Upload
pizzagod
View
215
Download
0
Embed Size (px)
Citation preview
8/7/2019 FE1001 Part 2 Lectures 22-27
1/41
12-1
Chapter 12:Static Equilibrium and Elasticity
.
12.2 Centre of Gravity
12.3 Rigid Objects in Static Equilibrium
12.4 Elastic Pro erties of Solids
12-2
8/7/2019 FE1001 Part 2 Lectures 22-27
2/41
Over 13 Weeks (39 hrs):
I. Vectors and D namics22 hrs
II . Static Equilibrium and Elasticity (6 hrs)
Chapter 12: Static Equilibrium and Elasticity
III. F ui Mec anics3 rs
IV. Thermod namics8 hrs
12-3
CONTENTS of PART II LECTURES
L23 Conditions for Equilibrium
L24 Equivalent Force Systems, Centre of Gravity
L25 Rigid Objects in Static Equilibrium (1)
L27 Analysis of Trusses.
g ec s n a c qu r um
L28 Elastic Properties of Solids: Youngs Modulus, ShearModulus and Bulk Modulus
12-4
8/7/2019 FE1001 Part 2 Lectures 22-27
3/41
an
Textbook:
R. A. Serway and J W Jewett, Jr, Physics for
Scientists and Engineers, 6th Ed, Vol 1, Thomson,2004.
R. C. Hibbeler, Engineering Mechanics Statics, 2nd
, , .
A. Gupta and C. K. Soh, Intelligent Interactive Tutoring
12-5
- , , .
12.1 Conditions for E uilibrium
Equilibrium - General
Equilibrium Equations End Supports
12-6
8/7/2019 FE1001 Part 2 Lectures 22-27
4/41
Static E uil ibrium
qu r um mp es t e o ect s at rest stat c
or its center of massmoves with a constant
Static equilibrium is a common situation in
Principles involved are of particular interest
eng neer ng
to civil engineers, architects, and mechanical
engineers
12-7
.
In this chapter, we will concentrate on Static
Equilibrium
The object will not be moving
D namic E uilibrium is also ossible The object would be translating with a constant
velocity and/or rotating with a constant angular
velocity
In either case F = 0 and = 0
12-8
8/7/2019 FE1001 Part 2 Lectures 22-27
5/41
Conditions for Equilibrium
Fi = 0 e o ec s mo e e as apar c e,
then this is the only condition that must
Net torque (or moment) equals zero
i = i =
This is needed if the object cannot be
12-9
. .
= r F (or = r F) -
rule to determine the
direction of the torque The tendency of the force
to cause a rotation about
moment arm d
12-10
8/7/2019 FE1001 Part 2 Lectures 22-27
6/41
Vector Re resentation of Moment 2
12-11
Moment of a Force
12-12
8/7/2019 FE1001 Part 2 Lectures 22-27
7/41
Vector Representation of Moment
12-13
Cross Product of Two Vectors
12-14
8/7/2019 FE1001 Part 2 Lectures 22-27
8/41
12-15
Coordinate Independence of Couples
F
12
FA
12-16
8/7/2019 FE1001 Part 2 Lectures 22-27
9/41
Need the angular acceleration
(vs.translational acceleration aof the ob ect to
be zero
= For rotational equilibrium,
i = i = This must be true for any axis of rotation
12-17
Ax is of Rotation for Torque Equation
e ne orque s a ou an ax s roug anypoint in the xyplane
If an object is in translational equilibrium and
,the net torque must be zero about any otheraxis i.e.
Fx = 0
Fy = 0
12-18
z = 0 (or Mz = 0), about any other axis
8/7/2019 FE1001 Part 2 Lectures 22-27
10/41
Axis of Rotation for Torque Equation, 2
Given that: F = 0 an
O = 0 (or MO = 0)022110 =++= LFrFr
( ) ( )
( )'
''
212211
2211'0
++++=
++=LL
L
FFrFrFr
FrrFrr
0=
--- If an object is in translational equilibrium and the
12-19
net torque is zero a out one axis, t en t e nettorque must be zero about any other axis
E uil ibrium E uations
.
