Chapter 5

INFLUENCE LINES

Noorli Ismail

Bambang Prihartanto

5.1 INTRODUCTION

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

In this chapter, will be discussed the moving load from one point to another point such as the live loads for bridge. This moving load maybe because of the moving vehicles. An influence lines represents the variation of 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 loads move 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 to right.

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

Signs convention;

(+ve) V M

1

5.2.1 Variations of Reactions, Ra And Rb 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 RA RB 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, RA and B, RB.

x 1kN 5m RA RB F↑ = F↓

RA + RB = 1 ……..(1) + ∑MA = 0; 1(x) – RB(5) = 0 RB = x/5

2

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

0.8 1.0 0.4 0.6 0.2 + (kN)

x(m) RB0 0 1 0.2 2 0.4 3 0.6 4 0.8 5 1.0

0 m 5 m

Influence line for RB From the equation of (1) and RB, find the support reaction at A.

RA + RB = 1

51

15

xR

xR

A

A

−=

=+

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

1.0

x(m) RA0 1

0

1 0.8 2 0.6 3 0.4 4 0.2 5 0

(kN) ( +)

Influence line for RA

3

5.2.2 Variations of Reactions, VC and Mc as Function of Load Position EXAMPLE 5.2 Construct the influence lines for RA, RB, VC and MC. A C B 2 m 6 m Solution; i) Influence lines for RA, RB.

1kN x 8-x RA RB

F↑ = F↓ RA + RB = 1 ……..(1)

+ ∑MB = 0; RA(8) -1(8-x) = 0 x(m) RA0 1 8 0

81 xRA −=

1.0 + Influence line for RA

From the equation of (1) and RA, find the support reaction at B. RA + RB = 1

8xR--- 1R)

8x1( BB =>=+−

x(m) RB0 0 8 1

1.0 + Influence line for RB

4

ii) Influence lines for VC and MC.

Refers to the sign convention

Mc

Vc Vc 2m 6m RA RB The beam will be separated to two segments;

1) 0< x <2 2) 2< x < 8

0< x <2 (unit load to the left of C) 1kN x 2-x Mc Vc

8

1 xRA −=

2< x < 8 (unit load to the right of C) 2 Mc Vc

8

1 xRA −=

F↑ = F↓ RA = 1 + Vc (1-x/8) = 1+ Vc

Vc = -x/8 + ∑Mc = 0;

(8

1 x− ) (2) – 1(2-x) –Mc = 0

Mc = xx+−− 2)

42(

x(m) VC0 0 2 -0.25

x(m) MC0 0 2 1.5 = 3x/4

F↑ = F↓ RA = Vc

Vc = 1- x/8

Mc = 0;

(

x(m) VC2 0.75 8 0

+ ∑

81 x− )(2)–Mc = 0

Mc = 4

2 x−

x(m) MC2 1.5 8 0

5

A C B 2 m 6 m 0.75

+ _

0.25 Influence lines for VC 1.5 + Influence lines for MC

EXAMPLE 5.3 Construct the influence lines for bending moment at C.

C

A B 10 m 3 m Solution;

Mc

Vc Vc 3m 7m

Refers to the sign convention

6

RA RB The beam will be separated to two segments;

1) 0< x <3 2) 3< x < 10

0< x ≤3 (unit load to the left of C) 1kN x 3-x Mc Vc

10

1 xRA −=

+ ∑Mc = 0;

(10

1 x− ) (3) – 1(3-x) –Mc = 0

Mc = xx+−− 3)

1033(

= 0.7x

x(m) MC0 0 3 2.1

3< x < 10 (unit load to the right of C)

3 Mc Vc

10

1 xRA −=

C

A B 10 m 3 m

2.1

+ ∑Mc = 0;

(10

1 x− )(3)–Mc = 0

Mc = 1033 x

−x(m) MC

3 2.1 10 0

Influence lines for MC

(+)

7

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 m 5 m c) Figure below shows a beam which is pinned at support B and supported by

a roller at D. The beam is loaded with a set of moving concentrated loads. Construct the influence lines for the;

(i) reaction at D, RD(ii) reaction at B, RB(iii) shear force at C, VC(iv) bending moment at C, MC

8

5.3 QUALITATIVE INFLUENCE LINES - MULLER-BRESLAU PRINCIPLE

Developed by Heinrich Müller-Breslau in 1886

The principle provides a quick method for establishing the shape of the influence 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 be determined, the pin is first replaced by a roller guide in Figure 5.1(b). When the positive(upward) force RA is applied at A, the beam deflects to the dashed position which represents the general shape of the influence line for RA.

