20 January 2014
Professor J. A. Caton
MEEN 410 INTERNAL COMBUSTION ENGINES
Problem Assignment Number 2
Due: 28 January 2014 (Tuesday)
(1.) A CFR, single cylinder, engine was operated at the following conditions. The resulting
cylinder pressures were plott
14 January 2014
Professor J. A. Caton
MEEN 410 INTERNAL COMBUSTION ENGINES
Problem Assignment Number 1
Due: 21 January 2014 (Tuesday)
(1.) For an engine you own, answer the following questions:
a.) List the manufacturer of the engine and the specific appl
09 February 2014
Professor J. A. Caton
MEEN 410 INTERNAL COMBUSTION ENGINES
Problem Assignment Number 5
Due: 18 February 2014 (Tuesday)
The following three problems involve computing flow rates past poppet valves. These problems will be
based on the same
CI Engine Combustion
Types of engines and general comments
Fuel systems and spray characteristics
Ignition and cetane no.
Combustion
Cold start and emissions
Other Combustion Modes HCCI,
Comparison of CI and SI Engines
Advantages relative to SI eng
Let us consider an element of length dx located at a distance y1 above the
neutral axis. Because of the shear force, the bending moment is not constant
along the x axis. Let us denote by M the bending moment on the near side of
the section and by M dM the
131
1. Stress
Mechanics
moment are internal for the whole system, but they become external when
applied to the isolated part. The interface forces, for example, are represented symbolically by the force vectors F1 ; F2 , and Fi in Fig. 1.7. The
isolated p
163
2. Deection and Stiffness
Solution
If we consider a cut at an arbitrary distance x from the origin and take only
the left part of the beam as a free body, the shear force can be expressed as
V R1 wx
wl
wx :
2
EXAMPLE 2.2
A concentrated load F is app
81
3. Dynamics of a Particle
The term
is the work where F is the total external force acting on the particle of mass m
and d r is the innitesimal displacement of the particle. Integrating Eq. (3.19),
one may obtain
v22
r2
1
1
1
m d v2 mv22 mv21 ;
F dr
233
4. Kinetostatics
Figure 4.8
Mechanisms
Used with
permission from
Ref. 15.
If the path I is followed (Fig. 4.8a), for the rotation kinematic pair at E (ER ) a
moment equation can be written as
P 4
ME rC rE F32 rC 4 rD F4 M4 0;
or
i
xC xE
F
34x
j
yC y
85
3. Dynamics of a Particle
If we use the function V , the integral dening the work is
r2
V2
F dr
dV V2 V1 ;
U12
r1
V1
3:28
1 2
1
mv V1 mv22 V2 ;
2 1
2
3:29
which means that the sum of the kinetic energy and the potential energy V is
constant:
1 2
mv
26
Statics
Statics
where the distance y y 0 b. Carrying out the operations,
Ixx y 02 dA 2b y 0 dA Ab 2 :
A
A
The rst term of the right-hand side is by denition Ix 0 x 0 ,
Ix 0 x 0 y 02 dA:
A
The second term involves the rst moment of area about the x 0 ax
20
Statics
Statics
Figure 2.8
The position x is actually the centroid coordinate of the loading curve area.
Thus, the simplest resultant force of a distributed load acts at the centroid of
the area under the loading curve.
Example 1
For the triangular loa
101
4. Planar Kinematics of a Rigid Body
Figure 4.6 shows two points A and B of a rigid body and their directions
of motion DA amd DB ,
and
v B kDB ;
Dynamics
v A kDA
Figure 4.6
where v A is the velocity of point A, and v B is the velocity of point B.
Thr
147
1. Stress
For the arm BC , the bending stress will reach a maximum near the shaft at B.
The bending stress for the rectangular cross-section of the arm is
s
M
6M
64000
61;440 psi:
I =c bh 2 0:251:252
The torsional stress is
T
t
1000
0:25
3 1:8
43;0
185
Mechanics
3. Fatigue
Figure 3.8
Goodman
diagram. Used
with permission
from Ref. 16.