Translational equilibrium => resultant external
Fi = 0
be zero when viewed from an inertial frame of reference
Rotational equilibrium => resultant external
torque about anyaxis must be zero:
i = 0 (or Mi = 0 )
12-20
The angular acceleration must equal zero
8/7/2019 FE1001 Part 2 Lectures 22-27
11/41
,
We will restrict the applications to situations
in which all the forces lie in the xyplane
These are called coplanarforces since they lie in
the same plane
here are three resulting equations
Fx = 0
Fy = 0
z = 0 or Mz = 0
12-21
12-22
8/7/2019 FE1001 Part 2 Lectures 22-27
12/41
MRy
-
Can move Can rotate in Fixed
12-23
horizontally androtate in the plane
the plane
Support Provided by a Smooth oroug ur ace
12-24
8/7/2019 FE1001 Part 2 Lectures 22-27
13/41
Support Reactions: Simply Supported Beam
for finding Reactions
RAxRAx
y
y RAx
y
y
12-25
FX = 0 RAx=0
Princi le of Transmissibil it
12-26
8/7/2019 FE1001 Part 2 Lectures 22-27
14/41
E uivalent Force S stems
body form a force system
Two force systems (applied to the same rigid
),,,( 121
1
1
2
1
1 Lvv
Lvv
MMFF ),,,( 222
1
2
2
2
1 Lvv
Lvv
MMFF
body) are equivalent when:
he resultant forces of the two force s stems are
the same;
he net torques of the two force systems (about
=j
j
i
i FF21vv
12-27
the same point) are the same )()( 21 OO = vv
E uivalent Force S stem
12-28
8/7/2019 FE1001 Part 2 Lectures 22-27
15/41
Force and Moment Vectors
12-29
Simplification of Force Systems
equivalent to a force applied at the given
. ., .
=
iA FF
1vv
= )(AMMAvv
(or, any force system can be simplified to be
12-30
8/7/2019 FE1001 Part 2 Lectures 22-27
16/41
12.2 Center of Gravit
force applied to a rigid body (or.
Find a point (i.e., A) such that therA
r
A
simplified moment is just 0.
( ) 0)( === WdrrAMM AAvvvvv
V
WrWdrWdr AV
A
V
vvvvvv==
12-31
--- e str ute grav tat ona orce s equ va ent to t e
concentrated simplified force applied at the point A.
t e var ous grav tat onaforces acting on all the
equivalent to a single
ravitational force actinthrough a single pointcalled the Center of Gravity(CG)
L++= 222111
xgmxgmx ii
xm=
12-32
L++ 2211 gmgm i
m
8/7/2019 FE1001 Part 2 Lectures 22-27
17/41
Center of Gravit 2
e orque ue o e grav a ona orce on
an object of mass Mis the force Mgacting at
Ifg is uniform over the object, then the
en er o rav y o e o ec co nc es
with its Center of Mass(CM)
t e o ect s omogeneous an symmetr ca ,
the Center of Gravity(CG) coincides with its
12-33
eome r c en er
Centre of Gravity and Mass for Particles
~~~
~~~
for CG===
i
ii
i
ii
i
ii
W
Wzz
W
Wyy
W
Wxx
for CM
===i
ii
i
ii
i
ii
mz
m
myy
m
mxx
:coordinates of
each particleyx ~,~,~
coor na es o ecentre of gravity
G of particles
:,, yx
12-34
8/7/2019 FE1001 Part 2 Lectures 22-27
