+t1=1
+v1=integrate('sin(W*t)*cos(W*t-%pi/6)','t',t0,t1)
+WD1=Po*Xo*W*v1
+//case 2
+t0=0
+t1=1/40
+v2=integrate('sin(W*t)*cos(W*t-%pi/6)','t',t0,t1)
+WD2=Po*Xo*W*v2
+//output
+mprintf(' (i)work done during the first second is %f N-m\n (ii)work done during the first 1/40th of second is %f N-m',WD1,WD2)
diff --git a/3532/CH1/EX1.7/Ex1_7.sce b/3532/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..fe16f356e --- /dev/null +++ b/3532/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.6.2\n')
+//given data
+//Z=r*e^(i*theta) is represented as Z=r*cos(theta) + i*r*sin(theta)= x +i*y
+//where r*cos(theta)=x and r*sin(theta)=y
+//case 1
+//V=5*e^(j*0.10)
+r1=5
+theta1=0.1
+x1=r1*cos(theta1)
+y1=r1*sin(theta1)
+v1=complex(x1,y1)
+//case 2
+//V=17*e^(-j*3.74)
+r2=17
+theta2=-3.74
+x2=r2*cos(theta2)
+y2=r2*sin(theta2)
+v2=complex(x2,y2)
+//output
+mprintf('case(i):V=5*e^(j*0.10) is represented as')
+disp(v1)
+mprintf('\ncase(ii):V=17*e^(-j*3.74) is represented as')
+disp(v2)
+mprintf('\nNOTE:complex number is represented as x+y*i in SCILAB')
diff --git a/3532/CH1/EX1.8.2/Ex1_10.sce b/3532/CH1/EX1.8.2/Ex1_10.sce new file mode 100644 index 000000000..3c869fd4c --- /dev/null +++ b/3532/CH1/EX1.8.2/Ex1_10.sce @@ -0,0 +1,93 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.8.2\n')
+//given data
+T=0.1//time period of periodic motion in sec
+W=2*%pi/T
+k=12/2//number of elements in half cycle
+mprintf('\tNo of elements in one cycle 2k=12,t(j) in degrees\n')
+mprintf('t(j) f(j) cos(t(j)) f(j)*cos(t(j)) sin(t(j)) f(j)*sin(t(j) cos(2*t(j)) f(j)*cos(2*t(j)) sin(2*t(j)) f(j)*sin(2*t(j)) cos(3*t(j)) f(j)*cos(3*t(j) sin(3*t(j)) f(j)*sin(3*t(j)\n')
+f(1)=10/6
+for j=1:6
+ t(j)=j*(%pi/k)
+ m(j)=cos(t(j))
+ n(j)=f(j)*m(j)
+ o(j)=sin(t(j))
+ p(j)=f(j)*o(j)
+ q(j)=cos(2*t(j))
+ r(j)=f(j)*q(j)
+ s(j)=sin(2*t(j))
+ u(j)=f(j)*s(j)
+ v(j)=cos(3*t(j))
+ x(j)=f(j)*v(j)
+ y(j)=sin(3*t(j))
+ z(j)=f(j)*y(j)
+ f(j+1)=f(j)+f(1)
+mprintf('%3.0f\t',t(j)*(180/%pi))
+mprintf('%3.4f\t\t',f(j))
+mprintf('%3.4f\t\t',m(j))
+mprintf('%3.4f\t\t',n(j))
+mprintf('%3.4f\t\t',o(j))
+mprintf('%3.4f\t\t',p(j))
+mprintf('%3.4f\t\t',q(j))
+mprintf('%3.4f\t\t',r(j))
+mprintf('%3.4f\t\t',s(j))
+mprintf('%3.4f\t\t',u(j))
+mprintf('%3.4f\t\t',v(j))
+mprintf('%3.4f\t\t',x(j))
+mprintf('%3.4f\t\t',y(j))
+mprintf('%3.4f\n',z(j))
+end
+f(7)=f(j)-f(1)
+for j=7:12
+ t(j)=j*(%pi/k)
+ m(j)=cos(t(j))
+ n(j)=f(j)*m(j)
+ o(j)=sin(t(j))
+ p(j)=f(j)*o(j)
+ q(j)=cos(2*t(j))
+ r(j)=f(j)*q(j)
+ s(j)=sin(2*t(j))
+ u(j)=f(j)*s(j)
+ v(j)=cos(3*t(j))
+ x(j)=f(j)*v(j)
+ y(j)=sin(3*t(j))
+ z(j)=f(j)*y(j)
+ f(j+1)=f(j)-f(1)
+ mprintf('%3.0f\t',t(j)*(180/%pi))
+mprintf('%3.4f\t\t',f(j))
+mprintf('%3.4f\t\t',m(j))
+mprintf('%3.4f\t\t',n(j))
+mprintf('%3.4f\t\t',o(j))
+mprintf('%3.4f\t\t',p(j))
+mprintf('%3.4f\t\t',q(j))
+mprintf('%3.4f\t\t',r(j))
+mprintf('%3.4f\t\t',s(j))
+mprintf('%3.4f\t\t',u(j))
+mprintf('%3.4f\t\t',v(j))
+mprintf('%3.4f\t\t',x(j))
+mprintf('%3.4f\t\t',y(j))
+mprintf('%3.4f\n',z(j))
+end
+sumf(j)=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)
+sumcos(t(j))=m(1)+m(2)+m(3)+m(4)+m(5)+m(6)+m(7)+m(8)+m(9)+m(10)+m(11)+m(12)
+sumfjcos(t(j))=n(1)+n(2)+n(3)+n(4)+n(5)+n(6)+n(7)+n(8)+n(9)+n(10)+n(11)+n(12)
+sumsin(t(j))=o(1)+o(2)+o(3)+o(4)+o(5)+o(6)+o(7)+o(8)+o(9)+o(10)+o(11)+o(12)
+sumfjsin(t(j))=p(1)+p(2)+p(3)+p(4)+p(5)+p(6)+p(7)+p(8)+p(9)+p(10)+p(11)+p(12)
+sumcos2(t(j))=q(1)+q(2)+q(3)+q(4)+q(5)+q(6)+q(7)+q(8)+q(9)+q(10)+q(11)+q(12)
+sumfjcos2(t(j))=r(1)+r(2)+r(3)+r(4)+r(5)+r(6)+r(7)+r(8)+r(9)+r(10)+r(11)+r(12)
+sumsin2(t(j))=s(1)+s(2)+s(3)+s(4)+s(5)+s(6)+s(7)+s(8)+s(9)+s(10)+s(11)+s(12)
+sumfjsin2(t(j))=u(1)+u(2)+u(3)+u(4)+u(5)+u(6)+u(7)+u(8)+u(9)+u(10)+u(11)+u(12)
+sumcos3(t(j))=v(1)+v(2)+v(3)+v(4)+v(5)+v(6)+v(7)+v(8)+v(9)+v(10)+v(11)+v(12)
+sumfjcos3(t(j))=x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)+x(12)
+sumsin3(t(j))=y(1)+y(2)+y(3)+y(4)+y(5)+y(6)+y(7)+y(8)+y(9)+y(10)+y(11)+y(12)
+sumfjsin3(t(j))=z(1)+z(2)+z(3)+z(4)+z(5)+z(6)+z(7)+z(8)+z(9)+z(10)+z(11)+z(12)
+a0=sumf(j)/(2*k)
+a1=sumfjcos(t(j))/k
+b1=sumfjsin(t(j))/k
+a2=sumfjcos2(t(j))/k
+b2=sumfjsin2(t(j))/k
+a3=sumfjcos3(t(j))/k
+b3=sumfjsin3(t(j))/k
+disp('The fourier components of periodic motion shown in example 1.8.1 are as follows')
+mprintf('\nao=%f\na1=%f\nb1=%f\na2=%f\nb2=%f\na3=%f\nb3=%f\n',a0,a1,b1,a2,b2,a3,b3)
diff --git a/3532/CH1/EX1.8/Ex1_8.sce b/3532/CH1/EX1.8/Ex1_8.sce new file mode 100644 index 000000000..5d4c7f118 --- /dev/null +++ b/3532/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 1.7.1\n')
+//given data
+Po=25//amplitude of force in N
+Xo=0.05//ampliyude of displacement in m
+W=20*%pi
+//calculations
+//case 1
+t0=0
+t1=1
+v1=integrate('sin(W*t)*cos(W*t-%pi/6)','t',t0,t1)
+WD1=Po*Xo*W*v1
+//case 2
+t0=0
+t1=1/40
+v2=integrate('sin(W*t)*cos(W*t-%pi/6)','t',t0,t1)
+WD2=Po*Xo*W*v2
+//output
+mprintf(' (i)work done during the first second is %f N-m\n (ii)work done during the first 1/40th of second is %f N-m',WD1,WD2)
diff --git a/3532/CH2/EX2.10/Ex2_10.sce b/3532/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..26547987e --- /dev/null +++ b/3532/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.5.2\n')
+//given data
+G=0.83*10^11//rigidity modulus in N/m^2
+J=14.7 //mass moment of inertia in kg-m^2
+l1=0.6 //lenght of section 1 in m
+l2=1.8 //lenght of section 2 in m
+l3=0.25 //lenght of section 3 in m
+d1=0.05 //dia of section 1 in m
+d2=0.08 //dia of section 2 in m
+d3=0.03 //dia of section 3 in m
+//calculations
+Kt1=(G/l1)*(%pi/32)*d1^4 //(%pi/32)*d^4 is the section modulus
+Kt2=(G/l2)*(%pi/32)*d2^4
+Kt3=(G/l3)*(%pi/32)*d3^4
+Kt=1/((1/Kt1)+(1/Kt2)+(1/Kt3)) //total effective stiffness of the torsional system
+Wn=sqrt(Kt/J)//natural freq in rad/sec
+fn=Wn/(2*%pi) //natural freq in Hz
+//output
+mprintf(' The natural frequency of torsional oscillation for the given system is\n %4.4f rad/sec or %4.4f Hz.',Wn,fn)
+mprintf('\nNOTE:Since the value of Kt in the textbook has been rounded of\n to 3 decimal places,the final answer varies slightly.')
diff --git a/3532/CH2/EX2.3.4/Ex2_6.sce b/3532/CH2/EX2.3.4/Ex2_6.sce new file mode 100644 index 000000000..c58b1adb1 --- /dev/null +++ b/3532/CH2/EX2.3.4/Ex2_6.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.3.4\n')
+//given data
+M=35//mass of flywheel in Kgs
+r=0.3/2 //distance of centre of mass from pivot in m
+T=1.22 //time period of oscillation in sec
+g=9.81//accelaration due to gravity in m/(sec^2)
+//concept is as follows
+//Jo=mass moment of inertia about pivot, Wn=natural freqency
+//thetadd=theta double dot(double differentiation)
+//Jo*thetadd=-M*g*r*theta ....sum of moments is = to zero
+//Jo*thetadd +(M*g*r*theta)=0
+//Wn=sqrt((M*g*r*)/Jo)=2*pi/T
+//calculations
+Jo=M*g*r/((2*%pi/T)^2)
+Jg=Jo-M*r^2 //mass moment of inertia about geometric axis
+//output
+mprintf('Mass moment of inertia about pivot is %4.4f Kg-m^2\n Mass moment of inertia about geometric axis is %4.4f Kg-m^2',Jo,Jg)
diff --git a/3532/CH2/EX2.4.1/Ex2_8.sce b/3532/CH2/EX2.4.1/Ex2_8.sce new file mode 100644 index 000000000..d05c64876 --- /dev/null +++ b/3532/CH2/EX2.4.1/Ex2_8.sce @@ -0,0 +1,18 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.4.1\n')
+//given data
+l=1 //lenght in m
+d=0.005 //dia of rod im m
+D=0.2 //dia of dotor in m
+M=2 //mass of motor in Kg
+G=0.83 *10^11 //modulus of rigidity in N/m^2
+//calculations
+J=M*((D/2)^2)/2 //mass moment of inertia in Kg-m^2
+Ip=(%pi/32)*d^4 //section modulus in m^4
+Kt=G*Ip/l //stiffness in N-m/rad
+Wn=sqrt(Kt/J) //natural freqency in rad/sec
+fn=Wn/(2*%pi) //natural freq in Hz
+//output
+mprintf(' The natural freqency of vibration of torsional pendulum is %4.4f rad/sec\n or %4.4f Hz',Wn,fn)
+mprintf('\nNOTE:In book the natural freqency of vibration of torsional pendulum\nis given as 36 Hz which is wrong.')
