diff options
Diffstat (limited to '1325')
168 files changed, 1782 insertions, 0 deletions
diff --git a/1325/CH10/EX10.1/10_1.PNG b/1325/CH10/EX10.1/10_1.PNG Binary files differnew file mode 100644 index 000000000..cff240e51 --- /dev/null +++ b/1325/CH10/EX10.1/10_1.PNG diff --git a/1325/CH10/EX10.1/10_1.sce b/1325/CH10/EX10.1/10_1.sce new file mode 100644 index 000000000..2afff4e9d --- /dev/null +++ b/1325/CH10/EX10.1/10_1.sce @@ -0,0 +1,7 @@ +//To find pitch diameter
+clc
+//given
+Teeth=48
+pitch=.75 //in
+D=Teeth*pitch/%pi
+printf("The pitch diameter is %.3f in",D)
diff --git a/1325/CH10/EX10.2/10_2.PNG b/1325/CH10/EX10.2/10_2.PNG Binary files differnew file mode 100644 index 000000000..d3c208707 --- /dev/null +++ b/1325/CH10/EX10.2/10_2.PNG diff --git a/1325/CH10/EX10.2/10_2.sce b/1325/CH10/EX10.2/10_2.sce new file mode 100644 index 000000000..e639b50f2 --- /dev/null +++ b/1325/CH10/EX10.2/10_2.sce @@ -0,0 +1,8 @@ +//To find pitch diameter and the circular pitch
+clc
+//given
+T=48//teeth
+pd=4//diametral pitch
+D=T/pd//pitch diameter
+p=%pi/pd//the circular pitch
+printf("\nThe pitch diameter = %.f in\nThe circular pitch = %.4f in\n",D,p)
diff --git a/1325/CH10/EX10.3/10_3.PNG b/1325/CH10/EX10.3/10_3.PNG Binary files differnew file mode 100644 index 000000000..937c65dcc --- /dev/null +++ b/1325/CH10/EX10.3/10_3.PNG diff --git a/1325/CH10/EX10.3/10_3.sce b/1325/CH10/EX10.3/10_3.sce new file mode 100644 index 000000000..a82ca058e --- /dev/null +++ b/1325/CH10/EX10.3/10_3.sce @@ -0,0 +1,10 @@ +//Find pitch diameter and pitch module
+clc
+//given
+T=48
+m=6//mm ; module
+D=m*T
+p=%pi*m
+dia=D/10//cm
+P=p*0.0393700787//inches
+printf("\nPitch diameter = %.1f cm\nCircular pitch = %.4f in\n",dia,P)
diff --git a/1325/CH10/EX10.4/10_4.PNG b/1325/CH10/EX10.4/10_4.PNG Binary files differnew file mode 100644 index 000000000..d15cb9cfa --- /dev/null +++ b/1325/CH10/EX10.4/10_4.PNG diff --git a/1325/CH10/EX10.4/10_4.sce b/1325/CH10/EX10.4/10_4.sce new file mode 100644 index 000000000..de06f8dc8 --- /dev/null +++ b/1325/CH10/EX10.4/10_4.sce @@ -0,0 +1,19 @@ +//To find the smallest number of teeth
+clc
+//given
+phi=20*%pi/180
+//Solution a)
+ar=1
+t1=2*ar/sin(phi)^2//from equation 10.7
+T1=ceil(t1)
+//Solution b)
+aw=1
+t2=2*aw/((1+3*sin(phi)^2)^(1/2)-1)//from euation 10.6
+T2=ceil(t2)
+//solution c)
+t=1
+T=3
+A=(t/T)*(t/T+2)
+t3=2*aw*(t/T)/((1+A*sin(phi)^2)^(1/2)-1)//from 10.5
+T3=ceil(t3)
+printf("\nSmallest number of teeth theoretically required in order to avoid interference on a pinion which is to gear with\na) A rack , t= %.f\nb) An equal pinion , t= %.f\nc) A wheel to give a ratio of 3 to 1 , t= %.f\n",T1,T2,T3)
diff --git a/1325/CH10/EX10.5/10_5.PNG b/1325/CH10/EX10.5/10_5.PNG Binary files differnew file mode 100644 index 000000000..5de659026 --- /dev/null +++ b/1325/CH10/EX10.5/10_5.PNG diff --git a/1325/CH10/EX10.5/10_5.sce b/1325/CH10/EX10.5/10_5.sce new file mode 100644 index 000000000..d427f8609 --- /dev/null +++ b/1325/CH10/EX10.5/10_5.sce @@ -0,0 +1,14 @@ +//To find the addendum required
+clc
+//given
+t=25
+phi=20*%pi/180
+//let pitch be 1
+R=t/(2*%pi)//R=t*p/(2*%pi)
+Larc=1.6//1.6*p
+//AB=Larc*cos(phi)
+AB=Larc*cos(phi)
+Ra=(4.47+13.97)^(1/2)//by simplifying AB+2{(Ra^2-R^2*cos(phi)^2)-R*sin(phi)} and using p =1
+Addendum=Ra-R
+//writing p in place of p=1
+printf("\nAddendum required = %.2fp",Addendum)
diff --git a/1325/CH10/EX10.6/10_6.PNG b/1325/CH10/EX10.6/10_6.PNG Binary files differnew file mode 100644 index 000000000..40f390ef7 --- /dev/null +++ b/1325/CH10/EX10.6/10_6.PNG diff --git a/1325/CH10/EX10.6/10_6.sce b/1325/CH10/EX10.6/10_6.sce new file mode 100644 index 000000000..3d93319c1 --- /dev/null +++ b/1325/CH10/EX10.6/10_6.sce @@ -0,0 +1,29 @@ +//to find thelength of path of contact and the length of arc of contact
+clc
+//let module be 1
+m=1
+t1=28
+t2=45
+r=t1*m/2
+R=t2*m/2
+ra=r+m
+Ra=R+m
+phi1=14.5*%pi/180
+//10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
+//AB=A+B-C
+A=m*(ra^2-r^2*cos(phi1)^2)^(1/2)
+B=m*(Ra^2-R^2*cos(phi1)^2)^(1/2)
+C=m*(r+R)*sin(phi1)
+AB=A+B-C
+p=%pi*m
+ABp=AB/%pi
+arc1=ABp/cos(phi1)//length of arc of contact
+phi2=20*%pi/180
+//10.8 => AB =(ra^2-r^2*cos(phi)^2)^(1/2)+(Ra^2-R^2*cos(phi)^2)^(1/2)-(r+R)*sin(phi)
+a=m*(ra^2-r^2*cos(phi2)^2)^(1/2)
+b=m*(Ra^2-R^2*cos(phi2)^2)^(1/2)
+c=m*(r+R)*sin(phi2)
+ab=a+b-c
+abp=ab/%pi
+arc2=abp/cos(phi2)//length of arc of contact
+printf("\nLength of path of contact\nWhen phi = 14.5 degrees = %.3fm\nWhen phi = 20 degrees = %.2fm\nLength of arc of contact\nWhen phi = 14.5 degrees = %.2fp\nWhen phi = 20 degrees = %.3fp\n",AB,ab,arc1,arc2)
diff --git a/1325/CH11/EX11.1/11_1.PNG b/1325/CH11/EX11.1/11_1.PNG Binary files differnew file mode 100644 index 000000000..1168950fb --- /dev/null +++ b/1325/CH11/EX11.1/11_1.PNG diff --git a/1325/CH11/EX11.1/11_1.sce b/1325/CH11/EX11.1/11_1.sce new file mode 100644 index 000000000..07a5c1348 --- /dev/null +++ b/1325/CH11/EX11.1/11_1.sce @@ -0,0 +1,18 @@ +//to find gear train suitable for connecting the spindle
+clc
+//given
+Ns=26//rpm of spindle
+N1=4//rpm of lead screw
+//the only wheel in the set of which 13 is a factor is that with 65 teeth
+T1=65
+T2=25//to satisfy the Ns/n1 ratio and to select from given set
+T3=75//to satisfy the Ns/n1 ratio and to select from given set
+T4=T1*T3*N1/(Ns*T2)
+//solution b
+Ns1=35
+N1=4
+Tb1=105//to satisfy the Ns/n1 ratio and to select from given set
+Tb2=30//to satisfy the Ns/n1 ratio and to select from given set
+Tb3=100//to satisfy the Ns/n1 ratio and to select from given set
+Tb4=Tb1*Tb3*N1/(Ns1*Tb2)
+printf("\na) The change wheel used will have %.f, %.f, %.f and %.f teeths\nb) The change wheel used will have %.f, %.f, %.f and %.f teeths",T1,T2,T3,T4,Tb1,Tb2,Tb3,Tb4)
diff --git a/1325/CH11/EX11.10/11_10.PNG b/1325/CH11/EX11.10/11_10.PNG Binary files differnew file mode 100644 index 000000000..6bc036ff9 --- /dev/null +++ b/1325/CH11/EX11.10/11_10.PNG diff --git a/1325/CH11/EX11.10/11_10.sce b/1325/CH11/EX11.10/11_10.sce new file mode 100644 index 000000000..c2a7cab17 --- /dev/null +++ b/1325/CH11/EX11.10/11_10.sce @@ -0,0 +1,20 @@ +//To find the ratio of engine speed to propeller shaft speed and the tooth loads for the third gear
+clc
+//given
+s1=26
+s2=24
+s3=23
+sr=31
+i1=70
+i2=72
+i3=61
+ir=71
+t=1500//lb in
+k1=-i3/s3//Ns3-Ni2/(Ni3-Ni2)=k
+//S3 is fixed thus
+k2=1-(1/k1)//k2=Ni3/Ni2
+k3=-i2/s2//k3=Ns2-Ni3/(Ni2-Ni3)
+k4=(1/k2-1)*k3+1//k4=Ns2/Ni3 ; reducing using k2 and k3
+k5=-i1/s1//Ns1-Nf/(Ni1-Nf)
+k6=(1-k5)/(1-k5/k4)//k6=Ns1/Nf
+printf("\n Ns1/Nf = %.2f",k6)
diff --git a/1325/CH11/EX11.2/11_2.PNG b/1325/CH11/EX11.2/11_2.PNG Binary files differnew file mode 100644 index 000000000..7d49b6579 --- /dev/null +++ b/1325/CH11/EX11.2/11_2.PNG diff --git a/1325/CH11/EX11.2/11_2.sce b/1325/CH11/EX11.2/11_2.sce new file mode 100644 index 000000000..be690e308 --- /dev/null +++ b/1325/CH11/EX11.2/11_2.sce @@ -0,0 +1,13 @@ +//To find the overall speed ratio
+clc
+//given
+v=15//ft/min
+d=2//ft
+N=450//rpm
+N1=d*v/(2*%pi)//rpm of barrel
+s=N/N1//total reduction speed required
+//With a minimum number of teeth = 20
+T=20
+T1=T*(s)^(1/3)
+R=(T1/T)^3
+printf("\nIf the minimum number of teeth is fixed at 20, the might be as follow ( %.f / 20 )^3 = %.1f\nThis is sufficiently close to the required ratio\n",T1,R)
diff --git a/1325/CH11/EX11.3/11_3.PNG b/1325/CH11/EX11.3/11_3.PNG Binary files differnew file mode 100644 index 000000000..cb97df3fc --- /dev/null +++ b/1325/CH11/EX11.3/11_3.PNG diff --git a/1325/CH11/EX11.3/11_3.sce b/1325/CH11/EX11.3/11_3.sce new file mode 100644 index 000000000..941a40c6d --- /dev/null +++ b/1325/CH11/EX11.3/11_3.sce @@ -0,0 +1,30 @@ +//To find number of teeth on each of the four wheels
+clc
+//given
+d=7//in; central distance
+k1=2*7*7//T1+t1/(2*7)=7
+k2=2*7*5//T2+t2/(2*5)=7
+G=9/1
+t1=(-(k1+k2)+((k1+k2)^2+4*(G-1)*(k1*k2))^(1/2))/(2*(G-1))
+a=ceil(t1)
+b=floor(t1)
+T1=k1-a
+T2=k2-a
+T3=k2-b
+G1=T1*T2/(a*a)
+G2=T1*T3/(a*b)
+dp=a/d
+//case b)
+tb1=23//let t1 = 23
+Tb1=k1-tb1
+Gb1=Tb1/tb1
+Gb2=G/Gb1
+tb2=k2/(Gb2+1)
+p=ceil(tb2)
+Tb2=k2-p
+l=Tb1-1
+m=tb1+1
+n=Tb2+1
+o=p-1
+Gb2=l*n/(m*o)
+printf("\na) No of teeth = %.f, %.f, %.f, %.f\nG = %.2f\n\nb) No of teeth = %.f, %.f, %.f, %.f\nG = %.2f\n\n",T1,T2,a,b,G2,l,m,n,o,Gb2)
diff --git a/1325/CH11/EX11.5/11_5.PNG b/1325/CH11/EX11.5/11_5.PNG Binary files differnew file mode 100644 index 000000000..7c0b9ff14 --- /dev/null +++ b/1325/CH11/EX11.5/11_5.PNG diff --git a/1325/CH11/EX11.5/11_5.sce b/1325/CH11/EX11.5/11_5.sce new file mode 100644 index 000000000..0fb9e55a4 --- /dev/null +++ b/1325/CH11/EX11.5/11_5.sce @@ -0,0 +1,12 @@ +//to find ratio of the speed of the driving shaft to the speed of the driven shaft
+clc
+//given
+Tb=27
+Tc=30
+Td=24
+Te=21
+k=Te*Tb/(Tc*Td)//k=Nd/Ne
+//by applying componendo and dividendo, using Ne=0 and reducing we get
+a=(1-k)//where a = Nd/Na
+b=1/a
+printf("\nThe ratio of the speed of driving shaft to the speed of driven shaft\n\nNa/Nd = %.2f",b)
diff --git a/1325/CH11/EX11.6/11_6.PNG b/1325/CH11/EX11.6/11_6.PNG Binary files differnew file mode 100644 index 000000000..ab73021a3 --- /dev/null +++ b/1325/CH11/EX11.6/11_6.PNG diff --git a/1325/CH11/EX11.6/11_6.sce b/1325/CH11/EX11.6/11_6.sce new file mode 100644 index 000000000..07ed9a83f --- /dev/null +++ b/1325/CH11/EX11.6/11_6.sce @@ -0,0 +1,19 @@ +//to find the speed of driven shaft
+clc
+//given
+Tb=75
+Tc=18
+Td=17
+Te=71
+N1=500//rpm
+k=Tb*Td/(Tc*Te)//k=Ne/Nb
+//case a)
+//using componendo and dividendo , Nb=0 and reducing we get
+a=1-k//a=Ne/Na
+Na=N1
+Ne=Na*a
+//case b)
+Na1=500//given
+Nb1=100//given
+Ne1=k*(Nb1-Na1)+Na1
+printf("\ncase a) Ne= %.3f rpm\ncase b) Ne= %.1f rpm\n",Ne,Ne1)
