This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Structural Analysis IV Chapter 1 – Course Introduction
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 3
2.1 Introduction
2.1.1 Purpose
Previously we only used virtual work to analyse structures whose members primarily
behaved in flexure or in axial forces. Many real structures are comprised of a mixture
of such members. Cable-stay and suspension bridges area good examples: the deck-
level carries load primarily through bending whilst the cable and pylon elements
carry load through axial forces mainly. A simple example is a trussed beam:
Other structures carry load through a mixture of bending, axial force, torsion, etc. Our
knowledge of virtual work to-date is sufficient to analyse such structures.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 4
2.2 Virtual Work Development
2.2.1 The Principle of Virtual Work
This states that:
A body is in equilibrium if, and only if, the virtual work of all forces acting on
the body is zero.
In this context, the word ‘virtual’ means ‘having the effect of, but not the actual form
of, what is specified’.
There are two ways to define virtual work, as follows.
1. Virtual Displacement:
Virtual work is the work done by the actual forces acting on the body moving
through a virtual displacement.
2. Virtual Force:
Virtual work is the work done by a virtual force acting on the body moving
through the actual displacements.
Virtual Displacements
A virtual displacement is a displacement that is only imagined to occur:
• virtual displacements must be small enough such that the force directions are
maintained.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 5
• virtual displacements within a body must be geometrically compatible with
the original structure. That is, geometrical constraints (i.e. supports) and
member continuity must be maintained.
Virtual Forces
A virtual force is a force imagined to be applied and is then moved through the actual
deformations of the body, thus causing virtual work.
Virtual forces must form an equilibrium set of their own.
Internal and External Virtual Work
When a structures deforms, work is done both by the applied loads moving through a
displacement, as well as by the increase in strain energy in the structure. Thus when
virtual displacements or forces are causing virtual work, we have:
00I E
E I
WW W
W W
δδ δ
δ δ
=− =
=
where
• Virtual work is denoted Wδ and is zero for a body in equilibrium;
• External virtual work is EWδ , and;
• Internal virtual work is IWδ .
And so the external virtual work must equal the internal virtual work. It is in this
form that the Principle of Virtual Work finds most use.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 6
Application of Virtual Displacements
For a virtual displacement we have:
0
E I
i i i i
WW W
F y P e
δδ δ
δ δ
==
⋅ = ⋅∑ ∑
In which, for the external virtual work, iF represents an externally applied force (or
moment) and iyδ its virtual displacement. And for the internal virtual work, iP
represents the internal force (or moment) in member i and ieδ its virtual deformation.
The summations reflect the fact that all work done must be accounted for.
Remember in the above, each the displacements must be compatible and the forces
must be in equilibrium, summarized as:
Set of forces in
equilibrium
Set of compatible
displacements
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 7
Application of Virtual Forces
When virtual forces are applied, we have:
0
E I
i i i i
WW W
y F e P
δδ δ
δ δ
==
⋅ = ⋅∑ ∑
And again note that we have an equilibrium set of forces and a compatible set of
displacements:
In this case the displacements are the real displacements that occur when the structure
is in equilibrium and the virtual forces are any set of arbitrary forces that are in
equilibrium.
Set of compatible
displacements
Set of forces in
equilibrium
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 8
2.2.2 Virtual Work for Deflections
Deflections in Beams and Frames
For a beam we proceed as:
1. Write the virtual work equation for bending:
0
E I
i i
WW W
y F M
δδ δ
δ θ δ
==
⋅ = ⋅∑
2. Place a unit load, Fδ , at the point at which deflection is required;
3. Find the real bending moment diagram, xM , since the real curvatures are given
by:
xx
x
MEI
θ =
4. Solve for the virtual bending moment diagram (the virtual force equilibrium
set), Mδ , caused by the virtual unit load.
5. Solve the virtual work equation:
0
1L
xx
My M dxEI
δ ⋅ = ⋅ ∫
6. Note that the integration tables can be used for this step.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 9
2.2.3 Virtual Work for Indeterminate Structures
General Approach
Using compatibility of displacement, we have:
Final = Primary + Reactant
Next, further break up the reactant structure, using linear superposition:
Reactant = Multiplier × Unit Reactant
We summarize this process as:
0 1M M Mα= +
• M is the force system in the original structure (in this case moments);
• 0M is the primary structure force system;
• 1M is the unit reactant structure force system.
The primary structure can be analysed, as can the unit reactant structure. Thus, the
only unknown is the multiplier, α , for which we use virtual work to calculate.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 10
Finding the Multiplier
For beams and frames, we have:
( )210 1
0 0
0L L
ii
i i
MM M dx dxEI EI
δδ α⋅= + ⋅∑ ∑∫ ∫
Thus:
( )
0 1
021
0
Li
i
Li
i
M M dxEI
Mdx
EI
δ
αδ
⋅−=∑∫
∑∫
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 11
2.2.4 Virtual Work for Compound Structures
Basis
In the general equation for Virtual Work:
i i i iy F e Pδ δ⋅ = ⋅∑ ∑
We note that the summation on the right hand side is over all forms of real
displacement and virtual force combinations. For example, if a member is in
combined bending and axial force, then we must include the work done by both
effects:
( ) ( ) ( )Axial BendingMemberiW e P e P
PL MP M dxEA EI
δ δ δ
δ δ
= ⋅ + ⋅
= ⋅ + ⋅∫
The total Virtual Work done by any member is:
( )Memberiv
PL M T VW P M dx T VEA EI GJ GA
δ δ δ δ δ= ⋅ + ⋅ + ⋅ + ⋅∫
In which Virtual Work done by axial, bending, torsion, and shear, respectively, is
accounted for. However, most members primarily act through only one of these stress
resultants, and so we commonly have only one term per member. A typical example
is when axial deformation of frame (bending) members is neglected; since the area is
large the contribution to virtual work is small.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 12
At the level of the structure as a whole, we must account for all such sources of
Virtual Work. For the typical structures we study here, we account for the Virtual
Work done by axial and flexural members separately:
0
E I
i i i i i i
WW W
y F e P M
δδ δ
δ δ θ δ
==
⋅ = ⋅ + ⋅∑ ∑ ∑
In which the first term on the RHS is the internal virtual work done by axial members
and the second term is that done by flexural members.
Again considering only axial and bending members, if a deflection is sought:
0
1
i i i i
Lx
i xi
y F e P M
PL My P M dxEA EI
δ δ θ δ
δ δ
⋅ = ⋅ + ⋅
⋅ = ⋅ + ⋅
∑ ∑
∑ ∑∫
To solve such an indeterminate structure, we have the contributions to Virtual Work:
0 1M M Mα= +
0 1P P Pα= +
for the structure as a whole. Hence we have:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 13
( ) ( )
( )
1
0
0 1 0 11
0
210 1 0 11 1
0
0
0 1
0
0
E I
i i i i i i
Lx
i xi
Lx x
i x
i
Lxx x
i ii i
WW W
y F e P M
PL MP M dxEA EI
P P L M MP M dx
EA EI
MP L P L M MP P dxEA EA EI
δδ δ
δ δ θ δ
δ δ
α δ αδ δ
δδ δδ α δ α
==
⋅ = ⋅ + ⋅
⋅ = ⋅ + ⋅
+ ⋅ + = ⋅ + ⋅
⋅= ⋅ + ⋅ ⋅ + + ⋅
∑ ∑ ∑
∑ ∑∫
∑ ∑∫
∑ ∑ ∑∫0
L
dxEI∑∫
Hence the multiplier can be found as:
( ) ( )
0 1 0 1
02 21 1
0
Li i i
i i
Li i i
i i
P P L M M dxEA EI
P L Mdx
EA EI
δ δ
αδ δ
⋅ ⋅ ⋅+= −
+
∑ ∑∫
∑ ∑∫
Note the negative sign!
Though these expressions are cumbersome, the ideas and the algebra are both simple.
Integration of Diagrams
We are often faced with the integration of various diagrams when using virtual work
to calculate the deflections, etc. As such diagrams only have a limited number of
shapes, a table of ‘volume’ integrals is used.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 14
2.3 Basic Examples
2.3.1 Example 1
Problem
For the following structure, find:
(a) The force in the cable BC and the bending moment diagram;
(b) The vertical deflection at D.
Take 3 28 10 kNmEI = × and 316 10 kNEA = × .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 15
Solution – Part (a)
This is a one degree indeterminate structure and so we must release one redundant.
We could choose many, but the most obvious is the cable, BC. We next analyze the
primary structure for the actual loads, and the unit virtual force placed in lieu of the
redundant:
From the derivation of Virtual Work for indeterminate structures, we have:
( )210 1 0 1
1 1
0 0
0L L
xx xi i
i i
MP L P L M MP P dx dxEA EA EI EI
δδ δδ α δ α ⋅
= ⋅ + ⋅ ⋅ + + ⋅
∑ ∑ ∑ ∑∫ ∫
We evaluate each term separately to simplify the calculations and to minimize
potential calculation error.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 16
Term 1:
This term is zero since 0P is zero.
Term 2:
Only member BC contributes to this term and so it is:
1
1 1 2 21ii
P L PEA EA EAδ δ ⋅
⋅ = ⋅ =
∑
Term 3:
Here we must integrate the bending moment diagrams. We use the volume integral
for the portion AD of both diagrams. Thus we multiply a triangle by a trapezoid:
( ) ( )( )( )
0 1
0
1 1 40 2 2 4 26
400 3
Lx xM M dxEI EI
EI
δ⋅ = − + −
= −
∑∫
Term 4:
Here we multiply the virtual BMD by itself so it is a triangle by a triangle:
( ) ( )( )( )
21
0
1 1 64 34 4 43
LxM
dxEI EI EI
δ = − − = ∑∫
With all terms evaluated the Virtual Work equation becomes:
2 400 3 64 30 0EA EI EI
α α= + ⋅ − + ⋅
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 17
Which gives:
400 3400
2 64 3 6 64EI
EIEA EI EA
α = =+ +
Given that 3 38 10 16 10 0.5EI EA = × × = , we have:
( )
400 5.976 0.5 64
α = =+
Thus there is a tension (positive answer) in the cable of 5.97 kN, giving the BMD as:
Note that this comes from:
( )( )( )( )
0
0
40 5.97 4 16.1 kN
0 5.97 2 11.9 kNA
D
M M M
M M M
α δ
α δ
= + ⋅ = + − =
= + ⋅ = + − = −
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 18
Solution – Part (b)
Recalling that the only requirement on applying virtual forces to calculate real
displacements is that an equilibrium system results, we can apply a vertical unit force
at D to the primary structure only:
The Virtual Work equation useful for deflection is:
0
1
i i i i
Lx
Dy i xi
y F e P M
PL MP M dxEA EI
δ δ θ δ
δ δ δ
⋅ = ⋅ + ⋅
⋅ = ⋅ + ⋅
∑ ∑
∑ ∑∫
Since 0Pδ = , we need only calculate the term involving the Virtual Work done by
the beam bending. This involves the volume integral of the two diagrams:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 19
Note that only the portion AD will count as there is no virtual moment on DB. Thus
we have:
However, this shape is not easy to work with, given the table to hand. Therefore we
recall that the real BMD came about as the superposition of two BMD shapes that are
easier to work with, and so we have:
A further benefit of this approach is that an equation of deflection in terms of the
multiplier α is got. This could then be used to determine α for a particular design
requirement, and in turn this could inform the choice of EI EA ratio. Thus:
( )( )( ) ( ) ( )( )( )
0
1 1 12 40 2 2 2 2 4 23 6
160 203
Lx
Dy xM M dxEI
EI
EI
δ δ
α
α
= ⋅
= + ⋅ − + − −
=
∑∫
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 20
Given 5.97α = , we then have:
( ) 33
160 20 5.97 13.9 13.9 10 1.7 mm3 8 10Dy EI EI
δ−
= = = × =×
The positive answer indicates that the deflection is in the direction of the applied
virtual vertical force and so is downwards as expected.
We can also easily work out the deflection at B, since it is the same as the elongation
of the cable:
( )( ) 33
5.97 210 0.75 mm
16 10ByPLEA
δ = = × =×
Draw the deflected shape of the structure.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 21
2.3.2 Example 2
Problem
For the following structure, find:
(a) The force in the cable CD and the bending moment diagram;
(b) Determine the optimum EA of the cable for maximum efficiency of the beam.
Take 3 28 10 kNmEI = × and 348 10 kNEA = × .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 22
Solution – Part (a)
Choose the cable CD as the redundant to give:
The equation of Virtual Work relevant is:
( )210 1 0 1
1 1
0 0
0L L
xx xi i
i i
MP L P L M MP P dx dxEA EA EI EI
δδ δδ α δ α ⋅
= ⋅ + ⋅ ⋅ + + ⋅
∑ ∑ ∑ ∑∫ ∫
We evaluate each term separately:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 23
Term 1:
This term is zero since 0P is zero.
Term 2:
Only member CD contributes to this term and so it is:
1
1 1 2 21ii
P L PEA EA EAδ δ ⋅
⋅ = ⋅ =
∑
Term 3:
Here we must integrate the bending moment diagrams. We use the volume integral
for each half of the diagram, and multiply by 2, since we have two such halves.
( )( )( )
0 1
0
2 5 1 10 212
50 3
Lx xM M dxEI EI
EI
δ⋅ = −
= −
∑∫
Term 4:
Here we multiply the virtual BMD by itself:
( ) ( )( )( )
21
0
2 1 4 31 1 23
LxM
dxEI EI EI
δ = − − = ∑∫
Thus the Virtual Work equation becomes:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 24
2 50 3 4 30 0EA EI EI
α α= + ⋅ − + ⋅
Which gives:
50 350
2 4 3 6 4EI
EIEA EI EA
α = =+ +
Given that 3 38 10 48 10 0.167EI EA = × × = , we have:
( )
50 106 0.167 4
α = =+
Thus there is a tension (positive answer) in the cable of 10 kN, giving:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 25
As designers, we want to control the flow of forces. In this example we can see that
by changing the ratio EI EA we can control the force in the cable, and the resulting
bending moments. We can plot the cable force and maximum sagging bending
moment against the stiffness ratio to see the behaviour for different relative
stiffnesses:
0
2
4
6
8
10
12
14
0.0001 0.001 0.01 0.1 1 10 100 1000 10000
Ratio EI/EA
Cable Tension (kN)Sagging Moment (kNm)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 26
Solution – Part (b)
Efficiency of the beam means that the moments are resisted by the smallest possible
beam. Thus the largest moment anywhere in the beam must be made as small as
possible. Therefore the hogging and sagging moments should be equal:
We know that the largest hogging moment will occur at 2L . However, we do not
know where the largest sagging moment will occur. Lastly, we will consider sagging
moments positive and hogging moments negative. Consider the portion of the net
bending moment diagram, ( )M x , from 0 to 2L :
The equations of these bending moments are:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 27
( )2P
PM x x= −
( ) 2
2 2W
w wLM x x x= − +
Thus:
( ) ( ) ( )
2
2 2 2
W PM x M x M xwL w Px x x
= +
= − −
The moment at 2L is:
( )2
2 2
2
22 2 2 2 2 2
4 8 4
8 4
wL L w L P LM L
wL wL PL
wL PL
= − −
= − −
= −
Which is as we expected. The maximum sagging moment between 0 and 2L is
found at:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 28
( )
max
max
0
02 2
2 2
dM xdx
wL Pwx
L Pxw
=
− − =
= −
Thus the maximum sagging moment has a value:
( )2
max
2 2 2 2
2 2
2 2 2 2 2 2 2 2 22
4 4 2 4 4 4 4 4
8 4 8
wL L P w L P P L PM xw w w
wL PL w L PL P PL Pw w w
wL PL Pw
= − − − − −
= − − − + − +
= − +
Since we have assigned a sign convention, the sum of the hogging and sagging
moments should be zero, if we are to achieve the optimum BMD. Thus:
( ) ( )max
2 2 2
2 2
22
2 0
08 4 8 8 4
04 2 8
1 08 2 4
M x M L
wL PL P wL PLw
wL PL Pw
L wLP Pw
+ =
− + + − =
− + =
+ − + =
This is a quadratic equation in P and so we solve for P using the usual method:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 29
( )
2 2
82 4 82 2 2 8
82 2
L L Lw L LP
wwL
± − = = ±
= ±
Since the load in the cable must be less than the total amount of load in the beam, that
is, P wL< , we have:
( )2 2 0.586P wL wL= − =
With this value for P we can determine the hogging and sagging moments:
( )( )2
2
2
2 22
8 42 2 3
8
0.0214
wL LwLM L
wL
wL
−= −
−=
= −
And:
( )
( )
2 2
max
2
2
2
2
8 4 8
2 22 2 38 8
3 2 28
0.0214
wL PL PM xw
wLwL
w
wL
wL
= − +
− − = + −
=
= +
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 30
Lastly, the location of the maximum sagging moment is given by:
( )
( )
max 2 22 2
2 2
2 120.207
L Pxw
wLLw
L
L
= −
−= −
= −
=
For our particular problem, 5 kN/mw = , 4 mL = , giving:
( )0.586 5 4 11.72 kNP = × =
( ) ( )2max 0.0214 5 4 1.71 kNmM x = × =
Thus, as we expected, 10 kNP > , the value obtained from Part (a) of the problem.
Now since, we know P we now also know the required value of the multiplier, α .
Hence, we write the virtual work equations again, but this time keeping Term 2 in
terms of L, since that is what we wish to solve for:
50 11.726 4
1 50 4 0.0446 11.72
EIEA
EIEA
α = =+
∴ = − =
Giving 3 38 10 0.044 180.3 10 kNEA = × = × . This is 3.75 times the original cable area
– a lot of extra material just to change the cable force by 17%. However, there is a
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 31
large saving by reducing the overall moment in the beam from 10 kNm (simply-
supported) or 2.5 kNm (two-span beam) to 1.71 kNm.
