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Structural Analysis II Analysis of Statically Indeterminate Structures by the Force Method

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Outline 1 2 3 4 5 6 7

Statically Indeterminate Structures General Procedure of Force Method Effect of Support Settlements Beam on Elastic Supports Frames and Trusses Composite Structures Several Degrees of Indeterminacy

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Statically Indeterminate Structures A statically indeterminate structure is a structure of which the number of unknown reactions or internal forces exceeds the number of equilibrium equations. The indeterminacy may arise as a result of added supports or members, or by the general form of the structure.

Statically Indeterminate Structures (cont’d) Most of the structures designed today are statically indeterminate. The benefits of statically indeterminate structures include:

1. For a given loading, the maximum stress and deflection of an indeterminate structure are generally smaller than those of its statically determinate counterpart.

 max 

M max c I

Disadvantage: Fixed support is more expensive to construct that pin or roller.

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Statically Indeterminate Structures (cont’d) 2. Statically indeterminate structure redistributes its load to its redundant supports or members in case of overloading (due to earthquake or wind). The structure maintains its stability and collapse is prevented. Plastic hinge

Stable after forming plastic hinge

Plastic hinge

Unstable after forming plastic hinge

Disadvantage: In a statically indeterminate structure, differential displacement of the supports, changes in member lengths caused by temperature or fabrication errors will introduce stresses in the structure.

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Method of Analysis When analyzing an indeterminate structure, it is necessary to satisfy:  Equilibrium  Compatibility

There are two methods to analyze a statically indeterminate structure: 1.

Force Method (also called: Flexibility Method, Compatibility Method, Method of Consistent Displacements. The unknowns are forces.)

This method is is discussed in this topic. 2.

Displacement Method (also called Stiffness Method. The unknowns are displacements. )

The two topics (Slope-Deflection Equations and Moment Distribution) are discussed in Part II. Further discussions are covered in Structural Analysis III. 6

General Procedure of Force Method A statically indeterminate structure can be equivalent to a statically determinate structure subjected to certain conditions.

Q

P

P

By

Q

The two structures are equivalent if Q=By or the displacement at B in the determinate structure δB=0. The latter is called compatibility equation.

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General Procedure of Force Method Consider the beam shown, which is indeterminate to the first degree. One additional equation is necessary.

(a)

=

(b)

Choose one of the support reactions as redundant ( in this case By ) and temporarily remove it.

+

The compatibility equation for displacement at B:

(c)

(d)

fBB = deflection at B ( the first subscript) caused by a unit load at B ( the second subscript) [unit: m/N, mm/kN, etc.]. It is called flexibility coefficient.

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General Procedure of Force Method (cont’d)

P

MA Ay

SFD

Once By is obtained, the three reactions at A can be calculated using the equations of equilibrium.

By

The shear force and bending moment diagram of the actual beam can be obtained using the equations of equilibrium. Since the force method depends on superposition of displacements, it is necessary that the material remain linearelastic when loaded.

BMD

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General Procedure of Force Method (cont’d) The choice of the redundant is arbitrary. (a)

+

(b)

(c)

(d)

The compatibility equation:

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Example 1 Determine the reactions and draw the bending moment diagram. EI is constant.

Solution: Take By as the redundant. Assume that By acts upward.

= The compatibility equation:

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Example 1 (cont’d) =

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Example 1 (cont’d) 112.8 kN.m

V(kN)

93.6

Bending moment at C can be computed using equilibrium equation.

-112.8

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Example 2 Determine the reactions and draw the shear and moment diagram. E is constant.

Solution:

=

The structure is indeterminate to the first degree. Select Cy as the redundant. The compatibility equation for displacement at C:

+

Computing ∆C and fCC using the method of virtual work. Cy 14

Example 2 (cont’d) Take x1 from B to the left, x2 from B to the right, x3 from D to the left, and x4 from the 18 kN point to the left.

1

15

Example 2 (cont’d) Compatibility equation

V (kN)

16.93

12.54

-7.07

M (kN.m)

16.48

6.23

-29.58

-5.46

-21.21 16

Effect of Support Settlements The effects of support settlements can easily be included in the force method by modifying certain terms of the compatibility equations.

Assume that RB is upward.

If support B does not settle when the beam is loaded, the compatibility equation can be written as:

If support B settles 1cm when the beam is loaded, the compatibility equation can be written as:

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Example 3 Draw the shear and bending moment diagrams of the beam. The support B settles 40mm. E = 200 GPa, I = 500 × 106 mm4.

Solution: Take the center support B as the redundant, so that the roller at B is removed. Assume that By acts downwards.