diff --git a/3532/CH2/EX2.5.1/Ex2_9.sce b/3532/CH2/EX2.5.1/Ex2_9.sce new file mode 100644 index 000000000..ffd2b201b --- /dev/null +++ b/3532/CH2/EX2.5.1/Ex2_9.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.5.1\n')
+//given data
+k1=2000 //stiffness of spring 1 in N/m
+k2=1500 //stiffness of spring 2 in N/m
+k3=3000 //stiffness of spring 3 in N/m
+k4=500 //stiffness of spring 4 in N/m
+k5=500 //stiffness of spring 5 in N/m
+fn =10 //natural frequency of system in Hz
+//calculations
+Ke1=1/((1/k1)+(1/k2)+(1/k3)) // effective stiffness of top 3 springs in series in N/m
+Ke2=k4+k5 // effective stiffness of lower 2 springs in parallel in N/m
+Ke=Ke1+Ke2 // total effective stiffness of sring system
+M=Ke/(2*%pi*fn)^2 //reqired mass such that the natural frequency of system is 10 Hz (in Kg)
+//output
+mprintf(' The mass required such that the natural frequency of system is 10 Hz\n is %4.4f Kg',M)
diff --git a/3532/CH2/EX2.5.2/Ex2_10.sce b/3532/CH2/EX2.5.2/Ex2_10.sce new file mode 100644 index 000000000..26547987e --- /dev/null +++ b/3532/CH2/EX2.5.2/Ex2_10.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.5.2\n')
+//given data
+G=0.83*10^11//rigidity modulus in N/m^2
+J=14.7 //mass moment of inertia in kg-m^2
+l1=0.6 //lenght of section 1 in m
+l2=1.8 //lenght of section 2 in m
+l3=0.25 //lenght of section 3 in m
+d1=0.05 //dia of section 1 in m
+d2=0.08 //dia of section 2 in m
+d3=0.03 //dia of section 3 in m
+//calculations
+Kt1=(G/l1)*(%pi/32)*d1^4 //(%pi/32)*d^4 is the section modulus
+Kt2=(G/l2)*(%pi/32)*d2^4
+Kt3=(G/l3)*(%pi/32)*d3^4
+Kt=1/((1/Kt1)+(1/Kt2)+(1/Kt3)) //total effective stiffness of the torsional system
+Wn=sqrt(Kt/J)//natural freq in rad/sec
+fn=Wn/(2*%pi) //natural freq in Hz
+//output
+mprintf(' The natural frequency of torsional oscillation for the given system is\n %4.4f rad/sec or %4.4f Hz.',Wn,fn)
+mprintf('\nNOTE:Since the value of Kt in the textbook has been rounded of\n to 3 decimal places,the final answer varies slightly.')
diff --git a/3532/CH2/EX2.6/Ex2_6.sce b/3532/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..c58b1adb1 --- /dev/null +++ b/3532/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.3.4\n')
+//given data
+M=35//mass of flywheel in Kgs
+r=0.3/2 //distance of centre of mass from pivot in m
+T=1.22 //time period of oscillation in sec
+g=9.81//accelaration due to gravity in m/(sec^2)
+//concept is as follows
+//Jo=mass moment of inertia about pivot, Wn=natural freqency
+//thetadd=theta double dot(double differentiation)
+//Jo*thetadd=-M*g*r*theta ....sum of moments is = to zero
+//Jo*thetadd +(M*g*r*theta)=0
+//Wn=sqrt((M*g*r*)/Jo)=2*pi/T
+//calculations
+Jo=M*g*r/((2*%pi/T)^2)
+Jg=Jo-M*r^2 //mass moment of inertia about geometric axis
+//output
+mprintf('Mass moment of inertia about pivot is %4.4f Kg-m^2\n Mass moment of inertia about geometric axis is %4.4f Kg-m^2',Jo,Jg)
diff --git a/3532/CH2/EX2.8/Ex2_8.sce b/3532/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..d05c64876 --- /dev/null +++ b/3532/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,18 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.4.1\n')
+//given data
+l=1 //lenght in m
+d=0.005 //dia of rod im m
+D=0.2 //dia of dotor in m
+M=2 //mass of motor in Kg
+G=0.83 *10^11 //modulus of rigidity in N/m^2
+//calculations
+J=M*((D/2)^2)/2 //mass moment of inertia in Kg-m^2
+Ip=(%pi/32)*d^4 //section modulus in m^4
+Kt=G*Ip/l //stiffness in N-m/rad
+Wn=sqrt(Kt/J) //natural freqency in rad/sec
+fn=Wn/(2*%pi) //natural freq in Hz
+//output
+mprintf(' The natural freqency of vibration of torsional pendulum is %4.4f rad/sec\n or %4.4f Hz',Wn,fn)
+mprintf('\nNOTE:In book the natural freqency of vibration of torsional pendulum\nis given as 36 Hz which is wrong.')
diff --git a/3532/CH2/EX2.9/Ex2_9.sce b/3532/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..ffd2b201b --- /dev/null +++ b/3532/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 2.5.1\n')
+//given data
+k1=2000 //stiffness of spring 1 in N/m
+k2=1500 //stiffness of spring 2 in N/m
+k3=3000 //stiffness of spring 3 in N/m
+k4=500 //stiffness of spring 4 in N/m
+k5=500 //stiffness of spring 5 in N/m
+fn =10 //natural frequency of system in Hz
+//calculations
+Ke1=1/((1/k1)+(1/k2)+(1/k3)) // effective stiffness of top 3 springs in series in N/m
+Ke2=k4+k5 // effective stiffness of lower 2 springs in parallel in N/m
+Ke=Ke1+Ke2 // total effective stiffness of sring system
+M=Ke/(2*%pi*fn)^2 //reqired mass such that the natural frequency of system is 10 Hz (in Kg)
+//output
+mprintf(' The mass required such that the natural frequency of system is 10 Hz\n is %4.4f Kg',M)
diff --git a/3532/CH3/EX3.2/Ex3_2.sce b/3532/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..b009c42d4 --- /dev/null +++ b/3532/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.3.2\n')
+//given data
+m=10 //mass of solid in Kg
+Kr=3000 //stiffness of natural rubber in N/m
+Kf=12000 //stiffness of felt in N/m
+Cr=100 //damping coefficient of natural rubber in N-sec/m
+Cf=330 //damping coefficient of felt in N-sec/m
+//calculations
+Ke=1/((1/Kf)+(1/Kr)) //equivalent stiffness in N/m
+Ce=1/((1/Cf)+(1/Cr)) //equivalent damping coefficient N-sec/m
+Wn=sqrt(Ke/m) // undamped natural freq in rad/sec
+fn=Wn/(2*%pi) // undamped natural freq in Hz
+zeta=Ce/(2*sqrt(Ke*m)) //damping factor
+Wd=sqrt(1-zeta^2)*Wn //damped natural freuency in rad/sec(eqn 3.3.16)
+fd=Wd/(2*%pi) // damped natural frequency in Hz
+//output
+mprintf(' The undamped natural frequency is %4.4f rad/sec or %4.4f Hz\n The damped natural freuency is %4.4f rad/sec or %4.4f Hz',Wn,fn,Wd,fd)
diff --git a/3532/CH3/EX3.3.2/Ex3_2.sce b/3532/CH3/EX3.3.2/Ex3_2.sce new file mode 100644 index 000000000..b009c42d4 --- /dev/null +++ b/3532/CH3/EX3.3.2/Ex3_2.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.3.2\n')
+//given data
+m=10 //mass of solid in Kg
+Kr=3000 //stiffness of natural rubber in N/m
+Kf=12000 //stiffness of felt in N/m
+Cr=100 //damping coefficient of natural rubber in N-sec/m
+Cf=330 //damping coefficient of felt in N-sec/m
+//calculations
+Ke=1/((1/Kf)+(1/Kr)) //equivalent stiffness in N/m
+Ce=1/((1/Cf)+(1/Cr)) //equivalent damping coefficient N-sec/m
+Wn=sqrt(Ke/m) // undamped natural freq in rad/sec
+fn=Wn/(2*%pi) // undamped natural freq in Hz
+zeta=Ce/(2*sqrt(Ke*m)) //damping factor
+Wd=sqrt(1-zeta^2)*Wn //damped natural freuency in rad/sec(eqn 3.3.16)
+fd=Wd/(2*%pi) // damped natural frequency in Hz
+//output
+mprintf(' The undamped natural frequency is %4.4f rad/sec or %4.4f Hz\n The damped natural freuency is %4.4f rad/sec or %4.4f Hz',Wn,fn,Wd,fd)
diff --git a/3532/CH3/EX3.3.3/Ex3_3.sce b/3532/CH3/EX3.3.3/Ex3_3.sce new file mode 100644 index 000000000..0c30824a7 --- /dev/null +++ b/3532/CH3/EX3.3.3/Ex3_3.sce @@ -0,0 +1,26 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.3.3\n')
+//given data
+m=600//mass of gun barrel in Kgs
+k=294000//stiffness in N/m
+x=1.3//recoil of gun in meters
+//calculations
+E=0.5*k*x^2//energy stored at the end of recoil
+Vo=sqrt(2*E/m)//velocity of recoil
+Cc=2*sqrt(k*m)//critical damping in N-sec/m
+Wn=sqrt(k/m)//natural frequency of undamped vibration in rad/sec
+T=2*%pi/Wn//time period of undamped vibration in sec
+Trecoil=(1/4)*T//time period for recoil or outward stroke in sec
+//x=(1.3+28.8*t)*e^(-22.1*t) from eqn 3.3.24
+mprintf('a)the initial recoil velocity of barrel is %f m/s\nb)critical damping co-efficient of the dashpot which is engaged at\nthe end of recoil stroke is %f N-sec/m\n\nsubstituting the value for t in eqn 3.3.24,starting from t=0.1 sec\nwith an increment of 0.01sec we get the following observations\n',Vo,Cc)
+t=0.1
+for i=1:20
+ x=(1.3 +28.8*t)*exp(-22.1*t)
+ mprintf('x=%f at t=%f\n',x,t)
+ t=t+0.01
+end
+mprintf('As x approaches the value of 0.05m,the value of t=0.22sec')
+Trec=0.22
+Tret=Trecoil+Trec
+mprintf('\nc)Therefore time required for barrel to return to position 5cm from\n the initial position is %f sec',Tret)
diff --git a/3532/CH3/EX3.4.1/Ex3_5.sce b/3532/CH3/EX3.4.1/Ex3_5.sce new file mode 100644 index 000000000..f5f515bd1 --- /dev/null +++ b/3532/CH3/EX3.4.1/Ex3_5.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.4.1\n')
+//given data
+J=0.06 //moment of inertia of disc of pendulum in Kg-m^2
+G=4.4*10^10 //rigidity modulus in N/m^2
+l=0.4 //lenght of shaft in m
+d=0.1 //diametre of shaft in m
+a1=9 //amplitude of first oscillation in degrees
+a2=6 //amplitude of second oscillation in degrees
+a3=4 //amplitude of third oscillation in degrees
+//calculations
+delta=log(a1/a2) //logarithmic decrement eqn 3.4.1 explained in sec 3.4
+zeta=delta/sqrt(4*%pi^2+delta^2) //representing zeta from eqn 3.4.1 in sec 3.4
+Kt=(G/l)*(%pi/32)*d^4 //(%pi/32)*d^4 is the section modulus
+C=zeta*2*sqrt(Kt*J) // torsional damping coefficient which is the damping torque at unit velocity (similar to eqn 3.3.6 in sec 3.3)
+Wn=sqrt(Kt/J) // undamped natural freq in rad/sec
+T=2*%pi/(sqrt(1-zeta^2)*Wn) //periodic time of vibration
+fn=Wn/(2*%pi) //natural freq of undamped vibration
+//output
+mprintf(' a)logarithmic decrement is %4.4f\n b)damping torque at unit velocity is %4.4f N-m/rad\n c)periodic time of vibration is %4.5f sec\n frequency of vibration if the disc is removed from viscous fluid is %4.4f Hz',delta,C,T,fn)
diff --git a/3532/CH3/EX3.5/Ex3_5.sce b/3532/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..f5f515bd1 --- /dev/null +++ b/3532/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.4.1\n')
+//given data
+J=0.06 //moment of inertia of disc of pendulum in Kg-m^2
+G=4.4*10^10 //rigidity modulus in N/m^2
+l=0.4 //lenght of shaft in m
+d=0.1 //diametre of shaft in m
+a1=9 //amplitude of first oscillation in degrees
+a2=6 //amplitude of second oscillation in degrees
+a3=4 //amplitude of third oscillation in degrees
+//calculations
+delta=log(a1/a2) //logarithmic decrement eqn 3.4.1 explained in sec 3.4
+zeta=delta/sqrt(4*%pi^2+delta^2) //representing zeta from eqn 3.4.1 in sec 3.4
+Kt=(G/l)*(%pi/32)*d^4 //(%pi/32)*d^4 is the section modulus
+C=zeta*2*sqrt(Kt*J) // torsional damping coefficient which is the damping torque at unit velocity (similar to eqn 3.3.6 in sec 3.3)
+Wn=sqrt(Kt/J) // undamped natural freq in rad/sec
+T=2*%pi/(sqrt(1-zeta^2)*Wn) //periodic time of vibration
+fn=Wn/(2*%pi) //natural freq of undamped vibration
+//output
+mprintf(' a)logarithmic decrement is %4.4f\n b)damping torque at unit velocity is %4.4f N-m/rad\n c)periodic time of vibration is %4.5f sec\n frequency of vibration if the disc is removed from viscous fluid is %4.4f Hz',delta,C,T,fn)
diff --git a/3532/CH3/EX3.6.1/Ex3_6.sce b/3532/CH3/EX3.6.1/Ex3_6.sce new file mode 100644 index 000000000..9fedb6e9c --- /dev/null +++ b/3532/CH3/EX3.6.1/Ex3_6.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.6.1\n')
+//given data
+m=5 //mass in spring mass system )in kg)
+k=980//stiffnes of spring in N/m
+u=0.025//coefficient of friction
+g=9.81//acceleration due to gravity
+//calculations
+F=u*m*g//frictional force in N
+Wn=sqrt(k/m)// freq of free oscillations in rad/sec
+fn=Wn/(2*%pi)// freq of free oscillations in Hz
+Ai=0.05//initial amplitude in m
+Ar=0.5*Ai//reduced amplitude in m
+totAreduc=Ai-Ar//total reduction in amp in m