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 32
2.3.3 Example 3
Problem
For the following structure:
1. Determine the tension in the cable AB;
2. Draw the bending moment diagram;
3. Determine the vertical deflection at D with and without the cable AB.
Take 3 2120 10 kNmEI = × and 360 10 kNEA = × .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 33
Solution
As is usual, we choose the cable to be the redundant member and split the frame up
as follows:
Primary Structure Redundant Structure
We must examine the BMDs carefully, and identify expressions for the moments
around the arch. However, since we will be using virtual work and integrating one
diagram against another, we immediately see that we are only interested in the
portion of the structure CB. Further, we will use the anti-clockwise angle from
vertical as the basis for our integration.
Primary BMD
Drawing the BMD and identify the relevant distances:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 34
Hence the expression for 0M is:
( ) ( )0 20 10 2sin 20 1 sinMθ θ θ= + = +
Reactant BMD
This calculation is slightly easier:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 35
( ) ( )1 1 2 2cos 2 1 cosMθ θ θ= ⋅ − = −
Virtual Work Equation
As before, we have the equation:
( )210 1 0 11 1
0 0
0L L
xx xi i
i i
MP L P L M MP P dx dxEA EA EI EI
δδ δδ α δ α ⋅
= ⋅ + ⋅ ⋅ + + ⋅
∑ ∑ ∑ ∑∫ ∫
Term 1 is zero since there are no axial forces in the primary structure. We take each
other term in turn.
Term 2
Since only member AB has axial force:
( )21 2 2Term 2EA EA
= =
Term 3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 36
Since we want to integrate around the member – an integrand ds - but only have the
moment expressed according to θ , we must change the integration limits by
substituting:
2ds R d dθ θ= ⋅ =
Hence:
( ) ( )
( )( )
( )
20 1
0 0
2
0
2
0
1 2 1 cos 20 1 sin 2
80 1 cos 1 sin
80 1 sin cos cos sin
Lx xM M dx dEI EI
dEI
dEI
π
π
π
δ θ θ θ
θ θ θ
θ θ θ θ θ
⋅= − − +
= − + +
= − − + +
∑∫ ∫
∫
∫
To integrate this expression we refer to the appendix of integrals to get each of the
terms, which then give:
( )
20 1
00
80 1cos sin cos24
80 1 10 1 1 0 1 02 4 4
80 1 11 12 4 4
80 12
Lx xM M dxEI EI
EI
EI
EI
πδ θ θ θ θ
π
π
π
⋅ = − + + −
= − + + − − − − + + − = − + + − + − =
∑∫
Term 4
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 37
Proceeding similarly to Term 3, we have:
( ) ( ) ( )
( )
21 2
0 0
22
0
1 2 1 cos 2 1 cos 2
8 1 2cos cos
LxM
dx dEI EI
dEI
π
π
δθ θ θ
θ θ θ
= − −
= − +
∑∫ ∫
∫
Again we refer to the integrals appendix, and so for Term 4 we then have:
( ) ( )
[ ]
21 22
0 0
2
0
8 1 2cos cos
8 12sin sin 22 4
8 12 0 0 0 02 4 4
8 3 74
LxM
dx dEI EI
EI
EI
EI
π
π
δθ θ θ
θθ θ θ
π π
π
= − +
= − + +
= − + + − − + + − =
∑∫ ∫
Solution
Substituting the calculated values into the virtual work equation gives:
2 80 1 8 3 70 02 4EA EI EIπ πα α− − = + ⋅ + + ⋅
And so:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 38
80 12
2 8 3 74
EI
EA EI
π
απ
− − =
− +
Simplifying:
20 20
3 7 EIEA
παπ
−=
− +
In this problem, 2EI EA = and so:
20 20 9.68 kN3 5παπ−
= =−
We can examine the effect of different ratios of EI EA on the structure from our
algebraic solution for α . We show this, as well as a point representing the solution
for this particular EI EA ratio on the following graph:
0
2
4
6
8
10
12
14
16
18
20
0.0001 0.001 0.01 0.1 1 10 100 1000 10000
Ratio EI/EA
α F
acto
r
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 39
As can be seen, by choosing a stiffer frame member (increasing EI) or by reducing
the area of the cable, we can reduce the force in the cable (which is just 1 α⋅ ).
However this will have the effect of increasing the moment at A, for example:
Deflections and shear would also be affected.
Draw the final BMD and determine the deflection at D.
0
5
10
15
20
25
30
35
40
45
0.0001 0.001 0.01 0.1 1 10 100 1000 10000
Ratio EI/EA
Ben
ding
Mom
ent a
t A (k
Nm
)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 40
2.3.4 Example 4
Problem
For the following structure:
1. draw the bending moment diagram;
2. Find the vertical deflection at E.
Take 3 2120 10 kNmEI = × and 360 10 kNEA = × .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 41
Solution
To begin we choose the cable BF as the obvious redundant, yielding:
Virtual Work Equation
The Virtual Work equation is as before:
( )210 1 0 11 1
0 0
0L L
xx xi i
i i
MP L P L M MP P dx dxEA EA EI EI
δδ δδ α δ α ⋅
= ⋅ + ⋅ ⋅ + + ⋅
∑ ∑ ∑ ∑∫ ∫
Term 1 is zero since there are no axial forces in the primary structure. As we have
done previously, we take each other term in turn.
Term 2
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 42
Though member AB has axial force, it is primarily a flexural member and so we only
take account of the axial force in the cable BF:
1
1 1 2 2 2 21ii
P L PEA EA EAδ δ
⋅ ⋅ = ⋅ =
∑
Term 3
Since only the portion AB has moment on both diagrams, it is the only section that
requires integration here. Thus:
( )( )( )0 1
0
1 1 220 2200 2 22
Lx xM M dxEI EI EIδ⋅ − = − =
∑∫
Term 3
Similar to Term 3, we have:
( ) ( )( )( )
21
0
1 1 4 32 2 23
LxM
dxEI EI EI
δ = − − = ∑∫
Solution
Substituting the calculated values into the virtual work equation gives:
2 2 220 2 4 30 0EA EI EI
α α= + ⋅ − + ⋅
Thus:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 43
220 22 2 4 3
EI
EA EI
α =+
And so:
220 242 23
EIEA
α =+
Since:
3
3
120 10 260 10
EIEA
×= =
×
We have:
( )220 2 40.4642 2 2
3
α = = ++
Thus the force in the cable BF is 40.46 kN tension, as assumed.
The bending moment diagram follows from superposition of the two previous
diagrams:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 44
To find the vertical deflection at E, we must apply a unit vertical load at E. We will
apply a downwards load since we think the deflection is downwards. Therefore we
should get a positive result to confirm our expectation.
We need not apply the unit vertical force to the whole structure, as it is sufficient to
apply it to a statically determinate sub-structure. Thus we apply the force as follows:
For the deflection, we have the following equation:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 45
0
1
i i i i
Lx
Ey i xi
y F e P M
PL MP M dxEA EI
δ δ θ δ
δ δ δ
⋅ = ⋅ + ⋅
⋅ = ⋅ + ⋅
∑ ∑
∑ ∑∫
However, since 0Pδ = , we only need calculate the second term:
For AB we have:
( )( )( )1 1 1371.2200 142.8 4 22
Bx
xA
M M dxEI EI EI
δ ⋅ = + = ∫
For BC we have:
( )( )( )1 1600200 4 2C
xx
B
M M dxEI EI EI
δ ⋅ = = ∫
For CD, we have the following equations for the bending moments:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 46
( ) ( )( )100 2sin200sin
M θ θθ
=
= ( ) ( )( )2 1 2sin
2 2sinMδ θ θ
θ= +
= +
Also note that we want to integrate around the member – an integrand ds - but only
have the moment expressed according to θ , we must change the integration limits by
substituting:
2ds R d dθ θ= ⋅ =
Thus we have:
( )( )
( )
2
0
22
0
2 22
0 0
1 200sin 2 2sin 2
800 sin sin
800 sin sin
Dx
xC
M M dx dEI EI
dEI
d dEI
π
π
π π
δ θ θ θ
θ θ θ
θ θ θ θ
⋅ = + ⋅
= +
= +
∫ ∫
∫
∫ ∫
Taking each term in turn:
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 47
[ ] ( )2
2
00
sin cos 0 1 1dπ
πθ θ θ= − = − − − = +∫
( ) ( )22
2 22 2
00
1 1 1 1sin sin 1 0 02 4 4 4 4 4
dππ θ π πθ θ θ − = − = − − − = ∫
Thus:
800 1 200 60014
Dx
xC
M M dxEI EI EI
π πδ − + ⋅ = + = ∫
Thus:
1371.2 1600 200 600 4200Ey EI EI EI EI
πδ += + + = +
Thus we get a downwards deflection as expected. Also, since 3 2120 10 kNmEI = × ,
we have:
3
4200 35 mm120 10Eyδ = = ↓
×
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 48
2.3.5 Problems
Problem 1
For the following structure, find the BMD and the vertical deflection at D. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .
(Ans. 7.8α = for BC, 1.93 mmByδ = ↓ )
Problem 2
For the following structure, find the BMD and the vertical deflection at C. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .
(Ans. 25.7α = for BD, 25 mmCvδ = ↓ )
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 49
Problem 3
For the following structure, find the BMD and the horizontal deflection at C. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .
(Ans. 47.8α = for BD, 44.8 mmCxδ = → )
Problem 4
For the following structure, find the BMD and the vertical deflection at B. Take P =
20 kN, 3 28 10 kNmEI = × and 316 10 kNEA = × .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 50
(Ans. 14.8α = for CD, 14.7 mmByδ = ↓ )
Problem 5
For the following structure, find the BMD and the vertical deflection at C. Take 3 250 10 kNmEI = × and 320 10 kNEA = × .
(Ans. 100.5α = for BC, 55.6 mmCyδ = ↓)
Problem 6
Analyze the following structure and determine the BMD and the vertical deflection at
D. For ABCD, take 210 kN/mmE = , 4 212 10 mmA = × and 8 436 10 mmI = × , and for
AEBFC take 2200 kN/mmE = and 3 22 10 mmA = × .
(Ans. 109.3α = for BF, 54.4 mmCyδ = ↓)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 51
Problem 7
Analyze the following structure. For all members, take 210 kN/mmE = , for ABC, 4 26 10 mmA = × and 7 4125 10 mmI = × ; for all other members 21000 mmA = .
(Ans. 72.5α = for DE)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 52
2.4 Past Exam Questions
2.4.1 Sample Paper 2007
3. For the rigidly jointed frame shown in Fig. Q3, using Virtual Work:
(i) Determine the bending moment moments due to the loads as shown; (15 marks)
(ii) Draw the bending moment diagram, showing all important values;
(4 marks)
(iii) Determine the reactions at A and E; (3 marks)
(iv) Draw the deflected shape of the frame.
(3 marks) Neglect axial effects in the flexural members. Take the following values: I for the frame = 150×106 mm4; Area of the stay EB = 100 mm2; Take E = 200 kN/mm2 for all members.
FIG. Q3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 53
2.4.2 Semester 1 Exam 2007
3. For the rigidly jointed frame shown in Fig. Q3, using Virtual Work:
(i) Determine the bending moment moments due to the loads as shown; (15 marks)
(ii) Draw the bending moment diagram, showing all important values;
(4 marks)
(iii) Determine the reactions at A and E; (3 marks)
(iv) Draw the deflected shape of the frame.
(3 marks) Neglect axial effects in the flexural members. Take the following values: I for the frame = 150×106 mm4; Area of the stay EF = 200 mm2; Take E = 200 kN/mm2 for all members.
Ans. 35.0α = .
FIG. Q3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 54
2.4.3 Semester 1 Exam 2008
QUESTION 3
For the frame shown in Fig. Q3, using Virtual Work:
(i) Determine the force in the tie; (ii) Draw the bending moment diagram, showing all important values; (iii) Determine the deflection at C; (iv) Determine an area of the tie such that the bending moments in the beam are minimized; (v) For this new area of tie, determine the deflection at C; (vi) Draw the deflected shape of the structure.
(25 marks)
Note:
Neglect axial effects in the flexural members and take the following values:
• For the frame, 6 4600 10 mmI = × ; • For the tie, 2300 mmA = ; • For all members, 2200 kN/mmE = .
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 55
2.4.4 Semester 1 Exam 2009
QUESTION 3
For the frame shown in Fig. Q3, using Virtual Work:
(i) Determine the axial forces in the members; (ii) Draw the bending moment diagram, showing all important values; (iii) Determine the reactions; (iv) Determine the vertical deflection at D; (v) Draw the deflected shape of the structure.
(25 marks)
Note:
Neglect axial effects in the flexural members and take the following values:
• For the beam ABCD, 6 4600 10 mmI = × ; • For members BF and CE, 2300 mmA = ; • For all members, 2200 kN/mmE = .
Ans. 113.7α = (for CE); 55 mmDyδ = ↓
FIG. Q3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 56
2.4.5 Semester 1 Exam 2010 QUESTION 3 For the frame shown in Fig. Q3, using Virtual Work:
(i) Draw the bending moment diagram, showing all important values;
(ii) Determine the horizontal displacement at C;
(iii) Determine the vertical deflection at C;
(iv) Draw the deflected shape of the structure.
(25 marks) Note: Neglect axial effects in the flexural members and take the following values:
• For the beam ABC, 3 25 10 kNmEI = × ; • For member BD, 2200 kN/mmE = and 2200 mmA = ; • The following integral results may assist in your solution:
sin cosdθ θ θ= −∫ 1cos sin cos 24
dθ θ θ θ= −∫ 2 1sin sin 22 4
d θθ θ θ= −∫
Ans. 37.1α = (for BD); 104 mmCxδ = ← 83 mmCyδ = ↓
FIG. Q3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 57
2.4.6 Semester 1 Exam 2011 QUESTION 3 For the frame shown in Fig. Q3, using Virtual Work: (i) Draw the bending moment diagram, showing all important values; (ii) Draw the axial force diagram; (iii) Determine the vertical deflection at D; (iv) Draw the deflected shape of the structure.
(25 marks) Note: Neglect axial effects in the flexural members and take the following values:
• For the member ABCD, 3 25 10 kNmEI = × ; • For members BF and CE, 2200 kN/mmE = and 2200 mmA = ; • The following integral result may assist in your solution:
2 1sin sin 2
2 4d θθ θ θ= −∫
Ans. 48.63α = (for BF); 108.4 mmDyδ = ↓
FIG. Q3
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 58
2.5 Appendix – Trigonometric Integrals
2.5.1 Useful Identities
In the following derivations, use is made of the trigonometric identities:
1cos sin sin 22
θ θ θ= (1)
( )2 1cos 1 cos22
θ θ= + (2)
( )2 1sin 1 cos22
θ θ= − (3)
Integration by parts is also used:
u dx ux x du C= − +∫ ∫ (4)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 59
2.5.2 Basic Results
Neglecting the constant of integration, some useful results are:
cos sindθ θ θ=∫ (5)
sin cosdθ θ θ= −∫ (6)
1sin cosa d aa
θ θ θ= −∫ (7)
1cos sina d aa
θ θ θ=∫ (8)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 60
2.5.3 Common Integrals
The more involved integrals commonly appearing in structural analysis problems are:
cos sin dθ θ θ∫
Using identity (1) gives:
1cos sin sin 22
d dθ θ θ θ θ=∫ ∫
Next using (7), we have:
1 1 1sin 2 cos22 2 2
1 cos24
dθ θ θ
θ
= −
= −
∫
And so:
1cos sin cos24
dθ θ θ θ= −∫ (9)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 61
2cos dθ θ∫
Using (2), we have:
( )2 1cos 1 cos2
21 1 cos22
d d
d d
θ θ θ θ
θ θ θ
= +
= +
∫ ∫
∫ ∫
Next using (8):
1 1 11 cos2 sin 22 2 2
1 sin 22 4
d dθ θ θ θ θ
θ θ
+ = +
= +
∫ ∫
And so:
2 1cos sin 22 4
d θθ θ θ= +∫ (10)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 62
2sin dθ θ∫
Using (3), we have:
( )2 1sin 1 cos2
21 1 cos22
d d
d d
θ θ θ θ
θ θ θ
= −
= −
∫ ∫
∫ ∫
Next using (8):
1 1 11 cos2 sin 22 2 2
1 sin 22 4
d dθ θ θ θ θ
θ θ
− = −
= −
∫ ∫
And so:
2 1sin sin 22 4
d θθ θ θ= −∫ (11)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 63
cos dθ θ θ∫
Using integration by parts write:
cos d u dxθ θ θ =∫ ∫
Where:
cosu dx dθ θ θ= =
To give:
du dθ=
And
cos
sin
dx d
x
θ θ
θ
=
=∫ ∫
Which uses (5). Thus, from (4), we have:
cos sin sin
u dx ux x du
d dθ θ θ θ θ θ θ
= −
= −∫ ∫
∫ ∫
And so, using (6) we have:
cos sin cosdθ θ θ θ θ θ= +∫ (12)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 64
sin dθ θ θ∫
Using integration by parts write:
sin d u dxθ θ θ =∫ ∫
Where:
sinu dx dθ θ θ= =
To give:
du dθ=
And
sin
cos
dx d
x
θ θ
θ
=
= −∫ ∫
Which uses (6). Thus, from (4), we have:
( ) ( )sin cos cos
u dx ux x du
d dθ θ θ θ θ θ θ
= −
= − − −∫ ∫
∫ ∫
And so, using (5) we have:
sin cos sindθ θ θ θ θ θ= − +∫ (13)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 65
( )cos A dθ θ−∫
Using integration by substitution, we write u A θ= − to give:
1duddu dθ
θ
= −
= −
Thus:
( ) ( )cos cosA d u duθ θ− = −∫ ∫
And since, using (5):
cos sinu du u− = −∫
We have:
( ) ( )cos sinA d Aθ θ θ− = − −∫ (14)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 66
( )sin A dθ θ−∫
Using integration by substitution, we write u A θ= − to give:
1duddu dθ
θ
= −
= −
Thus:
( ) ( )sin sinA d u duθ θ− = −∫ ∫
And since, using (6):
( )sin cosu du u− = − −∫
We have:
( ) ( )sin cosA d Aθ θ θ− = −∫ (15)
Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures
Dr. C. Caprani 67
2.6 Appendix – Volume Integrals
13
jkl 16
jkl ( )1 21 26
j j kl+ 12
jkl
16
jkl 13
jkl ( )1 21 26
j j kl+ 12
jkl
( )1 2
1 26
j k k l+ ( )1 21 26
j k k l+ ( )
( )
1 1 2
2 1 2
1 26
2
j k k
j k k l
+ +
+
( )1 212
j k k l+
12
jkl 12
jkl ( )1 212
j j kl+ jkl
( )1
6jk l a+ ( )1
6jk l b+
( )
( )
1
2
16
j l b
j l a k
+ +
+
12
jkl
512
jkl 14
jkl ( )1 21 3 5
12j j kl+ 2
3jkl
14
jkl 512
jkl ( )1 21 5 3
12j j kl+ 2
3jkl
14
jkl 112
jkl ( )1 21 3
12j j kl+ 1
3jkl
112
jkl 14
jkl ( )1 21 3
12j j kl+ 1
3jkl
13
jkl 13
jkl ( )1 213
j j kl+ 23
jkl
1 2
1 2
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
3.3.1 Example 1 .................................................................................................... 64
3.3.2 Example 2 .................................................................................................... 70
3.3.3 Example 3 .................................................................................................... 79
Rev. 1
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 2
3.1 Introduction
3.1.1 General
To further illustrate the virtual work method applied to more complex structures, the
following sets of examples are given. The examples build upon each other to
illustrate how the analysis of a complex structure can be broken down.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 3
3.2 Ring Beam Examples
3.2.1 Example 1
Problem
For the quarter-circle beam shown, which has flexural and torsional rigidities of EI
and GJ respectively, show that the deflection at A due to the point load, P, at A is:
3 3 3 8
4 4Ay
PR PREI GJ
π πδ − = ⋅ +
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 4
Solution
The point load will cause both bending and torsion in the beam member. Therefore
both effects must be accounted for in the deflection calculations. Shear effects are
ignored.
