Notes InfluenceLines

Structural Mechanics 4 CIE3109

MODULE : INFLUENCE LINES HANS WELLEMAN Civil Engineering TU-Delft

date, 18 March 2014

Structural Mechanics 4 Influence Lines

Ir J.W. Welleman March 2014 ii

TABLE OF CONTENT

1. INFLUENCE LINES ................................................................................................................................... 1

1.1 PROBLEM SKETCH AND ASSUMPTIONS ....................................................................................................... 1

1.1.1 Introduction of influence lines for force quantities. ........................................................................ 2

1.1.2 Work and reciprocity, Clapeyron and the theorem of Betti and Maxwell. ...................................... 6

1.1.2.1 Reciprocity, theorem of Betti and Maxwell’s law ...................................................................................... 7

1.1.2.2 example 2 : Illustration of Maxwell’s reciprocal theorem ........................................................................ 10

1.1.3 Influence lines for displacement quantities. .................................................................................. 11

1.2 STATICALLY DETERMINATE SYSTEMS ...................................................................................................... 14

1.2.1 Qualitative approach..................................................................................................................... 14

1.2.2 example 3 : influence lines for simply supported beam with a cantilever ..................................... 17

1.2.3 example 4 : Beam with a hinge ..................................................................................................... 20

1.3 STATICALLY INDETERMINATE SYSTEMS .................................................................................................. 21

1.3.1 Qualitative approach..................................................................................................................... 23

1.3.2 Quantitative calculation of influence lines .................................................................................... 25

1.3.3 Example 6 : Nth

order statically indeterminate structures ............................................................ 32

1.4 STIFFNESS DIFFERENCES IN STRUCTURES ................................................................................................. 35

1.5 SPRING SUPPORTS* .................................................................................................................................. 36

1.6 DISTRIBUTED LOADS AND INFLUENCE LINES ........................................................................................... 37

1.7 THE MOST UNFAVOURABLE POSITION OF THE LOAD ................................................................................ 38

1.8 EXERCISES ............................................................................................................................................... 41

Exercise 1 ................................................................................................................................................................. 41

Exercise 2 ................................................................................................................................................................. 41

Exercise 4 ................................................................................................................................................................. 42

1.9 APPENDIX ............................................................................................................................................ 47


Ir J.W. Welleman March 2014 iii

STUDY GUIDELINES

These lecture notes are part of the study material of Structural Mechanics 4. The theory and

examples in these lecture notes are presented in such a way that this topic can be mastered

through self-study. Apart from these notes with theory and examples there is additional

practice material available with extra examples, old exam questions and additional material.

An introduction to the topic of influence lines can be found in the book “Engineering

mechanics”, Volume 1, “Equilibrium” by Coen Hartsuijker and Hans Welleman, referred to

from this point on as EM-1-E. In EM-1-E chapter 16 an explanation on influence lines for

force quantities can be found for statically determinate systems. After a short introduction, the

method described in EM-1-E will be expanded and applied on influence lines for force- and

displacement quantities on statically determinate and statically indeterminate systems.

The powerpoint sheets used in the lectures will be available online on the following website:

http://icozct.tudelft.nl/TUD_CT/index.shtml

Despite the necessary precaution in putting together these lecture notes mistakes and flaws

will be inevitable. It is highly appreciated if mistakes are reported.

The teacher,

Hans Welleman

pdf-version, June 2013

[email protected]


Ir J.W. Welleman March 2014 1

1. INFLUENCE LINES

In this module the concept of influence lines will be introduced. First the concept of influence lines will be explained followed by the description of tools to construct influence lines of statically determinate and indeterminate systems for both force and displacement quantities. The theory will be applied to numerous examples.

1.1 Problem sketch and assumptions

Until now we have looked at the force distribution in a structure as a result of a static load at a

fixed location. This force distribution can be visualised using normal-, shear- and moment-

diagrams. If the position of the load is variable there is a problem in visualising the force

distribution.

Figure 1 : unit load that moves over the structure.

To find the magnitude of a force or displacement at a fixed position depending on the location

of the external load the concept of influence line is introduced.

Definition:

An influence line gives the value of a quantity in a specific point as a function of the

location of a moving unit load.

The unit load is a concentrated load with magnitude 1,0. Influence lines will be used to find

the most unfavourable position of the load for a certain quantity in statically determinate and

statically indeterminate systems. Quantities of interest could be:

- a support reaction

- a rotation in a certain point

- an internal moment in a certain cross-section

- a deflection in a specific point

- a shear force in a specific cross-section

In this chapter we will discuss how to determine influence lines for statically determinate and

indeterminate systems. In the last part the most unfavourable position of the load for a certain

quantity will be determined. For some applications only the shape of the influence line is

important. This is referred to as the qualitative aspect of the influence line. In other cases also

the exact magnitude of the influence factor is relevant. This latter aspect is referred to as the

quantitative aspect of the influence line.

1,0 kN

l

B A

AV BV



1.1.1 Introduction of influence lines for force quantities.

The concept of influence lines is introduced with the following example.

The beam in figure 2 is loaded with a concentrated load of 1,0 kN that moves from the left

support to the right support. This moving load is denoted with the horizontal dotted arrow.

Figure 2 : Example 1, moving unit load on a simply supported beam

The dynamic effects caused by the moving load are neglected completely. The problem is

considered tob e a static problem. Using this example we will introduce the influence lines for

force quantities:

• Influence lines for support reactions Av and Bv (positive reactions assumed upwards)

• Influence lines for internal forces VC and MC

Influence line for support reactions at A and B

The question is how the support reactions in A and B are related to the position of the moving

load. Equilibrium yields to:

V V

( ) 1,0 1,0l x xA B

l l

− × ×= =

The distribution of the support reaction at A and B depending on the position of the load x is

given below.

Figure 3 : Influence lines for Av and Bv.

1,0 kN

l=10 m

z

x

1,0

1,0

Influence line

AV

Influence line

BV

0,75

2,5 m

A B C

1,0 kN

l=10 m

B A C 2,5 m

AV BV



These graphs are the influence lines for Av and Bv. To determine the support reaction Av for a

load F placed somewhere on the structure we have to multiply the influence factor found from

the influence line at that location with the magnitude of the concentrated load. A load of 50

kN at 2.5 meters from the left support results in a support reaction at A of:

kN5,375075,0V =×=A

This support reaction acts upwards. It is important to take care of the signs and to pay

attention to the definition of the chosen coordinate system. Here we have chosen to plot the

positive values (upwards acting) of the support reaction downwards in the graph. Different

sign conventions are possible but must always be apparent from the assumed positive

direction of the quantity and positive coordinate directions should be depicted with an arrow.

The reader should pay attention to this. In these lecture notes we will apply the sign

convention that influence factors that correspond to upwards acting support reactions are

plotted below the x-axis.

Using the influence lines for the support reactions it is now possible to draw the influence

lines for a shear force and bending moment in a certain cross-section. It is important to realise

that a diagram will thus be drawn from which you can see the magnitude of a shear force or

moment in that specific cross-section if a unit load moves over the structure. This will be

demonstrated with two simple examples for the shear force and bending moment in a cross

section at point C (2,5 meters from the left support).

Influence line for the shear force in point C

The influence line for the shear force can be determined by calculating the influence factor for

a number of characteristic points. It seems obvious that following positions of the unit load

should be calculated:

x=0, x=2,5 (-), x=2,5 (+) en x=10,0

With (-) and (+) a cross-section just left and right of the given location is meant. Based on

equilibrium the shear force can be determined. If for example the moving unit load is located

at 2,5 meters from the left support the situation as seen in figure 4 is reached. At C the

magnitude of the shear force is 0,25 kN.

Figure 4 : Vertical equilibrium for a concentrated load at x = 2,5 (-) meters

1,0 kN

l=10 m

2,5(-) m

0,75 kN

0,25 kN

1,0 kN

0,75 kN 0,25 kN

B A C



For the characteristic points we can thus find the following values for the influence factor:

0)0,5(75,0)5,2(25,0)5,2(0)0( ==−== +− VVVV

Using these results we can create the influence line for the shear force at C. This is shown in

figure 5.