diff --git a/1325/CH11/EX11.8/11_8.PNG b/1325/CH11/EX11.8/11_8.PNG Binary files differnew file mode 100644 index 000000000..7fba1490d --- /dev/null +++ b/1325/CH11/EX11.8/11_8.PNG diff --git a/1325/CH11/EX11.8/11_8.sce b/1325/CH11/EX11.8/11_8.sce new file mode 100644 index 000000000..9be1d8ec3 --- /dev/null +++ b/1325/CH11/EX11.8/11_8.sce @@ -0,0 +1,15 @@ +//To find diameter of bicycle wheel
+clc
+//given
+Td=23
+Ta=19
+Tb=20
+Tc=22
+k=Td*Ta/(Tb*Tc)
+//using componendo and dividendo, Nc=0 and reducing we get
+a=1/k-1//a=Nd/Ne
+b=1/a//- denotes opposite direction
+d=5280*12/(%pi*5*b)
+p=ceil(d)
+printf("\nThe diameter must be = %.1f in\nThe numbers of teeths are therefore suitable for a cyclometer for bicycle with %.f inches wheels",d,p)
+
diff --git a/1325/CH12/EX12.10/12_10.PNG b/1325/CH12/EX12.10/12_10.PNG Binary files differnew file mode 100644 index 000000000..c9d1428c4 --- /dev/null +++ b/1325/CH12/EX12.10/12_10.PNG diff --git a/1325/CH12/EX12.10/12_10.sce b/1325/CH12/EX12.10/12_10.sce new file mode 100644 index 000000000..934336068 --- /dev/null +++ b/1325/CH12/EX12.10/12_10.sce @@ -0,0 +1,21 @@ +//to find the moment of inertia of the flying wheel
+clc
+//given
+ihp=25
+N=300//rpm
+Ks=2/100//given
+u=2.3//work done by gases during expansion is u(2.3) times that during compression
+E=ihp*33000/N//indicated work done per revolution
+E1=E*2//indicated work done per cycle
+We=E1/(1-1/u)//work done by gases during expansion
+AB=We*2/%pi//the maximum torque from fig 290
+AC=E/(2*%pi)//mean turning moment
+CB=AB-AC//maximum excess turning moment
+Ef=(CB/AB)^2*We//fluctuation of energy
+Ke=Ef/E
+w=%pi*N/30//angular speed
+g=32.2//ft/s^2
+moi=g*Ef/(w^2*Ks)//moment of inertia
+printf("Moment of inertia of the flywheel = %.f lb ft^2",moi)
+
+//answer is not EXACT due to the approximations in calculations done by the author of the book
diff --git a/1325/CH12/EX12.11/12_11.PNG b/1325/CH12/EX12.11/12_11.PNG Binary files differnew file mode 100644 index 000000000..dc3d38b1c --- /dev/null +++ b/1325/CH12/EX12.11/12_11.PNG diff --git a/1325/CH12/EX12.11/12_11.sce b/1325/CH12/EX12.11/12_11.sce new file mode 100644 index 000000000..f427e81a6 --- /dev/null +++ b/1325/CH12/EX12.11/12_11.sce @@ -0,0 +1,15 @@ +//To estimate the percentage variation from the mean speed
+clc
+//given
+N=100//rpm
+ke=1.93//As per given figure
+l=15//1 inch of fig = 15 ton ft
+x=40//degrees; 1 inch = 40 degree
+I=150//ton ft^2
+w=%pi*N/30//angular speed
+E=l*x*%pi/180//energy
+Ef=E*ke//fluctuation energy
+Ks=Ef*g/(w^2*I)//from equation 12.14
+p=Ks*100/2//dummy variables
+q=p*2//dummy variables
+printf("The total fluctuation of speed is %.2f percent and the variation in speed is %.2f percent on either side of the mean speed",q,p)
diff --git a/1325/CH12/EX12.2/12_2.PNG b/1325/CH12/EX12.2/12_2.PNG Binary files differnew file mode 100644 index 000000000..db0b465ed --- /dev/null +++ b/1325/CH12/EX12.2/12_2.PNG diff --git a/1325/CH12/EX12.2/12_2.sce b/1325/CH12/EX12.2/12_2.sce new file mode 100644 index 000000000..17e61b4c7 --- /dev/null +++ b/1325/CH12/EX12.2/12_2.sce @@ -0,0 +1,16 @@ +//To find torque exerted on driven shaft
+clc
+//given
+ne=31
+na=25
+nb=90
+nc=83
+Ta=10 //lbft
+//Ne-Nf/(Nc-Nf)=-83/31
+k=114/83//k=Nc/Nf As Ne = 0, on simplification we get Nc/Nf= 114/83
+j=-90/25//j=Na/Nb
+//Nc=Nb, Thus Na/Nc=-90/25
+//Na/Nf=(Na/Nc)*(Nc/Nf) ie Na/Nf=k*j
+//Tf*Nf=Ta*Na
+Tf=Ta*k*j
+printf("\nTorque exerted on driven shaft = %.1f lb.ft\n",Tf)
diff --git a/1325/CH12/EX12.3/12_3.PNG b/1325/CH12/EX12.3/12_3.PNG Binary files differnew file mode 100644 index 000000000..5ebcfb8d7 --- /dev/null +++ b/1325/CH12/EX12.3/12_3.PNG diff --git a/1325/CH12/EX12.3/12_3.sce b/1325/CH12/EX12.3/12_3.sce new file mode 100644 index 000000000..67f54f1ec --- /dev/null +++ b/1325/CH12/EX12.3/12_3.sce @@ -0,0 +1,22 @@ +//Find the torque exerted on the crankshaft
+clc
+//given
+D=9//in
+stroke=24//in
+d=2//in
+l=60//in
+CP=l
+N=120
+theta=40//degrees
+x=theta*%pi/180
+P1=160//lb/in^2
+P2=32//lb/in^2
+OC=stroke/2
+F=%pi*(D/2)^2*P1-%pi*(D/2)^2*P2+%pi*(d/2)^2*P2
+//Ft*Vc=F*Vp; Where Vc and Vp are velocities of crank and pin respectively
+//Vp/Vc=IP/IC=OM/OC - From similar triangles ; fig 274
+n=CP/OC
+OM=OC*(sin(x) + (sin(2*x)/(2*n)))//from 3.11
+T=F*OM/12//torque exerted on crankshaft
+Torque=floor(T)
+printf("The torque exerted on crankshaft= F*OM = %.f lb ft",Torque)
diff --git a/1325/CH12/EX12.4/12_4.PNG b/1325/CH12/EX12.4/12_4.PNG Binary files differnew file mode 100644 index 000000000..e04d2b2c1 --- /dev/null +++ b/1325/CH12/EX12.4/12_4.PNG diff --git a/1325/CH12/EX12.4/12_4.sce b/1325/CH12/EX12.4/12_4.sce new file mode 100644 index 000000000..5981534c0 --- /dev/null +++ b/1325/CH12/EX12.4/12_4.sce @@ -0,0 +1,18 @@ +//To find the total forces applied at point A and B
+clc
+//given
+AB=12.5//in
+IB=10.15//in
+IA=10.75//in
+IX=2.92//in
+IY=5.5//in
+w=3//lb
+Fi=5//lb
+Fa1=9//lb
+Fb1=(Fa1*IA-w*IY-Fi*IX)/IB
+//From the polygon of forces
+Fa2=7.66//lb
+Fb2=3.0//lb
+Fa=(Fa1^2+Fa2^2)^(1/2)
+Fb=(Fb1^2+Fb2^2)^(1/2)
+printf("\nThe total force applied to the link AB at the pin A = Fa = %.2f lb\nThe total force applied to the link AB at the pin B = Fb = %.2f lb\n",Fa,Fb)
diff --git a/1325/CH12/EX12.5/12_5.PNG b/1325/CH12/EX12.5/12_5.PNG Binary files differnew file mode 100644 index 000000000..9316f4168 --- /dev/null +++ b/1325/CH12/EX12.5/12_5.PNG diff --git a/1325/CH12/EX12.5/12_5.sce b/1325/CH12/EX12.5/12_5.sce new file mode 100644 index 000000000..852af91b0 --- /dev/null +++ b/1325/CH12/EX12.5/12_5.sce @@ -0,0 +1,27 @@ +//To find the inertia torque on the crankshaft
+clc
+//given
+CP=60//in
+l=CP/12
+a=41
+cg=19
+g=32.2//ft/s^2
+m1=580//lb
+Mr=500//lb
+n=5//from example 12.3
+x=40*%pi/180
+N=120
+r=1//ft
+k=25
+w=N*%pi/30
+Rm=m1+(cg/CP)*Mr
+fp=w^2*r*(cos(x)+cos(2*x)/n)
+Fp=-Rm*fp/g
+OM=0.7413//ft -from example 12.3
+Tp=Fp*OM//from 12.6
+L=a+k^2/a//length for simple equivalent pendulum
+L1=L/12
+Tc=-Mr*(a/12)*(l-L1)*w^2*sin(2*x)/(g*2*n^2)//from 12.10
+Tw=-Mr*a*cos(x)/(n*12)
+T=Tp+Tc+Tw
+printf("\nTp= %.f lbft\nTc = %.1f lbft\nTw = %.1f lbft\nTotal torque exerted on the crankshaft due to the inertia of the moving parts = Tp+Tc+tw = %.1f lbft",Tp,Tc,Tw,T)
diff --git a/1325/CH12/EX12.6/12_6.PNG b/1325/CH12/EX12.6/12_6.PNG Binary files differnew file mode 100644 index 000000000..51baf2958 --- /dev/null +++ b/1325/CH12/EX12.6/12_6.PNG diff --git a/1325/CH12/EX12.6/12_6.sce b/1325/CH12/EX12.6/12_6.sce new file mode 100644 index 000000000..06b3948e0 --- /dev/null +++ b/1325/CH12/EX12.6/12_6.sce @@ -0,0 +1,41 @@ +//To find the torque exerted on AB to overcome the inertia of the links and the forces which act on the pins B and C
+clc
+//given
+AB=2.5//in
+BC=7//in
+CD=4.5//in
+AD=8//in
+ED=2.3//from figure
+N=180
+w=N*%pi/30
+m=3//lb
+k=3.5//radius of gyration
+g=32.2//ft/s^2
+QT=1.35//inches from figure
+alpha=w^2*(QT/CD)
+Torque=m*(k/12)^2*alpha/g
+Torque1=Torque*12
+Tadd=m*ED//additional torque
+Tc=Tadd+Torque1//total torque
+Fc1=Tc/CD
+//link BC
+M=5//lb
+gA=1.8//in
+fg=w^2*(gA/12)
+F=M*fg/g
+OaG=5.6//in
+Kg=2.9//in
+GZ=Kg^2/OaG
+//scaled from figure
+IB=9//in
+IC=5.8//in
+IX=2.49//in
+IY=1.93//in
+Fb1=(Fc1*IC+F*IX+M*IY)/IB
+Tor=Fb1*AB
+//from force polygon
+Fc2=1//lb
+Fb2=15.2//lb
+Fb=(Fb1^2+Fb2^2)^(1/2)
+Fc=(Fc1^2+Fc2^2)^(1/2)
+printf("\nThe torque which must be exerted on AB in order to overcome the inertia of the links = Fb1*AB = %.1f lb.in\nThe total force applied to the link BC \nAt pin C = %.2f lb\nAt pin B = %.1f lb\n",Tor,Fc,Fb)
diff --git a/1325/CH12/EX12.7/12_7.PNG b/1325/CH12/EX12.7/12_7.PNG Binary files differnew file mode 100644 index 000000000..1d62a37f2 --- /dev/null +++ b/1325/CH12/EX12.7/12_7.PNG diff --git a/1325/CH12/EX12.7/12_7.sce b/1325/CH12/EX12.7/12_7.sce new file mode 100644 index 000000000..5ad67dd87 --- /dev/null +++ b/1325/CH12/EX12.7/12_7.sce @@ -0,0 +1,13 @@ +//To find the actual speed, the number of poles on the alternator and the required value of Ks
+clc
+//given
+N=210//rpm
+w=N*%pi/30
+F=50
+p1=F*120/(N*2)//N*p=F*120
+p2=floor(p1)//no of poles must be a whole number ; P2=P/2
+p=2*p2
+N1=F*120/p
+n=3//no of impulse per second
+Ks=n/(6*p)//equation 12.13
+printf("\nKs = %.4f\n\nActual speed = %.1f rpm\nNumber of poles = %.f",Ks,N1,p)
diff --git a/1325/CH12/EX12.8/12_8.PNG b/1325/CH12/EX12.8/12_8.PNG Binary files differnew file mode 100644 index 000000000..311e3b6e9 --- /dev/null +++ b/1325/CH12/EX12.8/12_8.PNG diff --git a/1325/CH12/EX12.8/12_8.sce b/1325/CH12/EX12.8/12_8.sce new file mode 100644 index 000000000..163de9066 --- /dev/null +++ b/1325/CH12/EX12.8/12_8.sce @@ -0,0 +1,11 @@ +//to find the weight of flywheel
+clc
+//given
+N=120//rpm
+k=3.5//ft
+Ef=2500//ft lb
+Ks=.01
+g=32.2//ft/s^2
+w=%pi*N/30//angular velocity
+W=g*Ef/(w^2*k^2*Ks*2240)//Weight of flying wheel
+printf("\nWeight of flying wheel, W = %.2f tons",W)
diff --git a/1325/CH12/EX12.9/12_9.PNG b/1325/CH12/EX12.9/12_9.PNG Binary files differnew file mode 100644 index 000000000..73aa6020d --- /dev/null +++ b/1325/CH12/EX12.9/12_9.PNG diff --git a/1325/CH12/EX12.9/12_9.sce b/1325/CH12/EX12.9/12_9.sce new file mode 100644 index 000000000..300e9e1cc --- /dev/null +++ b/1325/CH12/EX12.9/12_9.sce @@ -0,0 +1,16 @@ +//To find the fluctuation of speed
+clc
+//given
+N=270//rpm
+ihp=35.8
+k=2.25//ft
+g=32.2//ft/s^2
+ke=1.93//from table on p 440
+E=ihp*33000/N
+Ef=ke*E
+w=%pi*N/30
+W=1000//lb
+MOI=2*W*k^2//moment of inertia of both wheel
+ks=Ef*g/(MOI*w^2)//formula for ks
+p=ks/2
+printf("The fluctuation speed is therefore %.4f or %.3f on either side of the mean speed",ks,p)
diff --git a/1325/CH13/EX13.1/13_1.PNG b/1325/CH13/EX13.1/13_1.PNG Binary files differnew file mode 100644 index 000000000..055cc2089 --- /dev/null +++ b/1325/CH13/EX13.1/13_1.PNG diff --git a/1325/CH13/EX13.1/13_1.sce b/1325/CH13/EX13.1/13_1.sce new file mode 100644 index 000000000..0160eb230 --- /dev/null +++ b/1325/CH13/EX13.1/13_1.sce @@ -0,0 +1,24 @@ +//to find equilibrium speed
