lec1 introduction.unlocked

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Structural Analysis-I

Dr. Anis Shatnawi

Second Semester

2013-2014

Lecture 1

Structural Analysis

Introduction tothe Course

2


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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.


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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.


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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


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Intro 9

Very Big BridgeLongest single arch bridge in the country

New River Gorge- West Virginia

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Very Big Bridge

Steel bridge


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Very Big Bridge Note how three-dimensional the

bridge is!

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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:


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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?

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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


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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.

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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.


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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.

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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.


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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

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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.


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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?

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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?


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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.

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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!


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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

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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


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Example 2

This beam has 3 unknown reactions,R A, RB, and RC.

PP

RA RB RC


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.

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Example 3

This beam has 4 unknown reactions, R A, RB,RC, & M A.

PP

RB RCRA

MA


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.


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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.

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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.


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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 .

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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.


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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

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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

+


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Beam Sign Convention

Axial Force:

P P

P P

Compression is negative

Tension is positive

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Shear Force

Take a section:

+V makes the right FBD tends to rotate ClockWise

+V


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Shear Force

+V

+V makes both FBD’s tend to rotate Clock Wise.

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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)


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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)

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Moment

+M

+M causes Compression to occur

on the top of the beam

Take a section:


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Moment

+M



Notice the deformed shape.It holds water.

Take a section:

The deformed shape smiles at you

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Moment

-M causes Compression to occur

on the bottom of the beam

-M

This deformed shape sheds water.

+M



Smileyface

Frowneyface


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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.

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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.


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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.

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Alwayscheck

reactions!

OKso 0P2

P

2

PF:check

2

PR

LR2

LP0M

2

PR

LR

2

LP0M

y

A

AB

B

BA


Look firstat the

load of P

RBRA

P

L

2

P

2

P


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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!

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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


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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

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Structural Analysis

End ofIntroduction

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