Structure Analysis IChapter 6

Structure Analysis IChapter 6

Influence Line for Statically Determinate Structures

Influence lines

Influence lines provide a systematic procedure

of how force in a given part of structure varies

as the applied loads moves along the structure

Procedure for Analysis

• Place 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 the specified point.

• 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 beam

• If 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 moment diagram.

Structural Analysis IDr. Mohammed Arafa

• All statically determinate beams will have IL that

consist of straight line segments. After some practice

one should be able to minimize computations and

locate the unit load only at points representing the end

points of each line segment

Example 1

Draw the IL for the Reaction at A

xA

xA

M

y

y

B

1011

0)1)(10()10(

0

IL Equation

Example 2

Draw the IL for the Reaction at B

Structural Analysis IDr. Mohammed Arafa

Example 3

Draw the IL for the Shear at C

Load from A to C

Load > C

Example 4

Draw the IL for the Moment at C

Load from A to C

Load > C

Example 5

Draw the IL for the Shear & Moment at B

Structural Analysis IDr. Mohammed Arafa

Example 6

Draw the IL for the Shear & Moment at C

0.5

Example 7Internal Hinge Example

Determine the shear & moment at point D

Influence Lines for Beams

For Concentrated Force For any concentrated force F acting on the beam at any

position x, the value of the function can be found by multiplying

the ordinate of the influence line at the position x by the

magnitude of F.

1 1

2 2yA F F lb

Influence Lines for Beams

For Uniform LoadConsider a portion of a beam subjected to a uniform load w0

0

0

0 0

dF w dx

dF y w dx y

ydF yw dx w ydx

:

The area under the influence line

Where

ydx

Influence Lines for Beams

For Uniform LoadIn general, the value of a function caused by a uniform distributed

load is simply the area under the influence line multiplied by the

intensity of the uniform load.

0

0 0

1 11

2 2

yA area w

L w Lw

Example 7-Continue

Structural Analysis IDr. Mohammed Arafa

Example 8

Determine the moment at C in the two cases

cMLI .

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

Case 1

Example 8

cMLI .

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

Case 2

Example 8

Example 9

Determine the maximum positive shear that can be developed at

point C in the beam shown due to a concentrated moving

load of 4000 lb and a uniform moving load of 2000 lb/ft

Due to Concenterated Load

0.75 4000 3000CV lb lb

Due to

0.5 10 2.5 0.75 2000

5625

C

Uniform Load

V

lb

Total MaximumShear

3000 5625 8625CV lb

Qualitative Influence LinesMüller-Breslau Principle

Example 10

Sketch the influence line for the vertical reaction at A

Sketch the influence line for the vertical reaction at A

Sketch the influence line for the vertical reaction at A

Example 11Sketch the influence line for the shear at B

Sketch the influence line for the shear at B

Sketch the influence line for the shear at B

Sketch the influence line for the moment at B

Example 11-bSketch the influence line for the moment at B

Sketch the influence line for the moment at B

Example 11-CSketch the influence line for the moment at B

- the reaction at A and C

- shear at D,

- the moment at D, E and F

Example 11-C

Example 11-C

Example 11-C

Example 11-C

Example 11-C

Example 11-C

Influence Line for Floor Girders

Structural Analysis IDr. Mohammed Arafa

Draw the influence line for the shear at panel CD

Example 12

Draw the influence line for the moment at F

Example 13

Example 14Truss Example

Determine The force in member GB

EGB

EGB

RF

RF

41.1

45cos

AGB

AGB

RF

RF

41.1

45cos

part second for the

x FGB

0 0

6 0.354

12 -0.707

18 -0.354

24 0

Example 15Truss Example

Determine The force in member GF, BF

0 40 17.3

2.3

cos30 1.15

B D GF

GF D

BF D BF D

M R F

F R

F R F R

0 20 17.3

1.15

cos30 1.15

B A GF

GF A

BF A BF A

M R F

F R

F R F R

Example 16Truss Example

Determine The force in member CG

Example 17

Structural Analysis IDr. Mohammed Arafa

IL due to Series of

Concentrated Loads

1

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

2

1( 0.125) 4(0.75) 4(0.625) 5.375CV k

3

1(0) 4( 1.25) 4(0.75) 2.5CV k

1

2

3

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

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

2(0) 4(3.0) 3(7.5) 34.5 .

C

C

C

M k ft

M k ft

M k ft

Example 18

Example 19

max

8(1.2) 3(0.4) 10.8 .BM kN m

Structural Analysis IDr. Mohammed Arafa

Determine the maximum positive moment that created at point B

Example 19

Absolute Maximum Shear and Moment

Absolute Maximum Moment

• The absolute maximum moment in a simply

supported beam occurs under one of the

concentrated forces, such that this force is

positioned on the beam so that it and the

resultant force of the system are equidistant

from the beam’s centerline.

The location of Maximum Moment is at

2

xx

Absolute Maximum Moment

Structural Analysis IDr. Mohammed Arafa

Example 20

The truck has a mass of 2 Mg and a center of gravity at G.

Determine the absolute maximum moment developed in the

simply supported bridge deck due to the truck’s weight. The

bridge has a length of 10 m.

Example 21

3The truck weight= 2 10 kg 9.8

=19.62kN

Example 21

The maximum moment occurs under the front wheel loading.

Using the right section of the bridge deck,