วย.321 การวิเคราะห์โครงสร้าง 1 ce321 structural...

วย.321 การวิเคราะหโ์ครงสรา้ง 1

CE321 Structural Analysis 1

นคร ภูว่โรดม

คณะวิศวกรรมศาสตร ์มหาวิทยาลยัธรรมศาสตร ์

Faculty of Engineering, Thammasat University CE321 Structural Analysis I

Prerequisite: CE 221 Objective: The objective of this course begins with an introduction to structural analysis for civil engineering students. The fundamental characteristics such as stresses or stress resultants, deflection and support reactions are studied. The analysis of structures involves determination of these quantities as cause by a given loading condition. This course presents the methods for analysis of structures with linear elastic behaviors in static equilibrium. Course outlines: 1. Introduction to structural analysis 2. Loads on structures 3. Equilibrium and support reactions 4. Plane and space trusses 5. Beams and frames: shear and bending moment 6. Deflections of beams: Geometric methods 7. Deflections of trusses, beams and frames: work-energy method 8. Influence lines 9. Analysis of statically indeterminate structures: method of consistent deformation

สัปดาห์ที ่ หัวข้อ/รายละเอยีด จํานวนช่ัวโมง 1 หลกัการเบ้ืองตน้ของการวิเคราะห์โครงสร้าง (1-3) 3

2-3 การวิเคราะห์แรงภายในโครงถกั (4) 6 4-6 แรงปฏิกิริยา แรงเฉือน และโมเมนตด์ดัในโครงสร้างดีเทอร์

มิเนทเชิงสถิต (5) 9

7 การวิเคราะห์โครงสร้างโดยวิธีทางเรขาคณิต (6) 3 8-10 การวิเคราะห์การเสียรูปของโครงสร้างดีเทอร์มิเนทเชิงสถิต

โดยวิธีงานเสมือนและพลงังานความเครียด (7) 9

11-12 เสน้อิทธิพลสาํหรับโครงสร้างดีเทอร์มิเนทเชิงสถิต (8) 6 13-15 การวิเคราะห์โครงสร้างอินดีเทอร์มิเนทเชิงสถิตโดยวิธีการ

เสียรูปท่ีคงตวั (9) 9

Lecturer: Nakhorn Poovarodom, Ph.D. Textbooks: 1. A. Kassimali, “Structural Analysis”, 2nd ed., PWS Publishing, 1999 2. R. C. Hibbeler, “Structural Analysis”, 4th ed., Prentice-Hall Inc., 1999 References:

1. ปณธิาน ลักคุณะประสทิธิ์, “การวิเคราะห์โครงสร้าง”, สมาคมวิศวกรรมสถานแห่งประเทศไทย, กรกฎาคม ๒๕๓๘

2. R. C. Coates, M. G. Coutie, F. K. Kong, “Structural Analysis”, 3rd ed., Chapman & Hall, 1988

3. C. K. Wang, “Intermediate Structural Analysis”, McGraw-Hill, 1983 Conducts of Course Lecture 3 hours/week. Grade will be based on (tentative) Mid-term examination I 20 % Test 10 % Mid-term examination II 30 % Final examination 30 % Homework/Quiz/Project 10 %

Contents Chapter 1: Introduction to Structural Analysis 1.1 Role of Structural Analysis in Structural Engineering Projects 1 1.2 Classification of Structures 1 1.3 Analytical Model 2 Chapter 2: Loads on Structures 2.1 Dead Loads 4 2.2 Live Loads 4 2.3 Impact 4 2.4 Environmental Loads 4 2.5 Hydrostatic and Soil Pressure 5 2.6 Thermal and other effects 5 2.7 Load Combinations 5 Chapter 3: Equilibrium and Support Reactions 3.1 Equilibrium of structure 6 3.2 External and Internal Forces 6 3.3 Types of Supports for Plane Structures 6 3.4 Static Determinacy, Indeterminacy and Instability (Plane Structure) 7 3.5 Principle of Superposition 10 Chapter 4: Plane and Space Trusses 4.1 Assumptions for analysis of trusses 11 4.2 Arrangement of members of plane trusses: Internal stability 12 4.3 Equation of condition for plane trusses 12 4.4 Static Determinacy, Indeterminacy and Instability (plane truss) 13 4.5 Analysis of Plane truss by the method of joints 13 4.6 Analysis of Plane Truss by the Method of Sections 14 4.7 Analysis of Compound truss 15 4.8 Complex truss 15 4.9 Space Trusses 15 Examples 18 Chapter 5: Beams and Frames: Shear and Bending Moment 5.1 Axial force, Shear and Bending moment 27 5.2 Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) 27 5.3 Qualitative Deflected Shapes 27 5.4 Relationship between Loads, Shear and Bending Moments 28 5.5 Static Determinacy, Indeterminacy and Instability of Plan Frame 30 Examples 31

Chapter 6: Deflections of beams: Geometric methods 6.1 Differential equation for beam deflection 43 6.2 Direct integration method 44 6.3 Moment area method 45 6.4 Bending moment diagram by parts 46 Chapter 7: Deflections of trusses, beams and frames: Work-Energy method 7.1 Work 48 7.2 Principle of Virtual Work 49 7.3 Deflections of Trusses by the Virtual Work Method 51 7.4 Deflections of Beams by the Virtual Work Method 52 7.5 Deflections of Frames by the Virtual Work Method 54 7.6 Conservation of Energy and Strain Energy 55 7.7 Castigliano’s second Theorem 56 7.8 Betti’s Law and Maxwell’s Law of reciprocal deflections 59 Examples 60 Chapter 8: Influence Lines 8.1 Influence line for beams and frames by equilibrium method 84 8.2 Müller–Breslau’s principle and qualitative Influence Lines. 85 8.3 Influence line for Deflections 86 Examples 88 Chapter 9: Method of Consistent Deformation: Force Method 9.1 Introduction 95 9.2 Structure with a single degree of indeterminacy 95 9.3 Internal Forces and Moments as redundants 98 9.4 Structures with Multiple Degrees of Indeterminacy 100 9.5 Support settlements 101 9.6 Temperature change and Fabrication Errors 103 9.7 Method of Least Work 103 Examples 104 Appendix

Structural Analysis I

N. Poovarodom 1

Chapter 1 Introduction to Structural Analysis

1.1 Role of Structural Analysis in Structural Engineering Projects a) Planning Phase: To establish the functional requirements of the proposed structure - Aesthetics - Environmental impact b) Preliminary Structural Design: To estimate sizes of members by approximate analysis, past experiences or code requirements c) Estimation of Loads: Dead load, live load, environmental loads, etc. d) Structural Analysis: Loads Structure Stresses, deflections, support reactions e) Safety and serviceability checks: by building code requirements if the design is not satisfied, repeat step b to e f) Construction phase: Drawing, specification construction 1.2 Classification of Structures a) Tension structure; cable (no bending stiffness, no compression)

b) Compression structure

Column Arch


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c) Truss; members are in pure tension or compression, hinge joint d) Shear structure; for example, reinforced concrete shear wall e) Bending structure; beam, rigid frame, slab, plate 1.3 Analytical Model To simplify the analysis of a complicate structure Plane / Space Structure Plane Structure: all the members of structure and the applied loads lie in a single plane (2 Dimensions) Space structure: (3 Dimensions) Example Bridge


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Building

Connections a) Rigid; all members have the same translation and rotation

Transmit force & moment b) Hinge; all members have the same translation Transmit only force

Supports a) Fixed support b) Hinged support c) Roller


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Chapter 2 Loads on Structures

Types of Load Load estimation

Reference: ASCE 7 Standard

2.1 Dead Loads: Gravity load of constant magnitude and fixed position that act permanently on structure.

= weight of structure + equipments

Unit weight 3ft/lb 3m/kN 3m/kg - Concrete (RC) 150 23.6 2400 - Structural steel 490 77.0 7850 - Wood 40 6.3 640

2.2 Live Loads: load of varying magnitudes and/or positions caused by the use of the structure. LL for Building uniformly distributed surface load (load per unit area) LL for Bridge by “The American Association of state Highway and Transportation Officials” (AASHTO) Use the maximum effect of truck load (a) or lane load (b) 2.3 Impact: The dynamic effect of load

Impact factor 3.0125

50

LI

L = span length (ft) 2.4 Environmental Loads Example - Wind load - Earthquake load

W (hinge) (hi )


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2.5 Hydrostatic and Soil Pressure

h = unit weight of media (liquid / soil) 2.6 Thermal and other effects - Temperature changes (elongation / shortening) - Fabrication errors - Support movements 2.7 Load Combinations The structure must be designed to have adequate strength to resist the most unfavorable of all the load combination * Serviceability requirement - Deflections - Vibrations - Cracking - Corrosion - Fatigue


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Chapter 3 Equilibrium and Support Reactions

3.1 Equilibrium of structure

Plane 2D 0,0,0 zyx MFF

Space 3D 0,0,0 zyx FFF

and 0,0,0 zyx MMM

3.2 External and Internal Forces External Force = the actions of other bodies on the structure under consideration - Applied forces (load)

- Reactions Internal Force = the forces and couples exerted on a member or portion of the structure by the rest of the structure 3.3 Types of Supports for Plane Structures

x

y

z

x

y


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3.4 Static Determinacy, Indeterminacy and Instability (Plane Structure) Internal stability (rigid); structure can maintain its shape and remains a rigid body when detached from the supports. Internally Stable Structures Internally Unstable Structure Static Determinacy of Internally Stable Structures *Statically Determinate Externally = all support reaction can be determined by solving the equilibrium ( 3r ) *Statically Indeterminate Externally = all reactions cannot be determined from equilibrium ( 3r ) * Statically Unstable Externally = structure cannot remain equilibrium under any load ( 3r ) r = number of reactions For 3r , 3 ri e

= degree of external indeterminacy


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Externally Statically Determinate Structures Externally Statically Indeterminate Structures Externally Statically Unstable Structures


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Special case: geometrically unstable externally Static Determinacy of Internal Unstable Structures ( Equation of condition) Fig a) Internally unstable structure 3r , external unstable Fig b) 4r , stable (unknown = 4) Equation Equilibrium 3 + condition at B 1 Number of unknown = equation statically determinate externally Fig c) 5r

Equilibrium 3 5 Condition 0Fx at B 1

Condition 0M at B 1 statically determinate externally cer 3 = statically unstable externally ce = 1 for hinge

cer 3 = statically determinate externally = 2 for roller

cer 3 = statically indeterminate externally

“Degree of external indeterminacy” )3( ce eri


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Alternative Approach r = number of support reactions

if = number of internal forces that can be transmitted through internal hinge

)2( if or internal roller ( 1if )

rn = number of rigid members Unknown = ifr

Equation = rn3 ri nfr 3 Statically unstable externally

ri nfr 3 Statically determinate externally

ri nfr 3 Statically indeterminate externally

and rie nfri 3)(

3.5 Principle of Superposition

“On a linear elastic structure, the combined effect of several loads acting simultaneously is equal to the algebraic sum of the effects of each load acting individually” Condition (1) Small deformations (2) Linear elastic materials


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Chapter 4 Plane and Space Trusses

Truss = assemblage of straight members connected at their ends by flexible connections to form a rigid configuration Example Bridge truss Roof truss

… 4.1 Assumptions for analysis of trusses 1. All members are connected at their ends by frictionless connections; 2D hinges

3D ball and socket joints 2. All loads and support reactions are applied only at the joints 3. The centroidal axis of each member coincides with the line connecting the centers of the adjacent joints only axial force in members


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4.2 Arrangement of members of plane trusses : Internal stability Internally stable = Truss does not change its shape when detached from supports

Basic Truss Element

3 members and 3 joint Simple Truss By enlarging basic truss element; adding two additional members for each additional joint m = total number of members j = total number of joints 32)3(23 jjm Compound Truss By connecting 2 or more simple trusses and each truss is connected to

others by connections capable of transmitting at least 3 force components.

Example Internal stability (Simple and compound truss) 32 jm Internally unstable 32 jm Internally stable (with proper arrangement of members) 4.3 Equation of condition for plane trusses


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4.4 Static Determinacy, Indeterminacy and Instability (plane truss) Example Unknown Reaction 3 (r) Member force (axial only) 5 (m)

* force at joint = force in member (Newton rd3 law) Total unknown U = 3 + 5 = 8 Equation Equilibrium at joint; 1 joint 0,0 yx FF (j)

Total equation from j joint; E = 4 joint × 2 = 8 Equations 8E for 8U Statically Determinate for r = number of reactions m = number of members j = number of joints

1) jrm 2 Statically Unstable Truss

2) jrm 2 Statically Determinate Truss

3) jrm 2 Statically Indeterminate Truss Condition 1) is necessary and sufficient (true for all cases) Condition 2) and 3) are necessary but not sufficient (dependent of truss arrangement) Degree of static indeterminacy

jrmi 2)( 4.5 Analysis of Plane truss by the method of joints Determine the axial forces by considering the equilibrium of its joints

- Start from a joint with no more than 2 unknown forces or - Start from determining reactions by equilibrium of system


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Identification of zero-force members 1) If only 2 non-collinear members are connected to a joint that has no external loads or reactions applied to it

2 members are zero-force member 2) If three members, two of which are collinear, are connected to a joint that has no external load or reaction applied to it

The non-collinear member is zero force member

Procedure for analysis 1) Check the truss for static determinacy 2) Identify any zero-force member 3) Determine the slope of the inclined members (except zero-force member) 4) Draw FBD of the whole truss 5) Select a joint that has no more than 2 unknown If no such joint, determine support reaction first, then select a joint 6) a: Draw FBD of the selected joint Tensile force: arrow pulling away the joint Compressive force: arrow pushing into the joint b: Determine unknown by 0 xF and 0 yF at the joint

7) Carry on the analysis for the remaining joints 4.6 Analysis of Plane Truss by the Method of Sections This method is efficient for determining only certain members of a truss “Cutting the truss into two portions, pass through the members whose forces are desired” In general, sections should not pass through more than 3 unknown forces Procedure for Analysis

1) Select a section that passes through unknown ( 3 unknown) 2) Select the portion that requires the least amount of computational effort in determining the unknown forces 3) Draw FBD of truss, determine unknowns by equilibrium equation


