From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 3308/CH22/EX22.1/Ex22_1.sce | 36 ++++++++++++++++++++++++++++++++++++
 3308/CH22/EX22.1/Ex22_1.txt | 12 ++++++++++++
 3308/CH22/EX22.2/Ex22_2.sce | 20 ++++++++++++++++++++
 3308/CH22/EX22.2/Ex22_2.txt |  7 +++++++
 4 files changed, 75 insertions(+)
 create mode 100755 3308/CH22/EX22.1/Ex22_1.sce
 create mode 100755 3308/CH22/EX22.1/Ex22_1.txt
 create mode 100755 3308/CH22/EX22.2/Ex22_2.sce
 create mode 100755 3308/CH22/EX22.2/Ex22_2.txt

(limited to '3308/CH22')

diff --git a/3308/CH22/EX22.1/Ex22_1.sce b/3308/CH22/EX22.1/Ex22_1.sce
new file mode 100755
index 000000000..4f24bf3be
--- /dev/null
+++ b/3308/CH22/EX22.1/Ex22_1.sce
@@ -0,0 +1,36 @@
+clc 
+// Given that
+l=6//in inch Length of rod  
+di=1/2//in inch initial diameter of rod
+df=0.480//in inch final diameter of rod
+N=400//in rpm spindle rotation
+Vt=8//in inch/minute axial speed of the tool
+
+// Sample Problem on page no. 600
+
+printf("\n # Material Removal Rate and Cutting Force in Turning # \n")
+
+V=3.14*di*N
+printf("\n\n Cutting speed=%d in/min",V)
+
+v1=3.14*df*N//cutting speed from machined diameter
+d=(di-df)/2//depth of cut
+f=Vt/N//feed
+Davg=(di+df)/2
+MRR=3.14*Davg*d*f*N 
+printf("\n\n Material Removal Rate %f=in^3/min",MRR)
+
+t=l/(f*N)
+printf("\n\n Cutting time=%f min",t)
+
+P=(4/2.73)*MRR//average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3
+printf("\n\n Cutting power=%f hp",P)
+
+Fc=((P*396000)/(N*2*3.14))/(Davg/2)
+printf("\n\n Cutting force=%d lb",Fc)
+
+//answer in the book is given 118 lb due to approximation
+
+
+
+
diff --git a/3308/CH22/EX22.1/Ex22_1.txt b/3308/CH22/EX22.1/Ex22_1.txt
new file mode 100755
index 000000000..9b0ca6b2e
--- /dev/null
+++ b/3308/CH22/EX22.1/Ex22_1.txt
@@ -0,0 +1,12 @@
+ # Material Removal Rate and Cutting Force in Turning # 
+
+
+ Cutting speed=628 in/min
+
+ Material Removal Rate 0.123088=in^3/min
+
+ Cutting time=0.750000 min
+
+ Cutting power=0.180349 hp
+
+ Cutting force=116 lb 
\ No newline at end of file
diff --git a/3308/CH22/EX22.2/Ex22_2.sce b/3308/CH22/EX22.2/Ex22_2.sce
new file mode 100755
index 000000000..d2794ff75
--- /dev/null
+++ b/3308/CH22/EX22.2/Ex22_2.sce
@@ -0,0 +1,20 @@
+clc 
+// Given that  
+d=10//in mm diameter of drill bit
+f=0.2//in mm/rev feed
+N=800//in rpm spindle rotation
+
+// Sample Problem on page no. 632
+
+printf("\n # Material Removal Rate and Torque in Drilling # \n")
+
+MRR=[((3.14*(d^2))/4)*f*N]/60
+printf("\n\n Material Removal Rate %d=mm^3/sec",MRR)
+
+//Answer in the book is given 210 mm^3/sec
+
+//from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken
+T=(MRR*0.5)/((N*2*3.14)/60)
+printf("\n\n Torque on the drill %f=Nm",T)
+
+
diff --git a/3308/CH22/EX22.2/Ex22_2.txt b/3308/CH22/EX22.2/Ex22_2.txt
new file mode 100755
index 000000000..c0063062b
--- /dev/null
+++ b/3308/CH22/EX22.2/Ex22_2.txt
@@ -0,0 +1,7 @@
+
+ # Material Removal Rate and Torque in Drilling # 
+
+
+ Material Removal Rate 209=mm^3/sec
+
+ Torque on the drill 1.250000=Nm 
\ No newline at end of file
-- 
cgit