Drawing a plan view of the structure, we can identify the perpendicular distance of
the force, P, from the section of consideration, which we locate by the angle θ from
the y-axis:
The bending moment at C is P times the perpendicular distance AC , called m. The
torsion at C is the force times the transverse perpendicular distance CD , called t.
Using the triangle ODA, we have:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 5
sin sin
cos cos
m m RROD
OD RR
θ θ
θ θ
= ∴ =
= ∴ =
The distance CD , or t, is R OD− , thus:
( )cos
1 cos
t R ODR RR
θθ
= −
= −
= −
Thus the bending moment at point C is:
( )sin
M PmPR
θθ
=
= (1)
The torsion at C is:
( )
( )1 cos
T Pt
PR
θ
θ
=
= − (2)
Using virtual work, we have:
0
E I
Ay
WW W
M TF M ds T dsEI GJ
δδ δ
δ δ δ δ
==
⋅ = ⋅ + ⋅∫ ∫
(3)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 6
This equation represents the virtual work done by the application of a virtual force,
Fδ , in the vertical direction at A, with its internal equilibrium virtual moments and
torques, Mδ and Tδ and so is the equilibrium system. The compatible
displacements system is that of the actual deformations of the structure, externally at
A, and internally by the curvatures and twists, M EI and T GJ .
Taking the virtual force, 1Fδ = , and since it is applied at the same location and
direction as the actual force P, we have, from equations (1) and (2):
( ) sinM Rδ θ θ= (4)
( ) ( )1 cosT Rδ θ θ= − (5)
Thus, the virtual work equation, (3), becomes:
[ ][ ] ( ) ( )
2 2
0 0
1 11
1 1sin sin 1 cos 1 cos
Ay M M ds T T dsEI GJ
PR R Rd PR R RdEI GJ
π π
δ δ δ
θ θ θ θ θ θ
⋅ = ⋅ + ⋅
= + − −
∫ ∫
∫ ∫ (6)
In which we have related the curve distance, ds , to the arc distance, ds Rdθ= , which
allows us to integrate round the angle rather than along the curve. Multiplying out:
( )2 23 3
22
0 0
sin 1 cosAyPR PRd dEI GJ
π π
δ θ θ θ θ= + −∫ ∫ (7)
Considering the first term, from the integrals’ appendix, we have:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 7
( )
222
00
1sin sin 22 4
1 0 0 04 4
4
dππ θθ θ θ
π
π
= −
= − ⋅ − −
=
∫
(8)
The second term is:
( ) ( )
2 22 2
0 0
2 2 22
0 0 0
1 cos 1 2cos cos
1 2 cos cos
d d
d d d
π π
π π π
θ θ θ θ θ
θ θ θ θ θ
− = − +
= − +
∫ ∫
∫ ∫ ∫ (9)
Thus, from the integrals in the appendix:
( ) [ ] [ ]
( ) ( ) ( ) ( )
222 22
0 000
11 cos 2 sin sin 22 4
10 2 1 0 0 0 02 4 4
22 43 8
4
dππ
π π θθ θ θ θ θ
π π
π π
π
− = − + +
= − − − + + ⋅ − +
= − +
−=
∫
(10)
Substituting these results back into equation (7) gives the desired result:
3 3 3 8
4 4AyPR PREI GJ
π πδ − = +
(11)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 8
3.2.2 Example 2
Problem
For the quarter-circle beam shown, which has flexural and torsional rigidities of EI
and GJ respectively, show that the deflection at A due to the uniformly distributed
load, w, shown is:
( )24 4 212 8Ay
wR wREI GJ
πδ
−= ⋅ + ⋅
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 9
Solution
The UDL will cause both bending and torsion in the beam member and both effects
must be accounted for. Again, shear effects are ignored.
Drawing a plan view of the structure, we must identify the moment and torsion at
some point C, as defined by the angle θ from the y-axis, caused by the elemental
load at E, located at φ from the y-axis. The load is given by:
Force UDL length
w dsw R dφ
= ×= ⋅= ⋅
(12)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 10
The bending moment at C is the load at E times the perpendicular distance DE ,
labelled m. The torsion at C is the force times the transverse perpendicular distance
CD , labelled t. Using the triangle ODE, we have:
( ) ( )
( ) ( )
sin sin
cos cos
m m RROD
OD RR
θ φ θ φ
θ φ θ φ
− = ∴ = −
− = ∴ = −
The distance t is thus:
( )( )
cos
1 cos
t R OD
R R
R
θ φ
θ φ
= −
= − −
= − −
The differential bending moment at point C, caused by the elemental load at E is
thus:
( )[ ][ ] ( )
( )2
Force Distance
sin
sin
dM
wRd m
wRd R
wR d
θ
φ
φ θ φ
θ φ φ
= ×
= ×
= − = −
Integrating to find the total moment at C caused by the UDL from A to C around the
angle 0 to θ gives:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 11
( ) ( )
( )
( )
2
0
2
0
sin
sin
M dM
wR d
wR d
φ θ
φ
φ θ
φ
θ θ
θ φ φ
θ φ φ
=
=
=
=
=
= −
= −
∫
∫
∫
In this integral θ is a constant and only φ is considered a variable. Using the identity
from the integral table gives:
( ) ( )
( )
2
0
2
cos
cos0 cos
M wR
wR
φ θ
φθ θ φ
θ
=
== −
= −
And so:
( ) ( )2 1 cosM wRθ θ= − (13)
Along similar lines, the torsion at C caused by the load at E is:
( ) [ ][ ] ( ){ }
( )2
1 cos
1 cos
dT wRd t
wRd R
wR d
θ φ
φ θ φ
θ φ φ
= ×
= − −
= − −
And integrating for the total torsion at C:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 12
( ) ( )
( )
( )
( )
2
0
2
0
2
0 0
1 cos
1 cos
1 cos
T dT
wR d
wR d
wR d d
φ θ
φ
φ θ
φ
φ θ φ θ
φ φ
θ θ
θ φ φ
θ φ φ
φ θ φ φ
=
=
=
=
= =
= =
=
= − −
= − −
= − −
∫
∫
∫
∫ ∫
Using the integral identity for ( )cos θ φ− gives:
( ) [ ] ( ){ }
[ ]{ }
2
0 0
2
sin
sin 0 sin
T wR
wR
φ θφ θ
φ φθ φ θ φ
θ θ
==
= == − − −
= + −
And so the total torsion at C is:
( ) ( )2 sinT wRθ θ θ= − (14)
To determine the deflection at A, we apply a virtual force, Fδ , in the vertical
direction at A. Along with its internal equilibrium virtual moments and torques, Mδ
and Tδ and this set forms the equilibrium system. The compatible displacements
system is that of the actual deformations of the structure, externally at A, and
internally by the curvatures and twists, M EI and T GJ . Therefore, using virtual
work, we have:
0
E I
Ay
WW W
M TF M ds T dsEI GJ
δδ δ
δ δ δ δ
==
⋅ = ⋅ + ⋅∫ ∫
(15)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 13
Taking the virtual force, 1Fδ = , and using the equation for moment and torque at
any angle θ from Example 1, we have:
( ) sinM Rδ θ θ= (16)
( ) ( )1 cosT Rδ θ θ= − (17)
Thus, the virtual work equation, (15), using equations (13) and (14), becomes:
( ) [ ]
( ) ( )
22
0
22
0
1 11
1 1 cos sin
1 sin 1 cos
Ay M M ds T T dsEI GJ
wR R RdEI
wR R RdGJ
π
π
δ δ δ
θ θ θ
θ θ θ θ
⋅ = ⋅ + ⋅
= −
+ − −
∫ ∫
∫
∫
(18)
In which we have related the curve distance, ds , to the arc distance, ds Rdθ=
allowing us to integrate round the angle rather than along the curve. Multiplying out:
( )
( )
24
0
24
0
sin sin cos
sin cos cos sin
AywR dEI
wR dGJ
π
π
δ θ θ θ θ
θ θ θ θ θ θ θ
= −
+ − − +
∫
∫ (19)
Using the respective integrals from the appendix yields:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 14
( )
( ) ( )
24
0
24 2
0
4
4 2
4
4 2
1cos cos24
1cos sin cos cos22 4
1 10 14 4
1 10 1 0 1 0 1 0 18 2 4 4
12
1 18 2 4 4
AywREI
wRGJ
wREI
wRGJ
wREI
wRGJ
π
π
δ θ θ
θ θ θ θ θ θ
π π
π π
= − +
+ + − + −
= − − − − + + + − ⋅ + − − − + − + −
=
+ − + +
Writing the second term as a common fraction:
4 4 21 4 4
2 8AywR wREI GJ
π πδ − +
= ⋅ +
And then factorising, gives the required deflection at A:
( )224 4 21
2 8AywR wREI GJ
πδ
−= ⋅ + ⋅ (20)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 15
3.2.3 Example 3
Problem
For the quarter-circle beam shown, which has flexural and torsional rigidities of EI
and GJ respectively, show that the vertical reaction at A due to the uniformly
distributed load, w, shown is:
( )( )
24 22 2 3 8AV wR
β πβπ π
+ −=
+ −
where GJEI
β = .
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 16
Solution
This problem can be solved using two apparently different methods, but which are
equivalent. Indeed, examining how they are equivalent leads to insights that make
more difficult problems easier, as we shall see in subsequent problems. For both
approaches we will make use of the results obtained thus far:
• Deflection at A due to UDL:
( )24 4 212 8Ay
wR wREI GJ
πδ
−= ⋅ + ⋅ (21)
• Deflection at A due to point load at A:
3 3 3 8
4 4Ay
PR PREI GJ
π πδ − = ⋅ +
(22)
Using Compatibility of Displacement
The basic approach, which does not require virtual work, is to use compatibility of
displacement in conjunction with superposition. If we imagine the support at A
removed, we will have a downwards deflection at A caused by the UDL, which
equation (21) gives us as:
( )24 40 21
2 8Ay
wR wREI GJ
πδ
−= ⋅ + ⋅ (23)
As illustrated in the following diagram.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 17
Since in the original structure we will have a support at A we know there is actually
no displacement at A. The vertical reaction associated with the support at A, called V,
must therefore be such that it causes an exactly equal and opposite deflection, VAyδ , to
that of the UDL, 0Ayδ , so that we are left with no deflection at A:
0 0VAy Ayδ δ+ = (24)
Of course we don’t yet know the value of V, but from equation (22), we know the
deflection caused by a unit load placed in lieu of V:
3 3
1 1 1 3 84 4Ay
R REI GJ
π πδ ⋅ ⋅ − = ⋅ +
(25)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 18
This is shown in the following diagram:
Using superposition, we know that the deflection caused by the reaction, V, is V times
the deflection caused by a unit load:
1VAy AyVδ δ= ⋅ (26)
Thus equation (24) becomes:
0 1 0Ay AyVδ δ+ ⋅ = (27)
Which we can solve for V:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 19
0
1Ay
Ay
Vδδ
= − (28)
If we take downwards deflections to be positive, we then have, from equations(23),
(25), and (28):
( )24 4
3 3
212 8
1 1 3 84 4
wR wREI GJ
VR R
EI GJ
π
π π
−⋅ + ⋅
= − ⋅ ⋅ − − ⋅ +
(29)
The two negative signs cancel, leaving us with a positive value for V indicating that it
is in the same direction as the unit load, and so is upwards as expected. Introducing
GJEI
β = and doing some algebra on equation (29) gives:
( )
( )
( ) ( )
( )( )
12
12
12
2
21 1 1 1 1 3 82 8 4 4
21 1 1 3 82 8 4 4
4 2 3 88 4
4 2 88 2 2 3 8
V wREI EI EI EI
wR
wR
wR
π π πβ β
π π πβ β
β π βπ πβ β
β π ββ βπ π
−
−
−
− − = ⋅ + ⋅ × ⋅ + − − = + ⋅ × + + − + −
= × + −
= × + −
And so we finally have the required reaction at A as:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 20
( )( )
24 22 2 3 8AV wR
β πβπ π
+ −= + −
(30)
Using Virtual Work
To calculate the reaction at A using virtual work, we use the following:
• Equilibrium system: the external and internal virtual forces corresponding to a
unit virtual force applied in lieu of the required reaction;
• Compatible system: the real external and internal displacements of the original
structure subject to the real applied loads.
Thus the virtual work equations are:
0
E I
Ay
WW W
F M ds T ds
δδ δ
δ δ κ δ φ δ
==
⋅ = ⋅ + ⋅∫ ∫ (31)
At this point we introduce some points:
• The real external deflection at A is zero: 0Ayδ = ;
• The virtual force, 1Fδ = ;
• The real curvatures can be expressed using the real bending moments, MEI
κ = ;
• The real twists are expressed from the torque, TGJ
φ = .
These combine to give, from equation (31):
0 0
0 1L LM TM ds T ds
EI GJδ δ ⋅ = ⋅ + ⋅ ∫ ∫ (32)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 21
Next, we use superposition to express the real internal ‘forces’ as those due to the real
loading applied to the primary structure plus a multiplier times those due to the unit
virtual load applied in lieu of the reaction:
0 1 0 1M M M T T Tα α= + = + (33)
Notice that 1M Mδ = and 1T Tδ = , but they are still written with separate notation to
keep the ideas clear. Thus equation (32) becomes:
( ) ( )0 1 0 1
0 0
0 1 0 1
0 0 0 0
0
0
L L
L L L L
M M T TM ds T ds
EI GJ
M M T TM ds M ds T ds T dsEI EI GJ GJ
α αδ δ
δ α δ δ α δ
+ += ⋅ + ⋅
= ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅
∫ ∫
∫ ∫ ∫ ∫
(34)
And so finally:
0 0
0 01 1
0 0
L L
L L
M TM ds T dsEI GJ
M TM ds T dsEI GJ
δ δα
δ δ
⋅ + ⋅
= −
⋅ + ⋅
∫ ∫
∫ ∫ (35)
At this point we must note the similarity between equations (35) and (28). From
equation (3), it is clear that the numerator in equation (35) is the deflection at A of the
primary structure subject to the real loads. Further, from equation (15), the
denominator in equation (35) is the deflection at A due to a unit (virtual) load at A.
Neglecting signs, and generalizing somewhat, we can arrive at an ‘empirical’
equation for the calculation of redundants:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 22
of primary structure along due to actual loads line of action of redundant due to unit redundant
δαδ
=
(36)
Using this form we will quickly be able to determine the solutions to further ring-
beam problems.
The solution for α follows directly from the previous examples:
• The numerator is determined as per Example 1;
• The denominator is determined as per Example 2, with 1P = .
Of course, these two steps give the results of equations (23) and (25) which were
used in equation (28) to obtain equation (29), and leading to the solution, equation
(30).
From this it can be seen that compatibility of displacement and virtual work are
equivalent ways of looking at the problem. Also it is apparent that the virtual work
framework inherently calculates the displacements required in a compatibility
analysis. Lastly, equation (36) provides a means for quickly calculating the redundant
for other arrangements of the structure from the existing solutions, as will be seen in
the next example.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 23
3.2.4 Example 4
Problem
For the structure shown, the quarter-circle beam has flexural and torsional rigidities
of EI and GJ respectively and the cable has axial rigidity EA, show that the tension in
the cable due to the uniformly distributed load, w, shown is:
( ) ( )1
2
34 2 2 2 3 8 8 LT wRR
ββ π πβ πγ
− = + − + − + ⋅
where GJEI
β = and EAEI
γ = .
