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Statically Indeterminate Structures

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• Statically Indeterminate Structures

[email protected]

• • Statically Determinate Problems（静定问题）

• Statically Indeterminate Problems（超静定问题）

• Degrees of Indeterminacy（超静定次数）

• Advantages & Disadvantages of Indeterminate Structures

（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 FF F

FM

Statically Determinate Problems

4

A B

• A BC

• 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

FF F

F F

FMF

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

• • Advantages:

Advantages & Disadvantages of Indeterminate Structures

- Redistribution of reaction forces / internal forces

- Smaller deformation

- Greater stiffness as a whole structure

• Disadvantages:

- 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

restraints with extra constraining loads

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

subjected to external loads

• 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

bending11

• Statically Indeterminate Bars

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

the reaction forces.

12

1

2

0BB

B

A

P R LR LL

EA EA

LR P

L

LR P

L

• Solution:

12

• • Determine the reaction forces at the bar ends.

6

3

3 64

31

6

3

6

400 10

600 10

150 10 400 100

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 LL

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.

43

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 LL

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 21 2

1 1 2 2

1 1 2 21 2

1 1 2 2 1 1 2 2

1 1 2 2

,

P P P

PL P LL L

E A E A

E A P E A PP P

E A E A E A E A

PLL

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 FN3F

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

AF

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

11

cos

NF llEA

33

33

AE

lFl N

A

1 23

l

A*

17

• 5. Complementary equation

21 31 3 1 3

1 1 3 3 3 3

cos cos coscos

N NN N

F l F l EAl l F F

E A E A E A

6. Joint of force equilibrium and complementary equations

1 23 3

1 2 2

1 2 3

323

1 3

3 33 3

2coscos

( )cos

cos 1 2 cos

N N

N N

N N N

N

N N

FF F

E AF F

EAF F F F

FFEA EAF F

E A E A

18

• • Given: tension/compression rigidities E1A1, E2A2, E3A3,

external load F. Find: internal forces in bars.

F

A

1 23

l

Sample Problem

• Solution:

1. Make the truss determinate by

replacing extension/contraction

restraint along Bar 3 with FN3

1 2

1 2 3

sin sin 1

cos cos 0 2

N N

N N N

F F F

F F F

2. Static equilibrium at joint A

FN3FN1 FN2

FA

19

• 3. Displacement compatibility at joint A

1l

2

A*2l

3l

32 1

2 1 3

2sin sin tan

2 cos

ll l

l l l

4. Complementary equation

1 2 31 2 3

1 1 2 2 3 3

2 1 3

21 2 3

1

2 2 1 1 3

1 2 2 3 3

3

2 cos 0

, , cos cos

2 coscos cos

3

N N N

N

N

N N

N N

F l F l F ll l l

E A E A E A

F l F l F l

E A E A E A

F F F

E A E A E A

5. Joint of force equilibrium and complementary equations

A

1 23

l

• • Residual stresses exist in statically indeterminate

structures, originated from assembling / fabrication errors

of structural elements

A

1l 2l

3l

A

FN1 FN3 FN2

A

1 23

l

• Force equilibrium:

• Displacement compatibility:

N1 N2

N3 N1 N2( )cos

F F

F F F

cos

1

3

ll

• Bar 3 is δ shorter than required

Assembly Stress

21

• • Given: AB rigid; E1A1 = E2A2 = EA; Bar 1 is δ shorter than required. Find: internal forces in Bar 1 and 2 after assembly.

• Solution:

Bar 1 lengthened & Bar 2

shortened:

2. Displacement compatibility:

1 1 22

22

l al l

l a

BA

21

1l

a a

2l

BA

F1 F2FA

1 20 2AM aF aF

1. Basic determinate structure and force equilibrium of AB:

Sample Problem

22

• 3. Displacement-force relationship:

1 21 2;

Fl F ll l

EA EA

4. Complementary equation:

1 21 22( ) 2

Fl F ll l

EA EA

1 2 1

1 2

2

42

5

2 2

5

EAaF aF F

lFl F l

EAFEA EA

l

5. Joint of force equilibrium and complementary equations

23

• • A temperature change results in a change in length or

thermal strain. There is no stress associated with the

thermal strain unless the elongation is restrained by

the supports.

thermal expansion coefficient

T P

PLT L

AE

• Treat the additional support as redundant and apply

the principle of superposition.

0

0

AE

PLLT

PT

P AE T

PE T

A

• The thermal deformation and the deformation from

the redundant support must be compatible.

Thermo Stress

• Example (low-carbon steel)

6 0200 GPa 12.5 10 2.5 MPaC T T 24

• Coefficient of Thermal Expansion

25

• • Extension / contraction knots:

• Extension / contraction slot:

Addressing Thermo Stresses

26

• Statically Indeterminate Shafts

• Determine (a) the reactive torques at the ends, (b) the maximum shearing stresses in each segment of the bar, and (c) the angle of rotation at the cross section where the load T0 is applied.