N. Poovarodom 15

4.7 Analysis of Compound truss Using a combination of the method of joint and method of section can solve some problems of compound trusses 4.8 Complex truss Complex truss is not a simple truss and not a compound truss. It can not be solved by method of joint or method of section. Member forces can be determined by writing two equilibrium equations in terms of unknown member forces for each joint and then solving the system of 2j equations simultaneously. 4.9 Space Trusses Simple internally stable space truss

A simple space truss is formed by enlarging the tetrahedron element (Six members + four joints) by adding 3 members for one more joint The total number of members “m” for “j ” joint 63)4(36 jjm Reactions


N. Poovarodom 16

Equilibrium equations; 0,0,0 zyx FFF , 0,0,0 zyx MMM

For a space structure 6r statically unstable externally 6r statically determinate externally 6r statically indeterminate externally (r = number of reaction) Static Determinate, Indeterminacy and Instability Unknown m (force in each member) and r (reaction) Equation 3, )0,0,0( zyx FFF for each joint, total equation = 3j

For jrm 3 statically unstable space truss jrm 3 statically determinate space truss* jrm 3 statically indeterminate space truss*

* necessary but not sufficient Analysis of member forces Method of Joints Equilibrium at joint

0,0 yx FF and 0 zF

Use to determine joint with unknown 3 Rectangular coordinate for space member For length ABL with projections

ABx in x -axis

ABy in y - axis

ABz in z - axis 222 )()()( ABABABAB zyxL

For force component

AB

ABABxAB L

xFF in x direction

AB

ABAByAB L

yFF in y direction

AB

ABABzAB L

zFF in z direction

Then 222 )()()( zAByABxABAB FFFF


N. Poovarodom 17

Zero force member

a) if only AEF is not in the same b) if ACF and ABF are not collinear,

plane AEF = 0 they are zero force member Method of Sections - Pass an imaginary section through the truss - Divide truss into portion that no more than 6 unknowns can be determined


N. Poovarodom 18

Example 4.1: Determine force in all members of the truss shown.

m = 7 , r = 3 , j = 5, m + r = 2j (statically determinate) Determine reaction by 0, MF

Equilibrium at Joint C

60cos CEF ( 1.53 )

100CEF kN (T)

0sin CECB FF

80CBF kN

= 80 kN (C)

Joint E

0

cos10060cos

EB

EB

F

F

sin100EDF

80EDF kN (T)

Joint B 60cos BDF

100BDF kN 100 kN (C) 0sin10080 BAF

Joint A

100cos

60

ADF kN (T)

4m 4m 4m

3m

E D

C B A 60 kN

80 kN

60 kN 60 kN

60kN

80kN

4m

FCE

FCB

60

100

FED

FEB

60

80

60

FBD

FBA

80 0

60 FAD


N. Poovarodom 19

Member force;

Example 4.2: Determine force in all members of the truss shown. m = 10 , r = 4 , j = 7 m+r=2j (determinate) Determine reaction

;0XF 10 XX BA

;0YF 60 YY BA

;0BM + 012610320620920 YA

kNAY 25 Condition equation ;0GM +

032066 YX AA

kNB

kNB

kNA

kNA

Y

X

Y

X

35

25

25

15

3m

3m

3m 3m 3m 3m

20 20

BY

Bx

AY

Ax

20kN

10kN20kN

20kN

G

F

ED

C

BA

AX

AY

20

2010

G

14.14 61.85

20.67

47.7133.53

20.6720.67

14.14 47.71

20.67 15

25 35

25

20

1020

20

E D

C B A 60

80

60 60

60kN

80kN

100

0

0 100

80

80

100


N. Poovarodom 20

Example 4.3: Determine reactions of the truss shown.

m = 14 , r = 4 , j = 9 m + r = 2j (determinate)

0XF Ax = -30 = 30 k

+ ;0 ABBM 02030603060 yA

kAy 20

+ ;0 DM 02030120203060609060 yC

kC y 200

;0 yF kkDy 100100200206060

Example 4.4: Determine FGH and FBC, for h=3 ft and h=6 ft.

Soln ;0xF 0xA

0EM + 06412418424224 yA

kAy 8

;0yF Ey = 8k

20 ft

D C A

Ax

Ay Dy Cy

60k60k

30k

4 @ 30 ft = 120 ft

B

4 @ 6 ft = 24 ft

J I H

G F

E D C B A

2k 4k 4k 4k 2k

h

Ay

Ax

1

1


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Section 1-1 + ;0HM

hF

hF

BC

BC

48

012864122

+ ;0BM

hF

hF

GH

GH

36

06268

For h = 3 ft FBC = 16 k (T) FGH = -12 k = 12 k (C)

For h = 6 ft FBC = 8 k (T) FGH = -6 k = 6 k (C)

For h =3

For h = 6

To compare the effectiveness of system in terms of the least weight of the structure, consider the weight from the volume of the structure. For Volume = LA

The required area of section is dependent on the internal force;all

FA

Therefore, Volume LF For h=3 )247.424.13(71.6)2428(3)216412(6ii LF

= 2159.7 unit h = 6 )283.2248.8(48.8)2428(6)2846(6ii LF

= 575.8 unit

8.48 8.48 2.83 2.83 4 4 8 8

6 6 6 6

8 8

0

0 0

13.4 13.4 4.47 4.47 4 4 8 8

12 12 12 12

16 16 0

0

0

h

4 2

8

B

FGH H

FBH

FBC


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Example 4.5: Determine force in member FGJ, FJH and FHL

Soln

46.912

2tan 1

56.264

2tan 1

0IM +

0)tan4(50tan445085041004sintan44cos GJGJ FF

FGJ = 101.38 kN (T)

0yF 0coscos5050 ILGJ FF

FIL = -202.76 = 202.76 kN (C)

4m

4m

4m

4m

4m 4m

2m

4m 50 kN 50 kN

100 kN

50 kN

a a

b b L

K J

I H G

100 kN

100 kN

FGJ FIL

FHI FGH

50 50

50

100

Section a-a


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Section b-b

0JM +

0)4(50)4(50)4(cos76.202)4(cos HLFFHL = 111.80 kN (T)

0LM +

0)4(50)4(50)4(cos38.101)4(cos HJF

FHJ = -111.80 kN = 111.80 kN (C)

Example 4.8: Determine member forces of the space truss shown.

member projection (ft)

length (ft) x y z

AD 3 10 5 11.58 BD 8 10 7 14.59 CD 2 10 4 10.95

Joint D ;

;0XF 0695.10

2

59.14

8

58.11

3

CDBDAD FFF

;0YF 01295.10

10

59.14

10

58.11

10

CDBDAD FFF

;0ZF 095.10

4

59.14

7

58.11

5

CDBDAD FFF

Solve 3 equations FAD = 1.12 k (T) FBD = 7.56 k (C) FCD = 8.54 k (C)

50 50

50

100

FHL FHJ

202.76 101.38

J L

FGJ FIL


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Example 4.9: Determine member forces of the space truss shown.

member projection (ft)

length (ft) x y z

AB 20 0 0 20 AC 14 0 10 17.2 BC 6 0 10 11.66 AD 10 20 5 22.91 BD 10 20 5 22.91 CD 4 20 5 21

Joint D

;0XF 02521

4

91.22

10

91.22

10

CDBDAD FFF

;0YF 03021

20

91.22

20

91.22

20

CDBDAD FFF


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;0ZF 021

5

91.22

5

91.22

5

CDBDAD FFF

Solve FAD = 16.63 k (T) FBD = 33.84 k (C) FCD = 15.74 k (C) Joint A

;0ZF 02.17

1063.16

91.22

5

ACF 24.6 ACF k (C)

;0XF 024.62.17

1463.16

91.22

10

ABF 18.2 ABF k (C)

Joint B

;0ZF 066.11

1084.33

91.22

5

BCF 60.8 BCF k (T)

Exercise 1. Classify each of the structures shown as unstable, statically determinate, or statically indeterminate and determine the degree of indeterminacy.

(a) Truss

(b) Beam

2. Determine force in all members and reactions of the truss shown.

(hinge)

40 40 60

A B

C D E F

G

H

I

2 m

2 m

4 m 3 m


N. Poovarodom 26

3. Determine force in members AD, BD, CD, AB and AC of the space truss shown.

y

z

A

C

D

25 kN 20 kN

4 m

B x

40 kN

3 m

2.5 m 3 m

1.5 (CX, CY , CZ)

(BX, BZ )

(AY)

2 m

1 m


N. Poovarodom 27

Chapter 5 Beams and Frames: Shear and Bending Moment

5.1 Axial force, Shear and Bending moment Axial force “Q” Shear “S”, Moment: “M” Sign Convention (Beam Convention) * For inclined and vertical members, employ yx coordinate system y x-axis is in the direction of the member axis x y-axis is in the direction that make right – handed system with z-axis point out of paper y-axis should be upward direction and near the left portion of member 5.2 Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) They depict the variations of shear and bending moment quantities along the length of the members 5.3 Qualitative Deflected Shapes “A rough sketch of the neutral surface of structure, in the deformed position, under the action of a given loading condition” (Deformation due to bending moment only) * Positive M bends a beam concave upward * Negative M bends a beam concave downward * Note From example, A and D must contain no deflection E may deflect upward or downward (can not indicate)


N. Poovarodom 28

5.4 Relationship between Loads, Shear and Bending Moments For small element dx load w is assumed as uniform From b) 0 yF 0)( dSSwdxS wdxdS

wdx

dS

Slope of shear diagram at a point = intensity of distributed load at that point

B

AA

B

AB wdxSSdS

Change in shear between = area under the distributed load point A and B diagram between point A and B

0 aM 0)()()2/)(( dMMdxdSSdxdxwM

Neglecting higher order terms 2)(dx , dxds SdxdM

Sdx

dM

Slope of bending moment diagram at a point = Shear at that point

B

AA

B

AB SdxMMdM

Change in bending moment = Area under the shear diagram between point A and B between points A and B


N. Poovarodom 29

For concentrated loads Fig c) 0 yF ; 0)( dSSPS

For couples or concentrated moments

Fig d) 0 aM ; 0)( dMMMM

MdM Change in bending moment at the = magnitude of the point of application of a couple moment of the couple

Procedure for analysis to construct SFD and BMD 1. Calculate the support reactions. 2. Construct SFD as follows;

a) Determine Shear at the left end of beam. Upward concentrate load cause shear to increase. Downward concentrate load cause shear to decrease. b) Identify the next point to calculate shear.

(Ex. the other end or point of concentrate load) c) Determine the shear at point from (b), from the area under the load diagram. d) Determine the shape of the shear diagram; the slope of SFD is equal to the

load intensity at that point. e) If there is a concentrated load at this end, calculate the ordinate of shear diagram caused by the load. f) Return to step (b) until the right end is reached. Shear must be zero at the end. (include reaction) 3 Construct BMD as follows; a) Determine moment at left end. Clockwise couple cause bending moment to increase. Counterclockwise couple cause bending moment to decrease.

b) Identify the next point to calculate moment. Usually at point where shear was computed, where the couples are applied and where max/min moments occur and where shear is zero.

c) Determine bending moment at point from (b), from the area under the shear diagram.

d) Determine the shape of the BMD; the slope of BMD is equal to the shear at that point. e) If there is a couple at this end, calculate the ordinate of BMD from the couple. f) Return to (b) until the right end is reached. Moment at the end must be zero. (include reaction)

PdS Change in shear at that point of = magnitude of the load application of a concentrated load


N. Poovarodom 30

5.5 Static Determinacy, Indeterminacy and Instability of Plane Frame Statically Determinate = the bending moments, shears, and axial forces in all members, as well as the external reaction can be determined by using the equation of equilibrium and conditions. Example; Unknowns

6 per member + number of reactions

= (6 × 3) + 3 = 21

Equations 3 equations per member + 3 equations per joint = (3 × 3) + (3 × 4) = 21 Statically Determinate

Total Unknowns rmU 6

Total Equation

cejmE )(3

cejmrm )(36

or cejrm 33

necessary but not sufficient

“ ce ” Hinge; The number of equations of condition at the hinge joint is equal to the number

of members meeting at the joint minus one

0EHHM , 0GH

HM , 0 IHHM

They are not independent, one equation is automatically satisfied ( 1M trivial ) Roller: ce = 2 × (number of member at joint) – 2

(1 M , 1F trivial )

cejrm 33 Statically Unstable Frame

cejrm 33 Statically Determinate Frame

cejrm 33 Statically Indeterminate Frame


N. Poovarodom 31

2

P2

P

L/2 L/2

x

P

2

P

2

P

4

PL

BMD

SFD

Example 5.1: Write the equation for shear and bending moment, then draw SFD, BMD a)

2

2P

P

S

LxL

Lx

2

20

22

2xL

P

xP

M

LxL

Lx

2

20

b)

L

MS 0 < x < L

ML

xM

L

xM

M

LxL

Lx

2

20

2

M

2

M

L

M

L

M L

M

L/2 L/2

M

SFD

BMD

M

s L

M

M M

s L

M


N. Poovarodom 32

c)

;0 ABBM

3

303033.1

2

1

2

3030230 yA

kAy 67.36

;0yF 67.3645)42(2

1yC

kCy 33.98

;0AM +

0453

2452

2

1

2

454524533.98

CM

ftkM C .1050

45

2

x

z Then xz

45

2

xxxS

45

2

2

1267.36

45

267.362x

xS 0<x<L

3452267.36

2 xxxxxM

135

67.363

2 xxxM 0<x<L

For maximum M occurring at dx

dM = 0 = S

045

267.362

x

x x = 15.62 ft M = 300.6 k.ft

B A

-1050

300.6 -98.33

36.67

15.62

98.33 36.67

15 ft 30 ft

3.33 4k/ft

2k/ft

hinge C

1050 Ay

2 3.33

A B

x

z

2

36.67

S

M


N. Poovarodom 33

Example 5.2 Draw SFD, BMD and qualitative deflection shape a) b)

c) d)

15kN/m

195.2 128

-24.4

5.625.6

12 ft 8 ft5 ft

BMD

SFD

24.425.6

30k 20k

Deflected shape

105110 110.2

-2.5

-52.5

57.552.557.5

2m2m 4m

50kN

3.83

75

-125

4 ft6 ft 10 ft

20

20

125k-ft 75k-ft20 k

203.6 226.8

176.8

-24.82

-2.32

17.68

1.5k-ft20k 50k-ft

15 ft10 ft

24.8217.68

10 ft


N. Poovarodom 34

e) f)

To determine reaction in (f), consider part CE

0ACAM

13602

1313158 yB

kNBy 94.255

0yF 94.0yA kN94.0

0EGGM

2

13131013608 yF

kNFy 13.203

0yF kNkNGy 13.1313.13

35

-65 -31.25

23.2

8.4 ft 10

-21

16.5

-12.5

29 31

35k-ft

2.5k/ft

10 ft 5 ft 15 ft

10k

13.13203.13

-120.94

0.94

5m 5m 5m 5m 8m8m

GFE D CBA

120kN 10kN/m15kN/m

-0.94

255.94

135

60

-60 -110

93.1313.13

300

- 487.5 - 425

60 60

120

By Ay

15

60

60

Gy Fy

10


N. Poovarodom 35

Example 5.3: Arrange the support of the beam shown such that the maximum moment in the beam is the least.