18/41
=
=
dmxMx
m
~=
=
dWxWx ~
= dmyMy~
~= dWyWy~
~,
= mz=
12-35
-
( ) WxdWxxdWWx ==
=
12-36
( ) WydWyydWWy ==
8/7/2019 FE1001 Part 2 Lectures 22-27
19/41
Ex 12 .1: Locate the CM of theara o c o , x = y
2
( ) dyydy
xdydydxdL 22 211 +=
+=+=
m479.1411
21
=+== dyydLL
( ) 21
0
2m8484.021 =+== dyyydLyydm
m574.0== Mydmy
:density per length:
( ) ( ) 21
0
222 m6063.02121 =+=+==
dyyydyyxdLxxdm
12-37
m410.0== Mxdmx
Ex: Locate the Center of Geometry of
y2 2
1
2= - -
x
Parts Ai xi Aixi yi Aiyi2.14Ax
1 12 2 24 1.5 18
2 -3.14 1 -3.14 2 -6.28
07.286.6
.===
iAx
51.110.39
=== ii Ay
12-38
3 -2 3.33 -6.66 0.67 -1.33
6.86 --- 14.2 --- 10.39
6.86iA
8/7/2019 FE1001 Part 2 Lectures 22-27
20/41
12.3 Rigid Objects in Stat ic Equilibrium
Problem-Solving Strategy
External and Internal Forces
Support Provided by a Surface
x . : e g e an xamp e
Ex 12.3: Standing on a Horizontal Beam
Ex 12.4: Leaning Ladder
Ex 12.5: Negotiating a Curb
12-39
Truss
Problem-Solving Strategy
Draw a diagram of the system
Draw a free-body diagram(FBD)
object
Indicate the locations ofallthe forces
For systems with multiple objects, draw a
separate free-body diagramfor each object
12-40
8/7/2019 FE1001 Part 2 Lectures 22-27
21/41
Problem-Solving Strategy, 2
Find the components of the forces along the two axes = Be careful of signs
torque (moment) on the object
Remember that the choice of the axis is arbitrary
Choose an origin that simplifies the calculations as
much as possible
12-41
A force that acts along a line passing through the
origin produces a zero torque
Problem-Solving Strategy, 3
Apply the second condition for equilibrium,
i = 0 (or Mi = 0 ) The two conditions of equilibrium will give a
Solve the simultaneous equations
,
the direction opposite what you drew in the free-
12-42
8/7/2019 FE1001 Part 2 Lectures 22-27
22/41
External and Internal Forces
12-43
External and Internal Forces 2
12-44
8/7/2019 FE1001 Part 2 Lectures 22-27
23/41
External and Internal Forces 3
section.
12-45
Ex 12.2: Weighted Hand Example
Model the forearm as arigid bar
e weig t o t e orearmis ignored
the x-direction
conditions from the free-body diagram
12-46
8/7/2019 FE1001 Part 2 Lectures 22-27
24/41
Ex 12.2: Weighted Hand Example, 2
o ve or t e un nown
forces (F and R) pp y e con on or
torque equilibrium using the
rotation (o =0)mgo =
0)35)(50()3(F ==
12-47
N583F =
Ex 12.2: Weighted Hand Example, 3
Apply the condition for forceequilibrium (Fy = 0)
mgy =
0N0.50RF ==
50FR =
N53350583 ==
12-48
8/7/2019 FE1001 Part 2 Lectures 22-27
25/41
Ex 12.3: Standing on a Horizontal Beam
uniform => the center of ravit
is at the geometric centerof the beam
he person of 600 N isstanding on the beam
a are e ens on
in the cable and the
12-49
wall on the beam?
