structural statics

8/10/2019 structural statics

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Chapter 2

Fundamentals


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General Equations of Equilibrium

Supports

Member Forces

Connections

Stability and Determinacy of a Structure with Respect to

Supports

General Stability and Determinacy of Structures

Methods of Analysis

Chapter 2Fundamentals


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2-1 General

Fundamental Concepts of Structural Analysis

External

forces

Internal

forces

Internal

deformation

External

deformation

Equilibrium

Force-deformation Relationship

Compatibility


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2-2 Equations of Equilibrium

Static Equi li br ium:

A structure is said to be in equilibrium if, under the action ofexternal forces, it remains at rest relative to the earth.

Also, each part of the structure, if taken as a free bodyisolated

from the whole, must be at rest relative to the earth under the

action of the internal forces at the cut sections and of the

external forces thereabout.

If such is the case, the force system is balanced, or in

equilibrium, or in equilibrium, which implies that imposed on

the structure, or segment thereof, must be zero.

R 0


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Remark: In fact, there are always some small deformations that

may cause some small changes of dimension in a structure and ashifting of the action lines of the forces. In structural statics, such

effects are neglected and all force systems are assumed to act on

a rigid body. That is, the structural system is considered as a

rigid body when constructing the equations of equilibrium.

Non-conservative forces


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Alternative Forms

and ab

y a bF 0, M 0, M 0

( If a b y axis )

a

x

yb

and are not collinear

a b cM 0, M 0, M 0

( If a b c )

a

x

yb

c

x y aF 0, F 0, M 0

Coplanar System:

a

x

y


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Special Cases:Concurrent system and Parallel system

Concurrent system

(If is not on the line through the concurrent

( is satisfied automatica

point of forces

and perperdicular to y-axis)

(

lly)

or

or

I

x y o

y a

a b

F 0, F 0 M 0

F 0, M 0

M 0, M

a

0

f and does not pass through the concurrent point of forces)a b ab

a

x

y b

o


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Special Cases:

Parallel system

(If all forces parallel to y-axis, is satisfied automatical

or

ly)

(If a b and

ab //

x

y a

a b

F 0, M 0

M 0,

F 0

M 0

the forces of the system)

a

x

y b


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Two-force member

Not in equilibrium.


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Three-force member

Not in equilibrium.


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Conclusions: In general, there exists 3 equilibrium equations for a

coplanar system. However, there are only 2 equilibrium

equations for concurrent and parallel systems.

In space structural systems, there exists 6 equilibriumequations:

SFx=0, SFy=0, SFz=0, SMx=0, SMy=0, SMz=0

Exceptions: for concurrent and parallel systems the

number of equilibrium equations will be reduced. Forexample, in concurrent systems:

SFx=0, SFy=0, SFz=0


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2-3 Supports

Functions of supports

KinematicProvide constraints for a structural

system such that the structure can not be moved

freely.

Statics

Provide reactions such that equilibriumconditions can be preserved.

Types of Supports

Hinge supportLink support

Roller support

Fixed support


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2-3 Supports

Hinge Support

2 2

x yR R R


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2-3 Supports

Link Support

The constraints provided by any two concurrent and non-

parallel link supports is the similar to that provided by a

hinge support.


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2-3 Supports

Roller Support


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2-3 Supports

Fixed Support

x o

2 2

x y

R d M

R R R

Equivalent to a fixed support


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2-4 Member Forces

Truss


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2-4 Member Forces

Beams and Rigid Frames


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2-5 Connections

Type of Connections

Hinge connection(Do not transfer moment from onemember to the connected members, i.e., M=0 at hingeend of members)

Roller connection(Do not transfer moment and axialforce from one member to the connected members)

Rigid Connection(It can transfer moment, axial forceand shear force from a member to the connectedmembers)


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2-5 Connections

Semi-rigid Connections

It can transfer axial force, shear force and part ofmoment from a member to the connected members.

Link Connections(linker)

a a


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2-6 Stability and Determinacy of a Structure

With Respect to Supports

Stability

A structure remains in static equilibrium state whenit is acted on by a system of general loads, the structure isstable. (The structure is considered as a rigid body, i.e., thedeformations of structural members are not considered.)

Main reasons that caused a structure unstable: statically unstable without adequate number of

constraints

geometrically unstable the movement of a structure isnot well restrained by the supports, i.e., the geometricalarrangement of supports is not correct.


