Lecture notes, lectures 1-4 PDF

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Topics

Section I (Week 1 to Week 7)

Scope of Structures 1

Section I: Basic structural analysis Types of structures and loadings Basic concepts and analysis techniques Analysis of Trusses Analysis of beams and frames

Section II: Deflection of structures Deflection of beams Integration method Moment area method Deflection of structures using energy methods Principle of virtual work

Lecture 1

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Lecture 1

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Why we study structural analysis?

Structural Engineering is about the conception, design and construction of the structural systems that are needed in support of human activities. Civil engineering projects include bridges, building, dams, storage facilities etc. involves structural engineering.

Answer: To prevent this to happen!

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CV

Compassvale School, Singapore

•Structural analysis is rarely an end in itself. Rather, it is a tool used by structural designers to assist in the creation of design concepts and to demonstrate that these concepts satisfy project requirements.

Lecture 1

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2

Introduction

Structural forms and classifications What is a structure?

• Structural forms and classifications

• Structures refer to a system of connected parts used to support a load (e.g. se weight, occupants/furniture for building, traffics for bridges)

– Structural members and their characteristics – Different forms of structures and their properties

• Factors to consider – Safety (most important!) – Serviceability (e.g. extensive deflection or vibration is not allowed) – Esthetics – Economic and environmental constraints (cost to build and use, energy consumption etc.)

• Loads – Different loadings type and their characteristics

• Structural analysis and design – An introduction to the basic concepts

Lecture 1

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Lecture 1

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Structural members and their characteristics

Tie Rod and Struts • Tie rod normally refer to structural members subjected to tensile force only

• For the ease of design and construction, structures are often composing of a number o structural element o members • In this course, the following commonly encountered member types shall be discussed: – Tie rods – Struts – Beams – Columns

• Normally tie rods are rather slender • Strut are members that could subject to both tensile and compressive forces • They are commonly used in trusses and bridges More on tie rod in Wikipedia http://en.wikipedia.org/wiki/Tie_rod More on tie strut in Wikipedia http://en.wikipedia.org/wiki/Strut

Lecture 1

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Lecture 1

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•Compression Structures Strut

in an arch

•Stable 3-hinge arch continues to support the load

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•Tension Structures

Columns

Beams • Beam are usually straight horizonta members used primarily to carry vertical loads • They are classified to how they are supported • Design to resis bending momen and shear forces • In steel beam, wide flange cross section are commonly used

• Column usually straight vertical members to resist vertica compression • Tubes and wide flange cross section are often used • A column is called a beam column if it is subjected to both vertical compression and bending More about beam in Wikipedia: http://en.wikipedia.org/wiki/Beam_% 28structure%29

Lect

More about column in Wikipedia http://en.wikipedia.org/wiki/Column 19

Lecture 1

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Structural forms and classifications

Trusses • Used when long span is needed with not much depth constraints • Consist of slender struts/ties arranged in triangular or other fashions • Planer trusse (all members lie in same plane) used in bridges and roof • Space trus (members lie in different planes) used in derricks and towers

• The combination o structural elements (rods, strut, beam, column) and materials (concrete, steel, timber) generate different structural forms o structural systems • In this course, focus shall be given to the following two most popular types o skeleton structural systems – Trusses – Frames (including simple beams) Lecture 1

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Lecture 1

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Structural Analysis and Design Find out the loads on structures

Rechecking

Selection of structural form (trusses, frames) and material (structural steel, concrete)

Main study objective of this course

Structural analysis (calculate the forces in the structure)

Rechecking

Design (to provide strong enough members to resist the forces Build (to construct the structures)

Lecture 1

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Next Lecture • Some essential tools for structural analysis – Principle of superposition – Equilibrium – Internal forces – Free body diagrams

Lecture 1

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Idealization structures • In real sense, an exact analysis is almost impossible due to uncertainty in members dimensions and loading location (especially before the structure was built!) • Hence, in practice, in order to carry out force analysis, structural engineers must know how to idealize a structures to a simple form/geometry for analysis • Structural prismatic elements like rod/beam/column members are normally idealized as “line elements” with appropriate support conditions

Lecture 2 Idealization structures Geometry Loading Supports and connections

Principle of superposition Equilibrium Internal forces Free body diagrams

Lecture 2

• At the same time an appropriate idealized loading and supports and connections should also be applied to the structure 1

Lecture 2

2

•Simplification for the purpose of analysis

•All structures have to be simplified in a convenient form to carry out the structural analysis; actual structures often have complicated details •Experience and good judgement of structural behaviour are required in carrying out such simplification

Support connections Real and actual supports/connections could come with many different details In structural analysis, they are idealized into a few support and connection types