Figure 5 : Influence line for the shear force at C

In this figure we have shown to possible ways to draw the influence line. The first is based on

the deformation symbols. The second graph is based on the sign convention used in these

notes to plot positive values of the influence line always below the x-axis. Characteristic for

the influence line is that both segments are parallel and the jump in the influence line at point

C has a magnitude of 1,0.

1,0 kN

l = 10 m

z

x

Influence line

VC

0,75

2,5 m

-0,25

B A C

Influence line

VC

0,75

0,25

1,0



Influence line for the moment in point C

The influence line for the moment can also be determined using the influence line for the

support reactions. An equilibrium method will result in an influence line as shown in figure 6.

Figure 6 : Influence line for the moment in point C

The influence factor for the maximum moment at C follows from this diagram and is equal to

1,875. This is not a dimensionless number since the length is used in the calculation of this

value. Check this yourself!

Characteristic for this influence line is that the angle between both segments is equal to 1,0:

1,875 1,875

1,02,5 7,5

θ = + =

The concept of influence lines is introduced based on this example. Obviously we can also

draw influence lines for displacement quantities like deflections and rotations. For this we

need an additional tool which will be discussed in the next paragraph.

1,0 kN

l=10 m

z

x

Influence line

MC

2,5 m

1,875

B A C

1,875/2,5 1,875/7,5

θ = 1,0



1.1.2 Work and reciprocity, Clapeyron and the theorem of Betti and Maxwell.

Work is the product of a force and associated displacement. The associated displacement is

the component of the displacement in to the direction of the force as shown in figure 7. Using

vector notation work can be expressed as the dot product of the load vector F��

and

displacement vector u�

.

Figure 7 : Work generated by a force.

Obviously the work done can also be negative. In that case the force and corresponding

displacement are opposite towards each other. It is obvious that a displacement perpendicular

to the force does not contribute to the amount of work.

If on a linear elastic structure a force is applied slowly up to a value of F the structure will

deform and the deflection will grow up to a value of u. In figure 8a this situation is shown.

The force-displacement diagram is shown in figure 8b.

Figure 8 : work done by a concentrated load

If the force increases a little bit the deflection will increase with a value du. The load F will

generate additional work:

uFA dd ×=

When the final load is reached the total work is equal to:

∫ ×=u

uFA0

d

For the linear elastic structure we can easily see that the total work generated, is equal to the

area under the force-displacement curve:

FuA21=

(a) loaded situation

(b) force-displacementdiagram

u

force ukF ×=

du

F

deflection

u

F Linear elastic structure

uFA dd =

F

u

uF Work :

F orA F u A F u= × =��

i



In EM-1-E paragraph 15.1 the concept of strain energy is introduced. The amount of work

generated by the load is used to deform the structure. If the load is applied very slowly no

dynamics is involved and thus no kinenetic energy is involved. All generated work must be

converted into strain energy Ev.

VEA =

This direct relation between work and strain energy is also called the theorem of B.P.E.

Clapeyron (1799-1864). Clapeyron stated that for elastic systems the order in which the

system is loaded does not affect the total amount of work generated. This is because the work

is the sum of the work performed by each individual load. This principle of superposition

results in a total amount of work which must be stored in the deformed structure as strain

energy.

An application of the law of Clapeyron on a linear elastic system is shown below. The loads

are applied slowly and simultaneously to the given final values. The resulting final deflections

are shown as well.

Figure 9 : Clapeyron

The total amount of work that has to be stored as strain energy equals to:

iiuFuFuFuFEA ∑=++==

21

3321

2221

1121

V

1.1.2.1 Reciprocity, theorem of Betti and Maxwell’s law

When a structure is loaded the external forces generate work. Consider a beam loaded by two

concentrated loads applied one after the other. First the left concentrated load is slowly

applied until its final value of Fa is reached. Then the right load is applied slowly from 0 to its

final value Fb. This is shown in the left picture of figure 10.

Figuur 10 : Reciprocity

u1 u2 u3

F1 F2 F3

uaa uba

Fa Fb

ubb

uab

uaa uba

Fa Fb

ubb uab

A B A B



The displacements are shown with the first index representing the location of the

displacement and the second index indicates the point of application of the load. A

displacement uab is therefore a displacement at point A due to application of a load at point B.

The work generated by these loads for the left system equals to:

ababbb21

aaa21 uFuFuFA ×+×+×=

Please note that the load at A is applied completely before the load at B is applied. The force Fa remains constant while the force

Fb is applied.

Now we reverse the order of application of the loads. First we apply the force at B followed

by the application of the load at A. This is shown in the right picture of figure 10. The work

these point loads generate is equal to:

babaaa21

bbb21 uFuFuFA ×+×+×=

In the final situation we cannot discern in what order the loads were applied. The work

generated for both systems must be equal. This results in:

bababa uFuF =

This expression is known as the theorem of Betti. Maxwell has introduced the concept of

influence factor based on this theorem which we can use to define the deflection due to a load.

bbbbbababa

bababaaaaa

FcuFcu

FcuFcu

==

==

If we substitute these influence factors in the equation we found we obtain:

baab

ababbaba

cc

FcFFcF

=

=

This result reveals that the displacement at A due to a unit load at B is equal to the

displacement at B due to a unit load at point A. This is known as Maxwell’s reciprocal law.

The total deflection at A and B can now also be written as:

bbbababbbab

babaaaabaaa

FcFcuuu

FcFcuuu

+=+=

+=+=

Applying Maxwell and using a matrix-notation finally results in:

=

b

a

bbab

abaa

b

a

F

F

cc

cc

u

u

This relation is known as the flexibility relation and by using Maxwell’s reciprocal law this

matrix must be a symmetrical matrix.



Maxwell’s reciprocal law is derived here using forces and the displacements at different

positions. However this law is not restricted to this case only. The force vector can also be

comprised of couples and the degrees of freedom can also contain rotations thus resulting in

the following symmetrical matrix:

=

b

b

a

a

44342414

34332313

24232212

14131211

b

b

a

a

T

F

T

F

cccc

cccc

cccc

cccc

u

u

ϕ

ϕ

Two reciprocal relations are of particular interest here:

• The displacement at A due to a unit couple at A is equal to the rotation at A due a unit

load at A.

• The displacement at A due to a unit couple at B is equal to the rotation at B due to a

unit load at A.

These properties will be used later on influence lines.

The inverse of the flexibility relation is called the stiffness relation which can be written as:

−

−=

=

−

b

a

aaab

abbb

b

a

1

bbab

abaa

b

a

det

1

u

u

cc

cc

u

u

cc

cc

F

F

2

abbbaadet ccc −=

This stiffness relation can also be written as:

=

b

a

bbab

abaa

b

a

u

u

kk

kk

F

F

The stifness components can be expressed in terms of Maxwell’s influence factors:

2

abbbaa

aa

bb

2

abbbaa

abab

2

abbbaa

bb

aa

ccc

ck

ccc

ck

ccc

ck

−=

−−=

−=

The stiffness matrix is thus also a symmetrical matrix. Without elaborating on the proof behind this we

note that the determinant of the flexibility matrix is positive which means that the diagonal terms in the

stiffness matrix must always be positive. It can be proven that the eigenvalues of this matrix are all

positive and not equal to zero. This will result in a so-called positive-definite stiffness matrix.



1.1.2.2 example 2 : Illustration of Maxwell’s reciprocal theorem

In figure 11 a cantilever beam is shown that is loaded by a concentrated load and a couple at

point B.

Figure 11 : Cantilever beam loaded with a concentrated load and couple at the free end.

