Moving Load

2101-301 Structural Analysis I Dr. Jaroon Rungamornrat

Influence Line 1

INFLUENCE LINES PRELIMINARIES Moving loads -- Loads applied to a structure with points of application (including their magnitude) can vary as a function of positions on the structure. Examples of moving loads include live load on buildings, traffic or vehicle loads on bridges, loads induced by wind and earthquake, etc. In the analysis, the moving loads can be modeled as varying distributed loads, a series of concentrated loads, or the combination of distributed loads and concentrated loads.


Influence Line 2

Responses due to a moving unit load -- Quantities of interest at a particular point within a given structure, e.g. internal forces, support reactions, deformations, displacements and rotations, due to an applied moving unit load. The quantities are given in terms of functions of a position of a moving unit load on the structure; these response functions are termed as the influence functions and their graphical representations are known as the influence lines. Application of the influence functions (lines) Let fA be a quantity of interest at a point A within a given structure due to applied distributed load q and a series of concentrated loads {P1, P2, …, PN} and fAI denote the influence function of the corresponding quantity at point A. By a method of superposition, we obtain the relation of fA and fAI as

N

³ fAI q dx ¦ fAI (x i )Pi

fA

(1)

i 1

Figure 1

where the integral is to be taken over the region on which load q is applied and xi indicates the location on which the load Pi is applied. For instance, assume that the influence line of the support reaction at point A (RAI) of the beam is given as shown in the Figure 3a. The support reaction at point A (RA) due to applied loads as shown in Figure 3b can then be obtained using Eqn. (1) as follow: L/2

A moving unit load -- a concentrated load of unit magnitude with its point of application

RA

varies as a function of position on the structure.

³R

AI

q dx R AI (L/4 )P R AI (3L/4 )2P

0

§ L/2 · q ¨¨ ³1 - x/L dx ¸¸ 3/4 P 1/4 2P 3qL/8 5P/4 ©0 ¹ 1 1

q

x

A

2P

B RA L/4

RAI L

L/4

3/4

1-x/L

Figure 2

P

Area = 3L/8

Responses due to moving loads -- Quantities of interest that indicate the effect of the moving loads on a structure, e.g. internal forces, support reactions, displacements and rotations, deformations, etc.

RAI

0

L Figure 3a

x

L/4

L/4

1/2 1/4

RAI

L Figure 3b

x


Influence Line 3

In addition, the influence lines can also be used to predict the load pattern that maximizes responses at a particular point of the structure. For instance, let consider a two-span continuous beam subjected to both dead load (fixed load) and live load (varying load) as shown in the figure below.


Influence Line 4

INFLUENCE LINES FOR DETERMINATE BEAMS

1

x Live load Dead load A

B GBI RAI

B

To determine the maximum positive bending moment at points A, the maximum negative moment B, and the maximum positive shear at point A due to these applied dead and live loads, we construct first the influence lines MAI, MBI, and VAI as shown below.

MCI

C

TBI

D

Deformed state Undeformed state

RDI

¾ Support reactions (e.g. RAI, RDI) ¾ Bending moment at a particular section (e.g. MCI) ¾ Shear force at a particular section (e.g. VCI)

1

x

A

VCI

¾ Deflection at a particular point (e.g. GBI) A

B

¾ Rotation at a particular point (e.g. TBI)

Direct Methods for Constructing Influence Lines ¾ Treat a structure subjected to a moving unit load (as function of positions) MAI

x

¾ Influence functions are obtained by considering all possible load locations ¾ Support reactions

MBI

x

-- Equilibrium equations of the entire structure ¾ Internal forces -- Method of sections

VAI

x

-- Equilibrium equations of parts of the structure ¾ Displacement and rotations -- Determining support reactions and internal forces from equilibrium

It is evident from the influence lines that the maximum positive bending moment at point A occurs when the live load is placed only on the first span; the maximum negative moment at point B occurs when the live load is placed on both spans; and the maximum positive shear occurs when the live load is placed on the first half of the first span and on the second span. The maximum value of the responses can then be obtained using Eqn.(1) for each corresponding loading pattern. It is noted that the dead load is fixed and therefore it is applied to both spans of the beam for all cases.