=

The compatibility equation: 18

Example 3 (cont’d)

300

M

m

2

200

4

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Example 3 (cont’d) Find other reactions using the equations of equilibrium:

Draw the shear and bending moment diagrams:

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Example 4 12mm

18mm

5m

5m

Determine the reactions induced in the continuous beam if support B settles 18mm and support C settles 12mm. E = 200 GPa, I = 120 × 106 mm4.

Solution:

6mm 12mm

fBB

0.5

1

6.9 kN 13.8 kN

0.5

6.9 kN 21

Example 5 Determine the reaction and all bar forces if support A settles vertically 20mm. E = 200 GPa, A = 1000mm2 for all bars.

Solution: The structure is indeterminate to the first degree . Take the reaction at A as the redundant. Define RA positive downward. The compatibility equation:

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Example 5 (cont’d)

m/kN

The compatibility equation:

Member forces:

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Beam on Elastic Supports RB

RB Q A

B

If beam CD behaves elastically, the support B of beam AB can be idealized as a spring. The spring constant K indicates the force required at B (of beam CBD) for point B to displace one unit displacement (kN/m) When analyze beam AB, the compatibility equation at support B should consider ∆. Considering point A is fixed, the compatibility equation is

-|∆ B|+fBBRB= -RB/K 24

Example 6 w

Determine the deflection at point B. K = 1750 kN/m, w = 30 kN/m, I = 120 × 106 mm4 and E = 200 GPa. K=1750kN/m

5.5m

Solution: ∆B

∆B0

fBB 1 25

Example 7 (Frame) Determinate the support reactions on the frame. EI is constant.

Solution: Choose the horizontal reaction at B, Bx, as the redundant. The pin at B is replaced by a roller.

Compatibility Equations: Δ B  Bx f BB  0

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Example 7 (cont’d) Use the virtual work method to find ∆B and fBB.

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Example 7 (cont’d) Compatibility Equation:

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Example 8 (Frame) Determinate the support reactions on the frame. Point A is fixed and point B is supported by a roller. EI is constant.

Solution: Choose the end moment at A as the redundant. Compatibility Equations:

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Example 8 (cont’d) Use the virtual work method to find θA and αAA. Mm dx EI

L

A   0 A 

1 2.4 [ (0.13 x1)(1  0.278 x1) dx1 EI 0 1.5

  (1.34 x2  0.75x22 )(0.222 x2 )dx 2 ] 0

A 

1 0.333 (0.209  0.124)  EI EI

 AA  

L

0

m m dx EI

1.5 1 2.4 [ (1  0.278x1 )2 dx1   (0.222x 2 )2 dx2 ] 0 EI 0 1 1.210 (1.155  0.555)   EI EI

 AA   AA 1

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Example 8 (cont’d) The compatibility equation:

.m

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Degree of Indeterminacy of Trusses The unknowns:  The number of member forces (m)  The number of support reactions (r) Two equations can be written for each joint. The number of joint is j. Thus the number of available equilibrium equations is 2j.

Indeterminate truss: m + r > 2j Depending on m and r, a truss can be statically indeterminate internally or statically indeterminate externally or both. The redundant should be chosen correspondingly.

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Degree of Indeterminacy of Trusses m = 5, r = 4, j = 4 The truss is statically indeterminate externally to the first degree.

Choose one reaction as redundant.

m = 6, r = 3, j = 4 The truss is statically indeterminate internally to the first degree.

m = 6, r = 4, j = 4 The truss is statically indeterminate internally to the first degree and externally to the first degree.

Choose one member force as redundant.

Choose one reaction and one member force as redundant. 33

Example 9 Determinate the member force of the truss. The axial rigidity of AE is the same for all the members.

Solution: The truss is indeterminate internally to the first degree. Choose member AC as the redundant.

The Compatibility Equations:

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Example 9 (cont’d) Use the virtual work method to calculate the relative displacement ∆AC and fAC, AC.

1

1

35

Example 9 (cont’d) The compatibility equation:

The other member forces can be obtained by equilibrium at each joint:

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Composite Structures Composite structures are composed of some members subjected only to axial force while other members are subjected to bending. If the structure is statically indeterminate, the force method can conveniently be used for its analysis.

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Example 10 60 kN/m 3m C

Determine the horizontal and vertical reactions at A and the forces in member BC and BD. Take E = 200 GPa, I = 300 × 1.8m 106 mm4 for the beam and A = 1800 mm2 for each bar

Solution: The structure is indeterminate to the first degree. The force in member BD is chosen as the redundant. This member is cut to eliminate its capacity to sustain a force. 60 kN/m

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Example 10 (cont’d) 1.12

1 1

No need to calculate M. Why?

1 1.12

Consider only the bending strain energy in the beam and the axial strain energy in the bars.

