Analysis of Statically Indeterminate...

Analysis of Statically Indeterminate Structures

Force Method of Analysis

Chapter 10

Methods of AnalysisTwo different Methods are available

– Force method • known as consistent deformation, unit load method,

flexibility method, or the superposition equations method.

• The primary unknowns in this way of analysis are forces

– Displacement method• Known as stiffness method

• The primary unknowns are displacements

2

Force method of analysisThe deflection or slope at any point on a structure as a result

of a number of forces, including the reactions, is equal to the algebraic sum of the deflections or slopes at this particular point as a result of these loads acting individually

3

Force Method of Analysis• General Procedure

– Indeterminate to the first degree

– 1 Compatibility equation is needed

– Choosing one of the support reaction as a redundant

– The structure become statically determinate and stable

– Downward displacement B at B calculated (load action)

– BB upward deflection per unit force at B

– Compatibility equation

0 = B + ByBB

– Reaction By known

– Now the structure is statically determinate

4


– Indeterminate to the first degree

– 1 Compatibility equation is needed

– Choosing MA at A as a redundant

– The structure become statically determinate & stable

– Rotation A at A caused by load P is determined

– AA rotation caused by unit couple moment applied at A


0 = A + MA AA

– Moment MA known

– Now the structure is statically determinate

5


– Indeterminate to the 2nd degree

– 2 Compatibility equations needed

– Redundant reaction B & C

– Displacement B &C caused by load P1 & P2 are determined

6

Force Method of Analysis

• General Procedure – BB & BC Deflection per unit force

at B are determined

– CC & CB Deflection per unit force at C are determined

– Compatibility equations

0 = B + ByBB + CyBC

0 = C + ByCB + CyCC

– Reactions at B & C are known

– Statically determinate structure

7

Maxwell’s Theorem • The displacement of a point B on a

structure due to a unit load acting at a point C is equal to the displacement of point C when the unit load is acting at point B the is

fBC=fCB

• The rotation of a point B on a structure due to a unit moment acting at a point C is equal to the rotation of point C when the unit moment is acting at point B the is

BC=CB

Force Method of Analysis• Procedure for Analysis

– Determine the degree of statically indeterminacy

– Identify the redundants, whether it’s a force or a moment, that would be treated as unknown in order to form the structure statically determinate & stable

– Calculate the displacements of the determinate structure at the points where the redundants have been removed

– Calculate the displacements at these same points in the determinate structure due to the unit force or moment of each redundants individually

– Workout the compatibility equation at each point where there is a redundant & solve for the unknown redundants

– Knowing the value of the redundants, use equilibrium to determine the remaining reactions

9

Beam Deflections and Slopes

3 2

3 2

max max

36

3 2

Px Lx

EI

PL PLatx L

EI EI

20

2

0 0max max

2

2

Mx

EI

M L M Latx L

EI EI

Positive (+)


4 3 2 2

4 3

max max

4 624

8 6

wx Lx L x

EI

wL wLatx L

EI EI

3 2

3

max

2

max

4 3 , 048 2

2 48

0 or 2 16

P Lv x L x x

EI

L PLatx

EI

L PLatx

EI

Positive (+)


3 2 3

4

max

3

max

224

5

2 384

24

wxv x Lx L

EI

L wLatx

EI

wL

EI

2 2 2 , 06

6 6L R

Pbxv L b x x a

LEI

Pab L b Pab L a

LEI LEI

Positive (+)


3 2 3

3 2 2 3

3 3

9 24 16 02384

8 24 17384

2

3 7

128 384L R

wx LL Lx x xEI

wLx Lx L x L

EI

L x L

wL wL

EI EI

2 20

2

0max

0 0

3 26

9 2

6 3L R

M xv x Lx L

EIL

M L

EI

M L M L

EI EI

Positive (+)

Example 1• Determine the reaction at B

– Indeterminate to the 1st degree thus one additional equation needed

– Lets take B as a redundant

– Determine the deflection at point B in the absence of support B. Using the moment-area method

– Determine the deflection caused by the unit load at point B

39000 .B

kN m

EI

14

506=

300kN.m

B.M.D

BA

50kN

1 6 2300 . 6 6

2 3B

EI

Example 1


0 = B + ByBB

By = 15.6kN

– The reaction at B is known now so the structure is statically determinate & equilibrium equations can be applied to get the rest of the unknowns

3576BB

mf

EI

15

1 12 212 12

2 3BBf

EI

112=

12m

B.M.D

BA

19000 576

0 yBEI EI

B.M.D

Example 2• Determine the moment at A

– Indeterminate to the 1st degree thus one additional equation needed

– Lets take MA as a redundant

– Determine the slope A at point A ignoring the fixation at A. Using the moment-area method

1

1 10 10 333.3320

2 3d

EI EI

A

20k.ft

B.M.D

B1 1 333.33

.10

A

d

L EI

233.33 .A

k ft

EI

B

A A

d1

Elastic Curve

Example 2– Determine rotation caused by the unit

moment applied at A


0 = A +MAAA

MA = -10k.ft

– The moment at A is known now so the structure is statically determinate

2

1 10 2 10 33.331

2 3d

EI EI

17

33.33 3.330 AM

EI EI

B

A AA

d2

Elastic Curve

1 B.M.D

BA

12 1 33.33 3.33.