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 24
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 25
Solution
For this solution, we will use the insights gained from Example 3, in particular
equation (36). We will then verify this approach using the usual application of virtual
work. We will be choosing the cable as the redundant throughout.
Empirical Form
Repeating our ‘empirical’ equation here:
of primary structure along due to actual loads line of action of redundant due to unit redundant
δαδ
=
(37)
We see that we already know the numerator: the deflection at A in the primary
structure, along the line of the redundant (vertical, since the cable is vertical), due to
the actual loads on the structure is just the deflection of Example 1:
( )24 40 21
2 8Ay
wR wREI GJ
πδ
−= ⋅ + ⋅ (38)
This is shown below:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 26
Next we need to identify the deflection of the primary structure due to a unit
redundant, as shown below:
The components that make up this deflection are:
• Deflection of curved beam caused by unit load (bending and torsion);
• Deflection of the cable AC caused by the unit tension.
The first of these is simply the unit deflection of Example 3, equation (25):
( )3 3
1 1 1 3 8beam4 4Ay
R REI GJ
π πδ ⋅ ⋅ − = ⋅ +
(39)
The second of these is not intuitive, but does feature in the virtual work equations, as
we shall see. The elongation of the cable due to a unit tension is:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 27
( )1 1cableAy
LEA
δ ⋅= (40)
Thus the total deflection along the line of the redundant, of the primary structure, due
to a unit redundant is:
( ) ( )1 1 1
3 3
beam cable
1 1 3 8 14 4
Ay Ay Ay
R R LEI GJ EA
δ δ δ
π π
= +
⋅ ⋅ − ⋅ = ⋅ + +
(41)
Both sets of deflections (equations (39) and (41)) are figuratively summarized as:
And by making 0 1Ay AyTδ δ= , where T is the tension in the cable, we obtain our
compatibility equation for the redundant. Thus, from equations (37), (38) and (41) we
have:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 28
( )24 4
3 3
212 8
1 1 3 8 14 4
wR wREI GJ
TR R L
EI GJ EA
π
π π
−⋅ + ⋅
= ⋅ ⋅ − ⋅ ⋅ + +
(42)
Setting GJEI
β = and EAEI
γ = , and performing some algebra gives:
( )
( ) ( )
( ) ( )
12
3
12
3
1
2 3
21 1 1 1 1 3 82 8 4 4
4 2 3 88 4
82 2 3 84 28 8
LT wREI EI EI EI R EI
LwRR
LRwR
π π πβ β γ
β π βπ πβ β γ
ββπ πβ π γβ β
−
−
−
− − = ⋅ + ⋅ ⋅ + + + − + −
= +
+ − + + − =
(43)
Which finally gives the required tension as:
( ) ( )1
2
34 2 2 2 3 8 8 LT wRR
ββ π πβ πγ
− = + − + − + ⋅
(44)
Comparing this result to the previous result, equation (30), for a pinned support at A,
we can see that the only difference is the term related to the cable: 38 LR
βγ⋅ . Thus the
‘reaction’ (or tension in the cable) at A depends on the relative stiffnesses of the beam
and cable (through the 3R
EI,
3RGJ
and LEA
terms inherent through γ and β ). This
dependence on relative stiffness is to be expected.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 29
Formal Virtual Work Approach
Without the use of the insight that equation (37) gives, the more formal application of
virtual work will, of course, yield the same result. To calculate the tension in the
cable using virtual work, we use the following:
• Equilibrium system: the external and internal virtual forces corresponding to a
unit virtual force applied in lieu of the redundant;
• Compatible system: the real external and internal displacements of the original
structure subject to the real applied loads.
Thus the virtual work equations are:
0
E I
Ay
WW W
F M ds T ds e P
δδ δ
δ δ κ δ φ δ δ
==
⋅ = ⋅ + ⋅ + ⋅∑∫ ∫ (45)
In this equation we have accounted for all the major sources of displacement (and
thus virtual work). At this point we acknowledge:
• There is no external virtual force applied, only an internal tension, thus 0Fδ = ;
• The real curvatures and twists are expressed using the real bending moments and
torques as MEI
κ = and TGJ
φ = respectively;
• The elongation of the cable is the only source of axial displacement and is
written in terms of the real tension in the cable, P, as PLeEA
= .
These combine to give, from equation (45):
0 0
0L L
AyM T PLM ds T ds PEI GJ EA
δ δ δ δ ⋅ = ⋅ + ⋅ + ⋅ ∫ ∫ (46)
As was done in Example 3, using superposition, we write:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 30
0 1 0 1 0 1M M M T T T P P Pα α α= + = + = + (47)
However, we know that there is no tension in the cable in the primary structure, since
it is the cable that is the redundant and is thus removed, hence 0 0P = . Using this and
equation (47) in equation (46) gives:
( ) ( ) ( )0 1 0 1 1
0 0
0L LM M T T P L
M ds T ds PEI GJ EAα α α
δ δ δ + +
= ⋅ + ⋅ + ⋅ ∫ ∫ (48)
Hence:
0 1
0 0
0 1
0 0
1
0L L
L L
M MM ds M dsEI EI
T TT ds T dsGJ GJ
P L PEA
δ α δ
δ α δ
α δ
= ⋅ + ⋅ ⋅
+ ⋅ + ⋅ ⋅
+ ⋅ ⋅
∫ ∫
∫ ∫ (49)
And so finally:
0 0
0 01 1 1
0 0
L L
L L
M TM ds T dsEI GJ
M T P LM ds T ds PEI GJ EA
δ δα
δ δ δ
⋅ + ⋅
= −
⋅ + ⋅ + ⋅
∫ ∫
∫ ∫ (50)
Equation (50) matches equation (35) except for the term relating to the cable. Thus
the other four terms are evaluated exactly as per Example 3. The cable term,
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 31
1P L PEA
δ⋅ , is easily found once it is recognized that 1 1P Pδ= = as was the case for
the moment and torsion in Example 3. With all the terms thus evaluated, equation
(50) becomes the same as equation (42) and the solution progresses as before.
The virtual work approach yields the same solution, but without the added insight of
the source of each of the terms in equation (50) represented by equation (37).
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 32
3.2.5 Example 5
Problem
For the structure shown, the quarter-circle beam has the properties:
• torsional rigidity of GJ;
• flexural rigidity about the local y-y axis YEI ;
• flexural rigidity about the local z-z axis ZEI .
The cable has axial rigidity EA. Show that the tension in the cable due to the
uniformly distributed load, w, shown is:
( ) ( )12
2
4 2 1 1 8 21 3 82
T wRR
β ππ π
λ β γβ
− + − = + + − +
where Y
GJEI
β = , Y
EAEI
γ = and Z
Y
EIEI
λ = .
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 33
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 34
Solution
We will carry out this solution using both the empirical and virtual work approaches
as was done for Example 4. However, it is in this example that the empirical
approach will lead to savings in effort over the virtual work approach, as will be seen.
Empirical Form
Repeating our empirical equation:
of primary structure along due to actual loads line of action of redundant due to unit redundant
δαδ
=
(51)
We first examine the numerator with the following y-z axis elevation of the primary
structure loaded with the actual loads:
Noting that it is the deflection along the line of the redundant that is of interest, we
can draw the following:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 35
The deflection Azδ , which is the distance 'AA is known from Example 2 to be:
( )24 4 212 8Az
wR wREI GJ
πδ
−= ⋅ + ⋅ (52)
It is the deflection ''AA that is of interest here. Since the triangle A-A’-A’’ is a 1-1-
2 triangle, we have:
, 4 2Az
A π
δδ = (53)
And so the numerator is thus:
( )24 40 2
2 2 8 2A
wR wRGJEI
πδ
−= + ⋅ (54)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 36
To determine the denominator of equation (51) we must apply a unit load in lieu of
the redundant (the cable) and determine the deflection in the direction of the cable.
Firstly we will consider the beam. We can determine the deflection in the z- and y-
axes separately and combine, by examining the deflections that the components of the
unit load cause:
To find the deflection that a force of 12
causes in the z- and y-axes directions, we
will instead find the deflections that unit loads cause in these directions, and then
divide by 2 .
Since we are now calculating deflections in two orthogonal planes of bending, we
must consider the different flexural rigidities the beam will have in these two
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 37
directions: YEI for the horizontal plane of bending (vertical loads), and ZEI for loads
in the x-y plane, as shown in the figure:
First, consider the deflection at A in the z-direction, caused by a unit load in the z-
direction, as shown in the following diagram. This is the same as the deflection
calculated in Example 1 and used in later examples:
3 3
1 1 1 3 84 4Az
Y
R REI GJ
π πδ ⋅ ⋅ − = ⋅ +
(55)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 38
Considering the deflection at A in the y-direction next, we see from the following
diagram that we do not have this result to hand, and so must calculate it:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 39
Looking at the elevation of the x-y plane, we have:
The lever arm, m, is:
sinm R θ= (56)
Thus the moment at point C is:
( ) 1 1 sinM m Rθ θ= ⋅ = ⋅ (57)
Using virtual work:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 40
0
1E I
Ay
WW W
M ds
δδ δ
δ κ δ
==
⋅ = ⋅∫ (58)
In which we note that there is no torsion term, as the unit load in the x-y plane does
not cause torsion in the structure. Using ZM EIκ = and ds Rdθ= :
2
0
1 AyZ
M M RdEI
π
δ δ θ⋅ = ∫ (59)
Since sinM M Rδ θ= = , and assuming the beam is prismatic, we have:
3 2
2
0
1 sinAyz
R dEI
π
δ θ θ⋅ = ∫ (60)
This is the same as the first term in equation (7) and so immediately we obtain the
solution as that of the first term of equation (11):
3
1
4Ayz
REI
πδ = ⋅ (61)
In other words, the bending deflection at A in the x-y plane is the same as that in the
z-y plane. This is apparent given that the lever arm is the same in both cases.
However, the overall deflections are not the same due to the presence of torsion in the
z-y plane.
Now that we have the deflections in the two orthogonal planes due to the units loads,
we can determine the deflections in these planes due to the load 12
:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 41
3
1 2 1 1 3 84 42Az
Y
REI GJ
π πδ − = ⋅ +
(62)
3
1 2 142Ay
z
REI
πδ
= ⋅
(63)
The deflection along the line of action of the redundant is what is of interest:
Looking at the contributions of each of these deflections along the line of action of
the redundant:
From this we have:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 42
1 2
3
3
12
1 1 1 3 84 42 2
1 1 3 82 4 4
Az Az
Y
Y
AE
REI GJ
REI GJ
δ δ
π π
π π
= ⋅
− = ⋅ ⋅ +
− = ⋅ +
(64)
1 2
3
3
12
1 142 2
12 4
Ay Ay
z
z
AD
REI
REI
δ δ
π
π
=
= ⋅ ⋅
= ⋅
(65)
Thus the total deflection along the line of action of the redundant is:
1, 4
3 31 1 3 8 12 4 4 2 4
A Az Ay
Y z
AE AD
R REI GJ EI
πδ δ δ
π π π
= +
− = ⋅ + + ⋅
(66)
This gives, finally:
3
1, 4
1 1 1 3 82 4 4A
Y z
REI EI GJπ
π πδ − = + +
(67)
To complete the denominator of equation (51), we must include the deflection that
the cable undergoes due to the unit tension that is the redundant:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 43
1
2
LeEAREA
⋅=
= (68)
The relationship between R and L is due to the geometry of the problem – the cable is
at an angle of 45°.
Thus the denominator of equation (51) is finally:
3
1, 4 2
1 1 1 3 8 2 22 4 4A
Y z
REI EI GJ R EAπ
π πδ − = + + +
(69)
The solution for the tension in the cable becomes, from equations (51), (54) and (69):
( )2
4
3
2
21 12 2 8 2
1 1 1 3 8 2 22 4 4Y z
wRGJEI
TR
EI EI GJ R EA
π
π π
−+ ⋅
= − + + +
(70)
Using Y
GJEI
β = , Y
EAEI
γ = and Z
Y
EIEI
λ = , we have:
( )2
1
2
21 12 2 8 2
1 1 1 3 8 28 8
YY
Y Y Y Y
T wREIEI
EI EI EI R EI
πβ
π πλ β γ
−
−= + ⋅
− × + + +
(71)
Continuing the algebra:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 44
( )
( ) ( )
12
2
12
2
21 1 1 1 3 8 218 82 2 8 2
4 2 1 1 8 21 3 88 8 88 2
T wRR
wRR
π π πβ λ β γ
β π π πλ β γβ
−
−
− − = + ⋅ + + +
+ − = + + − +
(72)
Which finally gives the desired result:
( ) ( )12
2
4 2 1 1 8 21 3 82
T wRR
β ππ π
λ β γβ
− + − = + + − +
(73)
Formal Virtual Work Approach
In the empirical approach carried out above there were some steps that are not
obvious. Within a formal application of virtual work we will see how the results of
the empirical approach are obtained ‘naturally’.
Following the methodology of the formal virtual work approach of Example 4, we
can immediately jump to equation (46):
0 0
0L L
AyM T PLM ds T ds PEI GJ EA
δ δ δ δ ⋅ = ⋅ + ⋅ + ⋅ ∫ ∫ (74)
For the next step we need to recognize that the unit redundant causes bending about
both axes of bending and so the first term in equation (74) must become:
0 0 0
L L LY Z
Y ZY Z
M M MM ds M ds M dsEI EI EI
δ δ δ ⋅ = ⋅ + ⋅
∫ ∫ ∫ (75)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 45
In which the notation YM and ZM indicate the final bending moments of the actual
structure about the Y-Y and Z-Z axes of bending respectively. Again we use
superposition for the moments, torques and axial forces:
0 1
0 1
0 1
0 1
Y Y Y
Z Z Z
M M MM M M
T T TP P P
αααα
= +
= +
= +
= +
(76)
We do not require more torsion terms since there is only torsion in the z-y plane. With
equations (75) and (76), equation (74) becomes:
( ) ( )
( ) ( )
0 1 0 1
0 0
0 1 0 1
0
0L L
Y Y Z ZY Z
Y Z
L
M M M MM ds M ds
EI EI
T T P P LT ds P
GJ EA
α αδ δ
α αδ δ
+ += ⋅ + ⋅
+ +
+ ⋅ + ⋅
∫ ∫
∫
(77)
Multiplying out gives:
0 1
0 0
0 1
0 0
0 1
0 0
0 1
0L L
Y YY Y
Y Y
L LZ Z
Z ZZ Z
L L
M MM ds M dsEI EI
M MM ds M dsEI EI
T TT ds T dsGJ GJ
P L P LP PEA EA
δ α δ
δ α δ
δ α δ
δ α δ
= ⋅ + ⋅ ⋅
+ ⋅ + ⋅ ⋅
+ ⋅ + ⋅ ⋅
+ ⋅ + ⋅ ⋅
∫ ∫
∫ ∫
∫ ∫
(78)
At this point we recognize that some of the terms are zero:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 46
• There is no axial force in the primary structure since the cable is ‘cut’, and so 0 0P = ;
• There is no bending in the x-y plane (about the z-z axis of the beam) in the
primary structure as the loading is purely vertical, thus 0 0ZM = .
Including these points, and solving for α gives:
0 0
0 01 1 1 1
0 0 0
L LY
YY
L L LY Z
Y ZY Z
M TM ds T dsEI GJ
M M T P LM ds M ds T ds PEI EI GJ EA
δ δα
δ δ δ δ
⋅ + ⋅
= −
⋅ + ⋅ + ⋅ + ⋅
∫ ∫
∫ ∫ ∫ (79)
We will next examine this expression term-by-term.
0
0
LY
YY
M M dsEI
δ⋅∫
For this term, 0YM are the moments caused by the UDL about the y-y axis of bending,
as per equation (13):
( ) ( )0 2 1 cosYM wRθ θ= − (80)
YMδ are the moments about the same axis caused by the unit redundant. Since this
redundant acts at an angle of 45° to the plane of interest, these moments are caused
by its vertical component of 12
. From equation (4), we thus have:
( ) 1 sin2YM Rδ θ θ= − (81)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 47
Notice that we have taken it that downwards loading causes positive bending
moments. Thus we have:
( )
( )
02
0 0
23
0
1 11 cos sin2
sin sin cos2
L LY
YY Y
Y
M M ds wR R dsEI EI
wR RdEI
π
δ θ θ
θ θ θ θ
⋅ = − −
= − −
∫ ∫
∫ (82)
In which we have used the relation ds Rdθ= . From the integral appendix we thus
have:
[ ]
( ) ( ) ( ) ( )
20 42
000
3
1cos cos242
10 1 1 142
LY
YY Y
Y
M wRM dsEI EI
wREI
ππδ θ θ
⋅ = − − − −
= − − − + − −
∫ (83)
And so finally:
0 4
0 2 2
LY
YY Y
M wRM dsEI EI
δ⋅ = −∫ (84)
0
0
L T T dsGJ
δ⋅∫
The torsion caused by the UDL in the primary structure is the same as that from
equation (14):
( ) ( )0 2 sinT wRθ θ θ= − (85)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 48
Similarly to the bending term, the torsion caused by the unit redundant is 12
that of
the unit load of equation (17):
( ) ( )1 1 cos2
T Rδ θ θ= − − (86)
Again note that we take the downwards loads as causing positive torsion. Noting
ds Rdθ= we thus have:
( ) ( )
( )( )
202
0 0
24
0
1 1sin 1 cos2
sin 1 cos2
L T T ds wR R RdGJ GJ
wR dGJ
π
π
δ θ θ θ θ
θ θ θ θ
⋅ = − − −
= − − −
∫ ∫
∫ (87)
This integral is exactly that of the second term in equation (19). Hence we can take its
result from equation (20) to give:
( )220 4
0
282
L T wRT dsGJ GJ
πδ
−⋅ = − ⋅∫ (88)
1
0
LY
YY
M M dsEI
δ⋅∫
For this term we recognize that 1Y YM Mδ= and are the moments caused by the 1
2
component of the unit redundant in the vertical direction and are thus given by
equation (1):
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 49
( )1 1 sin2Y YM M Rδ θ θ= = (89)
Hence this term becomes:
1 2
0 0
3 22
0
1 1 1sin sin2 2
sin2
LY
YY Y
Y
M M ds R R RdEI EI
R dEI
π
π
δ θ θ θ
θ θ
⋅ =
=
∫ ∫
∫ (90)
From the integral tables we thus have:
( )
21 3
0 0
3
1 sin 22 2 4
1 0 0 02 4 4
LY
YY Y
Y
M RM dsEI EI
REI
πθδ θ
π
⋅ = −
= − ⋅ − −
∫ (91)
And so we finally have:
1 3
0 8
LY
YY Y
M RM dsEI EI
πδ⋅ = ⋅∫ (92)
1
0
LZ
ZZ
M M dsEI
δ⋅∫
Again we recognize that 1Z ZM Mδ= and are the moments caused by the 1
2
component of the unit redundant in the x-y plane and are thus given by equation (57).