2

22

22

2

2

0214.0

207.0;4

4

42

1

2

1

wLM

LaL

aLa

aaLL

MM

aLL

wM

waM

2

2

1wa

2

2

1wa

M-

M+

a

Lw

2

a

Lw

2 wa

wa

2

wL

2

wL

L a

w

a


N. Poovarodom 36

Example 5.4: Draw SFD, BMD and Axial force diagram, sketch the qualitative deflected shape.

-7

132.25

120

120 3.5 ft

-23

7

20

23 7

120

120

7

20

20

7

2k/ft 2k/ft

C B

A

6 ft

6 ft

23

15 ft

7

20

20 k FBD

SFD BMD

N-Diagram Deflected shape

x

y

x

y

x

y


N. Poovarodom 37

Example 5.5: Draw SFD, BMD, Axial force diagram, sketch the qualitative deflected shape.

Reaction kNAx 60

;0AM

494.8108204408 yD

kNDy 7.84

kNAy 7.4

Ax

Ay

4m

4m

20 kN10 kN/m

40 kN

8 m

D

C

B

A

LBC=8.94 Dy

1.47

-66.813.15

60

4.47

8.9410

84.7

2020

84.7

84.7

66.8

(84.7)55.8

(20)

(4.7)

15.8

8.94 kN/m

240

(20)

60

40 20

4.7

4.7

4.7

4.7

240

60

60

56.26

84.7

SFD

240

240 4.47 kN/m

13.15

x

y


N. Poovarodom 38

-84.7

+55.8

+15.8

-4.7

250

240

240

BMD

N-Diagram

Deflected shape

x

y

x

y

x

y


N. Poovarodom 39


Reaction

kAx 10

;0AM

kA

kD

D

y

y

y

83.8

17.27

1515202515302.130

1.2k/ft

10

25k

15 k

C

B

15 ft

15 ft

8.83 30 ft

DA

27.17

20 ft

-30.52

3.36

3.63

10

15

27.17

0.36

1.138

1.08

225

15

15

27.17

225

27.17

27.17

1515

225 225

(15)

5.64 (27.17)

30.52

3.6317.02

(8.83)

(15)

200

1.08k/ft15 25

10

8.83

8.83

200k-ft

8.83

10

10

8.83

LBC=31.62’

43.18

/138.162.31

302.1

ftkw

SFD

0.36 k/ft

x

y

+

x

y


N. Poovarodom 40

-27.17

-5.64

-17.02

-8.83

206.1

200

-225 -225 200

BMD

Axial

Note: the angle at rigid connection remain unchanged.

x

y

y

x

y

x

y

x


N. Poovarodom 41

Bx

10

25k

D C

1.25

A

ftkM A .75.181525.1

23.75

G

ftkM G .25.356

25

1.25

23.75 18.75

23.75

10 35

23.75

356.25

356.25

23.75

23.75

35

35

356.25

18.75

18.75

1.25

1.25

10

10


Reaction 0BM kGy 35 kAy 10

0DM

kB

B

x

x

25.12

155.1151015

2

kFx 75.23

1.5k/ft

25k

15 ft

G

F

E D C

B

A

15 ft

15 ft 15 ft

Gy Ay

1.5k/ft

25k

F

E D C

B


N. Poovarodom 42

-35

-23.75

-10

356.25

-356.25

-356.25

52.1 18.75

-18.75

18.75

23.75 -12.5 6.67

-35

10

1.25

SFD

BMD

Axial

y

x y

x

y

x

y

x


N. Poovarodom 43

Chapter 6 Deflections of Beams: Geometric Methods

Cause of deformation: load, temperature change or support movement. Structures are usually designed so that their deflections under normal service conditions will not exceed the allowable values. 6.1 Differential equation for beam deflection Fig a: an initially straight elastic beam subjected to transverse load Elastic curve = the neutral surface of the beam in the deformed state. Fig b: a differential element dx

of the beam in its deformed position. Assume plane section remains plane and perpendicular to the neutral axis

The slope of elastic curve dx

dy

Assume is small

Then 02 , sin , 1cos d = change in slope over the differential

length dx Consider the deformation of a fiber ab located at a distance y from the neutral surface ''ba

yd

dyabbad

22 (Shortening)

The strain in ab is

R

y

Rd

yd

ds

d

(R = radius of curvature)

For E

, E = modulus of elasticity, = stress in ab

R

yE

From fig c, c maximum stress at extreme fiber, “c” from the neutral axis.

cc

y

For moment M = the summation of the moments about the neutral axis throughout cross sectional area.

A

ydAM for + (tension) , and - (compression)


N. Poovarodom 44

Ic

dAyc

M C

A

C 2

I

McC A dAyI 2 = the moment of inertia

I

My

Then EI

M

R

1

From calculus;

232

22

1

1

dxdy

dxyd

R (Redefine y to be the vertical deflection.)

For small deformation 02 dxdy

2

21

dx

yd

R

The differential equations for the deflection of beams are; “Bernoulli–Euler beam equation”

dx

dy

, y can be solved.

Note: for small dxdy / the arc of elastic curve of beam “ ds ” is;

dxdxdxdydydxds 222 1

6.2 Direct Integration Method

- Writing EI

M in term of x

- Integrating this equation for , y. - Determining the constants of integration from boundary conditions.

This method is convenient for beams with continuous EI

M equation.

Note For linearly elastic beams subjected to several loads; The combined effect of loads = the algebraic sum of effect due to each of the loads acting individually on the beam.

EI

M

dx

yd

2

2

EI

M

dx

d


N. Poovarodom 45

6.3 Moment-Area Method Moment-area theorem: Relations between the geometry of the elastic curve of

a beam to its EI

M diagram.

Consider a beam shown, we have

dxEI

Md

= infinitesimal area under EI

M

diagram Take integration

B

A

B

Adx

EI

Md

B

AABBA dx

EI

M ******

BA , slope of elastic curve at A and B, respectively with respect to the un-deformed axis of the beam

BA the angle between the tangents to the elastic curve at A and B

The first moment-area theorem: “The change in slope between the tangents to the elastic curve at any two points is

equal to the area under EI

M diagram between the two points, provided that the elastic curve

is continuous between the two points.” (There must be no discontinuity from internal hinge.)

* For positiveEI

M the angle from the tangent at the left point to the tangent

at the right point will be counterclockwise the positive change in slope.

For negativeEI

M vise versa

Consider the deviation "" d between the tangent at two ends of dx on a line perpendicular to the un-deformed axis from point B;

dxd

x distance from B to dx

dxxEI

Md

Take integration

dxxEI

Md

B

A

B

A

, Then dxx

EI

MB

ABA ****


N. Poovarodom 46

BA The tangential deviation of B from the tangent at A = The deflection of B in the direction perpendicular to the un-deformed axis from the tangential at A

dxxEI

MB

A The moment of the area of

EI

M diagram between A to B, about point B

The second moment-area theorem: “The tangential deviation in the direction perpendicular to the un-deformed axis of the beam of a point on the elastic curve from the tangent to the elastic curve at another

point is equal to the moment of area under the EI

Mdiagram between the two points about

the point at which the deviation is desired, provided that the elastic curve is continuous between the two points.”

For positive EI

M between A to B, then BA is positive or B is above the tangent from A

6.4 Bending Moment Diagram by Parts For a beam that is subjected to different types of loads (point load and distributed loads);

the EI

M diagram becomes irregular shape.

Solution: constructing a separate BMD or each load Simple geometric shape BMD The total effect is obtained by superposition. Approach 1) Apply each of the loads separately


N. Poovarodom 47

2) BMD by cantilever parts:

Step for analysis; - Selecting a point on the beam (usually

at support or end) - Assuming that point is fixed - Applying each of the loads and

reaction separately on this imaginary cantilever beam

- Constructing the corresponding BMD


N. Poovarodom 48

Chapter 7 Deflections of trusses, beams and frames:

Work-Energy method To analyze the deflections or displacements of statically determinate structures by

using basic principles of work and energy. This method is more general than other methods since it can be applied to various types of structures, such as trusses, beams and frames. However, only one deflection component at one point can be computed from one calculation.

7.1 Work

Work = force × displacement of its point of application in the direction of the force.

* Positive: When force and displacement have the same sense * Negative: When force and displacement have the opposite sense

Consider the work done by force P during the deformation of a structure under the action of a system of force. The magnitude of P may vary as its point of application displace from the original position to the final position.

The small work done dW from force P with infinitesimal displacement, d , is

dPdW

The total work done W of force P performs over the entire displacement is

dPW0


N. Poovarodom 49

Consider 3 cases; From figure b): For general force-displacement relationship, W is equal to the area under the force-displacement diagram. From figure c): For linearly elastic structure, the force varies linearly with displacement from zero to its final value .Thus

PW2

1

From figure d): (special case), the force remains constant at P while its point of application undergoes a displacement cause by some other actions, then PW Also, the expressions for the work of couples are dMdW

- General case

0

MdW

- Linear elastic system

MW2

1

- Constant M during rotation MW

7.2 Principle of Virtual Work 7.2.1) Principle of Virtual Displacements for Rigid Bodies “If a rigid body is in equilibrium under a system of forces and if it is subjected to any small virtual rigid body displacements, the work done by the external force is zero” (Virtual = imaginary, not real) a) Consider a simple beam ABC subjected to force P b) The free body diagram yx PP ,

and reaction yx AA , , yx CC ,

c) Assume the beam displaces with small virtual rigid body motion to position A’B’C’

vx = Virtual disp. in x direction

vy = Virtual disp. in y direction

v = Virtual rotation about A

( v for virtual)


N. Poovarodom 50

When the beam deforms, all forces acting on it perform work called “Virtual Work” The total virtual work done by external forces veW

vrvyvxve WWWW vxW = Virtual work done during translation in x vyW = Virtual work done during translation in y vrW = Virtual work done during rotation

vxxvxxxvxxvxxvx FPAPAW

vyyvyyyyvyyvyyvyyvy FCPACPAW

vAvyyvyvyvr MLCaPLCaPW

vAvyyvxxve MFFW Since for equilibrium 0 xF , 0 yF , 0 AM , then ** 0veW ** 7.2.2) Principle of Virtual Forces for Deformable Bodies “If a deformable structure is in equilibrium under a virtual system of forces (and couples) and if it is subjected to any small real deformation consistent with the support and continuity conditions of the structure, then the virtual external work done by the virtual external forces (and couples) acting through the real external displacement (and rotations) is equal to the virtual internal work done by the virtual internal forces (and couples) acting through the real internal displacements (and rotations).”

(Virtual ext. force) × (real ext. disp.) = (Virtual int. force) × (real int. disp.) To derive this principle, consider the two member truss shown a) Virtual external force vP b) FBD of joint C

Equilibrium of C ( vBCvAC FF , = Virtual int. force)

0 xF , 0coscos 21 vBCvACv FFP

0 yF , 0sinsin 21 vBCvAC FF

Assume that C displaces with a small real (Note: A and B are supports and not displace, then no work done at A and B)


N. Poovarodom 51

The total virtual work done vW is from forces at C

21 coscos vBCvACvv FFPW

or 21 coscos vBCvACvv FFPW

Since from 0 xF at C 0W , Then

vive WW The principle of virtual forces

*Any cause of real deformations (loads, temperature changes, …etc.) can be analyzed for their effects by this principle Limitation for small deformation and small rotation

Virtual System virtual ext. force virtual int. force = real ext. displacement real int. displacement

Real System * Note: Since the virtual forces are independent of the actions causing the real deformation and remain constant during the real deformation, the expressions of the external and internal virtual work are in the form of P (no factor of 2

1 )

To determine displacement by virtual force principle, a virtual force system is selected so that the desired displacement is an unknown in the virtual work equation. 7.3 Deflections of Trusses by the Virtual Work Method Consider a statically determinate truss to determine the vertical deflection at B due to external load 1P and 2P The internal axial forces can be obtained by the method of joints. For member under axial force F, the axial deformation , is

AE

FL (real internal disp.*)

)cos()cos( 21 vBCvACv FFP

Virtual Ext. Force

Real Ext. Disp.

Virtual Int. Force

Real Int. Disp.