-- Displacement and rotations are obtained from 9 Direct integration method 9 Moment area and conjugate structure methods 9 Energy methods, etc.


Influence Line 5

Example1: Construct influence lines RAI, RBI, VCI, MCI, GCI, TCI of a simply supported beam

C

A


Influence Line 6

Influence lines for shear and bending moment VCI, MCI

B EI

L/3

2L/3

x L/3

1

A Solution Consider the beam subjected to a moving unit load as shown below.

B

C

A

> 6FY

B

C L/3

0

@

+

> 6M C

1

x

0

@

+

B

A

@

+

MCI

L- x L

+

R AI 1

(R AI 1)L x 3

2x 3

A

VCI B

C

C

A RAI

RBI

> 6FY

x L

0

@

+

R AI VCI VCI

0 1

R AI

1 RAI L

x

> 6M C

0

@

+

(R AI )(L/3) MCI

1

MCI

RBI 0

0

(RBI )(L) (1)(x) 0 RBI

0

x L

1

x L/3 RAI

@

0

(R AI )(L) (1)(L - x) 0 R AI

0

MCI

RAI

(R AI )(L/3) (1)(L/3 - x) MCI

RBI

RAI

> 6M A

R AI 1 VCI VCI

2L/3

Influence lines for reactions RAI, RBI

0

C

A

RBI

RAI

VCI

1

x

> 6M B

1

x L/3

L

x

x L

0

R AI L 3

L§ x· ¨1 ¸ 3© L¹

MCI


Influence Line 7

1


The deflection GCI for x L/3 can be obtained using the unit load method along with the actual system I and the virtual system I; i.e.

2/3

RAI 0

x

L

L

2/3

³

GCI

VCI 0

Influence Line 8

L/3 -1/3

0

x

L

1 § L x ·§ L · ª 2 2L º 1 § L ·§ L · ª§ 2 x · 2L º ¨ ¸¨ ¸ « ¨ x ¸¨ x ¸ «¨ ¸ » » 2EI © 3 ¹© 3 ¹ ¬ 3 9 ¼ 2EI © 3 ¹© 3 ¹ ¬© 3 L ¹ 9 ¼

2L/9 MCI 0

L/3

x

L

Influence lines for deflection and rotation GCI, TCI x L/3 A RAI

x L/3 C

A

B RBI

L

BMD M

³

GCI

x

0

L/3-x

Actual System I

1 B 1/3

A 1/L

C

MGM dx EI

1 § L x ·§ L · ª 2 2L º 1 § L L · ª§ 7 x · 2L º ·§ ¨ ¸¨ ¸ « ¨ x ¸¨ x ¸ «¨ ¸ » 2EI © 3 ¹© 3 ¹ ¬ 3 9 ¼ 2EI © 3 3 ¹ ¬© 6 2L ¹ 9 »¼ ¹©

Actual System II

1 C

The deflection GCI for x L/3 can be obtained using the unit load method along with the actual system II and the virtual system I; i.e.

L/3-x/3

x-L/3

2/3

B RBI

2x/3

x

A

1

RAI

L/3-x/3 2x/3 BMD M

1 § 2x ·§ 2L · ª 2 2L º ¨ ¸¨ ¸ 2EI © 3 ¹© 3 ¹ «¬ 3 9 »¼ x 5L2 9x 2 81EI

1 C

MGM dx EI

B 1/L

1 § 2x ·§ 2L · ª 2 2L º ¨ ¸¨ ¸ 2EI © 3 ¹© 3 ¹ «¬ 3 9 »¼ x-L 2 L 18Lx 9x 2 162EI

2L/9

1/3 BMD GM

x

BMD GM

x -2/3

Virtual System I

Virtual System II

4L3/243EI GCI 0

L/3

L

x


Influence Line 9

The rotation TCI for x L/3 can be obtained using the unit load method along with the actual system I and the virtual system II; i.e.

