1

INTRODUCTION TO INFLUENCE LINES

Why do we need the influence lines? For instance, when loads pass over a structure, say a bridge, one needs to know when the maximum values of shear/reaction/bending-moment will occur at a point so that the section may be designedNotations:

Normal Forces - +ve forces cause +ve displacements in +ve directionsShear Forces - +ve shear forces cause counter clockwise rotation & - ve shear force causes clockwise rotationBending Moments: +ve bending moments cause “cup holding water” deformed shape

• Influence lines describe the variation of an analysis variable (reaction, shear force, bending moment, twisting moment, deflection, etc.) at a point (say at C in Figure below)

A BC

A BC

INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES

2

• Introduction - What is an influence line?Dead loads are fixed in positions ---- Live loads can change their positions

Our objective is to establish how to position live loads (a truck or a train) to maximize the value of certain type of force (Shear or moment in a beam or axial force in a truss) at a designated section of a structure.To establish maximum design forces at critical sections produced by moving loads, we frequently construct influence lines. An influence line is a diagram whose ordinates, which are plotted as a function of distance along the span give the value of an influence an internal force, a reaction, or a displacement at a particular point in a structure as the unit load of 1 kN moves across the structure.

We can use the influence lines as- To determine where to place live load on a structure to maximize the force for which the influence line is drawn.- To evaluate the magnitude of the force produced by the live load.

INFLUENCE LINES FOR BEAMS

• Procedure:(1) Allow a unit load (1N, 1kN) to move over beam from left to right(2) Find the values of shear force or bending moment, at the point under consideration, as the unit load moves

over the beam from left to right(3) Plot the values of the shear force or bending moment, over the length of the beam, computed for the

point under consideration

3

MOVING CONCENTRATED LOADVariation of Reactions RA and RB as functions of load position

ΣMA =0(RB)(10) – (1)(x) = 0

RB = x/10RA = 1-RB

= 1-x/10

x1

A B

C10 m

3 m

x 1

A BC

RA=1-x/10 RB = x/10

x

A C

RA=1-x/10 RB = x/10

We want to draw I-line for the reaction at A of the simply supported beam.

We can establish the ordinates of the I-lines for the reaction at A by computing the value of RA for the successive positions of a unit load as it moves across the span.

4

RA occurs only at A; RB occurs only at B

Influence line

for RB

1-x/10

1

Influence

line for RA

x 10-x

x 10-x

x/10 1.0

1 ( )A BR R Applied load+ =

The several points to remember can be summarized as follows

1 All ordinates of the I-lines represent values of RA

2 Each value of RA is plotted directly below the position of the unit load that produced it.

3 The maximum value of RA occurs when the unit load acts at A

4 Since all ordinates are positive, reaction at A directed upward.

5 Influence line is a straight line.

5

Variation of Shear Force at C as a function of load position

0 < x < 3 m (unit load to the left of C)

Shear force at C is + ve, SC =+x/10

C

x 1.0

RA = 1-x/10 RB = x/10

3 m

10 m

A B

x/10

C

6

3 < x < 10 m (unit load to the right of C)

Shear force at C is -ve = 1-x/10

Influence line for shear at C

C

x

3 mA

RA = 1-x/10 RB = x/10

B

C

1

1

+ve

-ve

0.3

0.7

RA = 1-x/10

7

x/10

Variation of Bending Moment at C as a function of load position

0 < x < 3.0 m (Unit load to the left of C)

Bending moment is + ve at C

C

x

3 mA B

RA = 1-x/10RA = x/10

10 m

C

x/10

x/10

x/10

(x/10)(7)(x/10)(7)

8

3 < x < 10 m (Unit load to the right of C)

Moment at C is + ve

Influence line for bendingMoment at C

C

x m

3 mA

RA = 1-x/10

10 m

C

1-x/10

1-x/10

(1-x/10)(3)

(1-x/10)(3)

1 kN

RA = x/10

B

1-x/10(1-x/10)(3)

+ve

(1-7/10)(3)=2.1 kNm

9

MULLER-BRESLAU’S PRINCIPLE

This principle provides a simple procedure for establishing the shape of influence lines (I.L.) for reactions or the shear and moment in beams. The procedure does not produce numerical values of I.L. ordinates.