+clc
+//given
+//all lengths are in inches
+W=120//lb
+w=15//lb
+AB=12
+BF=8
+BC=12
+BE=6.5
+g=35230//inches rpm
+//at Minimum radius
+AF=(AB^2-BF^2)^(1/2)
+CE=(BC^2-BE^2)^(1/2)
+k2=(BE*AF)/(CE*BF)
+N2=(((W/2)*(1+k2)+w)*g/(w*AF))^(1/2)
+//At MAximum radius
+BF1=10
+BE1=8.5
+AF1=(AB^2-BF1^2)^(1/2)
+CE1=(BC^2-BE1^2)^(1/2)
+k1=(BE1*AF1)/(CE1*BF1)
+N1=(((W/2)*(1+k1)+w)*g/(w*AF1))^(1/2)
+printf("\nN1 (corresponding maximum radius) = %.1f rpm\nN2 (corresponding minimum radius) = %.1f rpm",N1,N2)
diff --git a/1325/CH13/EX13.10/13_10.PNG b/1325/CH13/EX13.10/13_10.PNG Binary files differnew file mode 100644 index 000000000..5f9350e7e --- /dev/null +++ b/1325/CH13/EX13.10/13_10.PNG diff --git a/1325/CH13/EX13.10/13_10.sce b/1325/CH13/EX13.10/13_10.sce new file mode 100644 index 000000000..54d8fbba4 --- /dev/null +++ b/1325/CH13/EX13.10/13_10.sce @@ -0,0 +1,15 @@ +//Find the coefficient of insensitiveness at the extreme radii of rotaion
+clc
+//given
+fs=3//lb
+W=90//lb
+w=15//lb
+//fb=(fs/2)*(1+k)*(r/h) From equation 13.31
+k=1//All the arms are of equal length
+//fb=fs*(r/h)
+//comparing the above result with the one obtained from example 8 , F=(W+w)*(r/h), we get coefficient of insensitiveness = k = (N1-N2)/N = fs/(W+w)
+k=fs/(W+w)
+K=k*100
+printf("Coefficient of insensitiveness = %.3f",k)
+
+
diff --git a/1325/CH13/EX13.11/13_11.PNG b/1325/CH13/EX13.11/13_11.PNG Binary files differnew file mode 100644 index 000000000..73d51e5bd --- /dev/null +++ b/1325/CH13/EX13.11/13_11.PNG diff --git a/1325/CH13/EX13.11/13_11.sce b/1325/CH13/EX13.11/13_11.sce new file mode 100644 index 000000000..2dea2af6a --- /dev/null +++ b/1325/CH13/EX13.11/13_11.sce @@ -0,0 +1,16 @@ +//find the coefficient of insensitiveness at their extereme radii of rotation
+clc
+//given
+a=4.5//in
+b=2//in
+r1=2.5//in
+r2=4.5//in
+F2=12.25//lb
+F1=25.4//lb
+fs=1.5//lb
+fb=(fs/2)*(b/a)
+//At minimum radii
+k1=fb/F2
+//At maximum radii
+k2=fb/F1
+printf("Coefficient of insensitiveness\nAt minimum radii = %.4f\nAt maximum radii = %.4f\n",k1,k2)
diff --git a/1325/CH13/EX13.2/13_2.PNG b/1325/CH13/EX13.2/13_2.PNG Binary files differnew file mode 100644 index 000000000..d556f0b80 --- /dev/null +++ b/1325/CH13/EX13.2/13_2.PNG diff --git a/1325/CH13/EX13.2/13_2.sce b/1325/CH13/EX13.2/13_2.sce new file mode 100644 index 000000000..2635e037b --- /dev/null +++ b/1325/CH13/EX13.2/13_2.sce @@ -0,0 +1,30 @@ +//to find the weight of ball required and maximum equilibrium speed
+clc
+//given
+BG=4//in
+//solution a
+w=15//lb
+W=120//lb
+k=.720
+BD=10.08//in
+CE=BD
+DG=BD+BG
+//by equating quations 13.2 and 13.10 and reducing, we get
+w1=(W/2*(1+k))/(((W/2*(1+k)+w)*DG/(BD*w))-1)
+printf("\nWeight of ball = %.3f lb\n",w1)
+//solution b
+CD=6.5//in
+BC=12//in
+BF=10//in
+AB=12//in
+CG=(DG^2+CD^2)^(1/2)
+gama=atan(CD/DG)
+bita=asin(CD/BC)
+alpha1=asin(BF/AB)
+bita1=asin(8.5/BC)
+gama1=gama+bita1-bita
+F=((w1+W/2)*8.471*(tan(alpha1)+tan(bita1)))/(CG*cos(gama1))-(w1*tan(gama1))
+printf("F1= %.1f lb",F)
+r1=CG*sin(gama1)+1.5//radius of rotation
+N1=(30/%pi)*(F*32.2*12/(w1*r1))^(1/2)
+printf("\nr1= %.2f in\nN1= %.1f rpm",r1,N1)
diff --git a/1325/CH13/EX13.3/13_3.PNG b/1325/CH13/EX13.3/13_3.PNG Binary files differnew file mode 100644 index 000000000..265ad2f67 --- /dev/null +++ b/1325/CH13/EX13.3/13_3.PNG diff --git a/1325/CH13/EX13.3/13_3.sce b/1325/CH13/EX13.3/13_3.sce new file mode 100644 index 000000000..e69add3ec --- /dev/null +++ b/1325/CH13/EX13.3/13_3.sce @@ -0,0 +1,20 @@ +//to find the rate of stiffness of the spring and the equilibrium speed
+clc
+//given
+w=3//lb
+g=32.2
+N2=300
+w2=(N2*%pi/30)
+r2=3/12//ft
+N1=1.06*N2
+r1=4.5/12//ft
+a=4//in
+b=2//in
+ro=3.5/12//ft
+F2=w*w2^2*r2/g
+F1=F2*N1^2*r1/(N2^2*r2)
+p=2*a^2*(F1-F2)/(b^2*(r1-r2))
+Fc=F2+(F1-F2)*(.5/1.5)
+N=(Fc*g/(ro*w))^(1/2)*30/%pi
+Ns=ceil(N)
+printf("N = %.f rpm",Ns)
diff --git a/1325/CH13/EX13.4/13_4.PNG b/1325/CH13/EX13.4/13_4.PNG Binary files differnew file mode 100644 index 000000000..e2fa74f9a --- /dev/null +++ b/1325/CH13/EX13.4/13_4.PNG diff --git a/1325/CH13/EX13.4/13_4.sce b/1325/CH13/EX13.4/13_4.sce new file mode 100644 index 000000000..f56373e6c --- /dev/null +++ b/1325/CH13/EX13.4/13_4.sce @@ -0,0 +1,17 @@ +//to find the equivalent stiffness of the auxiliary spring referred to the sleeve
+clc
+//given
+w=5//lb
+g=32.2
+N2=240//rpm
+w2=(N2*%pi/30)
+r2=5/12//ft
+N1=1.05*N2
+r1=7/12//ft
+a=6//in
+b=4//in
+pb=3/2
+F2=w*w2^2*r2/g
+F1=F2*N1^2*r1/(N2^2*r2)
+p=2*(a/b)^2*((F1-F2)/(r1*12-r2*12)-4*pb)
+printf("Equivalent stiffness; p = %.f lb/in",p)
diff --git a/1325/CH13/EX13.5/13_5.PNG b/1325/CH13/EX13.5/13_5.PNG Binary files differnew file mode 100644 index 000000000..35d5c9be6 --- /dev/null +++ b/1325/CH13/EX13.5/13_5.PNG diff --git a/1325/CH13/EX13.5/13_5.sce b/1325/CH13/EX13.5/13_5.sce new file mode 100644 index 000000000..7c7f18d22 --- /dev/null +++ b/1325/CH13/EX13.5/13_5.sce @@ -0,0 +1,26 @@ +//to find the stiffness of the governor spring
+clc
+//given
+w=3//lb
+W=15//lb
+g=32.2
+r2=2.5/12//ft
+N2=240//rpm
+w2=N*%pi/30
+F2=w*w2^2*r2/g
+a=4.5//in
+b=2//in
+sleevelift=0.5
+r1=r2*12+a*sleevelift/b//the increase of radius for a scleeve lift is 0.5 in
+N1=1.05*N2
+F1=(N1/N2)^2*(r1/(r2*12))*F2
+//a) at minimum radius
+S2=(F2*a/b-w)*2-W
+//b) At maximum radius
+DB=r1-r2*12
+BI=1.936//in
+AD=a
+BI=b
+S1=2*(F1*AD/BI-w*(DB+BI)/BI)-W
+k=(S1-S2)/sleevelift
+printf("Stiffness of the spring is %.1f lb/in",k)
diff --git a/1325/CH13/EX13.6/13_6.PNG b/1325/CH13/EX13.6/13_6.PNG Binary files differnew file mode 100644 index 000000000..aff25f473 --- /dev/null +++ b/1325/CH13/EX13.6/13_6.PNG diff --git a/1325/CH13/EX13.6/13_6.sce b/1325/CH13/EX13.6/13_6.sce new file mode 100644 index 000000000..d44cf551a --- /dev/null +++ b/1325/CH13/EX13.6/13_6.sce @@ -0,0 +1,13 @@ +//To find governor effort and power
+clc
+//given
+c=0.01
+W=120//lb
+w=15//lb
+k=.720
+h=8.944//in
+Q=c*(W+2*w/(1+k))
+x=(2*c/(1+2*c))*(1+k)*h
+P=Q*x
+printf("Governor power = Q*x = %.3f in lb",P)
+
diff --git a/1325/CH13/EX13.7/13_7.PNG b/1325/CH13/EX13.7/13_7.PNG Binary files differnew file mode 100644 index 000000000..736032191 --- /dev/null +++ b/1325/CH13/EX13.7/13_7.PNG diff --git a/1325/CH13/EX13.7/13_7.sce b/1325/CH13/EX13.7/13_7.sce new file mode 100644 index 000000000..2b3eae35a --- /dev/null +++ b/1325/CH13/EX13.7/13_7.sce @@ -0,0 +1,18 @@ +//to find governor power
+clc
+//given
+r=6//in
+a=6//in
+b=4//in
+//from example 4(using conditions and calculating constants A and B) we get F=11.1r-14.6
+//when r=6 , F= 52
+F=52//lb
+inc=2*.01*52//increase neglecting very small values
+F1=F+inc
+F2=2*a*inc/b//Force required to prevent the sleeve from rising
+F3=F2/2//Force is uniformly distributed
+r2=-14.6/(F1/r-11.1)//from equation 1
+x=r2-r//increase in radius of rotation
+lift=b*x/a//sleeve lift
+P=F3*lift//governor power
+printf("Governor power = %.3f in lb",P)
diff --git a/1325/CH14/EX14.1/14_1.PNG b/1325/CH14/EX14.1/14_1.PNG Binary files differnew file mode 100644 index 000000000..e26609002 --- /dev/null +++ b/1325/CH14/EX14.1/14_1.PNG diff --git a/1325/CH14/EX14.1/14_1.sce b/1325/CH14/EX14.1/14_1.sce new file mode 100644 index 000000000..1dd2dbd28 --- /dev/null +++ b/1325/CH14/EX14.1/14_1.sce @@ -0,0 +1,18 @@ +//to find balance weights
+clc
+//given
+W=200//lb
+r=9//in
+b1=15//in
+bm=b1
+l=10//in
+d=50//in
+//case a
+ma=d+l
+Bm1=W*r*l/(d*bm)//From 14.2
+B11=W*r*ma/(d*b1)//from 14.3
+//case b
+mb=d-l
+Bm2=W*r*l/(d*bm)//from 14.2
+B12=W*r*mb/(d*b1)//from 14.3
+printf("\na) Bm= %.f lb ; B1= %.f lb\nb) Bm= %.f lb ; B1= %.f lb",Bm1,B11,Bm2,B12)
diff --git a/1325/CH14/EX14.12/14_12.PNG b/1325/CH14/EX14.12/14_12.PNG Binary files differnew file mode 100644 index 000000000..866428c5b --- /dev/null +++ b/1325/CH14/EX14.12/14_12.PNG diff --git a/1325/CH14/EX14.12/14_12.sce b/1325/CH14/EX14.12/14_12.sce new file mode 100644 index 000000000..b327e23cd --- /dev/null +++ b/1325/CH14/EX14.12/14_12.sce @@ -0,0 +1,15 @@ +//To find the resultant primary and secondary force
+clc
+//given
+N=1500 //rpm
+R=4//lb
+g=32.2//ft/s^2
+w=%pi*N/30
+stroke=5//in
+r=stroke/2
+l=9//in
+b=3.5//in
+B=(3/2)*R*r/b//primary force
+n=l/r
+F=(3/2)*R*w^2*r/(g*12*n)//secondary force
+printf("\nResultant primary force = %.2f lb\nResultant secondary force = %.f lb",B,F)
diff --git a/1325/CH14/EX14.13/14_13.PNG b/1325/CH14/EX14.13/14_13.PNG Binary files differnew file mode 100644 index 000000000..013f7ac86 --- /dev/null +++ b/1325/CH14/EX14.13/14_13.PNG diff --git a/1325/CH14/EX14.13/14_13.sce b/1325/CH14/EX14.13/14_13.sce new file mode 100644 index 000000000..a2fa10375 --- /dev/null +++ b/1325/CH14/EX14.13/14_13.sce @@ -0,0 +1,17 @@ +//To find maximum and minimum secondary force
+clc
+//given
+g=32.2//ft/s^2
+n=2000//rpm
+R=6//lb
+r=3//in
+L=11//in
+w=%pi*n/30
+n=L/r
+//minimum secondary force
+F1=2*R*w^2*r/(g*n*12)
+a=floor(F1)
+//maximum secondary force
+F2=6*R*w^2*r/(g*n*12)
+b=floor(F2)
+printf("\nMinimum secondary force = %.f lb\nMaximum secondary force = %.f lb",a,b)
diff --git a/1325/CH14/EX14.2/14_2.PNG b/1325/CH14/EX14.2/14_2.PNG Binary files differnew file mode 100644 index 000000000..1ef9f1a8d --- /dev/null +++ b/1325/CH14/EX14.2/14_2.PNG diff --git a/1325/CH14/EX14.2/14_2.sce b/1325/CH14/EX14.2/14_2.sce new file mode 100644 index 000000000..22762df48 --- /dev/null +++ b/1325/CH14/EX14.2/14_2.sce @@ -0,0 +1,21 @@ +//to find the position and magnitude of the balance weight required
+clc
+//given
+Wa=200//lb
+Wb=300//lb
+Wc=240//lb
+W1=260//lb
+ra=9//in
+rb=7//in
+rc=10//in
+r1=12//in
+R=24//in
+alpha=45*%pi/180
+bita=75*%pi/180
+gama=135*%pi/180