9

Figure 5.1

Influence line for RA

(ii) If the support A is cantilever type, a double roller guide must be used at A

since this type of support will then transmit both a moment MA at the fixed support and axial load, RAx but will not transmit RAy.

RAy.

Deflected shape Influence line for RA.

RA

Figure 5.2 (iii) If the shape of influence line for the shear at C is to be determined, the

connection at C may be symbolized by a roller guide as shown in Figure 5.3(b). Applying a positive shear force, VC to the beam at C and allowing the beam to deflect to the dashed position.

10

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, an

internal hinge or pin is placed at C. Applying the positive moment MC to the beam, the beam then deflect to the dashed position as shown in Figure 5.5(b).

11

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 moving

load should be placed on the structure. For the concentrated load, the magnitude of associated reaction, shear or moment at the point can be calculated by multiplying the load and the ordinates of the influence line diagram.

12

EXAMPLE 5.5 The simply supported beam as shown in figure below has length of 15 m. If point load 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;

Find RA and RB.

Develop influence line for Vx.

6 m X

A B

0.6

+

- 0.4

Influence line for Vx

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

The maximum positive shear at X will occur when the point load of 105kN is located at X= (6→15)m, since this is the positive peak of the influence line, 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 the influence line, the ordinate peak is -0.4, so that;

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

15

1511

xBR

xAR

=

−=

15 m

13

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 moment at 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 50kN located at C.

50 kN A C B 1 m 3 m 2 m Solution;

The influence line for RA or RB can be drafted by following the previous procedures in Example-1 and 2.

6/5 x 1.0 = 1.2 1.0

+ 2/5 x 1.0 = 0.4 Influence line for RA

The positive live shear that can be developed at point load of 50kN.

R50kN = 50(0.4) = 20kN

x

xXM

x

xXM

1566

;15653

;60

−=

≤≤

=

≤≤

xRA 2.02.1 −=

14

EXAMPLE 5.7 The simply supported beam as shown in figure below has length of 8 m. If point load 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 15kN 1m A B C D E F

2 m 2 m 2 m 1 m 1 m

Solution; ¾ = 0.75 0.5 + + - - ¼ = 0.25 ¼ = 0.25 0.5

?

Influence line for VC

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

15

5.4.2 Uniform Load

The values of function caused by a uniform distributed load can be determined by multiplying the uniform load and area under influence line for 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 the influence line at the selected point.

(wdx)y = wydx

For overall,

L

= ∫ wydx

0

= 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 uniform distributed load of PQ located at 3m from A.

10 kN/m 5m X

P Q

2 m A

B

8 m

16

Solution ;

Construct the influence line for moment at X

x855M

;8x5

x83M

;5x0

X

X

−=

≤≤

=

≤≤

125.135

875.1

=

=

y

y

10 kN/m

2 m

5 mX

8 m

1.875

y = ?

3 m5

8xR

x811R

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

kNmxx 30)10)(875.1125.1(221

=+=

5.4.3 MOVING LOAD

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

If the structure undergoes one or two moving load, the position of the

maximum shear force and maximum bending moment can be determined easily 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 a structure, the maximum effect caused by a live concentrated force is determined by multiplying the peak ordinate of the influence line.

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

17

EXAMPLE 5.9 Determine maximum moment at point X on the deck due to the wheel loads of the moving car which is moving from A to B.

10 kN 15kN

4m X

A B

2.5m

16.5 m

Solution;

Find RA and RB.

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

0 4

0 ≤ x ≤4; Mx = 0.758x 4 ≤ x ≤16.5; Mx = 4 – 4x/16.5 +

0 1.5 4

3.03/4 = y/1.5 y = 1.14

18

CASE 2:

• Then, the load of 15kN will move and 10kN load acts at the point X

10kN 15kN

3.03

y = ?