+Areducpercycl=4*F/k //reduction in amplitude/cycle explained in section 3.6.2 in eqn 3.6.6
+n=round(totAreduc/Areducpercycl) //number of cycles for 50% reduction in amplitude
+Treduc=n*(2*%pi/Wn)//time taken to achieve 50%reduction
+//output
+mprintf(' a)The frequency of free oscillations is %4.4f rad/sec or %4.4f Hz\n b)number of cycles taken for 50 percent reduction in amplitude is %1.0f cycles\n c)time taken to achieve 50 percent reduction in amplitude is %4.4f sec',Wn,fn,n,Treduc)
diff --git a/3532/CH3/EX3.6.2/Ex3_7.sce b/3532/CH3/EX3.6.2/Ex3_7.sce new file mode 100644 index 000000000..4d3b71f4a --- /dev/null +++ b/3532/CH3/EX3.6.2/Ex3_7.sce @@ -0,0 +1,31 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.6.2\n')
+//given data
+k=9800//stiffnes of spring in N/m
+m=40 //mass in spring mass system )in kg)
+g=9.81//acceleration due to gravity
+F=49//frictional force in N
+x=0.126//total extension of spring in m
+xeq=m*g/k//extension of spring at equillibrium in m
+xi=x-xeq//initial extension of spring from equillibrium in m
+Alosspercycl=4*F/k//reduction in amplitude/cycle explained in section 3.6.2 in eqn 3.6.6
+n=int(xi/Alosspercycl)//number of complete cycles that system undergoes
+Af=xi-n*Alosspercycl//amplitude at the end of n cycles
+SF=k*Af//spring force acting on the upward direction for an extension of Af
+if F<SF then
+ disp('The spring will move up since spring force is greater than frictional force')
+ Xa=Af//assigning Af to a new variable Xa
+ Xb=0//assume Xb=0 at first
+ //solving the quadratic equation in Xb whose roots are Xb1 and Xb2
+ Xb1=(F+sqrt((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F*Xa))))/k
+ Xb2=(F-sqrt((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F*Xa))))/k
+ if int(Xb1-Xa)==0 then
+ Xb=Xb2
+ else
+ Xb=Xb1
+ end
+ finalext=xeq+Xb
+ mprintf(' The final extention of spring is %f m',finalext)
+ else disp('The spring will not move up since spring force is not greater than frictional force')
+end
diff --git a/3532/CH3/EX3.6/Ex3_6.sce b/3532/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..9fedb6e9c --- /dev/null +++ b/3532/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.6.1\n')
+//given data
+m=5 //mass in spring mass system )in kg)
+k=980//stiffnes of spring in N/m
+u=0.025//coefficient of friction
+g=9.81//acceleration due to gravity
+//calculations
+F=u*m*g//frictional force in N
+Wn=sqrt(k/m)// freq of free oscillations in rad/sec
+fn=Wn/(2*%pi)// freq of free oscillations in Hz
+Ai=0.05//initial amplitude in m
+Ar=0.5*Ai//reduced amplitude in m
+totAreduc=Ai-Ar//total reduction in amp in m
+Areducpercycl=4*F/k //reduction in amplitude/cycle explained in section 3.6.2 in eqn 3.6.6
+n=round(totAreduc/Areducpercycl) //number of cycles for 50% reduction in amplitude
+Treduc=n*(2*%pi/Wn)//time taken to achieve 50%reduction
+//output
+mprintf(' a)The frequency of free oscillations is %4.4f rad/sec or %4.4f Hz\n b)number of cycles taken for 50 percent reduction in amplitude is %1.0f cycles\n c)time taken to achieve 50 percent reduction in amplitude is %4.4f sec',Wn,fn,n,Treduc)
diff --git a/3532/CH3/EX3.7.1/Ex3_8.sce b/3532/CH3/EX3.7.1/Ex3_8.sce new file mode 100644 index 000000000..55624fded --- /dev/null +++ b/3532/CH3/EX3.7.1/Ex3_8.sce @@ -0,0 +1,11 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.7.1\n')
+//given data
+fnA=12 //frequency of free vibrations of system A in Hz
+fnB=15 //frequency of free vibrations of system B in Hz
+TdA=4.5 //time taken by system A to damp out completely in sec
+//calculations
+TdB=fnA*TdA/fnB //time taken by system B to damp out completely in sec
+//output
+mprintf(' The time taken by system B to damp out completely is %4.4f sec',TdB)
diff --git a/3532/CH3/EX3.7/Ex3_7.sce b/3532/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..4d3b71f4a --- /dev/null +++ b/3532/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,31 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.6.2\n')
+//given data
+k=9800//stiffnes of spring in N/m
+m=40 //mass in spring mass system )in kg)
+g=9.81//acceleration due to gravity
+F=49//frictional force in N
+x=0.126//total extension of spring in m
+xeq=m*g/k//extension of spring at equillibrium in m
+xi=x-xeq//initial extension of spring from equillibrium in m
+Alosspercycl=4*F/k//reduction in amplitude/cycle explained in section 3.6.2 in eqn 3.6.6
+n=int(xi/Alosspercycl)//number of complete cycles that system undergoes
+Af=xi-n*Alosspercycl//amplitude at the end of n cycles
+SF=k*Af//spring force acting on the upward direction for an extension of Af
+if F<SF then
+ disp('The spring will move up since spring force is greater than frictional force')
+ Xa=Af//assigning Af to a new variable Xa
+ Xb=0//assume Xb=0 at first
+ //solving the quadratic equation in Xb whose roots are Xb1 and Xb2
+ Xb1=(F+sqrt((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F*Xa))))/k
+ Xb2=(F-sqrt((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F*Xa))))/k
+ if int(Xb1-Xa)==0 then
+ Xb=Xb2
+ else
+ Xb=Xb1
+ end
+ finalext=xeq+Xb
+ mprintf(' The final extention of spring is %f m',finalext)
+ else disp('The spring will not move up since spring force is not greater than frictional force')
+end
diff --git a/3532/CH3/EX3.8/Ex3_8.sce b/3532/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..55624fded --- /dev/null +++ b/3532/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,11 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 3.7.1\n')
+//given data
+fnA=12 //frequency of free vibrations of system A in Hz
+fnB=15 //frequency of free vibrations of system B in Hz
+TdA=4.5 //time taken by system A to damp out completely in sec
+//calculations
+TdB=fnA*TdA/fnB //time taken by system B to damp out completely in sec
+//output
+mprintf(' The time taken by system B to damp out completely is %4.4f sec',TdB)
diff --git a/3532/CH4/EX4.1/Ex4_1.sce b/3532/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..8dfd86913 --- /dev/null +++ b/3532/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.2.1\n')
+//given data
+//T=To*sin(W*t)
+To=0.588 //maximum value of periodic torque in N-m
+W=4// freqency of applied force in rad/sec
+J=0.12//moment of inertia of wheel in kg-m^2
+Kt=1.176//stiffness of wire in N-m/rad
+Ct=0.392/1 //damping coefficient in N-m_sec/rad
+//calculations
+theta=To/sqrt((Kt-J*W^2)^2+(Ct*W)^2)//Equation for torsional vibration amplitude from Fig (4.2.2) and Eqn (4.2.5)
+MaxDcoup=Ct*W*theta//maximum damping couple in N-m
+if atan((Ct*W)/(Kt-J*W^2))>0 then
+ phiD=(180/%pi)*atan((Ct*W)/(Kt-J*W^2));//from eqn 4.2.6(in degrees)
+else
+ phiD=180+(180/%pi)*atan((Ct*W)/(Kt-J*W^2));
+
+end
+//output
+mprintf(' a)The maximum angular displacement from rest position is %4.4f radians\n b)The maximum couple applied to dashpot is %4.4f N-m\n c)angle by which the angular displacement lags the torque is %4.4f degrees',theta,MaxDcoup,phiD)
diff --git a/3532/CH4/EX4.10.1/Ex4_12.sce b/3532/CH4/EX4.10.1/Ex4_12.sce new file mode 100644 index 000000000..ccfe11d19 --- /dev/null +++ b/3532/CH4/EX4.10.1/Ex4_12.sce @@ -0,0 +1,28 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.1\n')
+//given data
+m=1000//mass of machine in kg
+Fo=490//amp of force in N
+f=180//freq inRPM
+//calculations
+//case a)
+K=1.96*10^6//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst1=Fo/K//amplitude of steady state
+X1=Xst1*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr1=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//case b)
+K=9.8*10^4//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst2=Fo/K//amplitude of steady state
+X2=Xst2*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr2=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//output
+mprintf(' a)The amplitude of motion of machine is %f m and the maximum force transmitted\n to the foundation because of the unbalanced force when\n K=1.96*10^6 N/m is %4.4f N\n b)for same case as in a)if K=9.8*10^4 N/m then\n the amplitude of motion of machine is %f m\n and the maximum force transmitted to the foundation because of\n the unbalanced force %4.4f N',X1,Ftr1,X2,Ftr2)
diff --git a/3532/CH4/EX4.10.2/Ex4_13.sce b/3532/CH4/EX4.10.2/Ex4_13.sce new file mode 100644 index 000000000..4d2ebbd0e --- /dev/null +++ b/3532/CH4/EX4.10.2/Ex4_13.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.2\n')
+//given data
+m=75//mass of machine in kg
+K=11.76*10^5//stiffness of springs in N/m
+zeta=0.2
+mo=2//mass of piston in kg
+stroke=0.08//in m
+e=stroke/2//in m
+N=3000//spee in c.p.m
+//calculations
+Wn=sqrt(K/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+y=(mo/m)
+Fo=mo*W^2*e//max force exerted
+X=y*e*bet^2/(sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn 4.3.2
+Ftr=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+mprintf(' a)The amplitude of vibration of machine is %f m and the \n the vibratory force Ftr transmitted to the foundation is %5.4f N',X,Ftr)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of values in textbook')
diff --git a/3532/CH4/EX4.10.3/Ex4_14.sce b/3532/CH4/EX4.10.3/Ex4_14.sce new file mode 100644 index 000000000..44a536569 --- /dev/null +++ b/3532/CH4/EX4.10.3/Ex4_14.sce @@ -0,0 +1,23 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.3\n')
+// given data
+m=20 //mass in kgs
+k=125600 //overall eqivalent stiffness i.e 4*31400 in N/m
+c=1568 //overall damping coefficient i.e 4*392 in N-sec/m
+n=500 //vibrating speed of machine in cpm
+//y=Ysin(w*t)
+Y=0.00005 //vibrating amplitude of machine in m
+W=2*%pi*n/60 //vibrating frequency in rad/sec
+Wn=sqrt(k/m) //natural frequency in rad/sec
+bet=(W/Wn) //speed ratio
+zeta=c/(2*sqrt(k*m)) //damping factor
+//calculations
+X=Y*sqrt((1+(2*zeta*bet)^2)/((1-bet^2)^2+(2*zeta*bet)^2)) //absolute amplitude of vibration of radio from eqn (4.4.6)
+Z=Y*((bet^2)/sqrt(((1-bet^2)^2+(2*zeta*bet)^2)))//from eqn 4.4.11
+FdynT=Z*sqrt((c*W)^2+k^2)//dynamic load total
+Fdyn=FdynT/4 //dynamic load on each isolator
+FdynTmax=m*W^2*X //max dynamic load on the isolators
+Fdynmax=FdynTmax/4 //max dynamic load on each isolator
+//output
+mprintf('a) The amplitude of vibration of radio is %f metres \n b)the dynamic load on each isolator due to vibration is %3.3f N',X,Fdyn)
diff --git a/3532/CH4/EX4.11.1/Ex4_15.sce b/3532/CH4/EX4.11.1/Ex4_15.sce new file mode 100644 index 000000000..4696ee57f --- /dev/null +++ b/3532/CH4/EX4.11.1/Ex4_15.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.1\n')
+//given data
+T=2//period of free vibration in sec
+f=1//vertical harmonic frequency of machine in in Hz
+Z=2.5//amplitude of vibrotometer mass relative to vibrotometer frame in mm
+//calculations
+Wn=2*%pi/T
+W=2*%pi*f
+bet=(W/Wn)
+zeta=0//for vibrotometers
+Y=Z*(sqrt((1-bet^2)^2+(2*zeta*bet)^2))/bet^2//amplitude of vibration of machine Eqn 4.4.11 in Sec 4.4.2
+//output
+mprintf(' The amplitude of vibration of support of machine is %4.4f mm',Y)
diff --git a/3532/CH4/EX4.11.2/Ex4_16.sce b/3532/CH4/EX4.11.2/Ex4_16.sce new file mode 100644 index 000000000..44826cf62 --- /dev/null +++ b/3532/CH4/EX4.11.2/Ex4_16.sce @@ -0,0 +1,35 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.2\n')
+//given data
+fn=5.75//natural frequency in Hz
+zeta=0.65
+ZbyY=1.01
+//case 1