Ex 12.3: Standing on a Horizontal Beam, 2
Draw a free-bodydiagram
Use the pivot in theproblem (at the wall)as e p vo
This will generally be
Note there are threeunknowns TR
12-50
8/7/2019 FE1001 Part 2 Lectures 22-27
26/41
Ex 12.3: Standing on a Horizontal Beam, 3
resolved into componentsin the free-bod dia ram
Apply the two conditionsof e uilibrium to obtainthree equations
Solve for the unknowns
026004200853sin
==
== TA=
=
N580
N313
R
T
12-51
020060053sinsin =+= TRFyx
= 1.71
he ladder is uniform
=> the weight of the ladder
center (its center of gravity)
here is static friction(s=0.4) between the
ladder and the ground
Find the minimum anglemin at which the ladder
12-52
oes not s ip
8/7/2019 FE1001 Part 2 Lectures 22-27
27/41
Ex 12.4: Leaning Ladder, 2
raw a ree- o y agram or
the ladder
The frictional force fis:
=s , s .static friction
Let Obe the axis of rotation
Apply the equations for the
two conditions of equilibrium
12-53
Solve the equations
Ex 12.4: Leanin Ladder 3
0cos2
sin == mgPo l
l
0
0
==
==
PfF
mgnF
x
y
from 1st and 2nd eqs:
mgnmg
P == andcot
from 3rd eq:
mgnmg
Pf ss === cot2
12-54
== 3.5125.12
1tan
min
s
Q
8/7/2019 FE1001 Part 2 Lectures 22-27
28/41
Ex 12.4: Leaning Ladder, Extended
at a distance dfrom the
The higher the personclimbs the lar er theangle at the base needsto be in order to remainin equilibrium
Eventually, the ladder
12-55
may s ip
Ex 12 .5: Negotiating a Curb
(A) n magn u e oforce F to be
(B) Magnitude and
reaction R
mg=700 N
r= 30 cm
h = 10 cm
12-56
8/7/2019 FE1001 Part 2 Lectures 22-27
29/41
Ex 12 .5: Negotiat ing a Curb, 2
be applied to roll up==
( ) 2 222 == hrhhrrd
N3132
=
=hr
mgdF
direction of reaction R
( )
==
=+=
66
N767
1
22
mgtan
FmgR
12-57
, ,form a triangle
F
Roof Truss
12-58
8/7/2019 FE1001 Part 2 Lectures 22-27
30/41
12-59
.(2). Connected by joints;(3). Loads are applied to joints.
Simpletruss:
(1). Start from a triangle;(2). Add every 2 members to form a new joint;
. .
12-60
8/7/2019 FE1001 Part 2 Lectures 22-27
31/41
(1). Find reactions;
A simple truss can be solved by Joint Method with 3 steps:
(2). Number the joints (from 0 to n);
(3). Solve the joints (from n to 0).
(a) Free body diagram of joint with tension convention;
(b) Fx = 0; Fy = 0;
(c) + is tension force, - indicates compression force.
12-61
(1)Statically Determinate Reactions
12-62
8/7/2019 FE1001 Part 2 Lectures 22-27
32/41
Determine forcesactin at the ins
Rockers / rollers atsu orts A &E
Fg= 7200 N
Note that FAB=FBAB D
FBA F
FBD FDB
F FDE
L = 50 mA C E
FAB
FAC FCA
FCB FCD
FCE
FED
FEC
12-63
nA nEFg
Anal sis of a Truss 2
B DF
FBD FDB
A C E
FAB
FAC FCA
FCB FCD
FCE
BC DC DE
FED
FECequilibrium
nA nEFg
( ) AEgEAgEAy nnFLLnMFnnF =====+= N360002,0
on)(compressiN7200030sin ==+= ABABAy FFnF
12-64
(tension)N6235030cos ==+= ACABACx FFFF
8/7/2019 FE1001 Part 2 Lectures 22-27
33/41
12.4 Elasticit
Deformation
Youngs Modulus
Shear Modulus
Moduli and Types of Materials
12-65
Deformation
So far we have assumed that objectsremain rigid when external forces act ont em
Except springs
ctua y, o ects are e orma e
It is possible to change the size and/or shape
Internal forces resist the deformation
12-66
8/7/2019 FE1001 Part 2 Lectures 22-27
34/41
Stress and Strain
Stress Is proportional to the force causing the
deformation
It is the external force acting on the object perun area
Strain
Is a measure of the degree of deformation
12-67
between stress and strain For sufficiently small stresses, strain is directly proportional
o s ress
It depends on the material being deformed
It also depends on the nature of the deformation Elastic modulus in general relates what is done to a
solid object to how that object responds
strain
s ressmoduluselastic
12-68
ar ous ypes o e orma on ave un que e as cmoduli
8/7/2019 FE1001 Part 2 Lectures 22-27
35/41
Youngs Modulus
Measures the resistance of a solid to a change in
Shear Modulus
within a solid parallelto each other
Bulk Modulus
Measures the resistance of solids or liquids to
changes in their volume
12-69
he bar is stretchedby an amount L
force F
ratio of the externalforce to the cross-sectional area A, i.e.