0 4 10

From the observation, position of CASE 2 will contribute the largest moment than CASE 1(from the slope) or the calculation can be made for both cases.

42.210)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 to wheel loads of moving truck.

19

Solution;

Construct the influence line for VB. 0.5 0 3 6

611

xR

xR

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 to the 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 7.5 0.5 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!

20

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 -0.5

Find the ordinates!

Due to the time constraints, the high peak of ordinate is +0.5 under the influence line of VB. The greatest wheel loads of the moving truck is 67.5kN. 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.

21

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 series of 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.

22

5.5 ABSOLUTE MAXIMUM SHEAR AND MOMENT

If the beam is cantilevered or simply supported, the problem in finding the absolute maximum shear and moment can be readily solved. Maximum moment on the beam subjected to the moving load is located on the beam’s centerline between the resultant force and the nearest point.

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

resultant force, FR

Nearest load C Lc A B

x xL−

2

L/2 Beam’s centerline L

EXAMPLE 5.11

Determine the absolute maximum moment in the simply supported beam.

23

Solution;

The magnitude and position of the resultant force of the system are determined first.

;∑=↓ FFR kNFR 18468 =++=

∑=+ CR MMC

mx

x

2

)5.4(4)3(618

=

+=

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

1st Trial; The position of FR and 6kN is located at the centre of beam.

FR

8kN 6kN 4kN

3m 1.5m

mx 2=

1m

FR

8kN 6kN 4kN

C

A B

2m 2m 1.5m

0.5 0.5

4.5m 4.5m

RA

Calculating RA with considering FR only;

kNR

RM

A

AB

10

)9()5(18

=

+−=+

24

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

Now using the left section of the beam;

8kN 6kN

Mcut

Vcut

kNmMM

cut

cut

26)3(8)5(10

=−+−=

kN10R A =

2nd Trial; The position of 8kN and FR is located at the centre of beam.

Calculating RA with considering FR only;

kN7R

)9(R)5.3(18M

A

AB

=

+−=+

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

Now using the left section of the beam;

8kN

Mcut

Vcut

kN7R A =

By comparison, the absolute maximum moment is; kNmM cut 26= , which occurs under 6kN load.

8 kN 6 kN 4 kN FR

3

8 kN 6 kN 4 kN FR

m

2 m

1.5 m 1.5 m 1 m 1 m1 m3.5 m

4.5 m 4.5 m

C

A B

kNmMM

cut

cut

5.24)15.4(7

=−+−=

25

EXAMPLE 5.12 Determine the absolute maximum shear subjected to the series of concentrated load. The loads can be moved from left to right or opposite direction. 5 20 20 15 15 (kN)

A B

10m 1m 1m 1m 1m

Solution; The magnitude and position of the resultant force of the system are determined first.

;∑=↓ FFR kNFR 75151520205 =++++=

∑=+ AR MMA

mx

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 = RA

∑MB = 0 RA(10) – 75(7.8) = 0

RA = 58.5 kN

RA

5 20 20 15

75

2.2 m

15

A B

10 m

RBB

26

Case 2

Consider the Absolute Maximum Shear occurs at Support B.

Vmax = RB

∑MA = 0 RB(10) – 75(8.2) = 0

RB = 61.5 kN

By comparison, the absolute maximum shear is, Vmax = 61.5 kN, which is occurs at support B.

5.6 INFLUENCE LINES FOR TRUSSES

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

RA

5 20 20 15

75

1.8 m

15

A B

10 m

RBB

27

Normally, the internal force members is the force axis whether its compression or tension. When the truss is subjected to the moving load, to determine 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 using the 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 FGB. Then, choose the section to be right or left system. Pick the right system. One unit load is not applied yet in the system.