L

³

TCI

0

Example2: Construct influence lines RAI, MAI, VBI, MBI, GBI, TBI of a cantilever beam

B EI

A L/2

x L2 3x 2 18EIL

L/2

Solution Consider the beam subjected to a moving unit load as shown below. 1

x

1 § 2x ·§ 2L · ª 2 2 º ¨ ¸¨ ¸ 2EI © 3 ¹© 3 ¹ «¬ 3 3 »¼

Influence Line 10

MGM dx EI

1 § L x ·§ L · ª 2 1 º 1 § L ·§ L · ª§ 2 x · 1 º ¨ ¸¨ ¸ « ¨ x ¸¨ x ¸ «¨ ¸ » » 2EI © 3 ¹© 3 ¹ ¬ 3 3 ¼ 2EI © 3 ¹© 3 ¹ ¬© 3 L ¹ 3 ¼


B EI

A

L/2

L/2

Influence lines for reactions RAI, MAI The rotation TCI for x L/3 can be obtained using the unit load method along with the actual system II and the virtual system I; i.e.

L

³ 0

RAI

MGM dx EI

> 6M A

1 § L x ·§ L · ª 2 1 º 1 § L L ·ª § 7 x · 2 º ·§ ¨ ¸¨ ¸ « ¨ x ¸¨ x ¸ « ¨ ¸ » 2EI © 3 ¹© 3 ¹ ¬ 3 3 ¼ 2EI © 3 3 ¹ ¬ © 6 2L ¹ 3 »¼ ¹©

> 6FY

L/3

@

+

M AI (1)(x) 0

0

@

+

x

R AI 1 0 R AI 1

0 0

0

M AI

1 § 2x ·§ 2L · ª 2 2 º ¨ ¸¨ ¸ 2EI © 3 ¹© 3 ¹ «¬ 3 3 »¼ L- x 2 L 6Lx 3x 2 18EIL

B

A

MAI

TCI

1

x

L

x

L

-L 1

TCI 2

-4L /162EI

x

MAI 1 RAI 0

L

x


Influence Line 11


Influence lines for shear and bending moment VCI, MCI

Influence Line 12

L

MAI 0

x

-L/2 1 x L/2 1 MAI

RAI

MBI VBI B

A

1 0

x

L 1

B

1

RAI VBI

> 6FY

0

> 6M B

0

@

@

+

+

VBI

0

VBI

0

M BI

MBI

0

L/2

L

0

L/2

L

x -L/2

Influence lines for deflection and rotation GBI, TBI

0

MAI

x L/2 1 A

1

x L/2 B

A

MAI

B

RAI

RAI

M BI

x

L/2

0 BMD M -x

L/2 x

BMD M -x

x L/2-x

-L/2 1

x L/2 MAI

A

MAI

B RAI

MBIVBI

A

Actual System I

Actual System II

B

RAI 1

> 6FY

0

@

+

R AI VBI

-L/2

0

B

A

A

1

0

1 1

VBI

> 6M B

0

@

+

R AI

1

(R AI )(L/2) M AI M BI M BI

BMD GM -L/2

0

R AI L M AI 2

L· § ¨ x ¸ 2 ¹ ©

x

1 B 1

BMD GM

x

-L/2+x

Virtual System I

Virtual System II


Influence Line 13

The deflection GBI for x L/2 can be obtained using the unit load method along with the actual system I and the virtual system I; i.e.

L

³

G BI

0


Influence Line 14

The rotation TBI for x L/2 can be obtained using the unit load method along with the actual system I and the virtual system II; i.e.

L

MGM dx EI

³

TCI

0

1 § L ·§ L · ª 1 L º 1 § L ·ª 1 L º - x ¨ ¸ « ¨ ¸¨ ¸ » 2EI © 2 ¹© 2 ¹ «¬ 3 2 »¼ EI © 2 ¹¬ 2 2 ¼