Quickly sketched I. Lines can be used in following three ways:

1- To verify that the shape of an I. L., produced by moving a unit load across the structure, is correct

2- To establish where to position the live load on a structure to maximize a particular force. Once the critical position of the load is established, it is simpler to analyze certain types of structures directly for the specified live load than to draw the I.L.

3- To determine the location of the maximum and minimum ordinates of an I.L. so that only a few positions of the unit load must be considered when the I.L. ordinates are computed.

The Muller-Breslau principle states:

The ordinates of an I.L. for any force are proportional to the deflected shape of the structure produced by removing the capacity of the structure to carry the force and than introducing into the released structure a displacement that corresponds to the restraint removed.

Remove the vertical restraint supplied by reaction at A

Released structure

Displace the left end of the beam upward in the direction of RA an arbitrary amount ∆

A B

A B

B

AR

∆A′

10

This deflected shape is proportional to the I.L. for RA . A simple way to produce the deflected shape is to imagine the force, associated with the restraint that has been removed, is applied to the released structure and displaces the member into its deflected position.

1 Place a unit load on the actual beam over support A and compute RA = 1 kN

Influence line for RA

Influence Line for a Determinate Beam by Muller-Breslau’s Method

Influence line for Reaction at A

11

Influence Lines for a Determinate Beam by Muller-Breslau’s Method

Influence Line for Shear at CInfluence Line for Bending Moment at C

12

USE OF INFLUENCE LINES

How to use an influence line to compute the maximum value of a function when the live load can act anywhere on the structure is either a single concentrated load or a uniformly distributed load of variable length.

Example: The beam in figure is to be designed to support its deadweight of 15 kN/m and a live load that consists of a 130 kN concentrated load and a variable length, uniformly distributed load of 40 kN/m . The live load can act anywhere on the span.

Draw the I.L. for moment at point C

Compute the maximum positive and negative values of the live load moment at section C and

The moment at C produced by the beam’s weight.

A B C D E

1.6 1.6 3 m 3 m

Variable

q=40 kN/m130

A B C D E

1.6 1.6 3 m 3 m

Mc

1 kN

Hinge

Deflected shape

Released structure

-0.8 -0.8

1.5

13

A B C D E

1.6 1.6 3 m 3 m

Mc

130 kN-0.8 -0.8

1.5

q=40 kN/m

130 40 kN/m 40 kN/m

12

max

130*1.5 40*(6*1.5* ) 375

positive moment

M kNm= + =

12

max

130*0.8 2*40*(1.6*0.8* ) 155.2

negative moment

M kNm− = + =

A B C D E

1.6 1.6 3 m 3 m

15 kN/m

1.5*6 1.6*0.82* *15 48.3

2 2

deadweigth moment

Mc kNm = − =

14

INFLUENCE LINES FOR TRUSSES

The unit load is positioned at successive points and the corresponding bar forces are plotted directly below the position of the load.

Example: Compute the ordinates of the influence lines for the reaction at A, and for bar forces BK, CK, and CD of the given truss. Assume that the loading is applied by the bottom chord members.

A

FEDCB

K J I H GL6 4.5X m

6 m

15

A

FEDCB

K J I H GL6 4 .5X m

6 m

1 56

46

36

[ ]AR kN

[ ]BKF kN

[ ]CKF kN

21.6m

2024

524

−

12

−

261

6

38

−68

−

98

− [ ]CDF kN

16

INCREASE-DECREASE METHOD

This method is used to maximize a function when the live load consists a set of concentrated loads whose relative position is fixed. For example the forces exerted by the wheels of a truck or a train.

Position the set of loads on the structure so that the leading load is located at the maximum ordinate of the influence line.

Now shift the entire set of loads forward a distance x1 so that the second wheel is located at the maximum ordinate of the I.L. As a result of this shift the value of the function changes

The contribution of the first wheel F1 to the function decreases the contribution of the F2 , F3 , F4 increases because they have moved to a position where the ordinates are larger.