+Hb=Wa*ra+Wb*rb*cos(alpha)-Wc*rc*cos(gama-bita)-W1*r1*cos(bita)//horizontal component after resolving
+Vb=Wb*rb*sin(alpha)+Wc*rc*sin(gama-bita)-W1*r1*sin(bita)//vertical component after resolving
+Bb=(Hb^2+Vb^2)^(1/2)
+B=Bb/R
+theta=atand(Vb/Hb)
+printf("\nBalance weight required = %.1f lb\ntheta = %.2f degrees",B,theta)
diff --git a/1325/CH14/EX14.5/14_5.PNG b/1325/CH14/EX14.5/14_5.PNG Binary files differnew file mode 100644 index 000000000..5f11d0015 --- /dev/null +++ b/1325/CH14/EX14.5/14_5.PNG diff --git a/1325/CH14/EX14.5/14_5.sce b/1325/CH14/EX14.5/14_5.sce new file mode 100644 index 000000000..26624ab7a --- /dev/null +++ b/1325/CH14/EX14.5/14_5.sce @@ -0,0 +1,15 @@ +//To find the balance weight required and the residual unbalanced force
+clc
+//given
+W=180//lb
+R=150//lb
+c=.5
+N=300//rpm
+r=7.5/12//ft
+Bb=(W+c*R)*r*12
+b=6//in
+B=Bb/b
+w=(%pi*N)/30
+Uf=(1/2)*(R/g)*w^2*r
+a=floor(Uf)
+printf("Balance weight required = %.1f lb\n The resultant unbalanced force = %.f lb\n",B,a)
diff --git a/1325/CH15/EX15.1/15_1.PNG b/1325/CH15/EX15.1/15_1.PNG Binary files differnew file mode 100644 index 000000000..ea5469f2d --- /dev/null +++ b/1325/CH15/EX15.1/15_1.PNG diff --git a/1325/CH15/EX15.1/15_1.sce b/1325/CH15/EX15.1/15_1.sce new file mode 100644 index 000000000..e983bc7c9 --- /dev/null +++ b/1325/CH15/EX15.1/15_1.sce @@ -0,0 +1,21 @@ +//to find the frequencies of the free longitudinal, transverse and torsional vibrations
+clc
+//given
+W=.3*2240//lb
+l=36//in
+D=3//in
+k=15//in
+A=%pi*(D/2)^2
+E=30*10^6//youngs modulus
+C=12*10^6
+g=32.2//ft/s^2
+d=W*l/(A*E)
+Fl=187.8/(d)^(1/2)
+I=%pi*(d/2)^4
+d1=W*(l^3)*64/(3*E*%pi*(3^4))
+Ft=187.8/(d1)^(1/2)
+j=%pi*3^4/32
+q=C*j/l
+Ftor=(1/(2*%pi))*(q*g*12/(W*k^2))^(1/2)
+F1=Ftor*60
+printf("\na) Frequency of Longitudinal vibrations = %.f per min\nb) Frequency of the transverse vibrations = %.f per min\nc) Frequency of the torsional vibration = %.f per min",Fl,Ft,F1)
diff --git a/1325/CH15/EX15.11/15_11.PNG b/1325/CH15/EX15.11/15_11.PNG Binary files differnew file mode 100644 index 000000000..4a94d2d2c --- /dev/null +++ b/1325/CH15/EX15.11/15_11.PNG diff --git a/1325/CH15/EX15.11/15_11.sce b/1325/CH15/EX15.11/15_11.sce new file mode 100644 index 000000000..9be8a056d --- /dev/null +++ b/1325/CH15/EX15.11/15_11.sce @@ -0,0 +1,28 @@ +//to find the natural frequencies of the torsional vibration of the system when inertia is neglected and when it is taken into account
+clc
+//given
+g=32.3//ft/s^2
+l2=25.5//in
+d1=2.75//in
+d2=3.5//in
+C=12*10^6//modulus of rigidity
+G=1/0.6//given speed ratio
+Ib=54//lb in^2
+Ic=850//lb in^2
+Id=50000//lb in^2
+Id1=Id/G^2//15.62
+Ia=1500//lb in^2
+la=Id1/(Id1+Ia)*66.5
+J=%pi*d1^4/32
+q=C*J/la//torsional stiffness
+n=(1/(2*%pi))*(q*g*12/Ia)^(1/2)
+nf=n*60//for minutes
+//case b)
+Ib1=Ib+Ic/(G^2)
+a=63.15//in; distance of the node from rotor A (given)
+b=3.661//in; distance of the node from rotor A (given)
+N1=n*(la/a)^(1/2)
+N2=n*(la/b)^(1/2)
+N1f=N1*60//for minutes
+N2f=N2*60//for minutes
+printf("\na) The frequency of torsional vibrations n = %.1f per sec or %.f per min\nb) The fundamental frquency = %.1f per sec or %.f per min\n and the two node frequency = %.f per sec or %.f per min",n,nf,N1,N1f,N2,N2f)
diff --git a/1325/CH15/EX15.2/15_2.PNG b/1325/CH15/EX15.2/15_2.PNG Binary files differnew file mode 100644 index 000000000..1913f0ec1 --- /dev/null +++ b/1325/CH15/EX15.2/15_2.PNG diff --git a/1325/CH15/EX15.2/15_2.sce b/1325/CH15/EX15.2/15_2.sce new file mode 100644 index 000000000..dc8e933c9 --- /dev/null +++ b/1325/CH15/EX15.2/15_2.sce @@ -0,0 +1,24 @@ +//To find the natural frequencies of the longitudinal, transverse and torsional vibration of the system
+clc
+//given
+l1=3//ft
+l2=2//ft
+l=l1+l2//ft
+W=.5*2240//lb
+k=20//in
+d=2//in
+Wa=2*W/5
+E=30*10^6
+A=%pi*(d/2)^2
+d1=Wa*l1*12/(A*E)
+N1=187.8/(d1)^(1/2)
+I=%pi*(d)^4/64
+d2=W*(l1*12)^3*(l2*12)^3/(3*E*(l*12)^3*I)
+N2=187.8/(d2)^(1/2)
+C=12*10^6//given
+g=32.2//given
+J=%pi*d^4/32
+q=C*J*((1/(l1*12))+(1/(l2*12)))
+n=(1/(2*%pi))*(q*g*12/(W*k^2))^(1/2)
+N3=n*60
+printf("\na)Longitudinal vibration = %.f per min\nb)Transverse Vibration = %.f per min\nc)Torsional Vibration = %.f per min\n",N1,N2,N3)
diff --git a/1325/CH15/EX15.3/15_3.PNG b/1325/CH15/EX15.3/15_3.PNG Binary files differnew file mode 100644 index 000000000..5ab75a7c1 --- /dev/null +++ b/1325/CH15/EX15.3/15_3.PNG diff --git a/1325/CH15/EX15.3/15_3.sce b/1325/CH15/EX15.3/15_3.sce new file mode 100644 index 000000000..03cf93309 --- /dev/null +++ b/1325/CH15/EX15.3/15_3.sce @@ -0,0 +1,10 @@ +//to find frequency of the natural transverse vibration
+clc
+//given
+l=10//ft
+d=4//in
+E=30*10^6//youngs modulus
+d1=0.0882//inches; maximum deflection as shown in the figure
+N=207/(d1)^(1/2)//From 15.20
+printf("\nFrequency of natural transverse vibration = %.f per min",N)
+
diff --git a/1325/CH15/EX15.4/15_4.PNG b/1325/CH15/EX15.4/15_4.PNG Binary files differnew file mode 100644 index 000000000..b204f02c5 --- /dev/null +++ b/1325/CH15/EX15.4/15_4.PNG diff --git a/1325/CH15/EX15.4/15_4.sce b/1325/CH15/EX15.4/15_4.sce new file mode 100644 index 000000000..517438f67 --- /dev/null +++ b/1325/CH15/EX15.4/15_4.sce @@ -0,0 +1,14 @@ +//To find the resistance offered by the dashpot
+clc
+//given
+m=50//lb
+k=100//lb/in
+g=32.2//ft/s
+d=m/k//static deflection
+n=(1/(2*%pi))*(g*12/d)^(1/2)
+//part 2
+b=g*12/d
+a=(b/20.79)^(1/2)
+nd=(1/(2*%pi))*((b-(a/2)^2))^(1/2)
+A=nd/n
+printf("\nFrequency of free vibrations = %.3f per sec\nFrequency of damped vibrations = %.3f per sec \nThe ratio of the frequencies of damped and free vibrationsis %.3f \n",n,nd,A)
diff --git a/1325/CH15/EX15.5/15_5.PNG b/1325/CH15/EX15.5/15_5.PNG Binary files differnew file mode 100644 index 000000000..78dd674bc --- /dev/null +++ b/1325/CH15/EX15.5/15_5.PNG diff --git a/1325/CH15/EX15.5/15_5.sce b/1325/CH15/EX15.5/15_5.sce new file mode 100644 index 000000000..6c525fa5b --- /dev/null +++ b/1325/CH15/EX15.5/15_5.sce @@ -0,0 +1,18 @@ +//To find the ratio nd/n
+clc
+//given
+//damping torque is directly proposrtional to the angular velocity
+C=12*10^6//Modulus of rigidity
+l=3//ft
+d=1//in
+g=32.2//ft/s^2
+I=500//lb ft^2 ; moment of inertia
+J=%pi*d^4/32
+q=C*J/(l*12)
+n=(1/(2*%pi))*(q*g*12/(I*12^2))^(1/2)
+//part 2
+b1=(q*g*12/(I*12^2))
+a1=(b1/10.15)^(1/2)//by reducing equation 15.28
+nd=(1/(2*%pi))*(b1-(a1/2)^2)^(1/2)
+A=nd/n
+printf("\nThe frequency of natural vibration = %.2f per sec\nThe frequency of damped vibration = %.2f per sec\nThe ratio nd/n = %.3f\n",n,nd,A)
diff --git a/1325/CH15/EX15.6/15_6.PNG b/1325/CH15/EX15.6/15_6.PNG Binary files differnew file mode 100644 index 000000000..85bdcde90 --- /dev/null +++ b/1325/CH15/EX15.6/15_6.PNG diff --git a/1325/CH15/EX15.6/15_6.sce b/1325/CH15/EX15.6/15_6.sce new file mode 100644 index 000000000..4e62ef86d --- /dev/null +++ b/1325/CH15/EX15.6/15_6.sce @@ -0,0 +1,18 @@ +//to find the amplitude if the period of the applied force coincided with the natural period of vibration of the system
+clc
+//given
+m=20//lb
+k=50//lb/in
+F=30//lb
+w=50//sec^-1
+g=32.2//ft/s^2
+d=m/k
+x=F/k//extension of the spring
+b=g*12/d
+a=(b/30.02)^(1/2)//from equation 15.28
+D=1/((1-w^2/b)^2+a^2*w^2/b^2)^(1/2)
+Af=D*x//amplitude of forced vibration
+D=(b/a^2)^(1/2)//At resonance
+A=D*x//amplitude at resonance
+printf("\nAmplitude of forced vibrations = %.3f in\nAmplitude of the forced vibrations at resonance = %.2f in",Af,A)
+
diff --git a/1325/CH15/EX15.7/15_7.PNG b/1325/CH15/EX15.7/15_7.PNG Binary files differnew file mode 100644 index 000000000..1f95a6e9b --- /dev/null +++ b/1325/CH15/EX15.7/15_7.PNG diff --git a/1325/CH15/EX15.7/15_7.sce b/1325/CH15/EX15.7/15_7.sce new file mode 100644 index 000000000..840abf0f0 --- /dev/null +++ b/1325/CH15/EX15.7/15_7.sce @@ -0,0 +1,27 @@ +//to find the fraction of the applied force transmitted at 1200 rpm and the amplitude of forced vibrations of the machines at resonance
+clc
+//given
+e=1/30
+n=1200//rpm
+w=%pi*n/30
+m=3//lb
+g=32.2//ft/s^2
+stroke=3.5//in
+r=stroke/2
+k=(1+1/e)^(1/2)//nf/n=k
+d=(k/187.7)^2
+W=200//lb ; given
+s=W/d//combined stiffness
+p=1/14.1//As a^2/b=1/198
+T=((1+p^2*k^2/((1-k^2)^2+p^2*k^2)))^(1/2)//actual value of transmissibility
+F=(m/g)*w^2*r/12//maximum unbalanced force transmitted on the machine
+Fmax=F*T//maximum force transmitted to the foundation
+//case b
+E=((1+p^2)/(p^2))^(1/2)
+Nreso=215.5//rpm
+Fub=F*(Nreso/n)^2
+Ftmax=E*Fub
+D=E//dynamic magnifier
+del=Fub/152//static deflection
+A=del*D
+printf("\na) Maximum force transmitted at 1200 rpm = %.f lb\nb) The amplitude of the forced vibrations of the machine at resonance = %.3f in\n Force transmitted = %.f lb\n",Fmax,A,Fub)
diff --git a/1325/CH15/EX15.8/15_8.PNG b/1325/CH15/EX15.8/15_8.PNG Binary files differnew file mode 100644 index 000000000..14a885501 --- /dev/null +++ b/1325/CH15/EX15.8/15_8.PNG diff --git a/1325/CH15/EX15.8/15_8.sce b/1325/CH15/EX15.8/15_8.sce new file mode 100644 index 000000000..a7b64e5a3 --- /dev/null +++ b/1325/CH15/EX15.8/15_8.sce @@ -0,0 +1,24 @@ +//To find the frequency of the natural torsional oscillations of the system
+clc
+//given
+l1=11//in
+l2=10//in
+l3=15//in
+l4=4//in
+l5=10//in
+d1=3//in
+d2=5//in
+d3=3.5//in
+d4=7//in
+d5=5//in
+I1=1500//lb ft^2
+I2=1000//lb ft^2
+leq=3//in from 15.49
+g=32.2//ft/s^2
+C=12*10^6
+J=%pi*leq^4/32
+l=l1+l2*(leq/d2)^4+l3*(leq/d3)^4+l4*(leq/d4)^4+l5*(leq/d5)^4
+la=I2*l/(I1+I2)
+qa=C*J/la
+n=(1/(2*%pi))*(qa*g*12/(I1*12^2))^(1/2)
+printf("\nThe frequency of the natural torsional oscillation of the system = %.1f per sec",n)
diff --git a/1325/CH15/EX15.9/15_9.PNG b/1325/CH15/EX15.9/15_9.PNG Binary files differnew file mode 100644 index 000000000..b3c67de9a --- /dev/null +++ b/1325/CH15/EX15.9/15_9.PNG diff --git a/1325/CH15/EX15.9/15_9.sce b/1325/CH15/EX15.9/15_9.sce new file mode 100644 index 000000000..84df93a92 --- /dev/null +++ b/1325/CH15/EX15.9/15_9.sce @@ -0,0 +1,25 @@ +//To find the frequencies of the free torsional vibrations of the system