1 § L ·§ L · 1 §L· ¨ ¸¨ ¸>1@ - x ¨ ¸>1@ 2EI © 2 ¹© 2 ¹ EI ©2¹

1 § L ·§ L · ª § 1 2x · L º ¨ x ¸¨ x ¸ « ¨ ¸ » 2EI © 2 ¹© 2 ¹ ¬ © 3 3L ¹ 2 ¼

x2 3L 2x 12EI

L

³ 0

L

MGM dx EI

TCI

³ 0

MGM dx EI

1 § L ·§ L · ª 1 L º 1 § L ·ª 1 L º - x ¨ ¸ « ¨ ¸¨ ¸ « » » 2EI © 2 ¹© 2 ¹ ¬ 3 2 ¼ EI © 2 ¹¬ 2 2 ¼

1 § L ·§ L · 1 §L· ¨ ¸¨ ¸>1@ - x ¨ ¸>1@ 2EI © 2 ¹© 2 ¹ EI ©2¹

L2 6x L 48EI

L L 4x 8EI

L /24EI L/2

0 TBI

3

0

x2 2EI

The rotation TBI for x L/2 can be obtained using the unit load method along with the actual system II and the virtual system I; i.e.

5L3/48EI GBI

1 § L ·§ L · ¨ x ¸¨ x ¸>1@ 2EI © 2 2 ¹© ¹

The deflection GBI for x L/2 can be obtained using the unit load method along with the actual system II and the virtual system I; i.e.

GCI

MGM dx EI

L

L/2

L

x

-L2/8EI

x -3L2/8EI


Influence Line 15


Example3: Construct influence lines RAI, MAI, RBI, VCI, VBLI, MBLI, VBRI, MBRI, VDI, and MDI of a beam shown below

From FBD I, we obtain

> 6FY

C

D

A

Influence Line 16

0

@

+

R AI RBI 1 0

B

R AI L/2

L/2

L

0

@

+

Solution Consider the beam subjected to a moving unit load as shown below.

