Chapter 21

Chapter 2

Analysis of Statically

Determinate Structures

Chapter 22

Idealized Structure

• In real sense exact analysis of a structure

can never be carried out.

• Estimates have always to be made of the

loadings and strength of materials.

• Furthermore, points of application for

the loadings must be estimated.

• Models or idealization should be made.

Chapter 23

Support connection

• Pin support and pin connection

Chapter 24

• Fixed support & fixed connection

Chapter 25

• Roller support

Chapter 26

Chapter 27

Chapter 28

Chapter 29

Chapter 210

Chapter 211

Support for coplanar structures

Chapter 212

Idealized Structure

Chapter 213

Chapter 214

Chapter 215

Chapter 216

Chapter 217

Chapter 218

Chapter 219

Chapter 220

Chapter 221

Chapter 222

Chapter 223

Principle of superposition

• The total displacement or internal loadings

(stress) at a point in a structure subjected to

several external loadings can be determined by

adding together the displacements or internal

loadings caused by each of the external loads

acting separately.

• Linear relationships among loads, stresses and

displacements

Chapter 224

Superposition requirements

• Material must be behave in a linear elastic

manner.

• The geometry of the structure must not

undergo significant change. Small

displacement theory applies.

Chapter 225

Chapter 226

Equation of Equilibrium

• In x-y plane

0

0

0

o

y

x

M

F

F

Internal loading

Chapter 227

Determinacy & Stability

• Determinacy: when all the forces in structure can

be determined from equilibrium equation , the

structure is referred to as statically determinate.

Structure having more unknown forces than

available equilibrium equations called statically

indeterminate

• If n is number of structure parts & r is number of

unknown forces:

r = 3n, statically determinate

r > 3n, statically indeterminate

Chapter 228

Classify determinate & indeterminate structure

eDeterminat Statically )1(33

1

3

n

r

degree 2

ateindetermin Statically )1(35

1

5

nd

n

r

edeterminat Statically )2(36

2

6

n

r

Chapter 229

degree 1

ateindetermin Statically )3(310

3

10

st

n

r

degree 1 ate,indetermin Statically )2(37

2

7

st

n

r

Chapter 230

degree 4

ateindetermin Statically )2(310

2

10

th

n

r

Chapter 231

Chapter 232

Chapter 233

Chapter 234

• Stability

Partial Constraints

case loading with thisunstablesatisfied benot will0 xF

Chapter 235

• Improper Constraints

This can occur if all the support reactions are concurrent at a

point.

0dP

Chapter 236

• This can occur also when the reactive forces are all parallel

In General

r < 3n, Then the structure is Unstable

r >= 3n, Also, Unstable if member reactions are

concurrent or parallel or some of the components

form a collapsible mechanism

Chapter 237

Classify The structure Stable or

Unstable

Sable )1(33

1

3

cases no special

n

r

Sable )2(38

2

8

cases no special

n

r

Unsable Bar concurrent are reactions three the)1(33

1

3

n

r

Chapter 238

Unsable )3(37

3

7

n

r

Chapter 239

Chapter 241

Application of

Equilibrium Equation

Chapter 242

Application of Equation of Equilibrium

1. If not given, establish a suitable x - y coordinate system.

2. Draw a free body diagram (FBD) of the object under

analysis.

3. Apply the three equations of equilibrium to solve for the

unknowns.

Procedure steps

Chapter 243

Example 1

Determine the Reactions

kA

A

kB

BM

kA

AF

y

y

y

yA

x

xx

4.13

05.3860sin60 0F

5.38

050)14()1(60cos60)10(60sin60 0

30

060cos60 0

y

Chapter 244

Chapter 245

Example 2

Determine the Reactions

600

0)6(60)4(60 0

120

06060 0F

0 0

y

kNM

MM

kNA

A

AF

A

AA

x

x

xx

Chapter 246

Chapter 247

Example 3

Determine the Reactions

IbA

A

IbA

AF

IbN

NNM

y

y

x

xx

B

BBA

2700

05.13313500 0F

1070

05.1331 0

5.1331

0)10()4()5.3(3500 0

53

y

54

53

54

Chapter 248

Chapter 249

Example 4

Determine the Reactions

IbA

A

AF

ftIbM

MM

IbC

CM

y

y

xx

A

AAat

y

yRightB

7600

08000400 0F

0 0

.72000

0)10(80006000)35(400 0

400

06000150

y

point

Chapter 250

Chapter 251

Example 4

Determine the Reactions

kNA

A

kNA

AF

kNC

CM

kNC

CM

y

y

x

xx

x

xA

y

yRightB

4.9

0)(863 0F

87.9

0)(87.14 0

7.14

0)2(8)3(6)4(3)5.1( 0

3

0)1(620

54

y

53

Chapter 252

Chapter 253

Example 5

Determine the Reactions

mkNM

MM

kNE

EF

EF

kNC

CM

kNA

AM

E

ErightD

y

y

xx

y

yLeftD

y

yLeftB

.33.5

0)4(33.5)2(8 0

33.5

067.6488 0

0 0

67.6

0)8(8)11(43 0

4

0)3(860

)(

y

)(

Chapter 254

Copyright © 2009 Pearson Prentice Hall Inc.

Chapter 257

Example 6

Determine the Reactions

kNA

A

kNA

AF

kNC

CM

kNC

CM

y

y

x

xx

x

xRightB

y

yA

120

045sin9.8445sin6.254240 0F

285

045cos9.8445cos6.25460180195 0

195

0)(9.84)5.4(60)3(240)6(0

240

0)5.4(45cos9.84)5.4(45cos9.84)5.1(45sin6.254

)5.4(45cos6.254)5.1(60)5.1(180)6( 0

y

2

23

Chapter 258

Chapter 259

Problem 1Determine the Reactions

Chapter 260

Problem 2Determine the Reactions

Chapter 261

Problem 3Determine the Reactions