AB

1x 2x 3x

1F 4F3F2F

1x 2x 3x

1F 4F3F2F

1x 2x 3x

1F 4F3F2F

1position

2position

3position

1

2my1 1m

y ′

If the net change is a decrease in the value of the function, the first position of the loads is more critical than the second position.

The change in value of the function produced by the movement of a particular wheel equals the difference between the product of the wheel load and the ordinate of the I.L. in the two positions.

( )1 1 1 1 1 1 1f F y F y F y y F y F m x′ ′∆ = − = − = ∆ =

Using similar triangles 1

1 1

my

x

∆= Let m1 is the slope of the influence line in the region of the shift.

17

Example: The 24 m bridge in figure must be designed to support the wheel loads shown below. Using the increase-decrease method determine the maximum value of the moment at panel point B. The wheels can move in either direction.

A B C D E

4 * 6 m = 2434

14

1 kN

14

(135 135 90 90)* *3 337.5+ + + =

3 m 3 m1.5 1.5

45 kN 90 90 135 135

1 2 3 4 5

3 m 3 m1.5 1.5

45 kN 90 90 135 135

1 2 3 4 5

3 m 3 m1 .5 1 .5

45 kN 90 90 135 135

1 2 3 4 5

3 m 3 m1.5 1.5

45 kN 90 90 135 135

1 2 3 4 5

1 1

34

− 14

4.5 kNmInfluence line for

moment at B

Position 1

Position 2

Position 3

The change in moment as all the loads shift left 3 meters

Decrease in moment: 34

45*( )*3 101.25− = −

Increase in moment:

Net change 236.25=

Position 2 is more critical then position 1

Case I : The 45 kN load moves from right to left

18

14

(135 135 90)* *1.5 135+ + =

Shift the loads again to determine if the moment continues to increase. Load 3 moves up to point B.

Decrease in moment: 34

(45 90)*( )*1.5 151.875+ − = −

Increase in moment:

Net change 16.875= −

Position 2 is more critical then position 3

4.545*2.25 (90*18 90*16.5 135*13.5 135*12)*

18

1738.125

BM

kNm

= + + + + =

=

Case II : The 135 kN load moves from right to left

Begin with a 135 kN load at B

3 m 3 m1 .51.5

4590 kN90135135

1 1

34

− 14

4.5 kNm

3 m 3 m1.51.5

4590 kN90135135

Position 1

Position 2

14

(135 90 90 45)* *1.5 135+ + + =Increase in moment:

Net change 16.875= −

Decrease in moment: 34

135*( )*1.5 151.875− = −

Position 1 is more critical then position 2

4.5*(135*18 135*16.5 90*13.5 90*12 45*9)

18

1839.375

BM

kNm

= + + + + =

=

19

MAXIMUM SHEAR FORCE AND BENDING MOMENTUNDER A SERIES OF CONCENTRATED LOADS

Taking moment about A,RE × L = PR ×[L/2 - )]( xx −

( / 2 )RE

PR L x x

L= − +

a1 a2 a3

xPR= resultant load

a1 a2 a3

x

PR= resultant load

C.L.

xL/2

LRE

AB C D

E

P1P2 P3 P4

P1 P2P3 P4

RA

20

Taking moment about E,

2

2

02.,.

])2/()2/[(

)1)(2/()2/(0

0

)()()2/)(2/(

)()2/(

)2/(

)](2/[

22211

22211

xx

xx

xxei

xLxxLL

P

xLL

PxxL

L

P

dx

dM

aPaaPxLxxLL

P

aPaaPxLRM

xxLL

PR

xxLPLR

R

RR

D

R

AD

RA

RA

=

=

=−

−−−+=

−++−+=

=

×−+−+−+=

×−+−+×=

−+=

−+×=×

The centerline must divide the distance between the resultant of all the loads in the moving series of loads and the load considered under which maximum bending moment occurs.

The maximum value of shear (simply supported or continuous) typically occurs adjacent to a support. In a simply supported beam the shear at the end of a beam will be equal to the reaction; therefore, to maximize the shear, we position loads to maximize the reaction. The influence line for the for the reaction indicates that load should be placed as close to the support as possible and that the entire span should be loaded. If a simple beam carries a set of moving loads, the increase-decrease method can be used to establish the position of the loads on member to maximize the reaction.