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Statically Unstable

xF 0

oM 0

Special cases for statically stable


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()


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Static Stability :

a

No. of independent unknowns (r)

Equilibrium Eqs. (r )

No. of independent static equations= +Condition Eqs. (c)

< Statically Unstable

= Statically Stable and Determinate (SD)> Statically Stable and Indeterminate (SI)


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Static Stability :No. of independent unknowns (r):External (

) : reactions

External + Internal: reactions and member forces

No. Equilibrium Eqs. (for planar structure) (ra):

External: 3 eqs.(SFx=0, SFy=0, and SMo=0)

External + Internal: depends on the no. of joints and

structural type

Beam and Frames: 3 eqs. for each joint=3j

Truss : 2 eqs. for each joint=2jNo. of Condition Eqs. (c):compound type

structures

hinge

roller

linker

e.g.hingec=1roller

linker

c=2


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Condition Equations

Simple type structures


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Condition Equations (contl)

Compound type structuresroller

hinge

linker


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Condition Equations

simple type structure

compound type structure

(

)

compound type structure

x y aF 0, F 0, M 0


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No. of Condition Equations

orAO OB

O OM 0 M 0

Hinge Connection

OA B

Ohinge

Note:hingenn1

2 2 abcbdb1

a

cb

d


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or

or

AO OB

O O

AO OB

x x

M 0 M 0

F 0 F 0

Roller or Link Connection

OA

B Orollerconnection

Note:roller or link connection2

2A

Bd

c

2 6 S i i i f S


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Examples

hinge

c=3c=3


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Statically Unstable Externally: r


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Statically Stable and Determinate Externally (r=ra+c)

r=3, ra=3, c=0 SD externally


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Statically Stable and Indeterminate Externally(r>ra+c)

r=5, ra=3, c=0SI externally to the 2nddegree of indeterminacy

(degree of indeterminacy =r(ra+c)=53=2)

(

)

(

)


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External Geometric Unstable (

)r ra+c

xF 0 oM 0

2 6 S i i i f S


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External Geometrical Unstable

2 6 S bili d D i f S


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Geometrically UnstableCompound type structure

dcouple0

r=4, ra=3, c=1a b c


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2 6 St bilit d D t i f St t


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Conclusions for stability and determinacy of a structurewithrespect to supports(External stability and determinacy)

If the number of unknown reactions is less than 3, the eqs.of equilibrium are generally not satisfied, and the system issaid to be statically unstable externally.

If the number of unknown reactions is equal to 3 and if noexternal geometric instability is involved, the system is saidto be statically stable and determinate externally.

If the number of unknown reactions is greater than 3 and ifno external geometric instability is involved, the system issaid to be statically stable and indeterminate externally.

The excess number n of unknown elements designated then-thdegree of statically determinacy.


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2-7 General Stability and Determinacy of Structures

(= External + Internal)

No. of independent unknowns :

External + Internal: reactions and member forces

No. Equilibrium Eqs. (for planar structure):

External + Internal: depends on the no. of joints and

structural typeBeam and Frames: 3 eqs. for each joint=3j

Truss : 2 eqs. for each joint=2j

No. of Condition Eqs.

No. of independent unknowns> Equilibrium Eqs.

< No. of independent static equations= +

= Condition Eqs.


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Truss: Unknowns : One unknown internal force for each member

(comp. or tension)No. of unknown reactions

Equilibrium Equations: Taking each node as a free body and

each free body can provide two equilibrium equations.

In truss structures, members are connected to each other by

hinges. No condition equations exists in truss structures.

< statically unstable

r+b= 2j statically stable and determinate> statically stable and indeterminate

Without

geometrically

unstable

2 7 G l St bilit d D t i f St t


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Beams and Frames:

Each member has three independent unknown internal forces

(shear, axial force and bending moment)

unknown reactions

Taking each node as a free body which can provide 3

equilibrium equations. If members are connected by hinge, roller or linker, some

condition equations can be constructed.

