Lec1 Introduction.unlocked
Transcript of Lec1 Introduction.unlocked
-
8/21/2019 Lec1 Introduction.unlocked
1/25
9/14/2013
1
Structural Analysis-I
Dr. Anis Shatnawi
Second Semester
2013-2014
Lecture 1
Structural Analysis
Introduction tothe Course
2
-
8/21/2019 Lec1 Introduction.unlocked
2/25
9/14/2013
2
Intro 3
Introduction-Contents
Previous Structures Courses Stress Equations
Static Determinacy and Indeterminacy
This Course
Types of Structures
Getting Feedback
Loads
Units and Accuracy
Sign Conventions
Principle of Superposition
Intro 4
A structure refers to a system of connected
parts used to support a load.
Important examples related to civil engineering
include buildings, bridges, and towers; and in otherbranches of engineering, ship and aircraft frames,
tanks, pressure vessels, mechanical systems, and
electrical supporting structures are important.
-
8/21/2019 Lec1 Introduction.unlocked
3/25
9/14/2013
3
Intro 5
We examine Structures in this classBridge
RainbowBridge inFolsom
Intro 6
We examine Structures in this classAnother Bridge
Vancouver, B.C.
•Note it is composed of several smallerstructures, called substructures.
-
8/21/2019 Lec1 Introduction.unlocked
4/25
9/14/2013
4
Intro 7
We examine Structures in this classAnother Bridge-Closer View
Quick
view of
Tacoma
Narrows
Bridge
Note it is composed of several substructures.
How many different substructures can youidentify?
Suspension Cables
Tension Cables
Stiffening Truss
Support Frames
Roadway/Beams
Any Others?
Intro 8
We examine Structures in this classAnother Bridge-End View
Tower Support Frames
Foundation
-
8/21/2019 Lec1 Introduction.unlocked
5/25
9/14/2013
5
Intro 9
Very Big BridgeLongest single arch bridge in the country
New River Gorge- West Virginia
I nt ro 1 0
Very Big Bridge
Steel bridge
-
8/21/2019 Lec1 Introduction.unlocked
6/25
9/14/2013
6
I nt ro 11
Very Big Bridge Note how three-dimensional the
bridge is!
I nt ro 1 2
What is this class?
StaticsDeals with equilibrium of rigid bodies.Structures did NOT deform.
Mechanics of MaterialsLooked at what goes on inside a structure
when loads are applied. Structures deform,
that is, they change shape.
You have had two previous courses in structures:
-
8/21/2019 Lec1 Introduction.unlocked
7/25
9/14/2013
7
I nt ro 1 3
Statics
Statics is Equilibrium:
•These are vector equations.
•In 2D, there are 2 force and 1 momentequations, for a total of 3 equations.
•In 3D there are 3 force and 3 momentequations, for a total of 6 equations.
F=0
M=0
What did you do in Statics?
I nt ro 1 4
Statics
Equilibrium:
Equilibrium is very important for thiscourse.
WARNING: Most errors you make inthis course will be errors in applying
equilibrium.
F=0
M=0
-
8/21/2019 Lec1 Introduction.unlocked
8/25
9/14/2013
8
I nt ro 1 5
Free Body Diagrams (FBD)
The concept of a FBD is very important.
A FBD of the entire structure will show allforces acting on it. You replace the supports
with the reactive forces that are supplied by the
supports.
I nt ro 1 6
Free Body Diagrams (FBD)
The concept of a FBD is very important.
A FBD can be only a portion of the structureobtained by taking a section at one or more
points.
A FBD of the entire structure will show allforces acting on it. You replace the supports
with the reactive forces that are supplied by the
supports.
You must show all forces on the FBD.When a section is taken you must include
all internal forces acting at the section.
-
8/21/2019 Lec1 Introduction.unlocked
9/25
9/14/2013
9
I nt ro 1 7
External and Internal Forces
There are two basic types of forces,External and Internal:
External forces are those that act on the FBD.Loads and reactions are examples.
I nt ro 1 8
External and Internal Forces
There are two basic types of forces,External and Internal:
External forces are those that act on the FBD.Loads and reactions are examples.
Internal forces exist within the structure andare necessary to hold the structure together. Axial forces, shear forces and bendingmoments are examples.
Internal forces always occur in pairs, sincetaking a section will produce two FBD’s andeach FBD must have the internal force actingon it.
-
8/21/2019 Lec1 Introduction.unlocked
10/25
9/14/2013
10
I nt ro 1 9
Reactions at Supports
In 2 dimensions, there are basicallythree types of supports.
R
y
Roller -has only 1 force reaction to prevent
translation in 1 direction. The reaction can
act up or down.R
xR
y
Pin -supplies 2 force reactions to prevent
translation in 2 directions.
Fixed-provides the 2 force reactions likethe pin but also prevents rotation by
supplying a moment reaction.
Note that the
reactions arealways shown
in the positive
directions.