Leet & Uang, Fundamentals of structural analysis, Mc Graw Hill 2002

21

Example Determine the absolute maximum moment and shear produced in the simply supported beam with a span of 20 m. by the wheel loads.

20m

24 kN 24 8

6m 3

2 4 2 4 8

6 m

56 kN

3

3 .8 6 2.14 m

24 24 8 56

0

24*6 8*9

3.86

F kN

M About left end

R x

x m

Σ = + + =

Σ =

+ =

=

i

Assume that the maximum moment occurs under the left end force. Position the loads as the beam’s centerline divides the distance between the 24 kN load and the resultant.

8 .0 7 m

2 4 2 4 85 6 kN

3 .8 6 2.14 m

22.604

2 0 m

2 4 2 4 85 6 kN

3 .8 6 2.14 m

31

0 22 11.93 24 5.93 8 2.93 20

22.604

max 22.604*8.07 182.41

BM A

A kN

absolute M kNm

Σ = + + =

=

= =

i i i

Assume that the maximum moment occurs under the middle 24 kN force. Position the loads as the centerline of the beam is located halfway between the middle 24 kNload and the resultant.

0 24 14.93 24 8.93 8 5.93 20

31.004

max 31*11.07 24*6 224.89

BM A

A kN

absolute M kNm

Σ = + + =

=

= − =

i i i

A B

A B

22

-Compute the absolute maximum shear and moment produced in the simply supported beam by two concentrated live loads of 16 kips spaced 12 ft apart. The beam spans 24 ft.

Answer : S(max)=24 kips; M(max)=108 kip-ft

-Compute the absolute maximum value of live load shear and moment produced in a simply supported beam spanning 40 ft by the wheel loads shown below.

Answer : S(max)=43.2 kips; M(max)=365.4 kip-ft

24 kips 24 kips6 kips

12′ 12′

Leet & Uang, Fundamentals of structural analysis, Mc Graw Hill 2002. pages 289-290

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24

25

MÜHENDĐSLĐK MĐMARLIK FAKÜLTESĐ

Đnşaat Mühendisliği Bölümü Lisans Programı

ĐNŞ 313: Structural Analysis I (Yapı Statiği I)

(ECTS:5 Ders yükü:3+0)

Genel Amaç:

Dersin amacı, izostatik yapıların dış yükler etkisinde mesnet reaksiyonları, kesit zorları ve belirli noktalarındaki eğim ve yer değiştirme gibi performans karakteristiklerinin belirlenmesidir.

Öğrenme Çıktıları ve Alt Beceriler

Ders sonunda öğrenci,

1- Basit ve çıkmalı kirişlerle çerçevelerde mesnet reaksiyonlarını hesaplayabilir. Bu yapıtiplerinin kesit zorlarına ait fonksiyonları belirleyerek diyagramlarını çizebilir.

1.1. Denge denklemlerini kolaylıkla yazar, el ve/veya bilgisayar programı yardımıyla çözer.

1.2. Kesim yöntemini kullanır.

1.3. Đntegrasyon yöntemini kullanır.

1.4. Diferansiyel denge denklemlerinin fiziksel anlamından hareketle diyagram çizer.

2- Üç mafsallı kemerlerin kesit zorlarını hesaplayabilir.

2.1 Kesit zorlarını formülize eder el ve/veya bilgisayar yardımıyla sonuçlara ulaşır.

3- Kafes kirişlerin çubuk kuvvetlerini hesaplayabilir.

3.1 Kafes kirişleri sınıflandırır.

3.2 Basit, bileşik ve karmaşık kafes kirişlerde çubuk kuvvetlerini hesaplar.

3.3 Basit kafes kirişlerde deplasman diyagramı çizer.

3.4 Đş-enerji yöntemi ile yer değiştirme bulur.

4- Tesir çizgileri çizebilir.

4.1. Basit ve çıkmalı kirişlerle kafes kirişlerde tesir çizgisi çizer.

4.2 Müller-Breslau ilkesini kullanır.

5- Kirişlerde eğim ve yer değiştirme hesaplayabilir.

5.1. Moment-Area yöntemini kullanır.

5.2. Eşlenik kiriş yöntemini kullanır.