In this equation, a and b are the coordinates of the points of intersection of
the straight line with the x and y axes, respectively. For example, the
equation for the
where the vector rt Dt rt is the change in position, or displacement
of P ; during the interval of time Dt (Fig. 2.1). The velocity is the rate of
change of the position of the point P . The magnitude of the velocity v is the
speed v jvj. The dimensions o
57
2. Kinematics of a Point
rotates is Dy yt Dt yt . The triangle in Fig. 2.3a is isosceles, so the
magnitude of Du is
jDuj 2juj sinDy=2 2 sinDy=2:
The vector Du is
where n is a unit vector that points in the direction of Du (Fig. 2.3a). The time
derivati
251
1. Screws
Machine Components
For square threads the normal thread load, F, is parallel to the axis of the
screw (Figs 1.6 and 1.7). The preceding equations can be applied for square
threads.
For Acme threads (Figs 1.5) or other threads, the normal thr
191
1. Fundamentals
Mechanisms
uniquely dene its position in space at any instant of time. The number of
DOF is dened with respect to a reference frame.
Figure 1.2 shows a free rigid body, RB, in planar motion. The rigid body
is assumed to be incapable of
24
Statics
Statics
2.9 Second Moments and the Product of Area
The second moments of the area A about x and y axes (Fig. 2.14), denoted as
Ixx and Iyy , respectively, are
2:12
Ixx y 2 dA
Iyy
A
A
x 2 dA:
2:13
Figure 2.14
The second moment of area cannot be
157
2. Deection and Stiffness
The integration constants C1 to C4 can be evaluated by considering appropriate boundary conditions. Both EI y 0 and EIy 0 at x 0. This gives
C3 0 and C4 0. At x 0 the shear force is equal to R1 . Therefore,
Eq. (2.28) gives V
30
Statics
Statics
or
2a02 tan1
2Ixy
2Ixy
p ) a02 0:5 tan1
0:5p:
Iyy Ixx
Iyy Ixx
This means that there are two axes orthogonal to each other having extreme
values for the second moment of area at 0. On one of the axes is the
maximum second moment of are
139
1. Stress
The foregoing equation gives R1 600 lb. The force equation with respect to
the y axis is
R1 400 150 R2 0;
yielding R2 500 lb. The next step is to draw the shear force and the bending
moment diagrams shown in Figs. 1.13b and 1.13c. From the b
107
Dynamics
4. Planar Kinematics of a Rigid Body
Figure 4.10
frame, i.e., they are the velocity and acceleration measured by an observer
moving with the rigid body (Fig. 4.10).
If A is a point of the rigid body, A 2 RB, v Arel 0 and a Arel 0.
4.7.1 MOTIO
10
Statics
Statics
Figure 1.9
where n is a unit vector whose direction is the same as the direction of
advance of a right-handed screw rotated from a toward b, through the angle
(a, b), when the axis of the screw is perpendicular to both a and b. m
The ma
143
1. Stress
Substituting Eq. (1.54) into Eq. (1.53) gives
t
V 2
c y12 :
2I
1:55
This equation represents the general equation for shear stress in a beam of
rectangular cross-section. The expression for the second moment of area I
for a rectangular secti
151
Mechanics
2. Deection and Stiffness
Figure 2.1
Springs. (a)
Linear spring;
(b) stiffening
spring; (c) softening spring.
Used with
permission from
Ref. 16.
where F is the force applied on the bar, l the length of the bar, A the crosssectional area, and
vii
Table of Contents
4.7 Motion of a Point That Moves Relative to a Rigid Body
5. Dynamics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . .
5.1 Equation of Motion for the Center of Mass. . . . . . . . .
5.2 Angular Momentum Principle for a
95
4. Planar Kinematics of a Rigid Body
Integrating Eq. (3.62) with respect to time, one may obtain
t1
r F dt HO 2 HO 1 :
3:64
The integral on the left-hand side is called the angular impulse.
The principle of angular impulse and momentum: The angular imp
75
3. Dynamics of a Particle
3.2 Newtonian Gravitation
F
Gm1 m2
;
r2
3:3
Figure 3.1
where G is called the universal gravitational constant. Equation (3.3) may
be used to approximate the weight of a particle of mass m due to the
gravitational attraction o