Using the “forget-me-nots” we can determine the deflections at the free end. For the vertical

displacement and rotation at point B the following equations must hold:

EI

lT

EI

lF

EI

lT

EI

lFw

B

2

B

B

2

B

3

BB

2

23

+=

+=

ϕ

In Matrix-notation this can be written as:

=

B

B

2

23

B

B

2

23T

F

EI

l

EI

lEI

l

EI

lw

ϕ

At the end of the beam the rotation due to a unit load F is equal to the deflection caused by a

unit couple T. This is in full compliance with Maxwell’s reciprocal theorem.

l

FB

wB

EI

TB ϕB x

A B



1.1.3 Influence lines for displacement quantities.

Using the tools we developed in the last paragraph we can now determine the influence lines

for displacement quantities. This will be applied on the example of a simply supported beam.

First we will look at the influence line for the vertical displacement and then we will look at

the influence line for the rotation of a specific point on the beam axis.

Influence line for the vertical displacement at point C.

To determine the influence line of the deflection at C we will first investigate the deflection at

C due to different point of application of the load. In figure 12 the structure is drawn again

with a number of characteristic points.

Figure 12 : Determining the deflection at C for several positions of the unit load.

To be able to draw the influence line we need the influence factors for several characteristic

points of the moving load. In figure 12 three positions of the unit load are investigated. The

influence factors for the deflection at C due to the three load positions are:

cc cd ce, andw w w

It is not necessary to determine the influence factors by moving the unit load to a different

position since Maxwell’s reciprocal theorem holds:

ecce

dccd

ww

ww

=

=

This means that the deflected beam due to a unit load at C is equal to the influence line for the

deflection at C. Check this yourself thoroughly!

To determine the value of the influence factors the deflection has to be determined for a unit

load at point C. This can be done in several ways:

- By using the forget-me-nots

- Method of elastic weights

- By application of the moment-area theorem (M/EI-diagram)

- By using computer software like for example: MatrixFrame

1,0 kN

4 × 2,5 m

B A C D E

1,0 kN 1,0 kN

wcc wdc wec



Try to determine the deflection line as shown in figure 13 yourself using one of the above

mentioned methods.

Figure 13 : Influence line for the deflection at point C

Influence line for the rotation in point C

In the diagram below the requested influence factors for the rotation at C are indicated.

Figure 14 : Rotation of the cross-section at C due to a moving unit load

For different positions of the unit load we monitor the slope of the deflection line at point C.

To determine the influence factors of the influence line for a rotation at point C we can once

again make use of Maxwell’s reciprocal theorem.

1,0 kN

4 × 2,5 m

B A C D E

11,718/EI Influence line for the

deflection in point C 14,322/EI

9,114/EI

1,0 kN

4 × 2,5 m

B A C D E

Deflection due to F at C

1,0 kN 1,0 kN

Deflection due to F at D

Deflection due to F at E

ϕcc

ϕcd

ϕce



According to Maxwell the following holds:

=

d

d

c

c

44342414

34332313

24232212

14131211

d

d

c

c

T

F

T

F

cccc

cccc

cccc

cccc

w

w

ϕ

ϕ

• The displacement in C due to a unit load at D is equal to the displacement at D caused by

a unit load at C.

• The displacement at C due to a unit couple at C is equal to the rotation in C as a result of

the unit load at C.

• The displacement at C due to a unit couple at D is equal to the rotation at D caused by a

unit load at C.

Based on the reciprocal relations above and the influence factors can be obtained with:

Step 1: The rotation at C due to a unit load at C (=desired influence factor) is equal to the deflection at

C as a result of a unit couple at C.

12c12c

12c12c

)0,1(

)0,1(

cTcw

cFc

==×=

==×=ϕ

Step 2 The rotation at C due to a unit load at D (=desired influence factor) is equal to the deflection at

D due to a unit couple at C.

23c23d

23d23c

)0,1(

)0,1(

cTcw

cFc

==×=

==×=ϕ

The result of these steps is that the deflection line due to a unit couple in C is the requested

influence line for the rotation of the cross-section at C. In the figure below this approach is

applied to the simply supported beam.

Figure 15 : Influence line for the rotation at C

Determining the value of the influence factors is left to the reader. In an identical manner as

described for the influence line for the deflection these influence factors can be obtained for

characteristic points of the beam. From this introduction it is apparent that determining the

influence factors of displacement quantities is somewhat laborious and a general procedure

for force quantities is yet missing. In the next paragraph we develop a convenient and fast

method to draw influence lines for force quantities.

l=10 m

B A C D E

Influence line for the

rotation at C

TC=1,0 kNm



1.2 Statically determinate systems

In the previous paragraph the concept of influence lines has been introduced. For statically

determinate systems we will now present a fast method to determine influence lines for force

quantities based on a qualitative approach. To determine influence lines that correspond to

displacements the method shown in the previous paragraph can be used.

1.2.1 Qualitative approach

To determine the influence line in the previous paragraph we put the load at a specific point

and found the force distribution based up on equilibrium. Repeating this procedure for several

characteristic points of the load leads to the required influence line. Professor Heinrich

Müller-Breslau introduced a qualitative and fast way to determine influence lines of force

quantities that is based on virtual work.

Müller-Breslau The shape of the deformed structure is equal to the influence line of a force quantity (for example a support reaction, a shear force

or bending moment) if this quantity performs negative virtual work by applying a virtual unit associated displacement at the

location of the quantity. The influence factor is positive if the associated displacement has the same direction as the unit load.

In figure 16 this is once again explained using a simply supported beam.

Figure 16 : Müller-Breslau for the influence lines AV and BV

To determine the influence line for the support reaction at A according to Müller-Breslau’s

principle we will perform the following two steps:

1. Turn the structure into a mechanism and apply a unit displacement δw =1,0 in the direction of the

quantity (support reaction at A) such that the support reaction performs negative virtual work. For a

upward assumed positive support reaction therefore the corresponding unit displacement is downwards.

2. The shape of the deformed structure is equal to the influence line for the required force quantity

(support reaction at A). Check this yourself based on the previously derived influence lines. Since the

influence line has no change of sign, the support reaction will be positive for all possible positions of

the unit load.

+

+

1,0 kN

l=10 m

B A C 2,5 m

B

AV

δ w =1,0 Influence line

for AV

A

BV

δ w =1,0

Influence line

for BV

x

z



The prove for this approach can be shown using the principle of virtual work. Releasing one

of the degrees of freedom changes the structure into a mechanism. The virtual work generated

by all external loads due to a kinematically admissible virtual displacement δw must be zero.

The support reaction generates negative virtual work due to the imposed virtual

displacement. The unit load in this case will generate positive work. Applying the principle of

virtual work for the load at A results in:

V

V

load at A: 1,0 0

1,0

w A w

A

δ δ− × + × = ⇔

= ↑

V

V

load at : 1,0 0x

x

x w A w

A w

δ δ

δ

− × + × = ⇔

= ↑

This result is equal to the influence factor we found for the support reaction at A due to a unit

load at A. In a similar way we can derive an influence factor for a unit load at x which will

give an expression which is identical to the displaced mechanism (right expressions).

For the internal forces like the shear force and the bending moment we can also apply Müller-

Breslau’s principle. To determine the influence line for a bending moment we have to perform

the following steps:

1. The structure has to be turned into a mechanism. By placing a hinge at C we have

created a mechanism. This is shown in figure 17. The virtual unit displacement

corresponding to the internal moment is a rotation of δθ =1,0

2. The displaced structure is the influence line. The influence factor at C for the bending

moment follows from the displacement at point C. With the geometric relation

between the unit rotation and the deflection at C the influence factor can be found.

This is demonstrated in the figure below.

Figure 17 : Müller-Breslau for the influence line MC

The displaced structure is the influence line for MC . The influence factor at C is 1,875. The

positive axis of the influence line is indicated by the black arrow with the + at the origin. The

influence line is positive over the entire length of the beam since the displacement has the

same direction as the unit load for all positions of the load.