Without

geometrically

unstable


r+3b= 3j+c statically stable and determinate

> statically stable and indeterminate


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Truss:b=13, r=3, j=8 b+r=16=2jSD

b=13, r=3, j=8 b+r=16=2jUnstable



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:


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abcd



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Beams

Beams and Frames:


r+3b= 3j+c statically stable and determinate



r = 3+c statically stable and determinate


2 7 General Stability and Determinacy of Structures


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r=5, b=5, j=6, c=2, r+3b=20=3j+c

r=6, b=5, j=6, c=2, r+3b=21>3j+c=20

r=5, b=4, j=5, c=2, r+3b=17=3j+c

r=4, b=4, j=5, c=3, r+3b=163j+c=17



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b=14, r=9, j=13, c=4, 3b+r=51>3j+c=43

b=11, r=9, j=10, c=1, 3b+r=42>3j+c=31



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Remark:

Examples

a b(1) r=6,b=3, j=4, c=0

r+3b=15 > 3j+c=12

3

(2) r=6,b=2, j=3, c=0

r+3b=12 > 3j+c=9

3

abbabb


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A special method for portal frames

3x4x8+3x15x3=231



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Note:

3x41=11



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Truss:(A) Simple Rigid

(1) Formed from a basic rigid unit

3

Rigid Body

Non-rigid Body+additional constraints

stable

stable

Basic Rigid Unit for a Truss:

+3



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y y


Truss

:

3

simple type truss

rigid unit

Note:



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Truss

:



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Truss

:

(2) Formed from a basic stable unit

Basic Stable Unit for a Truss:

+4

Basic Stable Unit


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Truss

:(B) Compound Rigid

3Compound Rigid

Compound Rigid

3

Compound Type Truss

unstable unstable



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Truss:(B) Compound Rigid

r=6, b=40, j=23, r+b=2j



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Truss:(C) Non-rigid member arrangement

Non-rigid unit + 3+= stable



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(D) Complex Type Truss (

sec.3-1)

basic rigid trusssimple type trusscompound type trusscomplex truss



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(

)

n

n() (b+r=2j3b+r=3j+c)

11 1 12 2 1n n 1

21 1 22 2 2n n 2

n1 1 n 2 2 nn n n

a x a x ... a x b

a x a x ... a x b

a x a x ... a x b

xi

bi



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11 12 1 1n

21 22 2 2n

n1 n 2 n nn mm

11 12 1m 1n

21 22 2m 2n

n1 n 2 nm nn

a a b a

a a b a

a a b ax

a a a a

a a a a

a a a a

01.mxm

2.m=0xm=0

01.

mxm=2.

m=0xm



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Conclusions:

00

Example:

x

x 1 2 3

A 2 3

F 0 0 0 0 Q

F 0 R R R P

M 0 0 LR 2LR PL

0 0 0

1 1 1 0 unstable

0 L 2L

P

R1R2

R3



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()

s1 s2

s3

Example:

s1=xSFx=0 s2=s1=x

SFy=0 s2=s1=x

q q

s1 s2

s1cosq

s1sinq

s2cosq

s2sinqs2=s1=x=0

s3=0 Stable



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s1 s2

s3

Example:A double symmetric complex truss

s1=xJoint a: SFx=0 s2=s1=x

SFy=0 s9=6x/5

Joint b: SFx=0 s7=s1=xSFy=0 s3=6x/5

Joint f: SFx=0 s8=s2=x

SFy=0 s4=6x/5

Joint d: SFx=0 and SFy=0s4=s6=x

x

Unstable

s4

s5 s6

s7s8

s9A=3

B=3

4 4

6

a

b

c

d

e

f

Note:AB


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2-8 Methods of Analysis


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Force method ( , EA=const.)

1. Joint equilibrium

Ra=P1, P1=P+P2, P2=Rc

2. Member flexibility

Elongation of member 1:

e1= (P+Rc)(1.5L)/EA

Elongation of member 2:

e2=RcL/EA

3. Joint displacementua=0

ub=e1= (P+Rc)(1.5L)/EA

uc=e1+e2=(P+Rc)(1.5L)/EA

+RcL/EA= 0Rc=0.6P

P2=0.6P, P1=0.4P

ub=0.6PL/EA

2-8 Method of Analysis


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Displacement method

1. Joint displacement

ua=uc=0, ub=unknown2. Member stiffness

Elongation of members:

e1=ubua=ub

e2=uc

ub=

ubMember forces:

P1=ubEA/(1.5L)

P2=ubEA/L

3. Joint equilibrium

P=P1P2=ubEA/(1.5L)+ubEA/L

ub=0.6PL/EAP1=0.4P, P2=0.6P

2-8 Method of Analysis


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8 et od o a ys s

Force methods (Flexibility methods)

Chapter 3 Structural statics

Chapter 5 Consistent deformation methods

Chapter 6 Matrix force method

others

Displacement method (Stiffness methods) Chapter 8 Slope deflection method

Chapter 9 Matrix displacement method

Chapter 7 Moment distribution method (A special version

of displacement methods)

structural statics

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