+//substituting for (Z/Y)=1.01 and (W/Wn)=r^2 in Eqn 4.4.11 we get the quadratic eqn as follows
+//0.02*r^4-0.31*r^2+1=0
+//solving for r in above eqn whose rootes are r1 and r2
+r1=sqrt(((0.31)+sqrt(((-0.31)^2)-4*0.02*1))/(2*0.02))
+r2=sqrt(((0.31)-sqrt(((-0.31)^2)-4*0.02*1))/(2*0.02))
+if r1>r2 then
+ r=r1
+ else r=r2
+end
+bet=r//bet=(W/Wn)
+f1=bet*fn
+//case 2
+ZbyY=0.98
+//substituting for (Z/Y)=0.98 and (W/Wn)=r^2 in Eqn 4.4.11 we get the quadratic eqn as follows
+//0.04*r^4+0.31*r^2-1=0
+//solving for r in above eqn whose rootes are r3 and r4
+r3=sqrt((-0.31+sqrt(((0.31)^2)-4*0.04*-1))/(2*0.04))
+r4=sqrt((-0.31-sqrt(((0.31)^2)-4*0.04*-1))/(2*0.04))
+t1=real(r3)
+t2=real(r4)
+if t1>t2 then
+ r=r3
+ else r=r4
+end
+bet=r//bet=(W/Wn)
+f2=bet*fn
+mprintf('The lowest frequency beyond which the amplitude can be measured within\n (i)one percent error is %4.4f Hz\n (ii)two percent error is %4.4f Hz',f1,f2)
diff --git a/3532/CH4/EX4.11.3/Ex4_17.sce b/3532/CH4/EX4.11.3/Ex4_17.sce new file mode 100644 index 000000000..827e5825e --- /dev/null +++ b/3532/CH4/EX4.11.3/Ex4_17.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.3\n')
+//given data
+J=0.049//moment of inertia in kg-m^2
+Kt=0.98//stiffness in N-m/rad
+Ct=0.11//damping coefficient in N-m_sec/rad
+N=15//R.P.M
+thetaRD=2//relative amplitude between ring and shaft in degrees
+//calculations
+W=N*2*%pi/60 //frequency of vibrating shaft in rad/sec
+Wn=sqrt(Kt/J) //natural freqency in rad/sec
+zeta=(Ct/(2*sqrt(Kt*J))) //damping factor
+thetaRR=(thetaRD/(57.3)) //relative amplitude in radians
+bet=(W/Wn)
+thetamax=thetaRR*((sqrt((1-bet^2)^2+(2*zeta*bet)^2)/bet^2))
+maxacc=(W^2)*thetamax
+//output
+mprintf('The maximum acceleration of the shaft is %4.4f rad/(sec^2)',maxacc)
diff --git a/3532/CH4/EX4.11.4/Ex4_18.sce b/3532/CH4/EX4.11.4/Ex4_18.sce new file mode 100644 index 000000000..b568e3c24 --- /dev/null +++ b/3532/CH4/EX4.11.4/Ex4_18.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.4\n')
+//given data
+RF=1800//resonant frequency in rpm
+L=0.050//lenght of steel reed in metres
+B=0.006//width of steel reed in metres
+t=0.00075//thickness of steel reed in metres
+E=19.6*10^10//young's modulus in N/(m^2)
+//calculations
+Wn=2*%pi*RF/60//natural frequency in radians
+I=(B*t^3)/12//moment of inertia in (m^4)
+m=3*E*I/((Wn^2)*L^3)//required mass
+//output
+mprintf('The required mass M to be placed at the end of the reeds of Frahm tachometer is %f Kgs',m)
diff --git a/3532/CH4/EX4.12/Ex4_12.sce b/3532/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..ccfe11d19 --- /dev/null +++ b/3532/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,28 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.1\n')
+//given data
+m=1000//mass of machine in kg
+Fo=490//amp of force in N
+f=180//freq inRPM
+//calculations
+//case a)
+K=1.96*10^6//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst1=Fo/K//amplitude of steady state
+X1=Xst1*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr1=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//case b)
+K=9.8*10^4//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst2=Fo/K//amplitude of steady state
+X2=Xst2*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr2=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//output
+mprintf(' a)The amplitude of motion of machine is %f m and the maximum force transmitted\n to the foundation because of the unbalanced force when\n K=1.96*10^6 N/m is %4.4f N\n b)for same case as in a)if K=9.8*10^4 N/m then\n the amplitude of motion of machine is %f m\n and the maximum force transmitted to the foundation because of\n the unbalanced force %4.4f N',X1,Ftr1,X2,Ftr2)
diff --git a/3532/CH4/EX4.13/Ex4_13.sce b/3532/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..4d2ebbd0e --- /dev/null +++ b/3532/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.2\n')
+//given data
+m=75//mass of machine in kg
+K=11.76*10^5//stiffness of springs in N/m
+zeta=0.2
+mo=2//mass of piston in kg
+stroke=0.08//in m
+e=stroke/2//in m
+N=3000//spee in c.p.m
+//calculations
+Wn=sqrt(K/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+y=(mo/m)
+Fo=mo*W^2*e//max force exerted
+X=y*e*bet^2/(sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn 4.3.2
+Ftr=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+mprintf(' a)The amplitude of vibration of machine is %f m and the \n the vibratory force Ftr transmitted to the foundation is %5.4f N',X,Ftr)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of values in textbook')
diff --git a/3532/CH4/EX4.14/Ex4_14.sce b/3532/CH4/EX4.14/Ex4_14.sce new file mode 100644 index 000000000..44a536569 --- /dev/null +++ b/3532/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,23 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.3\n')
+// given data
+m=20 //mass in kgs
+k=125600 //overall eqivalent stiffness i.e 4*31400 in N/m
+c=1568 //overall damping coefficient i.e 4*392 in N-sec/m
+n=500 //vibrating speed of machine in cpm
+//y=Ysin(w*t)
+Y=0.00005 //vibrating amplitude of machine in m
+W=2*%pi*n/60 //vibrating frequency in rad/sec
+Wn=sqrt(k/m) //natural frequency in rad/sec
+bet=(W/Wn) //speed ratio
+zeta=c/(2*sqrt(k*m)) //damping factor
+//calculations
+X=Y*sqrt((1+(2*zeta*bet)^2)/((1-bet^2)^2+(2*zeta*bet)^2)) //absolute amplitude of vibration of radio from eqn (4.4.6)
+Z=Y*((bet^2)/sqrt(((1-bet^2)^2+(2*zeta*bet)^2)))//from eqn 4.4.11
+FdynT=Z*sqrt((c*W)^2+k^2)//dynamic load total
+Fdyn=FdynT/4 //dynamic load on each isolator
+FdynTmax=m*W^2*X //max dynamic load on the isolators
+Fdynmax=FdynTmax/4 //max dynamic load on each isolator
+//output
+mprintf('a) The amplitude of vibration of radio is %f metres \n b)the dynamic load on each isolator due to vibration is %3.3f N',X,Fdyn)
diff --git a/3532/CH4/EX4.15/Ex4_15.sce b/3532/CH4/EX4.15/Ex4_15.sce new file mode 100644 index 000000000..4696ee57f --- /dev/null +++ b/3532/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.1\n')
+//given data
+T=2//period of free vibration in sec
+f=1//vertical harmonic frequency of machine in in Hz
+Z=2.5//amplitude of vibrotometer mass relative to vibrotometer frame in mm
+//calculations
+Wn=2*%pi/T
+W=2*%pi*f
+bet=(W/Wn)
+zeta=0//for vibrotometers
+Y=Z*(sqrt((1-bet^2)^2+(2*zeta*bet)^2))/bet^2//amplitude of vibration of machine Eqn 4.4.11 in Sec 4.4.2
+//output
+mprintf(' The amplitude of vibration of support of machine is %4.4f mm',Y)
diff --git a/3532/CH4/EX4.16/Ex4_16.sce b/3532/CH4/EX4.16/Ex4_16.sce new file mode 100644 index 000000000..44826cf62 --- /dev/null +++ b/3532/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,35 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.2\n')
+//given data
+fn=5.75//natural frequency in Hz
+zeta=0.65
+ZbyY=1.01
+//case 1
+//substituting for (Z/Y)=1.01 and (W/Wn)=r^2 in Eqn 4.4.11 we get the quadratic eqn as follows
+//0.02*r^4-0.31*r^2+1=0
+//solving for r in above eqn whose rootes are r1 and r2
+r1=sqrt(((0.31)+sqrt(((-0.31)^2)-4*0.02*1))/(2*0.02))
+r2=sqrt(((0.31)-sqrt(((-0.31)^2)-4*0.02*1))/(2*0.02))
+if r1>r2 then
+ r=r1
+ else r=r2
+end
+bet=r//bet=(W/Wn)
+f1=bet*fn
+//case 2
+ZbyY=0.98
+//substituting for (Z/Y)=0.98 and (W/Wn)=r^2 in Eqn 4.4.11 we get the quadratic eqn as follows
+//0.04*r^4+0.31*r^2-1=0
+//solving for r in above eqn whose rootes are r3 and r4
+r3=sqrt((-0.31+sqrt(((0.31)^2)-4*0.04*-1))/(2*0.04))
+r4=sqrt((-0.31-sqrt(((0.31)^2)-4*0.04*-1))/(2*0.04))
+t1=real(r3)
+t2=real(r4)
+if t1>t2 then
+ r=r3
+ else r=r4
+end
+bet=r//bet=(W/Wn)
+f2=bet*fn
+mprintf('The lowest frequency beyond which the amplitude can be measured within\n (i)one percent error is %4.4f Hz\n (ii)two percent error is %4.4f Hz',f1,f2)
diff --git a/3532/CH4/EX4.17/Ex4_17.sce b/3532/CH4/EX4.17/Ex4_17.sce new file mode 100644 index 000000000..827e5825e --- /dev/null +++ b/3532/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.3\n')
+//given data
+J=0.049//moment of inertia in kg-m^2
+Kt=0.98//stiffness in N-m/rad
+Ct=0.11//damping coefficient in N-m_sec/rad
+N=15//R.P.M
+thetaRD=2//relative amplitude between ring and shaft in degrees
+//calculations
+W=N*2*%pi/60 //frequency of vibrating shaft in rad/sec
+Wn=sqrt(Kt/J) //natural freqency in rad/sec
+zeta=(Ct/(2*sqrt(Kt*J))) //damping factor
+thetaRR=(thetaRD/(57.3)) //relative amplitude in radians
+bet=(W/Wn)
+thetamax=thetaRR*((sqrt((1-bet^2)^2+(2*zeta*bet)^2)/bet^2))
+maxacc=(W^2)*thetamax
+//output
+mprintf('The maximum acceleration of the shaft is %4.4f rad/(sec^2)',maxacc)
diff --git a/3532/CH4/EX4.18/Ex4_18.sce b/3532/CH4/EX4.18/Ex4_18.sce new file mode 100644 index 000000000..b568e3c24 --- /dev/null +++ b/3532/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.11.4\n')
+//given data
+RF=1800//resonant frequency in rpm
+L=0.050//lenght of steel reed in metres
+B=0.006//width of steel reed in metres
+t=0.00075//thickness of steel reed in metres
+E=19.6*10^10//young's modulus in N/(m^2)
+//calculations
+Wn=2*%pi*RF/60//natural frequency in radians
+I=(B*t^3)/12//moment of inertia in (m^4)
+m=3*E*I/((Wn^2)*L^3)//required mass
+//output
+mprintf('The required mass M to be placed at the end of the reeds of Frahm tachometer is %f Kgs',m)
diff --git a/3532/CH4/EX4.2.1/Ex4_1.sce b/3532/CH4/EX4.2.1/Ex4_1.sce new file mode 100644 index 000000000..8dfd86913 --- /dev/null +++ b/3532/CH4/EX4.2.1/Ex4_1.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.2.1\n')
+//given data
+//T=To*sin(W*t)
+To=0.588 //maximum value of periodic torque in N-m
+W=4// freqency of applied force in rad/sec
+J=0.12//moment of inertia of wheel in kg-m^2
+Kt=1.176//stiffness of wire in N-m/rad
+Ct=0.392/1 //damping coefficient in N-m_sec/rad
+//calculations
+theta=To/sqrt((Kt-J*W^2)^2+(Ct*W)^2)//Equation for torsional vibration amplitude from Fig (4.2.2) and Eqn (4.2.5)
+MaxDcoup=Ct*W*theta//maximum damping couple in N-m
+if atan((Ct*W)/(Kt-J*W^2))>0 then
+ phiD=(180/%pi)*atan((Ct*W)/(Kt-J*W^2));//from eqn 4.2.6(in degrees)
+else
+ phiD=180+(180/%pi)*atan((Ct*W)/(Kt-J*W^2));
+
+end
+//output
+mprintf(' a)The maximum angular displacement from rest position is %4.4f radians\n b)The maximum couple applied to dashpot is %4.4f N-m\n c)angle by which the angular displacement lags the torque is %4.4f degrees',theta,MaxDcoup,phiD)
diff --git a/3532/CH4/EX4.2.2/Ex4_2.sce b/3532/CH4/EX4.2.2/Ex4_2.sce new file mode 100644 index 000000000..ebbfc6c90 --- /dev/null +++ b/3532/CH4/EX4.2.2/Ex4_2.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.2.2\n')
+//given data
+Wd=9.8*2*%pi// damped natural freqency in rad/sec
+Wp=9.6*2*%pi//freqency from forced vibration test in rad/sec
+//calculations
+//(Wp/Wn)=sqrt(1-2*zeta^2)...(1) from Eqn 4.2.18 from Sec 4.2.1
+//(Wd/Wn)=sqrt(1-zeta^2)...(2) from Eqn 4.2.19 from Sec 4.2.1
+//dividing (1) by (2)
+x=(Wp/Wd)
+//x=[sqrt(1-2*zeta^2)]/[sqrt(1-zeta^2)]
+zeta=sqrt((1-x)/(2-x))//damping factor obtained on simplifying the above eqn
+//substituting for zeta in eqn 2 above
+Wn=Wd/sqrt(1-zeta^2)//natural frequency of system in rad/sec
+fn=Wn/(2*%pi)//natural frequency of system in Hz
+//output
+mprintf('The damping factor for the system is %f and\n the natural frequency is %4.4f rad/sec or %4.2f Hz',zeta,Wn,fn)
+mprintf('\nNOTE:The damping factor zeta given in textbook is 0.196,which is wrong.')