AF
12-70
8/7/2019 FE1001 Part 2 Lectures 22-27
36/41
Youn s Modulus 2
ens on stra n s t e rat o o t e c ange n
length to the original length, i.e. LL oung s mo u us, , s e ra o o ose wo
ratios:
tensile stress
tensile strainY
L =
Units are: N / m2
(Pa)
i
12-71
Stress vs. Strain Curve
Ex eriments show that for certainstresses, the stress is directlyproportional to the strain
the curve
The elastic limitis themaximum stress that can beapplied to the substance before itbecomes ermanentl deformed
When the stress exceeds the elastic limit, the substance willbe permanently deformed
12-72
The curve is no longer a straight line
With additional stress, the material ultimately breaks
8/7/2019 FE1001 Part 2 Lectures 22-27
37/41
deformation occurs whena force acts parallel to oneof its faces while theopposite face is held fixedy ano er orce
This is called a shears ress
For small deformations, no change in volume occurs
12-73
A good first approximation
Shear Modulus, 2
Shear stressis / A Fis the tangential force
A is the area of the face being sheared
x x is the horizontal distance the sheared face moves h is the height of the object
Shear modulus is the ratio of the shear stress tothe shear strain:
shear stress
shear strainS
x= =
12-74
Units are: N / m2 (Pa)
8/7/2019 FE1001 Part 2 Lectures 22-27
38/41
occurs when a force of uniformma nitude is a liedperpendicularly over the entiresurface of the object.
The object will undergo achange in volume, but not ins ape.
Volume stressis defined as the ratio of the magnitude
12-75
, , A, of the surface.
Bulk Modulus 2
s s a so ca e epressure.
Volume strainis the ratio of the change in.
Bulk modulus is the ratio of the volume stress
to the volume strain:
volume stressF
PAB
= = =
The negative sign indicates that an increase in
i iV V
12-76
pressure wi resu t in a ecrease in vo ume.
8/7/2019 FE1001 Part 2 Lectures 22-27
39/41
Compressibility is the inverse of the bulkmodulus.
It is often used instead of the bulk modulus.
12-77
Both solids and liquids have a bulk modulus
Liquids cannot sustain a shearing stress or
a ens e s ress
If a shearing force or a tensile force is applied
to a liquid, the liquid simply flows in response.
12-78
8/7/2019 FE1001 Part 2 Lectures 22-27
40/41
Moduli Values
12-79
A rigid body is supported by 3 deformable rods, find
reactions N , N and N (all the self wei hts are ne li ible).
x
yP= 100 N
o u us: = m
Rod length: L=1 m
Cross area: A=0.1 m2
NA N N
Distances between rods: 1 m
01000 =++= CBAy NNNF :Forces are in equilibrium:
=+= .: CBA
Deformations are compatible: Points A, B and C are on the same line
12-80
+AE
LNA A10,:
+AE
LNB B11,:
+AE
LNC C12,:
8/7/2019 FE1001 Part 2 Lectures 22-27
41/41
Thank You !Thank You !
12-81