R
x R
y
MR
I nt ro 2 0
Mechanics of Materials
In this course we allowed the structures todeform, i.e. change in shape.
Strain is the quantitative measure of thisdeformation.
Stress is related to strain through Hooke’s law.
You learned to calculate stresses, strains, anddisplacements of structures in mechanics.
-
8/21/2019 Lec1 Introduction.unlocked
11/25
9/14/2013
11
I nt ro 2 1
Displacement and Deformation
Displacement is the movement from onelocation to another.
Deformation means a change in shape.
You can have one without the other.
They usually occur together.
Can you think of examples of them
occurring separately and together?
I nt ro 2 2
Stress Equations from Mechanics
These equations for stress are justfine. We really cannot improveupon them.
J
Tr
Ib
VQ
I
My
A
N
T
V
M
ax
How about A; I; and J? Or y or r? Q? Or N; M; V; T?
Which quantities in theseequations are most difficult to find?
-
8/21/2019 Lec1 Introduction.unlocked
12/25
9/14/2013
12
I nt ro 2 3
Statically Determinate Structures
A structure is Statically Determinate (SD) ifyou can calculate all reactions and internalforces just using the equations of statics, i.e.equations of equilibrium.
This means that you must have the samenumber of unknowns as you have equationsof equilibrium.
Number of unknowns=Number of equilibrium equations All problems we examined in Statics and most
in Mechanics were statically determinate.
I nt ro 2 4
Statically Indeterminate Structures
A structure is Statically Indeterminate(SI) if you cannot calculate all reactions andinternal forces just using the equations ofstatics, i.e. equations of equilibrium.
This means that you have more unknownsthan you have equations of equilibrium.
You solved a few SI problems in yourMechanics class.
Number of unknowns>Number of equilibrium equations
But not many!
In this class we solve SI problems!
-
8/21/2019 Lec1 Introduction.unlocked
13/25
9/14/2013
13
I nt ro 2 5
Example 1
P
This beam has 2 unknown reactions, R A and RB.
RA RB
In a beam, there are assumed to be no hor izontalforces and therefore we show no horizontalreactions on a beam.
There are 2 equilibr ium equations instead of theusual 3 because the FX=0 gives us no usefulinformation since there are no horizontal forces.
2 equations to solve for the 2 unknown reactionsmeans this problem is Statically Determinate (SD).Number of equations =Number of unknowns
I nt ro 2 6
Example 2
This beam has 3 unknown reactions,R A, RB, and RC.
PP
RA RB RC
There are still only 2 equilibrium equations.
2 equations to solve for the 3 unknownreactions means it is StaticallyIndeterminate (SI).
Why no R AX?
Number of equations < Number of unknowns
-
8/21/2019 Lec1 Introduction.unlocked
14/25
9/14/2013
14
I nt ro 2 7
Example 2
This beam has 3 unknown reactions,R A, RB, and RC.
PP
RA RB RC
There are still only 2 equilibrium equations.
2 equations to solve for the 3 unknownreactions means it is StaticallyIndeterminate (SI).
Why no R AX?
Since there is 1 more unknown thanequations, we say it is SI 1.
I nt ro 2 8
Example 3
This beam has 4 unknown reactions, R A, RB,RC, & M A.
PP
RB RCRA
MA
There are still only 2 equilibrium equations.
2 equations to solve for the 4 unknownreactions means this problem is StaticallyIndeterminate (SI). equations < unknowns
Since there are 2 more unknowns thanequations, we say it is SI 2.
-
8/21/2019 Lec1 Introduction.unlocked
15/25
9/14/2013
15
I nt ro 2 9
In the stress equations, the internal forces, Vand M for a beam, are the most difficult termsto evaluate if the structure is at all complicated.
Main Goal in this Course
For SI structures, evaluating these internalforces can be a formidable task.
Finding these internal forces is the main
goal of this course.
I nt ro 3 0
Once the internal forces are found, the stressequations from Mechanics can be used.
Completing the Analysis
Once the stresses are found, they are
compared to the material strength to see ifthe structure is safe.
In this class, we will usually stop after theinternal forces are found, but always rememberthat these next steps must always be performedto complete the analysis.
-
8/21/2019 Lec1 Introduction.unlocked
16/25
9/14/2013
16
I nt ro 3 1
The internal forces are best illustrated bydrawing the internal force diagrams.
Internal Force Diagrams
For beams, you are familiar with shear forceand bending moment diagrams.
Trusses have only axial forces, no shear ormoment.
For frames, we will have axial force, shearforce, and bending moment diagrams .
I nt ro 3 2
Sign Conventions
We must be very careful with our signs.
Remember that we always have two signconventions: equilibrium and beam.
They are used for different purposes.
They are not contradictory. They are complimentary.
We always need to use the two conventionstogether.
-
8/21/2019 Lec1 Introduction.unlocked
17/25
9/14/2013
17
I nt ro 3 3
Fy=0: In this equation a force will be positiveif it acts upward, the positive direction of the y-axis.