Hence this term becomes:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 50
1 2
0 0
1 1 1sin sin2 2
LY
YY Y
M M ds R R RdEI EI
π
δ θ θ θ ⋅ = ∫ ∫ (93)
This is the same as equation (90) except for the different flexural rigidity, and so the
solution is got from equation (92) to be:
1 3
0 8
LZ
ZZ Z
M RM dsEI EI
πδ⋅ = ⋅∫ (94)
1
0
L T T dsGJ
δ⋅∫
Once again note that 1T Tδ= and are the torques caused by the 12
vertical
component of the unit redundant. From equation (2), then we have:
( )1 1 1 cos2
T T Rδ θ= = − (95)
Thus:
( ) ( )
( )
21
0 0
232
0
1 1 11 cos 1 cos2 2
1 cos2
L T T ds R R RdGJ GJ
R dGJ
π
π
δ θ θ θ
θ θ
⋅ = − −
= −
∫ ∫
∫ (96)
This integral is that of equation (9) and so the solution is:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 51
1 3
0
3 88
L T RT dsGJ GJ
πδ − ⋅ = ∫ (97)
1P L P
EAδ⋅
Lastly then, since 1 1P Pδ= = and 2L R= , this term is easily calculated to be:
1 2P L RP
EA EAδ⋅ = (98)
With the values for all terms now worked out, we substitute these values into
equation (79) to determine the cable tension:
( )224 4
3 3 3
282 2 2
3 8 28 8 8
Y
Y Z
wR wREI GJ
R R R REI EI GJ EA
π
απ π π
− − − ⋅ = −
− ⋅ + ⋅ + +
(99)
Cancelling the negatives and re-arranging gives:
( )2
4
3
2
21 12 2 8 2
1 1 1 3 8 2 22 4 4
Y
Y z
wRGJEI
TR
EI EI GJ R EA
π
π π
−+ ⋅
= − + + +
(100)
And this is the same as equation (70) and so the solution can proceed as before to
obtain the tension in the cable as per equation (73).
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 52
Comparison of the virtual work with the empirical form illustrates the interpretation
of each of the terms in the virtual work equation that is inherent in the empirical view
of such problems.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 53
3.2.6 Review of Examples 1 – 5
Example 1
For a radius of 2 m and a point load of 10 kN, the bending and torsion moment
diagrams are:
Using the equations derived in Example 1, the Matlab script for this is:
function RingBeam_Ex1 % Example 1 R = 2; % m P = 10; % kN theta = 0:(pi/2)/50:pi/2; M = P*R*sin(theta); T = P*R*(1-cos(theta)); hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off;
0 10 20 30 40 50 60 70 80 900
5
10
15
20
Mom
ent (
kNm
)
Degrees from Y-axis
BendingTorsion
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 54
Example 2
For a radius of 2 m and a UDL of 10 kN/m, the bending and torsion moment
diagrams are:
Using the equations derived in Example 2, the Matlab script for this is:
function RingBeam_Ex2 % Example 2 R = 2; % m w = 10; % kN/m theta = 0:(pi/2)/50:pi/2; M = w*R^2*(1-cos(theta)); T = w*R^2*(theta-sin(theta)); hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off;
0 10 20 30 40 50 60 70 80 900
5
10
15
20
25
30
35
40
Mom
ent (
kNm
)
Degrees from Y-axis
BendingTorsion
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 55
Example 3
For the parameters given below, the bending and torsion moment diagrams are:
Using the equations derived in Example 3, the Matlab script for this is:
function [M T alpha] = RingBeam_Ex3(beta) % Example 3 R = 2; % m w = 10; % kN/m I = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EI = E*I/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 1 beta = GJ/EI; % Torsion stiffness ratio end alpha = w*R*(4*beta+(pi-2)^2)/(2*beta*pi+2*(3*pi-8)); theta = 0:(pi/2)/50:pi/2;
0 10 20 30 40 50 60 70 80 90-10
-5
0
5
10
15
20
X: 90Y: 17.19
X: 59.4Y: -4.157
X: 90Y: 0.02678M
omen
t (kN
m)
Degrees from Y-axis
X: 28.8Y: -6.039
BendingTorsion
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 56
M0 = w*R^2*(1-cos(theta)); T0 = w*R^2*(theta-sin(theta)); M1 = -R*sin(theta); T1 = -R*(1-cos(theta)); M = M0 + alpha.*M1; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off; end
The vertical reaction at A is found to be 11.043 kN. Note that the torsion is
(essentially) zero at support B. Other relevant values for bending moment and torsion
are given in the graph.
By changing β , we can examine the effect of the relative stiffnesses on the vertical
reaction at A, and consequently the bending moments and torsions. In the following
plot, the reaction at A and the maximum and minimum bending and torsion moments
are given for a range of β values.
Very small values of β reflect little torsional rigidity and so the structure movements
will be dominated by bending solely. Conversely, large values of β reflect structures
with small bending stiffness in comparison to torsional stiffness. At either extreme
the variables converge to asymptotes of extreme behaviour. For 0.1 10β≤ ≤ the
variables are sensitive to the relative stiffnesses. Of course, this reflects the normal
range of values for β .
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 57
The Matlab code to produce this figure is:
% Variation with Beta beta = logspace(-3,3); n = length(beta); for i = 1:n [M T alpha] = RingBeam_Ex3(beta(i)); Eff(i,1) = alpha; Eff(i,2) = max(M); Eff(i,3) = min(M); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(beta,Eff(:,1),'b:'); plot(beta,Eff(:,2),'k-','LineWidth',2); plot(beta,Eff(:,3),'k-'); plot(beta,Eff(:,4),'r--','LineWidth',2); plot(beta,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('Va','Max M','Min M','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Beta'); ylabel('Load Effect (kN & kNm)');
10-3
10-2
10-1
100
101
102
103
-10
-5
0
5
10
15
20
25
Beta
Load
Effe
ct (k
N &
kN
m)
Va Max M Min M Max T Min T
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 58
Example 4
For a 20 mm diameter cable, and for the other parameters given below, the bending
and torsion moment diagrams are:
The values in the graph should be compared to those of Example 3, where the support
was rigid. The Matlab script, using Example 4’s equations, for this problem is:
function [M T alpha] = RingBeam_Ex4(gamma,beta) % Example 4 R = 2; % m - radius of beam L = 2; % m - length of cable w = 10; % kN/m - UDL A = 314; % mm2 - area of cable I = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EA = E*A; % kN - axial stiffness EI = E*I/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 2 beta = GJ/EI; % Torsion stiffness ratio end if nargin < 1 gamma = EA/EI; % Axial stiffness ratio end
0 10 20 30 40 50 60 70 80 90-10
-5
0
5
10
15
20
X: 90Y: 17.58
X: 59.4Y: -3.967X: 28.8
Y: -5.853
X: 90Y: 0.4128
Mom
ent (
kNm
)
Degrees from Y-axis
BendingTorsion
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 59
alpha = w*R*(4*beta+(pi-2)^2)/(2*beta*pi+2*(3*pi-8)+8*(beta/gamma)*(L/R^3)); theta = 0:(pi/2)/50:pi/2; M0 = w*R^2*(1-cos(theta)); T0 = w*R^2*(theta-sin(theta)); M1 = -R*sin(theta); T1 = -R*(1-cos(theta)); M = M0 + alpha.*M1; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off; end
Whist keeping the β constant, we can examine the effect of varying the cable
stiffness on the behaviour of the structure, by varying γ . Again we plot the reaction
at A and the maximum and minimum bending and torsion moments for the range of
γ values.
For small γ , the cable has little stiffness and so the primary behaviour will be that of
Example 1, where the beam was a pure cantilever. Conversely for high γ , the cable is
very stiff and so the beam behaves as in Example 3, where there was a pinned support
at A. Compare the maximum (hogging) bending moments for these two cases with
the graph. Lastly, for 0.01 3γ≤ ≤ , the cable and beam interact and the variables are
sensitive to the exact ratio of stiffness. Typical values in practice are towards the
lower end of this region.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 60
The Matlab code for this plot is:
% Variation with Gamma gamma = logspace(-3,3); n = length(gamma); for i = 1:n [M T alpha] = RingBeam_Ex4(gamma(i)); Eff(i,1) = alpha; Eff(i,2) = max(M); Eff(i,3) = min(M); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(gamma,Eff(:,1),'b:'); plot(gamma,Eff(:,2),'k-','LineWidth',2); plot(gamma,Eff(:,3),'k-'); plot(gamma,Eff(:,4),'r--','LineWidth',2); plot(gamma,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('T','Max M','Min M','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Gamma'); ylabel('Load Effect (kN & kNm)');
10-3
10-2
10-1
100
101
102
103
-10
0
10
20
30
40
Gamma
Load
Effe
ct (k
N &
kN
m)
T Max M Min M Max T Min T
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 61
Example 5
Again we consider a 20 mm diameter cable, and a doubly symmetric section, that is
Y ZEI EI= . For the parameters below the bending and torsion moment diagrams are:
The values in the graph should be compared to those of Example 4, where the cable
was vertical. The Matlab script, using Example 5’s equations, for this problem is:
function [My T alpha] = RingBeam_Ex5(lamda,gamma,beta) % Example 5 R = 2; % m - radius of beam w = 10; % kN/m - UDL A = 314; % mm2 - area of cable Iy = 2.7e7; % mm4 Iz = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EA = E*A; % kN - axial stiffness EIy = E*Iy/1e6; % kNm2 EIz = E*Iz/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 3 beta = GJ/EIy; % Torsion stiffness ratio end if nargin < 2 gamma = EA/EIy; % Axial stiffness ratio end if nargin < 1
0 10 20 30 40 50 60 70 80 90-20
-10
0
10
20
30
X: 90Y: -19.22
X: 52.2Y: -2.605
X: 90Y: 3.61
X: 90Y: 20.78
X: 25.2Y: -4.378
Mom
ent (
kNm
)
Degrees from Y-axis
YY BendingZZ BendingTorsion
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 62
lamda = EIy/EIz; % Bending stiffness ratio end numerator = (4*beta+(pi-2)^2)/(beta*sqrt(2)); denominator = (pi*(1+1/lamda)+(3*pi-8)/beta+8*sqrt(2)/(gamma*R^2)); alpha = w*R*numerator/denominator; theta = 0:(pi/2)/50:pi/2; M0y = w*R^2*(1-cos(theta)); M0z = 0; T0 = w*R^2*(theta-sin(theta)); M1y = -R*sin(theta); M1z = -R*sin(theta); T1 = -R*(1-cos(theta)); My = M0y + alpha.*M1y; Mz = M0z + alpha.*M1z; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,My,'k'); plot(theta.*180/pi,Mz,'k:'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('YY Bending','ZZ Bending','Torsion','location','NW'); hold off; end
Keep all parameters constant, but varying the ratio of the bending rigidities by
changing λ , the output variables are as shown below. For low λ (a tall slender
beam) the beam behaves as a cantilever. Thus the cable requires some transverse
bending stiffness to be mobilized. With high λ (a wide flat beam) the beam behaves
as if supported at A with a vertical roller. Only vertical movement takes place, and the
effect of the cable is solely its vertical stiffness at A. Usually 0.1 2λ≤ ≤ which means
that the output variables are usually quite sensitive to the input parameters.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 63
The Matlab code to produce this graph is:
% Variation with Lamda lamda = logspace(-3,3); n = length(lamda); for i = 1:n [My T alpha] = RingBeam_Ex5(lamda(i)); Eff(i,1) = alpha; Eff(i,2) = max(My); Eff(i,3) = min(My); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(lamda,Eff(:,1),'b:'); plot(lamda,Eff(:,2),'k-','LineWidth',2); plot(lamda,Eff(:,3),'k-'); plot(lamda,Eff(:,4),'r--','LineWidth',2); plot(lamda,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('T','Max My','Min My','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Lamda'); ylabel('Load Effect (kN & kNm)');
10-3
10-2
10-1
100
101
102
103
-20
-10
0
10
20
30
40
Lamda
Load
Effe
ct (k
N &
kN
m)
T Max My Min My Max T Min T
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 64
3.3 Grid Examples
3.3.1 Example 1
Problem
For the grid structure shown, which has flexural and torsional rigidities of EI and GJ
respectively, show that the vertical reaction at C is given by:
12 3CV P
β
= +
Where
EIGJ
β =
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 65
Solution
Using virtual work, we have:
0
0
E I
WW W
M TM ds T dsEI GJ
δδ δ
δ δ
==
= ⋅ + ⋅∫ ∫
(101)
Choosing the vertical reaction at C as the redundant gives the following diagrams:
And the free bending moment diagram is:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 66
But the superposition gives:
0 1M M Mα= + (102)
0 1T T Tα= + (103)
Substituting, we get:
( ) ( )0 1 0 10M M T T
M ds T dsEI GJα α
δ δ+ +
= ⋅ + ⋅∫ ∫ (104)
2 2
0 1 1 0 1 1 0M M M T T Tds ds ds dsEI EI GJ GJ
α α+ + + =∫ ∫ ∫ ∫ (105)
2 2
0 1 1 0 1 1 0M M M T T Tds ds ds dsEI EI GJ GJ
α α+ + + =∫ ∫ ∫ ∫ (106)
Taking the beam to be prismatic, and EIGJ
β = gives:
2 20 1 1 0 1 1 0M M ds M ds T T ds T dsα β αβ+ + + =∫ ∫ ∫ ∫ (107)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 67
From which:
0 1 0 1
2 21 1
M M ds T T ds
M ds T ds
βα
β
+ = − +
∫ ∫∫ ∫
(108)
From the various diagrams and volume integrals tables, the terms evaluate to:
( )( )( )
( )
( )( )( )
( )( )( )
3
0 1
0 1
2 31
2 31
13 30 0
1 223 3
PLM M ds L PL L
T T ds
M ds L L L L
T ds L L L L
β β
β β β
= − = −
= =
= =
= =
∫∫
∫
∫
(109)
Substituting gives:
( )
3
3 3
3
3 23
03
23
1 13
PL
L L
PLL
αβ
β
− + = − +
= ⋅ ⋅+
(110)
Which yields:
12 3CV Pα
β
≡ = + (111)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 68
Numerical Example
Using a 200 × 400 mm deep rectangular concrete section, gives the following:
3 4 3 41.067 10 m 0.732 10 mI J= × = ×
The material model used is for a 50N concrete with:
230 kN/mm 0.2E ν= =
Using the elastic relation, we have:
( ) ( )
66 230 10 12.5 10 kN/m
2 1 2 1 0.2EGν
×= = = ×
+ +
From the model, LUSAS gives: 0.809 kNCV = . Other results follow.