+……


N. Poovarodom 52

To determine (real ext. disp.*), we select a virtual system with a unit load (virtual ext. force*) acting at B in the same direction as assumed )( . The internal force vF (virtual

int. force*) can be obtained by the method of joints. Now 1veW

And the total internal work done (summation of all members) )( vvi FW

For vive WW

Then )(1 vF

For the real internal deformation due to external load AE

FL

AE

FLFv1

Temperature Changes and Fabrication Errors For the member axial deformations are caused by temperature change T LT )( is the coefficient of thermal expansion of member Then )(1 TLFv

For the effect of fabrication errors, substitute changes in member length into the virtual work equation. Procedure for Analysis 1) Real System - External load: Obtain the real internal force F by method of joints. ( F is positive for tension ) - Temperature change: Consider the effect of increase of temperature as + T . - Fabrication error: Positive effects for too long member. 2) Virtual System - Remove all loads then apply a unit force at the joint to determine the deflection, in the direction of the deflection. - Compute all virtual internal force vF ( positive for tension)

3) can be obtained by applying the virtual work equation vF1

Positive solution of means has the same direction as the unit load applied in the virtual system 7.4 Deflections of Beams by the Virtual Work Method Consider a beam shown, to determine the vertical deflection at B ( )


N. Poovarodom 53

A virtual system with a unit load acting at B and in the direction of is selected as shown in fig (b). The virtual external work by virtual unit load × real deflection is, )(1 veW

To obtain the virtual internal work, consider a small element dx of the beam at distance x from the left support, as fig(b). From (b), beam with virtual load, is subjected to the deformation due to the real load, fig. (a), then the virtual internal moment vM acting on the element dx performs

virtual internal work with the real rotation d as in fig (c). ( vM is obtained from a virtual unit load )

)( dMdW vvi

Note: since vM remains constant during the real internal rotation d , no factor 21 for

work done. From the beam equation, the change of slope d over length dx can be written as

dxEI

Md

M = bending moment due to real loading causing d

dxEI

MMdW vvi

The total work done

dxEI

MMW

Lv

vi 0

Then dxEI

MMLv

0

)(1

To determine slope at point C, then we use a virtual system consisting of a unit couple acting at the point C as shown in fig (d).

Then dxEI

MMLv

0

)(1


N. Poovarodom 54

In a similar manner The internal work done by axial force

dxEA

FFW

Lv

Fvi 0

The internal work done by shear

dxkGA

VVW

Lv

Vvi 0

k = shape factor, G = shear modulus The internal work done by torsion

dxGJ

TTW

Lv

Tvi 0

J = The polar moment of inertia Then the generation equation for beam is TviVviFviMvive WWWWW

However, for most beams , the effect from bending moment is highest. Procedure for analysis (Bending effect only) 1. Real System Draw a diagram of the beam showing all the real loads acting on it. 2. Virtual System Remove all real loads from the beam and apply a unit action on it. - To determine , apply a unit force at the position and the direction of . - To determine , apply a unit couple at the position and direction of the desired .

3. If there is discontinuity in dxEI

MM v , divide the beam into segment for continuous

function in the integration. The point to divide beam must be at; - Position of point load - Position of couple moment - Position of intermediate support - Position where distributed load changes - Position where EI change 4. For each segment, determine the equation for M due to real loads in terms of x. The origin of x can be anywhere. The sign convention should follow sign convent for beam. 5. Determine the equation for vM due to virtual action using the same procedure as step 4.

6. Determine the deflection or slope of the beam by the virtual work equation. 7.5 Deflections of Frames by the Virtual Work Method The application on frames is similar to on beams. The analysis may involve more than one member. The computation can be done on each member separately and the total work done is obtained by the summation at all members. If consider bending and axial deformation

or

1

1 =

dxEI

MM

AE

FLF v

v


N. Poovarodom 55

7.6 Conservation of Energy and Strain Energy Energy of Structure: The capacity of structure for doing work. Strain Energy “U”: The energy that a structure has because of its deformation. The principle of conservation of energy “The work performed on an elastic structure in equilibrium by statically (gradually) applied external forces is equal to the work done by the internal forces, or the strain energy of the structure.” ie WW

or UWe

Strain Energy of Trusses Consider a truss subjected to P P increases from zero to its final value For elastic structure, the deformation also increases linearly with the load

PWe 21

To write U, consider an arbitrary member j (for example, member CD)

If F = axial force in member due to load P Then = the axial deformation of this member

AE

FL

The internal work or strain energy stored in member j, jU is

AE

LFFU j

2

2

1

2

1

The strain energy for the entire truss is

AE

LFU

2

2

Strain Energy of Beams Consider a beam shown Force P increases from 0 P The internal bending moment M on element dx also increases from zero to its final value. The internal work or the strain energy in the element dx is

MddU2

1

For dxEI

Md


N. Poovarodom 56

dxEI

MdU

2

2

1

Then dxEI

MU

L

0

2

2

* If EI

M is not a continuous function of x, the beam must be divided into segments

of continuous EI

M.

Strain Energy of Frames The portions of frames may be subjected to axial forces and bending moment, then ba UUU

For the frame with divided segments of constant AE

F,

AE

LFU a 2

2

For the frame with divided segments of constant EI

M

dxEI

MU b

2

2

The total strain energy

dxEI

M

AE

LFU

22

22

7.7 Castigliano ’s second Theorem “For linearly elastic structures, the partial derivative of the strain energy with respect to an applied force (or couple) is equal to the displacement (or rotation) of the force (or couple) along its line of action.”

iiP

U

or iiM

U

U = strain energy i = deflection of the point of application of iP , in the direction of iP .

i = rotation of the point of application of iM , in the direction of iM

To prove the theorem, consider a beam under force 321 ,, PPP

321 ,, PPP increase from zero causing

321 ,, which vary linearly with

the applied force.


N. Poovarodom 57

dP2

dU

U

P2

dP2

dU

The total strain energy stored in the beam from the external work done ( eW ) is;

332211 2

1

2

1

2

1 PPPWU e (a)

or 321 ,, PPPfU (b)

To determine the deflection, for example, 2 increase 2P by a small quantity 2dP . Then the

increase in strain energy due to 2dP is;

22

.dPP

UdU

(c)

The total strain energy TU is now given by;

22

dPP

UUdUUU T

(d)

Now, consider a new sequence of loading. First apply only 2dP gradually and then

apply 21 , PP and 3P gradually from zero.

For linear elastic system, the strain energy must be the same and the deflection at each point must be the same. Thus,

3322112222 2

1

2

1

2

1

2

1 PPPdPddPU T (e)

(* Since 2dP remains constant during the additional deflection 2 , no factor 21 for the

work of this term.)

neglect the second order of small quantity, 02

122 ddP

33221122 2

1

2

1

2

1 PPPdPU T (f)

Substituting (a) into (f) UdPU T 22 (g) From (d) = (g)

UdPdPP

UU

2222

2

2 P

U

***

Application to trusses

From AE

LFU

2

2

, iiP

U

AE

LF

P 2

2

P

FF

P

F

22

AE

FL

P

F


N. Poovarodom 58

Application to breams

L

dxEI

M

P 0

2

2 and

L

dxEI

M

M 0

2

2

or and Application to Frames

dxEI

M

P

M

AE

FL

P

F

dxEI

M

M

M

AE

FL

M

F

Procedure for analysis 1. 1.1) If there is an external force (or couple) acting on the given structure at the point and in the direction of the desired deflection (or rotation), then that force (or couple) may be used as P or M . Then, go to step 2 1.2) Otherwise, apply a fictitious load P or couple M correspondent to the desired unknown displacement or rotation. 2. Determine - the axial force F - the equations for bending moment )(xM in each member in terms of P and M 3. Determine

- P

F

- P

M

or - M

F

- M

M

4. 4.1) For a case of 1.1, substitute the numerical value of P or M into the expression for F and )(xM and derivatives. 4.2) For a case of 1.2, the numerical value is zero. 5. Determine , by the Castigliano’s second theorem

The positive results indicate the same direction as P or M

dxEI

M

P

ML

0

dxEI

M

M

ML

0


N. Poovarodom 59

7.8 Betti’s Law and Maxwell’s Law of reciprocal deflections Betti’s Law “For a linearly elastic structure, the virtual work done by a P system of forces and couples acting through the deformation caused by a Q system of forces and couples is equal to the virtual work done of the Q system acting through the deformation due to the P system.” Consider a beam under two different systems of forces , P and Q The virtual external work done ( veW ) due to

force P system acting through deformation from Q is

nQnQQve PPPW ...21 21

n

iQi i

P1

From vive WW , using the internal work done of beam

dxEI

MMP

LQP

Q

n

ii i

01

Next, consider the virtual external work done by Q through the deformation from P, Then equating vive WW , obtain

dxEI

MMQ

LPQ

m

jPj j

01

The order of term in integral does not affect, 2 equations are identical as;

m

jPj

n

iQi ji

QP11

* Betti’s Law *

Maxwell’s Law of reciprocal deflections “For a linearly elastic structure, the deflection at a point i due to a unit load applied at a point j is equal to the deflection at j due to a unit load at i ” *deflection can be rotation and load can be a couple. Consider a beam subjected to P and Q separately, consisting of the unit load at point i and j , respectively

ijf = the deflection at i due to the unit load at j

jif = the deflection at j due to the unit load at i

= flexibility coefficient (deflection per unit load) From Betti’s Law, jiij ff 11

* Maxwell’s Law *

jiij ff


N. Poovarodom 60

Example 7.1: Determine the vertical and horizontal deflection at B.

E = 200 GPa A = 1000 mm2 EA = 2 × 105 kN

Sol n Real System: analyze the reaction and

member axial forces Virtual System:

1) for BH : apply a unit load in the horizontal direction at B

2) for BV : apply a unit load in the vertical

direction at B

Calculation for B : form a table contain the real forces F and virtual force Fv with EA , L Member L(m) F (kN) Fv1 Fv2 Fv1FL Fv2FL

AB 4.24 -5.66 0.849 -0.566 -20.37 13.58 AC 5 24 0.4 0.4 48.0 48.0 BC 3.61 -43.27 -0.721 -0.721 112.62 112.62

∑ 140.25 174.2

From Wvi = )(AE

FLFv = )(

1FLF

EA v

And Wve = )(1 Wve = Wvi

BH = EA

25.140 =

5102

25.140

= 0.000701 m = 0.701 mm

BV = EA

2.174 = 0.000871 m = 0.871 mm

40kN

A C

B 20kN

3 m

3 m 2 m

36

5.66 43.27

24 4

20

40

20 F-Force

0.6

1

0.849 0.721

0.4 0.6

1

Fv1 - Force

0.6

0.566 0.721

0.4 0.4

1 Fv2 - Force


N. Poovarodom 61

Example 7.2: Obtain BH and BV , E= 29000 ksi

Soln Real system

F

Virtual 1 Virtual 2 Fv1 Fv2

Member L(in) A(in2) F (k) Fv1 (k/k) Fv1

A

FLFv2 (k/k) Fv2

A

FL

AB 240 4 33.33 0.889 1777.8 0 0 BC 240 4 33.33 0.444 887.9 -1 -1999.8 CD 240 4 33.33 0.444 887.9 -1 -1999.8 EF 240 4 -33.33 -0.889 1777.8 0 0 AE 300 4 -41.67 -1.110 3469.0 0 0 BF 300 3 0 0.556 0 0 0 DF 300 4 -41.67 -0.556 1737.6 0 0 BE 180 3 25 0.667 1000.5 0 0 CF 180 3 25 0 0 0 0

11538.5 -3999.6

BV = 29000

5.11538 = 0.398 in

BH = 29000

6.3999 = -0.138 in = 0.138 in

D C B A

(4in2)

(4) (3)

E F (4)

(3)

(4)

(3)

(4)

(4)

25k 25k

20 ft 20 ft 20 ft

15 ft

25

41.67 41.67 0 25 25

33.33 33.33 33.33

25

33.33

25 25

0.556 0.556 0.667 0

0.889 0.444 0.444

2/3 1/3

0.889

1

1.11 0

0

0

0

0

1

0 0

1 1


N. Poovarodom 62

Example 7.3: Obtain EH E = 29000 ksi A = 6 in2

Soln m=6, j = 5, r = 4 m + r = 2j Statically determinate

5

10tan 1

1

AC sin 1 + 6 =0 AC = -6.71 CD = -7.0

Member L (ft) F (k) Fv (k/k) Fv(FL) (k-ft) AD 14.14 -1.41 0.707 -14.1 BD 14.14 -7.07 -0.707 70.68 DE 5 -3 1 -15 AC 0 CD 0 BE 0

58.41

in 0029.0629000

1258.41EH

ED C

BA

6k 6k 6k

4k

10 ft

5 ft 5 ft 5 ft 5 ft

37

6.71 1.41 7.07

6.71

8

11

4

6 6 6 4

7

F-Force

1

0.707

1

0.5

0.5 0.5

0.5

0.707

Fv-Force

CD

AC

1

6 4

C


N. Poovarodom 63

50 50

100

50 50

100

Example 7.4: Determine the smallest A required for the structure shown so that the horizontal displacement at D (DH) does not exceed 10 mm, E = 70 GPa a)

Member L (m) F (kN) Fv Fv(FL) AB 3 0 0 0 CD 3 -100 0 0 AC 4 -50 0 0 BD 4 -183.3 -1.33 975.32 AD 5 166.7 1.67 1391.67

2367

1 (DH) = kN.m 2367

EA

For DH = 0.01 m = A61070

2367

A = 0.00338 m2 = 3380 mm2

b) c)

Arequired = 3050 2mm Arequired = 2621 2mm

183.3

50 kN

D C

B A

83.3

166.7

50kN 100 kN

50

100 kN

4 m 183.3

10

3 m

F-Force

1.33 1.33

1.33 1.67

1

1

Fv-Force


N. Poovarodom 64

Example 7.5: Determine the vertical displacement at G ( vG ) due to temperature increase

of 65 F in AB, BC, CD and DE and a temperature drop of 20 F in FG and GH

= 6.5x 610 F

1

vF - Force

Member )( FT vF (k/k) vF )( T (k. F)

AB 65 -0.5 -32.5 BC 65 -1 -65 CD 65 -1 -65 and L are DE 65 -0.5 -32.5 identical FG -20 0.5 -10 GH -20 0.5 -10

215

TFLG vv

= )215(12105.6 6 = -0.0168 ft

= in 202.0 Example 7.6: Obtain vG if member BC and CG are made too short of 0.5 in.

member in Fv vF

BC -0.5 -1 0.5 CG -0.5 0 0

5.0

G = vF

= in 5.0

-20 -20

+65 +65 +65 +65

12 ft

C D E

F G H

B A

Real System

12 ft 12 ft 12 ft

12 ft

1

0.5

0.707 0.707

0.707 0.707

0.5 0.5

0.5 0.5

0.5 1 1 0.5

0.5


N. Poovarodom 65

x

1

Example 7.7: Obtain and at A

EI = constant.