- M AI (RBI )(2L) (1)(x) 0 M AI

2RBI L x

-x

1

x C

D

L/2

1

L

> 6M A

A

1 RBI

B

L/2

L

xL

L MAI

Influence lines for reactions RAI, MAI, RBI and shear force VCI

A

1

VCI

B

C

RAI

1 B

C

RBI

RBI

FBD III

xL MAI

A

FBD IV

From FBD IV, we obtain

1 VCI

B

C

RAI

B

C

RBI

> 6MC

RBI

FBD I

0

@

+

RBI

FBD II

From FBD II, we obtain

> 6FY

> 6MC

0

@

+

0

@

+

(RBI )(L) 0 RBI

(RBI )(L) (1)(x L) 0 x 1 L

VCI RBI 1 0 VCI

1 RBI

0

2

x L

From FBD III, we obtain

> 6FY

0

@

+

VCI RBI VCI

> 6FY

0 RBI

0

0

@

+

R AI RBI 1 0 R AI

1 RBI

2

x L


Influence Line 17


Influence Line 18

Influence lines for shear and bending moment VBLI, MBLI

> 6M A

0

@

+

- M AI (RBI )(2L) (1)(x) 0 M AI

2RBI L - x

x 2L xL MAI

A

1 C

RAI MAI 0

L

MBLI VBLI B

B RBI

RBI

L

2L

x

3L

> 6FY

-L

0

@

VBLI RBI

+

VBLI

> 6M B RAI

1

1

0

L

2L

3L

0

L

@

M BLI

+

M BLI

x

-RBI 0 0

-1

2

1 RBI

0

0

2L

MAI x

3L

A

1

x 2L C

MBLIVBLI B

B

RAI

RBI

> 6FY

0

@

+

RBI

VBLI RBI 1 0

1 2L

VCI 0

3L

VBLI

x

L -1

> 6M B

0

@

+

1 RBI

M BLI (1)(x 2L) 0 M BLI

2L x

1


Influence Line 19

0

2L

L

Influence Line 20

2

1 RBI


x

3L

MAI

A

1

x 2L C

B

RAI

VBLI

0

L

3L

2L

MBLI

0

L

RBI

x

> 6FY

-1

-1 3L

2L

MBRIVBRI

0

@

+

VBRI 1 0 VBRI

x

1

-L

> 6M B

0

@

+

M BRI (1)(x 2L) 0 M BRI

2L x

Influence lines for shear and bending moment VBRI, MBRI

xL MAI

A

1

MBRI VBRI

C

1

B

RAI

RBI

> 6FY

> 6M B

0

0

@

@

+

+

VBRI

0

VBRI

0

M BRI

0

M BRI

0

1

VBRI 0

L

2L

3L

0

L

2L

3L

MBRI

x

x -L

1


Influence Line 21


Influence Line 22

> 6M B

Influence lines for shear and bending moment VDI, MDI

0

@

+

M DI (1)(x L/2) (RBI )(3L/2) 0 M DI

3LRBI /2 L/2 x

x L/21 MAI

D

A

C

B

RAI

RBI

RBI MDI VDI

C

0

B

D

0

@

+

VDI RBI VDI

> 6M B

0

@

+

MAI

A

L/2

L

3L

x

2L

MDI

0

L/2

L/2 L

2L 3L

x

-L/2

M DI (RBI )(3L/2) 0 3LRBI /2

Remarks

C

1. The influences lines of support reactions and internal forces (shear force and bending moment) for statically determinate beams are piecewise linear; i.e. they consists of only straight line segments.

B

RAI

RBI MDI VDI

1 C

B

D

> 6FY

1

x

3L

-1

1 D

1

2L

0

-RBI

M DI

x L/2

L

RBI VDI 0

> 6FY

2

1

0

@

+

RBI

VDI RBI 1 0 VBLI

1 RBI

2. The influence functions of the internal forces can be obtained in terms of the influence functions of the support reactions; therefore, the influence lines of internal forces can be readily obtained from those for support reactions. 3. The influence lines of the deflection and rotation at any points of the statically determinate beam generally consist of curve segments.


Influence Line 23


Influence Line 24

¾ Virtual displacement due to release of shear constraint.

Muller-Breslau Principle Actual Structure. Consider a statically determinate beam subjected to a moving unit load as shown in the figure below. x

1

1. Remove the shear constraint by introducing a shear release at point of interest 2. The beam becomes statically unstable (partially or completely) 3. Introduce unit relative virtual displacement between the two ends of the shear release with their slope remaining the same (provided that the moment constraint exists at that point) 4. The virtual displacement at all other points results from the development of the mechanism (or rigid body motion) of the entire beam. 1 Virtual System 2a

Virtual Displacement -- The fictitious and arbitrary displacement that is introduced to the structure. For use further below, the following three types of virtual displacement for the beam structure are considered: RELEASE shear constraint

¾ Virtual displacement due to release of a support constraint. 1. Release a support constraint in the direction of interest 2. The beam becomes statically unstable (partially or completely) 3. Introduce unit virtual displacement (or unit virtual rotation if the rotational constraint is released) in the direction that the support constraint is released. 4. The virtual displacement at all other points results from the development of the mechanism (or rigid body motion) of the entire beam

1 Virtual System 2b

RELEASE shear constraint ¾ Virtual displacement due to release of bending moment constraint.

1

Virtual System 1a RELEASE displacement constraint

1. Remove the moment constraint by introducing a hinge at point of interest 2. The beam becomes statically unstable (partially or completely) 3. Introduce unit relative virtual rotation at the hinge without separation (provided that the shear constraint exists at that point). 4. The virtual displacement at all other points results from the development of the mechanism (or rigid body motion) of the entire beam. 1

1

Virtual System 1b

RELEASE rotational constraint

Virtual System 3a

RELEASE moment constraint 1 Virtual System 1c 1

RELEASE displacement constraint

RELEASE moment constraint

Virtual System 3b


Influence Line 25

Principle of Virtual Work: Consider a system or structure subjected to external applied loads. The support reactions and internal forces at any locations within the structure are in equilibrium with the applied loads if and only if the external virtual work (work done by the external applied loads) is the same as the internal virtual work (work done by the internal forces) for all admissible virtual displacements, i.e.