To determine FGB use equation ΣFy = 0(so FGH and FCB can neglected)

45G F

E

DC FCB

FGH

FGB

RE

28

↑ ΣFy = ↓ ΣFy RE = FBG sin 45 ……..(1)

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

1 x A E 24 m RERA

RA+ RE = 1

LOCATION OF UNIT

LOAD

x RE

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

+ MA = 0 - RE (24) + 1 (x) = 0 RE = x /24

RA = 1- x/24 Influence lines for RE

1 0.75 0.5 0.25 0 A B C D E Determine FBG using the equation FBG sin 45 = RE ……..(1). Put one unit load at joint A,B,C,D and E in truss and relate to I.L for RE.

i) 1 unit load at A;

RE = 0 …(determine from I.L for VE), so FBG =0…(determine from Equation 1)

ii) 1 unit load move to B; RE = 0.25, so FBG = 0.354

iii) 1 unit load move to C; =when 1 unit load move at C, the unit load is inclusive in the right system. Refer to the next figure. So, the new equation is FBG sin 45 +1 = RE ……..(2)

29

when 1 unit load at C, RE = 0.5 so FBG = -0.707

VE

FGB

45° G F

E VEDC

FGH

FCB

1

↑ ΣFy = ↓ ΣFy FBG sin 45 +1= RE..(2)

iv) 1 unit load move to D;

RE = 0.75, so FBG= -0.354 ….(determine from Equation 2)

v) 1 unit load move to E; RE = 1, so FBG=0 ….(determine from Equation 2)

Draw the influence lines for member FBG.

LOCATION OF UNIT

LOAD

FBG

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

30

G F H

E A C DB

6 m @ 4

0.354 0 0 -0.354 -0.707

Influence lines for member FBG 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 FHG, use Method of Section and pick left system (or right system). One unit load is not applied yet in the system.

31

Take the moment at B, because FBG and FBC can be neglected. + ΣMB = 0

= FHG (6) + RA (6) ……..(1)

The left system is selected, so influence line for RA will be developed. How? (assume the system (A→E) is a beam in order to determine influence line for RA). 1 x A

E 24 m

RA = 1- x/24 Influence lines for RA 1 0.75 0.5 0.25 0 A B C D E

LOCATION OF UNIT

LOAD

x RA

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

H

A

FHG

FBG

FBC

BRA

RERA

32

Put one unit load at joint A,B,C,D and E in the truss. Determine FHG using equation (1). i) 1 unit load at A;

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

FHG (6) + RA (6) – 1(6) =0 ……..(2)

H FHG

Then, FHG = 0

ii) 1 unit load move to B; RA = 0.75

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

FHG (6) + RA (6) – 1(0) =0 ……..(2)

Then, FHG = -0.75 NOTE: This example is differing from the previous example in determining 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 unit load is not inclusive in the selected system);

RA = 0.5 ….use equation (1) to determine FHG. FHG = -0.5

iv) 1 unit load move to D;

RA = 0.25 FHG = -0.25

v) 1 unit load move to E;

RA = 0 FHG = 0

FBG

FBCBRA

A

+ ΣMB = 0 = FHG (6) + RA (6) -1(6)…..(2)

1

33

Draw the influence lines for FHG.

F G

LOCATION OF UNIT

LOAD

FHG

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

H

A

0 0 -0.25 -0.5 -0.75 EXAMPLE 5.15 Draw the influence line for member EF, ED and CD for the truss which is subjected to the uniform distributed load of 2kN/m. Determine the maximum compression and tensile load for the three members.

Solution;

6 m @ 4DB C

2 kN/m

2 m

E F

A BC D3 mm 3 3 m 3 m

34

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

F

Ө=53.13º A

BC D

To determine FED use equation ΣFy = 0(so FEF and FCD can be neglected)

↑ ΣFy = ↓ ΣFy RB + FED sin 53.13=0 ……..(1)

To determine FEF, take the moment at D, because FED and FCD can be neglected.

+ ΣMD = 0

= -FEF (2) – RB (6) ……..(1)

To determine FCD, take the moment at E, because FED and FEF can be neglected.

+ ΣME = 0

= -FCD (2) – RB (7.5) ……..(1)

Influence line for RB

3 m 3 m

FEF

FED

FCD

E

2 m

A

3 m 3 m

B

10.5

0.25

35

If the left system is selected, the influence line for RA is constructed; One unit load is not applied yet in the system.