+clc
+//given
+Ia=2.5//ton ft^2
+Ib=7.5//ton ft^2
+Ic=3//ton ft^2
+g=32.2//ft/s^2
+AB=9.5//ft
+BC=25//ft
+d=8.5//in
+C=11.8*10^6//lb/in^2
+k=Ic/Ia//la/lc=k
+lc1=(25.6+(25.6^2-4*114.1)^(1/2))/2//from 1 and 2 , reducing using quadratic formula
+lc2=(25.6-(25.6^2-4*114.1)^(1/2))/2//from 1 and 2 , reducing using quadratic formula
+la1=lc1*k
+la2=lc2*k
+J=%pi*d^4/32
+q=C*J/(lc1*12)//torsional stiffness
+IC=Ic*2240*12^2/(g*12)//moment of inertia
+nc=(1/(2*%pi))*(q/IC)^(1/2)//fundamental frequency of vibration
+a1=nc*60
+a=floor(a1)
+n=16*(lc1/lc2)^(1/2)
+b=n*60
+printf("\nFundamental frequency of vibration = %.f per min\nTwo node frequency = %.f per min\n",a,b)
diff --git a/1325/CH2/EX2.1/2_1.PNG b/1325/CH2/EX2.1/2_1.PNG Binary files differnew file mode 100644 index 000000000..1382aabc4 --- /dev/null +++ b/1325/CH2/EX2.1/2_1.PNG diff --git a/1325/CH2/EX2.1/ex2_1.sce b/1325/CH2/EX2.1/ex2_1.sce new file mode 100644 index 000000000..130dbf689 --- /dev/null +++ b/1325/CH2/EX2.1/ex2_1.sce @@ -0,0 +1,49 @@ +//to find velocity and change in kinetic energy when impact between two spheres moving in a same line is a) INELASTIC , b) ELASTIC , c) e = 0.6
+clc
+//a) INELASTIC
+//for sphere 1 ,mass=m1 and initial velocity=u1
+//for sphere 2 ,mass=m2 and initial velocity=u2
+m1=100//lb
+u1=10//ft/s
+m2=50//lb
+u2=5//ft/s
+v=(m1*u1+m2*u2)/(m1+m2)
+//change in kinetic energy
+//initial kinetic energy = ke1
+ke1=(m1*(u1^2)+m2*(u2^2))/(2*32.2)
+//Kinetic Energy after inelastic colision = ke2
+ke2=((m1+m2)*8.333^2)/(2*32.2)
+//Change in Kinetic Energy =l
+l=ke1-ke2
+//b) Elastic
+// for a very short time bodies will have a common velocity given by v=8.333 ft/s
+// for a very short time bodies will have a common velocity given by v=8.333 ft/s
+//immidiately after impact ends the velocities for both the bodies are given by v1 and v2
+v1=2*v-u1
+v2=2*v-u2
+//c) Coeeficient of Restitution=0.6
+e=0.6
+ve1=(1+e)*v-e*u1
+ve2=(1+e)*v-e*u2
+ke3=(m1*(ve1^2)+m2*(ve2^2))/(2*32.2)
+loss=ke1-ke3
+printf("kinetic energy before collisio0n is %f ft lb\n",ke1)
+printf("\n")
+printf("a) INELASTIC\n")
+printf("\n")
+printf("velocity after collision is %f ft/s\n",v)
+printf("the Kinetic Energy after collision is %f ft lb\n",ke2)
+printf("the change in Kinetic Energy after collision is %f ft lb\n",l)
+printf("\n")
+printf("b) ELASTIC\n")
+printf("\n")
+printf("velocity of 1 after collision is %f ft/s\n",v1)
+printf("velocity of 2 after collision is %f ft/s\n",v2)
+printf("there is no loss of kinetic energy in case of elastic collision\n")
+printf("\n")
+printf("c) e=0.6\n")
+printf("\n")
+printf("velocity of 1 after collision is %f ft/s\n",ve1)
+printf("velocity of 2 after collision is %f ft/s\n",ve2)
+printf("the Kinetic Energy after collision is %f ft lb\n",ke3)
+printf("the change in Kinetic Energy after collision is %f ft lb\n",loss)
diff --git a/1325/CH2/EX2.10/2_10.PNG b/1325/CH2/EX2.10/2_10.PNG Binary files differnew file mode 100644 index 000000000..f1e993b2e --- /dev/null +++ b/1325/CH2/EX2.10/2_10.PNG diff --git a/1325/CH2/EX2.10/ex2_10.sce b/1325/CH2/EX2.10/ex2_10.sce new file mode 100644 index 000000000..bfbf4826a --- /dev/null +++ b/1325/CH2/EX2.10/ex2_10.sce @@ -0,0 +1,22 @@ +//to find the acceleration if mass is allowed to fall freely and when efficiency of the gearing were 90%
+//gravitaional force (g)=32.2 ft/s^2
+clc
+//given
+Ia=200//lb ft2
+Ib=15//lb ft2
+G=5//wb==5*wa
+m=150//lb
+r=8//in
+printf("\n")
+//the equivalent mass of the geared system referred to the circumference of the drum is given by
+//Me=(1/r)^2*(Ia+(G^2*Ib))
+Me=(12/r)^2*(Ia+(G^2*Ib))
+M=m+Me
+a=(m/M)*32.2//acceleration
+//if efficiency of gearing is 90% then Me=(1/r^2)*(Ia+(G^2*Ib)/n)
+n=.9
+Me1=(12/r)^2*(Ia+(G^2*Ib)/n)
+M1=Me1+m
+a1=(m/M1)*32.2
+printf("acceleration = %.2f ft/s2\n",a)
+printf("acceleration when gear efficiency is 0.9= %.2f ft/s2\n",a1)
diff --git a/1325/CH2/EX2.11/2_11.PNG b/1325/CH2/EX2.11/2_11.PNG Binary files differnew file mode 100644 index 000000000..c238d61a0 --- /dev/null +++ b/1325/CH2/EX2.11/2_11.PNG diff --git a/1325/CH2/EX2.11/ex2_11.sce b/1325/CH2/EX2.11/ex2_11.sce new file mode 100644 index 000000000..359a2b030 --- /dev/null +++ b/1325/CH2/EX2.11/ex2_11.sce @@ -0,0 +1,37 @@ +//to find the maximum acceleration of car on each gear
+//gravitaional force (g)=32.2 ft/s^2
+clc
+printf("\n")
+//let
+//S=displacement of car from rest with uniform acceleration a, the engine torque T assumed to remain ocnstant
+//v=final speed ofcar
+//G=gear ratio
+//r=effective radius
+//n=efficiency of transmission
+//M=mass of the car
+//Ia and Ib=moments of inertia of road whels and engine
+//formulas => F=29.5nG ; Me= 1648+$.54nG^2 ; a=32.2 F/Me
+//given
+G1=22.5
+G2=12.5
+G3=7.3
+G4=5.4
+n=.82//for 1st ,2nd and 3rd gear
+n4=.9//for 4th gear
+F1=29.5*n*G1
+F2=29.5*n*G2
+F3=29.5*n*G3
+F4=29.5*n4*G4
+//on reduction and putting values we get
+Me1=1648+4.54*n*G1^2
+Me2=1648+4.54*n*G2^2
+Me3=1648+4.54*n*G3^2
+Me4=1648+4.54*n4*G4^2
+a1=32.2*F1/Me1
+a2=32.2*F2/Me2
+a3=32.2*F3/Me3
+a4=32.2*F4/Me4
+printf("Maximum acceleration of car on top gear is %.2f ft/s^2 \n",a4)
+printf("Maximum acceleration of car on third gear is %.2f ft/s^2 \n",a3)
+printf("Maximum acceleration of car on second gear is %.2f ft/s^2 \n",a2)
+printf("Maximum acceleration of car on first gear is %.2f ft/s^2 \n",a1)
diff --git a/1325/CH2/EX2.12/2_12.PNG b/1325/CH2/EX2.12/2_12.PNG Binary files differnew file mode 100644 index 000000000..eade451e8 --- /dev/null +++ b/1325/CH2/EX2.12/2_12.PNG diff --git a/1325/CH2/EX2.12/ex2_12.sce b/1325/CH2/EX2.12/ex2_12.sce new file mode 100644 index 000000000..8282be87b --- /dev/null +++ b/1325/CH2/EX2.12/ex2_12.sce @@ -0,0 +1,12 @@ +//to find the couple supplied to shaft
+//gravitaional force (g)=32.2 ft/s^2
+clc
+printf("\n")
+//given
+I=40//lb ft2
+n=500//rpm
+w=%pi*n/30//angular velocity
+wp=2*%pi/5//angular velocity of precession
+I1=I/32.2
+T=I1*w*wp//gyroscopic couple
+printf("the couple supplied to the shaft= %.2f lb ft\n",T)
diff --git a/1325/CH2/EX2.13/2_13.PNG b/1325/CH2/EX2.13/2_13.PNG Binary files differnew file mode 100644 index 000000000..442f75d2f --- /dev/null +++ b/1325/CH2/EX2.13/2_13.PNG diff --git a/1325/CH2/EX2.13/ex2_13.sce b/1325/CH2/EX2.13/ex2_13.sce new file mode 100644 index 000000000..3e764705d --- /dev/null +++ b/1325/CH2/EX2.13/ex2_13.sce @@ -0,0 +1,22 @@ +//to find the gyroscopic reaction of the airscrew on the aeroplane when it has a) three blades and b)two blades
+//gravitaional force (g)=32.2 ft/s^2
+clc
+//given
+printf("\n")
+I=250//lb ft2
+n=1600//rpm
+v=150//mph
+r=500//ft
+w=%pi*160/3//angular velocity of rotation
+wp=(150*88)/(60*500)//angular velocity of precession
+//a) with three bladed screw
+//T=I*w*wp
+T=(250/32.2)*%pi*(160/3)*wp
+//b)with two bladed air screw
+//T1=2*I*w*wp*sin(o)
+printf("The magnitude of gyroscopic couple is given by %.0f lb ft\n",T)
+//Tix=T(1-cos(2o)) lb ft
+//T1y=Tsin(2o)) lb ft
+printf("The component gyroscopic couple in the vertical plane =%.0f(1-cos(2x)) lb ft\n",T)
+printf("The component gyroscopic couple in the horizontal plane =%.0f(sin(2x)) lb ft\n",T)
+// for direction refer the book example
diff --git a/1325/CH2/EX2.2/2_2.PNG b/1325/CH2/EX2.2/2_2.PNG Binary files differnew file mode 100644 index 000000000..52b96bdab --- /dev/null +++ b/1325/CH2/EX2.2/2_2.PNG diff --git a/1325/CH2/EX2.2/ex2_2.sce b/1325/CH2/EX2.2/ex2_2.sce new file mode 100644 index 000000000..d4c24527f --- /dev/null +++ b/1325/CH2/EX2.2/ex2_2.sce @@ -0,0 +1,42 @@ +//to find speed of truck immidiately after collision and the maximum deflection of spring during impact. Moreover if k=0.5 then determine hoe the final speedsw will be affected and amount of dissipated energy
+clc
+//given
+m1=15//tons
+u1=12//m/h
+m2=5//tons
+u2=8//m/h
+k=2//ton/in
+e1=0.5//coefficient of restitution
+printf("\n")
+//conservation of linear momentum
+v=(m1*u1+m2*u2)/(m1+m2)
+printf("velocity at the instant of collision is %.2f mph",v)
+e=(m1*m2*(88/60)^2*(u1-u2)^2)/(2*32.2*(u1+u2))
+printf("\n")
+printf("The difference between the kinetic energy before and during the impact is %.2f ft tons\n",e)
+//energy stored in spring equals energy dissipated
+//s=(1/2)*k*x^2
+//s=e
+//since there are 4 buffer springs ,4x^2=24 inches (2 ft=24 inches)
+x=((e*12)/4)^.5
+printf("Maximum deflection of the spring is %.2f in\n",x)
+// maximum force acting between pair of buffer = stiffness of spring*deflection
+f=k*x
+printf("Maximum force acting between each buffer is %.2f tons\n",f)
+//assuming perfectly elastic collision
+//for loaded truck
+v1=2*11-12
+//for unloaded truck
+v2=2*11-8
+printf("Speed of loaded truck after impact %.2f mph\n",v1)
+printf("speed of unloaded truck after impact %.2f mph\n",v2)
+//if coefficient of restitution =o.5
+//for loaded truck
+ve1=(1+.5)*11-.5*12
+//for unloaded truck
+ve2=(1+.5)*11-.5*8
+printf("Speed of loaded truck after impact when e=0.5 %.2f mph\n",ve1)
+printf("Speed of unloaded truck after impact when e=0.5 %.2f mph\n",ve2)
+//net loss of kinetic energy=(1-e^2)*energy stored in spring
+l=(1-(e1^2))*2//ft tons
+printf("Net loss of kinetic energy is %.2f ft tons\n",l)
diff --git a/1325/CH2/EX2.3/2_3.PNG b/1325/CH2/EX2.3/2_3.PNG Binary files differnew file mode 100644 index 000000000..4f2e09edc --- /dev/null +++ b/1325/CH2/EX2.3/2_3.PNG diff --git a/1325/CH2/EX2.3/ex2_3.sce b/1325/CH2/EX2.3/ex2_3.sce new file mode 100644 index 000000000..091910bcb --- /dev/null +++ b/1325/CH2/EX2.3/ex2_3.sce @@ -0,0 +1,29 @@ +//to find maximum twist,apeed of flywheels when twist is maximum and when springs regains its shape