10AA

d ft

L EI EI

Example 3

• Neglect the axial load

• The end moments at A&B will be considered as redundants

• From Table inside the front cover

1(375)A

EI

1(291.7)B

EI

Example 4

-

3.33BA

EI

13.33

20(2/3)10

6.67

6.67AA

EI

6.67BB

EI

3.33AB

EI

375 6.67 3.330 A BM M

EI EI EI

291.7 3.33 6.670 A BM M

EI EI EI

45.8 .AM k ft 20.8 .BM k ft

Example 4

24Example 4

Example 5Determine the reaction at the support for the beam shown, EI is Constant.Choose the internal moment at internal support as the redundant

Compatibility EquationsThe relative rotation of one end of one beam

with respect to the end of the other beam to be zero

Example 5

Example 6Determine the support reactions on the frame shown EI is Constant.

Example 6

Example (Additional)The frame, shown in the photo is used to support the bridge deck.

Assuming EI is constant, a drawing of it along with the dimensions and

loading is shown. Determine the support reactions.

Example 7Determine the moment at the fixed support A for the frame shown EI is

Constant.

Example 7

Problem 1

Problem 2

Problem 3

Force Method of Analysis: TrussExample8

Determine the force in member AC.

Assume EA is the same for all the members

44

Example 8

Force Method of Analysis: Truss

46

Example 9

• Determine the force in each member if the turnbuckle on member AC is used to shorten the member by 0.5in. Each member has a cross-section area of 0.2 in2 & E=29(106)psi

47

Example 9

Example 9

Example 10The beam shown is supported by a pin at A and two pin-connected bar at B.

Determine the force in member BD. Take E=29(103), I=800 in4 for the beam

and A=3 in2 for each bar.

Example 10

Example 11The simply supported beam shown in the photo is to be designed to support a

uniform load of 2 kN/m. Determine the force developed in member CE. Neglect

the thickness of the beam and assume the truss members are pin connected to the

beam. Also, neglect the effect of axial compression and shear in the beam. The

cross-sectional area of each strut is 400 mm2, and for the beam I=20(106)mm4 .

Take E=200 GPa

Example 11

Symmetric Structures

Anti-symmetric Structures

Transformation of Loading

Influence Lines for Statically Indeterminate Beams

Reaction at A

1

Scale Factor 1

y DA

AA

AA

A ff

f

Reaction at A


1

Scale Factor 1

E DE

EE

EE

V ff

f

Shear at E


Moment at E

1

Scale Factor 1

E DE

EE

EE

M f


Example 11

Influence Line of a continuous Beam

4m 8m

A B C

G D E F

Draw the Influence Line of1. Reaction at A, B and C2. Shear at G and E3. Moment at G and E

Influence Line of RA

1

3

1 1

2

3

2 2

0 8

5.33

12

0 4

18.67 64

6

xA

xA

x

x xf

EI EI

x

x xf

EI EI EI

4m 8mA

B C

D

16/EI8/EI 4/EI

5.33/EI

x1y=x1/2x2

y=x2

18.67/EI

64/EI


1

3

1 1

2

3

2 2

0 8

5.33

12

0 4

18.67 64

6

xA

xA

x

x xf

EI EI

x

x xf

EI EI EI

Point fxA fxA/fAA

A 64/EI 1.0

G 28/EI 0.4375

B 0 0

D -14/EI -0.2188

E -16/EI -0.25

F -10/EI -0.1562

C 0 0


0

8 12 1 0

3

2 8

c

B A

AB

M

R R x

R xR

Point x RA RB

A 12 1.0 0

G 10 0.4375 0.5939

B 8 0 1

D 6 -0.2188 1.078

E 4 -0.25 0.875

F 2 -0.1562 0.485

C 0 0 0

Influence Line of RBUsing equilibrium conditions for the influence line of RB

0

4 8 1 8 0

18 2

B

A C

AC

M

R R x

x RR

Point x RA RC

A 12 1.0 0

G 10 0.4375 -0.0312

B 8 0 0

D 6 -0.2188 0.1406

E 4 -0.25 0.375

F 2 -0.1562 0.6719

C 0 0 1

Influence Line of RCUsing equilibrium conditions for the influence line of RC

1

Check

0y

A B C

F

R R R

Influence Line of VG

Influence Line of VE

Influence Line of MG

Influence Line of ME

Live Load Pattern in Continuous Beams

LL

DL

Support Reactions

Influence Line for positive reaction at support 1

Load pattern for maximum positive reaction at support 1

Span Positive Moment

Influence line for positive moment at 7

Load pattern for maximum positive moment at 7

Support Negative Moment

Influence line for negative moment at support 2

Load pattern for maximum negative moment at support 2

Internal Shear

Influence line for positive shear at 7

Load pattern for maximum positive shear at 7

Moment EnvelopesThe moment envelope curve defines the extreme boundary values of

bending moment along the beam due to critical placements of design

live loading.

Example of Three Span Continuous Beam

Moment Envelopes

Dead Load Only

BMD due to DL

Moment Envelopes

Live Load Arrangement

Max. Positive Moment at span 1&3(Min. Moment at span 2)

Max. Positive Moment at span 2

(Min. Moment at span 1 & 3)

Moment Envelopes


Max. negative Moment at support 2

Max. negative Moment at support 3

Moment Envelopes


Max. positive Moment at support 2

Max. positive Moment at support 3

Moment Envelopes

Max. +M at span 2Max. +M at spans 1&3

DL only

Max. -M at support 2 Max. -M at support 3

Max. +M at support 2 Max. +M at support 3

Summarize all critical cases

Moment EnvelopesMoment Diagram for All Cases

Solving each of these cases and combining the results with the dead

only case, results in six different moment diagrams.

The dead load only case has been left as a heavier line.

Moment EnvelopesMoment Diagram for All Cases

Analysis of Statically Indeterminate...

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