With virtual work the result found can be checked. Use the mechanism from figure 17 with a

virtual displacement at C equal to δw. Zero total virtual work results in:

1,0 kN

l=10 m

z

x

Influence line

MC

2,5 m

B A C

S MC MC

δ w

δ θ =1,0 875,1

0,15,75,2

=

=+=

w

ww

δ

δδδθ

+



{ }kNm875,1

00,1

0

C

5,71

5,21

C

C

=

+×=

=×−×

=

M

wMw

Mw

A

δδ

δθδ

δ

This result is indeed the same as the previously found result.

remark : The bending moment generates negative work. This can be seen by splitting the

mechanism in two parts at point C and checking how the bending moments act on the

two beam ends and how these beams rotate. The figure below shows for both beam

ends opposite direction of the beam axis rotation and the bending moments at the

beam ends. The generated work is therefore negative.

Figure 18 : Work generated by bending moments at C

For the shear force we also need to find a mechanism in which the shear force generates

negative work due to a kinematically admissible unit displacement. This can be achieved by

placing a sliding hinge at the position of the cross section for which we have to determine the

influence line. This can be seen in figure 19. The shear force is drawn in its positive direction,

the sliding hinge is moved in the direction opposite to the direction of the force. The shear

force thus generates negative work.

Figure 19 : Müller-Breslau for the influence line VC

From the geometric demands1 of the mechanism we can find the values of the influence

factors. In this way we directly find the influence line as we derived earlier in figure 5. The

influence factor is positive when the displacement has the same direction as the unit load.

With these examples it has been shown that the principle of Müller-Breslau complies with the

virtual work method. Ensure that the required force quantity generates negative work in order

to obtain the correct influence line including the sign. The influence line will be equal to the

displaced mechanism.

1 Only the shear force has to generate work not the bending moment. The beam axis is therefore not allowed to

rotate at the sliding hinge.

1,0 kN

l=10 m

z

x

Influence line

VC

2,5 m

B A C

“sliding hinge”

VC

VC

δ w=1,0

+

δw

MC MC 5,72

wδθ =

5,21

wδθ = x

z

ϕ



1.2.2 example 3 : influence lines for simply supported beam with a cantilever

By means of an example Müller-Breslau’s principle will be demonstrated. In figure 20 a

statically determinate beam is shown that has a cantilever on the left side.

Figure 20 : example 3, beam with cantilever

Find the influence lines for the following quantities.

� Support reaction at A

� Support reaction at B

� Internal moment at C

� Shear force at C

� deflection at C

� the rotation at C

Influence lines for the support reactions

To determine the influence lines for the support reactions we apply a unit displacement at the

location of the supports in such a way that the support reaction performs negative work. The

mechanism that results has a displacement equal to the influence line that corresponds to the

support reaction. In figure 21 this is shown, the unit displacements are shown in bold.

Figure 21 : Influence lines for the support reactions

The value of the influence factors can also be derived using the virtual work equation.

A

8 m 3,5 m

B δ f

δ f

BV

AV

1,0 kN

1,0 kN

δ w

δ w

+

+

1,0 (0,5)

1,44

1,0 (0,5)

0,44

1,0 kN

8 m

B A C 4,0 m

x-as

z-as 3,5 m

D

EI



By applying a unit load at the location of interest we can use virtual work to determine this

influence factor, e.g. for point C the following results:

1V 2

1V 2

1,0 0 with :8 4

1,0 0,5

f wA f A w w f

wA

f

δ δδ δ δ δ δ

δδ

= − × + × = = ⇒ =

= = × =

These influence factors are shown in brackets in figure 21. The support reaction at A is

always upwards, but this does not hold for the support reaction at B!

Influence lines for internal forces

To find the influence line for the bending moment in the cross-section at point C we apply:

• Create a mechanism by placing a hinge at C.

• Apply a unit rotation at C such that the bending moment MC generates negative work.

• The displaced shape of the mechanism is the required influence line.

This is shown in figure 22.

Figure 22 : Influence line for MC.

From the geometrical demands of the mechanism can be seen that the deflection at C is equal

to 2,0. This is the influence factor for the bending moment at C due to a unit load at C:

kNm0,210,2CC =×=M

This same value can be found based on equilibrium: 0,280,125,041

CC =××== FlM kNm

To find the shear force at C we also have to create a mechanism such that a unit displacement

that corresponds to the force quantity (shear force) can be applied. In figure 23 a sliding hinge

is placed at C to create the mechanism.

Figure 23 : Influence line for the shear force at point C

8 m 3,5 m

δ w

+ δδδδθθθθ = 1,0

(2,0)

1,75

8 m 3,5 m +

1,0

0,4375

0,5

0,5



Influence lines for the displacements

To determine the influence lines for displacements we will make use of the deflection line due

to a unit load as derived in paragraph 1.1.3. To determine the influence line for the vertical

deflection at C we must place the unit load at point C. The resulting deflection line is the

influence line we are looking for.

Figure 24 : Influence line for the deflection at C

The maximum deflection at the C can be found using the forget-me-nots:

EIEI

Fw

EIEI

Fw

145,3

16

85,3

667,10

48

8 2

CD

3

CC −=××

−=×−==×

= θ

To find the influence line for the rotation at C we have to apply a unit couple at C. In this case

the influence line we are looking for is also equal to the deflection line that results from this

unit couple. This deflection line is shown in figure 25.

Figure 25 : Influence line for the rotation at C

The displacement at C is zero. This means that the rotation at C due to a unit load at C is

equal to zero. For point D the influence factor can be derived using the forget-me-nots :

EIEI

Tw

1667,15,3

24

8CD −=×

×−=

8 m 3,5 m

1,0

+

1,1667/EI

D C

8 m 3,5 m

1,0

+ 10,667/EI

14,0/EI

D C A B θ



1.2.3 example 4 : Beam with a hinge

In figure 26 we have shown a hinged beam.

Figure 26 : example 5, a hinged beam

For this beam the following influence lines are requested:

• All the support reactions (upwards is the positive convention)

• Internal moment in E, halfway across the 2nd

span.

• Shear force in E, halfway across the 2nd

span.

• Displacement in S2

In the figures below the influence lines are given. Determine the values marked with y

yourself.

Figure 27 : Influence lines corresponding to example 4

Influence line

for AV

Influence line

for BV

Influence line

for CV

Influence line

for DV

Influence line

for ME

Influence line

for VE +

Influence line

for wS2

+

+

+

+

+

+

1,0

1,0

1,0

y

y

1,0

y

y

1,0

1,0

y y

1,0

5,0 m 5,0 m 10,0 m

4,0 m 2,5 m

S2 S1 EI

EI EI

z

x

1,0

A B C D E



1.3 Statically indeterminate systems

For statically indeterminate structures a similar approach as for statically determinate

structures can be applied to find the influence lines for force quantities. There is one

important difference though. The influence lines for all the force quantities are no longer

linear functions but they are curved. This will be illustrated with a statically indeterminate

beam as shown in figure 28. This structure has one redundant since it is statically

indeterminate to the degree of one.

Figuur 28 : Statically indeterminate beam

The clamping moment TB (negative bending moment) at B can be determined using the

forget-me-nots. The compatibility equation with respect to the deformation is that the rotation

in point B has to be equal to zero:

BB2

22

B

B

22B

AB

en2

)(

036

)(

TMl

xlFxT

EI

lT

EIl

xlFx

−=−

=

=−−

=ϕ

The clamping moment works in the shown direction for all values of x between 0 and l. The

vertical support reactions in A and B can be found from equilibrium. Check for yourself that

the results you find for these are equal to:

VV

3

323

V

1

2

23)(

AB

Fl

lxlx

l

T

l

xlFA B

−=

×+−

=−−

=

The support reactions act upwards for all values of x between 0 and l. With the functions we

have derived above the influence lines are now determined. For a number of values in

between the influence factors can be determined. In the table below these values are given for

points A, B, C and D.