+clc
+//given
+m1=500//lb ft^2
+m2=1500//lb ft^2
+k=150//lb ft^2
+w1=150//rpm
+//angular momentum will be conserved as net external force is zero
+//let final angular velocity be N then (m1+m2)N=w1*m1
+N=(w1*m1)/(m1+m2)
+printf("Angular velocity at the instant when speeds of the flywheels are equalised is given by %.2f r.p.m\n",N)
+//kinetic energy at this instance
+ke1=(1/2)*((m1+m2)/32.2)*((%pi*N)/30)^2
+printf("The kinetic energy of the system at this instance is %.2f ft lb\n",ke1)
+printf("which is almost equal to 480 ft lb \n")
+//initial kinetic energy
+ke0=(1/2)*((m1)/32.2)*((%pi*w1)/30)^2
+printf("The initial kinetic energy of the system is %.2f ft lb\n",ke0)
+printf("which is almost equal to 1915 ft lb \n")
+//strain energy = s
+s=ke0-ke1
+printf("strain energy stored in the spring is %.2f ft lb which is approximately 1435 ft lb\n",s)
+//if x is the maximum anglular displacement of wheel and the mean torque applied by spring is i/2*k*x then work done or strain energy is given by 1/2 *k*x^2
+x=((1435*2)/150)^.5
+printf("Maximum angular displacement is %.2f in radians which is equal to 250 degrees\n",x)
+//na1 and na are initial and final speeds of the flywheel 1 and same nb1 and nb for flywheel 2
+na=2*N-w1//w1=na1
+nb=2*N-0//nb1=0
+printf ("Speed of flywheel a and b when spring regains its unstrained position are %.2f rpm and %.2f rpm respectively\n",na,nb)
diff --git a/1325/CH2/EX2.4/2_4.PNG b/1325/CH2/EX2.4/2_4.PNG Binary files differnew file mode 100644 index 000000000..10921a480 --- /dev/null +++ b/1325/CH2/EX2.4/2_4.PNG diff --git a/1325/CH2/EX2.4/ex2_4.sce b/1325/CH2/EX2.4/ex2_4.sce new file mode 100644 index 000000000..f0f9e39dd --- /dev/null +++ b/1325/CH2/EX2.4/ex2_4.sce @@ -0,0 +1,18 @@ +//to find length of equivalent simple pendulum
+//gravitaional force (g)=32.2 ft/s^2
+clc
+//given
+m1=150 //lb
+l=3//ft
+//number of oscillation per second is given by n
+printf("\n")
+n=(50/92.5)
+printf ("number of oscillation per second = %.2f\n",n)
+//length of simple pendulum is given by L=g/(2*%pi*n)^2
+L=32.2/(2*%pi*n)^2
+printf ("length of simple pendulum = %.2f ft\n",L)
+// distance of cg from point of suspension is given by a
+a=25/12
+k=(a*(L-a))^.5//radius of gyration
+moi=m1*k^2
+printf("The moment of inertia of rod is %.2f lb ft^2",moi)
diff --git a/1325/CH2/EX2.5/2_5.PNG b/1325/CH2/EX2.5/2_5.PNG Binary files differnew file mode 100644 index 000000000..93d992588 --- /dev/null +++ b/1325/CH2/EX2.5/2_5.PNG diff --git a/1325/CH2/EX2.5/ex2_5.sce b/1325/CH2/EX2.5/ex2_5.sce new file mode 100644 index 000000000..a075a31cd --- /dev/null +++ b/1325/CH2/EX2.5/ex2_5.sce @@ -0,0 +1,24 @@ +//to find moment of inertia and distance of cg from small end centre
+clc
+//let l1 and l2 be length of equivalent simple pendulum when axis coincides with small end and big end respectively
+//n1 and n2 =corresponding frequencies of oscillation per second
+n1=50/84.4
+n2=50/80.3
+//let a1 and a2 = distances of cg from small end and big end centers respectively
+//gravitaional force (g)=32.2 ft/s^2
+//L=g/(2*%pi*n)
+L1=(32.2*12)*(84.4/(100*%pi))^2
+L2=(32.2*12)*(80.3/(100*%pi))^2
+//a1(L1-a1)=k^2=a2(L2-a2) and a1+a2=30 inches
+//substituting and solving for a we get
+a1=141/6.8
+a2=30-a1
+k=(a1*(L1-a1))^.5
+moi=90*(149/144)//moi=m*k^2
+printf("length of equivalent simple pendulum when axis coincides with small end and big end respectively-\n")
+printf("L1=%.1f in\n",L1)
+printf("L2=%.1f in\n",L2)
+printf("distances of cg from small end and big end centers respectively are-\n")
+printf("a1=%.1f in\n",a1)
+printf("a2=%.1f in\n",a2)
+printf("Moment of inertia of rod =%.2f lb ft^2",moi)
diff --git a/1325/CH2/EX2.6/2_6.PNG b/1325/CH2/EX2.6/2_6.PNG Binary files differnew file mode 100644 index 000000000..06f59b237 --- /dev/null +++ b/1325/CH2/EX2.6/2_6.PNG diff --git a/1325/CH2/EX2.6/ex2_6.sce b/1325/CH2/EX2.6/ex2_6.sce new file mode 100644 index 000000000..800f4f6be --- /dev/null +++ b/1325/CH2/EX2.6/ex2_6.sce @@ -0,0 +1,16 @@ +//to find radius of gyration about the mass centre
+//gravitaional force (g)=32.2 ft/s^2
+clc
+//given
+printf("\n")
+m1=150
+l=8.5
+g=32.2
+a=83.2
+n=25
+//k=(a/2*%pi*n)*(g/l)^0.5
+k=(14*a*((g)^0.5))/(2*%pi*n*(l^0.5))
+k1=14.5/12
+printf("radius of gyration is %.2f inches which is equal to %.2f ft \n",k,k1)
+moi=m1*(k1^2)
+printf("moment of inertia=%.2f lb ft^2",moi)
diff --git a/1325/CH2/EX2.7/2_7.PNG b/1325/CH2/EX2.7/2_7.PNG Binary files differnew file mode 100644 index 000000000..2eb72ebde --- /dev/null +++ b/1325/CH2/EX2.7/2_7.PNG diff --git a/1325/CH2/EX2.7/ex2_7.sce b/1325/CH2/EX2.7/ex2_7.sce new file mode 100644 index 000000000..eeb018df4 --- /dev/null +++ b/1325/CH2/EX2.7/ex2_7.sce @@ -0,0 +1,26 @@ +//to find the equivalent dynamical system
+//gravitaional force (g)=32.2 ft/s^2
+clc
+printf("\n")
+//given
+m=2.5//lb
+a=6//in
+k=3.8//in
+l=9//in
+c=3//in
+w=22500
+//k^2=ab
+//case a) to find equivalent dynamic system
+b=(k^2)/a
+ma=(2.5*6)/8.42//m*a/a+b
+mb=m-ma
+printf("Mass ma =%.2f lb will be situated at 6 inches from cg and mb =%.2f lb will be situated at %.2f inches from cg in the equivalent dynamical system",ma,mb,b)
+printf("\n")
+//if two masses are situated at the bearing centres
+ma1=(2.5*6)/9
+mb1=m-ma1
+k1=(a*c)^.5
+//t=m*((k1^2)-(k^2))*w
+t=((2.5*(18-3.8^2))*22500)/(32.2*12*12)
+printf("correction couple which must be applied in order that the two mass system is dynamically equivalent to the rod is given by %.2f lb ft\n",t)
+
diff --git a/1325/CH2/EX2.8/2_8.PNG b/1325/CH2/EX2.8/2_8.PNG Binary files differnew file mode 100644 index 000000000..a36f35853 --- /dev/null +++ b/1325/CH2/EX2.8/2_8.PNG diff --git a/1325/CH2/EX2.8/ex2_8.sce b/1325/CH2/EX2.8/ex2_8.sce new file mode 100644 index 000000000..9a7fb5f5e --- /dev/null +++ b/1325/CH2/EX2.8/ex2_8.sce @@ -0,0 +1,17 @@ +//to find forces throught pin A and B in order to accelerate the link
+//gravitaional force (g)=32.2 ft/s^2
+clc
+printf("\n")
+m=20//lb
+g=32.2
+a=200//ft/s^2
+w=120//rad/s^2
+k=7//in
+f=(m/g)*a//effective force appllied to the link
+//this force acts parallel to the acceleration fg
+t=(m/g)*(k/12)^2*w//couple required in order to provide the angular acceleration
+//the line of action of F is therefore at a distance from G given by
+x=t/f
+printf("Effective force applied to the link is %.3f lb and the line of action of F is therefore at a distance from G given by %.3f ft \n",f,x)
+printf("F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved along the appropriate lines of action to give the magnitudes of Fa and Fb\n")
+printf("From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb\n")
diff --git a/1325/CH2/EX2.9/2_9.PNG b/1325/CH2/EX2.9/2_9.PNG Binary files differnew file mode 100644 index 000000000..29081ea64 --- /dev/null +++ b/1325/CH2/EX2.9/2_9.PNG diff --git a/1325/CH2/EX2.9/ex2_9.sce b/1325/CH2/EX2.9/ex2_9.sce new file mode 100644 index 000000000..02a79a83e --- /dev/null +++ b/1325/CH2/EX2.9/ex2_9.sce @@ -0,0 +1,20 @@ +//to find force that must be exerted in oeder to give an acceleration of 3ft/s^2 and smallest value of u(friction coefficient)
+//gravitaional force (g)=32.2 ft/s^2
+clc
+printf("\n")
+//given
+m=10//ton
+m2=1000//lb
+a=3//ft/s^2
+//the addition to actual mass in order to allow for the rotational inertia of the wheels and axles
+m1=2*(1000/2240)*(15/21)^2//m1=m2*k^2/r^2 and 1 ton=2240 lbs
+M=m+m1
+F=3*(10.46/32.2)//F=M.a
+f=F*2240//lb
+Fa=(2*1000/2240)*(3/32.2)*(15/21)^2//total tangential force required in order to provide the angular acceleration of the wheels and axles
+//Limiting friction force =uW
+//u*10>0.042
+u=0.042/10
+printf("The total tangential force required in order to provide the angular acceleration of the wheels and axles is %.4f ton\n",Fa)
+printf("If there is to be pure rolling ,u>%.4f",u)
+
diff --git a/1325/CH3/EX3.3/3_3.PNG b/1325/CH3/EX3.3/3_3.PNG Binary files differnew file mode 100644 index 000000000..9b300293c --- /dev/null +++ b/1325/CH3/EX3.3/3_3.PNG diff --git a/1325/CH3/EX3.3/3_3.sce b/1325/CH3/EX3.3/3_3.sce new file mode 100644 index 000000000..eb7bd3682 --- /dev/null +++ b/1325/CH3/EX3.3/3_3.sce @@ -0,0 +1,25 @@ +//To find velocities of point p, x and y
+clc
+//Given
+OC=6//in
+CP=24//in
+N=240//rpm
+X=45//degrees
+XP=19//in
+XC=6//in
+YP=32//in
+YC=9//in
+//Scalling off lenghts from fig , we have
+CI=2.77//in
+PI=2.33//in
+XI=2.33//in
+YI=3.48//in
+//Solution
+Vc=((%pi*N)/30)*(OC/12)//changing OP into feets
+printf("\nw=%.2f ft/s\n",Vc)
+//w=Vc/CI=Vp/PI=Vx/XI=Vy/YI
+w=Vc/CI
+Vp=w*PI
+Vx=w*XI
+Vy=w*YI
+printf("velocity of points P, X and Y are %.2f ft/s, %.2f ft/s and %.1f ft/s respectively",Vp,Vx,Vy)
diff --git a/1325/CH3/EX3.4/3_4.PNG b/1325/CH3/EX3.4/3_4.PNG Binary files differnew file mode 100644 index 000000000..dd3210507 --- /dev/null +++ b/1325/CH3/EX3.4/3_4.PNG diff --git a/1325/CH3/EX3.4/3_4.sce b/1325/CH3/EX3.4/3_4.sce new file mode 100644 index 000000000..b06a8be98 --- /dev/null +++ b/1325/CH3/EX3.4/3_4.sce @@ -0,0 +1,40 @@ +//To find accelerations of point p and x and angular acceleration of rod
+clc
+printf("\n")
+//Given
+OC=9//inches
+CP=36//inches
+XC=12//inches
+X=40//degrees
+CM=6.98//from the scaled figure
+N1=240//rpm
+N2=240//rpm (instantaneous) with angular aceleration (ao) 100 rad/s^2
+ao=100 //rad/s^2
+w=(%pi*N1/30)
+a=w^2*(OC/12)
+printf("Centripetal acceleration = %.f ft/s^2\n",a)
+Wr=w*CM/CP//rad/s^2
+f1=Wr^2*(CP/12)//centripetal component of acceleration of p realtive to C
+//Solution a)
+//given from fig 58(a)
+tp=296
+cp=306
+ox=422
+f2=tp //Tangential component of acceleration of p realtive to C
+f3=cp//acceleration of p realtive to C
+fx=ox//acce;eration of x
+ar=f2/(CP/12)//angular acceleration of rod
