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Chapter 5INFLUENCE LINES

Noorli IsmailBambang Prihartanto

5.1 INTRODUCTION

Influence lines have important application for the design of structures thatresist large live loads. The structures is acted by load at certain locationsare discussed in the previous chapters.

In this chapter, will be discussed the moving load from one point toanother point such as the live loads for bridge. This moving load maybe

because of the moving vehicles. An influence lines represents the variationof the reaction, shear, moment or deflection at a specific point in a member as a concentrated force moves over the member.

For this reasons, influence lines play an important part in the design of bridges, industrial crane rails, conveyors and other structures where loadsmove across their span.

5.2 INFLUENCE LINES FOR BEAMS

Procedures in constructing the influence lines:

a) allow a one unit load (either 1kN, 1N, 1kip or 1 ton) to move over the beam from left to right.

b) find the values of shear force or bending moment at the point under consideration as the unit load moves over the beam from left toright.

c) plot the values of the shear force or bending moment over thelength of the beam computed for the point under consideration.

Signs convention;

(+ve) V

M

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5.2.1 Variations of Reactions, R a And R b As Function of Load Position

EXAMPLE 5.1

Construct the influence lines for vertical reaction at A and B.

A 5m B

Solution ;

Allow one unit load (1kN) to move over the beam from left to right.

x 1 kN

5m

R A R B

Find the value of shear forces at the point under the support reaction at A and B.

Plot the values in drafted of influence lines for vertical reaction at A, R A and B,R B.

x 1kN

5m

R A R B

F = F R A + R B = 1 ..(1)

+ MA = 0; 1(x) R B(5) = 0R B = x/5

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From the equation of R B, substitute the value of 0 m to 5 m in the table.

0.8 1.00.4 0.6

0.2 +(kN)

x(m) R B0 01 0.22 0.43 0.64 0.85 1.0

0 m 5 m

Influence line for R B

From the equation of (1) and R B, find the support reaction at A.

R A + R B = 1

51

15

x R

x R

A

A

=

=+

From the equation of R A, substitute the value of 0m to 5m in the table.

1.0

x(m) R A0 1

0

1 0.82 0.63 0.44 0.25 0

(kN) (+)

Influence line for R A

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5.2.2 Variations of Reactions, V C and Mc as Function of Load Position

EXAMPLE 5.2

Construct the influence lines for R A

, R B, V

Cand M

C.

A C B

2 m 6 m

Solution;

i) Influence lines for R A, R B.

1kNx 8-x

R A R BF = F

R A + R B = 1 ..(1)

+ MB = 0; R A(8) -1(8-x) = 0 x(m) R A0 18 0

81

x R A =

1.0

+


From the equation of (1) and R A

, find the support reaction at B.R A + R B = 1

8x

R --- 1R )8x

1( BB =>=+ x(m) R B0 08 1

1.0

+


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ii) Influence lines for V C and M C.

Refers to the sign convention

Mc

Vc Vc2m 6m

R A R B

The beam will be separated to two segments;1) 0< x


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A C B

2 m 6 m

0.75

+

_

0.25

Influence lines for V C

1.5

+

Influence lines for M C

EXAMPLE 5.3

Construct the influence lines for bending moment at C.

C

A B

10 m3 m

Solution;

Mc

Vc Vc3m 7m

Refers to the sign convention

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R A R B

The beam will be separated to two segments;1) 0< x


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EXERCISE 5.1

a) The given figure of beams, construct;i) the influence line for vertical reaction at A and B.ii) the influence line for shear at C.iii) the influence line for bending moment at C.

C

A B

5 m 5 m

2.5 m

b) The given figure of beams, construct;i) the influence line for vertical reaction at A and B.ii) the influence line for shear at C.iii) the influence line for bending moment at C.

C

A B

15 m5 m

c) Figure below shows a beam which is pinned at support B and supported bya roller at D. The beam is loaded with a set of moving concentrated loads.Construct the influence lines for the;

(i) reaction at D, R D(ii) reaction at B, R B(iii) shear force at C, V C(iv) bending moment at C, M C

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5.3 QUALITATIVE INFLUENCE LINES - MULLER-BRESLAUPRINCIPLE

Developed by Heinrich Mller-Breslau in 1886

The principle provides a quick method for establishing the shape of theinfluence line.