diff --git a/3532/CH4/EX4.2/Ex4_2.sce b/3532/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..ebbfc6c90 --- /dev/null +++ b/3532/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.2.2\n')
+//given data
+Wd=9.8*2*%pi// damped natural freqency in rad/sec
+Wp=9.6*2*%pi//freqency from forced vibration test in rad/sec
+//calculations
+//(Wp/Wn)=sqrt(1-2*zeta^2)...(1) from Eqn 4.2.18 from Sec 4.2.1
+//(Wd/Wn)=sqrt(1-zeta^2)...(2) from Eqn 4.2.19 from Sec 4.2.1
+//dividing (1) by (2)
+x=(Wp/Wd)
+//x=[sqrt(1-2*zeta^2)]/[sqrt(1-zeta^2)]
+zeta=sqrt((1-x)/(2-x))//damping factor obtained on simplifying the above eqn
+//substituting for zeta in eqn 2 above
+Wn=Wd/sqrt(1-zeta^2)//natural frequency of system in rad/sec
+fn=Wn/(2*%pi)//natural frequency of system in Hz
+//output
+mprintf('The damping factor for the system is %f and\n the natural frequency is %4.4f rad/sec or %4.2f Hz',zeta,Wn,fn)
+mprintf('\nNOTE:The damping factor zeta given in textbook is 0.196,which is wrong.')
diff --git a/3532/CH4/EX4.3.1/Ex4_3.sce b/3532/CH4/EX4.3.1/Ex4_3.sce new file mode 100644 index 000000000..43d47e288 --- /dev/null +++ b/3532/CH4/EX4.3.1/Ex4_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations b G.K.Grover\n Example 4.3.1\n')
+//given data
+m=1200//mass of motor in kg
+mo=1//unbalanced mass on motor in kg
+e=0.06//location of unbalanced mass from motor in m
+Wn=2210*(2*%pi/60)//resonant freq in rad/sec
+W=1440*(2*%pi/60)//operating freq
+//calculations
+//case 1
+zeta=0.1
+bet=(W/Wn)
+y=(mo/m)//from eqn 4.3.2
+X1=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//case 2
+zeta=0
+X2=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//output
+mprintf('If the damping is less than 0.1 then the amplitude of \n vibration will be between %f m and %f m',X1,X2)
diff --git a/3532/CH4/EX4.3.2/Ex4_4.sce b/3532/CH4/EX4.3.2/Ex4_4.sce new file mode 100644 index 000000000..d02fc736f --- /dev/null +++ b/3532/CH4/EX4.3.2/Ex4_4.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.3.2\n')
+//given data
+m=320//mass of engine in kg
+mo=24//reciprocating mass on motor in kg
+r=0.15//vertical stroke in m
+e=r/2
+delst=0.002//stati defln in m
+C=490/(0.3)//damping recistance in N-sec/m
+g=9.81// gravity in m/sec^2
+N=480//speed in rpm in case b)
+//calculation
+Wn=sqrt(g/delst) //natural freqency in rad/sec
+Nr=Wn/(2*%pi)*60 //resonant speed in rpm
+W=(2*%pi*N/60)
+bet=(W/Wn)
+zeta=(C/(2*m*Wn)) //damping factor
+y=(mo/m)//from eqn 4.3.2
+X=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//output
+mprintf(' a)speed of driving shaft at which esonance occurs is %4.4f RPM\n b)The amplitude of steady state forced vibrations when the driving shaft \n of the engine rotates at 480 RPM is %f m',Nr,X)
diff --git a/3532/CH4/EX4.3/Ex4_3.sce b/3532/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..43d47e288 --- /dev/null +++ b/3532/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations b G.K.Grover\n Example 4.3.1\n')
+//given data
+m=1200//mass of motor in kg
+mo=1//unbalanced mass on motor in kg
+e=0.06//location of unbalanced mass from motor in m
+Wn=2210*(2*%pi/60)//resonant freq in rad/sec
+W=1440*(2*%pi/60)//operating freq
+//calculations
+//case 1
+zeta=0.1
+bet=(W/Wn)
+y=(mo/m)//from eqn 4.3.2
+X1=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//case 2
+zeta=0
+X2=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//output
+mprintf('If the damping is less than 0.1 then the amplitude of \n vibration will be between %f m and %f m',X1,X2)
diff --git a/3532/CH4/EX4.4.1/Ex4_5.sce b/3532/CH4/EX4.4.1/Ex4_5.sce new file mode 100644 index 000000000..6b19a1838 --- /dev/null +++ b/3532/CH4/EX4.4.1/Ex4_5.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.1\n')
+//given data
+T=0.8//time period of free vibration in sec
+t=0.3//time for which the vertical distance has to be calculated
+//y=18*sin(2*pi*t)
+Y=18//max amplitude in mm
+//calculations
+W=2*%pi
+Wn=(2*%pi/T)
+bet=(W/Wn)
+x=(Y/(1-bet^2))*(sin(W*t)-bet*sin(Wn*t))// from eqn 4.4.17 explained in the same problem
+//output
+mprintf('The vertical distance moved by mass in the first 0.3 sec is %4.4f mm',x)
diff --git a/3532/CH4/EX4.4.2/Ex4_6.sce b/3532/CH4/EX4.4.2/Ex4_6.sce new file mode 100644 index 000000000..9a40f5466 --- /dev/null +++ b/3532/CH4/EX4.4.2/Ex4_6.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.2\n')
+//given data
+m=0.9//mass in kg
+K=1960//stiffness in N/m
+Y=5//amp of vibration of support in m
+N=1150//frequency in cycles per min
+//calculations
+Wn=sqrt(K/m)
+W=N*2*%pi/60//frequency of vibration of support
+bet=(W/Wn)
+//case 1
+zeta=0
+X1=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//case 2
+zeta =0.2
+X2=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//output
+mprintf('The amplitude of vibration when damping factor=0 is %4.4f mm \n If damping factor=0.2,then amplitude of vibration is %4.4f mm',X1,X2)
diff --git a/3532/CH4/EX4.4.3/Ex4_7.sce b/3532/CH4/EX4.4.3/Ex4_7.sce new file mode 100644 index 000000000..975af1a15 --- /dev/null +++ b/3532/CH4/EX4.4.3/Ex4_7.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.3\n')
+//given data
+delst=0.1//steady state defln in m
+g=9.81//acceleration due to gravity
+Y=0.08//amp of vibration of automobile in m
+lambda=14//wavelenght of profile in m
+//calculations
+Wn=sqrt(g/delst)
+fn=Wn/(2*%pi)//frequency of vibration of automobile in Hz
+Vc=(3600/1000)*lambda*fn//critical speed in km/hr
+V=60 //speed in km/hr
+W=V*(1000/3600)*(2*%pi/lambda)
+bet=(W/Wn)
+zeta=0
+X=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//output
+mprintf(' The critical speed of automobile %4.4f km/hr\n The amplitude of vibration at 60 Km/Hr is %4.4f m',Vc,X)
diff --git a/3532/CH4/EX4.4/Ex4_4.sce b/3532/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..d02fc736f --- /dev/null +++ b/3532/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.3.2\n')
+//given data
+m=320//mass of engine in kg
+mo=24//reciprocating mass on motor in kg
+r=0.15//vertical stroke in m
+e=r/2
+delst=0.002//stati defln in m
+C=490/(0.3)//damping recistance in N-sec/m
+g=9.81// gravity in m/sec^2
+N=480//speed in rpm in case b)
+//calculation
+Wn=sqrt(g/delst) //natural freqency in rad/sec
+Nr=Wn/(2*%pi)*60 //resonant speed in rpm
+W=(2*%pi*N/60)
+bet=(W/Wn)
+zeta=(C/(2*m*Wn)) //damping factor
+y=(mo/m)//from eqn 4.3.2
+X=(y*e)*(bet)^2/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//from eqn 4.3.2
+//output
+mprintf(' a)speed of driving shaft at which esonance occurs is %4.4f RPM\n b)The amplitude of steady state forced vibrations when the driving shaft \n of the engine rotates at 480 RPM is %f m',Nr,X)
diff --git a/3532/CH4/EX4.5.1/Ex4_8.sce b/3532/CH4/EX4.5.1/Ex4_8.sce new file mode 100644 index 000000000..b7f52cdb5 --- /dev/null +++ b/3532/CH4/EX4.5.1/Ex4_8.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.5.1\n')
+//given data
+X=0.015//amplitude of vibration of spring mass dashpot system in m
+f=100//frquency of vibration of spring mass dashpot system in Hz
+zeta=0.05
+fnD=22//damped natural frequency in Hz
+m=0.5//mass in kg
+//calculations
+W=2*%pi*fnD
+c=2*m*W*zeta// from Eqn 3.3.6 and Eqn 3.3.7
+Epercycl=%pi*c*(2*%pi*f)*X^2//Eqn 4.5.1...energy dissipated per cycle
+Epersec=Epercycl*f//energy dissipated per sec
+//output
+mprintf(' The power required to vibrate spring mass dashpot system with \n an amplitude of 1.5 cm and at frequency of 100 Hz is %4.4f Watts',Epersec)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of value of pi')
diff --git a/3532/CH4/EX4.5/Ex4_5.sce b/3532/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..6b19a1838 --- /dev/null +++ b/3532/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,15 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.1\n')
+//given data
+T=0.8//time period of free vibration in sec
+t=0.3//time for which the vertical distance has to be calculated
+//y=18*sin(2*pi*t)
+Y=18//max amplitude in mm
+//calculations
+W=2*%pi
+Wn=(2*%pi/T)
+bet=(W/Wn)
+x=(Y/(1-bet^2))*(sin(W*t)-bet*sin(Wn*t))// from eqn 4.4.17 explained in the same problem
+//output
+mprintf('The vertical distance moved by mass in the first 0.3 sec is %4.4f mm',x)
diff --git a/3532/CH4/EX4.6.1/Ex4_9.sce b/3532/CH4/EX4.6.1/Ex4_9.sce new file mode 100644 index 000000000..5e07c87c7 --- /dev/null +++ b/3532/CH4/EX4.6.1/Ex4_9.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.6.1\n')
+//given data
+mprintf('NOTE:The mass given in textbook should be equal\n to 3.7 kgs and not 8.7 Kgs')
+m=3.7//mass in kg
+g=9.81// gravity
+K=7550////stiffness of in N/m
+u=0.22//coefficient of friction
+Fo=19.6//amp of force in N
+f=5//frequency of force
+//calculations
+F=u*m*g//frictional force
+W=2*%pi*f
+Wn=sqrt(K/m)
+bet=(W/Wn)
+X=(Fo/K)*sqrt(1-(4*F/(%pi*Fo))^2)/(1-bet^2)//Eqn 4.6.2 in Sec 4.6
+Ceq=4*F/(%pi*W*X)//equivalent viscous damping Eqn 4.6.1 in Sec 4.6
+//output
+mprintf('\nThe amplitude of vibration of mass is %f m\n The equivalent viscous damping is %f N-sec/m',X,Ceq)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of value of pi in the taxtbook')
diff --git a/3532/CH4/EX4.6/Ex4_6.sce b/3532/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..9a40f5466 --- /dev/null +++ b/3532/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.2\n')
+//given data
+m=0.9//mass in kg
+K=1960//stiffness in N/m
+Y=5//amp of vibration of support in m
+N=1150//frequency in cycles per min
+//calculations
+Wn=sqrt(K/m)
+W=N*2*%pi/60//frequency of vibration of support
+bet=(W/Wn)
+//case 1
+zeta=0
+X1=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//case 2
+zeta =0.2
+X2=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//output
+mprintf('The amplitude of vibration when damping factor=0 is %4.4f mm \n If damping factor=0.2,then amplitude of vibration is %4.4f mm',X1,X2)
diff --git a/3532/CH4/EX4.7/Ex4_7.sce b/3532/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..975af1a15 --- /dev/null +++ b/3532/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.3\n')
+//given data
+delst=0.1//steady state defln in m
+g=9.81//acceleration due to gravity
+Y=0.08//amp of vibration of automobile in m
+lambda=14//wavelenght of profile in m
+//calculations
+Wn=sqrt(g/delst)
+fn=Wn/(2*%pi)//frequency of vibration of automobile in Hz
+Vc=(3600/1000)*lambda*fn//critical speed in km/hr
+V=60 //speed in km/hr
+W=V*(1000/3600)*(2*%pi/lambda)
+bet=(W/Wn)
+zeta=0
+X=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//output
+mprintf(' The critical speed of automobile %4.4f km/hr\n The amplitude of vibration at 60 Km/Hr is %4.4f m',Vc,X)
diff --git a/3532/CH4/EX4.8/Ex4_8.sce b/3532/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..b7f52cdb5 --- /dev/null +++ b/3532/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,17 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.5.1\n')
+//given data
+X=0.015//amplitude of vibration of spring mass dashpot system in m
+f=100//frquency of vibration of spring mass dashpot system in Hz
+zeta=0.05
+fnD=22//damped natural frequency in Hz
+m=0.5//mass in kg
+//calculations
+W=2*%pi*fnD