+printf("Case a) \nap= %.f ft/s^2,\nax= %.f ft/s^2 and\nar= %.1f rad/s^2 \n",f3,fx,ar)
+//Solution b)
+//given from fig 58(b)
+oc1=474
+oc=480
+pt=238
+pc=246
+xo=452
+f4=pt//Tangential component of acceleration of p realtive to C
+f5=pc//acceleration of p realtive to C
+Ar=f4/(CP/12)//angular acceleration of rod
+f6=ao*(OC/12)//tangential component of acceleration realtive to C
+Fx=xo//acce;eration of x
+printf("Case b) \nap= %.f ft/s^2,\nax= %.f ft/s^2 and\nar= %.1f rad/s^2 \n",f4,Fx,Ar)
diff --git a/1325/CH3/EX3.5/3_5.PNG b/1325/CH3/EX3.5/3_5.PNG Binary files differnew file mode 100644 index 000000000..58a01f4e8 --- /dev/null +++ b/1325/CH3/EX3.5/3_5.PNG diff --git a/1325/CH3/EX3.5/3_5.sce b/1325/CH3/EX3.5/3_5.sce new file mode 100644 index 000000000..91d122645 --- /dev/null +++ b/1325/CH3/EX3.5/3_5.sce @@ -0,0 +1,28 @@ +//To find angular acceleration of CD and BC
+clc
+//Given
+AB=2.5//inches
+BC=7//inches
+CD=4.5//inches
+DA=8//inches
+N=100//rpm
+X=60//degrees
+w=(%pi*N)/30
+//From triangle ABM we have
+AM=0.14//feet
+BM=0.12//feet
+Vb=w*AB/12//ft/s
+Vc=w*AM//ft/s
+Vcb=w*BM//ft/s
+fb=w^2*(AB/12)//ft/s^2
+bt=Vcb^2/(BC/12)//ft/s^2
+os=Vc^2/(CD/12)//ft/s^2
+//By measurement from acceleration diagram
+sc=19.1//ft/s^2
+tq=14.4//ft/s^2
+Acd=sc/(CD/12)
+Abc=tq/(BC/12)
+printf("\n")
+printf("Vb=%.2f ft/s \nVc=%.2f ft/s\nVcb=%.2f ft/s\nfb=%.2f ft/s^2\nbt=%.2f ft/s^2\nos=%.2f ft/s^2\n",Vb,Vc,Vcb,fb,bt,os)
+printf("Angular acceleration of CD(counter-clockwise)= %.1f rad/s^2 \n",Acd)
+printf("Angular acceleration of BC(counter-clockwise)= %.1f rad/s^2 \n",Abc)
diff --git a/1325/CH3/EX3.6/3_6.PNG b/1325/CH3/EX3.6/3_6.PNG Binary files differnew file mode 100644 index 000000000..2b4f0b611 --- /dev/null +++ b/1325/CH3/EX3.6/3_6.PNG diff --git a/1325/CH3/EX3.6/3_6.sce b/1325/CH3/EX3.6/3_6.sce new file mode 100644 index 000000000..d6143ffdf --- /dev/null +++ b/1325/CH3/EX3.6/3_6.sce @@ -0,0 +1,19 @@ +//To find the acceleration of P realative to the fixed point O
+clc
+//Given
+printf("\n")
+OP=2//ft
+f=4//ft/s^2
+w=2 //rad/s (anticlockwise)
+a=5 //rad/s^2 (anticlockwise)
+Vpq=3 //ft/s
+r=OP
+os=w^2*r//component 1
+sq=a*r//component 2
+qt=f//component 3
+tp=2*w*Vpq//component 4
+Aqo=(os^2+sq^2)^1/2//vector addition of component(a,b)
+Apq=(qt^2+tp^2)^1/2//vector addition of component(c,d)
+//Apo=Apq+Aqo (vector addition)
+Apo=((os-qt)^2+(sq+tp)^2)^(1/2)
+printf("Acceleration of P realative to fixed point O is %.1f ft/s^2",Apo)
diff --git a/1325/CH3/EX3.7/3_7.PNG b/1325/CH3/EX3.7/3_7.PNG Binary files differnew file mode 100644 index 000000000..3c2dea696 --- /dev/null +++ b/1325/CH3/EX3.7/3_7.PNG diff --git a/1325/CH3/EX3.7/3_7.sce b/1325/CH3/EX3.7/3_7.sce new file mode 100644 index 000000000..31fdca8a3 --- /dev/null +++ b/1325/CH3/EX3.7/3_7.sce @@ -0,0 +1,26 @@ +//to find velocity and acceleration of ram R
+clc
+printf("\n")
+//GIVEN
+OC=8//inches
+CP=4//inches
+N=60//inches
+ON=15//inches
+RN=6//inches
+X=120//degrees
+OP=10.6
+OQ=OP
+//from fig 65(a)
+Vq=1.56//ft/s
+Vrn=0.74//ft/s
+//from fig 65(b)
+ftq=3.74//ft/s^2
+ftrn=2.03//ft/s^2
+w1=(%pi*N)/30
+w=Vq/(OQ/12)
+wrn=Vrn/(RN/12)
+a=ftq/(OP/12)//Angular acceleration of ON
+a1=ftrn/(RN/12)//angular acceleration of RN
+printf("W=%.2f rad/s\nWrn=%.2f rad/s\n",w,wrn)
+printf("Angular acceleration of ON= %.2f rad/s^2\nAngular acceleration of RN=%.2f rad/s^2\n",a,a1)
+
diff --git a/1325/CH3/EX3.8/3_8.PNG b/1325/CH3/EX3.8/3_8.PNG Binary files differnew file mode 100644 index 000000000..73a34b777 --- /dev/null +++ b/1325/CH3/EX3.8/3_8.PNG diff --git a/1325/CH3/EX3.8/3_8.sce b/1325/CH3/EX3.8/3_8.sce new file mode 100644 index 000000000..639bd9610 --- /dev/null +++ b/1325/CH3/EX3.8/3_8.sce @@ -0,0 +1,34 @@ +//to find the velocity and acceleration of the piston along the cylinder, the angular velocity and angular acceleration of the connecting rod cp and the coriolis component of the acceleration of P
+clc
+//given
+OC=3//inches
+CP=9//inches
+N=1200 //rpm (clockwise)
+X=55 //degrees
+//from the figure 66
+OP=10.35//inches
+PM=10.74//inches
+OM=2.95//inches
+PC=12.84//inches
+PR=PC
+RV=2.49//inches
+UV=1.29//inches
+OU=5.90//inches
+PV=13.05//inches
+OV=6.06//inches
+OQ=OP
+//Solution
+w=(%pi*N)/30//the angular velocity of the cylinder line OP
+Vq=w*(OP/12)//the velocity of Q
+Vp=w*(PM/12)//The velocity of P
+w1=Vp/(CP/12)//The angular velocity of CP
+Vpq=w*(OM/12)//the velocity of sliding of the piston along the cylinder
+fq=w^2*(OQ/12)//the centripetal acceleration of Q
+Acp=w1^2*(PC/12)//The centripetal component of acceleration of P
+Atp=w^2*(RV/12)//The tangential component of acceleration of P
+acp=Atp/(CP/12)// The angular acceleration of the connecting rod CP
+f=w^2*(UV/12)//component c
+d=2*w*Vpq//component d
+Ap=w^2*PV//the resultant acceleration of P
+Apq=w^2*OV//the acceleration of P realative to Q
+printf("\nThe velocity and acceleration of the piston along the cylinder are %.1f ft/s and %.f ft/s^2 respectively\nThe angular velocity and angular acceleration of the connecting rod cp are %.1f rad/s and %.f rad/s^2 respectively\nAnd the coriolis component of the acceleration of P is %.f ft/s^2\n",Vpq,f,w1,acp,d)
diff --git a/1325/CH4/EX4.1/4_1.PNG b/1325/CH4/EX4.1/4_1.PNG Binary files differnew file mode 100644 index 000000000..c74e936c9 --- /dev/null +++ b/1325/CH4/EX4.1/4_1.PNG diff --git a/1325/CH4/EX4.1/4_1.sce b/1325/CH4/EX4.1/4_1.sce new file mode 100644 index 000000000..2d3094bdb --- /dev/null +++ b/1325/CH4/EX4.1/4_1.sce @@ -0,0 +1,23 @@ +//To find the extreme angular velocities of the driven shaft and its maximum acceleration
+clc
+//given
+rpm=1000
+angle=20//degree
+ang=(angle*%pi)/180
+printf("\n")
+w=2*%pi*rpm/60
+printf("The angular velocity of the driving shaft is %.1f rad/s \n",w)
+//maximum value of w1=w/cos(angle) and minimum value w2=w*cos(angle)
+w1=w/cos(ang)
+w2=w*cos(ang)
+printf("Extreme angular velocities :-\n")
+printf("maximum value of angular velocity w1=%.1f rad/s \nminimum value of angular velocity w2=%.1f rad/s\n",w1,w2)
+//using equation 4.11, cos(2x)=(2*sin(angle)^2)/(2-sin(angle)^2)
+x=acos((2*sin(ang)^2)/(2-sin(ang)^2))*180/(%pi)
+y=360-x//for cosine inverse, angle and 360-angle are same and must be considered
+x1=x/2
+y1=y/2
+printf("The acceleration of driven shaft is a maximum when theta =%.2f or %.2f degrees\n",x1,y1)
+amax=(w^2*cos(ang)*(sin(ang)^2)*sin(x*%pi/180))/((1-((cos(x1*%pi/180)^2)*(sin(ang)^2)))^2)//maximum angular acceleration, numerically
+printf("Maximum angular acceleration is %.f rad/s^2\n",amax)
+
diff --git a/1325/CH5/EX5.1/5_1.PNG b/1325/CH5/EX5.1/5_1.PNG Binary files differnew file mode 100644 index 000000000..2007adcb9 --- /dev/null +++ b/1325/CH5/EX5.1/5_1.PNG diff --git a/1325/CH5/EX5.1/5_1.sce b/1325/CH5/EX5.1/5_1.sce new file mode 100644 index 000000000..d7ae85202 --- /dev/null +++ b/1325/CH5/EX5.1/5_1.sce @@ -0,0 +1,26 @@ +//to find theta at admission, cut-off, release and compression
+clc
+//given
+s=1.125//inch
+e=0.25//inch
+t=2.25//inch
+alpha=35//degrees
+//from 5.2, we know theta+alpha=sininverse(s/t)
+x=asind(s/t)
+y=180-x//sin(x)=sin(180-x)=sin(y)
+//at admission
+p=x-alpha
+//at cutoff
+q=y-alpha
+//from 5.3, theta+alpha=sininnverse(-e/t)
+ang=asind(-e/t)
+angle=abs(ang)
+a=180+angle//lies in the negative region of sine curve
+b=360-angle//lies in hte negative region of sine curve
+//at release
+r=a-alpha
+//at compression
+s=b-alpha
+printf("Angle theta at admission, cut-off, release and compression are %.2f, %.2f, %.2f and %.2f degrees respectively",p,q,r,s)
+
+
diff --git a/1325/CH6/EX6.1/6_1.PNG b/1325/CH6/EX6.1/6_1.PNG Binary files differnew file mode 100644 index 000000000..8907ace28 --- /dev/null +++ b/1325/CH6/EX6.1/6_1.PNG diff --git a/1325/CH6/EX6.1/6_1.sce b/1325/CH6/EX6.1/6_1.sce new file mode 100644 index 000000000..c2179abc2 --- /dev/null +++ b/1325/CH6/EX6.1/6_1.sce @@ -0,0 +1,13 @@ +//to find the maximum efficiency
+clc
+//given
+theta=60//degrees
+u1=0.15//between surfaces A annd B
+u2=0.10//for the guides
+phi=atand(u1)
+phi1=atand(u2)
+alpha=(theta+phi+phi1)/2//from 6.22, maximum efficiency is obtained at alpha
+//from 6.23, maximum efficiency is given by nmax=(cos(theta+phi+phi1)+1)/(cos(theta-phi-phi1)+1)
+nmax=(cos((theta+phi+phi1)*%pi/180)+1)/(cos((theta-phi-phi1)*%pi/180)+1)
+printf("Maximum efficiency = %.4f and it is obtained when alpha = %.2f degrees",nmax,alpha)
+
diff --git a/1325/CH6/EX6.3/6_3.PNG b/1325/CH6/EX6.3/6_3.PNG Binary files differnew file mode 100644 index 000000000..b2c041891 --- /dev/null +++ b/1325/CH6/EX6.3/6_3.PNG diff --git a/1325/CH6/EX6.3/6_3.sce b/1325/CH6/EX6.3/6_3.sce new file mode 100644 index 000000000..c3c32d224 --- /dev/null +++ b/1325/CH6/EX6.3/6_3.sce @@ -0,0 +1,19 @@ +//to find power absorbed and number of collars required
+clc
+//from equation 6.36 we know, M=(2/3)*u*W*(ri^3-r2^3)/(r1^2-r2^2)
+//given
+u=0.04
+W=16//tons
+w=W*2240//lbs
+r1=8//in
+r2=6//in
+N=120
+P=50//lb/in^2
+M=(2/3)*u*w*(r1^3-r2^3)/(r1^2-r2^2)
+hp=M*2*%pi*N/(12*33000)//horse power absorbed
+//from fig 137,effective bearing surface per pad is calsulate from the dimensions to be 58.5 in^2
+A=58.5//in^2
+n=w/(A*P)
+x=floor(n)
+printf("\n")
+printf("Horsepower absorbed = %.2f\nNumber of collars required = %.f\n",hp,x)
diff --git a/1325/CH6/EX6.4/6_4.PNG b/1325/CH6/EX6.4/6_4.PNG Binary files differnew file mode 100644 index 000000000..a3274ea05 --- /dev/null +++ b/1325/CH6/EX6.4/6_4.PNG diff --git a/1325/CH6/EX6.4/6_4.sce b/1325/CH6/EX6.4/6_4.sce new file mode 100644 index 000000000..15f2ec96e --- /dev/null +++ b/1325/CH6/EX6.4/6_4.sce @@ -0,0 +1,16 @@ +//To find the dimensions of the clutch plate and the total axial pressure which must be exerted by the springs.
+clc
+//given
+ratio=1.25
+u=.675
+P=12//hp
+//W=P*%pi*(r1^2-r2^2); Total axal thrust.