Deflected Shape
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 69
Bending Moment Diagram
Torsion Moment Diagram
Shear Force Diagram
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 70
3.3.2 Example 2
Problem
For the grid structure shown, which has flexural and torsional rigidities of EI and GJ
respectively, show that the reactions at C are given by:
4 4 4 28 5 8 5C CV P M PLβ ββ β
+ += = + +
Where
EIGJ
β =
(Note that the support symbol at C indicates a moment and vertical support at C, but
no torsional restraint.)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 71
Solution
The general virtual work equations are:
0
0
E I
WW W
M TM ds T dsEI GJ
δδ δ
δ δ
==
= ⋅ + ⋅∫ ∫
(112)
We choose the moment and vertical restraints at C as the redundants. The vertical
redundant gives the same diagrams as before:
And, for the moment restraint, we apply a unit moment:
Which yields the following:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 72
Again the free bending moment diagram is:
Since there are two redundants, there are two possible equilibrium sets to use as the
virtual moments and torques. Thus there are two equations that can be used:
1 10 M TM ds T dsEI GJ
= ⋅ + ⋅∫ ∫ (113)
2 20 M TM ds T dsEI GJ
= ⋅ + ⋅∫ ∫ (114)
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 73
Superposition gives:
0 1 1 2 2M M M Mα α= + + (115)
0 1 1 2 2T T T Tα α= + + (116)
Substituting, we get from equation (113):
( ) ( )0 1 1 2 2 0 1 1 2 21 10
M M M T T TM ds T ds
EI GJα α α α+ + + +
= ⋅ + ⋅∫ ∫ (117)
20 1 1 2 1
1 2
20 1 1 2 1
1 2 0
M M M M Mds ds dsEI EI EI
T T T T Tds ds dsGJ GJ GJ
α α
α α
+ +
+ + + =
∫ ∫ ∫
∫ ∫ ∫ (118)
Taking the beam to be prismatic, and EIGJ
β = gives:
2
0 1 1 1 2 2 1
20 1 1 1 1 2 1 0
M M ds M ds M M ds
T T ds T ds T T ds
α α
β α β α β
+ +
+ + + =
∫ ∫ ∫∫ ∫ ∫
(119)
Similarly, substituting equations (115) and (116) into equation (114) gives:
2
0 2 1 1 2 2 2
20 2 1 1 2 2 2 0
M M ds M M ds M ds
T T ds TT ds T ds
α α
β α β α β
+ +
+ + + =
∫ ∫ ∫∫ ∫ ∫
(120)
We can write equations (119) and (120) in matrix form for clarity:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 74
0 1 0 1
0 2 0 2
2 21 1 2 1 2 1 1
2 221 2 1 2 2 2
0
M M ds T T ds
M M ds T T ds
M ds T ds M M ds T T ds
M M ds TT ds M ds T ds
β
β
β β ααβ β
+ + +
+ + = + +
∫ ∫∫ ∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(121)
Evaluating the integrals for the first equation gives:
3
0 1 0 1
32 2 3
1 1
2 22 1 2 1
03
23
12
PLM M ds T T ds
LM ds T ds L
M M ds L T T ds L
β
β β
β β
−= =
= =
= − = −
∫ ∫
∫ ∫
∫ ∫
(122)
And for the second:
0 2 0 2
2 21 2 1 2
2 22 2
0 0
12
M M ds T T ds
M M ds L TT ds L
M ds L T ds L
β
β β
β β
= =
= − = −
= =
∫ ∫
∫ ∫∫ ∫
(123)
Substituting these into equation (121), we have:
( )
3 23
1
22
23 2
030 1
2
L LPL
L L
β βαα
β β
1 + − + − + = 1 − + +
(124)
Giving:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 75
( )
13 2 3
1
2 2
23 2
301
2
L L PL
L L
β βαα
β β
− 1 + − + = 1 − + +
(125)
Inverting the matrix gives:
( ) ( )
( ) ( )
33 2
1
22
12 61 1 21
35 8 6 4 01 2 2 3
PLL L
L L
β βαα β
β β
+ + = + + +
(126)
Thus:
( )
( )
( )( )
3
31
32
2
12 13 4 11
2 1 25 8 5 86 1 23
PLL P
LPLL
ββαβα β β
β
+ + = = ++ + +
(127)
Thus, since 1
2
C
C
VM
αα
≡
, we have:
4 4 4 28 5 8 5C CV P M PLβ ββ β
+ += = + +
(128)
And this is the requested result.
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 76
Some useful Matlab symbolic computation script appropriate to this problem is:
syms beta L P A = [ L^3*(2/3+beta) -L^2*(0.5+beta); -L^2*(0.5+beta) L*(1+beta)]; A0 = [P*L^3/3; 0]; invA = inv(A); invA = simplify(invA); disp(simplify(det(A))); disp(invA); alpha = invA*A0; alpha = simplify(alpha);
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 77
Numerical Example
For the numerical model previously considered, for these support conditions, LUSAS
gives us:
5.45 kN 14.5 kNmC CV M= =
Deflected Shape
Shear Force Diagram
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 78
Torsion Moment Diagram
Bending Moment Diagram
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 79
3.3.3 Example 3
Problem
For the grid structure shown, which has flexural and torsional rigidities of EI and GJ
respectively, show that the reactions at C are given by:
( )( ) ( )2 1 1
2 4 1 4 1C C C
P PL PLV M Tββ β
+= = ⋅ = ⋅
+ +
Where
EIGJ
β =
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 80
Solution
The general virtual work equations are:
0
0
E I
WW W
M TM ds T dsEI GJ
δδ δ
δ δ
==
= ⋅ + ⋅∫ ∫
(129)
We choose the moment, vertical, and torsional restraints at C as the redundants. The
vertical and moment redundants give (as before):
Applying the unit torsional moment gives:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 81
Again the free bending moment diagram is:
Since there are three redundants, there are three possible equilibrium sets to use. Thus
we have the following three equations:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 82
1 10 M TM ds T dsEI GJ
= ⋅ + ⋅∫ ∫ (130)
2 20 M TM ds T dsEI GJ
= ⋅ + ⋅∫ ∫ (131)
3 30 M TM ds T dsEI GJ
= ⋅ + ⋅∫ ∫ (132)
Superposition of the structures gives:
0 1 1 2 2 3 3M M M M Mα α α= + + + (133)
0 1 1 2 2 3 3T T T T Tα α α= + + + (134)
Substituting, we get from equation (113):
( ) ( )0 1 1 2 2 3 3 0 1 1 2 2 3 31 10
M M M M T T T TM ds T ds
EI GJα α α α α α+ + + + + +
= ⋅ + ⋅∫ ∫ (135)
20 1 1 2 1 3 1
1 2 3
20 1 1 2 1 3 1
1 2 3 0
M M M M M M Mds ds ds dsEI EI EI EI
T T T T T T Tds ds ds dsGJ GJ GJ GJ
α α α
α α α
+ + +
+ + + + =
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫ (136)
Taking the beam to be prismatic, and EIGJ
β = gives:
2
0 1 1 1 2 2 1 3 3 1
20 1 1 1 2 2 1 3 3 1 0
M M ds M ds M M ds M M ds
T T ds T ds T T ds T T ds
α α α
β α β α β α β
+ + +
+ + + + =
∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(137)
Similarly, substituting equations (115) and (116) into equations (114) and (132)
gives:
Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples
Dr. C. Caprani 83
2
0 2 1 1 2 2 2 3 3 2
20 2 1 1 2 2 2 3 3 2 0
M M ds M M ds M ds M M ds
T T ds TT ds T ds T T ds
α α α
β α β α β α β
+ + +
+ + + + =
∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(138)
2
0 3 1 1 3 2 2 3 3 3
20 3 1 1 3 2 2 3 3 3 0
M M ds M M ds M M ds M ds
T T ds TT ds T T ds T ds
α α α
β α β α β α β
+ + +
+ + + + =
∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(139)
We can write equations (119), (120), and (139) in matrix form for clarity:
To find the forces in the bars, we can now use the member stiffness matrices, since
we know the end displacements:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 18
Member 1
1 3 3
2
100 100 0 4.810 10
100 100 0.048 4.8FF
−− − = × = −
(4.2.21)
Thus Member 1 has a tension of 4.8 kN, since the directions of the member forces are
interpreted by our sign convention:
Also note that it is in equilibrium (as we might expect).
Member 2
2 3 3
3
200 200 0.048 54.810 10
200 200 0.322 54.8FF
−− − = × = −
(4.2.22)
Member 2 thus has tension of 54.8 kN.
Member 3
3 3 3
4
140 140 0.322 45.0810 10
140 140 0 45.08FF
−− = × = − −
(4.2.23)
Thus Member 3 has a compression of 45.08 kN applied to it.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 19
Problem
Find the displacements of the connections and the forces in each member for the
following structure:
Ans. 0.22 mm, 2.11 mm
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 20
4.2.4 General Methodology
Steps
The general steps in Matrix Stiffness Method are:
1. Calculate the member stiffness matrices
2. Assemble the global stiffness matrix
3. Restrict the global stiffness matrix and force vector
4. Solve for the unknown displacements
5. Determine member forces from the known displacements and member stiffness
matrices
6. Determine the reactions knowing member end forces.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 21
Matlab Program - Implementation
These steps are implemented in the Matlab Program as follows:
function [D F R] = AnalyzeTruss(nData,eData) % This function analyzes the truss defined by nData and eData: % nData = [x, y, xLoad, yLoad, xRestraint, yRestraint] % eData = [iNode, jNode, E, A]; kg = AssembleTrussK(nData, eData); % Assemble global stiffness matrix fv = AssembleForceVector(nData); % And the force vector [kgr fv] = Restrict(kg, fv, nData); % Impose restraints D = fv/kgr; % Solve for displacements F = ElementForces(nData,eData,D); % Get the element forces R = D*kg; % Get the reactions
The output from the function AnalyzeTruss is:
• D: vector of nodal deflections;
• F: vector of element forces;
• R: vector of nodal forces (indicating the reactions and applied loads).
The input data required (nData and eData) will be explained later.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 22
4.2.5 Member contribution to global stiffness matrix
Consider a member, ij, which links node i to node j. Its member stiffness matrix will
be:
Node i Node j
Node i k11ij k12ij
Node j k21ij k22ij
Its entries must then contribute to the corresponding entries in the global stiffness
matrix:
… Node i … Node j …
… … … … … …
Node i … k11ij … k12ij …
… … … … … …
Node j … k21ij … k22ij …
… … … … … …
If we now consider another member, jl, which links node j to node l. Its member
stiffness matrix will be:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 23
Node j Node l
Node j k11jl k12jl
Node l k21jl k22jl
And now the global stiffness matrix becomes:
… Node i … Node j … Node l …
… … … … … … … …
Node i … k11ij … k12ij … … …
… … … … … … … …
Node j … k21ij … k22ij + k11jl
… k12jl …
… … … … … … … …
Node l … … k21lj … k22jl …
… … … … … … … …
In the above, the identifiers k11 etc are sub-matrices of dimension:
ndof × ndof
where ndof refers to the number of degrees of freedom that each node has.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 24
Matlab Program – Element Contribution
Considering trusses, we have 2 degrees of freedom (DOFs) per node, the x direction
and the y direction. Thus, for a truss with nn number of nodes, there are 2nn DOFs in
total. The x-DOF for any node i is thus located at 2i-1 and the y-DOF at 2i.
Consider a truss member connecting nodes i and j. To add the 4×4 truss element
stiffness matrix into the truss global stiffness matrix, we see that each row adds into
the following matrix columns:
2i-1 2i 2j-1 2j
The rows in the global stiffness matrix corresponding to the rows of the element
stiffness matrix are:
1. Row 1: Adds to row 2i-1 of the global stiffness matrix;
2. Row 2: Adds to row 2i;
3. Row 3: adds to row 2j-1;
4. Row 4: adds to row 2j.
Note of course that the column and row entries occur in the same order.
These rules are implemented for our Truss Analysis Program as follows:
function kg = AddElement(iEle,eData,ke,kg) % This function adds member iEle stiffness matrix ke to the global % stiffness matrix kg. % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % The DOFs in kg to enter the properties into DOFs = [2*iNode-1 2*iNode 2*jNode-1 2*jNode]; % For each row of ke for i = 1:4 % Add the row to the correct entries in kg kg(DOFs(i),DOFs) = kg(DOFs(i),DOFs) + ke(i,:); end
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 25
Matlab Program – Global Stiffness Matrix Assembly
The function that assembles the truss global stiffness matrix for the truss is as
follows:
function kg = AssembleTrussK(nData, eData) % This function assembles the global stiffness matrix for a truss from the % joint and member data matrices % How many nodes and elements are there? [ne ~] = size(eData); [nn ~] = size(nData); % Set up a blank global stiffness matrix kg = zeros(2*nn,2*nn); % For each element for i = 1:ne E = eData(i,3); % Get its E and A A = eData(i,4); [L c s] = TrussElementGeom(i,nData,eData); % Geometric Properties ke = TrussElementK(E,A,L,c,s); % Stiffness matrix kg = AddElement(i,eData,ke,kg); % Enter it into kg end
Note that we have not yet covered the calculation of the truss element stiffness
matrix. However, the point here is to see that each element stiffness matrix is
calculated and then added to the global stiffness matrix.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 26
Matlab Program – Force Vector
Examine again the overall equation (4.2.10) to be solved:
{ } [ ]{ }=F K u
We now have the global stiffness matrix, we aim to calculate the deflections thus we
need to have a force vector representing the applied nodal loads. Again remember
that each node as two DOFs (x- and y-loads). The code for the force vector is thus:
function f = AssembleForceVector(nData) % This function assembles the force vector % How may nodes are there? [nn ~] = size(nData); % Set up a blank force vector f = zeros(1,2*nn); % For each node for i = 1:nn f(2*i - 1) = nData(i, 3); % x-load into x-DOF f(2*i) = nData(i, 4); % y-load into y-DOF end
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 27
4.2.6 Interpretation of Stiffness Matrix
It is useful to understand what each term in a stiffness matrix represents. If we
consider a simple example structure:
We saw that the global stiffness matrix for this is:
11 12 13 1 1
21 22 23 1 1 2 2
31 32 33 2 2
0
0
K K K k kK K K k k k kK K K k k
− = = − + − −
K
If we imagine that all nodes are fixed against displacement except for node 2, then we
have the following:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 28
From our general equation:
1 11 12 13 12
2 21 22 23 22
3 31 32 33 32
010
F K K K KF K K K KF K K K K
= =
(4.2.24)
Thus:
1 12 1
2 22 1 2
3 32 2
F K kF K k kF K k
− = = + −
(4.2.25)
These forces are illustrated in the above diagram, along with a free-body diagram of
node 2.
Thus we see that each column in a stiffness matrix represents the forces required to
maintain equilibrium when the column’s DOF has been given a unit displacement.
This provides a very useful way to derive member stiffness matrices.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 29
4.2.7 Restricting a Matrix – Imposing Restraints
In Example 1 we solved the structure by applying the known supports into the global
stiffness matrix. We did this because otherwise the system is unsolvable; technically
the determinant of the stiffness matrix is zero. This mathematically represents the fact
that until we apply boundary conditions, the structure is floating in space.
To impose known displacements (i.e. supports) on the structure equations we modify
the global stiffness matrix and the force vector so that we get back the zero
displacement result we know.
Considering our two-element example again, if node 1 is supported, 1 0u = . Consider
the system equation:
1 11 12 13 1
2 21 22 23 2
3 31 32 33 3
F K K K uF K K K uF K K K u
=
(4.2.26)
Therefore to obtain 1 0u = from this, we change K and F as follows:
1
2 22 23 2
3 32 33 3
0 1 0 000
uF K K uF K K u
=
(4.2.27)
Now when we solve for 1u we will get the answer we want: 1 0u = . In fact, since we
now do not need this first equation, we could just consider the remaining equations:
2 22 23 2
3 32 33 3
F K K uF K K u
=
(4.2.28)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 30
And these are perfectly solvable.
Thus to summarize:
To impose a support condition at degree of freedom i:
1. Make the force vector element of DOF i zero;
2. Make the i column and row entries of the stiffness matrix all zero;
3. Make the diagonal entry ( ),i i of the stiffness matrix 1.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 31
Matlab Program – Imposing Restraints
To implement these rules for our Truss Analysis Program, we will first create of
vector which tells us whether or not a DOF is restrained. This vector will have a zero
if the DOF is not restrained, and a 1 if it is.
Once we have this vector of restraints, we can go through each DOF and modify the
force vector and global stiffness matrix as described before. The implementation of
this is as follows:
function [kg f] = Restrict(kg, f, nData) % This function imposes the restraints on the global stiffness matrix and % the force vector % How may nodes are there? [nn ~] = size(nData); % Store each restrained DOF in a vector RestrainedDOFs = zeros(2*nn,1); % For each node, store if there is a restraint for i = 1:nn % x-direction if nData(i,5) ~= 0 % if there is a non-zero entry (i.e. supported) RestrainedDOFs(2*i-1) = 1; end % y-direction if nData(i,6) ~= 0 % if there is a support RestrainedDOFs(2*i) = 1; end end % for each DOF for i = 1:2*nn if RestrainedDOFs(i) == 1 % if it is restrained f(i) = 0; % Ensure force zero at this DOF kg(i,:) = 0; % make entire row zero kg(:,i) = 0; % make entire column zero kg(i,i) = 1; % put 1 on the diagonal end end
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 32
4.3 Plane Trusses
4.3.1 Introduction
Trusses are assemblies of members whose actions can be linked directly to that of the
simple spring studied already:
EAkL
= (4.3.1)
There is one main difference, however: truss members may be oriented at any angle
in the xy coordinate system (Cartesian) plane:
Thus we must account for the coordinate transformations from the local member axis
system to the global axis system.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 33
Matlab Program – Data Preparation
In the following sections we will put the final pieces of code together for our Truss
Analysis Program. At this point we must identify what information is required as
input to the program, and in what format it will be delivered.
The node data is stored in a matrix nData. Each node of the truss is represented by a
row of data. In the row, we put the following information in consecutive order in
columns:
1. x-coordinate;
2. y-coordinate;
3. x-load: 0 or the value of load;
4. y-load: 0 or the value of load;
5. x-restraint: 0 if unrestrained, any other number if restrained;
6. y-restraint: 0 if unrestrained, any other number if restrained.
