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Influence Lines Page 1 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.
Influence Lines Page 2 of 14 1/27/2006
• 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
Influence Lines Page 3 of 14 1/27/2006
-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.
Influence Lines Page 4 of 14 1/27/2006
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.
Influence Lines Page 5 of 14 1/27/2006
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.
Influence Lines Page 6 of 14 1/27/2006
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.
Influence Lines Page 7 of 14 1/27/2006
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
Influence Lines Page 8 of 14 1/27/2006
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
Influence Lines Page 9 of 14 1/27/2006
-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
Influence Lines Page 10 of 14 1/27/2006
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
2−+−++−= xRxRxRM
dxvdEI CBAAB
1
20
22
214
8128
2C
xRxEI
xRxRxM
dxdvEI CBAAB +
−+−−
−++−=
21
3332
614
868
62CxC
xRxEI
xRxRxMEIv CBAAB ++
−+−−
−++−=
x = 0, v = 0 therefore C2 = 0
x = 0 , 0=dxdv therefore C1 = 0
614
06051,0868
1030,06
04955,02
1322,03332 −
+−−−
−+−=x
xEIx
EIxEIxEIEIv
Influence Lines Page 11 of 14 1/27/2006
-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
2−+−++−= xRxRxRM
dxvdEI CBAAB
1
222
214
28
2C
xR
xRxRxM
dxdvEI CBAAB +
−+
−++−=
21
33032
614
68
462
CxCx
Rx
RxEIxRxMEIv CBAAB ++−
+−
+−++−=
x = 0, v = 0 therefore C2 = 0
x = 0 , 0=dxdv therefore C1 = 0
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)
Influence Lines Page 12 of 14 1/27/2006
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
Influence Lines Page 13 of 14 1/27/2006
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
2−+−++−= xRxRxRM
dxvdEI CBAAB
1
222
214
28
2C
xR
xRxRxM
dxdvEI CBAAB +
−+
−++−=
21
33032
614
68
862
CxCx
Rx
RxEIxRxMEIv CBAAB ++−
+−
+−++−=
x = 0, v = 0 therefore C2 = 0
x = 0 , 0=dxdv therefore C1 = 0
614
0113502,068
0251695,086
015146,02
071641,033
032 −+
−−−++−=
xxEIxEIxEIxEIEIv
Influence Lines Page 14 of 14 1/27/2006
-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