influence lines

Influence Lines of 14 1/27/2006

Influence Lines Definition An influence line is a graphical representation of a function that describes a force (axial, shear, moment) at a point due to a unit force anywhere on the member. Example 1: Draw the influence lines for reaction at A, Shear at B and moment at B of the following member.

When the unit load is at A the reaction will be equal to 1,0 and when it is at B the reaction will be equal to 0. The reduction in the reaction is linear with the distance away from A. When the unit load is to the left of B, the shear force at B is equal to the reaction at A minus the unit load, hence negative. When the unit load is to the right of B, the shear force is equal to the reaction at A. For the bending moment at B, when the unit load is at A the reaction at C is equal to 0 with the lever arm equal to 4 m. The bending moment at B is then equal to 0. As the load moves towards B the reaction at C increase linearly until the load is at B, then the reaction at C is equal to 0,3333 kN. Bending moment at B is 0,333 x 4 m = 1,333 kN.m. As the unit load moves past B, the bending moment is equal to the reaction at A times 2 m. Unit load at C, reaction at A = 0. Bending moment = 0. Use of influence lines.

• To obtain the maximum value of a function due to a single concentrated live load, the live load should be placed where the ordinate to the influence line is a maximum.

• The value of the function due to the action of a single live point load equals the product of

the magnitude of the live load and the value of the ordinate at that point.


• To obtain the maximum value of a function due to a uniformly distributed live load, the load should be placed over all those portions of the structure for which the ordinates to the influence line have the same sign.

• The value of a function due to a uniformly distributed live load is equal to the product of the

loading and the net area under the portion of the influence line covered by the load. Example 2: For the following structure draw the influence lines for,

a) The shear to the left of A, b) Shear at D, c) Moment at A d) Moment at D

Shear to the left of A: The shear stays constant at –1,0 until the unit load reaches A. Thereafter there is no shear force in the section CA. Shear at D: When the unit load is at C the reaction at A is equal to 1,5. The shear at D is then equal to –1,0 + 1,5 = +0,5. When the unit load is at A the reaction at A is equal to 1,0 with no shear force between A and B. When the unit load is just to the left of D, the reaction at A is equal to 0,7 with the shear at D equal to 0,7 – 1,0 = -0,3. As the load moves to the right of D the shear force at D is equal to the reaction at A = 0,7. Moment at A: When the load is at C, the moment at A is equal to –5,0. When the load is at A the moment is equal to 0. Moment at D: When the unit load is at C the reaction at B is –0,5 and the moment at D is equal to


-0,5 x 7 = -3,5. When the load is at A the reaction is 0 and the moment at D is then also 0. When the load is at D the reaction at at B is 1*3/10 = 0,3 and the moment at D is equal to 0,3 x 7 = 2,1. When the load is at B the moment at D will be equal to 0. If a 10 kN load moves between C and B, the following maximum values will occur.

e) The shear to the left of A, will be –1,0 x 10 = -10 kN. f) Shear at D, Load just to the right of D, shear = 0,7 x 10 = 7 kN g) Moment at A, load at C, moment = -5 x 10 = -50 kN.m h) Moment at D, load at C, moment = -3,5 x 10 = -35,00 kN.m, load at D moment = 2,1 x 10 =

21 kN.m. Alternative method of construction influence lines. Another way of constructing influence lines is to use the Müller-Breslau method: The ordinates of the influence line for any stress element (such as axial force, shear moment or reaction) of any structure are proportional to those of the deflection curve, which is obtained by removing the constraint corresponding to the that element from the structure and introducing in its place a corresponding deformation into the primary structure that remains. In the case of indeterminate structures this method is restricted to structures the material of which is elastic and follows Hooke’s law. Example 3: Use the Müller-Breslau method and determine the influence line for the reaction at A, reaction at C, shear at B, moment at B, shear to left of C and shear to right of C.


One can also construct influence for more complex statically determinate structures. Example 4: The unit load moves across DEFGH. Construct the influence lines for the reaction at A, and the moment at B.

In each case construct the influence line for the member EABC. Member DEF rests on the support at D and on the deflected member EABC at E. Draw in the member DEF so that it is supported at D and E on the deflected member EABC. In a similar fashion, member FGH rests on member EABC at G an H. Draw the member so that it is supported at G and H on the deflected shape. Calculate the coordinates of member DEF and FGH. As a self study construct the influence lines for the shear force to the left of A, shear force to the right of A, shear force at B and the bending moment at G. Absolute Maximum Moment as a Result of Live Point Loads. Consider a simply supported beam that carries a series of moving live loads. One is aware that the maximum moment will occur in the region of the centre of the member. The maximum moment for point loads will occur under one of the point loads. Two questions must be answered:

a) Under which load does absolute maximum live moment occur b) What is the position of this load when maximum moment occurs.


