Structure Analysis IStructure Analysis IChapter 6Chapter 6

Influence Line for Statically yDeterminate Structures

Influence linesInfluence lines

• Influence lines provide a systematic procedure of how force in a given part ofprocedure of how force in a given part of structure varies as the applied loads moves l halong the structure

Procedure for AnalysisProcedure for Analysis• Place a unit load at various locations x along thePlace a unit load at various locations, x along the

member, and at each location use static to determine the value of the function ( Reaction, Shear, Moment) at h ifi d ithe specified point.

If h IL f i l f i i• If the IL for a vertical force reaction at a point on a beam is to be constructed, consider the reaction to be positive at the point when it acts upward on the beampositive at the point when it acts upward on the beam

• If a shear or moment IL is to be drawn for a point, takeIf a shear or moment IL is to be drawn for a point, take the shear or moment at the point as +ve according to the same sign convention used for drawing shear and

dimoment diagram.

• All statically determinate beams will have IL that consist of straight line segments. After some practice g g pone should be able to minimize computations and locate the unit load only at points representing the end y p p gpoints of each line segment

Example 1Example 1Draw the IL for the Reaction at A

IL Equation

M B 0=∑

xA

xAy

B

1011

0)1)(10()10(

−=

=−+−∑

xAy 101

Example 2Example 2Draw the IL for the Reaction at B

Example 3Example 3Draw the IL for the Shear at C

Load from A to C

Load > C

Example 4Example 4Draw the IL for the Moment at C

Load from A to C

Load > C

Example 1Example 1Draw the IL for the Shear & Moment at B

Example 2Example 2Draw the IL for the Shear & Moment at C

0.5

Example 3Example 3Internal Hinge ExampleDetermine the shear & moment at point D

Example 3 ContinueExample 3-Continue

Example 4Example 4Determine the moment at C in the two cases

Case 1

cMLI .

mkNM c . 5.32)5.2(10)25.1(6 =+=

Case 2

MLI cMLI.

mkNM c . 5.27)25.1(10)5.2(6 =+=

Qualitative Influence LinesQualitative Influence Lines

Example 5Example 5Sketch the influence line for the vertical reaction at A

Example 6Example 6Sketch the influence line for the shear at B

Example 6-bExample 6 bSketch the influence line for the moment at B

Influence Line for Floor GirdersInfluence Line for Floor Girders

Example 7Draw the influence line for the shear at panel CD

Example 7

Example 8Draw the influence line for the moment at F

Example 8

Example 9Example 9Truss ExampleDetermine The force in member GB

EGB

EGB

RFRF

41.145cos

==

df h

AGB RF 45cospartsecondfor the

−=

AGB RF 41.1−=

Example 10Example 10Truss ExampleDetermine The force in member GF, BF

( )0 40 17.32.330 1 15

B D GF

GF D

M R FF RF R F R

= ⇒ = −

= −∑

cos30 1.15BF D BF DF R F R= →→ =

( )0 20 17.31 15

F A GFM R FF R

= ⇒ = −∑1.15

cos30 1.15GF A

BF A BF A

F RF R F R

= −= − →→ = −

Example 11pTruss ExampleDetermine The force in member CG

Example 11p

IL due to Series of Concentrated LoadsConcentrated Loads

( )1 1(0.75) 4(0.625) 4(0.5) 5.25CV k= + + =

( )21( 0.125) 4(0.75) 4(0.625) 5.375CV k= − + + =( )2( ) ( ) ( )C

( )31(0) 4( 1.25) 4(0.75) 2.5CV k= + − + =( )3( ) ( ) ( )C

( ) 2(7 5) 4(6 5) 3(5 0) 56 0M k ft+ +( )( )( )

1

2

2(7.5) 4(6.5) 3(5.0) 56.0 .

2(4.5) 4(7.5) 3(6.0) 57.0 .C

C

M k ft

M k ft

= + + =

= + + =

( )32(0) 4(3.0) 3(7.5) 34.5 .CM k ft= + + =

Example 12p

Example 13p

( )max8(1.2) 3(0.4) 10.8 .BM kN m= + =

Example 14p

Absolute Maximum Shear and Moment

Simply Supported Beamp pp

The location of Maximum Moment is at

2xx =

Example 15p