The principle gives only a procedure to determine of the influence line of a parameter for a determinate or indeterminate structure.

But using the basic understanding of the influence lines, the magnitudes of

the influence lines also can be computed.

In order to draw the shape of the influence lines properly, the capacity of the beam to resist the parameter investigated (reaction, bending moment,shear force, etc.), at that point, must be removed.

The principle states;

The influence line for a parameter (say, reaction, shear or bending moment),at a point, is to the same scale as the deflected shape of the beam, when the

beam is acted upon by that parameter.

The capacity of the beam to resist that parameter, at that point,must be removed.

Then allow the beam to deflect under that parameter

Positive directions of the forces are the same as before

EXAMPLE 5.4

(i) If the shape of influence line for the vertical reaction at A is to bedetermined, the pin is first replaced by a roller guide in Figure 5.1(b).When the positive(upward) force R A is applied at A, the beam deflectsto the dashed position which represents the general shape of theinfluence line for R A.

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Figure 5.1


(ii) If the support A is cantilever type, a double roller guide must be used at Asince this type of support will then transmit both a moment M A at the fixedsupport and axial load, R Ax but will not transmit R Ay .

R Ay .

Deflected shape Influence line for R A.

R A

Figure 5.2

(iii) If the shape of influence line for the shear at C is to be determined, theconnection at C may be symbolized by a roller guide as shown in Figure5.3(b). Applying a positive shear force, V C to the beam at C and allowingthe beam to deflect to the dashed position.

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Figure 5.3

(iv) If the support is a cantilever type, a roller guide must be placed at B,the positive shear is applied and corresponding influence line is shown.The left segment of the beam does not deflect due to the fixed support.

Figure 5.4

(v) If the shape of influence line for the moment at C is to be determined, aninternal hinge or pin is placed at C. Applying the positive moment M C tothe beam, the beam then deflect to the dashed position as shown in Figure5.5(b).

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Figure 5.5

5.4 THE APPLICATION OF INFLUENCE LINES

5.4.1 Concentrated/Point/Axial Load

Influence lines represent the variation of functions either the reaction,shear or moment at a specific point due to the unit load.

Once the line is constructed, one can tell at a glance where the movingload should be placed on the structure. For the concentrated load, themagnitude of associated reaction, shear or moment at the point can becalculated by multiplying the load and the ordinates of the influence linediagram.

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EXAMPLE 5.5

The simply supported beam as shown in figure below has length of 15 m. If pointload of 105 kN moving from A to B. Determine :

a) maximum positive shear that can be developed at point X. b) maximum negative shear that can be developed at point X.c) maximum moment at X.

Solution;

9 Find R A and R B.

Develop influence line for V x.

6 m X

A B

0.6

+

- 0.4

Influence line for V x

a) Maximum positive shear that can be developed at point X.

The maximum positive shear at X will occur when the point load of 105kNis located at X= (6 15)m, since this is the positive peak of the influenceline, the ordinate peak is +0.6, so that;

Max +ve shear : 105 x 0.6 = 63 kN

b) Maximum negative shear that can be developed at point X.

The maximum negative shear at X will occur when the point load of 105kN is located at X= (0 6)m, since this is the negative peak of theinfluence line, the ordinate peak is -0.4, so that;

Max -ve shear: 105 x (-0.4) = -42 kN

15

151

1

x B R

x A R

=

=

15 m

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c) Maximum moment at X.

Construct the influence line for moment at X.

6 m

3.6

The ordinate peak of influence line for Mx is +3.6. The maximum momentat X due to the concentrated load of 105kN is;

= 105 x 3.6 = 378 kNm

EXAMPLE 5.6

Determine the positive live shear that can be developed at point load of 50kNlocated at C.