The element data is stored in a matrix called eData. Each element has a row of data
and for each element the information stored in the columns in order is:
1. i-Node number: the node number at the start of the element;
2. j-Node number: the other node the element connects to;
3. E: the Modulus of Elasticity of the element material;
4. A: the element area;
We will prepare input data matrices in the above formats for some of the examples
that follow so that the concepts are clear. In doing so we keep the units consistent:
• Dimensions are in m;
• Forces in kN
• Elastic modulus is in kN/mm2;
• Area is mm2.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 34
Matlab Program – Data Entry
To enter the required data, one way is:
1. Create a new variable in the workspace (click on New Variable);
2. Name it eData for example;
3. Double click on the new variable to open the Matlab Variable Editor;
4. Enter the necessary input data (can paste in from MS Excel, or type in);
5. Repeat for the nodal data.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 35
4.3.2 Truss Element Stiffness Matrix
For many element types it is very difficult to express the element stiffness matrix in
global coordinates. However, this is not so for truss elements. Firstly we note that the
local axis system element stiffness matrix is given by equation (4.2.3):
[ ] 1 11 1
k kk
k k− −
= = − − k (4.3.2)
Next, introducing equation (4.3.1), we have:
[ ] 1 11 1
EAL
− = −
k (4.3.3)
However, this equation was written for a 1-dimensional element. Expanding this to a
two-dimensional axis system is straightforward since there are no y-axis values:
[ ]
1 0 1 00 0 0 01 0 1 0
0 0 0 0
i
i
j
j
xyEAxLy
←− ← =
← − ←
k (4.3.4)
Next, using the general element stiffness transformation equation (See the Appendix):
[ ] [ ] [ ][ ]Tk = T k T (4.3.5)
And noting the transformation matrix for a plane truss element from the Appendix:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 36
cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos
P
P
α αα α
α αα α
− = −
T 0T =
0 T (4.3.6)
We have:
[ ]
1cos sin 0 0 1 0 1 0sin cos 0 0 0 0 0 00 0 cos sin 1 0 1 00 0 sin cos 0 0 0 0
cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos
EAL
α αα α
α αα α
α αα α
α αα α
−−
− = ⋅ − −
− −
k
(4.3.7)
Carrying out the multiplication gives:
2 2
2 2
2 2
2 2
cos cos sin cos cos sincos sin sin cos sin sin
cos cos sin cos cos sincos sin sin cos sin sin
EAL
α α α α α αα α α α α α
α α α α α αα α α α α α
− − − − = − − − −
k (4.3.8)
If we examine the nodal sub-matrices and write cosc α≡ , sins α≡ :
[ ]
2 2
2 2
2 2
2 2
c cs c cscs s cs sEA
L c cs c cscs s cs s
− − − − = − − − −
k (4.3.9)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 37
Labelling the nodal sub-matrices as:
[ ] =
k11 k12k
k21 k22 (4.3.10)
Then we see that the sub-matrices are of dimension 2 × 2 (No. DOF × No. DOF) and
are:
2
2
c csEAL cs s
=
k11 (4.3.11)
And also note:
k11 = k22 = -k12 = -k21 (4.3.12)
Therefore, we need only evaluate a single nodal sub-matrix (k11) in order to find the
total element stiffness matrix in global coordinates.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 38
Matlab Program – Element Stiffness Matrix
Calculating the element stiffness matrix for our Truss Analysis Program is easy. The
only complexity is extracting the relevant data from the input node and element data
matrices. Rather than try determine the angle that the truss member is at (remember
we only have the nodal coordinates), we can calculate cosα and sinα directly (e.g.
adjacent/hypotenuse). Further, the element length can be found using Pythagoras,
given the nodal coordinates. These element properties are found in the script below:
function [L c s] = TrussElementGeom(iEle,nData,eData); % This function returns the element length % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % What are the coordinates of these nodes? iNodeX = nData(iNode,1); iNodeY = nData(iNode,2); jNodeX = nData(jNode,1); jNodeY = nData(jNode,2); % Use Pythagoras to work out the member length L = sqrt((jNodeX - iNodeX)^ 2 + (jNodeY - iNodeY)^ 2); % Cos is adjacent over hyp, sin is opp over hyp c = (jNodeX - iNodeX)/L; s = (jNodeY - iNodeY)/L;
The E and A values for each element are directly found from the input data element
matrix as follows:
E = eData(i,3); % Get its E and A A = eData(i,4);
Thus, with all the relevant data assembled, we can calculate the truss element
stiffness matrix. In the following Matlab function, note that we make use of the fact
that each nodal sub-matrix can be determined from the nodal sub-matrix k11 :
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 39
function k = TrussElementK(E,A,L,c,s) % This function returns the stiffness matrix for a truss element k11 = [ c^2 c*s; c*s s^2]; k = (E*A/L) * [ k11 -k11; -k11 k11];
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 40
4.3.3 Element Forces
The forces applied to a member’s ends are got from the element equation:
{ } [ ]{ }e e=F k u (4.3.13)
Expanding this in terms of nodal equations we have:
i i
j j
=
F δk11 k12F δk21 k22
(4.3.14)
Thus we know:
j i j= ⋅ + ⋅F k21 δ k22 δ (4.3.15)
From which we could determine the member’s axial force. However, for truss
members, we can determine a simple expression to use if we consider the change in
length in terms of the member end displacements:
x jx ixL δ δ∆ = − (4.3.16)
y jy iyL δ δ∆ = − (4.3.17)
And using the coordinate transforms idea:
cos sinx yL L Lα α∆ = ∆ + ∆ (4.3.18)
Also we know that the member force is related to the member elongation by:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 41
EAF LL
= ⋅∆ (4.3.19)
Thus we have:
cos sinx y
EAF L LL
α α = ⋅ ∆ + ∆ (4.3.20)
And introducing equations (4.3.16) and (4.3.17) gives:
[ ]cos sin jx ix
jy iy
EAFL
δ δα α
δ δ−
= ⋅ − (4.3.21)
A positive result from this means tension and negative compression.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 42
Matlab Program – Element Force
Once the element nodal deflections are known, the element forces are found as
described above. Most of the programming effort is dedicated to extracting the nodal
deflections that are relevant for the particular member under consideration:
function F = TrussElementForce(nData, eData, d, iEle) % This function returns the element force for iEle given the global % displacement vector, d, and the node and element data matrices. % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % Get the element properties E = eData(iEle,3); % Get its E and A A = eData(iEle,4); [L c s] = TrussElementGeom(iEle,nData,eData); % Geometric Properties dix = d(2*iNode-1); % x-displacement at node i diy = d(2*iNode); % y-displacement at node i djx = d(2*jNode-1); % x-displacement at node j djy = d(2*jNode); % y-displacement at node j F = (E*A/L) * (c*(djx-dix) + s*(djy-diy));
Note also that the way the program is written assumes that tension is positive and
compression is negative. We also want to return all of the element forces, so we use
the function just described to calculate all the truss elements’ forces:
function F = ElementForces(nData,eData,d) % This function returns a vector of the element forces % How many elements are there? [ne ~] = size(eData); % Set up a blank element force vector F = zeros(ne,1); % For each element for i = 1:ne % Get its force and enter into vector F(i) = TrussElementForce(nData, eData, d, i); end
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 43
4.3.4 Example 2: Basic Truss
Problem
Analyse the following truss using the stiffness matrix method.
Note that:
• 2200 kN/mmE = ;
• The reference area is 2100mmA = .
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 44
Solution
STEP 1: Determine the member stiffness matrices:
Member 12
The angle this member makes to the global axis system and the relevant values are:
21 1cos cos4522
c cα≡ = = ⇒ =
21 1 1sin sin 452 22
s s csα≡ = = ⇒ = ⇒ =
Therefore:
2
12 212
0.5 0.5200 100 20.5 0.510 2
c csEAL cs s
⋅ = = k11
Thus:
312
0.5 0.510
0.5 0.5
=
k11 (4.3.22)
Notice that the matrix is symmetrical as it should be.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 45
Member 23
The angle this member makes to the global axis system and the relevant values are:
21 1cos cos31522
c cα≡ = = ⇒ =
21 1 1sin sin3152 22
s s csα≡ = = − ⇒ = ⇒ = −
Therefore:
2
23 223
0.5 0.5200 100 20.5 0.510 2
c csEAL cs s
− ⋅ = = − k11
Thus:
323
0.5 0.510
0.5 0.5−
= − k11 (4.3.23)
Again the matrix is symmetrical.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 46
STEP 2: Assemble the global stiffness matrix
For 3 nodes, the unrestricted global stiffness matrix will look as follows:
11 12 13
21 22 23
31 32 33
Node 1 Node 2 Node 3
← = ← ←
K K KK K K K
K K K (4.3.24)
Note that each of the sub-matrices is a 2×2 matrix, e.g.:
11 12
21 22
Node 1 Node 1
xx xy
yx yy
k k xk k y
← = ←
11K (4.3.25)
The member stiffness nodal sub-matrices contribute to the global stiffness nodal sub-
matrices as follows:
11 12 13 12 12
21 22 23 12 12 23 23
31 32 33 23 23
= =
K K K k11 k12 0K K K K k21 k22 + k11 k12
K K K 0 k21 k22 (4.3.26)
Expanding this out and filling in the relevant entries from equations (4.3.22) and
(4.3.23) whilst using equation (4.3.12) gives:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 91
As can be seen, the load is split between the two members in a way that depends on
their relative stiffness.
The total solution is thus:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 92
4.4.5 Problems
Problem 1
Determine the bending moment diagram and rotation of joint 2. Take 3 210 10 kNmEI = × .
Problem 2
Determine the bending moment diagram and the rotations of joints 1 and 2. Take 3 220 10 kNmEI = × .
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 93
Problem 3
Determine the bending moment diagram and the displacements of joints 2 and 3.
Take 3 220 10 kNmEI = × .
Problem 4
Determine the bending moment diagram and the vertical displacement under the 100
kN point load. Take 3 210 10 kNmEI = × .
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 94
Problem 5
Determine the bending moment diagram and the rotations of joints 2 and 3. Take 3 220 10 kNmEI = × .
Problem 6
Determine the bending moment diagram and the rotations of all joints. Take 3 240 10 kNmEI = × . You may use Excel or Matlab to perform some of the numerical
calculations. Check your member stiffness and global stiffness matrices with LinPro,
and your final results. Identify and explain discrepancies. Verify with LUSAS.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 95
4.5 Plane Frames
4.5.1 Plane Frame Element Stiffness Matrix
A plane frame element is similar to a beam element except for some differences:
• The presence of axial forces;
• The member may be oriented at any angle in the global axis system;
• The inter-nodal loads may be applied in the local or global coordinates.
These points are illustrated in the following:
Lastly, an easy way to deal with inter-nodal point loads ( GP , LP ) is to introduce a
node under the point load (splitting the member in two), then it is no longer inter-
nodal and so no transformations or equivalent load analysis is required. The downside
to this is that the number of equations increases (which is only really a problem for
analysis by hand).
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 96
Axial Forces
To include axial forces, we can simply expand the beam element stiffness matrix to
allow for the extra degree of freedom of x-displacement at each node in the member
local coordinates. Thus expanding equation (4.4.8) to allow for the extra DOFs gives:
[ ]
11 14
3 2 3 2
2 2
41 44
3 2 3 2
2 2
0 0 0 012 6 12 60 0
6 4 6 20 0
0 0 0 012 6 12 60 0
6 2 6 40 0
X XEI EI EI EI
L L L LEI EI EI EIL L L L
X XEI EI EI EI
L L L LEI EI EI EIL L L L
−
− = − − −
−
k (4.5.1)
However, these terms that account for axial force are simply those of a plane truss
element in its local coordinate system:
[ ] 1 11 1
EAL
− = −
k (4.5.2)
Thus equation (4.5.1) becomes:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 97
[ ]
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0
12 6 12 60 0
6 4 6 20 0
0 0 0 0
12 6 12 60 0
6 2 6 40 0
EA EAL L
EI EI EI EIL L L LEI EI EI EIL L L L
EA EAL L
EI EI EI EIL L L LEI EI EI EIL L L L
− − − = − − − − −
k (4.5.3)
This is the stiffness matrix for a plane frame element in its local coordinate system
and can also be written in terms of nodal sub-matrices as:
[ ] =
k11 k12k
k21 k22 (4.5.4)
Where the nodal sub-matrices are as delineated in equation (4.5.3).
Note that if axial forces are neglected, we can just use the regular beam element
stiffness matrix instead, though coordinate transformation may be required.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 98
Transformation to Global Coordinates
From the Appendix, the plane frame element stiffness matrix in global coordinates is:
[ ] [ ][ ]Te K = T k T (4.5.5)
As a consequence, note that we do not need to perform the transformation when:
1. The member local axis and global axis system coincide;
2. The only unrestrained DOFs are rotations/moments.
Again from the Appendix, the transformation matrix for a plane frame element is:
cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1
α αα α
α αα α
−
= −
T (4.5.6)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 99
Inter-nodal Loads
In plane frames, loads can be applied in the global axis system, or the local axis
system. For example, if we consider a member representing a roof beam, we can have
the following laods:
• Case 1: Gravity loads representing the weight of the roof itself;
• Case 2: Horizontal loads representing a horizontal wind;
• Case 3: Net pressure loads caused by outside wind and inside pressures.
Case 1 Case 2 Case 3
Most structural analysis software will allow you to choose the axis system of your
loads. However, in order to deal with these loads for simple hand analysis we must
know how it works and so we consider each case separately.
In the following the member local axis system has a prime (e.g. x’) and the global
axis system does not (e.g. x).
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 100
Case 1: Vertically Applied Loads
In this case we can consider an equivalent beam which is the projection of the load
onto a horizontal beam of length XL :
Since the resulting nodal forces and moments are in the global axis system no further
work is required.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 101
Case 2: Horizontally Applied Loads:
Similarly to vertically applied loads, we can consider the horizontal projection of load
onto an equivalent member of length YL .
Again the resulting nodal loads are in the global axis system and do not require any
modification.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 102
Case 3: Loads Applied in Local Member Axis System
In this case there is no need for an equivalent beam and the fixed-fixed reactions are
worked out as normal:
However, there is a complication here since the reactions are now not all in the global
axis system. Thus the forces (not moments) must be transformed from the local axis
to the global axis system. Thus there is a simple case:
If axial forces are neglected, only moments are relevant and so no transformations are
required.
For generality though we can use the transformations given in the Appendix:
{ } [ ] { }'TF = T F (4.5.7)
Writing this out in full for clarity, we have:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 103
'
'
'
'
cos sin 0 0 0 0sin cos 0 0 0 0
0 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1
ij ijix ixij ij
iy iyij iji iij ijjx jxij ijjy jyij ijj j
F FF FM MF FF FM M
α αα α
α αα α
− = −
(4.5.8)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 104
4.5.2 Example 9 – Simple Plane Frame
Problem
For the following frame, determine the rotation of the joints and the bending moment
The fact that we can neglect axial deformation makes this problem much simpler. As
a consequence, the only possible displacements are the rotations of joints 1 and 2.
Since node 3 is fully restricted out, we have the following partially-restricted set of
equations in terms of nodal sub-matrices:
=
1 1
2 2
F δK11 K12F δK21 K22
(4.5.9)
If we expand this further, we will be able to restrict out all but the rotational DOFs:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 105
1 33 36 1
2 63 66 2
M k k
M k k
θ
θ
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅
∑
∑
(4.5.10)
The member contributions to each of these terms are:
• 33 Term 33 of Member 12k = ;
• 36 Term 36 of Member 12k = ;
• 63 Term 63 of Member 12k = ;
• 66 Term 66 of Member 12 Term 33 of Member 23k = + .
• Member 12:
Looking at equation (4.5.3):
3
3
12
4 4 10Term 33 4 101
EIL
⋅ = = = ×
(4.5.11)
3
3
12
2 2 10Term 36 2 101
EIL
⋅ = = = ×
(4.5.12)
3
3
12
2 2 10Term 63 2 101
EIL
⋅ = = = ×
(4.5.13)
3
3
12
4 4 10Term 66 4 101
EIL
⋅ = = = ×
(4.5.14)
• Member 23:
Again, from equation (4.5.3):
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 106
3
3
23
4 4 10Term 33 4 101
EIL
⋅ = = = ×
(4.5.15)
Thus the system equation becomes:
1 13
2 2
4 210
2 8MM
θθ
=
∑∑
(4.5.16)
Next we must find the net moments applied to each node. There are no directly
applied nodal moment loads, so the ‘force’ vector is, from equation (4.4.13):
{ } { }= − FF F (4.5.17)
• Member 12 Moments:
2 2121
2 2122
12 1 1 kNm12 12
12 1 1 kNm12 12
wLM
wLM
⋅= = = +
⋅= − = − = −
(4.5.18)
• Member 23 Moments:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 107
232
233
16 1 2 kNm8 8
16 1 2 kNm8 8
PLM
PLM
⋅= = = +
⋅= − = − = −
(4.5.19)
Thus the net nodal loads become:
{ } { } 1
2
1 1 kNm
1 2 1MM
+ − = − = − = − = − + −
∑∑FF F (4.5.20)
And so equation (4.5.16) is thus:
13
2
1 4 210
1 2 8θθ
− = −
(4.5.21)
Which is solved to get:
( )
1 33
2
8 2 1 3 141 1 10 rads2 4 1 1 1410 4 8 2 2
θθ
−− − − = = × − − −⋅ − ⋅
(4.5.22)
The negative results indicate both rotations are clockwise.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 108
Lastly, we must find the member end forces. Since we only need to draw the bending
moment diagram so we need only consider the terms of the member stiffness matrix
relating to the moments/rotations (similar to equation (4.4.9)). Also, we must account
for the equivalent nodal loads as per equation (4.4.11):
• Member 12:
121 3 3122
1 4 2 3 14 010 10 kNm
1 2 4 1 14 12 7MM
−+ − = + = − − −
(4.5.23)
• Member 23:
232 3 3233
2 4 2 1 14 12 710 10 kNm
2 2 4 0 17 7MM
−+ − + = + = − −
(4.5.24)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 109
4.5.3 Example 10 –Plane Frame Using Symmetry
Problem
For the following frame, determine the rotation of the joints, the displacement under
the 8 kN point load and the bending moment diagram. Neglect axial deformations.
Take 3 21 10 kNmEI = × .
Solution
Again, the fact that we can neglect axial deformation makes this problem much
simpler. Since the structure is symmetrical and it is symmetrically loaded, it will not
sway. Further, because of this symmetry, we can adopt the following model for
analysis:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 110
Notice two things from this model:
• we have renumber the joints – there is no need to retain the old numbering system;
• The remaining DOFs are 2θ and 3 yδ - we can restrict all other DOFs. Thus in terms
of nodal sub-matrices we immediately have:
2 2
3 3
=
F δK11 K12F δK21 K22
(4.5.25)
And expanding this further, we restrict out all other restrained DOFs:
2 66 68 2
3 86 88 3y y
M k k
F k k
θ
δ
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
∑
∑
(4.5.26)
The member contributions to each of these terms are:
• 66 Term 66 of Member 12 Term 33 of Member 23k = + ;
• 68 Term 35 of Member 23k = ;
• 86 Term 53 of Member 23k = ;
• 66 Term 55 of Member 23k = .