Table : Influence factors

x/l AV ][ F× BV ][ F× TB ][ Fl× MB ][ Fl×

0,00 1,0000 0,0000 0,0000 -0,0000

0,25 0,6328 0,3672 0,1172 -0,1172

0,50 0,3125 0,6875 0,1875 -0,1875

0,75 0,0859 0,9141 0,1641 -0,1641

1,00 0,0000 1,0000 0,0000 -0,0000

1,0 kN

l

B A C

x=0,25 l

D

x=0,75 l

TB

BV AV



The influence lines for the force quantities can be plotted with this data. In figure 29 this

result is shown. These influence lines are no longer straight. Check for yourself that the

maxim value of the clamping moment is found if the unit load is placed at a distance

lx 331= .

Figure 29 : Influence lines for a statically indeterminate beam.

The consequence of this “discovery” is that determining the exact shape of the influence line

becomes a more laborious process than in statically determinate cases. Especially drawing the

influence lines for shear forces and bending moments in a cross section is a bit more work.

That is the reason we have not looked at them at first. In the next paragraph we will explain

that it is not always necessary to determine the exact shape of the influence line and that we

can suffice with the following:

• Values at the characteristic points,

• combined with the knowledge that the influence line is curved,

• and the shape can be determined with the principle of Müller-Breslau as described

before.

1,0 kN

l

B A C

Influence line

for MB

D

Influence line

for AV

Influence line

for BV

0,12l 0,19 l

0,16 l

1,00

0,31

0,63

0,09

1,0 0,37

0,69 0,91



1.3.1 Qualitative approach

When determining influence lines for statically determinate systems we have used the

following approach:

Displacement quantitites:

• Apply a unit load that corresponds to the requested displacement quantitiy, the

deformed shape is the influence line

Force quantities:

• Apply a unit displacement that corresponds to the force quantity you are interested in

such that the quantity performs negative work. The deformed shape is the desired

influence line. The quantity is positive for displacements in the direction of the unit

load.

To apply the unit displacement we added a degree of freedom by introducing a sliding hinge

or hinge (shear force, bending moment) or by removing one of the supports. For a statically

determinate structure this approach results in a mechanism. For statically indeterminate

systems the same approach will be used. The difference is that because the systems is

indeterminate the released structure will not be a mechanism. This principle will be applied on

the indeterminate structure of figure 28. The proof for this approach is found in the

APPENDIX.

Influence lines for the support reactions

To determine the influence line for the support reactions we have to apply a unit displacement

at the point of interest. In figure 30 this is illustrated for both support reactions.

Figure 30 : Influence lines for the support reactions according to Müller-Breslau

Without immediately determining the influence factors we quickly have a good view of the

influence line. Pay attention with this method that only the displacement that corresponds to

the force quantity you are looking for is given a unit displacement. This can be seen at the

right support, the clamped end results in a horizontal tangent for both influence lines at the

right end.

1,0 kN

l

B A

1,0

1,0

Influence line

for AV

Influence line

for BV

BV AV



The clamping moment at support B is of course also a support reaction. Here we can also

apply the same principle. The displacement that corresponds with the clamping moment is the

rotation of point B. Assume a positive moment in the beam and apply at B a unit rotation

such that TB performs negative work. The result is the displacement field as shown in figure

31. This is the influence line we are looking for.

Figure 31 : Influence line for the clamping moment at B.

The determination of the influence factor for characteristic points will be addressed later.

Influence lines for the internal forces

To determine the influence line for the shear force and the internal moment we can also apply

Müller-Breslau’s principle. The shape of the influence line for the shear force at point C is

sketched in figure 32. Pay attention: we have applied a sliding connection, the slope left and

right of C has to be the same! (otherwise the bending moment would also perform work.)

Figure 32 : Influence line for the shear force at point C

1,0 kN

l

B A

Influence line

for VC

C

1,0

1,0 kN

l

B A

θ = 1,0

0,1875 l

Influence line

for MB

TB

TB

MB



The influence line for the internal moment at point C can also be drawn in the same manner.

This is shown in figure 33.

Figure 33 : Influence line for the internal moment in point C

An important difference between the influence lines for statically determinate and

indeterminate systems is that the influence lines are not linear. In figure 33 for example the

part AC cannot be straight since the connecting parts at C have to make an angle equal to a

unit rotation!

1.3.2 Quantitative calculation of influence lines

Next to the previously shown qualitative approach it is necessary to determine the influence

factors at some characteristic points. This will be shown based upon the previous example.

Support reaction in A

From the influence line as shown in figure 30 we want to find the influence factor at mid

span. There are again multiple methods at our disposal to determine this influence factor. One

option is to use the force method based on the forget-me-nots to determine the force

distribution in the beam. To do so we apply a unit load at mid span. Then we create a

statically determinate system by removing the support at A. As a compatibility equation we

demand zero deflection at point A (there is a support there). This approach is further clarified

in figure 34.

Figure 34 : Support reaction at A due to a unit load at E

The demand for zero deflection at A using the forget-me-nots results in the following

expression for the vertical support reaction at A:

1,0 kN

l

B A

Influence line

for MC

C

1,0

B E

½l AV

½l

z-as

x-as

1,0



( ) ( )

3125,016

5

2

0,1

3

0,1

30

V

21

2

213

213

A

==

××

+×

+×

−==

A

lEI

l

EI

l

EI

lAw V

This influence factor for the support reaction at A due to a unit load at E was already

indicated in figure 29.

Internal Moment in B

To determine the influence line for the clamping moment at B we will apply a unit rotation at

point B. The deflection line we obtain using this method is the requested influence line. This

load case is shown in figure 35. The deflection at mid span can be determined using for

example the forget-me-nots or the moment-area theorem.

Figure 35 : Deflection line due to a unit roation at B.

The rotation at B is equal to -1,0. Using a forget-me-not we can thus determine the clamping

moment at B:

l

EIM

EI

lM 30,1

3B

BB −=⇒−==ϕ

Using the second moment-area theorem we can also determine the deflection at mid span. The

reader is asked to check these results:

lwE 163−=

This deflection is the requested influence factor for the clamping moment at point B due to a

unit load at E:

lM 1875,0B −=

This value corresponds to the value indicated in figure 29.

kwispel-effect

1,0 Deflection line

for a unit

rotation in B

l

3 5,0A =ϕ

83

23

21

21 ** ==

llθ

la61=

M/EI – diagram

(moment-area

theorem)

x

z

M/EI

0,5

E

MB



Influence line for the bending moment at C

To determine the characteristic values of the influence line for the bending moment at C we

can use the influence line for the support reaction at A. The influence factor can be

determined using the following:

1

C V 4

1C V 4

0 0,25 : 1,0 (0,25 )

0, 25 :

x l M A l l x

l x l M A l

≤ ≤ = × − × −

≤ ≤ = ×

In the table below the influence factor for several values of x have been collected.

Table : Influence factors for MC

x MC [*Fl] 0,00 0,0000

0,10 0,0626

0,20 0,1260

0,25 0,1582

0,30 0,1409

0,40 0,1080

0,50 0,0781

0,75 0,0214

1,00 0,0000

The influence line for MC is shown in figure 36. The line appears linear for the part AC but it

actually isn’t, to see this check the values in the table above and the earlier remark about this

that corresponds to figure 33.

Figure 36 : Influence line for MC.

The influence line for the shear force can be determined in the same manner. Determine for

yourself several characteristic values of the influence factor and draw the influence line based

on figure 36.

1,0 kN

l B A

Influence line

for MC

C

0,158 l



Example 5 : Continuous beam with three supports

In figure 37 a continuous beam with three supports is drawn.

Figure 37 : Example 5, continuous beam with three supports

The question is to determine the following influence lines:

o Support reactions at A, B and C

o Bending moment MD

o deflection at D

Before we will determine the influence lines precisely we will first use a qualitative approach

to sketch the correct shape of the influence lines. Based on these sketches we can then

determine for which characteristic points we have to calculate the exact values of the

influence factors to get a sufficiently accurate influence line.