50 kN

A C B1 m 3 m 2 m

Solution;

9 The influence line for R A or R B can be drafted by following the previous procedures in Example-1 and 2.

6/5 x 1.0 = 1.21.0

+ 2/5 x 1.0 = 0.4

Influence line for R A The positive live shear that can be developed at point load of 50kN.

R 50kN = 50(0.4)= 20kN

x

x X M

x

x X M

156

6

;15653

;60

=

=

x R A 2.02.1 =

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EXAMPLE 5.7

The simply supported beam as shown in figure below has length of 8 m. If pointload of 40 kN, 25 kN and 15 kN stop at the selected position as shown in figure

below. Determine the shear force at C.

40kN 25kN 15kN1m

A B C D E F

2 m 2 m 2 m 1 m 1 m

Solution;

= 0.750.5

+ +- -

= 0.25 = 0.25 0.5

?

Influence line for V C

Shear force at C, V C = 40(+0.5) + 25(+0.50) + 15(-0.25)= 28.75kN

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5.4.2 Uniform Load

The values of function caused by a uniform distributed load can bedetermined by multiplying the uniform load and area under influence linefor the function.

Prove;w wdx

dx

L

As shown, (wdx) can be assumed as the concentrated force acts at the

point, dx while function value is (wdx)y where y is ordinate under theinfluence line at the selected point.

(wdx)y = wydx

For overall,L

= wydx0

= w ydx

where; ydx is area under the influence line

EXAMPLE 5.8

Using the influence line, determine the bending moment at point X when uniformdistributed load of PQ located at 3m from A.

10 kN/m 5m X

P Q

2 m A

B

8 m

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Solution ;

Construct the influence line for moment at X

x

8

55M

;8x5

x83

M

;5x0

X

X

=

=

125.135

875.1

=

=

y

y

10 kN/m

2 m

5 mX

8 m

1.875

y = ?

3 m5

8x

R

x81

1R

B

A

=

=

The bending moment at point X when uniform distributed load of PQ located at

3m from A is the total of shaded area@ trapezoid area

kNm x x 30)10)(875.1125.1(221 =+=

5.4.3 MOVING LOAD

In designing the structure, the positions of the maximum shear force andmaximum bending moment should be determined.

If the structure undergoes one or two moving load, the position of themaximum shear force and maximum bending moment can be determinedeasily from the influence line.

5.4.3.1 Maximum Influence at a point due to a Series of Concentrated Loads.

Once the influence line of function has been drafted for a point in astructure, the maximum effect caused by a live concentrated force isdetermined by multiplying the peak ordinate of the influence line.

However, several concentrated forces must be placed on the structure byusing trial and error procedure to determine the maximum shear or moment. The series of concentrated load would be the wheel loading of atruck, train or car.

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EXAMPLE 5.9

Determine maximum moment at point X on the deck due to the wheel loads of themoving car which is moving from A to B.

10 kN 15kN

4m X

A B

2.5m

16.5 m

Solution;

Find R A and R B.

Construct the influence line for mom

Using trial and error procedure;

CASE 1:

Imagine that the 15kN will arrive first at the point X

10kN 15kN

3.03

y = ?

Mmax = 15(3.03) + 10(1.14)

= 56.85 kNm

3.03

04

0 x 4;Mx = 0.758x

4 x 16.5;Mx = 4 4x/16.5+

0 1.5 4

3.03/4 = y/1.5y = 1.14

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CASE 2:

Then, the load of 15kN will move and 10kN load acts at the pointX

10kN 15kN

3.03

y = ?

0 4 10

From the observation, position of CASE 2 will contribute the largestmoment than CASE 1(from the slope) or the calculation can be made for

both cases.

42.2

10)45.16(03.3

=

=

y

y

Mmax = 15(2.42) + 10(3.03)

= 66.6 kNm

It is confirm that the maximum moment occurred at position of CASE 2.

Mmax = 66.6 kNm

EXAMPLE 5.10

Determine the maximum positive shear created at point B in the beam due towheel loads of moving truck.

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Solution;

Construct the influence line for V B.