0<x<L 2

2wxM

0<x<L xM v 11

0<x<L 12 vM

for A

1 ( A ) = L

v dxEI

MM

0

1 = dxEI

wxxL

0

2

)2

)((

= L

EI

wx

0

4

8 =

EI

wL

8

4

For Ä

1 )( A = dxwx

EI

L

o

)2

)(1(1 2

= EI

wL

6

3

= EI

wL

6

3

w

L

A B

x

x

1

Real system

Virtual System 1 (for A )

Virtual System 2 (for A )


N. Poovarodom 66

Example 7.8: Obtain C

E=250 GPa I= 4610600 mm Divide beam into Segment at location of point load (or point where EI change)

Segment X coordinate

EI M (kN-m) vM

(kN.m/kN)Origin Limit

AB A 0-5 EI 175x 0.25x BC A 5-15 2EI 175x – 200(x-5) 0.25x CD D 0-5 EI 125x 0.75x

c = 5

0

15

5

)5(20017525.02

1)175)(25.0(

1dxxxx

EIdxxx

EI 5

0

)125)(75.0(1

dxxxEI

= 1

14844

E 3mkN

= 310150

14844

x = m 099.0

5 m 10m 5 m B C D A

100kN 200kN

I I 2I

x x

x B C D A

125 175

Real 200 100

B C D A x

x x

0.75 0.25

1 Virtual

EI = 43

6

3

9

10

10600

10

10250

= 310150 kN.m2


N. Poovarodom 67


E= 29000 ksi I= 3500 in4

Segment x-coordinate M (k-ft)

vM

(k-ft/k) origin limit

AB A 0-15 12.6x-x 2 -0.4x CB C 0-6 -x 2 -x

1 )( C = EI

1

15

0

6

0

22 ))(()6.12)(4.0( dxxxdxxxx

= EI

5.283 k-ft3

= 350029000

125.283 3

= -0.00483 in = in 00483.0

2k/ft

29.4 12.6

6 ft 15 ft B A

C

1

1.4 k 0.4 k

x x

Virtual system


N. Poovarodom 68

Example 7.10: Obtain D and D E = 30,000 ksi I = 3,000 in4


M (k-ft) 1vM

ftk

ftk 2vM (k-ft/k) Origin limit

DC D 0-8 -35x -1 -x

CB D 8-16 -35x+70(x-8) -1+8

1(x-8) x-16

AB A 0-16 240+5x-1.25x 2

2- 8x 16-x

D =

16

8

8

0

)56035)(28

()35)(1(1

dxxx

xdxEI

dxxx

x16

0

2 )25.15240)(8

2(4

3

= EI

4427 k-ft2 =

300030000

124427 2

x

= 0.00708 rad

(follow sense of unit couple 1 )

D =

16

8

8

0

)56035)(16()35)((1

dxxxdxxxEI

16

0

2 )25.15240)(16(4

3dxxxx

= EI

32427 k-ft3

= 300030000

1232427 3

= 0.623 in

2

2

D 70

x x x

8 ft 8 ft 16 ft

240

5

C B A

35 k 2.5 k/ft hinge

I3

4 I

1

8

1 8

1

1vM 21

16 1

2vM


N. Poovarodom 69

Example 7.11: Obtain D , consider only bending effect. E = 2000 ksi

I = 10000 in 4

Real M DE = 29x – 1.5x 2

Virtual M VDE = 16

x

D = dxxxx

EI)5.129(

162

1 216

0

= EI

3.469 k-ft2 =

)10000(2000

123.469 2

= 0.00338 rad

8 ft

8 ft

16 ft 8 ft

x

I 2I E

D C

B

A

30k

49

40

40k 29

3 k/ft

2I

M-system

16

1

16

1

1

vM only in member DE

Mv-system


N. Poovarodom 70

Example 7.12: Obtain B , consider only bending effect. E = 200 GPa, I = 500 x 10 6 mm 4

Segment X – coordinate M

(kN.m) vM

(kN.m/kN.m) Origin Limit AD A 0-2.5 91.875x -0.075x DB A 2.5-5 1.875x + 225 -0.075x CB C 0-5 96.875x-10x2 0.125x

B =

5.2

0

)875.91)(075.0(1

dxxxEI

5

5.2

)225875.1()075.0( dxxx +

dxxxx

5

0

2 )10875.96(125.0

=EI

01.110 kN.m2

= )500(200

01.110

= 0.00110 rad

5m

C B

A

D 96.875

xx

4m

1.5 1.5

(91.875)

(90) (120)

150

20kN/m

M-system x

153.125 (122.5)

D

1

0.125

0.125

(0.1)

(0.075) Mv-system


N. Poovarodom 71

Example 7.13: Determine E (in vertical direction)

E = 200 GPa, I = 350x10 6 mm4

A = 6.5 x 104 mm2

(Approximately = square section)

5/2

1/2

5/12

5/12

1/2 1/2

5/12

5/12

1/2

5/2 5/2 5/2 5/12

5/12

1

1/2 1/2

5/12

1/2 1/2

5/12

1 kN

GFED

C

BA

Virtual System

(I,A)

(2I) (2I,2A) (2I,2A) (2I,2A)

(I,A)

H

G FE (hinge)

D C

BA

5m 5m 2m 3m

120.75

39.375

104.25

25.625

65 kN

3m

3m

15kN/m

104.25

30

30 39.375

120.75

236.25

x

39.375

x x

39.375

59.25 90.75

39.375

266.25 108.75 41.25

x

x

25.625

65

104.25

45

67.5

120.75

39.375


N. Poovarodom 72


M (kN.m) Mv (kN.m/kN) origin Limits(m)

AH A 0-3 25.625x -5x/12 HD A 3-6 25.625x-65(x-3) -5x/12 BF B 0-6 39.375x 5x/12 DE D 0-5 59.25x-108.75-7.5x 2 2/52/ x FE F 0-5 90.75x – 266.25-7.5x 2 2/52/ x

1. E = EI

1

3

0

6

3

)19565625.25)(12

5()625.25)(

12

5( dxxx

xdxx

x

dxxxx

dxxx

)5.775.10825.59)(2

5

2(

2

1)375.39(

12

5 25

0

6

0

+

5

0

2 )5.725.26675.90)(2

5

2(

2

1dxxx

x

= EI

81.1607 kN- m 3 = m 0230.0 = mm 0.23

Axial deformation effect Segment F

vF

AD -104.25 -1/2 DF -39.375 -5/12 BF -120.75 -1/2

Effect of axial force F

= EA

LFF v

=

)10)(12

5)(375.39(

2

1)6)(

2

1)(75.120(6)

2

1)(25.104(

1

EAEA

= EA

757 kN.m

= m 0000582.0 (very small) Total E = 0.0230 + 0.0000582 m.


N. Poovarodom 73

Example 7.14: Determine C , consider effect of bending and axial deformation.

E = 70 GPa I = 1030 10 6 mm 4 A = 1.925 10 4 mm 2 EI = 72100 kN.m 2 EA = 1.3475 10 6 kN (WF 588 x 300)


M (kN.m) Mv

(kN.m/kN) F (kN)

Fv (kN/kN) Origin Limit (m)

AB A 0-5 76x-530 -4 -48 -1 CB C 0-5 2/12 2x x)5/4( 0 -3/5

C1 = 5)1)(48(1

)2

12)(5

4()53076)(4(

1 5

0

25

0

EA

dxx

xdxxEI

= EAEI

240 kN.m

7550 3 kN.m = 0.10471 + 1.781 x 10 4 m

= 0.1049 m

1

C

A

B

3 m

4 m

530

48

76

40kN

x

x

12kN/m

5 m

1 (4/5) 1

4

1

4 1

4

11

x

4

60 150

150

36

48

40

530

48

76

xReal Virtual

(3/5)


N. Poovarodom 74

Example 7.15: Determine the maximum deflection of the beam shown.

Segment Origin Limit M Mv

AC A 0-z 121 Mx

L

MM

xL

z

1

BC C 0 to (L-z) 221 Mx

L

MM

xL

z

z = EI

1

dxMxL

MMx

L

zdxMx

L

MMx

L

z Lz

o 0

221

121 ))(())(()())(1(

= EI

1

L6

1 )2)()(( 2121 zMzMLMLMzzL

For dz

d =

31LM

+ 6

2LM + zM 1 -

L

zM

2

21 -

L

zM

2

22 = 0

z = )(3 21 MM

L

2/12

2212

11 ))(3(3 MMMMM

and z = 2/12221

211

21

))(3(3)(3

MMMMMMM

L

For 1M = -20 k-ft, 2M = -80 k-ft, L = 10 ft. z = -2.1633 and 6.163 ft. Only z = 6.163 ft. is a possible solution

Then z = EI

5.400

L

z

L

z1

xx

BC

A

z

L L

MM 21 L

MM 21

M1 M2

Real Virtual


N. Poovarodom 75

Example 7.16: Determine BH and BV .

E = 200 GPa A = 1000 mm 2

Soln P1 and P2 act in the direction of BH and BV ,

then use them for variable of the analysis

Member L(m) F 1P

F

2P

F

P1 = 20, P 2 = 40

FLP

F

1

FLP

F

2

AB 4.24 0.849P1 -

0.566P 2 0.849 -0.566 -20.37 13.58

AC 5 0.4(P1 +P 2 ) 0.4 0.4 48 48

BC 3.61 -0.721(P1 +P 2 ) -0.721 -0.721 112.6 112.6

140.23 174.18

BH = EA

23.140

= 0.000701 m = mm 701.0

BV = EA

18.174

= 0.000871 m = mm 871.0

20kN

40kN

3m

2 m 3 m

C

B

A

P1 0.6P1 0.6P1

0.4P1

0.721P1 0.849P1

P1(=20kN)

0.721P0.566P2

0.4P2

0.6P2 0.4P2

P2(=40kN)


N. Poovarodom 76

Example 7.17: Obtain BH and BV

E=29000 ksi

Member L(in) A(in 2 ) F 1P

F

2P

F

P1 = 0, P 2 = 25k

1P

F

)(A

FL

2P

F

)(A

FL

AB 240 4 11.11 + 0.889 P 2 0 0.889 0 1778.1

BC 240 4 22.22 - P1 +

0.444 P 2

-1 0.444 -1999.2 887.6

CD 240 4 22.22 - P1 +

0.444 P 2

-1 0.444 -1999.2 887.6

EF 240 4 -11.11 – 0.889P 2 0 -0.889 0 1778.1

AE 300 4 -13.89 – 1.11 P 2 0 -1.11 0 3466.5

BF 300 3 -13.89 + 0.556P 2 0 0.556 0 0

DF 300 4 -27.78 – 0.556P 2 0 -0.556 0 1738.1

BE 180 3 8.33 + 0.667P 2 0 0.667 0 1000.7

CF 180 3 25 0 0 0 0 = -3998.4 11536.7

367.16 2P

23

233.8 P

27.78+0.566P2 13.89- 0.556P2 25

8.33 + 0.667P2

13.89+1.11P2

11.11+0.889P2

11.11+0.889P2 22.22-P1+0.444P2 22.22-P1+0.444P2 P1

25k P2(=25k)

P1(=0)

D C B A

(4in2)

(4) (3)

E F (4)

(3)

(4)

(3)

(4)

(4)

25k 25k

20 ft 20 ft 20 ft

15 ft


N. Poovarodom 77

BH = 29000

4.3998 = -0.138 in = in 138.0

BV = 29000

7.11536 = in 398.0

Example 7.18: Obtain EH

E = 29000 ksi A = 6 in2

Member L (in) F (k) P

F

)(

0

FLP

FP

AC 134.16 -6.71 0 0 BE 134.16 -6.71 0 0 AD 169.71 -1.41 + 0.707P 0.707 -169.18 BD 169.71 -7.07 – 0.707P -0.707 848.29 CD 60 -7 0 0 DE 60 -3+P 1 -180

499.11

EH = EA

11.499 = in 00287.0

10 ft

5 ft

24

P

A B

E D C

P(=0) 3-P 7

6.71 1.41- 0.707P

7.07 + 0.707P

6.71

28

P

211

P

27

P

6k 6k 6k

4k

E

5 ft 5 ft 5 ft


N. Poovarodom 78

x

2k/ft

29.4+1.4P 12.6-0.4P

6 ft 15 ft B A

C

x

P (=0)


E = 29000 ksi I = 4000 in 4

Segment x-coordinate

M P

M

Origin Limits

AB A 0-9 (37.5 + 2/P )x 0.5x

BC A 9-18 -12.5x + 450 +

)2/(P x 0.5x

DC D 0-18 (12.5 + 2/P )x 0.5x Substitute P=0;

C =

18

0

9

0

18

9

)5.12)(5.0()4505.12)(5.0()5.37)(5.0(1

dxxxdxxxdxxxEI

= 3k.ft 33412

EI = in 498.0


E = 29000 ksi I = 3500 in 4

Segment x – coordinate

M P

M

Origin Limits

AB A 0-15 (12.6-0.4P) x –x 2 -0.4x CD C 0-6 -Px -x 2 -x

Substitute P=0;

C =

15

0

6

0

22 ))(()6.12)(4.0(1

dxxxdxxxxEI

= EI

5.283 k.ft 3

= -0.00483 in = in 00483.0

25.12

P

xx x

18 ft9 ft 9 ft

P(=0) 50k

DC

BA

25.37

P


N. Poovarodom 79

Example 7.21: Obtain D and D E = 30,000 ksi I BC = I CD = I

I AB = (4/3)I

Segment x– coordinate

M M

M

P

M

Origin Limits

DC D 0-8 -Px - -1 -x

CB D 8-16 -Px - +(2P+8

)(x-8) 2

8

x x-16

AB A 0-16 225.1)8

40()216320( xx

2-8

x 16-x

Substitute = 0 and P = 35 k

D =

8

0

16

8

16

0

2 )25.15240)(8

2(4

3)56035)(2

8()35)(1(

1dxxx

xdxx

xdxx

EI

= EI

7.4426 k.ft 2 = 0.00708 rad

D =

8

0

16

8

)56035)(16()35)((1

dxxxdxxxEI

16

0

2 )25.15240)(16(4

3dxxxx

= EI

7.32426 k.ft 3 = in 622.0

2P+8

x x x

8 ft 8 ft

320-16P-2

40-P-8

C B A

35 k 2.5 k/ft hinge

I=4000in4 I=3000in4

(P=35)

D

=0

16 ft


N. Poovarodom 80

Example 7.22: Obtain D , consider only effect of bending deformation. E = 10,000 ksi I = 3,000 in 4

Segment X – coordinate

M M

M

Origin Limits

AB A 0-20 x50)700( -1

BC B 0-15 300 -1

DC D 0-10 x30 1

Substitute = 0

D =

20

0

15

0

10

0

)30)(1()300)(1()50700)(1(2

11dxxdxdxx

EI

= EI

4000 k.ft 2

= -0.0192 rad = 0.0192 rad

10 ft

10 ft

15 ft

=0

50

20k

(I)

D

A

B C

30k

(I)

700+

(2I)

300-

30

30

x

30 30 300-300-

x

50

50 300-

700+

x


N. Poovarodom 81

Example 7.23: Obtain CH , consider only effect of bending deformation.