FX=0: In this equation a force will be positiveif it acts to the right, the positive direction ofthe x-axis.
Equilibrium Sign Convention
We use this for summing forces and moments ,e.g. to calculate reactions.
x
y
z
+
+
Positive horizontal force for equilibr ium
Positive vertical force for equilibr ium
I nt ro 3 4
Equilibrium Sign Convention
M=0. In this equation a moment will bepositive if it acts counterclockwise, (CCW),around the positive z-axis by the right handrule.
We will maintain this consistent sign convention
throughout the semester.
Positive moment for
applying equilibrium.
x
y
z
+
-
8/21/2019 Lec1 Introduction.unlocked
18/25
9/14/2013
18
I nt ro 3 5
Beam Sign Convention
Axial Force:
P P
P P
Compression is negative
Tension is positive
I nt ro 3 6
Shear Force
Take a section:
+V makes the right FBD tends to rotate ClockWise
+V
-
8/21/2019 Lec1 Introduction.unlocked
19/25
9/14/2013
19
I nt ro 3 7
Shear Force
+V
+V makes both FBD’s tend to rotate Clock Wise.
I nt ro 3 8
Shear Force
-V causes a Counter Clock Wise (CCW)
rotation of each Free Body Diagram (FBD)
-V
+V
+V causes a Clock Wise (CW) rotation
of each Free Body Diagram (FBD)
-
8/21/2019 Lec1 Introduction.unlocked
20/25
9/14/2013
20
I nt ro 3 9
Shear Force
+V
+V causes a Clock Wise (CW) rotation
of each Free Body Diagram (FBD)
-V
-V causes a Counter Clock Wise (CCW)
rotation of each Free Body Diagram (FBD)
I nt ro 4 0
Moment
+M
+M causes Compression to occur
on the top of the beam
Take a section:
-
8/21/2019 Lec1 Introduction.unlocked
21/25
9/14/2013
21
I nt ro 4 1
Moment
+M
+M causes Compression to occur
on the top of the beam
Notice the deformed shape.It holds water.
Take a section:
The deformed shape smiles at you
I nt ro 4 2
Moment
-M causes Compression to occur
on the bottom of the beam
-M
This deformed shape sheds water.
+M
+M causes Compression to occur
on the top of the beam
Smileyface
Frowneyface
-
8/21/2019 Lec1 Introduction.unlocked
22/25
9/14/2013
22
I nt ro 4 3
Linear Problems
Nearly all structures you have examined sofar are linear problems.
For a linear problem, if you plot load vs. anyresult of that load (e.g. a reaction,displacement, internal force, etc.), you get astraight line.
What is the only nonlinear problem you
looked at in Mechanics? For linear problems, the Principle ofSuperposition always applies.
I nt ro 4 4
Principle of Superposition
The Principle of Superposition says manythings :
If you double the loads then you double anyresult of the loads.
If you have several loads acting on thestructure, you can solve the problem for eachload separately and then add the results.
-
8/21/2019 Lec1 Introduction.unlocked
23/25
9/14/2013
23
I nt ro 4 5
Principle of Superposition-Loads
For example:
To solve this problem with both a concentratedload and a uniform load, we can solve the
problem first for the load of P and then for the
load of w0, and then add the results together.
RBRA
P w0
L
This beam has two loads: a concentrated load of
P and a uniform load of w0.
I nt ro 4 6
Alwayscheck
reactions!
OKso 0P2
P
2
PF:check
2
PR
LR2
LP0M
2
PR
LR
2
LP0M
y
A
AB
B
BA
Principle of Superposition-Loads
Look firstat the
load of P
RBRA
P
L
2
P
2
P
-
8/21/2019 Lec1 Introduction.unlocked
24/25
9/14/2013
24
I nt ro 4 7
Principle of Superposition-Loads
Now look
at the
uniform
load w0
RBRA
w0
L
2LwO
2LwO
0Lw2
Lw
2
LwF:check
2
LwR
LR2
L
Lw0M
2
LwR
LR2
LLw0M
OOO
y
OA
AOB
OB
BOA
Always check
reactions!
I nt ro 4 8
Principle of Superposition-Loads
So, we just
add together
the solutions
for these two
simplerproblems to
get the
answer to the
original
problem
RA RB
P
+ RBRA
w0
RBRA
Pw
0
=
2
P
2
P
2LwO
2LwO
2Lw
2
PO
2Lw
2
PO
-
8/21/2019 Lec1 Introduction.unlocked
25/25
9/14/2013
I nt ro 4 9
Principle of Superposition-Displacements
It also works
with
displacements.
Find those due
to P, DP, and
those due to
w0
, Dw
, and
add them
together:
D=DP+Dw.
P w0
D
=
w0
Dw
+
P
DP
I nt ro 5 0
Structural Analysis
End ofIntroduction