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
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•
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
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-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) f) g) h)
The shear to the left of A, will be –1,0 x 10 = -10 kN. Shear at D, Load just to the right of D, shear = 0,7 x 10 = 7 kN Moment at A, load at C, moment = -5 x 10 = -50 kN.m 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.
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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
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L + x − d 2
R R MY =
L
=
R Rx Rd 2
+
L
−
L
Calculate the moment under the load B L M B = R MY − x − A ⋅ a 2
R Rx Rd L = + − ⋅ − x − A ⋅ a 2 L L 2 =
RL Rd Rx 2 4
−
2
−
L
+
Rxd L
− A ⋅ a
For the maximum value of M B, dM B dx
=−
2 Rx Rd L
+
L
=0
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. x =
30 * 2 + 20 * 6 + 10 * 8 20 + 30 + 20 + 10
= 3,25 m
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
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Rleft *16 − R * (8 − 0,625) = 0 Rleft =
80 * 7,375 16
= 36,875 kN
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 patte rn loading should be applied to obtain maximum values for each of the cases.
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Example 7: Determine the equations for the influence lines for moment M AB 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, R A, R B, R C and R D Any number of methods may be used to determine the unknown forces. Use slope-deflection equations to solve M A, etc. M AB =
2 EI
M BA =
2 EI
M BC =
2 EI
M CB = M CD =
(2 ⋅ −1 + θ B ) = −0,5 EI + 0,25EI θ B
8 8
6 2 EI 6 3 EI 8
(− 1 + 2θ B ) = −0,25 EI + 0,5 EI θ B (2θ B + θ C ) = 0,6667 EI θ B + 0,3333EI θ C (θ B + 2θ C ) = 0,3333 EI θ B + 0,66667 EI θ C (θ C ) = 0,375EI θ C
ΣM B = 0 ∴ M BA + M BC = 0
1,16667 θ B + 0,33333θ C =+ 0,25
(1)
ΣM C = 0 ∴ M CB + M CD = 0
0,3333θ B + 1,04167 θ C = 0
(2)
θB = +0,235848 θC = -0,07547 MAB = -0,44104 EI Influence Lines
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MBA = -0,13208 EI MBC = +0,13208 EI MCB = +0,02830 EI MCD = -0,02830 EI VAB =
M AB + M BA 8
VBA = − VBC =
VCD =
M AB + M BA
8 M BC + M CB
6
VCB = −
= −0,07164 EI
= +0,02673 EI
M BC + M CB 6
M CD 8
= +0,07164 EI
= −0,02673 EI
= −0,0035375 EI
R A = VAB = - 0,07164 EI R B = VBA + VBC = +0,09837 EI R C = VCB + VCD = -0,0302675 EI For the influence line, use equation for deflected shape. EI
EI
d 2 v dx 2 dv dx
= − M AB + R A x + R B x − 8 + RC x − 14
= − M AB x + R A
EIv = − M AB
x 2 2
+ R A
x 2 2
x 3 6
+ R B + R B
x − 8
2
+ RC
2
x − 8 6
3
+ RC
x − 14
2
+ C 1
2
x − 14
3
6
+ C 1 x + C 2
x = 0, v = 0 therefore C 2 = 0 x=0,
dv dx
= −1,0 therefore C1 = -1,0 EI
x − 8 x 2 x 3 EIv = +0,44104 EI − 0,07164 EI + 0,0938 EI 2 6 6
Influence Lines
3
− 0,03027 EI
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x − 14 6
3
− EIx
1/27/2006
0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1.0 -1.1 -1.2 -1.3 -1.4 0
1
2
3
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5
6
7
8
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20
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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: 0 M BC = 0 M CB =
2 EI L 2 EI L
(2θ B + θ C ) = (θ B + 2θ C ) =
2 EI 6 2 EI 6
(2 ⋅ −1) = −0,6667 EI (− 1) = −0,3333EI
Stiffness at B k BA =
4 EI
k BC =
4 EI
8
6
0,5
= 0,5
D BA =
= 0,6667
D BC =
0,6667
= 0,6667
DCB =
0,6667
= 0,375
DCD =
1,1667
= 0,42856
1,1667
= 0,57144
Stiffness at C k CB =
4 EI
k CD =
3EI
6 8
Influence Lines
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1,04167 0,375 1,04167
= 0,640 = 0,360
1/27/2006
MAB
MBA 0,4286
+0,1429
+0,2858