įWE

įWI

(2)


Muller-Breslau Principle: “The influence line of a particular support reaction has an identical shape to the virtual displacement obtained from releasing the support constraint in the direction of the support reaction (under consideration) and introducing a rigid body motion with unit displacement/unit rotation in the direction of the released constraint.” Influence Line for Shear Force. Let assume that the influence line of the shear force at point C, VCI, is to be determined. By applying the principle of virtual work to the actual system with a special choice of the virtual displacement as indicated in the virtual system 2a (the virtual displacement associated with the rigid body motion of the beam resulting from the release of the shear constraint at C) , we obtain

It is important to note that the portion of the structure that undergoes virtual rigid body motion (virtual displacement that produces no deformation) produces zero internal virtual work. Influence Line for Support Reactions. To clearly illustrate the strategy, let assume that the influence line of the support reaction RAI is to be determined. By applying the principle of virtual work to the actual system with a special choice of the virtual displacement as indicated in the virtual system 1a (the virtual displacement associated with the rigid body motion of the beam resulting from the release of the displacement constraint at A) , we obtain

įWE įWE

R AI 1 1 įv ( x ) R AI įv ( x )

;

įWI

įWE

1 įv ( x ) įv ( x )

įWE

įWI

0 MAI

R AI

įWI

;

(4)

1 MCI

A

VCI B

Actual system

C

RAI

įv ( x )

VCI 1 VCI

įv ( x )

VCI

x

įWI

Influence Line 26

(3)

RBI 1 Gv(x)

x

1

A

MAI

Virtual System 2a B

Actual system

RELEASE shear constraint

RAI RBI 1

VCI

1

x

Gv(x) Virtual System 1a RELEASE displacement constraint

Muller-Breslau Principle: “The influence line of the shear force at a particular point has an identical shape to the virtual displacement obtained from releasing the shear constraint at that point and introducing a rigid body motion with unit relative virtual displacement between the two ends of the shear release with their slope remaining the same.”

1 RAI

x


Influence Line 27

Influence Line for Bending Moment. Let assume that the influence line of the bending at point C, MCI, is to be determined. By applying the principle of virtual work to the actual system with a special choice of the virtual displacement as indicated in the virtual system 3a (the virtual displacement associated with the rigid body motion of the beam resulting from the release of the bending moment constraint at C) , we obtain

įWE

1 įv ( x ) įv ( x )

įWE

įWI

įWI

;


Influence Line 28

Example4: Use Muller-Breslau principle to construct influence lines RAI, RDI, RFI, VBI, VCLI, VCRI, VDLI, VDRI,VEI, MBI, MDI, and MEI of a statically determinate beam shown below

MCI

E

D

F

MCI 1 MCI L/4

L/4

C

B

A

įv ( x )

(5)

L/2

L/2

L/2

Solution The influence line of the support reaction RDI is obtained as follow: 1) release the displacement constraint at point D, 2) introduce a rigid body motion, 3) impose unit displacement at point D, and 4) the resulting virtual displacement is the influence line of RDI.

1 x MAI

A

1 MCI

E

F

D

B

Actual system

RELEASE displacement constraint

C

RAI

C

B

A

VCI

RBI 1

L/4

L/2

L/4

L/2

L/2

Gv(x) Virtual System 3a

h2=3/2 1

h1=3/4

RELEASE moment constraint

h3=1/2

RDI

x

1 MCI

1

x

Muller-Breslau Principle: “The influence line of the shear force at a particular point has an identical shape to the virtual displacement obtained from releasing the shear constraint at that point and introducing a rigid body motion with unit relative virtual displacement between the two ends of the shear release with their slope remaining the same.”

The value of the influence line at other points can be readily determined from the geometry, for instance,

h2

(1)(3L/2 ) /(L) 3/2

h3

(1)(L/2 ) /(L) 1/2

h1

(3/2 )(L/4 ) /(L/2 ) 3/4


Influence Line 29

The influence line of the shear force VEI is obtained as follow: 1) release the shear constraint at point E, 2) introduce a rigid body motion, 3) impose unit relative displacement at point E and 4) the resulting virtual displacement is the influence line of VEI.