+c=2*m*W*zeta// from Eqn 3.3.6 and Eqn 3.3.7
+Epercycl=%pi*c*(2*%pi*f)*X^2//Eqn 4.5.1...energy dissipated per cycle
+Epersec=Epercycl*f//energy dissipated per sec
+//output
+mprintf(' The power required to vibrate spring mass dashpot system with \n an amplitude of 1.5 cm and at frequency of 100 Hz is %4.4f Watts',Epersec)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of value of pi')
diff --git a/3532/CH4/EX4.9/Ex4_9.sce b/3532/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..5e07c87c7 --- /dev/null +++ b/3532/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,21 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.6.1\n')
+//given data
+mprintf('NOTE:The mass given in textbook should be equal\n to 3.7 kgs and not 8.7 Kgs')
+m=3.7//mass in kg
+g=9.81// gravity
+K=7550////stiffness of in N/m
+u=0.22//coefficient of friction
+Fo=19.6//amp of force in N
+f=5//frequency of force
+//calculations
+F=u*m*g//frictional force
+W=2*%pi*f
+Wn=sqrt(K/m)
+bet=(W/Wn)
+X=(Fo/K)*sqrt(1-(4*F/(%pi*Fo))^2)/(1-bet^2)//Eqn 4.6.2 in Sec 4.6
+Ceq=4*F/(%pi*W*X)//equivalent viscous damping Eqn 4.6.1 in Sec 4.6
+//output
+mprintf('\nThe amplitude of vibration of mass is %f m\n The equivalent viscous damping is %f N-sec/m',X,Ceq)
+mprintf('\nNOTE: slight differnce in answer compared to textbook\n is due approximation of value of pi in the taxtbook')
diff --git a/3532/CH5/EX5.3.2/Ex5_3.sce b/3532/CH5/EX5.3.2/Ex5_3.sce new file mode 100644 index 000000000..4cfe37007 --- /dev/null +++ b/3532/CH5/EX5.3.2/Ex5_3.sce @@ -0,0 +1,18 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.3.2\n')
+//given data
+m1=5*0.75//mass of rod 1 in kgs
+m2=5*1.00//mass of rod 2 in kgs
+l1=0.75//lenght of rod 1 in m
+l2=1.00//lenght of rod 2 in m
+K=2940//stiffness of spring in N/m
+//calculations
+Wn=sqrt(3*(m1+m2)*K/(m1*m2))//natural frequency in rad/sec
+fn=Wn/(2*%pi)//natural frequency in Hz as solved in the textbook itself
+b1=(K*l2)
+b2=(K*l1-m1*l1*Wn^2/3)
+x=(b2/b1)
+Fmax=K*(l1*1-l2*x)/57.3//to convert into radians
+//output
+mprintf('The frequency of the resulting vibrations if the efect of gravity\n is neglected is %4.4f rad/sec or %4.4f Hz.\n The angular movement of CD is %3.3f degrees(out of phase) \n with the movement of AB.\n The maximum force in the spring is %4.4f N',Wn,fn,x,Fmax)
diff --git a/3532/CH5/EX5.3.3/Ex5_4.sce b/3532/CH5/EX5.3.3/Ex5_4.sce new file mode 100644 index 000000000..2a3cdc65b --- /dev/null +++ b/3532/CH5/EX5.3.3/Ex5_4.sce @@ -0,0 +1,24 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.3.3\n')
+//given data
+m1=500//mass of disc 1 in Kgs
+m2=1000//mass of disc 2 in Kgs
+D1=1.25//outer dia of disc 1 in m
+D2=1.9//outer dia of disc 2 in m
+l=3.0//lenght of shaft in m
+d=0.10//dia of shaft in m
+G=0.83*10^11//rigidity modulus in N/m^2
+//calculations
+J1=m1*(D1/2)^2/2//mass moment of inertia in kg-m^2
+J2=m2*(D2/2)^2/2//mass moment of inertia in kg-m^2
+Ip=(%pi/32)*d^4//section modulus of shaft in m^4
+Kt=G*Ip/l//stiffness in N-m/rad
+Wn=sqrt(Kt*(J1+J2)/(J1*J2))//from Eqn 5.3.28,Sec 5.3.3
+fn=Wn/(2*%pi)
+Kt1=2*Kt
+Kt2=2*Kt*2^4
+Kte=1/((1/Kt1)+(1/Kt2))
+x=sqrt(Kte/Kt)//ratio of modified natural freq to original natural frequency
+//output
+mprintf('The natural frequency of the torsional vibration is\n %4.4f rad/sec or %3.3f Hz.\n The ratio of modified natural frequency to original natural frequency\n is %3.3f.Which means stiffening a system increases its natural frequency',Wn,fn,x)
diff --git a/3532/CH5/EX5.3/Ex5_3.sce b/3532/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..4cfe37007 --- /dev/null +++ b/3532/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,18 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.3.2\n')
+//given data
+m1=5*0.75//mass of rod 1 in kgs
+m2=5*1.00//mass of rod 2 in kgs
+l1=0.75//lenght of rod 1 in m
+l2=1.00//lenght of rod 2 in m
+K=2940//stiffness of spring in N/m
+//calculations
+Wn=sqrt(3*(m1+m2)*K/(m1*m2))//natural frequency in rad/sec
+fn=Wn/(2*%pi)//natural frequency in Hz as solved in the textbook itself
+b1=(K*l2)
+b2=(K*l1-m1*l1*Wn^2/3)
+x=(b2/b1)
+Fmax=K*(l1*1-l2*x)/57.3//to convert into radians
+//output
+mprintf('The frequency of the resulting vibrations if the efect of gravity\n is neglected is %4.4f rad/sec or %4.4f Hz.\n The angular movement of CD is %3.3f degrees(out of phase) \n with the movement of AB.\n The maximum force in the spring is %4.4f N',Wn,fn,x,Fmax)
diff --git a/3532/CH5/EX5.4/Ex5_4.sce b/3532/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..2a3cdc65b --- /dev/null +++ b/3532/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,24 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.3.3\n')
+//given data
+m1=500//mass of disc 1 in Kgs
+m2=1000//mass of disc 2 in Kgs
+D1=1.25//outer dia of disc 1 in m
+D2=1.9//outer dia of disc 2 in m
+l=3.0//lenght of shaft in m
+d=0.10//dia of shaft in m
+G=0.83*10^11//rigidity modulus in N/m^2
+//calculations
+J1=m1*(D1/2)^2/2//mass moment of inertia in kg-m^2
+J2=m2*(D2/2)^2/2//mass moment of inertia in kg-m^2
+Ip=(%pi/32)*d^4//section modulus of shaft in m^4
+Kt=G*Ip/l//stiffness in N-m/rad
+Wn=sqrt(Kt*(J1+J2)/(J1*J2))//from Eqn 5.3.28,Sec 5.3.3
+fn=Wn/(2*%pi)
+Kt1=2*Kt
+Kt2=2*Kt*2^4
+Kte=1/((1/Kt1)+(1/Kt2))
+x=sqrt(Kte/Kt)//ratio of modified natural freq to original natural frequency
+//output
+mprintf('The natural frequency of the torsional vibration is\n %4.4f rad/sec or %3.3f Hz.\n The ratio of modified natural frequency to original natural frequency\n is %3.3f.Which means stiffening a system increases its natural frequency',Wn,fn,x)
diff --git a/3532/CH5/EX5.7.1/Ex5_8.sce b/3532/CH5/EX5.7.1/Ex5_8.sce new file mode 100644 index 000000000..cee0aee50 --- /dev/null +++ b/3532/CH5/EX5.7.1/Ex5_8.sce @@ -0,0 +1,29 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.7.1\n')
+//given data
+J1=0.735//moment of inertia of main system in Kg-m^2
+Kt1=7.35*10^5//torsional stiffness
+To=294//amplitude of applied torque
+W=10^3//frequency of applied torque
+//u=ratio of absorber mass to main mass i.e M2/M1
+//Wn is exitation frequency
+//calculations
+W1=sqrt(Kt1/J1)
+//case1
+x1=0.8//where x=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+x2=1.2//where x=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+if u1>u2 then
+ u=u1
+else
+ u=u2
+end
+J2=u*J1//moment of inertia of absorber in Kg-m^2
+Kt2=u*Kt1// total torsional stiffness of absorber
+K=Kt2/(4*0.1^2)//stiffness of each spring in N/m
+b2=-(To/Kt2)//amplitude of vibration in rad
+//output
+mprintf('The maximum moment of inertia of absorber(J2) is %4.4f Kg-m^2 and\n %f is the stiffness of each of the four absorber springs such that\n the resonant frequencies are at least 20 percent from exitation frequency.\n The amplitude of vibration of this absorber(b2) at exitation frequency\n is %f radians',J2,K,b2)
diff --git a/3532/CH5/EX5.7.2/Ex5_9.sce b/3532/CH5/EX5.7.2/Ex5_9.sce new file mode 100644 index 000000000..3c7585937 --- /dev/null +++ b/3532/CH5/EX5.7.2/Ex5_9.sce @@ -0,0 +1,46 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.7.2\n')
+//given data
+W1=220*2*%pi/60//vibrating frequency at 220 RPM (in rad/sec)
+W2=W1//frequency to which the spring mass system is tuned to.
+M2=1//mass in spring mass system in kgs
+N1=188//first resonant freq of spring mass system in cpm
+N2=258//second resonant freq of spring mass system in cpm
+//u=ratio of absorber mass to main mass i.e M2/M1
+//calculations
+K2=M2*W2^2
+Wn1=N1*2*%pi/60//first resonant freq of spring mass system in rad/sec
+Wn2=N2*2*%pi/60//second resonant freq of spring mass system in rad/sec
+//case 1
+W=Wn1
+x1=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+W=Wn2
+x2=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+//therefore
+u=(u1+u2)/2//which is equal to M2/M1
+M1=M2/u// mass of main system in kgs
+K1=K2/u//stiffness of main system in N/m
+//now
+Wn21=150*2*%pi/60//new first resonant frequency in rad/sec
+Wn22=310*2*%pi/60//new second resonant frequency in rad/sec
+W=Wn21
+x1=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+W=Wn22
+x2=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+//choosing the higher value
+if u1>u2 then
+ u=u1
+else
+ u=u2
+end
+M3=M1*u// mass of main system in kgs
+K3=K1*u//stiffness of main system in N/m
+//output
+mprintf(' The mass of main system required is %4.4f kgs\n stiffness of main system reqired is %5.5f N/m\n If the resonant frequencies lie outside the range of 150 to 310 rpm then\n mass of main system is %4.4f kgs\n stiffness of main system is %5.5f N/m',M1,K1,M3,K3)
diff --git a/3532/CH5/EX5.8/Ex5_8.sce b/3532/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..cee0aee50 --- /dev/null +++ b/3532/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,29 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.7.1\n')
+//given data
+J1=0.735//moment of inertia of main system in Kg-m^2
+Kt1=7.35*10^5//torsional stiffness
+To=294//amplitude of applied torque
+W=10^3//frequency of applied torque
+//u=ratio of absorber mass to main mass i.e M2/M1
+//Wn is exitation frequency
+//calculations
+W1=sqrt(Kt1/J1)
+//case1
+x1=0.8//where x=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+x2=1.2//where x=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+if u1>u2 then
+ u=u1
+else
+ u=u2
+end
+J2=u*J1//moment of inertia of absorber in Kg-m^2
+Kt2=u*Kt1// total torsional stiffness of absorber
+K=Kt2/(4*0.1^2)//stiffness of each spring in N/m
+b2=-(To/Kt2)//amplitude of vibration in rad
+//output
+mprintf('The maximum moment of inertia of absorber(J2) is %4.4f Kg-m^2 and\n %f is the stiffness of each of the four absorber springs such that\n the resonant frequencies are at least 20 percent from exitation frequency.\n The amplitude of vibration of this absorber(b2) at exitation frequency\n is %f radians',J2,K,b2)
diff --git a/3532/CH5/EX5.9/Ex5_9.sce b/3532/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..3c7585937 --- /dev/null +++ b/3532/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,46 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 5.7.2\n')
+//given data
+W1=220*2*%pi/60//vibrating frequency at 220 RPM (in rad/sec)
+W2=W1//frequency to which the spring mass system is tuned to.
+M2=1//mass in spring mass system in kgs
+N1=188//first resonant freq of spring mass system in cpm
+N2=258//second resonant freq of spring mass system in cpm
+//u=ratio of absorber mass to main mass i.e M2/M1
+//calculations
+K2=M2*W2^2
+Wn1=N1*2*%pi/60//first resonant freq of spring mass system in rad/sec
+Wn2=N2*2*%pi/60//second resonant freq of spring mass system in rad/sec
+//case 1
+W=Wn1
+x1=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+W=Wn2
+x2=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+//therefore
+u=(u1+u2)/2//which is equal to M2/M1
+M1=M2/u// mass of main system in kgs
+K1=K2/u//stiffness of main system in N/m
+//now
+Wn21=150*2*%pi/60//new first resonant frequency in rad/sec
+Wn22=310*2*%pi/60//new second resonant frequency in rad/sec
+W=Wn21
+x1=(W/W2)
+u1=[x1^2-1]^2/x1^2//from Eqn 5.7.9,Sec 5.7.1.