• Solution:

0

/

0 0

max max

0

,

2 2,

B AB BB A

pB pA

A pB B pA

B A

B pA A pB B pA A pB

B B A A B B A ABC AC C

pB pA pB pA

T T LT L

GI GI

L I L IT T T T

L I L I L I L I

T d T d T L T L

I I GI GI

27

• Statically Indeterminate Shafts

• For the special case of dA = dB:

0 0 0 0

0 0

max max

0

,

2 2,

2 2

A pB B pAA BB A

B pA A pB B pA A pB

B AA BBC AC

p p p p

A BB B A AC

p p p

L I L IL LT T T T T T

L I L I L L I L I L

T d T dL dT L dT

I LI I LI

L L TT L T L

GI GI LGI

28

• Sample Problem

• What is the deformation of the

rod and tube when a torque T is

exerted on a rigid end plate as

shown?

1 2

1 21 2

1 1 2 2

1 1 2 2

1 2

1 1 2 2 1 1 2 2

1 2

1 1 2 2

,

p p

p p

p p p p

p p

T T T

T L T L

G I G I

G I T G I TT T

G I G I G I G I

TL

G I G I

• Solution:

29

• Sample Problem

• Determine the maximum shearing stress in each of

the two shafts with a diameter of 1.5 in.

• Solution:

4max

max max

4 12 600

2 12 0

2 2 4 4

120 lb ft, 240 lb ft, 1440 lb

2 240 12 1.5 24.35 ksi

1.5 32

2 12.17 ksi

2

A

B

B AF E

p p

A B

B

BD

p

A

AC BD

p

T F

T F

T L T L

GI GI

T T F

T d

I

T d

I

30

• • Make the beam determinate by replacing the restraint at B with FB

l

q

A B

qA B

FB

qBA

MA• Make the beam determinate by replacing the rotational restraint at A with MA

• A propped cantilever beam

Statically Indeterminate Beams

31

• q

AB

l

• Known: q and l. Find: reaction forces at the supports.

Sample Problem

32

• • Method 1:

A B

q

FB

3 4

0

03 8

3

8

B

B

B

w

F l ql

EI EIql

F

• Deformation compatibility requires:

• Take the deflection restraint at B as the redundant restraint

33

A B

q

l

• • Take the rotational restraint at A as the redundant restraint

q

BA

MA

3

2

0

03 24

8

A

A

A

M l ql

EI EI

qlM

• Deformation compatibility requires:

A B

q

l

• Method 2:

34

• PA

B

C D

aa

• Cantilever beam AB is enhanced by another cantilever

beam CD with the same EI. Find (1) contact force at D;

(2) ratio of deflections at B with and without enhancement;

(3) ratio of the maximum bending moments in AB with and

without enhancement.

Sample Problem

35

• • Contact force at DP

A

B

CFD

FD

• Deflection compatibility

at D requires D Dw AB w CD

3

3

DD

F aw CD

EI

CD

FD

• Solution:

36

• 2 32 35

3 3 26 6 3 6

D DD

F a F aPa Paw AB a a a a

EI EI EI EI

P

AB

D1

FD

AB

D

FD

P

AB

D

3 335 5

3 6 3 4

D DD D D

F a F aPaw AB w CD F P

EI EI EI

37

• • Ratio of deflections at B with and without the enhancement

3 3

0

2 33 3 3

1

3 3

1

0

2 8

3 3

58 8 393 2

6 3 6 3 24

3939 8

24 3 64

B

D DB

B

B

P a Paw

EI EI

F a F aPa Pa Paw a a

EI EI EI EI EI

w Pa Pa

EI EIw

38

• • Ratio of the maximum bending moments in AB with and

without the enhancement.

0 0max 0 2AM x Px M M Pa

• Without:

• With:

1

1max 1

2D

D

Px x aM x

Px F x a a x a

M M Pa

PA B

x

P

ABD

FD

x

39

• Exercise

• Find the reaction forces for the following continuous beam.

MMqqa

a2a

AB D

C

a

q qa

a2a

AB D

C

a

• Solution

40

• Moment-Area Theorems with Indeterminate Beams

• Reactions at supports of statically indeterminate

beams are found by designating a redundant

constraint and treating it as an unknown load which

satisfies a displacement compatibility requirement.

• The (M/EI) diagram is drawn by parts. The

resulting tangential deviations are superposed and

related by the compatibility requirement.

• With reactions determined, the slope and deflection

are found from the moment-area method.

41

• • For the coplanar structure shown below, the flexural

rigidity of AB is EI, tension rigidity of CD is EA; P, L and

a are given. Find FCD.

P

A BC

D

a

2L

2L

Combined Indeterminate Structures

42

• • Solution

3

3

3

48

48

C CD

C CC CD

C

w l

P F L F aw l

EI EA

ALF P

Ia AL

D

a

C

P

A BC

2L

2L

CF

• Deformation compatibility requires:

43

Combined Indeterminate Structures

• More Examples

44

• More Examples

• More Examples

46

• More Examples

47

Consider

only the

effects of

bending.

• • Statically Determinate Problems（静定问题）

• Statically Indeterminate Problems（超静定问题）

• Degrees of Indeterminacy（超静定次数）

• Advantages & Disadvantages of Indeterminate Structures

（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

48

• • 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

49