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• Statically Indeterminate Structures

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• • Statically Determinate Problems（静定问题）

• Statically Indeterminate Problems（超静定问题）

• Degrees of Indeterminacy（超静定次数）

（超静定结构的优缺点）

• Redundancy & Basic Determinate System（冗余约束与 基本静定系）

• General Ideas for Analyzing Indeterminate Structures

（超静定问题一般解法）

Contents

2

• • Force Method Solution Procedure（力法）

• Statically Indeterminate Bars（超静定拉压杆）

• Assembly Stress（装配应力）

• Thermo Stress（热应力）

• Statically Indeterminate Shafts（超静定扭转轴）

• Statically Indeterminate Beams（超静定弯曲梁）

• Moment-Area Theorems with Indeterminate Beams（图乘 法求解超静定梁）

• Combined Indeterminate Structures （组合超静定结构）

Contents

3

• • Problems can be completely solved via static equilibrium

• Number of unknowns (forces/moments) = number of

independent equations from static equilibration

0 ? 0 ?

?0

x Ax

y Ay

ByA

F F F F

FM

     

       

  

Statically Determinate Problems

4

A B

• A B C

• Problems cannot be solved from static equilibrium alone

• Number of unknowns (forces/moments) > number of

independent equations from static equilibration

? 0 ?

0 ?

?0 ?

Ax

x Ay

y By

CxA

Cy

F F F

F F

FM F

    

     

    

  

Statically Indeterminate Problems

5

• • For a co-planer structure, there are at most three

equilibrium equations for each portion of the structure. If

there is a total of n portions and m unknown reaction

forces from supports:

3 Statically determinate

3 Statically

- 3 Degrees of indetermi

indet

n

erminate

acy

m n

m n

m n 

 

 

Statically determinate Degrees of Indeterminacy = 2

Degrees of Indeterminacy of Structures

6

- Redistribution of reaction forces / internal forces

- Smaller deformation

- Greater stiffness as a whole structure

- Thermal and residual stresses due to temperature change

and fabrication errors

7

• • Redundancy: unnecessary restraints without which the

static equilibrium of a structure still holds.

• Basic determinate system: the same structure as of a

statically indeterminate system after replacing redundant

Redundancy & Basic Determinate System

8

• • Basic determinate structure: obtained via replacing

redundant restraints with extra constraining loads.

How to Analyze Statically Indeterminate Structures?

• Equilibrium: is satisfied when the reaction forces at

supports hold the structure at rest, as the structure is

• Deformation compatibility: satisfied when the various

segments of the structure fit together without intentional

breaks or overlaps

• Deformation-load relationship: depends on the manner

the material of the structure responds to the applied loads,

which can be linear/nonlinear/viscous and elastic/inelastic;

for our study the behavior is assumed to be linearly elastic

9

• • The method of consistent deformations or force method

was originally developed by James Clerk Maxwell in 1874

and later refined by Otto Mohr and Heinrich Műller-

Breslau.

Christian Otto Mohr

(1835 – 1918)

James Clerk Maxwell

(1831 – 1879)

Heinrich Műller-Breslau

(1851 - 1925)

Force Method (Method of Consistent Deformations)

10

• • Making the indeterminate structure determinate

Force Method Solution Procedure

• Obtaining its equations of equilibrium

• Analyzing the deformation compatibility that must be

satisfied at selected redundant restraints

• Substituting the deformation-load relationship into

deformation compatibility, resulting in complementary

equations

• Joining of equilibrium equations and complementary

equations

• Types of Problems can be addressed include (a) bar

tension/compression; (b) shaft torsion; and (c) beam

bending 11

• Statically Indeterminate Bars

• For an axially loaded bar with both ends fixed, determine

the reaction forces.

  12

1

2

0 BB

B

A

P R LR L L

EA EA

L R P

L

L R P

L

     

 

 

• Solution:

12

• • Determine the reaction forces at the bar ends.

 

 

 

6

3

3 64

3 1

6

3

6

400 10

600 10

150 10 400 10 0

600 10

250 10

900 10

250 10

577kN

900 577 323kN

B

B

i i

i i B

B

B

A

R

R

F L L

EA E R

R

R

R

 

     

          

    

  

    

 

 

   

• Solution:

Sample Problem

• • Solution:

Sample Problem

• Determine the reaction forces, if any, at the bar ends. E = 200 Gpa.

 

 

 

4 3

1

6

3

3 6

9 3

6

3

6

4.5 10

400 10

600 10

150 10 400 10

200 10 600 10

250 10

900 10

250 10

115.4kN

900 115.4 784.6kN

i i

i i

B

B

B

B

B

A

F L L

EA

R

R

R

R

R

R

 

   

     

       

     

  

    

 

 

   

14

• Sample Problem

• What is the deformation of the

rod and tube when a force P is

exerted on a rigid end plate as

shown?

• Solution:

1 2

1 2 1 2

1 1 2 2

1 1 2 2 1 2

1 1 2 2 1 1 2 2

1 1 2 2

,

P P P

PL P L L L

E A E A

E A P E A P P P

E A E A E A E A

PL L

E A E A

         

    

   

15

• • Given: tension/compression rigidities E1A1 = E2A2 = EA,

E3A3, external load F. Find: internal forces in bars.

1. Make the truss determinate by

replacing extension/contraction

restraint along Bar 3 with FN3 F



A

1 23

l

1 2

1 2 3

0

0 ( )cos

x N N

y N N N

F F F

F F F F F

  

    

 A F

FN3FN1 FN2

Sample Problem

• Solution:

2. Static equilibrium at joint A

16

• 4. Displacement-force relationship

3. Displacement compatibility at joint A

Due to symmetry

21 ll  

 cos31 ll 

1 1

cos

NF ll EA 

  33

3 3

AE

lF l N



A

1 23

l

A*

17

• 5. Complementary equation

21 3 1 3 1 3

1 1 3 3 3 3

cos cos cos cos

N N N N

F l F l EA l l F F

E A E A E A   

       

6. Joint of force equilibrium and complementary equations

1 2 3 3

1 2 2

1 2 3

32 3

1 3

3 3 3 3

2cos cos

( )cos

cos 1 2 cos

N N

N N

N N N

N

N N

F F F

E A F F

EA F F F F

F FEA EAF F

E A E A

 

 

          

 