160.9.1 indet analysis force method...write the n compatibility equations at relevant points solve...

Analysis of Statically Indeterminate Structures Using the Force Method

StevenVukazichSanJoseStateUniversity

Statically Indeterminate Structures

At the beginning of the course, we learned that a stable structure that contains more unknowns than independent equations of equilibrium is Statically Indeterminate.

• Redundancy (several members must fail for the structure to become unstable);

• Often maximum stresses is certain members are reduced;

• Usually deflections are reduced.

Advantages Disadvantages

• Connections are often more expensive;

• Finding forces and deflections using hand analysis is much more complicated.

Steps in Solving an Indeterminate Structure using the Force Method

Determine degree of Indeterminacy Let n =degree of indeterminacy

(i.e. the structure is indeterminate to the nth degree)

Define Primary Structure and the n Redundants

Define the Primary Problem

Solve for the nRelevant

Deflections in Primary Problem

Define the nRedundant Problems

Solve for the nRelevant Deflections in each Redundant

Problem

Write the nCompatibility Equations at

Relevant Points

Solve the nCompatibility

Equations to find the n Redundants

Use the Equations of Equilibrium to

solve for the remaining unknowns

Chapter 3

Chapters 3,4,5 then 7 or 8

Construct Internal Force

Diagrams (if necessary)

Chapter 3

Chapters 3,4,5 then 7 or 8

Chapters 3,4,5

Force Method of Analysis

EBA C

w P

Consider the beam

FBD

EBA D

w P

Ax

Ay

MA

Dy

X = 5

3n = 3(1) = 3

Beam is stable

Statically Indeterminate to the 2nd degree

EI D

C

Cy

Define Primary Structure and Redundants• Remove all applied loads from the actual structure;• Remove support reactions or internal forces to define a primary structure;• Removed reactions or internal forces are called redundants;• Same number of redundants as degree of indeterminacy• Primary structure must be stable and statically determinate;• Primary structure is not unique – there are several choices.

EBA D

Primary Structure Redundants

MA

EBA DDy

EBA DMQ

C

C Cy

Cy

CyQ C

Define and Solve the Primary Problem

• Apply all loads on actual structure to the primary structure;• Define a reference coordinate system;• Calculate relevant deflections at points where redundants were

removed.

DBA C

w Py

x

Δ"

EIE

Δ#

Define and Solve the Redundant Problems• There are the same number of redundant problems as degrees of indeterminacy;• Define a reference coordinate system;• Apply only one redundant to the primary structure;• Write the redundant deflection in terms of the flexibility coefficient and the

redundant for each redundant problem.• Calculate the flexibility coefficient associated with the relevant deflections for

each redundant problem;

EBA

y

x

Δ##

EI D

Cy

EBA

y

x

𝛿##

EI D1

Δ## = 𝐶'𝛿##C

Δ"#

𝛿"#

C

Redundant Problem 1

Δ"# = 𝐶'𝛿"#

Define and Solve the Redundant Problems

EBA

y

x

Δ#"

EI D

Dy

EBA

y

x

𝛿#"

EI D1

Δ#" = 𝐷'𝛿#"

C

Δ""

𝛿""

C

Redundant Problem 2

Δ"" = 𝐷'𝛿""

Compatibility Equations

Δ# + Δ## + Δ#" = 0

Compatibility at Point C

Δ" + Δ"# + Δ"" = 0

Compatibility at Point D

Compatibility Equations in terms of Redundants and Flexibility Coefficients

∆# + 𝐶'𝛿## + 𝐷'𝛿#" = 0∆" + 𝐶'𝛿"# + 𝐷'𝛿"" = 0

Solve for Cy and Dy

The Force Method is

Based on the Principle of

Superposition

IndeterminateProblem

PrimaryProblem

RedundantProblem 1

RedundantProblem 2

EBA C

w P

EI D

DBA C

w Py

xΔ"

EIEΔ#

EBA

y

x

Δ#"

EI DDy

C

Δ""

EBA

y

x

Δ##

EI DC

Δ"#

Cy

y

x

=

+

+

Example Problem

AB

P

EI C

𝐿2

𝐿2

y

x

For the indeterminate beam subject to the point load, P, find the support reactions at A and C. EI is constant.