LRd

LRxR

L

dxLRRMY −+=

−+

=2

2

Calculate the moment under the load B

aAxLRM MYB ⋅−

−=

2 aAxL

LRd

LRxR

⋅−

−⋅

−+=

22

aAL

RxdL

RxRdRL⋅−+−−=

2

24

For the maximum value of MB,

02=+−=

LRd

LRx

dxdM B whence x = d/2

The maximum moment directly beneath one of a series of concentrated live loads that are applied to a simply supported beam occurs when the centre of the span is halfway between that particular load and the resultant of all the loads on the span. Example 5: Determine the absolute maximum moment that will occur as a result of the four point loads that can move across the beam.

The position of the resultant force, relative to the first 20 kN load, can be calculated by taking moments about the first 20 kN force.

mx 25,310203020

8*106*202*30=

+++++

=

Position the resultant so that the centre line of the beam falls between the resultant and the 30 kN load, ie, x = d/2 with x = 0,625 m. The left reaction can be calculated from taking moments about N.


kNR

RR

left

left

875,3616

375,7*80

0)625,08(*16*

==

=−−

M under the 30 kN load = 36,875 * 7,375 – 20 * 2 = 231,953 kN.m Move the loads so that the resultant load falls on the left hand side and that the centre line falls between the resultant and the second 20 kN load. The resultant load should now be 1,375 m to the left of the centre line. The right reaction is now equal to 33,125 kN and the moment under the 20 kN load = 199,453 kN.m. The maximum moment will thus occur when the 30 kN load is 0,625 m to the left of the centre line of the beam. Influence lines for statically indeterminate structures The Müller-Breslau method may be used to draw the shape of the influence line. The influence line gives a good indication of where live loads should be applied to obtain the maximum value for the forces in the structure. Various methods may be used to determine the actual ordinates of the deflected shape. Example 6: Draw influence lines for, reaction at A, shear at E, Shear to left of B and moment at B of the following continuous beam and show where pattern loading should be applied to obtain maximum values for each of the cases.


Example 7: Determine the equations for the influence lines for moment MAB and MBC, shear at E and to left of B of the following statically indeterminate beam. The load moves from A to D.

Moment at A: Unknown forces are, MA, RA, RB, RC and RD Any number of methods may be used to determine the unknown forces. Use slope-deflection equations to solve MA, etc.

( ) BBAB EIEIEIM θθ 25,05,0128

2+−=+−⋅=

( ) BBBA EIEIEIM θθ 5,025,0218

2+−=+−=

( ) CBCBBC EIEIEIM θθθθ 3333,06667,026

2+=+=

( ) CBCBCB EIEIEIM θθθθ 66667,03333,026

2+=+=

( ) CCCD EIEIM θθ 375,08

3==

00 =+∴=Σ BCBAB MMM 1,16667θB + 0,33333θC =+ 0,25 (1)

00 =+∴=Σ CDCBC MMM 0,3333θB + 1,04167θC = 0 (2) θB = +0,235848 θC = -0,07547 MAB = -0,44104 EI


MBA = -0,13208 EI MBC = +0,13208 EI MCB = +0,02830 EI MCD = -0,02830 EI

VAB = EIMM BAAB 07164,0

8−=

+

VBA = EIMM BAAB 07164,0

8+=

+−

VBC = EIMM CBBC 02673,0

6+=

+

VCB = EIMM CBBC 02673,0

6−=

+−

VCD = EIM CD 0035375,0

8−=

RA = VAB = - 0,07164 EI RB = VBA + VBC = +0,09837 EI RC = VCB + VCD = -0,0302675 EI For the influence line, use equation for deflected shape.

1482

2−+−++−= xRxRxRM

dxvdEI CBAAB

1

222

214

28

2C

xR

xRxRxM

dxdvEI CBAAB +

−+

−++−=

21

3332

614

68

62CxC

xR

xRxRxMEIv CBAAB ++

−+

−++−=

x = 0, v = 0 therefore C2 = 0

x = 0 , 0,1−=dxdv therefore C1 = -1,0 EI

EIxx

EIx

EIxEIxEIEIv −−

−−

+−+=614

03027,068

0938,06

07164,02

44104,03332


-1.4 -1.3 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Influence line for the moment at A. Influence line for moment at B: Use moment distribution to obtain the moments and the reactions. Apply the unit rotation and the beam will have the following deflected shape prior to moment distribution.