Qualitative approach

Based on Müller-Breslau’s principle we can sketch the correct shape of the influence line. For

force quantities we apply a unit displacement, for displacement quantities we apply a unit

load. Check for yourself that the shown influence lines of figure 38 comply with this approach

in terms of shape.

Figure 38 : Example 5, qualitative influence lines

4,0 m 2,0 m

EI

C A B D

1,0 m 1,0

1,0

1,0

1,0

1,0

Influence line for AV

Influence line for BV

Influence line for CV

Influence line for MD

Influence line for wD

1,0



For the support reactions the influence lines we have found so far based on this approach are

quite accurate already. To be able to determine the influence line for the bending moment at

point D it is convenient if we first determine the influence line for the support reaction at C.

This is why we will discuss, for completeness, all the influence lines in a quantitative way.

Quantitative approach, support reactions

To determine the support reactions one can make use of several methods. The fastest method

would of course be to use some computer software to solve the problem. In these lecture notes

though we will once again make use of the forget-me-nots.

The structure is statically indeterminate to the first degree. If the support reaction at B is

chosen as the unknown redundant (so we remove the support at B) we need a compatibility

demand for zero deflection at B, since the beam is supported at B. The influence line for the

support reaction at B can then be determined in the same way as the previous example. This is

clarified in figure 39 in which a positive support reaction is assumed to act downwards. This

choice is made to make Maxwell’s formulas more recognizable in this example.

Figure 39 : Compatibility demand at B

The support reaction RBX at B due to a unit load at a distance x can be determined as follows:

bb

xb

bb

bxBX

BXbbbxbbbxB 00,10

c

c

c

cR

Rccwww

−=−=

=×+×=+==

The influence line can be determined by calculating the deflection at several points along the

beam due to a unit load at B. This means one single load case! These deflections are then

scaled by the deflection at B due to the unit load. This approach is basically a proof for the

method that was introduced in 1886 by Müller-Breslau. See for this proof, also the

APPENDIX.

4,0 m 2,0 m

EI

C A B

1,0

x

C A B

wbx

1,0

x

C A B

wbb

RBX

Please note we apply Maxwell’s law

here.



This calculation can be exectued in several ways. Here forget-me-nots will be used to

ccalculate the influence factor cxb once every meter along the beam.

determining cbx:

To determine the deflection at B due to a unit load at B using forget-me-not’s:

Figure 40 : forget-me-not for a point load

The deflection due to a concentrated load at B can be determined for every meter along the

beam using the moment-area theorem as has been done in a previous example. The results are

collected in the following table:

Table : Influence factors for cbx

x cbx 0 0

1

EI9

19

2

EI9

32

3

EI9

5,34

4

EI9

28

5

EI9

5,15

6 0

With this set of data we can determine the influence factors of the influence line, for several

characteristic points by making use of:

bb

xb

bb

bx

BXc

c

c

cR −=−=

Obviously because of the scaling the influence factor at B will be equal to -1,0 and at A and C

it’s must be equal to 0,0.

b = 4,0 a = 2,0

EI

A B

1,0

A

C

2 2 2 2

BB

( )

6

( )

6

32

3 3 9bb

F ab l b

lEI

F ab l a

lEI

Fa b a bw c

lEI lEI EI

ϕ

ϕ

× += −

× +=

= ⇔ = =

C x

z



With this we find the following results:

Table : influence factors for RB

x RB 0,0 0,000

1,0 -0,594

2,0 -1,000

3,0 -1,078

4,0 -0,875

5,0 -0,484

6,0 0,000

remark:

The influence factors can obviously also be found using computer software. By placing a unit load at

point B and determining the deflection for every meter we can quickly determine the influence line.

This means that we can find the influence line based on one load case with one calculation!

The influence line for RB can now be plotted accurately. Based on equilibrium all values of

the other support reactions can be found. In the table below the influence factors of all the

support reactions have been listed. In figure 41 all influence lines have been drawn again

making use of the standard conventions for directions and signs of the support reactions.

Table : Influence factors for the support reactions

x RA RB RC 0,0 -1,000 0,000 0,000

1,0 -0,437 -0,594 0,031

2,0 0,000 -1,000 0,000

3,0 0,219 -1,078 -0,141

4,0 0,250 -0,875 -0,375

5,0 0,156 -0,484 -0,672

6,0 0,000 -0,000 -1,000

Figure 41 : Influence line for the vertical support reactions at A, B and C

For these influence lines the standard conventions for vertical support reactions is used:

upwards acting reactions are positive which is plotted downwards in the influence line. These

lines show a clear resemblance with the expectations from the qualitative approach.

1,0

1,0

1,0

Influence line for AV



1,078

0,141

Note:

The assumed direction of the support reaction

B was wrong. The influence factors we found

are all negative! The correct direction is

upwards.

For the drawing of the influence lines the

standard conventions for signs and directions

is used. The vertical support reactions are

now AV, BV and CV. These support reactions

are positive if they act upwards. For all

influence lines holds that positive values are

plotted downwards.




The influence line for the bending moment at D can be determined using the found influence

line for CV. Check for yourself that following influence factors are found:

Table : Influence factors for MD

x MD ][ l×

0,0 0,000

1,0 -0,031

2,0 0,000

3,0 0,141

4,0 0,375

5,0 0,672

6,0 0,000

The influence line is shown in figure 42 (not to scale).

Figure 42 : Influence line for MD

1.3.3 Example 6 : Nth order statically indeterminate structures

The structures discussed so far have been statically indeterminate with one single redundant.

For nth

-times statically indeterminate structures the same solution strategy can be applied. For

these cases n unknowns are chosen accompanied by n compatibility equations. In figure 43 a

2-times statically indeterminate beam is shown together with the statically determinate

auxiliary system with the two unknowns, the vertical support reactions at B and C. Positive

support reaction are in this case assumed to act downwards. This choice is made to make

Maxwell’s formulas more recognizable in this example.

Figure 43 : 2-times statically indeterminate system

The accompanying compatibility equations are the demands for zero deflection at B and C.

This approach is identical to the outlined procedure in the previous example, use a statically

determinate auxiliary structure and elaborate the compatibility demand.


0,67 l

6,0 m 4,0 m

EI

C A B D

1,0

5,0 m

x

1,0

A

RB

D

RC



To find the influence lines, Maxwell’s reciprocal theorem is used which states that the

deflection at B due to a load at x is equal to the deflection at x due to the same load at B.

o Deflection at B due to a load at x : wbx = xb

o Deflection at B due to the unknown at B : wbb

o Deflection at B due to the unknown at C : wbc = cb

o Deflection at C due to a load at x : wcx = xc

o Deflection at C due to the unknown at B : wcb = bc

o Deflection at C due to the unknown at C : wcc

Using these results the compatibility equations can be rewritten as:

0

0

cccbxc

bcbbxb

=++

=++

www

www

These conditions can be written in terms of influence factors:

00,1

00,1

CccBbcxc

CbcBbbxb

=×+×+×

=×+×+×

RcRcc

RcRcc

Solving this system of equations results in the expressions for the two unknown (redundant)

support reactions:

2

bcccbb

xcbbxbbc

2

bcccbb

xbccxcbc

B enccc

ccccR

ccc

ccccR C −

−=

−

−=

To conclude the general procedure to follow is:

• First determine once the coefficients cbb , ccc and cbc for the auxiliary system,

• Then determine for a number of points the coefficients cxb and cxc

( perform two calculations, first with a load of 1,0 kN at B and then with 1,0 kN at C )

• Then solve the influence factors RB and RC for every position of x.

All other influence lines can be found based upon equilibrium by using the previously found

influence lines for the support reactions. This also holds for the bending moment and shear

force at a certain cross-section of the beam.

It should be obvious from the above example that determining an influence line of a statically

indeterminate structure is quite a laborious process. Based on the principle of Müller-Breslau

however a quick sketch of the influence line (qualitative) can be obtained without

calculations. In most cases this will give enough info for the most unfavourable position of

the load. If necessary more precise calculations will give the specific values of the influence

factors (quantitative) at characteristic points. For this frame analysis programs can be used or

in simple cases classical methods based on the forget-me-nots in combination with a

spreadsheet.