Pin connection

Lecture 2

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Fixed connection

Lecture 2

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1

Idealized connection types

Idealized structures

Roller suppor : no relative vertical movement, allows some freedoms for slight rotation tal and vertical movement, allows some freedoms for slight rotation Fixed join : no relative horizontal, vertical movement and rotation 9  Roller support

• When selecting the model for each support, the engineer must be aware how the assumptions will affect the actua performance • The assumptions must be reasonable for the structural design • The simplified loadings should give results that closely approximate the actual loadings • Common types of connections on coplanar structures are given in Table 2.1 of Textbook (Page 37)

Pin support Pin connection =9 

Fixed connection Lecture 2

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Lecture 2

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Reactions

Different types o idealised support conditions for planar structures:

Pin support with F and Fy

Tie support with Fn

Generally known as roller suppor with only one reaction force Fy

Partially fixed or fixed support with F F and M

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Quite often, a body or a combination of bodies is to be analysed and it is isolated as a single body from the surrounding bodies. Hence, a free body diagram is a diagrammatic representation of the isolated body or combination of bodies treated as a single body, showing ALL forces applied to it and by mechanical contact with other bodies that are imagined to be removed.

Example: Draw the free body diagram for the beam. Neglect the weight of the beam.

Idealized structures: An example

Lecture 2

Exampl : A cylinder is supported on a smooth inclined surface by a tw bar frame . Draw the free body diagram for the cylinder, the two bar frame and the pin at C.

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Example: Draw the free body diagram for the pulley, the post and the beam CD.

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Equations of Equilibrium • For the general 3D cases, six equations: F

x 0

M

x 0

F 

y0

M y 0

F 0 z  M 0 z

• For planar structures (and problems appear in this course), reduces to 3 equations:  Fx  0 ,  Fy  0 ,  M o  0

Lecture 2

Principle of Superposition

Principle of Superposition • Forms the basis for much theory of structural analysis in this course • The total displacements or internal loadings

P

2P



(stress) at a point in a structure subjected to several external loadings can be determined by adding together the displacements or internal loadings (stress) caused by each of the external loads acting separately • Linear relationship exist among loads, stresses and displacements

Lecture 2

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

• 2 requirements for the principle to apply: – Material must behave in a linea elastic manner, Hooke’s Law is valid – The geometry of the structure mus no undergo significant change when the loads are applied, small displacement theory

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Lecture 2

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More examples on FBD

D

More examples on FBD

P

P

1 C

2

F

y

A

1

FyA

P

x

FxB

FxA

B

2

FyB

FBD for the whole structure

FxA

Fxc

FyA Fxc

M

Fyc

Fyc

FxA + 0 = 0 FyA – F = 0 Moment above A M +

3=0

•K •1m •1m

•8kN/m

Lecture 2

•12kN

•10kN

•18kN.m

•18kN.m

•4m

•2m

•10kN

•22kN.m

•29kN

•8kN/m

•22kN.m

•10kN

•25kN

•17 • •

•x=17/8

•K

•29kN

•8kN/m

18

•K •1m•1m

•29 •12kN

FyB

17

•22kN.m

FxB

FBD for CB

FyA

Lecture 2

•12kN •18kN.m

FBD for ACD

FxA

•11

•28

•46.0625 •32

•Q(kN)

•15 •20

•18

•25kN

•10 •

•17

•M(kN·m)

19 / 47

5

•80kN.m

•160kN

•1m •1m

•40kN/m

•2m

•80kN.m

•40kN

•130 •80kN.m

•160kN

•40kN/m

•310kN

• 

Some hints on drawing FBD and equilibrium equations (EE) •

FBD should never contain any support

•

When removing the supports to the structures, replaced them by the corresponding reactions

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The number of reactions should always equal to the numbers of constrains removed.

•

Away fix the global x and y axes before writing down the EE, this shall fix the positive directions of forces

•

In general, one could always assume that all reactions are acting in the positive direction

•

When writing down the EE, always use the condition of “Sum T Zero . That is, instead of writing:

FxB + Pcos(4 Lecture 2

• 

•210

•  •190

•160

•340

•Q(kN)

•M(kN·m)

•280 哈工大 22 / 47 学院

Internal forces • Internal forces are forces developedinside a member under the action of external loadings and reactions from the supports • In structural analysis, to find out the internal forces of members are very important as such member forces information are critical to the design of the structures so that engineers could design a member section size strong enough to resist the internal forces • Hence, it is very important that in structural analysi all the member forces are calculated correctly. Otherwise, it could lead to error in design and eventually failure of members and structures

FxB = Pcos(4 ) one shall write:

•120 •30

•130 •130kN

•310kN

• 

•40kN

•40kN

•40kN/m

•130kN

•2m

•4m

•160kN

=0 23

Lecture 2

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Next Lecture • To obtain member force, one needs to make a “cut” to the member under concern and expose the corresponding internal forces P