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Example 10 (cont’d)

The compatibility equation:

Apply equilibrium equations to the structure:

The same results can be obtained by taking the horizontal reaction at A as the redundant. 40

Example 10 (cont’d) 60 kN/m

Ax

Alternate solution

3m

The horizontal reaction Ax is chosen as the redundant.

1.8m

=

A, x  f AA Ax  0 m0

1

+

30 kN

43.92 kN

A ,x   f AA  

31.06 kN

0.732

nNL ( 43.92)(0.732)(1.8 / sin 600 ) ( 31.06)( 0.897)(1.8 / sin 450 )   AE AE AE

n L (0.732) (1.8 / sin 60 ) (0.897) (1.8 / sin 45 )   AE AE AE 2

2

0

2

0.897

Ax 

0

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Example 11 200 kN

1.5m 3.6m

3.6m 200 kN

FCD

FCD

Determine the force in each of the three struts and the maximum moment in the beam. Each of the struts has a crosssectional area of 1250 mm2 . Neglect the depth of the beam and the effect of axial compression in the beam. E = 200 GPa, I for the beam is 150 × 106 mm4 .

Solution: The structure is indeterminate to the first degree. Take force in member CD as the redundant. The compatibility equation for member CD:

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Example 11 (cont’d) 200 kN

Use the virtual work method:

1.5m 3.6m

3.6m

= 200 kN

N=0

α

N=0

+

cosα=5/13

1 1 α 43

Example 11 (cont’d) 200 kN

1.5m 3.6m

3.6m

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Several Degrees of Indeterminacy Procedures for indeterminate structure to the second degree.

= +

+

The compatibility equations:

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Several Degrees of Indeterminacy (cont’d) fBB = the deflection at B caused by a unit load at B fCB = the deflection at C caused by a unit load at B fBC = the deflection at B caused by a unit load at C fCC = the deflection at C caused by a unit load at C The compatibility equations:

By and Cy can be obtained. The reactions at A and D can be computed using the equations of equilibrium. 46

Several Degrees of Indeterminacy (cont’d) For a structure having n redundant reaction, n compatibility equations can be written.

Δ1  f11R1  f 12R2  ...  f 1nR n  0 Δ  f R  f R  ...  f R  0  2 21 1 22 2 2n n   Δn  fn 1 R1  fn 2 R2  ...  fnn Rn  0  f11 f  21     fn1

f12 ... f 22 ...  f n 2 ...

f 1n   R1   Δ1  Δ  f 2 n  R2      2            f nn  Rn  Δ n 

f is symmetric

fR  -Δ 47

Theorem of Reciprocal Displacements Principle of virtual work:

mA

mB

fBA= fAB

The displacement of point B on a structure due to a unit load acting at point A is equal to the displacement of point A due to a unit load acting at point B.

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Theorem of Reciprocal Displacements The rotation at point B on a structure due to a unit couple moment acting at point A is equal to the rotation at point A when the unit couple moment is acting at point B.

The rotation at point B on a structure due to a unit load acting at point A is equal to the displacement at point A when a unit couple moment is acting at point B.

This theorem of reciprocal displacement is valid for any elastic structure – be it a truss, a beam or a frame.

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General Procedure of Force Method 1.

Determinate the number of degrees n to which the structure is indeterminate.

2.

Remove n redundant forces or moments. Define the positive directions of the redundants.

3.

Write a compatibility equation for the displacement or rotation at each joint where there is redundant.

4.

Solve for unknown redundants.

5.

Solve for the remaining unknown reactions using the equations of equilibrium.

6.

Draw the shear and bending moment diagrams.

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Example 12 Draw the shear and bending moment diagrams for the beam. EI is constant. Neglect the effects of axial force

=

Solution: Consider the two end moments at A and B as the redundants.

+

The compatibility equation:

+

51

Example 12 (cont’d) Use the virtual work method to find the angle of rotations and the angular flexibility coefficients.

The compatibility equation:

52

Example 12 (cont’d) Find other reactions using the equations of equilibrium.

Draw the shear and bending moment diagrams:

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Table I: Deflections and Slopes of Cantilever Beams

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Table II: Deflections and Slopes of Simple Beams

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Table II: Deflections and Slopes of Simple Beams (2)

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Reference/Further Reading Hibbeler, R. C. “Structural Analysis,” 7th Edition, Pearson Prentice Hall, 2009, Chapter 6. (OR 8th Ed. 2012)

Leet, K. M., Uang, C. M., Gilbert, A. M. “Fundamentals of Structural Analysis,” 3rd Edition, McGraw-Hill, 2008, Chapter 8. (OR 4th Ed. 2011) Software for Structural Analysis RISA 2D - http://www.risatech.com/demos/demo_2d.html

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