Assignment:

Determine for the given beam the influence lines for RB and RC. Practically this means that

two calculations for the statically determinate auxiliary system have to be performed. First a

calculation with a unit load at B, and then a calculation with a unit load at C. Collect the

deflection for every meter and then, using the system of equations, solve the unknowns and

draw the influence lines for RB and RC. This can all be done using a frame analysis program

and a spreadsheet but also by using the forget-me-nots.



For the calculations the previous defined directions of the support reactions RB and RC are

used again (positive downwards). As a check the results of the calculations are gathered in the

following table. To draw the influence lines the standard conventions concerning signs and

directions is used (positive values are plotted downwards).

The result shown in figure 44 corresponds with the expected result from a qualitative

approach: The influence line for the vertical support reaction has the same shape as the

deflection line when the support is given a unit displacement.

Figure 44 : influence lines for BV and CV

Table : calculation results

x cxb [*10-3] cxc [*10-3] RB RC

0 0,00 0,00 0,00 0,00

1 12588,89 11055,56 -0,34 0,05

2 24444,45 21777,78 -0,65 0,09

3 34833,34 31833,33 -0,88 0,08

4 43022,22 40888,89 -1,00 0,00

5 48444,45 48611,11 -0,98 -0,15

6 51200,00 54666,67 -0,85 -0,36

7 51555,56 58722,22 -0,64 -0,58

8 49777,78 60444,45 -0,41 -0,79

9 46133,34 59500,00 -0,18 -0,94

10 40888,89 55555,56 0,00 -1,00

11 34311,11 48444,45 0,10 -0,95

12 26666,67 38666,67 0,14 -0,80

13 18222,22 26888,89 0,12 -0,57

14 9244,44 13777,78 0,07 -0,30

15 0,00 0,00 0,00 0,00

cbb

ccc

cbc

cbc

1,0 A D

1,0

+

+



B

C



1.4 Stiffness differences in structures

For structures with varying stiffness the principle of Müller-Breslau can also be applied to

find influence lines. A pronounced example is the theoretical case where one part of the

structure is infinitely stiff compared to the other parts of the structure. A typical example is

shown in figure 45.

Figure 45 : Statically indeterminate structure with varying stiffness

Applying Müller-Breslau has to be done with caution here. Since the infinitely stiff elements

will not bend the resulting influence lines will differ from the expected results. Check the

influence lines in figure 46 for correctness.

Figure 46 : Influence line for bending moment and shear force at D

For the shear force at B take into account that part AB will not bend. This result is shown in

figure 47.

Figure 47 : Influence lines for the shear force at B

Assignment :

Draw the influence lines for the deflection at mid span of the right span and

for the rotation at point C.

Influence

line for MD

Influence

line for VD

1,0

1,0

0,5

0,5

+

+

Influence line

for VB-left

Influence line

for VB-right

1,0

+ 1,0

+

l l

1,0

EI EI=∞

C B A D

½l



1.5 Spring supports*

To find the influence lines for structures with spring supports the previously outlined method

of example 5 will be used with a small modification due to the influence of the spring. The

modified model is shown in figure 48.

Figure 48 : Statically indeterminate structure with a spring support

The spring support with stiffness k may represent a bad foundation or for example a

supporting structure that deforms elastically.

The support reaction at B is again taken as the redundant. The compatibility equation

associated to this support reaction is the displacement at B which is equal to the shortening of

the spring. This is shown in figure 49.

Figure 49 : Displacements due to the spring support

Like in example 5 the positive support reaction is chosen in the positive z-direction. The

linear spring characteristic is also shown in this figure and will be used in the compatibility

equation. The compatibility equation for B can thus be written as follows:

kc

cR

kRuRcc

1

/0,1

bb

bxBX

BXveerBXbbbx

+−=

−=−=×+×

In the limit of zero or infinite spring stiffness k, this equation gives the expected results.

4,0 m 2,0 m

EI

C A B

1,0

k

z

x

x

C A B wbx

1,0

x

C A B wbb

RBX

B k

RBX

uspring

k

uspring

RBX

z

z

z



1.6 Distributed loads and influence lines

The influence line gives insight into the magnitude of a quantity due to a moving unit load. In

case of an applied distributed load also use can be made of influence lines. In figure 50 a

simply supported beam is shown with the influence line for the bending moment at mid span.

Figure 50 : Influence lines and distributed loads

The maximum bending moment at mid span due to a distributed load can be found by using

the influence line for a moving unit load. The structure with this uniformly distributed load is

shown in figure 51.

Figure 51 : Distributed load

The distributed load can be regarded as a series of concentrated loads each with a magnitude q

that are grouped close together. The influence of all these concentrated loads on the bending

moment at C can be found using the influence factors at the location of these small loads, this

is shown in figure 52.

Figure 52 : Distributed load

The bending moment at C thus becomes the sum of all the contributions:

qcqcqcqcM iC ∑=+×+×+×= ...321

For a continuous distributed load this summation changes into an integral:

2

81

21

41

21

0

21

0

22d)(2

1

qlqlldxxqxxcqM

ll

C =××××=×=×= ∫∫

If the distributed load is more complex it can be solved numerically using software like for

example MAPLE and DERIVE.

l

½l

EI

C A B

1,0

Influence line for MC

+

l

EI

A B

q

Influence line for MC

+

q q q

c1 c2 c3

¼l c(x)



1.7 The most unfavourable position of the load

Using influence lines it is possible to find for a specific quantity the most unfavourable

position of the load. If this is possible for a single load then this is also possible for a set of

loads or an infinite set of loads, the distributed load. In this paragraph we will first look at the

most unfavourable position of the above mentioned loads. As a starting point we will use the

statically determinate structure shown in figure 53. In this figure we have already drawn

several influence lines. Check the correctness of these influence lines.

Figure 53 : Statically determinate structure with influence lines

Unit load

Using these influence lines the position for which the quantity reaches its maximum value can

be read. To determine for example the maximum support reaction at A the load has to be

applied at the left end of the left cantilever.

Uniformly distributed load

A uniformly distributed load is

nothing more than a set of

concentrated loads packed

closely together. To determine

for example the maximum

support reaction at A due to a

distributed load the distributed

load has to be applied over the

entire part of the structure

where the influence line is

positive (below the axis). This

is shown in figure 54.

Figure 54 : Maximum AV due to a distributed load

4,0 m 2,0 m 2,0 m

1,0

EI

A B C

1,0

Influence

line for AV

1,5 +

q

Assumption:

Positive support reaction acts

upwards.

4,0 m 2,0 m 2,0 m

1,0

EI

A B C

1,0

1,0

0,5

0,5

Influence

line for AV

Influence

line for MC

Influence

line for VC

1,5 +

+

+

1,0

1,0

1,0



The vertical support reaction at A can be found with:

∫ ×=0,6

0

A d)( xxcqR

In this case the integral is equal to the area of the triangular area of the influence line

multiplied with the magnitude of the distributed:

qqR 5,45,10,65,0maxA =×××=

Check this answer also based up on equilibrium.

The use of influence lines to find the most unfavourable positions of distributed loads is also

illustrated with the example in figure 55 in which a braced frame is shown. To find the

unfavourable position of distributed loads for the bending moment at Z first the influence line

for the bending moment at Z is sketched. To find this influence line a unit rotation is applied

at a hinge which is positioned at Z and with a simple frame analysis the deflection xx can be

found which represents the influence factor. From this the spans with maximum loading

become clear as is indicated in figure 55.

Figure 55 : Most unfavourable position of the vertical distributed load for MZ

Load systems

Load systems are multiple (mobile) concentrated load which have a fixed relative position to

each other. These systems may represent a large moving truck or a train on a bridge. In figure

56 such a load system is shown.