0.5

0 3 6

61

1

x R

x R

B

A

=

=

-0.5

xV

x

xV

x

B

B

6

Using trial and error procedure;

From the figure, the wheel load of 18kN will move first and continue tothe next wheel.

CASE 1;

Imagine that the 18kN will arrive first at the point B.

18kN 40.5kN 67.5kN 45kN

3 3.9 5.7 7.50.5 0.35

0.05

0 3

-0.5

VB = 18(0.5)+40.5(0.35)+67.5(0.05) = 26.55 kN

11

;6361

;30

=

=

Wheel distance

Find the ordinates!

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CASE 2;

Then, the load of 18kN will move and 40.5kN load acts at the point B

18kN 40.5kN 67.5kN 45kN

2.1 3 4.8 6.6(distance)

0.5

0.2

0 3-0.35

Find the ordinates!

-0.5

VB = 18(-0.35)+40.5(0.5)+67.5(0.2) = 27.45 kN

CASE 3;

Then, the load of 40.5kN will move and 67.5kN load acts at the point B.

18kN 40.5kN 67.5kN 45kN

0.3 1.2 3 4.8(distance)0.5

0.2

0 -0.05 3-0.2

-0.5Find the ordinates!

Due to the time constraints, the high peak of ordinate is +0.5 under theinfluence line of V B. The greatest wheel loads of the moving truck is67.5kN. So, the highest shear is achieved when this wheel loads act the

point B.

VB = 18(-0.05)+40.5(-0.2)+67.5(0.5)+45(0.2) = 33. 8kN

Therefore, Case 3 is the maximum positive shear created at point B.

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EXERCISE 5.2

a) Determine the maximum moment +ve at point C when a series of concentrated load is moving from A to B with load of 10kN is leading.

5m

2kN 5kN 15kN 10 kN

C

A

B

1m 2m 1m 10m 5m

[Ans : 67kNm]

b) The simply supported beam with length of 80 m is subjected to the seriesof concentrated loads and moving from left to right of the beam. Determine;

i)

maximum positive shear that can be developed at 10m from the left of beam.ii) maximum negative shear that can be developed at 10m from the left of

beam.

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5.5 ABSOLUTE MAXIMUM SHEAR AND MOMENT

If the beam is cantilevered or simply supported, the problem in finding theabsolute maximum shear and moment can be readily solved. Maximummoment on the beam subjected to the moving load is located on the

beams centerline between the resultant force and the nearest point.

Absolute maximum moment in simply supported beam occurs under oneof the concentrated forces series. This principle will have to be applied toeach load in the series and the corresponding maximum moment iscomputed. By comparison, the largest moment is the absolute maximum.

resultant force, F R Nearest load C

Lc

AB

x x L 2

L/2

Beams centerline

L

EXAMPLE 5.11

Determine the absolute maximum moment in the simply supported beam.

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Solution;

The magnitude and position of the resultant force of the system are determinedfirst.

;

= F F R

kN F R

18468 =++=

=+ C R M M C m x x

2

)5.4(4)3(618

=

+=

The resultant force, F R is located between 8kN and 6kN.

1st Trial;

The position of F R and 6kN is located at the centre of beam.FR

8kN 6kN 4kN

3m 1.5m

m x 2= 1m

FR

8kN 6kN 4kN

C

A B

2m 2m 1.5m

0.5 0.5

4.5m 4.5m

R A

Calculating R A with considering F R only;

kN R

R M

A

A B

10

)9()5(18

=

+=+

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First assume the absolute maximum moment occurs under the 6 kN load.

Now using the left section of the beam;

8kN 6kN

M cut

VcutkNm M

M

cut

cut

26

)3(8)5(10=

+=

kN10R A =

2nd Trial;

The position of 8kN and F R is located at the centre of beam.

Calculating R A with considering F R only;

kN7R

)9(R )5.3(18M

A

AB

=

+=+

There is possibility that the absolute maximum moment may occur under 8kNload.

Now using the left section of the beam;

8kN

M cut

Vcut

kN7R A =

By comparison, the absolute maximum moment is; kNm M cut 26= , which occursunder 6kN load.