+//M=u*W*(r1+r2); Total friction moemnt
+//reducing the two equations and using ratio=1.25(r1=1.25*r2) we get, M=u*21.2*r2^3
+ReqM=65//lb ft
+RM=ReqM*12//lb in
+r2=(RM/(u*P*%pi*(1.25^2-1)))^(1/3)
+r1=1.25*r2
+d1=r1*2
+d2=r2*2
+printf("The dimensions of the friction surfaces are:\nOuter Diameter= %.1f in\nInner Diameter= %.1f in\n",d1,d2)
diff --git a/1325/CH6/EX6.5/6_5.PNG b/1325/CH6/EX6.5/6_5.PNG Binary files differnew file mode 100644 index 000000000..1eb586c11 --- /dev/null +++ b/1325/CH6/EX6.5/6_5.PNG diff --git a/1325/CH6/EX6.5/6_5.sce b/1325/CH6/EX6.5/6_5.sce new file mode 100644 index 000000000..4a8851954 --- /dev/null +++ b/1325/CH6/EX6.5/6_5.sce @@ -0,0 +1,15 @@ +//to find the number of plates required
+clc
+P=20//lb/in^2
+u=0.07//friction coefficient
+N=3600//rpm
+H=100//hp
+r1=5//in
+r2=0.8*r1//given
+A=%pi*(r1^2-r2^2)//the area of each friction surface
+W=A*P//total axial thrust on plates
+M=(1/2)*u*W*(r1+r2)//friction moment for each pair of contacts
+T=H*33000*12/(2*%pi*N)//total torque to be transmitted
+x=(T/M)//effective friction surfaces required
+printf("\nNumber of effective friction surfaces required= %.f\n",x)
+
diff --git a/1325/CH6/EX6.7/6_7.PNG b/1325/CH6/EX6.7/6_7.PNG Binary files differnew file mode 100644 index 000000000..ebe05086a --- /dev/null +++ b/1325/CH6/EX6.7/6_7.PNG diff --git a/1325/CH6/EX6.7/6_7.sce b/1325/CH6/EX6.7/6_7.sce new file mode 100644 index 000000000..d6a3e95dc --- /dev/null +++ b/1325/CH6/EX6.7/6_7.sce @@ -0,0 +1,29 @@ +//to find the turning moment on the cranckshaft when a) friction at the bearing is neglected b)when u=0.5
+clc
+//given
+P=6 //tons
+u=0.05
+theta=60//degrees
+CP=80
+Stroke=16//in
+OC=Stroke/2
+r1=7//in
+r2=15//in
+r3=4.4//in
+//Radius of friction circle
+ro=u*r1
+rc=u*r2
+rp=u*r3
+phi=asind(OC*sin((theta)*%pi/180)/CP)
+alpha=asind((rc+rp)/CP)
+//a) without friction
+Qa=P/cos((phi)*%pi/180)
+Xa=OC*cos((30-phi)*%pi/180)//tensile force transmitted along the eccentric rod when friction is NOT taken into account
+Ma=Qa*Xa/12
+//b) with friction
+Qb=P/cos((phi-alpha)*%pi/180)//tensile force transmitted along the eccentric rod when friction is taken into account
+Xb=OC*cos((30-(phi-alpha))*%pi/180)-(rc+ro)
+Mb=Qb*Xb/12
+n=Mb/Ma
+printf("Turning moment applied to OC:\na)Without friction= %.2f ton.ft\nb)With friction(u=0.05)= %.2f ton.ft",Ma,Mb)
+printf("\nThe efficiency of the mechanism is %.2f ",n)
diff --git a/1325/CH6/EX6.8/6_8.PNG b/1325/CH6/EX6.8/6_8.PNG Binary files differnew file mode 100644 index 000000000..cd9ab06a4 --- /dev/null +++ b/1325/CH6/EX6.8/6_8.PNG diff --git a/1325/CH6/EX6.8/6_8.sce b/1325/CH6/EX6.8/6_8.sce new file mode 100644 index 000000000..e3f0a0919 --- /dev/null +++ b/1325/CH6/EX6.8/6_8.sce @@ -0,0 +1,19 @@ +//To find maximum horizontal force that can be transmitted to the pump
+clc
+stroke=4//in
+d=11.5//in
+ds=4//in
+dp=14//in
+theta=%pi
+u1=.25
+T=350//lb
+u2=0.1
+k=%e^(u1*theta)
+T2=T1/k
+Tor=(T1-T2)*(dp/2)//total resisting torque
+//total resisting torque is also given by P*(r+2*(cos%pi/6))+u2*R*(ds/2)
+//equating and putting values we get the following quadratic equation
+p=[1 -1.163D3 3.342D5]
+a=roots(p)
+printf("\nP=%.1f",a)
+printf("\nThe larger of two values is inadmissible. It corresponds to a negative sign in front of the second term on the right hand side of equation (1)")
diff --git a/1325/CH7/EX7.1/7_1.PNG b/1325/CH7/EX7.1/7_1.PNG Binary files differnew file mode 100644 index 000000000..9112884df --- /dev/null +++ b/1325/CH7/EX7.1/7_1.PNG diff --git a/1325/CH7/EX7.1/7_1.sce b/1325/CH7/EX7.1/7_1.sce new file mode 100644 index 000000000..8cd0731f0 --- /dev/null +++ b/1325/CH7/EX7.1/7_1.sce @@ -0,0 +1,15 @@ +//To find maximum horsepower which belt can transmit
+clc
+//given-belt is perfectly elastic and massless
+u=0.3
+v=3600//ft/min
+V=v/60//ft/sec
+theta=165//degrees
+x=theta*%pi/180
+k=%e^(u*x)//k=T1/T2=e^(u*x)
+To=500//lb
+T1=2*k*To/(k+1)
+T2=T1/k
+T=T1-T2//effective tension
+H=T*V/550//horsepower transmitted
+printf("\nThe horse-power transmitted = %.2f\n",H)
diff --git a/1325/CH7/EX7.2/7_2.PNG b/1325/CH7/EX7.2/7_2.PNG Binary files differnew file mode 100644 index 000000000..c989cb9cd --- /dev/null +++ b/1325/CH7/EX7.2/7_2.PNG diff --git a/1325/CH7/EX7.2/7_2.sce b/1325/CH7/EX7.2/7_2.sce new file mode 100644 index 000000000..4a1e56237 --- /dev/null +++ b/1325/CH7/EX7.2/7_2.sce @@ -0,0 +1,28 @@ +//to find maximun horsepower when drive is a) Vertical b) Horizontal
+clc
+w=1.2//lb/ft^2
+u=0.3
+v=3600//ft/min
+V=v/60//ft/sec
+theta=165//degrees
+g=32.2//ft/s^2
+x=theta*%pi/180
+k=%e^(u*x)//k=T1/T2=e^(u*x)
+To=500//lb
+//Solution a)Vertical drive
+Tc=w*V^2/g//equation 7.5
+//solution a)
+H=2*(k-1)*(To-Tc)*V/((k+1)*550)
+Vmax=(To*g/(3*w))^(1/2)
+Hmax=2*(k-1)*(To-Tc)*Vmax/((k+1)*550)
+//Solution b)
+To1=To+Tc
+//from equation 7.15 2/To1^2=1/Tt^2+1/Ts^2
+//T1/T2=k
+T2=367 //lb - from trail and error
+T1=k*T2
+Tt=T1+Tc
+Ts=T2+Tc
+HP=(T1-T2)*V/550
+printf("\nSolution a)\nHorsepower transmitted= %.1f\nMaximum Horsepower transmitted= %.1f (at velocit = %.1f ft/s^2)Solution b)\nTt=%.f lb\nTs=%.f lb\nHorsepower transmitted= %.1f",H,Hmax,Vmax,Tt,Ts,HP)
+
diff --git a/1325/CH8/EX8.1/8_1.PNG b/1325/CH8/EX8.1/8_1.PNG Binary files differnew file mode 100644 index 000000000..bbae5cf83 --- /dev/null +++ b/1325/CH8/EX8.1/8_1.PNG diff --git a/1325/CH8/EX8.1/8_1.sce b/1325/CH8/EX8.1/8_1.sce new file mode 100644 index 000000000..a769fa12f --- /dev/null +++ b/1325/CH8/EX8.1/8_1.sce @@ -0,0 +1,15 @@ +//Find the braking torque applied to the drum
+clc
+//given
+dia=12//in
+r=dia/2
+CQ=7//in
+OC=6//in
+OH=15//in
+u=0.3
+P=100//lb
+phi=atan(u)
+x=r*sin(phi)//in inches;radius of friction circle
+a=5.82//from figure
+Tb=P*OH*x/a//braking torque
+printf("\nThe braking torque of the drum Tb= %.2f lb in\n",Tb)
diff --git a/1325/CH8/EX8.2/8_2.PNG b/1325/CH8/EX8.2/8_2.PNG Binary files differnew file mode 100644 index 000000000..2b196d539 --- /dev/null +++ b/1325/CH8/EX8.2/8_2.PNG diff --git a/1325/CH8/EX8.2/8_2.sce b/1325/CH8/EX8.2/8_2.sce new file mode 100644 index 000000000..d24caee55 --- /dev/null +++ b/1325/CH8/EX8.2/8_2.sce @@ -0,0 +1,32 @@ +//To find braking torque applied to the drum
+clc
+//given
+
+OH=15//in
+l=OH
+u=0.3
+P=100//lb
+phi=atan(u)
+//according to fig 170(b)
+//for clockwise rotation
+a=6//from figure
+x=r*sin(phi)//in inches;radius of friction circle
+Tb=P*l*x/a//braking torque on the drum
+//for counter clockwise rotation
+a1=5.5//in
+Tb1=P*l*x/a1//braking torque on the drum
+//according to figure 172(a)
+//for clockwise rotation
+a2=6.48//from figure
+x=r*sin(phi)//in inches;radius of friction circle
+Tb2=P*l*x/a2//braking torque on the drum
+//for counter clockwise rotation
+a3=6.38//in
+Tb3=P*l*x/a3//braking torque on the drum
+T1=ceil(Tb1)
+T2=ceil(Tb2)
+T3=ceil(Tb3)
+printf("\nbraking torque on drum\nWhen dimensions are measured from fig 170(b)\nFor clockwise rotation= %.f lb in\nFor counter clockwise rotation= %.f lb in\nWhen dimensions are measured from fig 171(a)\nFor clockwise rotation= %.f lb in\nFor counter clockwise rotation= %.f lb in",Tb,T1,T2,T3)
+
+
+
diff --git a/1325/CH8/EX8.3/8_3.PNG b/1325/CH8/EX8.3/8_3.PNG Binary files differnew file mode 100644 index 000000000..1123e050a --- /dev/null +++ b/1325/CH8/EX8.3/8_3.PNG diff --git a/1325/CH8/EX8.3/8_3.sce b/1325/CH8/EX8.3/8_3.sce new file mode 100644 index 000000000..017cc168a --- /dev/null +++ b/1325/CH8/EX8.3/8_3.sce @@ -0,0 +1,22 @@ +//To find a) magnitude of P b) magnitude of force at Hd
+clc
+//given
+u=.35
+Tb=500//lb.ft
+rd=10//in
+phi=atan(u)
+x=rd*sin(phi)
+//F*OD=R*a=R1*a
+//R=R1
+//2*R*x=Tb
+OD=24//in
+a=11.5//inches; From figure
+F=Tb*a*12/(OD*2*x)
+//from figure
+HG=4//in
+GK=12//in
+HL=12.22//in
+P=F*HG/GK
+Fhd=HL*P/HG
+printf("\na) Magnitude of P = %.f lb",P)
+printf("\nb) Magnitude of Fhd = %.f lb",Fhd)
diff --git a/1325/CH8/EX8.4/8_4.PNG b/1325/CH8/EX8.4/8_4.PNG Binary files differnew file mode 100644 index 000000000..605f9715f --- /dev/null +++ b/1325/CH8/EX8.4/8_4.PNG diff --git a/1325/CH8/EX8.4/8_4.sce b/1325/CH8/EX8.4/8_4.sce new file mode 100644 index 000000000..c052e3d18 --- /dev/null +++ b/1325/CH8/EX8.4/8_4.sce @@ -0,0 +1,15 @@ +//To find the least force required in order to support load of .5 tons
+clc
+//given
+u=.3
+theta=270*%pi/180
+l=18//in
+a=4//in
+Di=15//in
+Do=21//in
+w=.5//tons
+W=w*2204//lb
+Q=W*Di/Do//required tangential braking force on the drum
+k=%e^(u*theta)//k=T1/T2
+p=Q*a/(l*(k-1))
+printf("Least force required, P = %.f lb",p)
diff --git a/1325/CH8/EX8.5/8_5.PNG b/1325/CH8/EX8.5/8_5.PNG Binary files differnew file mode 100644 index 000000000..92c3fc658 --- /dev/null +++ b/1325/CH8/EX8.5/8_5.PNG diff --git a/1325/CH8/EX8.5/8_5.sce b/1325/CH8/EX8.5/8_5.sce new file mode 100644 index 000000000..aed6d06a6 --- /dev/null +++ b/1325/CH8/EX8.5/8_5.sce @@ -0,0 +1,16 @@ +//To find the least effort applied at the end of the lever which will provide a braking torque of 4000 lb ft
+clc
+//given
+n=12
+u=.28
+a=4.5//in
+b=1//in
+l=21//in
+r=15//in
+Tb=4000//lb
+theta=10*%pi/180
+//k=Tn/To
+k=((1+u*tan(theta))/(1-u*tan(theta)))^n
+Q=Tb*(12/r)
+P=Q*(a-b*k)/(l*(k-1))//from combining 8.6 with k=e^u*theta
+printf("The least effort required = P = %.1f lb",P)
diff --git a/1325/CH8/EX8.6/8_6.PNG b/1325/CH8/EX8.6/8_6.PNG Binary files differnew file mode 100644 index 000000000..7fce9b101 --- /dev/null +++ b/1325/CH8/EX8.6/8_6.PNG diff --git a/1325/CH8/EX8.6/8_6.sce b/1325/CH8/EX8.6/8_6.sce new file mode 100644 index 000000000..7630ffed8 --- /dev/null +++ b/1325/CH8/EX8.6/8_6.sce @@ -0,0 +1,30 @@ +//To find the minimum distance in which the car may be stopped
+clc
+//given
+w=9.5 //ft
+h= 2 //ft
+x=4 //ft
+v=30//mph
+V=1.46667*v//ft/s
+u1=.1
+u2=.6
+g=32.2//ft/s^2
+//a) rear wheels braked
+fa1=(u1*(w-x)*g)/(w+u1*h)
+fa2=(u2*(w-x)*g)/(w+u2*h)
+sa1=V^2/(2*fa1)
+sa2=V^2/(2*fa2)
+//b) front wheels braked
+fb1=u1*x*g/(w-u1*h)
+fb2=u2*x*g/(w-u2*h)
+sb1=V^2/(2*fb1)
+sb2=V^2/(2*fb2)
+//c) All wheels braked
+fc1=u1*g
+fc2=u2*g
+sc1=V^2/(2*fc1)
+sc2=V^2/(2*fc2)
+k1=(x+u1*h)/(w-x-u1*h)//Na/Nb
+k2=(x+u2*h)/(w-x-u2*h)//Na/Nb
+printf("\nCoefficient of friction = 0.1\na) Minimum distance in which car may be stopped when the rear brakes are applied = %.f ft\nb) Minimum distance in which car may be stopped when the front brakes are applied = %.f ft\nc) Minimum distance in which car may be stopped when all brakes are applied = %.f ft\nCoefficient of friction = 0.6\na) Minimum distance in which car may be stopped when the rear brakes are applied = %.f ft\nb) Minimum distance in which car may be stopped when the front brakes are applied = %.f ft\nc) Minimum distance in which car may be stopped when all brakes are applied = %.f ft\n",sa1,sb1,sc1,sa2,sb2,sc2)
+printf("Required ration of Na/Nb\nFor u1 = 0.1 -> %.3f\nFor u2 = 0.6 -> %.2f\n",k1,k2)
diff --git a/1325/CH9/EX9.5/9_5.PNG b/1325/CH9/EX9.5/9_5.PNG Binary files differnew file mode 100644 index 000000000..24f8ac2d0 --- /dev/null +++ b/1325/CH9/EX9.5/9_5.PNG diff --git a/1325/CH9/EX9.5/9_5.sce b/1325/CH9/EX9.5/9_5.sce new file mode 100644 index 000000000..bcb6aa297 --- /dev/null +++ b/1325/CH9/EX9.5/9_5.sce @@ -0,0 +1,35 @@ +//To draw complete displacement, velocity and acceleration diagrams
+clc
+//given
+alpha=55*%pi/180
+N=1200//rpm
+lift=.5//in
+rn=.125//in ; noseradius
+rmin=1.125//in ; minimum radius
+OQ=rmin+lift-rn
+OP=(OQ^2-1)/(2*(1-OQ*cos(alpha)))//from triangle opq fig 201(a)
+PQ=OP+rmin-rn
+phi=asin(OQ*sin(alpha)/PQ)
+x1=[0:.0001:phi]
+x2=[phi:.0001:alpha]
+y1=4.477*(1-cos(x1))//from 9.6
+y2=1.5*cos(alpha-x2)-1//from 9.9
+v1=%pi*N*4.477*sin(x1)/(30*12)//from 9.7
+v2=15.71*sin(alpha-x2)//from 9.10
+f1=(%pi*N/30)^2*(4.477/12)*cos(x1)//from 9.8
+f2=-1974*cos(alpha-x2)//from 9.11
+a=[0:.0001:phi]
+b=[phi:.0001:alpha]
+p=[0:.0001:phi]
+q=[phi:.0001:alpha]
+subplot(3,1,3)
+subplot(311)
+plot(x1,y1,x2,y2)
+xtitle("","angle","displacement")
+subplot(312)
+plot(a,v1,b,v2)
+xtitle("","angle","velocity")
+subplot(313)
+plot(p,f1,q,f2)
+xtitle("","angle","acceleration")
+
diff --git a/1325/CH9/EX9.7/9_7.PNG b/1325/CH9/EX9.7/9_7.PNG Binary files differnew file mode 100644 index 000000000..85a5336f1 --- /dev/null +++ b/1325/CH9/EX9.7/9_7.PNG diff --git a/1325/CH9/EX9.7/9_7.sce b/1325/CH9/EX9.7/9_7.sce new file mode 100644 index 000000000..a8677e50d --- /dev/null +++ b/1325/CH9/EX9.7/9_7.sce @@ -0,0 +1,20 @@ +//to find the angular velocity and the angular acceleration of the follower
+clc
+//given
+N=600//rpm
+BC=3//in
+rmin=1.125//in
+rf=39/8//in
+OP=rf-rmin
+OM1=0.79//in;given
+NZ1=2.66//in
+w=N*%pi/30
+vb=w*OM1
+Vang=vb/BC
+at=w^2*NZ1
+fBC=at/BC
+OM2=.52//in
+NZ2=3.24//in
+af=w*OM2/BC
+angf=w^2*NZ2/BC
+printf("\nWhen theta = 25 degrees\nangular velocity = %.1f rad/s\nangular acceleration = %.f rad/s^2\nWhen theta = 45 degrees\nangular velocity = %.1f rad/s\nangular acceleration = %.f rad/s^2",Vang,fBC,af,angf)
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