E = 10,000 ksi I = 1500 in 4

Substitute P=0

CH = 12

0

2 )5.118(5.11

dxxxxEI

= EI

3888 k-ft3

= in 448.0

Segment X – coordinate

M P

M

Origin Limits

AB A 0-18 Px x CB C 0-12 (18+1.5P)x-1.5x 2 1.5x

18 ft

12 ft

12 ft 18-1.5P

18+1.5P

(P=0)

P

x

x

D

A

B C

3k/ft


N. Poovarodom 82

Example 7.24: Determine E in vertical direction. E = 200 GPa I = 350 x 10 6 mm 4 A = 6.5 x 10 4 mm 2

Segment X-Coordinate

M P

M

F P

F

Origin Limits

AH A 0-3 25.625x - Px12

5 x

12

5

-104.25-2

2

1

HD A 3-6 25.625x – 65(x-3) - Px12

5 x

12

5

BF B 0-6 39.375x + Px12

5 x

12

5 -120.75-

2

P

2

1

DE D 0-5 59.25x – 108.75-7.5x 2

+ Px

)2

5

2( 2

5

2

x

-39.375-12

5P

12

5

FE F 0-5 90.75x-266.25-7.5x 2 + Px

)2

5

2(

2

5

2

x

25.625-12

5P

P=0

(I,A)

(2I,2A) (2I,2A)

(I,A)

G FE

(hinge) D C

BA

5m 5m 2m 3m

120.75+0.5P

39.375+12

5P

104.25+0.5P

65k

3m

3m

15kN/m

(2I,2A)

H

120.75+0.5P 104.25+0.5P

30

30

39.375+ P12

5

120.75+0.5P

236.25+ P2

5

x

39.375+ P12

5

x x

39.375+ P12

5

59.25+2

P

90.75+2

P

39.375+ P12

5

266.25+ P2

5

108.75+ P2

5

41.25+ P2

5

x

x

25.625 -5P/12

65

39.375+ P12

5

104.25+0.5P 45

67.5 P=0


N. Poovarodom 83

Substitute P=0 and calculate

E =

AE

FL

P

F+

dxEI

M

P

M

= )10)(12

5)(375.39(

2

1)6)(

2

1)(75.120(6)

2

1)(25.104(

1

EAEA

+

3

0

6

3

)19565625.25)(12

5()625.25)(

12

5(

1dxxx

xdxx

x

EI

+ 6

0

5

0

2 )5.775.10825.59)(2

5

2(

2

1)375.39)(

12

5( dxxx

xdxx

x

+

5

0

2 )5.725.26675.90)(2

5

2( dxxxx

= EA

757 +

EI

81.1607

= m 0230.00000582.0


N. Poovarodom 84

Chapter 8 Influence Lines

The analysis of structures for variable loads consists of 2 steps. 1.) Determining the position of the load at which the response function becomes maximum. 2.) Computing the maximum value of the response function. “An influence line (I.L.) is a graph of a response function of a structure (e.g. reaction, shear or bending moment) as a function of the position of a downward unit load moving across the structure.” 8.1 Influence line for beams and frames by equilibrium method Consider a simple beam subjected to a unit moving load, determine the influence line of the reaction at A and C and shear and moment at B. I.L. for reaction equilibrium

0 cM ;

L

x

L

xLAy

1

0 yF ; L

xC y

Figure b) shows I.L. of yA .

It shows how the movement of a unit load across the length of the beam influence the magnitude of yA .

The ordinate of the I.L. at any position x is equal to the magnitude of yA due to a unit load at x.

Figure c) shows I.L of yC .

*at any x, 0.1 yy CA

I.L. for shear and bending moment at B (Follow beam sign convention)

When the unit load is located to the left of B

L

xCS yB , ax 0

aLL

xaLCM yB , ax 0

When the unit load is located to the right of B

L

xAS yB 1 , Lxa

aL

xaAM yB

1 , Lxa


N. Poovarodom 85

Note: for statically determinate structures, I.L. is straight line. 8.2 Müller–Breslau’s principle and qualitative Influence Lines. “The I.L. for a force (or moment) response function is given by the deflected shape of the released structure obtained by removing the restraint corresponding to the response function from the original structure and by giving the released structure a unit displacement (or rotation) at the location and in the direction of the response function, so that only the response function and the unit load perform external work.” (This principle is not valid for I.L. of deflection) Consider the simple beam of previous example again I.L. for reaction For yA ( Fig b)

* remove the restraint corresponding to yA

Then support A is free in vertical direction. * assume a virtual unit displacement at A in vertical direction ( 1 , positive for upward). Principle of virtual displacement Virtual Work = Real force × Virtual displacement = 0 011 yAW yve

yAy

L

x 1

Figure c) shows the same procedure for yC


N. Poovarodom 86

I.L. for BS

- remove the restraint of BS By cutting the beam at B - Displace B vertically 211

1 and 2 make 2 segments parallel

( no work done from BM )

0121 ySSW BBe

yS B

From Fig .L

a

L

a 1, 21

I.L. for BM

- remove the restraint of BM by inserting a hinge at B - rotate the portions AB and AC - introduce virtual rotation 1 by rotating AB by 1 , AC by 2

121

0121 yMMW BBve

yM B

From figure (e) L

a

L

a 21 ,1

L

aa 1

8.3 Influence line for Deflections = The variation of a deflection of a structure as a unit load moves across the structure. By placing a unit load successively at arbitrary points, then determine an expression for the vertical deflection, and plot the expression. More efficient method by “ Maxwell’s law of reciprocal deflection” From a; BXf = at B when the load at x

BXf = the ordinate at x of the I.L. for B

From b; XBf = at x when the load at B


N. Poovarodom 87

From Maxwell’s law BXXB ff

XBf = the ordinate at x of the I.L. for B “ The deflected shape (elastic curve) of a structure due to a unit load applied at a point represents the I.L. for deflection at the point where the unit load is applied.” ( XBf can be obtained easily)


N. Poovarodom 88

Ex. 8.1 Draw Influence Line for Ay , Cy , SB and MB

a)

0CM 91 xAy

0AM 9xC y

Load is located at the left of B;

Consider the right FBD; SB = -Cy MB = Cy(6) Load is located at the right of B;

Consider the left FBD;

SB = Ay MB = Ay(3)

b)

0CM 81 xAy

0AM 8xC y

Load is located at the left of B;

Consider the right FBD; SB = -Cy MB = Cy(4) Load is located at the right of B;

Consider the left FBD; SB = +Ay MB = Ay(4)

-1.0

-0.25

-0.25

1.0 1.25

C

2m

2

0.5

-0.5

0.5

1/2 1.0

MB

4m 4m

B DA

SB

Cy

Ay

x

Ay Cy SB

MB 1

Ay Cy SB

MB 1

Ay Cy SB

MB 1

2

2/3

-1/3

1/3

2/3

1.0

1.0

MB

6m 3m

B CA

SB

Cy

Ay

x

Ay Cy SB

MB 1


N. Poovarodom 89

Ex. 8.2 Draw Influence Line for SC, MC and shear just to the left, SD,L and just to the right SD,R of support D

123

4 xBy ,

123

1 xDy

Load at the left of C SC = -Dy

MC = Dy(8)

Load at the right of C SC = By MC = By(4) …………………………….. SD,L Load at the left of D

SD,L = -Dy Load at the right of D SD,L = By

…………………… SD,R

Load at the left of D SD,R = 0

Load at the right of D SD,R = 1

+1.0

-0.5-1.0

0.33

-2

2.67

-2.67

-0.5

0.67

-0.33

0.33

1.0

0.33

-0.33

0.671.0

1.33

1.5

-0.5

4m

E

6m8m 4m

C DA

By

Dy

SC

MC

SD,L

SD,R

B x

D


N. Poovarodom 90

Ex. 8.3 Draw Influence Line for Ay, Dy, MD, SB and MB

120 x

0CM

121

xAy ,

12

xDy , )6(yD DM

1812 x

0yA 1yD

)18(1 xM D Load at the left of B

DyB

yB

MDM

DS

)12(

Load at the right of B

)6(yB

yB

AM

AS

3

-0.5

0.5

-6 -3

0.5 1.0

0.5

hinge D

6m

1.0

6m 6m

B A

Ay

Dy

MD

SB

MB

C

MD

Dy Ay

C

C

MD

A Dy


N. Poovarodom 91

Ex. 8.4 Draw Influence Line for Ay , By , SE , ME

0BM , 1010

13 xAy

0AM , 10

3

xBy

Load at the left of E SE = - By ME = By(4) Load at the right of E SE = Ay ME = Ay(6)

-3.0

2.4

-1.2

-0.5 -0.6

0.4 0.3

-0.3

0.6 1

1.5

-0.5

0.4 1 1.3

6m 4m 5m 3m

7m 5m

E F G D C

B

A

Ay

By

SE

ME


N. Poovarodom 92

Ex. 8.5 Draw Influence Line for Ay, MA, Ey, Gy, SC, MC, SF, and MF

+ MA

10’ 10’ 15’ 15’ 20’ 20’

G F E D C hinge hinge

B A

1

1

20

5/3

1

-2/3 1/2

-1/2

5

2/3 1/2

-1/2

7.5

-10

Ey

Gy

SC

MC

SF

MF

Ay

Ey

( =1, ccw)

MA

Ay

Released structure

1

1

1

Gy

SC 1

=1

1

=1


N. Poovarodom 93

Ex. 8.6 Draw Influence Line for Ay, MA, Fy, MF, SB, MB, SE, and ME

-3

-1

-5

1

1 -6

10

1

C D E F B A

5 m 6 m 3 m 3 m 5 m

Ay

MA

MF

AF

SB

MB

SE

ME


N. Poovarodom 94

Ex. 8.7 Draw Influence Line for MA, Ay, By, and SC

-1

1

1

10

-10

10 ft 10 ft 10 ft 10 ft

16 ft

F

E D C

B

A

MA

Ay

By

SC

MA

F

E D C

B

A


N. Poovarodom 95

Chapter 9 Method of Consistent Deformation: Force Method

9.1 Introduction

Force method or Flexibility method or method of consistent deformation involves; 1) Remove enough restraints from indeterminate structure to render it statically determinate (and stable). The system is called as primary structure. The removed restraints are called redundant restraints 2) Obtain the deformations, deflections of primary structure due to external loads. 3) Apply the redundants as unknown loads on primary structure, obtain deformations in terms of redundants. 4) Solve for redundants by the compatibility equations: “The deformations of primary structure due to the combined effect of the redundants and the given external loadings The deformations of the original indeterminate structure” 9.2 Structure with a single degree of indeterminacy Consider the propped cantilever beam subjected to a concentrated load P ( 1SI ) To analyze this beam one additional equation is required find one compatibility equation .


N. Poovarodom 96

Fig. a ) Select one reaction to be redundant, to establish the compatibility equation. For example, select YC , the compatibility condition is

0C

Fig b) Remove the roller C from the indeterminate beam. Then the beam become determinate and stable; “primary beam”

* Reaction YC is now considered as the unknown action on the primary beam. * The deflection at C = The deflection at C + The deflection of primary beam due to external load at C due to YC

CCCOC

Assume positive if it has the same direction as redundant YC (+ upward). Since YC is unknown, to determine CC by applying unit value of YC , then

multiplying the deflection by the unknown magnitude. YCCCC Cf

CCf the deflection at C of the primary beam due to the unit

value of redundant YC = flexibility coefficient Then the compatibility equation YCCCOC Cf

CC

COY f

C

Using the beam deflection formulars

EI

PLLL

EI

P L

CO 48

5

23

6

)( 322

5123000048

)1220(325 3

-3 in.

512300003

)1220(

3

33

EI

LfCC = 0.3 in./k

3.0

)3(

CC

COY f

C 10 k

or

P

EI

L

EI

PLCY 16

5

348

5 33

10 k

Then calculate the remaining reaction;

,0XF ; 0XA

,0YF ; 22YA k

,0 AM ; 120AM ftk

Bending moment diagram (BMD) can be obtained by superimposing the BMD of the primary beam due to external load only and due to YC .

+


N. Poovarodom 97

Moment as the Redundant In force method, the redundants which render statically determinate and stable can be arbitrary selected. a) Consider the previous beam again. b) Remove the rotation restraint AM fixed support hinge * Compatibility condition “Slope at A due to external forces and redundants must be zero.”

AO is the slope at A due to P

c) AAf is the slope at A due to a

unit value of AM

AAAAA Mf

* assume all , AM Compatibility condition 0 AAAAO Mf

From beam-deflection formular,

EI

PLAO 16

2

opposite to AM direction

EI

Lf AA 3

12016

33

16

2

PL

L

EI

EI

PL

fM

AA

AOA

k-ft.

Then all other remaining redundants can be obtained. Summary of the Analysis Procedures ( for single degree of external indeterminacy ) 1) - Select one of the support reactions as the redundant. - The structure after removing this redundant must be determinate and stable. - The sense of the redundant can be arbitrarily assumed. - Remove the restraint corresponding to the redundant primary structure 2) Draw a diagram of the primary structure

a) with external loads only. b) with only the unit value of redundant applied in the positive direction of redundant.