8 kN 6 kN 4 kNFR

3

8 kN 6 kN 4 kNFR

m

2 m

1.5 m1.5 m1 m1 m1 m3.5 m

4.5 m 4.5 m

C

A B

kNm M

M

cut

cut

5.24

)15.4(7=

+=

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EXAMPLE 5.12

Determine the absolute maximum shear subjected to the series of concentratedload. The loads can be moved from left to right or opposite direction.

5 20 20 15 15 (kN)

A B

10m1m 1m 1m 1m

Solution;

The magnitude and position of the resultant force of the system are determinedfirst.

;= F F R kN F R 75151520205 =++++=

=+ A R M M A m x x

2.2

)4(15)3(15)2(20)1(2075

=

+++=

The resultant force, FR

is located between 20kN and 15kN.

Case 1

Consider the Absolute Maximum Shear occurs at Support A.

Vmax = R A

MB = 0R A(10) 75(7.8) = 0

R A = 58.5 kN

R A

5 20 20 15

75

2.2 m

15

A B

10 m

R B B

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Case 2

Consider the Absolute Maximum Shear occurs at Support B.

Vmax = R B

MA = 0R B(10) 75(8.2) = 0

R B = 61.5 kN

By comparison, the absolute maximum shear is, V max = 61.5 kN, which is occursat support B.

5.6 INFLUENCE LINES FOR TRUSSES

Trusses are often used as primary load-carrying elements for bridges. Theloading on the bridge deck is transmitted to stringers and then transmit theloading to floor beams and lastly to the joints along the bottom cord of thetruss. For the moving load, the loads maybe not acted at the joint of thetruss. The loads will transmit to the truss member.

R A

5 20 20 15

75

1.8 m

15

A B

10 m

R B B

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Normally, the internal force members is the force axis whether itscompression or tension. When the truss is subjected to the moving load, todetermine the maximum or minimum force axis within the truss member,influence lines can be drafted.

Influence lines for truss member can be determined by located one unit load at node or truss joint . Determine the force of truss member by usingthe method of section.

EXAMPLE 5.13

Construct the influence lines for force in member GB from the given problem.

6m @ 4

E

H

A

G F

B C D

6m

Solution;

Use the Method of Section to determine the force of F GB . Then, choose the sectionto be right or left system.

Pick the right system. One unit load is not applied yet in the system.

To determine F GB use equation Fy = 0(so F GH and F CB can neglected)

45G

F

E

DCFCB

F

FGB

R E

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Fy = FyR E = F BG sin 45 ..(1)

The right system is selected, so influence line for R E will be constructed.How? (assume the system (A E) is a beam in order to determine influence linefor R E).

1x

A E

24 m R ER A

R A+ R E = 1

LOCATIONOF UNIT

LOAD

x R E

A 0 0B 6 0.25C 12 0.5D 18 0.75E 24 1

+ M A = 0

- R E (24) + 1 (x) = 0R E = x /24R A = 1- x/24

Influence lines for R E1

0.75

0.50.25

0A B C D E

Determine F BG using the equation FBG sin 45 = R E ..(1). Put one unit load at joint A,B,C,D and E in truss and relate to I.L for R E.

i) 1 unit load at A;R E = 0 (determine from I.L for V E), so F BG =0(determine fromEquation 1)

ii) 1 unit load move to B;R E = 0.25, so F BG = 0.354

iii) 1 unit load move to C;=when 1 unit load move at C, the unit load is inclusive in the rightsystem. Refer to the next figure. So, the new equation is F BG sin 45 +1= R E ..(2)

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when 1 unit load at C, R E = 0.5 so F BG = -0.707

VE

FGB

45G F

EVEDC

FGH

FCB1

Fy = FyFBG sin 45 +1= R E..(2)

iv)

1 unit load move to D;R E = 0.75 , so F BG = -0.354 .(determine from Equation 2)

v) 1 unit load move to E;R E = 1 , so F BG =0 .(determine from Equation 2)

Draw the influence lines for member F BG .