+//case 2
+W=Wn22
+x2=(W/W2)
+u2=[x2^2-1]^2/x2^2//from Eqn 5.7.9,Sec 5.7.1.
+//choosing the higher value
+if u1>u2 then
+ u=u1
+else
+ u=u2
+end
+M3=M1*u// mass of main system in kgs
+K3=K1*u//stiffness of main system in N/m
+//output
+mprintf(' The mass of main system required is %4.4f kgs\n stiffness of main system reqired is %5.5f N/m\n If the resonant frequencies lie outside the range of 150 to 310 rpm then\n mass of main system is %4.4f kgs\n stiffness of main system is %5.5f N/m',M1,K1,M3,K3)
diff --git a/3532/CH6/EX6.15/Ex6_15.sce b/3532/CH6/EX6.15/Ex6_15.sce new file mode 100644 index 000000000..0c16e4d46 --- /dev/null +++ b/3532/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 6.8.2\n')
+//given data
+m1=250;m2=100//mass of two blocks in Kgs
+c1=80;c2=60,c=20//damping coefficients in N-sec/m
+F1=1000;F2=1500//amplitude of force acting on block 1 and 2 rsptly
+k=250000//stiffness of spring in N/m
+W=60//frequency of applied force in rad/sec
+//calculations
+M=[m1,0;0,m2];
+K=[k,-k;-k,k];
+C=[c+c1,-c;-c,c+c2];
+R=[F1;F2;0;0];
+X=K-(W^2)*M
+Y=W*C
+G=[X,-Y;Y,X]
+AB=inv(G) *R//from Eqn6.8.4 in Sec 6.8
+X1=sqrt(AB(1,1)^2 +AB(3,1)^2)
+X2=sqrt(AB(2,1)^2 +AB(4,1)^2)
+//output
+mprintf('The amplitude of vibrations are %fm for mass 1 and %fm for mass 2',X1,X2)
diff --git a/3532/CH6/EX6.8.2/Ex6_15.sce b/3532/CH6/EX6.8.2/Ex6_15.sce new file mode 100644 index 000000000..0c16e4d46 --- /dev/null +++ b/3532/CH6/EX6.8.2/Ex6_15.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 6.8.2\n')
+//given data
+m1=250;m2=100//mass of two blocks in Kgs
+c1=80;c2=60,c=20//damping coefficients in N-sec/m
+F1=1000;F2=1500//amplitude of force acting on block 1 and 2 rsptly
+k=250000//stiffness of spring in N/m
+W=60//frequency of applied force in rad/sec
+//calculations
+M=[m1,0;0,m2];
+K=[k,-k;-k,k];
+C=[c+c1,-c;-c,c+c2];
+R=[F1;F2;0;0];
+X=K-(W^2)*M
+Y=W*C
+G=[X,-Y;Y,X]
+AB=inv(G) *R//from Eqn6.8.4 in Sec 6.8
+X1=sqrt(AB(1,1)^2 +AB(3,1)^2)
+X2=sqrt(AB(2,1)^2 +AB(4,1)^2)
+//output
+mprintf('The amplitude of vibrations are %fm for mass 1 and %fm for mass 2',X1,X2)
diff --git a/3532/CH7/EX7.1/Ex7_1.sce b/3532/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..8ad1c9349 --- /dev/null +++ b/3532/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,23 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.2.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+y1=g*(M1*a(1,1)+M2*a(1,2))
+y2=g*(M1*a(2,1)+M2*a(2,2))
+Wn=sqrt(g*(M1*y1+M2*y2)/(M1*y1^2+M2*y2^2))
+//now to find out lower natural frequency
+F1=M1*y1*Wn^2
+F2=M2*y2*Wn^2
+y1new=F1*a(1,1)+F2*a(1,2)
+y2new=F1*a(2,1)+F2*a(2,2)
+Wnnew=sqrt((F1*y1new+F2*y2new)/(M1*y1new^2+M2*y2new^2))//actual natural frequency in rad/sec
+//output
+mprintf(' The practical natural frequency Wn is %4.4f rad/sec,but the lower \n natural frequency Wn`is %4.4f rad/sec which is closer to the actual\n natural frequency',Wn,Wnnew)
diff --git a/3532/CH7/EX7.10/Ex7_10.sce b/3532/CH7/EX7.10/Ex7_10.sce new file mode 100644 index 000000000..1d7ff7f30 --- /dev/null +++ b/3532/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,37 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.7.2\n')
+//given data
+J(1)=100//moment of inertia of first rotor in Kg-m^2
+J(2)=50//moment of inertia of second rotor in Kg-m^2
+J(3)=10//moment of inertia of third rotor in Kg-m^2
+J(4)=50//moment of inertia of fourth rotor in Kg-m^2
+Kt(1)=10^4//stiffness of shaft between 1 and 2 in N-m/rad
+Kt(2)=10^4//stiffness of shaft between 2 and 3 in N-m/rad
+Kt(3)=2*10^4//stiffness of shaft between 3 and 4 in N-m/rad
+To=10000//amplitude of applied torque in N-m
+W=5//frequency of applied torque in rad/sec
+//calculations
+b(1)=-(0.789*To)/3825//twist of shaft 1 in rad
+P(1)=J(1)*W^2
+Q(1)=P(1)*b(1)//twisting moment of shaft 1 in N-m
+R(1)=Q(1)
+S(1)=R(1)/Kt(1)//twist of shaft 1 in radians
+b(2)=b(1)-S(1)//twist of shaft 2 in rad
+P(2)=J(2)*W^2
+Q(2)=P(2)*b(2)
+R(2)=Q(1)+Q(2)+To//twisting moment of shaft 2 in N-m
+S(2)=R(2)/Kt(2)//twist of shaft 2 in radians
+b(3)=b(2)-S(2)//twist of shaft 3 in rad
+P(3)=J(3)*W^2
+Q(3)=P(3)*b(3)
+R(3)=Q(2)+Q(3)//twisting moment of shaft 3 in N-m
+S(3)=R(3)/Kt(3)//twist of shaft 3 in radians
+b(4)=b(3)-S(3)//twist of shaft 4 in rad
+P(4)=J(4)*W^2
+Q(4)=P(4)*b(4)
+R(4)=Q(3)+Q(4)//twisting moment of shaft 4 in N-m
+mprintf('The amplitudes of discs are as follows\n disc1=%4.4f rad\n disc2=%4.4f rad\n disc3=%4.4f rad\n disc4=%4.4f rad',b(1),b(2),b(3),b(4))
+mprintf('\nThe twists of shaft are as follows\nfirst shaft=%5.5f rad\nsecond shaft=%5.5f rad\nthird shaft=%5.5f rad',S(1),S(2),S(3))
+mprintf('\nThe twisting moments of shafts are as follows\nfirst shaft=%5.5f N-m\nsecond shaft=%5.5f N-m\nthird shaft=%5.5f N-m',R(1),R(2),R(3))
+mprintf('\nNOTE:The slight difference in values are due to the more accurate values\ncalculated by SCILAB')
diff --git a/3532/CH7/EX7.2.1/Ex7_1.sce b/3532/CH7/EX7.2.1/Ex7_1.sce new file mode 100644 index 000000000..8ad1c9349 --- /dev/null +++ b/3532/CH7/EX7.2.1/Ex7_1.sce @@ -0,0 +1,23 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.2.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+y1=g*(M1*a(1,1)+M2*a(1,2))
+y2=g*(M1*a(2,1)+M2*a(2,2))
+Wn=sqrt(g*(M1*y1+M2*y2)/(M1*y1^2+M2*y2^2))
+//now to find out lower natural frequency
+F1=M1*y1*Wn^2
+F2=M2*y2*Wn^2
+y1new=F1*a(1,1)+F2*a(1,2)
+y2new=F1*a(2,1)+F2*a(2,2)
+Wnnew=sqrt((F1*y1new+F2*y2new)/(M1*y1new^2+M2*y2new^2))//actual natural frequency in rad/sec
+//output
+mprintf(' The practical natural frequency Wn is %4.4f rad/sec,but the lower \n natural frequency Wn`is %4.4f rad/sec which is closer to the actual\n natural frequency',Wn,Wnnew)
diff --git a/3532/CH7/EX7.3.1/Ex7_4.sce b/3532/CH7/EX7.3.1/Ex7_4.sce new file mode 100644 index 000000000..2b5f13fc7 --- /dev/null +++ b/3532/CH7/EX7.3.1/Ex7_4.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.3.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+y1=g*M1*a(1,1)//considering only M1 to be acting
+y2=g*M2*a(2,2)//considering only M2 to be acting
+W1=sqrt(g/y1)
+W2=sqrt(g/y2)
+Wn=sqrt(1/((1/W1^2)+(1/W2^2)))//applying Eqn 7.3.7,Sec7.3
+//output
+mprintf(' The natural frequency of transverse vibration obtained from \n Dunkerly method is %4.4f rad/sec which is slightly lower\n than the correct value',Wn)
diff --git a/3532/CH7/EX7.4.1/Ex7_5.sce b/3532/CH7/EX7.4.1/Ex7_5.sce new file mode 100644 index 000000000..c14b212b0 --- /dev/null +++ b/3532/CH7/EX7.4.1/Ex7_5.sce @@ -0,0 +1,28 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.4.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+x1(1)=1;x2(1)=1
+for i=1:10//upto 10th iteration for more perfect answer
+F1(i)=100*x1(i)//'i' represents the dash(')
+F2(i)=50*x2(i)
+x1(i)=F1(i)*a(1,1)+F2(i)*a(1,2)
+x2(i)=F1(i)*a(2,1)+F2(i)*a(2,2)
+r=(x2(i)/x1(i))
+x2(i+1)=r
+x1(i+1)=1
+end
+x1dd=1
+W1=(x1dd/x1(10))
+W2=(r/x2(10))
+Wn=sqrt((W1+W2)/2)//natural frequency in rad/sec
+mprintf('The natural frequency of system in iilustrative example 7.2.1 obtained by\nStodala method is Wn=%f rad/sec',Wn)
+mprintf('\nNOTE:The obtained answer is more near to the perfect answer \since 10 iterations/trials\nhas been carried out.In textbook only upto 3rd iteration has been carried out')
diff --git a/3532/CH7/EX7.4/Ex7_4.sce b/3532/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..2b5f13fc7 --- /dev/null +++ b/3532/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,19 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.3.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+y1=g*M1*a(1,1)//considering only M1 to be acting
+y2=g*M2*a(2,2)//considering only M2 to be acting
+W1=sqrt(g/y1)
+W2=sqrt(g/y2)
+Wn=sqrt(1/((1/W1^2)+(1/W2^2)))//applying Eqn 7.3.7,Sec7.3
+//output
+mprintf(' The natural frequency of transverse vibration obtained from \n Dunkerly method is %4.4f rad/sec which is slightly lower\n than the correct value',Wn)
diff --git a/3532/CH7/EX7.5/Ex7_5.sce b/3532/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..c14b212b0 --- /dev/null +++ b/3532/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,28 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.4.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+x1(1)=1;x2(1)=1
+for i=1:10//upto 10th iteration for more perfect answer
+F1(i)=100*x1(i)//'i' represents the dash(')
+F2(i)=50*x2(i)
+x1(i)=F1(i)*a(1,1)+F2(i)*a(1,2)
+x2(i)=F1(i)*a(2,1)+F2(i)*a(2,2)
+r=(x2(i)/x1(i))
+x2(i+1)=r
+x1(i+1)=1
+end
+x1dd=1
+W1=(x1dd/x1(10))
+W2=(r/x2(10))
+Wn=sqrt((W1+W2)/2)//natural frequency in rad/sec
+mprintf('The natural frequency of system in iilustrative example 7.2.1 obtained by\nStodala method is Wn=%f rad/sec',Wn)
+mprintf('\nNOTE:The obtained answer is more near to the perfect answer \since 10 iterations/trials\nhas been carried out.In textbook only upto 3rd iteration has been carried out')
diff --git a/3532/CH7/EX7.7.2/Ex7_10.sce b/3532/CH7/EX7.7.2/Ex7_10.sce new file mode 100644 index 000000000..1d7ff7f30 --- /dev/null +++ b/3532/CH7/EX7.7.2/Ex7_10.sce @@ -0,0 +1,37 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 7.7.2\n')
+//given data
+J(1)=100//moment of inertia of first rotor in Kg-m^2
+J(2)=50//moment of inertia of second rotor in Kg-m^2
+J(3)=10//moment of inertia of third rotor in Kg-m^2
+J(4)=50//moment of inertia of fourth rotor in Kg-m^2
+Kt(1)=10^4//stiffness of shaft between 1 and 2 in N-m/rad
+Kt(2)=10^4//stiffness of shaft between 2 and 3 in N-m/rad
+Kt(3)=2*10^4//stiffness of shaft between 3 and 4 in N-m/rad
+To=10000//amplitude of applied torque in N-m
+W=5//frequency of applied torque in rad/sec
+//calculations
+b(1)=-(0.789*To)/3825//twist of shaft 1 in rad
+P(1)=J(1)*W^2
+Q(1)=P(1)*b(1)//twisting moment of shaft 1 in N-m
+R(1)=Q(1)
+S(1)=R(1)/Kt(1)//twist of shaft 1 in radians
+b(2)=b(1)-S(1)//twist of shaft 2 in rad
+P(2)=J(2)*W^2
+Q(2)=P(2)*b(2)
+R(2)=Q(1)+Q(2)+To//twisting moment of shaft 2 in N-m