A B

P

EI C

𝐿2

𝐿2

y

xAx

Ay

MA

Cy

FBD

X = 4

3n = 3(1) = 3

Beam is stable

Statically Indeterminate to the 1st degree

Define Primary Structure and Redundant• Remove all applied loads from the actual structure;• Remove support reactions or internal forces to define a primary structure;• Removed reactions or internal forces are called redundants;• Same number of redundants as degree of indeterminacy• Primary structure must be stable and statically determinate;• Primary structure is not unique – there are several choices.

BAC

Primary Structure Redundant

MA

BA CCy

Define and Solve the Primary Problem

• Apply all loads on actual structure to the primary structure;• Define a reference coordinate system;• Calculate relevant deflections at points where redundants were

removed.

B

A C

Py

x

EI𝜃/

𝐿2

𝐿2

𝜃/ = −𝑃𝐿2

16𝐸𝐼

From Tabulated Solutions

+ Counter-clockwiserotations positive

Define and Solve the Redundant Problem• There are the same number of redundant problems as degrees of indeterminacy;• Define a reference coordinate system;• Apply only one redundant to the primary structure;• Write the redundant deflection in terms of the flexibility coefficient and the

redundant for each redundant problem.• Calculate the flexibility coefficient associated with the relevant deflections for

each redundant problem;

𝜃// = 𝑀/𝛼//

Redundant ProblemBA C

y

x

EI𝜃//MA

BA Cy

x

EI𝛼//1

𝛼// = −𝐿3𝐸𝐼

From Tabulated Solutions

L

L

Compatibility Equation at Point A

𝜃/ + 𝜃// = 0

Compatibility at Point A

Compatibility Equation in terms of Redundant and Flexibility Coefficient

𝜃/ +𝑀/𝛼// = 0

−𝑃𝐿2

16𝐸𝐼+ 𝑀/ −

𝐿3𝐸𝐼

= 0

𝑀/ =𝑃𝐿2

16𝐸𝐼−3𝐸𝐼𝐿

𝑀/ = −316𝑃𝐿

Solve for MA

A B

P

EI C

𝐿2

𝐿2

y

xAx

Ay

MA

Cy

Free Body Diagram

𝑀/ = −316𝑃𝐿

A B

P

EI C

𝐿2

𝐿2

y

xAx

Ay Cy

316𝑃𝐿

Can now use equilibrium equations to find the remaining three unknowns

Find Remaining Unknowns

:𝑀/

�

�

= 0+

:𝐹=

�

�

= 0+

:𝐹'

�

�

= 0+

Ax = 0

A B

P

EI C

𝐿2

𝐿2

y

xAx

Ay Cy

316𝑃𝐿

Can now use equilibrium equations to find the remaining three unknowns

𝐶' =516𝑃

𝐴' =1116𝑃

+

+V

M0

Draw V and M Diagrams of the Beam

A B

P

EI C

𝐿2

𝐿2

y

x

316𝑃𝐿

516𝑃

1116𝑃

1116𝑃

−516𝑃

𝑀@ = −316𝑃𝐿 +

1132𝑃𝐿 =

532𝑃𝐿

𝑀@ − 𝑀/ =1116𝑃

𝐿2

−3𝑃𝐿16

532𝑃𝐿

V

M 0

Superposition of Primary and Redundant Problems

A

C

P

B𝐿2

𝐿2

316𝑃𝐿

516𝑃11

16𝑃

1116𝑃

−316𝑃𝐿

532𝑃𝐿

P

12𝑃

14𝑃𝐿

316𝑃

−332𝑃𝐿

−12𝑃

12𝑃

12𝑃

316𝑃 3

16𝑃

316𝑃𝐿

−316𝑃𝐿

B B

0

IndeterminateProblem

PrimaryProblem

RedundantProblem= +

C CA A

0

−516𝑃

0