• Determinacy and stability • Simple applications of equilibrium equations

P

B

A

A

C A simple beam ACB

N

N

M

M

V

B V C

A cut before C exposes the internal forces N, V and M P

FxA

A

N

N

M

M

V

FyA

B V C

FxB

FyB

FBD for RHS

FBD for LHS Lecture 2

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Lecture 2

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Determinacy • Equilibrium equations (EE) provide sufficient conditions for equilibrium • When all forces (internal and reactions) can be determined strictly from the EE, the structure is referred as statically determine • If the structure has more unknown forces than the numbers of EE , the structure is a statically indeterminate structure • If a structure is statically indeterminate, addition equations, known a compatibility equation are needed to solve for the forces • In this course, we shall only work on statically determine planar (2D) structures

Lecture 3 • Determinacy • Stability • Simple application of equilibrium equations

Fo Forr s tability, a minimum o f 3 rea reaction ction co components mponents is needed: 1

Lecture 3

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3

Lecture 3

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Stability • To ensure equilibrium of a structure or its members: – Must satisfy EEs – Members must be properly “held” or constrained by their supports • If the structure is not properly supported, instability may be resulted and lead to collapse of the (partially or whole) structure • Instability could be caused by – Partial constraints (i.e. no enough numbers of constraints / support reactions) – Improper Constraints (i.e. enough numbers of constraints but put in wrong positions/ directions)

Lecture 3

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Lecture 3

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Lecture 3

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Lecture 3

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Lecture 3

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2

•Examples Examples o f external s ta tatical tical cl classification assification assification:

•Computation of reactions using Equation of Equilibrium

•Example: Determine the reactions for the structure.

Lecture 3

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3

•And And t hey mu must st be properly a rranged rranged:

•Example: A pin-connected two-bar frame is supported and loaded as shown below. Determine the reactions at supports A and B.

In some cases, the number of reaction forces may equal EEs. However, instability or movement of structure could still occur if support reactions are • Concurrent (i.e. pass through) at a point • All parallel and acting along one direction only

Simple application of EEs

Simple application of EEs Procedure of analysis using EEs

Some useful hints

(1) Identify all the supports and hinge connections (if any) (2) Remove the supports and r apply the reactions (3) Draw FBD for the whole structures or FBDs for separate parts of the structure. Remember that 3 EEs could by applied to each separated FBD (4) Apply EEs in turns (to different separate parts if needed) to solve for the unknown reaction and forces (5) Always try to solve one unknown at a time

(1) There is no need to know in advance the directions of the reaction before the analysis. One could always assuming the reactions are acting in the positive directions, if eventually the answer is negative, it simply implies that the reaction is acting in the opposite direction (2 When apply the EEs, very often the solutions will become very straightforward by take moment (i.e. using the M=0 condition) at appropriate point (3 Always write the EE in “sum t zero” form, this could reduce the chance of making errors

Lecture 3

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Lecture 3

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Numerical examples

 Fx  0; Ax  270 cos 600  0 Ax  135kN

The 6kN/m udl is converted to 36kN at its centroid

Note: the 6.8kNm moment at the free end should be included when taking moment at A

Segment BC :

With anti - clockwise in the  direction, 0 0  M A  0;  270sin 60 (3)  270 cos 60 (0.3)  By (4.2)  6.8  0

With anti - clockwise in the  direction,  M c  0;  8 B y (4.5)  0  By  1.78kN

B y  159kN

   Fy  0;  1.78 Cy  0  Cy  1.78kN

   F y  0;  270sin 60 0 159  A y 0

 F x  0; B x  0

A y  74.8 kN Lecture 3

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Lecture 3

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•Example : •Compute the reactions for the structure shown below.

Segment AB : With anti - clockwise in the  direction,  M A  0; M A  36(3)  (1.78)(6)  0 M A  97.3kN.m    Fy  0; Ay  36 1.78  0 Ay  34.2kN  Fx  0; Ax  0 Lecture 3

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5

•40kN/m •A

•120kN

•B

•D

•K

•C

•8m

•2m

•3m

•3m •120kN

•40kN/m

•145kN

•40kN/m

•235kN

•Home Works

•120kN

•20kN •A

•B

•C

•10kN

•D

•K

•2kN/m •A

•120 •(kN·m)

•3m

•B •3m

•E •2m •2m

•C •2m

•180 •145

•10kN

•263

•60

• 

• 

• 

•175

•60kN

•60kN

• 

•60 (kN)

•3m

•D •4m

•10kN

•5kN/m •2m •2m

•F •2m

•20kN·m •2m

•2m

•2m

•5m

6

•q •B

•A

•x

•D

•l

•C

•l - x

•l

•q •B

•A

•x

•D

...