+S(2)=R(2)/Kt(2)//twist of shaft 2 in radians
+b(3)=b(2)-S(2)//twist of shaft 3 in rad
+P(3)=J(3)*W^2
+Q(3)=P(3)*b(3)
+R(3)=Q(2)+Q(3)//twisting moment of shaft 3 in N-m
+S(3)=R(3)/Kt(3)//twist of shaft 3 in radians
+b(4)=b(3)-S(3)//twist of shaft 4 in rad
+P(4)=J(4)*W^2
+Q(4)=P(4)*b(4)
+R(4)=Q(3)+Q(4)//twisting moment of shaft 4 in N-m
+mprintf('The amplitudes of discs are as follows\n disc1=%4.4f rad\n disc2=%4.4f rad\n disc3=%4.4f rad\n disc4=%4.4f rad',b(1),b(2),b(3),b(4))
+mprintf('\nThe twists of shaft are as follows\nfirst shaft=%5.5f rad\nsecond shaft=%5.5f rad\nthird shaft=%5.5f rad',S(1),S(2),S(3))
+mprintf('\nThe twisting moments of shafts are as follows\nfirst shaft=%5.5f N-m\nsecond shaft=%5.5f N-m\nthird shaft=%5.5f N-m',R(1),R(2),R(3))
+mprintf('\nNOTE:The slight difference in values are due to the more accurate values\ncalculated by SCILAB')
diff --git a/3532/CH8/EX8.1/Ex8_1.sce b/3532/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..f91c1baa5 --- /dev/null +++ b/3532/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.2.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+m=5//mass of rotor in kg
+d=0.01//dia of shaft in m
+I=(%pi/64)*d^4///moment of area in m^4
+l=0.4//bearing span in m
+e=0.02//distance of CG away from geometric centre of rotor in mm
+N=3000//speed of shaft in RPM
+//calculations
+k=48*E*I/l^3//stiffness of shaft in N/m
+Wn=sqrt(k/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+r=(bet^2*e/(1-bet^2))//from Eqn 8.2.2 in Sec 8.2
+rabs=abs(r)//absolute value of displacement
+Rd=k*rabs/1000//total dynamic load in bearings in N(divide by 1000 since r is in mm)
+F=Rd/2//dynamic load on each bearings in N
+//output
+mprintf(' The amplitude of steady state vibration of shaft is %f mm\nNOTE:negetive sign shows that displacement is out of phase with centrifugal force\nThe dynamic force transmtted to the bearings is %4.4f N\n The dynamic load on each bearing is %4.4f N',r,Rd,F)
diff --git a/3532/CH8/EX8.2.1/Ex8_1.sce b/3532/CH8/EX8.2.1/Ex8_1.sce new file mode 100644 index 000000000..f91c1baa5 --- /dev/null +++ b/3532/CH8/EX8.2.1/Ex8_1.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.2.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+m=5//mass of rotor in kg
+d=0.01//dia of shaft in m
+I=(%pi/64)*d^4///moment of area in m^4
+l=0.4//bearing span in m
+e=0.02//distance of CG away from geometric centre of rotor in mm
+N=3000//speed of shaft in RPM
+//calculations
+k=48*E*I/l^3//stiffness of shaft in N/m
+Wn=sqrt(k/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+r=(bet^2*e/(1-bet^2))//from Eqn 8.2.2 in Sec 8.2
+rabs=abs(r)//absolute value of displacement
+Rd=k*rabs/1000//total dynamic load in bearings in N(divide by 1000 since r is in mm)
+F=Rd/2//dynamic load on each bearings in N
+//output
+mprintf(' The amplitude of steady state vibration of shaft is %f mm\nNOTE:negetive sign shows that displacement is out of phase with centrifugal force\nThe dynamic force transmtted to the bearings is %4.4f N\n The dynamic load on each bearing is %4.4f N',r,Rd,F)
diff --git a/3532/CH8/EX8.2/Ex8_2.sce b/3532/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..61792beb2 --- /dev/null +++ b/3532/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,30 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.3.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+m=4//mass of rotor in kg
+g=9.81//acc due to gravity in m/sec^2
+d=0.009//dia of shaft in m
+I=(%pi/64)*d^4///moment of area in m^4
+l=0.48//bearing span in m
+e=0.003//distance of CG away from geometric centre of rotor in mm
+N=760//speed of shaft in RPM
+c=49//equivalent viscous damping in N-sec/m
+//calculations
+K=48*E*I/l^3//stiffness of shaft in N/m
+Wn=sqrt(K/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+zeta=c/(2*sqrt(K*m))
+r=e*(bet^2/sqrt(((1-bet^2)^2+(2*zeta*bet)^2)))//from Eqn 8.3.4 ,Sec 8.3
+Fd=sqrt((K*r)^2+(c*W*r)^2)//dynamic load on bearing in N
+Fs=m*g//static load in N
+Fmax=Fd+Fs//maximum static load on the shaft under dynamic condition in N
+smax=(Fmax*l/4)*(d/2)/I//maximum stress under dynamic condition in N/m^2
+ss=(Fs*l/4)*(d/2)/I//maximum stress under dead load condition in N/m^2
+Fdamp=(c*W*r)//damping force in N
+Tdamp=Fdamp*r//damping torque in N-m
+P=2*%pi*N*Tdamp/60//power in Watts
+//output
+mprintf(' The mamximum stress in the shaft under dynamic condition is %.3f N/(m^2)\n The dead load stress is %.3f N/(m^2)\n The power required to drive the shaft at 760 RPM is %4.4f Watts',smax,ss,P)
diff --git a/3532/CH8/EX8.3.1/Ex8_2.sce b/3532/CH8/EX8.3.1/Ex8_2.sce new file mode 100644 index 000000000..61792beb2 --- /dev/null +++ b/3532/CH8/EX8.3.1/Ex8_2.sce @@ -0,0 +1,30 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.3.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+m=4//mass of rotor in kg
+g=9.81//acc due to gravity in m/sec^2
+d=0.009//dia of shaft in m
+I=(%pi/64)*d^4///moment of area in m^4
+l=0.48//bearing span in m
+e=0.003//distance of CG away from geometric centre of rotor in mm
+N=760//speed of shaft in RPM
+c=49//equivalent viscous damping in N-sec/m
+//calculations
+K=48*E*I/l^3//stiffness of shaft in N/m
+Wn=sqrt(K/m)
+W=2*%pi*N/60
+bet=(W/Wn)
+zeta=c/(2*sqrt(K*m))
+r=e*(bet^2/sqrt(((1-bet^2)^2+(2*zeta*bet)^2)))//from Eqn 8.3.4 ,Sec 8.3
+Fd=sqrt((K*r)^2+(c*W*r)^2)//dynamic load on bearing in N
+Fs=m*g//static load in N
+Fmax=Fd+Fs//maximum static load on the shaft under dynamic condition in N
+smax=(Fmax*l/4)*(d/2)/I//maximum stress under dynamic condition in N/m^2
+ss=(Fs*l/4)*(d/2)/I//maximum stress under dead load condition in N/m^2
+Fdamp=(c*W*r)//damping force in N
+Tdamp=Fdamp*r//damping torque in N-m
+P=2*%pi*N*Tdamp/60//power in Watts
+//output
+mprintf(' The mamximum stress in the shaft under dynamic condition is %.3f N/(m^2)\n The dead load stress is %.3f N/(m^2)\n The power required to drive the shaft at 760 RPM is %4.4f Watts',smax,ss,P)
diff --git a/3532/CH8/EX8.3/Ex8_3.sce b/3532/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..bb812c820 --- /dev/null +++ b/3532/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.4.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+p=M1*a(1,1)+M2*a(2,2)//from Eqn 8.4.6 ,Sec 8.4
+q=M1*M2*(a(1,1)*a(2,2)-(a(1,2)^2))//from Eqn 8.4.6 ,Sec 8.4
+Wn1=sqrt((p-sqrt(p^2-4*q))/(2*q))//from Eqn 8.4.6 ,Sec 8.4
+Wn2=sqrt((p+sqrt(p^2-4*q))/(2*q))//from Eqn 8.4.6 ,Sec 8.4
+Nc1=Wn1*60/(2*%pi)//critical speed in RPM
+Nc2=Wn2*60/(2*%pi)//critical speed in RPM
+//output
+mprintf(' The critical speeds for the system shown in fig 7.2.1 are %4.4f RPM and %4.4f RPM',Nc1,Nc2)
diff --git a/3532/CH8/EX8.4.1/Ex8_3.sce b/3532/CH8/EX8.4.1/Ex8_3.sce new file mode 100644 index 000000000..bb812c820 --- /dev/null +++ b/3532/CH8/EX8.4.1/Ex8_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.4.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+I=4*10^-7//moment of area in m^4
+M1=100;M2=50//mass of discs 1 and 2 in Kgs
+c=0.18//distance of disc 1 from support in m
+l=0.3//distance of disc 2 from support in m
+g=9.81//aceleration due to gravity in m/sec^2
+//calculations
+a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)*(3*l-c)),(l^3/(3*E*I))]//from SOM
+p=M1*a(1,1)+M2*a(2,2)//from Eqn 8.4.6 ,Sec 8.4
+q=M1*M2*(a(1,1)*a(2,2)-(a(1,2)^2))//from Eqn 8.4.6 ,Sec 8.4
+Wn1=sqrt((p-sqrt(p^2-4*q))/(2*q))//from Eqn 8.4.6 ,Sec 8.4
+Wn2=sqrt((p+sqrt(p^2-4*q))/(2*q))//from Eqn 8.4.6 ,Sec 8.4
+Nc1=Wn1*60/(2*%pi)//critical speed in RPM
+Nc2=Wn2*60/(2*%pi)//critical speed in RPM
+//output
+mprintf(' The critical speeds for the system shown in fig 7.2.1 are %4.4f RPM and %4.4f RPM',Nc1,Nc2)
diff --git a/3532/CH8/EX8.4/Ex8_4.sce b/3532/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..4659fd16d --- /dev/null +++ b/3532/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.6.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+M=10//mass of rotor in kg
+g=9.81//acc due to gravity in m/sec^2
+ra=0.12//radius of gyration in m
+l=0.3//lenght of steel shaft in m
+b=0.06//thickness of rotor in m
+I=10*10^-8//moment of inertia of section in m^4
+//calculations
+r=sqrt((ra^2/2)+(b^2/12))
+h=3*(r^2)/l^2//from Eqn 8.6.4 ,Sec 8.6
+g1=sqrt((2/h)*((h+1)-sqrt((h+1)^2-h)))//natural frequency,from Eqn 8.6.4 ,Sec 8.6
+g2=sqrt((2/h)*((h+1)+sqrt((h+1)^2-h)))//natural frequency,from Eqn 8.6.4 ,Sec 8.6
+W1=g1*sqrt(3*E*I/(M*l^3))//from Eqn 8.6.4 ,Sec 8.6
+W2=g2*sqrt(3*E*I/(M*l^3))//from Eqn 8.6.4 ,Sec 8.6
+Nc1=W1*60/(2*%pi)//critical speed in RPM
+Nc2=W2*60/(2*%pi)//critical speed in RPM
+//output
+mprintf(' The operating speed of 10000 RPM is not near to either of \n the critical speeds i.e %4.4f RPM or %4.4f RPM.\n Therefore the operating speed is safe.',Nc1,Nc2)
diff --git a/3532/CH8/EX8.6.1/Ex8_4.sce b/3532/CH8/EX8.6.1/Ex8_4.sce new file mode 100644 index 000000000..4659fd16d --- /dev/null +++ b/3532/CH8/EX8.6.1/Ex8_4.sce @@ -0,0 +1,22 @@ +clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 8.6.1\n')
+//given data
+E=1.96*10^11//youngs modulus in N/m^2
+M=10//mass of rotor in kg
+g=9.81//acc due to gravity in m/sec^2
+ra=0.12//radius of gyration in m
+l=0.3//lenght of steel shaft in m
+b=0.06//thickness of rotor in m
+I=10*10^-8//moment of inertia of section in m^4
+//calculations
+r=sqrt((ra^2/2)+(b^2/12))
+h=3*(r^2)/l^2//from Eqn 8.6.4 ,Sec 8.6
+g1=sqrt((2/h)*((h+1)-sqrt((h+1)^2-h)))//natural frequency,from Eqn 8.6.4 ,Sec 8.6
+g2=sqrt((2/h)*((h+1)+sqrt((h+1)^2-h)))//natural frequency,from Eqn 8.6.4 ,Sec 8.6
+W1=g1*sqrt(3*E*I/(M*l^3))//from Eqn 8.6.4 ,Sec 8.6
+W2=g2*sqrt(3*E*I/(M*l^3))//from Eqn 8.6.4 ,Sec 8.6
+Nc1=W1*60/(2*%pi)//critical speed in RPM
+Nc2=W2*60/(2*%pi)//critical speed in RPM
+//output
+mprintf(' The operating speed of 10000 RPM is not near to either of \n the critical speeds i.e %4.4f RPM or %4.4f RPM.\n Therefore the operating speed is safe.',Nc1,Nc2)
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