LOCATION

OF UNITLOAD

FBG

A 0B 0.354C -0.707D -0.354E 0

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G FH

EAC DB

6 m @ 4

0.354

0 0

-0.354

-0.707

Influence lines for member F BG

EXAMPLE 5.14

Construct the influence line for force in member HG from the same truss.

Solution;

6 m @ 4

D

G F

EB C

H

A

To determine F HG , use Method of Section and pick left system (or right system).One unit load is not applied yet in the system.

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Take the moment at B, because F BG and F BC can be neglected.

+ MB = 0= F HG (6) + R A (6) ..(1)

The left system is selected, so influence line for R A will be developed.How? (assume the system (A E) is a beam in order to determineinfluence line for R A).

1x

A

E

24 m

R A = 1- x/24

Influence lines for R A1

0.750.5

0.25

0A B C D E

LOCATIONOF UNIT

LOAD

x R A

A 0 1B 6 0.75C 12 0.5D 18 0.25E 24 0

H

A

FHG

FBG

FBCBR A

R ER A

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Put one unit load at joint A,B,C,D and E in the truss. Determine F HG usingequation (1).

i) 1 unit load at A;

when 1 unit load at A, the unit load is inclusive the left system asshown in the figure above. So, the new equation will be :

FHG (6) + R A (6) 1(6) =0 ..(2)

H FHG

Then, F HG = 0

ii) 1 unit load move to B;R A = 0.75

when 1 unit load at B, the unit load is inclusive the left system as thefigure above. So, the equation (1) will be:

FHG (6) + R A (6) 1(0) =0 ..(2)

Then, F HG = -0.75

NOTE: This example is differing from the previous example indetermining the equation because using the moment equation. So,the equation is different at the different point.

iii) 1 unit load move to C (when unit load move to C, D and E, this unitload is not inclusive in the selected system);

R A = 0.5 .use equation (1) to determine F HG.FHG = -0.5

iv) 1 unit load move to D;R A = 0.25FHG = -0.25

v) 1 unit load move to E;R A = 0FHG = 0

FBG

FBCBR AA

+ MB = 0= F HG (6) + R A (6) -1(6)..(2)1

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Draw the influence lines for F HG .

FG LOCATION

OF UNITLOAD

FHG

A 0B -0.75C -0.5D -0.25E 0

H

A

0 0-0.25

-0.5-0.75

EXAMPLE 5.15Draw the influence line for member EF, ED and CD for the truss which issubjected to the uniform distributed load of 2kN/m. Determine the maximumcompression and tensile load for the three members.

Solution;

DB C

2 kN/m

2 m

E F

A BC D

3 mm 33 m 3 m

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If the right system is selected, the influence line for R B is constructed;One unit load is not applied yet in the system.

F

= 53.13A

BC D

To determine F ED use equation Fy = 0(so F EF and F CD can be neglected)

Fy = FyR B + F ED sin 53.13=0 ..(1)

To determine F EF, take the moment at D, because F ED and F CD can beneglected.

+ MD = 0= -FEF (2) R B (6) ..(1)

To determine F CD , take the moment at E, because F ED and F EF can beneglected.

+ ME = 0= -FCD (2) R B (7.5) ..(1)


3 m 3 m

FEF

FED

FCD

E2 m

A

3 m 3 m

B

1

0.50.25

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If the left system is selected, the influence line for R A is constructed; Oneunit load is not applied yet in the system.

To determine F ED use equation Fy = 0(so F EF and F CD can be neglected)

Fy = FyR A = F ED sin 53.13 ..(1)

To determine F EF, take the moment at D, because F ED and F CD can beneglected.

+ MD = 0= -F EF (2) + R A (6) ..(1)

To determine F CD , take the moment at E, because F ED and F EF can beneglected.

+ ME = 0= -F CD (2) + R A (4.5) ..(1)

All the influence lines for three member forces are drafted by followingthe previous example.

= 53.13

E FFEF

2 m

C D

FEDFCD

AB

3 m 3 m3 m 3 m

10.75

0.5

AB


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To determine F ED;

RIGHTSYSTEM

LEFTSYSTEM

One unit load applied; R B FED One unit load applied; R A FED

From this table, it is confirm that the both system will indicate the samevalue of member force. Students can continue in determining the member forces of F EF and F CD with the same procedures.

