initial commit / add all books

6 files changed, 111 insertions, 0 deletions

diff --git a/3594/CH6/EX6.1/6_1.sce b/3594/CH6/EX6.1/6_1.sce
new file mode 100644
index 000000000..dad8be876
--- /dev/null
+++ b/3594/CH6/EX6.1/6_1.sce
@@ -0,0 +1,13 @@
+

+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/3594/CH6/EX6.3/6_3.sce b/3594/CH6/EX6.3/6_3.sce
new file mode 100644
index 000000000..4772e64ad
--- /dev/null
+++ b/3594/CH6/EX6.3/6_3.sce
@@ -0,0 +1,19 @@
+

+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/3594/CH6/EX6.4/6_4.sce b/3594/CH6/EX6.4/6_4.sce
new file mode 100644
index 000000000..f3710b2bf
--- /dev/null
+++ b/3594/CH6/EX6.4/6_4.sce
@@ -0,0 +1,16 @@
+

+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/3594/CH6/EX6.5/6_5.sce b/3594/CH6/EX6.5/6_5.sce
new file mode 100644
index 000000000..c9bcee920
--- /dev/null
+++ b/3594/CH6/EX6.5/6_5.sce
@@ -0,0 +1,15 @@
+

+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/3594/CH6/EX6.7/6_7.sce b/3594/CH6/EX6.7/6_7.sce
new file mode 100644
index 000000000..ebe996a3e
--- /dev/null
+++ b/3594/CH6/EX6.7/6_7.sce
@@ -0,0 +1,29 @@
+

+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/3594/CH6/EX6.8/6_8.sce b/3594/CH6/EX6.8/6_8.sce
new file mode 100644
index 000000000..349f5ac28
--- /dev/null
+++ b/3594/CH6/EX6.8/6_8.sce
@@ -0,0 +1,19 @@
+

+clc

+stroke=4//in

+d=11.5//in

+ds=4//in

+dp=14//in

+theta=%pi

+u1=.25

+T1=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. \n It corresponds to a negative sign in front of the second term on the \n right hand side of equation (1)")