-0,0098
VBC =
M AB + M BA 8
6
VCD =
M CD 8
-0,2642 EI
-0,0565 EI
+0,0565 EI
+0,0047 +0,0004
= −0,04955 EI
= −0,05345 EI
M BC + M CB 6
MCD 0,360
= +0,04955 EI
M AB + M BA
8 M BC + M CB
VCB = −
+0,2642 EI
-0,0018
+0,1322EI
VBA = −
+0,0514
-0,0196
-0,0009
VAB =
MCB 0,640 -0,3333 +0,1905 +0,0914 -0,0130 +0,0083 -0,0011 +0,0007
-0,0002
MBC 0,5714 -0,6667 +0,3809 +0,0457 -0,0261 +0,0041 -0,0023 +0,0004 -0,0002
= +0,05345 EI
= 0,00706 EI
R A = VAB = + 0,049555 EI R B = VBA + VBC = -0,1030 EI R C = VCB + VCD = 0,06051 EI EI
d 2 v dx
EI
dv dx
2
= − M AB + R A x + R B x − 8 + RC x − 14
= − M AB x + R A
EIv = − M AB
x 2 2
+ R A
x 2 2
x 3 6
+ R B + R B
x − 8
2
− 1 EI x − 8
2
x − 8 6
0
+ RC
3
− EI x − 8 + RC
x − 14 2
x − 14
2
+ C 1
3
6
+ C 1 x + C 2
x = 0, v = 0 therefore C 2 = 0 x=0,
dv dx
= 0 therefore C1 = 0
x − 8 x 2 x 3 EIv = −0,1322 EI + 0,04955 EI − 0,1030 EI 2 6 6
Influence Lines
3
− EI x − 8 + 0,06051
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x − 14
3
6
1/27/2006
0.4 0.3 0.2 0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 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: EI
d 2 v dx
EI
dv dx
2
= − M AB + R A x + R B x − 8 + RC x − 14
= − M AB x + R A
EIv = − M AB
x 2 2
+ R A
x 2 2
x 3 6
+ R B
x − 8 2
+ EI x − 4
0
2
+ RC + R B
x − 14 2
x − 8 6
2
+ C 1
3
+ RC
x − 14 6
3
+ C 1 x + C 2
x = 0, v = 0 therefore C 2 = 0 x=0,
dv dx
= 0 therefore C1 = 0
x = 8, v = 0 -32 MA + 85,333 R A +1,0 EI = 0
(1)
x = 14, v = 0 -98 MA + 457,333 R A + 36 R B +1,0 EI = 0
(2)
x = 22, v = 0 -242 MA + 1774,6667 R A + 457,333 R B + 85,3333 R C + 1,0 EI = 0
(3)
Take moments about R D -1 MA + 22 R A + 14 R B + 8 R C = 0
(4)
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MA R A R B R C
= 0,071639 EI = 0,015146 EI = -0,02517 EI = 0,01135 EI
x − 8 x 2 x 3 0 EIv = −0,071639 EI + 0,015146 EI + EI x − 4 + −0,02517 EI 2 6 6
3
+ 0,01135 EI
x − 14
3
6
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
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22
Influence line for the shear force at E
In this case use super-position to solve the unknown forces. Remove the unknown reactions R B, R C and R D.
∆ P = ∆ PS + P δ PP + Qδ PQ + Rδ PR = 0 ∆ Q = ∆ QS + P δ PQ + Qδ QQ + Rδ QR = 0 Influence Lines
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∆ R = ∆ RS + P δ PR + Qδ QR + Rδ RR = 0 h
δ PP =
8
3 h
δ QQ
=
δ RR
=
h
δ PQ
=
h
δ PR
=
3
6 h h
⋅ a1a 2 = ⋅ a1a 2 =
3
6
⋅ a1a 2 = ⋅ 8 ⋅ 8 = 170,6667 3 14 3 22 3
⋅14 ⋅14 = 914,6667 ⋅ 22 ⋅ 22 = 3549.3333 8
⋅ a1 (2a 2 + b2 ) =
6 8
⋅ a1 (2a 2 + b2 ) =
⋅ 8 ⋅ (2 * 22 + 14) = 618,6667
14
⋅14 ⋅ (2 * 22 + 8) = 1698,6667 6 6 1 + 170,6667 P + 362,6667Q + 618,6667 R = 0 δ QR
=
⋅ a1 (2a 2 + b2 ) =
6
⋅ 8 ⋅ (2 *14 + 6) = 362,6667
1 + 362,6667 P + 914,6667Q + 1698,6667 R = 0 1 + 618,6667 P + 1698,6667Q + 3549,3333R = 0
P = - 0,0251695EI Q = 0,0113502EI R = - 0,0013267EI R A = 0,015146EI MA = 0,071641EI Function for the influence line: EI
EI
d 2 v dx 2 dv dx
= − M AB + R A x + R B x − 8 + RC x − 14
= − M AB x + R A
EIv = − M AB
x 2 2
+ R A
x 2 2
x 3 6
+ R B
x − 8 2
+ EI x − 8
0
2
+ RC + R B
x − 14 2
x − 8 6
2
+ C 1
3
+ RC
x − 14 6
3
+ C 1 x + C 2
x = 0, v = 0 therefore C 2 = 0 x=0,
dv dx
= 0 therefore C1 = 0
x − 8 x 2 x 3 0 EIv = −0,071641 EI + 0,015146 EI + EI x − 8 − 0,0251695 EI 2 6 6
Influence Lines
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3
+ 0,0113502
x − 14
3
6
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0.2 0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1 0
1
2
3
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5
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7
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Influence line for the shear force to the left of B
Influence Lines
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