Influence Line 30

The influence line of the bending moment MEI is obtained as follow: 1) release the bending moment constraint at point E, 2) introduce a rigid body motion, 3) impose unit relative rotation at point E without separation and 4) the resulting virtual displacement is the influence line of MEI.

1 C

B

A

1

E

D

F

C

B

A

RELEASE shear constraint L/4

L/2

L/4

L/2

F

RELEASE moment constraint L/2 L/4

h2=1/2

E

D

L/2

L/4

L/2

L/2

h4=1/2

h1=1/4

h3=L/4

1

VEI

1

x MEI h3=-1/2

x h1=-L/8 h2=-L/4


h3 /(L / 2 )

h4 /(L / 2 ) h3

h4

h4 h3 1 h4 (h4 ) 2h4 1 h4 1/2 h3

h4

h2

(h3 )(L/2 ) /(L/2 ) 1/2

h1

(h2 )(L/4 ) /(L/2 ) 1/4


h3 /(L / 2 ) h3 /(L / 2 ) 1 h3 h2

(h3 )(L/2 ) /(L/2 )

h1

(h2 )(L/4 ) /(L/2 )

1/2

L/4

L/4 L/8

The rest of the influence lines can be determined in the same manner and results are given below.


Influence Line 31

x

E

D

Example5: Use Muller-Breslau principle to construct influence lines RAI, MAI, RDI, VBI, VCI, VDI,VELI,VERI, MBI, MDI, and MEI of a statically determinate beam shown below.

F

C L/4

Influence Line 32

1

B

A


L/2

L/4

L/2

L/2

C

B

A

E

D

F

1 1/2

L/4

L/4

RAI

L/2

L/2

L/2

x 1 1/2

RFI

x 1/4

1/2

Solution By Muller-Breslau principle, we obtain the influence lines as follow: 1) release the constraint associated with the quantity of interest, 2) introduce a rigid body motion, 3) impose unit virtual displacement/rotation in the direction of released constraint, and 4) the resulting virtual displacement is the influence line to be determined. It is noted that values at points on the influence line can be readily determined from the geometry.

1/2

1

x

VBI

x

C

B

A

E

D

F

-1/2 VCLI

x

L/4 1

1

-1/2

L/2

L/4

L/2

L/2

1

-1

1/2

VCRI

x

RAI

x

-1/2

-1/2

L/2

-1 VDLI

L/4

L/4

x MAI

x

-1/2

-L/4 -1

-1 1

1 1/2

1/4 VDRI

1/2 x

REI

x 1

1 1/2

L/8 MBI

x

VBI

x 1

L/4

-1/2 1/2

L/8 MDI

3/2

VCI x

x -1/2


Influence Line 33

A

C

B

Influence Line 34

Example6: Use Muller-Breslau principle to construct influence lines RAI, MAI, RDI, RFI, VBI, VCI, VDLI, VDRI,VEI, MBI, and MDI of a statically determinate beam shown below.

1

x


E

D

F

L/4

L/2

L/4

L/2

C

B

A

E

D

F

L/2 L/4

L/4

L/2

L/2

L/2

1/2 VDI

x -1/2

-1/2

VELI

x

Solution By Muller-Breslau principle, we obtain the influence lines as follow: 1) release the constraint associated with the quantity of interest, 2) introduce a rigid body motion, 3) impose unit virtual displacement/rotation in the direction of released constraint, and 4) the resulting virtual displacement is the influence line to be determined. It is noted that values at points on the influence line can be readily determined from the geometry.

-1/2

-1/2

C

B

A 1

1

x

-1

E

D

F

1

VERI

x

L/4 1

1

L/2

L/4

L/2

L/2

1

L/8 MBI

x

RAI

x

-L/8

L/2

-L/4 L/4 MDI

-1

L/4 MAI

x

x

-L/2

-L/4

2 1

MEI

REI x

x -L/2

1 REI x


Influence Line 35

1

x A

C

B

L/4

L/2

L/4 1

E

D

L/2

F

L/2

1

VBI

x 1 -1

VCI

x

-1 VDLI

x -1

-1

1

1

VDRI

x 1

VEI

x L/4

x

MBI -L/4

MBI

-L/2

Moving Load

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