+


N. Poovarodom 98

3) Write the compatibility equation in terms of the deflection of primary structure in 2a) and 2b).

4) Compute the deflection of the primary structure in 2a) and 2b). 5) Solve the compatibility equation for the unknown redundant. 6) Determine the remaining support reactions and draw SFD and BMD. 9.3 Internal Forces and Moments as redundants Choices of redundant are arbitrary, select the redundant in such the way that the analysis is simple. Consider a beam with 1SI , select BM as the redundant .

a) the slope of the elastic curve is continuous at B

b) remove restraint of rotation at B, point B becomes hinge. Then, a discontinuity at B is the change of slope between left and right of B due to external load

BRBLBOrel

c) Apply a unit value of BM by a pair of equal but opposite couples flexibility coefficient BBrelf

BBrelf = the change of slope between the tangents to the elastic curve at left and

right of B due to a unit value of BM .

= BBRBBL ff Now the continuous 2 span beam becomes 2 simple support beams and the elastic deflection can be computed easily.

EIEIEIBL

420)1420(

206

1410)620(

206

610 2222

EIEIBR

3.533)1030(

306

1012 22

EIEIBOrel

3.9533.533420

10 10

6 8 6

BL

12

20 10 BR


N. Poovarodom 99

Flexibility coefficient;

EIf BBL 3

20

EI

f BBR

10

EIEI

f BBrel

67.16)10

3

20(

1

Then solving the compatibility equation, 0 BBBrelBOrel Mf

067.163.953

BMEIEI

BM = -57.19 k-ft (opposite to assumption) Then the other support reactions and SFD and BMD can be determined. Internally indeterminate Structures If structure is internally indeterminate but externally determinate then the redundants must be internal forces. Consider a truss with internal indeterminacy )]32([)3( jmrSI ( jrm 2)( ) 110)]342(6[)33( Select the axial force in AD to be the redundant ( ADF ).

The restraint corresponding to ADF is then removed from the truss by cutting member AD as in Fig (b). There exists a gap ADO in member AD.

Then the redundant force ADF must close the gap.

To evaluate the effect of ADF and its magnitude, apply the equal and opposite unit axial loads to member AD. ADF causes the ends of member AD to overlap by ADADf ,

The compatibility equation; 0, ADADADADO Ff

1 1


N. Poovarodom 100

The axial deformation ADO and ADADf , can be determined by virtual work

AE

FLFv

Then the redundant ADF can be obtained, and the axial forces in the members are from superposition. 9.4 Structures with Multiple Degrees of Indeterminacy To extend the method of consistent deformations to the problems of multiple degree of indeterminacy. Consider a beam with 3SI

Require 3 selected redundants, YY CB , and YD 3 compatibility conditions that the deflections of the primary beam at B,

C and D must be zero - The deflections of primary beam at B, C and D due to the external load

are COBO , and DO respectively.

- The flexibility coefficients in Fig (c) though (e); ijf

* ijf the deflection corresponding to a redundant i due to a unit value of a redundant j *

( i location of the deflection and j location of the unit load causing the deflection) To write the compatibility equations

For example, the deflection at B; 1) Due to external load = BO

+ 2) Due to YB

= YBB Bf + 3) Due to YC

= YBC Cf

+ 4) Due to YD

= YBD Df


N. Poovarodom 101

Then; 0 YBDYBCYBBBO DfCfBf

and 0 YCDYCCYCBCO DfCfBf

0 YDDYDCYDBDO DfCfBf

* The compatibility equations must be solve simultaneously to determine the unknown redundants . * From this example, the total number of deflections to be computed is

3 (due external loads) + 9 (flexibility coefficients) = 12 = 2)( SISI However, according to “Maxwell’s law of reciprocal deflections”

“For a linear elastic structure, the deflection at a point i due to a unit load applied at a point j is equal to the deflection at j due to a unit load at i .”

jiij ff

The number of deflections to be computed is 9 or 2

)3( 2SISI

Procedure For the method of consistent deformations . 1) - Determine SI - Select redundants, number = SI , determinate structure and stable. - Sense can be arbitrary. - Remove redundants and structure becomes a primary structure. 2) Draw a diagram of the primary structure. a) Only external loads. b) Only unit value of redundant. Total number of diagram = SI +1 3) Write compatibility equations (number of equations = SI ). 4) Compute the deflections and flexibility coefficients using Maxwell’s law. 5) Solve the simultaneous compatibility equations. 6) Determine the remaining reactions, member forces and draw SFD, BMD. 9.5 Support settlements * For externally indeterminate structures, this causes stress in member. * For internally indeterminate structures, this causes only translation or rotation as rigid bodies, without changing their shapes. Consider a continuous beam with support settlement B and C


N. Poovarodom 102

Consider YB and YC as redundants. For primary beam in Fig (b), (c) and (d) The compatibility conditions are

CYCCYCBCO

BYBCYBBBO

CfBf

CfBf

* The magnitude of support settlements must

be measured from the chord connecting the deformed positions of the supports to the deformed positions of the redundant supports . For example, for A and D not settle, the settlement of B and C are B , C

However , if beam undergoes settlement the displacement BR and CR

relative to the chord of primary beam must be used. * Only the displacement relative to the chord causes stresses in the beam.*


N. Poovarodom 103

9.6 Temperature change and Fabrication Errors Temperature change and Fabrication errors may affect the stress conditions of externally and/or internally indeterminate structures. 9.7 Method of Least Work Consider a statically indeterminate beam Suppose select YB as the unknown redundant, the strain energy of this beam is a function of

external load w and redundant YB

),( YBwfU

Castigliano’s second theorem: “The partial derivative of the strain energy with respect to a force is equal tothe deflection of the point of application of the force along its line of action.”

at B , 0 BO we can write the compatibility equation as;

0

YB

U

minimum of the strain energy “The principle of Least Work” “The magnitudes of the redundants of a statically indeterminate structure must be such that the strain energy stored in the structure is minimum (internal work done is the least).” For multiple degree of indeterminacy, n redundants 1R , ….. nR

),.......,,,( 21 nRRRwfU

Apply the principle of least work separately for each iR

0

0

0

2

1

nR

U

R

U

R

U

Simultaneous equations of compatibility.

Note this principle cannot be used in problems of temperature change, support settlement or fabrication error.

A B C

W

By


N. Poovarodom 104

Ex. 9.1 Determine the reactions and draw the SFD and BMD of the beam shown. Select the reaction at the roller to be the redundant.

EI = Constant

Soln 1) Select Dy as the redundant 2) 3) The compatibility equation, at D 0 yDDDO Df

4) Using the beam-deflection formulas

EI

Lf

EIEI

L

LLEI

LLL

EI

L

DD

DO

DO

3

54

742500

54

220

3236

321533

6

330

3

3

22

5)

k

L

EI

EI

L

fD

DD

DOy 22.12

3

54

2203

3

6)

MA

DyAy

Ax

DCBA5 ft 5 ft 5 ft

15 k 30 k

DO

15 30

(a)

DDf

1 (b)

-116.7

47.261.1

-12.22

2.7832.78

116.7 k.ft

12.22k32.78k

1530Reaction

SFD

BMD


N. Poovarodom 105

Ex. 9.2 Determine the reactions and draw the SFD and BMD of the beam shown. Select the moment at support as redundant.

Soln 1) Select MC as the redundant 2)

3) 0 CCCCO Mf

4) EIEIEIEI

PL

EI

MLCO

1230

16

)12)(120(

6

)12)(75(

166

22

(opposite to MC)

EIEI

LfCC

4

3

5) mkNf

MCC

COC .5.307

4

1230

6)

MC

Cx 120kN

Cy Ay

75kN.m

C

6m6m

B

A

CO 12075

(a)

CCf 1

(b)

-307.5

243.875

-91.87

28.13

307.5

28.13

75120

91.87

Reaction

SFD

BMD


N. Poovarodom 106

Ex. 9.3 Determine the reactions and draw the SFD and BMD of the beam shown. Select the reaction at central support as redundant.

E = Constant

Solution 1) Select By as the redundant 2) Determine BO and fBB by Virtual Force Principle

3)

Segment Origin Limit EI M Mv

AB A 0-10 EI 2

15165

2xx x

2

1

BC D 5-10 2EI 2

15)5(60195

2xxx x

2

1

CD D 0-5 2EI 2

15195

2xx x

2

1

BO dxxx

xEI

dxEI

MM v

2

1

2

15165

1 10

0

2

5

0

210

5 2

1

2

15195

2

1

2

15)5(60195

2

1dxx

xxdxx

xxx

EI

EI

28125

fBB

dxxdxx

EIdx

EI

MM v

v

210

0

210

0 2

1

2

1

2

11

EI

125

4) By = BB

O

fB =

125

28125 EI

EI kN225

Dy By Ay

Ax D

C

B (2I)

(I)

60 kN15 kN/m

5 m5 m10 m

A

xx

B

195165

DC

6015

A

Actual system

M-system 1/21/2

1 D

Virtual system

Mv-system


N. Poovarodom 107

5)

-7.5

52.5

-97.5

127.5

52.5

-82.5

91.88

225

-225

225 82.552.5

D

6015

A

3.5 m


N. Poovarodom 108

Ex. 9.4 Determine the reactions and force in each member of the truss shown. EA = Constant E = 29000 ksi A = 6 in2

Soln 1) Select Cy as the redundant 2)

3) 0 yCCCO Cf

4) By virtual work principle for truss deformation

AE

FLFv

F are obtained from (a) => Actual System Fv are obtained from (b) => Virtual System Member L (in) F Fv FFvL FvFvL Fi = F+FvCy

AB 180 29.17 -0.25 -1313 11.25 18.8 BC 180 29.17 -0.25 -1313 11.25 18.8 CD 180 25.83 -0.5 -2325 45 5.1 EF 180 -25.83 0.5 -2325 45 -5.1 AE 300 -31.94 0.417 -3996 52.2 -14.6 CE 300 -5.56 -0.417 696 52.2 -22.9 DF 300 -43.06 0.833 -10760 208.2 -8.5 BE 240 30 0 0 0 30 CF 240 4.44 -0.667 -711 106.8 -23.2

-22047 532

AE

FLFvCO

AE

22047

20 ft

30k

10k

Ax

Ay Cy Dy

D C B A

E F

15 ft 15 ft 15 ft

30k

10 30

34.44

43.05 31.94 5.56

30 4.44

29.17 25.83 29.17

25.56

25.83

30 CO

(a)

F-system

fCC

0.667

0.833 0.417

0.417 0

0.667

0.25

0.5 0.25

0.333

0.5

1

(b)

Fv-system


N. Poovarodom 109

AE

LFFf vv

CC

AE

532

CC

COy f

C

532

)22047(

kC y 5.41

Reactions can be obtained by equilibrium of the system; Ay = 11.7 k , Ax = 10 k , Dy = 6.8 k Member axial forces are obtained by superposition Fig. (a) + Fig. (b) Cy Fi = F+FvCy


N. Poovarodom 110

Ex. 9.5 Determine reactions of the frame shown, consider axial and bending deformations. E , A , I are constant.

Section "" 1212

I = 1728 in4 = 0.0833 ft4 A = 144 in4 = 1ft4

Soln 1) Select Dx as redundant 2)

3) 0 xDDDO Df

4) By virtual work

15’

1k/ft

C

20k

15’ 15’ Ay

Ax

A

B

Dy

D Dx

DO

13.75

6.25

15

1k/ft

20k

(a)

xD1 fDD

0.5

0.5

1 (b)

x x

x

D C B

A

13.75

6.25

15

1k/ft

20k 1

D

0.5

0.5

1


N. Poovarodom 111

Segment Origin Limit M Mv F Fv

AB A 0-15 2

152x

x x -6.25 0.5

BC D 15-30 13.75x-20(x-15) 0.5x 0 1 CD D 0-15 13.75x 0.5x 0 1

15

0

30

15

15

0

2

)5.0)(75.13()5.0))(15(2075.13()(2

151

dxxxdxxxxdxxx

xEI

EA

FLFdx

EI

MM vvDO

EEEEAEI

dxEA

5317309.465317779.4644297

)5.0)(25.6(1 15

0

301155.01

)5.0(1 22

15

0

30

0

22

EAdxxdxx

EI

EA

LFFdx

EI

MMf v

vv

vDD

EEEEAEI

4055475.334052075.333375

5) kf

DDD

DOx 11.13

kDx 11.13

If not consider axial deformation; kDx 125.13

Other reactions are obtained by superposition Reactions = Fig (a) + Fig (b) Dx

1.9

12.8

7.2

13.1


N. Poovarodom 112

Ex. 9.6 Repeat Ex. 9.3, by selecting MB as the redundant.

Solution 1) Select MB, remove restraint (internal moment at B) 2)

3) BBRBBLBBrel

BRBLBOrel

fff

0 BBBrelBOrel Mf

4) By deflection formula, assume the direction of to be the same as redundant.