Initial moments:

( ) ( ) EIEILEIM CBBC 6667,012

62220 −=−⋅=+= θθ

( ) ( ) EIEILEIM CBCB 3333,01

62220 −=−=+= θθ

Stiffness at B

5,08

4==

EIk BA 42856,01667,1

5,0==BAD

6667,06

4==

EIkBC 57144,01667,16667,0

==BCD

Stiffness at C

6667,06

4==

EIkCB 640,004167,16667,0

==CBD

375,08

3==

EIkCD 360,004167,1

375,0==CDD


MAB MBA MBC MCB MCD

0,4286 0,5714 0,640 0,360 -0,6667 -0,3333

+0,1429 +0,2858 +0,3809 +0,1905 +0,0457 +0,0914 +0,0514

-0,0098 -0,0196 -0,0261 -0,0130 +0,0041 +0,0083 +0,0047

-0,0009 -0,0018 -0,0023 -0,0011 +0,0004 +0,0007 +0,0004 -0,0002 -0,0002

+0,1322EI +0,2642 EI -0,2642 EI -0,0565 EI +0,0565 EI

VAB = EIMM BAAB 04955,0

8+=

+

VBA = EIMM BAAB 04955,0

8−=

+−

VBC = EIMM CBBC 05345,0

6−=

+

VCB = EIMM CBBC 05345,0

6+=

+−

VCD = EIM CD 00706,0

8=

RA = VAB = + 0,049555 EI RB = VBA + VBC = -0,1030 EI RC = VCB + VCD = 0,06051 EI

1482


dxvdEI CBAAB

1

20

22

214

8128

2C

xRxEI

xRxRxM

dxdvEI CBAAB +

−+−−

−++−=

21

3332

614

868

62CxC

xRxEI

xRxRxMEIv CBAAB ++

−+−−

−++−=


x = 0 , 0=dxdv therefore C1 = 0

614

06051,0868

1030,06

04955,02

1322,03332 −

+−−−

−+−=x

xEIx

EIxEIxEIEIv


-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 2 4 6 8 10 12 14 16 18 20 22

Influence line for moment at B Influence line for the shear force at E. Use McCauley notation:

1482


dxvdEI CBAAB

1

222

214

28

2C

xR

xRxRxM

dxdvEI CBAAB +

−+

−++−=

21

33032

614

68

462

CxCx

Rx

RxEIxRxMEIv CBAAB ++−

+−

+−++−=



x = 8, v = 0 -32 MA + 85,333 RA +1,0 EI = 0 (1) x = 14, v = 0 -98 MA + 457,333 RA + 36 RB +1,0 EI = 0 (2) x = 22, v = 0 -242 MA + 1774,6667 RA + 457,333 RB + 85,3333 RC + 1,0 EI = 0 (3) Take moments about RD -1 MA + 22 RA + 14 RB + 8 RC = 0 (4)


MA = 0,071639 EI RA = 0,015146 EI RB = -0,02517 EI RC = 0,01135 EI

614

01135,068

02517,046

015146,02

071639,033

032 −+

−−+−++−=

xEI

xEIxEIxEIxEIEIv

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Influence line for the shear force at E In this case use super-position to solve the unknown forces. Remove the unknown reactions RB, RC and RD.

0=+++∆=∆ PRPQPPPSP RQP δδδ 0=+++∆=∆ QRQQPQQSQ RQP δδδ


0=+++∆=∆ RRQRPRRSR RQP δδδ

6667,1708838

3 21 =⋅⋅=⋅= aahPPδ

6667,91414143

143 21 =⋅⋅=⋅= aah

QQδ

3333.354922223

223 21 =⋅⋅=⋅= aah

RRδ

( ) ( ) 6667,362614*28682

6 221 =+⋅⋅=+⋅= baahPQδ

( ) ( ) 6667,6181422*28682

6 221 =+⋅⋅=+⋅= baahPRδ

( ) ( ) 6667,1698822*2146

1426 221 =+⋅⋅=+⋅= baah

QRδ

06667,6186667,3626667,1701 =+++ RQP 06667,16986667,9146667,3621 =+++ RQP 03333,35496667,16986667,6181 =+++ RQP

P = - 0,0251695EI Q = 0,0113502EI R = - 0,0013267EI RA = 0,015146EI MA = 0,071641EI Function for the influence line:

1482


dxvdEI CBAAB

1

222

214

28

2C

xR

xRxRxM

dxdvEI CBAAB +

−+

−++−=

21

33032

614

68

862

CxCx

Rx

RxEIxRxMEIv CBAAB ++−

+−

+−++−=



614

0113502,068

0251695,086

015146,02

071641,033

032 −+

−−−++−=

xxEIxEIxEIxEIEIv


-1

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Influence line for the shear force to the left of B

influence lines

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