At A;R B+FED sin 53.13=0

0 0 At this point, unitload is inclusive inthe left system.At A;R A=FEDsin 53.13 +1

1 0

At C;R B+FED sin 53.13=0

0.25 -0.313 At C;R A=FEDsin 53.13 +1

0.75 -0.313

At this point, unit loadis inclusive in the rightsystem.

At D;R B+FED sin 53.13=1

0.5 0.625At D;R A=FEDsin 53.13r

0.5 0.625

At B;R B+FED sin 53.13=1

1 0 At B;R A=FEDsin 53.13

0 0

0.625

AB

C

D

x

-0.313Influence line for F ED

m1x

x625.x313.939.0x3

625.0313.0

=x

00 ==

AB

C D

-0.75-1.5

Influence line for F EF

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A BC D

0.941.13

Influence line for F CD

To determine the maximum load at all influence lines, it can be observedthe influence line for F ED , F EF and F CD is located at the beams centerline.The load can be determined by multiplying the uniform load and the areain influence line.

Refer to the influence line for F ED;

)T(kN5)6)(625.0(21

)2)(625.0(21

2F

)C(kN25.1)1)(313.0(21

)3)(313.0(21

2F

)Tmax(

)Cmax(

=+=

=+=

Refer to the influence line for F EF;

Fmax = ( ) )(18265.121

2 C kN x x =

Refer to the influence line for F CD;

( ) ( )

)T(kN82.15

)39.311.341.1(2

6x13.121

3x)13.194.0(21

3x94.021

2Fmax

=

++=

+++=

The maximum compression load = 18 kN

The maximum tensile load = 15.82 kN

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TUTORIAL 5

1. Using the Principle of Muller Breslau, draft the influence lines for verticalreaction at A.

a)

b)

c)

A

A

A

hinged

2. Using the Principle of Muller Breslau, draft the influence lines for shear at

B.

a)

hinged

B

B

B

b)

c)

3. Figure Q1(a) shows a beam which is pinned at support B and supported bya roller at D. The beam is loaded with a set of moving concentrated loadsas shown in Figure Q1(b).

(a) Sketch the influence lines for:(i) the reaction at B, R B(ii) the reaction at D, R D(iii) the shear force at C, V C(iv) the bending moment at C, M C

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(b) Determine the maximum values V C and M C when the train loadsshown in Figure Q1(b) crosses the beam from right to left.

(c) Determine the absolute maximum shear force and absolute

maximum bending moment for span BD when the loads moveacross BD.

A B C D E

2 m 4 m 6 m 3 m

3 m 2

25 60 20 (kN)Movingdirection

(b)

(a) Figure Q1 OCTOBER 2006-UTHM

4. Figure Q2 shows a beam which is pinned at support A and roller at supportB.

(a) Construct the influence line for the vertical reaction at A and B byusing statics and the method of sections.

(b) If point X located 4 meter from A, determine the shear, Vx andmoment, Mx by using method of sections.

(c) Determine the maximum positive shear that can be develop at pointX for the beam due to the concentrated moving load of 10 kN and auniform moving load of 4 kN/m along the span.

(d) There is difference between constructing an influence line andconstructing a shear or bending moment diagram. What do youthink about that statement?

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B16 m 4 m

5. (a) Describe the procedure for drawing the influence line and givetwo(2) applications of the influence line.

(b) Figure Q3 shows a bridge truss of span 20m. A uniformlydistributed load of 60kN/m and longer than the bridge is moving onthe bridge deck from A to E.

i) Draw the influence line for member BC and BG.ii) Calculate the maximum force in member BC and BG.

Figure Q3

NOVEMBER 2005-UTHM

6. Figure shows a Parker truss supported with a hinge at A and a roller at K.Draw the influence lines for the axial force in member GH and HJ of the

plane truss as a unit load moves on the bottom chord from point A to L.

analisis struktur c5

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