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Structural Analysis CE2100Influence Lines11Influence Lines:Moment and shear force diagrams represent variation of these values along the length of the beam for a set of stationary loads.

Influence lines represent variation in the response* at a specific point due to a unit concentrated load moving along the beam.

*Response may mean shear force, bending moment, reaction or deflection.2Example 1:Influence line for R1.R1 + R2 = 1R1. L = 1.(L-x)R1 = 1(1-x/L)

Plot x v/s R1 (Upward force as positive).3P=1R1R2x1X=LX=0Example 2:Influence line for shear force at mid-span (C).R1 = (1-x/L)When load is acting on the left side of mid-span (x < L/2):Vc = 1-x/L -1 = -x/L

When load is acting on the left side of mid-span (x > L/2):Vc = 1-x/L

Plot x v/s Vc 41R1R2xC

+ve shear0.5X=0X=L-0.5Example 3:Influence line for bending moment at mid-span (C).R1 = (1-x/L); R2 = x/LWhen load is acting on the left side of mid-span (x < L/2):Mc = (x/L)(L/2) = x/2

When load is acting on the left side of mid-span (x > L/2):Mc = (1-x/L)(L/2) = 0.5(L-x)

Plot x v/s Mc 51R1R2xCX=0X=LL/4Influence line and distributed loadUnit concentrated load can be substituted by dF = w(dx), acting over a length dx at location x.

Response at C due to dF is given by (ordinate of influence line at x) (dF).

If we apply dF=w(dx) all along the beam, the response will be given by (area under the influence line)*(w).

w*Area under the influence line = w*(L/4)(L/2) = wL2/8

Bending moment at mid span due to uniformly distributed load = wL2/8

61R1R2xCX=0X=LL/4Influence line for bending moment at mid-span (C).Qualitative assessmentHeinrich Muller-Breslau (1886)Influence line for a function has the same shape as thedeflected shape of the beam when the beam is acted upon by the function.

7R1R2X=LX=0Influence line for R1.R1R2C0.5X=0X=L-0.5Influence line for Vc.Qualitative assessmentHeinrich Muller-Breslau (1886)8X=0X=LInfluence line for Vc.CCInfluence line for Mc.CCX=0X=LSeries of concentrated loads9

L/32/3L0.15L0.15L

0.15LInfluence line for shear force at C.Influence line for a function can tell us the maximum value of that function as a concentrated load moves along the member length.

For a series of concentrated loads:Net influence line can be developed by superimposing influence lines for individual loadsSeries of concentrated loads10

L/32L/30.15L0.15L

0.15LFor a series of concentrated loads:Response history can be developed by multiplying the ordinates of influence lines for individual loads and superimposing them.

Since the individual influence lines are multi-linear in nature, the net response history will be multi-linear as well. Therefore the maxima or minima will be located at one of the corners, i.e., when one of the load passes the point of interest (C).

Example: Shear force at a point due to a series of concentrated loads111(0.75)+4(0.625)+4(0.5)=5.25k

Or

1(-0.25)+4(0.625)+4(0.5) = 4.25 k

Example:continued.121(-0.125)+4(0.75)+4(0.625)=5.375k

Or

1(-0.125)+4(-0.25)+4(0.625) = 1.375k

1(0)+4(-0.125)+4(0.75)=2.5k

Or

1(0)+4(-0.125)+4(-0.25) = -1.5k

Therefore, maximum shear at C is equal to 5.375 k

Example: Bending moment at a point due to a series of concentrated loads13

Influence line for bending moment at C2(7.5)+4(6.5)+3(5.0)=56 k.ft

Or

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6.55.014 20

Example: continued14Influence line for bending moment at C2(4.5)+4(7.5)+3(6.0)=57 k.ft

2(0)+4(3.0)+3(7.5)=34.5 k.ft

Therefore, maximum moment at C is equal to 57.0 k.ft3.0414Absolute maximum due to a series of concentrated loads15Assuming that only vertically downward concentrated loads are actingCantilever beam: Maximum shear and maximum bending moments are always located near the support.

Simply supported beam:Maximum shear always takes place near one of the supports.Maximum bending moment will takes place under one of the loads in the series. Find the absolute maximum for each individual load and compare.How to find the location of the series that produces maximum bending moment under a particular load?

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Consider a special case when F2 and FR coincide.Example: Absolute maximum bending moment18

Resultant: FR = 4.5 k6.67 ft from F1

Configuration 1: Moment under load F1: Position the series of loads so that F1 and FR are equidistant from the beams centerline.Reaction Ay=FR (11.67)/30 =1.7505 kMaximum Moment under F1 = Ay. (15-3.33) = 20.42 k.ft

Example: Absolute maximum bending moment19Configuration 2: Moment under load F2: Position the series of loads so that F2 and FR are equidistant from the beams centerline.

Reaction Ay=FR (15+1.67)/30 =2.5 kMaximum Moment under F2 = Ay. (15+1.67) F1(10) = 21.675 k.ft

Configuration 3:Similarly check for moment under load F3.

Configuration 2 generates the absolute maximum bending moment = 21.67 k.ft