EIIEIE

EIEIEI

wL

BR

BL

500

)2(16

10160

)2(24

1015

625

24

1015

2423

33

EIEIBOrel

1125500625

)2(3

103

10

IEf

EIf

BBR

BBL

EIEI

fBBrel

5

3

15

mkNM B 2255

1125 (The same result as in Ex. 9.3)

BR BL 2I I

60 15

BM1

BBRf BBLf

BR BL

BL BOrel

BBRf BBLf

BBLf BBrelf

Dy By Ay

Ax D

C

B (2I)

(I)

60 kN15 kN/m

5 m5 m10 m

A


N. Poovarodom 113

Ex. 9.7 Determine reactions and force in each member. AE = Constant

Soln 10)]352(8[)33()]32([)3( jmrSI Internal indeterminacy 1 degree

1) Select FAD to be the redundant, assume tension. 2)

3) 0, ADADADADO Ff

4) By virtual work Member L(m) F Fv FvFL Fv FvL Fi = F+FvFAD

AB 2 0 -0.707 0 1 6.04 BC 2.83 14.14 0 0 0 14.14 CD 2 -10 0 0 0 -10 DE 2 -10 -0.707 14.14 1 -3.96 BE 2.83 14.14 1 40.02 2.83 5.60 AD 2.83 0 1 0 2.83 -8.54 AE 2 0 -0.707 0 1 6.04 BD 2 -20 -0.707 28.3 1 -13.96

82.46 9.66

46.82

AEADO and AE

f ADAD

66.9 ,

5) )(54.854.866.9

46.82CkNFAD

C

2m

2m2m

25

15kN20kN10kN

20

E D

BA

0

14.14

14.140

10 10

20 25

1520 10

20

E D

B A

0

11

0.707

1

00.707

0.707

0.707

0


N. Poovarodom 114

If select FAB as a redundant;

Member L(m) Fa Fb Fa FbL Fb2L

AE 2 0 1 0 2 AD 2.83 0 -1.414 0 5.66 BE 2.83 14.14 -1.414 -56.62 5.66 BD 2 -20 1 -40 2 BC 2.83 14.14 0 0 0 ED 2 -10 1 -20 2 DC 2 -10 0 0 0 AB 2 0 1 0 2

-116.62 19.32

)(04.632.19

)62.116(TkNFAB

20 15

0 10

(a) 1

0

(b)


N. Poovarodom 115

Ex. 9.8 Determine reactions and draw SFD and BMD of the beam shown.

Soln 1) Select Cy and Dy as redundant. 2) a) b1) b2) 3) Compatibility At C; 0 yCDyCCCO DfCf

At D; 0 yDDyDCDO DfCf

5m 5m 10m 5m

Ay Cy Dy

MA

Ax

20kN/m 80kN

E D C B A

DO CO

20kN/m 80

yC1

fDC fCC

fDD fCD

yD1


N. Poovarodom 116

4) Consider formula for calculation of deflection

calculate deflection y(x)

4223 6424

xxLLxEI

wy

+

xaEI

way

xxaaxEI

wy

424

6424

3

4223

Lxa

ax

0

xaaxxLLxEI

w

xaLxaLEI

w

y344223

2223

46424

)(6)(424

Lxa

ax

0

10356

58010410101025610254

24

20 2444223

EIEICO

EI

220833

20356

5802010410202025610254

24

20 2344223

EIEIDO

EI

676667

EIEI

xaEI

PafCC

3.333)3010(

6

10)3(

6

22

EIEI

f DD

7.2666)20320(

6

202

a

L

x

w

w

w


N. Poovarodom 117

EIEI

axxEI

PfCD

3.833

6

)1020310()3(

6

2323

EIEIEI

xaPaf DC

3.833

6

)20310(10

6

)3( 22

5) From compatibility equations,

07.26663.8336766671

03.8333.3332208331

yy

yy

DCEI

DCEI

Cy = 128.7 kN Dy = 213.5 kN

DCCD ff

-250

72

-115

96

-93

4.325 113.5

100

37.8

-42.2

86.5

37.8 128.7 213.5

93

20 80


N. Poovarodom 118

Ex. 9.9 Determine reaction and forces in members.

211

)]352(8[)34(

)]32([)3(

jmrSI

External indeterminate = 1 degree, and internal indeterminate = 1 degree

1) Select redundant as; Cy (for external redundant ) and FBD (for internal redundant, tension +) 2)

(6)

(6) (6)

(6)

(8in2)

(8) (8)

(8)

16 ft 16 ft Cy 20k Ay

12 ft

Ax

Dx E D

C B

A

26.67

O

33.3

20

26.67

0

0

20

26.67

a)

0

20

SystemFa BDO

CO

1

1

0.8

0.6 1 0.6

0.8

0

0 b) SystemF

F

b

BD

BDBDf ,

BDCf ,

2.67

1

1.33

O 1.67

0

2.67

1.33

1.67

1

2.67

c) SystemF

C

c

y

CBDf ,

CCf


N. Poovarodom 119

3) Compatibility equations; At BD; 0,, yCBDBDBDBDBDO CfFf

At C; 0, yCCBDBDCCO CfFf

4)

Member L

(in) A

(in2) Fa

(k) Fb Fc Fa FbL/A Fa FcL/A Fb

2L/A Fc2L/A Fb FcL/A

F=Fa+ FbFBD+ FcCy

AB 192 8 0 -0.8 1.33 0 0 15.36 42.45 -25.54 -1.44 BC 192 8 0 0 1.33 0 0 0 42.45 0 8.29 DE 192 8 26.7 -0.8 -2.67 -512 -1709 15.36 171.1 51.26 0.31 AD 144 6 0 -0.6 0 0 0 8.64 0 0 -7.3 BE 144 6 20 -0.6 0 -288 0 8.64 0 0 12.7 AE 240 6 -33.3 1 1.67 -1333 -2226 40 111.56 66.8 -10.77 CE 240 8 0 0 -1.67 0 0 0 83.67 0 -10.4 BD 240 6 0 1 0 0 0 40 0 0 12.16

-2133 -3935 128 451.2 92.52

EAE

LFF baBDO

2133 ,

EAE

LFF caCO

3935 ,

EAE

LFf b

BDBD

1282

, , EAE

LFf c

CC

2.4512

EAE

LFFff cb

BDCCBD

52.92,,

6) Solve for FBD , Cy FBD = 12.16 k (T) , Cy = 6.23 k Other reactions and member forces can be obtained; Ax = 10.04 k , Ay = 13.77 k , Dx = 10.04 k


N. Poovarodom 120

Ex. 9.10 Determine reactions of the frame shown by considering effect of bending deformation only, EI = constant. Solution

1) Select Cy , Dy as redundant

2)

10 ft 10 ft

15 ft

2.5 k/ft

MA

Ay

Ax

Cy Dy

DB

A

C

50 k-ft

DO CO 50

x

x

x2.5

37.5 231.25

Ma - System

1

1

DCf CCf

10

Mb - System

1

1

DDf CDf

20

Mc - System


N. Poovarodom 121

3) Compatibility equations; At C; 0 yCDyCCCO DfCf

At D; 0 yDDyDCDO DfCf

Segment Origin Limit Ma Mb Mc

AB A 0-15 25.23125.25.37 2 xx 10 20 BC D 10-20 50 x-10 x CD D 0-10 50 0 x

CO = dxEI

MM ba

=

15

0

20

10

2

)10(501025.2312

5.25.37

1dxxdx

xx

EI

= EI

5.4062

DO = dxEI

MM ca

=

15

0

20

0

2

502025.2312

5.25.37

1xdxdx

xx

EI

= EI

3125

CCf = dxEI

MM bb

=

15

0

20

10

2)10(1001

dxxdxEI

= EI

3.1833

DDf = dxEI

MM cc

=

15

0

20

0

22201

dxxdxEI

= EI

7.8666

CDf = DCf = dxEI

MM cb

=

15

0

20

10

)10(2001

xdxxdxEI

= EI

3.3833

4) Solve compatibility equations; kC y 45.19 , kkDy 24.824.8


N. Poovarodom 122

Ex. 9.11 Determine reactions for the beam due to a small downward settlement at C E = constant

1) Select Cy as a redundant 2) (no external force , CO =0)

3) 0+fCCCy = -

4) L

L

L

CC dxxEI

dxxEI

f2

22

0

2

2

11

EI

L3

16

3

5)

33

16

L

EIC y

33

16

L

EI

x x

(I)(2I)

L/2L/2

CB A

1

CCf

23

16

L

EI

33

16

L

EI

33

16

L

EI

Reaction


N. Poovarodom 123

Ex. 9.12 Analyze the system with loading shown and support displacements of ½ in at A , 4 in at C, 3 in at E, and 2 ½ in at G

Solution

1) Select MC and ME as redundant

2)

And settlement

30k 30k 15k

(I) (2I) (I)

15 ft 10 ft 10 ft 15 ft 10 ft 10 ft

G F E C A B D

E = 29000 ksi, I = 5000 in4

ER EL CR CL

1 1

ECLf CCRf CCLf

EERf

1 1

EELf CERf

CL

2 ½ 3 4

½

A C E G

is measured from chord

CR ER

EL


N. Poovarodom 124

3) Compatibility equations;

EEEErelCECEOrel

CECECCCrelCOrel

MfMf

MfMf

4) By Tables of Beam-Deflection formula;

EIEIff

EIEIf

EIEIf

EIEI

EIEI

ECCE

EErel

CCrel

EOrel

CRCLCOrel

08.2

26

25

5.12

3

25

23

251

83.10

23

25

3

201

18001525

256

15301525

256

1530

2

11

9001025

256

1030

2

1

16

20151

2222

222

From figure L

a

0146.01220

5.04

CL rad

00333.01225

34

CR rad

0179.0 CRCLC rad

00333.0300

43

EL rad

00167.0300

5.23

ER rad

00166.0 ERELE rad 5) Solving compatibility equation, EI = 29000 ksi 5000 in4 MC = 1682 k-ft ME = -559 k-ft

(For example, rad 0.000894 500029000

12900 2

COrel )

Then the reactions; Ay = 91.6 k Cy = 154.2 k Ey = 148 k Gy = 10.4 k

chord

is measured from chord

from settlement is also measured from chord

chord


N. Poovarodom 125

Ex. 9.13 Determine reactions and force in each member E = 200 GPa A = 3000 mm2 = 15102.1 C

a) Due to a temperature drop of 25 C in AB, BC and CD and increase of 60 C in EF Solution 1) Select FBF as redundant (Tension as +) 2)

3) 0, BFBFBFBFO Ff

4) Virtual Work aBFO FLT

AE

LFf a

BFBF

2

,

Member L (m) T Fa TLFa Fa2L F=0+FaFBF (kN)

AB 8 -25 0 0 0 0 BC 8 -25 -0.8 160 5.12 -37.33 CD 8 -25 0 0 0 0 EF 8 +60 -0.8 -384 5.12 -37.33 BF 10 0 1.0 0 10 46.67 CE 10 0 1.0 0 10 46.67 BE 6 0 -0.6 0 2.16 -28.0 CF 6 0 -0.6 0 2.16 -28.0

-224 34.56 *AE and DF has zero axial force.

8 m8 m 8 m

6m

FE

DCB A

BFO

+60

-25 -25-25

BFBFf ,

11

10.6 0.6

0

00.8

0.8

0

0

Fa-System


N. Poovarodom 126

m/kN 1076.5

300010200

100056.34

m 00269.0102.1224

56

2

,

5

BFBF

BFO

f

5) kN 7.46,

BFBF

BFOBF f

F (T)

b) If member EF is 30 mm too short

0

104.2)8.0(1000

30

,

2

BFBFBFBFO

aBFO

Ff

mFe

)(7.416

7.4161076.5

104.25

2

,

CkNF

kNf

F

BF

BFBF

BFOBF

Ex. 9.14 Analysis the beam shown by the method of least work

Soln Select By as redundant

0

2

1

0

2

L

yy

L

dxEI

M

B

M

B

U

dxEI

MU

Segment x-coordinate

M yBM

Origin Limit CB C 0-6 -100x 0

BA C 6-18 )6(2

)6(10100

2

xBx

x y x-6

0))6()6(5100)(6(18

6

2

dxxBxxxB

Uy

y

0576126720 yB

kNBy 220

(2I)(I) 6m12m

CB A

10kN/m 100kN

x By 220-By

2520-12By 100

x


N. Poovarodom 127

Ex. 9.15 Determine the reactions and member forces by the method of least work. E = 29,000 ksi

Soln Select FBD as redundant. (Tension = +) Then to determine FBD by minimizing the strain energy with respect to FBD.

0

2

2

AE

FL

F

F

F

U

AE

LFU

BDBD

Member L (in)

A (in2)

F F/ FBD ( F/ FBD)(FL/A) (Sub. FBD) F

AB 192 8 -26.67-0.8 FBD -0.8 512+15.36 FBD -51.6 BC 192 8 -26.67 0 0 -26.7 DE 192 8 80-0.8 FBD -0.8 -1536+15.36 FBD 55.1 AD 144 6 -0.6 FBD -0.6 8.64 FBD -18.7 BE 144 6 20-0.6 FBD -0.6 -288+8.64 FBD 1.3 AE 240 6 -66.67+ FBD 1 -2667+40 FBD -35.6 CE 240 8 33.33 0 0 33.3 BD 240 6 FBD 1 40 FBD 31.1

-3979+128 FBD

01

A

FL

F

F

EF

U

BDBD

-3979+128 FBD = 0 FBD = 31.1 k (T)

(6)

(6) (6)

(6)

(8)

(8 in2) (8)

(8)

16 ft 16 ft 20k

20k

12 ft

E D

C B

A

66.67- FBD 20-0.6 FBD

80-0.8 FBD

26.67+0.8 FBD

FBD

FBD

0.6FBD

80

80

40

26.67

33.33

20 20

C


N. Poovarodom 128

Ex. 9.16 Determine force in rod BC and BD of the system shown. A = 100 mm2 I = 200106 mm4 E = 200 GPa Consider only bending deformation in beam AB

Solution Select FBD and FBC as redundant. (Tension = +)

AE

FL

F

Fdx

EI

M

F

M

F

U

AE

LFdx

EI

MU

i

L

ii

AB

0

22

2

1

2

1

AB M = (FBC + 5

3FBD – 80)x

xF

M

BC

, xF

M

BD 5

3

BC F = FBC

1

BCF

F , 0

BDF

F

BD F = FBD

1

BDF

F , 0

BCF

F

0000,20

5

580

5

3

000,40

1

0000,20

3

380

5

3

000,40

1

005

380

5

31

00805

31

4

0

3

4

0

3

24

0

4

0

2

BDBDBC

BCBDBC

BDBDBDBC

BD

BCBCBDBC

BC

FxFF

FxFF

AE

LFdxxFF

EIF

U

AE

LFdxxFF

EIF

U

FBC = 53.43 kN (T) FBD = 19.23 kN (T)

80kN

3m

4m

(A)(A)

(I)

D C

B (hinge)

A

37

FBCFBD

x80


N. Poovarodom 129

Appendix


N. Poovarodom 130


N. Poovarodom 131


N. Poovarodom 132

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