Figure 56 : Load system on a statically determinate beam

The influence line for the bending moment at C is shown in figure 57.

x-as

1,0 kN

8,0 m

B A C 3,0 m

z-as 3,5 m

D

1,0 m

Z

xx

Influence

line for MZ

1,0



The question is to determine the most unfavourable position of the load system for the

bending moment at C.

Figure 57 : Starting position of the load system

The most unfavourable position of the system can be found by trying a few positions. Here

the sensitiveness of the position of the load system on the bending moment will be examined.

To do so the system is put at a starting position as is shown in figure 57. Then the system is

moved over a small distance x. This displacement will increase the contribution to the bending

moment of the left concentrated load and decrease the contribution of the right concentrated

load. This results in an total influence factor of the load system which depends on x:

0,10)0,5

875,1()

0,3

875,1()( 21 ≤≤−++= x

xc

xcxc

From this expression can be seen that moving the load system by 1,0 m tot the right results in

the maximum influence factor for the load system. For this case the result is easily seen. The

slope of the left part is steeper than the right part. By moving the system to the right we

increase the contribution of the left load more than we decrease the contribution of the right

load. The maximum influence of the left load is therefore the most unfavourable position of

the load system. This approach can be summarised as follows:

Increase of the influence = total load left × 1,875/3,0

Decrease of the influence = total load right × 1,875/5,0

For load systems consisting of two concentrated loads the most unfavourable position can

quite quickly be seen. It becomes slightly more difficult if the system consists of three

concentrated loads. The approach described here is still applicable though. For further

explanation of this we refer to EM-1-E paragraph 16.3.2.

8,0 m 3,5 m

+ δδδδθθθθ = 1,0

c2=1,875

2,1875

1,0 kN 1,0 m 2,0 m

x

c1=1,25



1.8 Exercises

Exercise 1

A hinged beam structure is shown in the figure below. The beam is fully clamped at C and

has two hinges S1 and S2.

a) Construct the influence line for the clamping moment at C

b) Construct the influence line for the shear force just right of B

c) Sketch the influence line for the deflection of hinge S2

remark : Construct means that the influence line must also be determined

quantitatively. Sketch means that the influence line must be determined

qualitatively where it also has to be clear whether the lines are straight or

curved.

Exercise 2

Another hinged beam structure is give in the figure below. The beam is fully clamped at C

and has one hinge S1.

a) Sketch the influence line for the moment at C

b) Sketch the influence line for the shear force just to the right of B

2,0 m 3,0 m 2,0 m 3,0 m

C S2 B A

S1

2,0 m 3,0 m 5,0 m

C B A

S1



Exercise 3

A hinged beam structure is shown in the figure below. The beam is clamped at A and B and

has two hinges S1 and S2. The influence of the axial deformation may be neglected.

a) Construct the influence line for the bending moment at A

b) Construct the influence line for the shear force halfway between S1-S2

c) Sketch the influence line for the deflection of hinge S2

remark : Construct means that the influence line must also be determined

quantitatively. Sketch means that the influence line must be determined

qualitatively where it also has to be clear whether the lines are straight or

curved.

Exercise 4

In the figure below a frame structure is given which consists of a beam on a column. The

beam is loaded with a downwards acting uniformly distributed load q.

a) Sketch the influence line for the bending moment at point G at mid span of AE.

b) Find the most unfavourable position of the distributed load q for the bending moment

at G.

c) Describe how the magnitude of this maximum bending moment can be determined

(qualitative approach)

3,0 m 4,0 m 3,0 m

B S2 A S1

C D

A

B

E

G



Exercise 5

A hinged beam structure is given in the figure below.

Find the influence lines for:

a) The vertical support reaction at B

b) The shear force at D

c) The bending moment at D

d) The most unfavourable position of the uniformly distributed load q for the above

quantities.

e) The corresponding extreme values of the quantities for this distributed load.

3a a a 3a

q kN/m

S

A B

C D



ANSWERS

2,0 m 3,0 m 2,0 m 3,0 m

C S2 B A

S1

1,0

Influence

line for MC

Influence line

for VB-right 1,0 1,5

Influence

line for wS-2

F=1,0 straight

straight

curved

2,0 m 3,0 m 5,0 m

C B A

S1

Influence

line for MC

1,0 curved

straight straight

Influence line

for VB-right

1,0

straight straight curved

3,0

4,5



3,0 m 4,0 m 3,0 m

B S2 A S1

Influence

line for MA

1,0

Influence

line for VS1-S2

0,5

0,5

Influence

line for wS2

F=1,0

straight

straight

curved

C D

A

B

E

G

Influence

line for MG

(top beam) 1,0

Total bending moment is equal to area

under the influence line times the

magnitude of the load q.



3a a a 3a

q kN/m

S

A B

C D

Influence

line for BV

BV

1,0

1,25

Influence

line for VD

0,25

0,75

0,25

Influence

line for MD 1,0 0,75 a

0,75 a Please note:

• Rewrite the unit rotation at D to a

deflection (position of the

mechanism)

• 2 options for the position of the

distributed load (left half or right

half since this has no influence on

the absolute value of the bending

moment at D) 231

max 2 24 0,75M a a q qa= × × × =

Maximum shear force: 1

max 2

1312 8

4 0,25

3 0,75

V a q

a q qa

= × × × +

× × × =



1.9 APPENDIX

Müller-Breslau’s method was introduced in 1886 to determine the influence lines of force

quantities. In this appendix the method is proven based on the continuous beam on three

supports for which we will determine the influence line of one of the support reactions. The

beam is shown in the figure below:

To find the support reaction at B due to a unit load at location x an auxiliary statically

determinate system is used in which the support at B is removed and the support reaction RB

is taken as an unknown load. The compatibility condition is a zero deflection at B from which

the unknown support reaction RB can be obtained.

Using Maxwell’s notation with influence factors this condition can be written as:

B

B B

= 0

1,0 0

bb bx

bb bx

w w w

w c R c

+ =

= + × =

The support reaction RB in B due to a unit load at x follows from:

B = 1,0bx

bb

cR

c− ×

Using Maxwell’s reciprocal theoreme bx xbc c= finally results in:

B = 1,0xb

bb

cR

c− ×

b a

EI

C A B

1,0

x

C A B

wbx

1,0

x

C A B

wbb

RB



This expression is in fact the influence line for the support reaction at B. The influence factor

depends on the variable x and is equal to:

xb

bb

c

c−

Apart from a scaling factor, this influence factor is equal to cxb which is the influence factor

for the deflection at x due to a load at B. In word, the deflection line of the statically

determinate system due to a unit load at B. The exact values of the influence factor have not

yet been determined but it must hold that:

xb xb

bb bb

c w

c w− = −

since:

B

B

bb bb

xb xb

w c R

w c R

= ×

= ×

The influence factor for the support reaction is a ratio that is independent of the value RB. If in

the above expression we choose for the scaling factor wbb a unit displacement the influence

factor becomes:

1,0

xb xbxb

bb

w ww

w

− −= = −

Expressed in words the above equation states that the influence factor for the support reaction

in B is equal to minus the deformed shape of the statically determinate system we have

chosen due to a unit displacement applied at the point of the force quantity we are looking at.

This is exactly what the principle of Müller-Breslau states; release the associated degree of

freedom and apply a unit displacement or rotation.

The minus sign in the expression arises from the assumed positive direction of the chosen

unknown RB. This is done deliberately to conform directly to the notation as introduced by

Maxwell. It should be clear that if the unknown (the support reaction at B) was chosen to be

positive if acting upwards combined with a positive (downwards) acting unit displacement the

minus sign would disappear from the above expression. In that case we let the requested force

quantity perform negative work which is in accordance to the approach we have presented

earlier for Müller-Breslau’s:

Apply a unit displacement that corresponds to the force quantity you are

investigating, such that the force quantity performs negative work. The deformed

structure is the influence line you are looking for. The quantity is positive for

displacements in the direction of the unit load.

Notes InfluenceLines

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