To determine FED use equation ΣFy = 0(so FEF and FCD can be neglected)

↑ ΣFy = ↓ ΣFy RA = FED sin 53.13 ……..(1)

To determine FEF, take the moment at D, because FED and FCD can be neglected.

+ ΣMD = 0

= -FEF (2) + RA (6) ……..(1)

To determine FCD, take the moment at E, because FED and FEF can be neglected.

+ ΣME = 0

= -FCD (2) + RA (4.5) ……..(1)

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

Ө=53.13º

E FFEF

2 m

C D

FED

FCDA B

3 m 3 m3 m 3 m

10.75

0.5

A B

Influence line for RA

36

To determine FED;

RIGHT SYSTEM

LEFT SYSTEM

One unit load applied; RB FED One unit load applied; RA FED

From this table, it is confirm that the both system will indicate the same value of member force. Students can continue in determining the member forces of FEF and FCD with the same procedures.

At A; RB+FED sin 53.13=0

0 0 At this point, unit load is inclusive in the left system. At A; RA=FEDsin 53.13 +1

1 0

At C; RB+FED sin 53.13=0

0.25 -0.313 At C; RA=FEDsin 53.13 +1

0.75 -0.313

At this point, unit load is inclusive in the right system. At D; RB+FED sin 53.13=1

0.5 0.625 At D; RA=FEDsin 53.13r

0.5 0.625

At B; RB+FED sin 53.13=1

1 0 At B; RA=FEDsin 53.13

0 0

0.625

A B

C D

x

-0.313

Influence line for FED

m 1xx625.x313.939.0

x3625.0313.0−

=x

00 =−=

A B

C D

-0.75-1.5

Influence line for FEF

37

A BC D

0.941.13

Influence line for FCD

To determine the maximum load at all influence lines, it can be observed the influence line for FED, FEF and FCD is located at the beam’s centerline. The load can be determined by multiplying the uniform load and the area in influence line. Refer to the influence line for FED;

)T(kN5)6)(625.0(

21)2)(625.0(

212F

)C(kN25.1)1)(313.0(21)3)(313.0(

212F

)Tmax(

)Cmax(

=⎥⎦⎤

⎢⎣⎡ +=

=⎥⎦⎤

⎢⎣⎡ −+−=

Refer to the influence line for FEF;

Fmax = ( ) )(18265.1212 CkNxx =−

Refer to the influence line for FCD;

( ) ( )

)T(kN82.15

)39.311.341.1(2

6x13.1213x)13.194.0(

213x94.0

212Fmax

=

++=

⎥⎦

⎤⎢⎣

⎡+⎥⎦

⎤⎢⎣⎡ ++=

The maximum compression load = 18 kN

The maximum tensile load = 15.82 kN

38

TUTORIAL 5

1. Using the Principle of Muller Breslau, draft the influence lines for vertical

reaction 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 by

a roller at D. The beam is loaded with a set of moving concentrated loads as shown in Figure Q1(b).

(a) Sketch the influence lines for:

(i) the reaction at B, RB(ii) the reaction at D, RD(iii) the shear force at C, VC(iv) the bending moment at C, MC

39

(b) Determine the maximum values VC and MC when the train loads shown 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 move across BD.

A B C D E

2 m 4 m 6 m 3 m

3 m 2

25 60 20 (kN)Moving direction

(b)

(a) Figure Q1 OCTOBER 2006-UTHM

4. Figure Q2 shows a beam which is pinned at support A and roller at support B. (a) Construct the influence line for the vertical reaction at A and B by

using statics and the method of sections.

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

(c) Determine the maximum positive shear that can be develop at point

X for the beam due to the concentrated moving load of 10 kN and a uniform moving load of 4 kN/m along the span.

(d) There is difference between constructing an influence line and

constructing a shear or bending moment diagram. What do you think about that statement?

40

A B C16 m 4 m

x

5. (a) Describe the procedure for drawing the influence line and give

two(2) applications of the influence line. (b) Figure Q3 shows a bridge truss of span 20m. A uniformly

distributed load of 60kN/m and longer than the bridge is moving on the 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.

APRIL 2003-UiTM

41