Transformation of the member stiffness matrices is not required. Member 12 only has
a rotational DOF and Member 23’s local member coordinate system is parallel to the
global axis coordinate system.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 111
• Member 12:
From equation (4.5.3):
3
3
12
4 4 10Term 66 4 101
EIL
⋅ = = = ×
(4.5.27)
• Member 23:
Again, from equation (4.5.3):
3
3
23
4 4 10Term 33 4 101
EIL
⋅ = = = ×
(4.5.28)
3
32 2
23
6 6 10Term 35 6 101
EIL
⋅ = − = − = − ×
(4.5.29)
3
32 2
23
6 6 10Term 53 6 101
EIL
⋅ = − = − = − ×
(4.5.30)
3
33 3
23
12 12 10Term 55 12 101
EIL
⋅ = = = ×
(4.5.31)
Thus the system equation becomes:
22 3
33
8 610
6 12 yy
MF
θδ
− = −
∑∑
(4.5.32)
The 4 kN point load is directly applied to node 3 so this causes no difficulty. The
equivalent nodal loads for the UDL are:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 112
2 2121
2 2122
12 1 1 kNm12 12
12 1 1 kNm12 12
wLM
wLM
⋅= = = +
⋅= − = − = −
(4.5.33)
Notice that we do not need to find the vertical reaction forces as there is no sway of
the frame and we are neglecting axial deformation.
The nodal load vector, from equation (4.4.13) is thus:
{ } { } { }0 1 14 0 4
− + = − = − = − −
∑ FF F F (4.5.34)
And so equation (4.5.32) is thus:
23
3
1 8 610
4 6 12 y
θδ
+ − = − −
(4.5.35)
Which is solved to get:
( )( )
2 33
3
12 6 1 0.2 rads1 1 106 8 4 0.43 m10 8 12 6 6y
θδ
−+ − = = × − −⋅ − − −
(4.5.36)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 113
The negative results indicate the rotation is clockwise and the displacement
downwards, as may be expected:
Lastly then we find the bending moments. For member 12 only the terms relating to
bending moments are relevant.
• Member 12:
121 3 3122
1 4 2 0 0.610 10 kNm
1 2 4 0.2 1.8MM
−+ + = + = − − −
(4.5.37)
However, for member 23, the downwards deflection also causes moments and so the
relevant DOFs are rotation of node i and vertical movement of node j (as calculated
earlier). It is easier to see this if we write the member equation in full:
• Member 23:
232 3 3
233
4 6 0.210 10
0.432 6
M
M
−
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ −
= × ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ −
⋅ ⋅ ⋅ − ⋅ ⋅
(4.5.38)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 114
Thus:
232233
1.8 kNm
2.2MM
+ = +
(4.5.39)
And so the BMD is:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 115
4.5.4 Problems
Problem 1
Determine the bending moment diagram and the rotation of joint 2. Take 3 210 10 kNmEI = × and neglect axial deformations.
(Ans. 2 5 6 mradsθ = )
Problem 2
Determine the bending moment diagram and the rotations of joints 1 and 2. Take 3 220 10 kNmEI = × and neglect axial deformations.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 116
Problem 3
Determine the bending moment diagram. Take 3 220 10 kNmEI = × and neglect axial
deformations.
Problem 4
Determine the bending moment diagram, the rotation of joint 2, and the horizontal
displacements of joints 2 and 3. Take 3 210 10 kNmEI = × and neglect axial
deformations.
(Ans. 2 211.33 mrads; 44.0mmxθ δ= − = )
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 117
Problem 5
Determine the bending moment diagram. Take 3 220 10 kNmEI = × and neglect axial
deformations.
Problem 6
Determine the bending moment diagram and the vertical displacement of joint 3.
Take 3 240 10 kNmEI = × and neglect axial deformations.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 118
Problem 7
Determine the bending moment diagram, the rotation of joint 2, and the vertical
displacement under the 80 kN point load. Take 3 210 10 kNmEI = × and neglect axial
deformations.
(Ans. 2 1.071 mrads; 1.93mmyθ δ= − = − )
Problem 8
For the frame of Problem 1, determine the bending moment diagram and the rotation
and vertical displacement of joint 2 if member 24 has 310 10 kNEA = × . Neglect axial
deformation in the other members.
(Ans. 2 0.833 mrads; 0.01mmyθ δ= = − )
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 119
Problem 9
Determine the bending moment diagram for the prismatic portal frame. Take 3 220 10 kNmEI = × and neglect axial deformations. You may use Excel or Matlab to
perform some of the numerical calculations. Check your member stiffness and global
stiffness matrices with LinPro, and your final results. Identify and explain
discrepancies. Verify with LUSAS.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 120
4.6 Appendix
4.6.1 Plane Truss Element Stiffness Matrix in Global Coordinates
Compatibility Conditions
Firstly we indentify the conditions of compatibility of a truss element nodal
deflections and the member elongation. We use the following notation for the
deflections at each node of the truss:
If we now consider the deflected position of the truss member, we have:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 121
Obviously the change in length of the truss will be related to the difference between
the nodal deflections. Hence, we define the changes in movements such that an
elongation gives positive changes:
x jx ix y jy iyδ δ δ δ δ δ∆ = − ∆ = −
Moving the deflected position of node i back to its original location gives:
Looking more closely at the triangle of displacements at node j, and remembering
that we are assuming small deflections—which in this case means the deflected
position of the member is still at a rotation of θ . Hence we have:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 122
And so the elongation is given by:
( ) ( )
cos sin
cos sinx y
jx ix jy iy
e δ θ δ θ
δ δ θ δ δ θ
= ∆ + ∆
= − + − (4.6.1)
Now multiply out and re-order to get:
cos sin cos sinix iy jx jye δ θ δ θ δ θ δ θ= − + − + + (4.6.2)
If we define a direction vector, α , and a displacement vector, δ , as:
cossin
cossin
ix
iy
jx
jy
δθδθδθδθ
− − = =
α δ (4.6.3)
Then, from (4.6.2) and (4.6.3), we can say:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 123
te = α δ (4.6.4)
Thus we have related the end displacements to the elongation of the member which
therefore maintain compatibility of displacement.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 124
Virtual Work for Element Forces
Looking at the forces acting on the nodes of the bar element, we have:
This is a force system in equilibrium—the external nodal loading is in equilibrium
with the internal bar force, N. If we consider a pattern of compatible displacements
such as the following:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 125
We can apply virtual work to this:
0
E I
WW Wδδ δ
==
And we have:
Substituting in our notations for the bar element:
ix ix iy iy jx jx jy jyeN F F F Fδ δ δ δ= + + + (4.6.5)
If we define the force vector, F , as:
ix
iy
jx
jy
FFFF
=
F (4.6.6)
Then we can write (4.6.5) as:
t eN=F δ (4.6.7)
Set of forces in
equilibrium
Set of compatible
displacements
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 126
If we use (4.6.4) we how have:
t t N=F δ α δ (4.6.8)
Post-multiply both sides by 1−δ , and noting that N is a scalar, gives:
t t N=F α
N=F α (4.6.9)
Expanding this out gives:
cossin
cossin
ix
iy
jx
jy
F NF NF NF N
θθθθ
− − =
(4.6.10)
Which are the equations of equilibrium of the bar element:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 127
Relating Forces to Displacements
Lastly, in order to relate the end forces to the element nodal displacements, we note
from the constitutive law:
EAN eL
= ⋅ (4.6.11)
And so from (4.6.9) we have:
EA eL
=F α (4.6.12)
And using equation (4.6.4) gives:
tEAL
=F α α δ (4.6.13)
Hence the term tEAL
α α relates force to displacement and is called the stiffness
matrix, k , which is evaluated by multiplying out terms:
[ ]
cossin
cos sin cos sincossin
tEAL
EAL
θθ
θ θ θ θθθ
=
− − = − −
k α α
(4.6.14)
And multiplying this out gives:
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 128
2 2
2 2
2 2
2 2
cos cos sin cos cos sincos sin sin cos sin sin
cos cos sin cos cos sincos sin sin cos sin sin
EAL
θ θ θ θ θ θθ θ θ θ θ θ
θ θ θ θ θ θθ θ θ θ θ θ
− − − − = − − − −
k (4.6.15)
And for clarity, we write out the final equation in matrix form and in full:
=F kδ (4.6.16)
2 2
2 2
2 2
2 2
cos cos sin cos cos sincos sin sin cos sin sin
cos cos sin cos cos sincos sin sin cos sin sin
ix ix
iy iy
jx jx
jy jy
FF EAF LF
δθ θ θ θ θ θδθ θ θ θ θ θδθ θ θ θ θ θδθ θ θ θ θ θ
− − − − = − − − −
(4.6.17)
So for example, the stiffness that relates a horizontal force at node j to the horizontal
displacement at node j is:
2cosjx jx
EAFL
θ δ =
And other relationships can be found similarly.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 129
4.6.2 Coordinate Transformations
Point Transformation
We consider the transformation of a single point P from one coordinate axis system
xy to another x’y’:
From the diagram, observe:
' coordinate of
' coordinate of
OC x P
PC y P
=
= (4.6.18)
Also:
coordinate of
coordinate of
OB x P
PB y P
=
= (4.6.19)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 130
Next we can say:
OC OA AC= + (4.6.20)
PC PD CD= − (4.6.21)
Introducing the relevant coordinates:
cos cosOA OA xα α= = (4.6.22)
sin sinAC BD PB yα α= = = (4.6.23)
Thus equation (4.6.20) becomes:
' cos sinOC x x yα α= = + (4.6.24)
Next we have:
cos cosPD PB yα α= = (4.6.25)
sin sinCD AB OB xα α= = = (4.6.26)
Thus equation (4.6.21) becomes:
' cos sinPC y y xα α= = − (4.6.27)
Writing equations (4.6.24) and (4.6.27) together:
' cos sinx x yα α= + (4.6.28)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 131
' sin cosy x yα α= − + (4.6.29)
And now in matrix form gives:
' cos sin' sin cos
x xy y
α αα α
= −
(4.6.30)
Often we write:
cossin
cs
αα
≡≡
(4.6.31)
To give:
''
x c s xy s c y
= −
(4.6.32)
Lastly, if we generically name the two coordinate systems as q and q’, we then have
in matrix form:
{ } [ ]{ }Nq' = T q (4.6.33)
Where [ ]NT is the nodal transformation matrix given by:
cos sinsin cosN
α αα α
= −
T (4.6.34)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 132
Force/Displacement Transformation
Forces and moments can be oriented in the local member axis system or in the global
structure axis system. In general we will need to transform the forces and
displacements of both nodes, thus we write:
''
i iN
j jN
F FT 0=
F F0 T (4.6.35)
And finally we can write:
{ } [ ]{ }'F = T F (4.6.36)
Where:
[ ] N
N
T 0T =
0 T (4.6.37)
Similarly for deflections:
{ } [ ]{ }'δ = T δ (4.6.38)
A very useful property of the transformation matrix (not derived here) is that it is
orthogonal. This means that its transpose is equal to its inverse:
[ ] [ ] 1T −=T T (4.6.39)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 133
Thus when either a force or displacement is known for the local axis system, it can be
found in the global axis system as follows:
{ } [ ] { }'TF = T F (4.6.40)
{ } [ ] { }'Tδ = T δ (4.6.41)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 134
Transformations for Plane Truss Element
For a plane truss member, there will be x and y components of force at each of its
nodes. Using the transformation for a point, we therefore have:
'
'
cos sinsin cos
x x
y y
F FF F
α αα α
= −
(4.6.42)
And so for a truss element, we have directly from equation (4.6.34):
cos sinsin cosN
α αα α
= −
T (4.6.43)
And so, from equation (4.6.37),
[ ]
cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos
α αα α
α αα α
− = −
T (4.6.44)
For clarity, we write the transformation out in full:
'
'
'
'
cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos
ix ix
iy iy
jx jx
jy jy
F FF FF FF F
α αα α
α αα α
− = −
(4.6.45)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 135
Transformations for Plane Frame Element
Based on the DOF transformation matrix for a plane truss member (in terms of
forces), we can determine the transformation matrix for a plane frame node quite
easily:
cos sin 0sin cos 00 0 1
ex xe
y ye
F FF FM M
α αα α
= −
(4.6.46)
This is because a moment remains a moment in the plane. So for a single node, and
both nodes, we have, respectively:
{ } [ ]{ }' NF = T F (4.6.47)
''
i iN
j jN
F FT 0=
F F0 T (4.6.48)
Thus, we can now write the final transformation matrix for a plane frame element as:
cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1
α αα α
α αα α
−
= −
T (4.6.49)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 136
Element Stiffness Matrix Transformation
Using the general expression for a single element:
e e eF = K δ (4.6.50)
Regardless of member type or the number of dimensions, we will always have some
coordinate transform from local to global coordinates such that:
eF = TF (4.6.51)
eδ = Tδ (4.6.52)
Hence from equation (4.6.50) we can write:
eTF = K Tδ (4.6.53)
And so the force-displacement relationship in the global axis system is:
1 e− F = T K T δ (4.6.54)
The term in brackets can now be referred to as the element stiffness matrix in global
coordinates. Thus, using equation (4.6.39), we write:
e T eG LK = T K T (4.6.55)
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 137
4.6.3 Past Exam Questions
Summer 2001
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 138
Summer 2002
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 139
Summer 2004
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 140
Summer 2006
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 141
Sample Paper 2006/7 1. (a) Using the stiffness method, determine the displacement of the joints of the pin-jointed truss shown in Fig.
Q1(a), under the load as shown. (10 marks)
(b) Members 15 and 16 are added to the truss of Fig. 1(a) to form the truss shown in Fig. Q1(b). However,
member 16 is found to be 15 mm too long and is forced into place. The same load of 100 kN is again to be applied. Using the stiffness method, determine the displacement of the joints and the force in member 16.
(15 marks) Take EA = 2×104 kN and the cross sectional areas of the members as:
Members 12, 13, and 16: 3A; Diagonal Members 14 and 15: 3√2A.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 142
Semester 1 2006/7 1. Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-
jointed truss shown in Fig. Q1, allowing for:
(i) The 100 kN vertical load as shown, and; (ii) A lack of fit of member 12, which was found to be 5 mm too short upon arrival at site, and
which was then forced into place.
Take EA = 2×104 kN and the cross sectional areas of the members as: • Members 12: 3A; • Members 13 and 14: 3√2A.
(25 marks)
Ans. 50 kN; -75√2 kN; -25√2 kN.
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 143
Semester 1 Repeat 2006/7 1. Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-
jointed truss shown in Fig. Q1, allowing for:
(ii) The 100 kN vertical load as shown, and; (ii) A lack of fit of member 12, which was found to be 10√2 mm too short upon arrival at site, and
which was then forced into place.
Take EA = 2×104 kN and the cross sectional areas of all members as 3√2A. (25 marks)
Ans. 125√2 kN; -50√2 kN; -75√2 kN.
FIG. Q1
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 144
Semester 1 2007/8 QUESTION 1 Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-jointed truss shown in Fig. Q1, allowing for: (i) The 100 kN load as shown, and; (ii) A lack of fit of member 13, which was found to be 4 mm too short upon arrival at site, and which was then
forced into place; (iii) A temperature rise of 20 ˚C in member 24. Note: Take 3125 10 kNEA = × and the coefficient of thermal expansion -5 -12 10 Cα = × ° .
(25 marks)
Ans. -55.7 kN; +69.7 kN; -55.3 kN.
FIG. Q1
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 145
Semester 1 2008/9 QUESTION 1 Using the stiffness method, for the continuous beam shown in Fig. Q1, do the following:
(i) determine the displacement of the joints; (ii) draw the bending moment diagram;
(iii) determine the reactions.
Note: Take 3 210 10 kNmEI = × .
(25 marks)
Ans. 98.7 kNm; 102.6 kNm; 60.9 kNm.
FIG. Q1
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 146
Semester 1 2009/10 QUESTION 1 Using the stiffness method, for the frame shown in Fig. Q1, do the following:
(i) determine the vertical displacement at the centre of the middle span; (ii) draw the bending moment diagram;
(iii) determine the reactions.
Note: Take 3 210 10 kNmEI = × .
(25 marks)
Ans. -11.88 mm
FIG. Q1
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 147
Semester 1 2010/11 QUESTION 1 Using the stiffness method, for the truss shown in Fig. Q1: (a) Determine:
(i) The displacement of the joints; (ii) The forces in the members; (iii) The deflected shape of the structure.
(15 marks) (b) Determine the lack of fit of member 23, which would result in no horizontal displacement of joint 2 under the
100 kN load shown. (10 marks)
Note: • Take 2200kN/mmE = for all members.
• Area for member 12 is 2400 10 mmA = . • Area for member 23 is 2960mmA = .
• Area for member 13 is 2640 2 mmA = .
Ans. 68.7 kN, 34.6 kN, 49.3 kN; 5.21 mm.
FIG. Q1
Structural Analysis IV Chapter 4 – Matrix Stiffness Method
Dr. C. Caprani 148
4.7 References • Alberty, J., Carstensen, C. and Funken, S.A. (1999), ‘Remarks around 50 lines of
Matlab: short finite element implementation’, Numerical Algorithms, 20, pp. 117-
137, available at: web address.
• Brown, D.K. (1990), An Introduction to the Finite Element Method using Basic
Programs, 2nd Edn., Taylor and Francis, London.
• Carroll, W.F. (1999), A Primer for Finite Elements in Elastic Structures, John
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 21
The input parameters (shown in red) are:
m – the mass;
k – the stiffness;
delta_t – the time step used in the response plot;
u_0 – the initial displacement, 0
u ;
v_0 – the initial velocity, 0
u .
The properties of the system are then found:
w, using equation (5.2.10);
f, using equation (5.2.26);
T, using equation (5.2.26);
, using equation (5.2.29);
, using equation (5.2.30).
A column vector of times is dragged down, adding delta_t to each previous time
value, and equation (5.2.24) (“Direct Eqn”), and equation (5.2.28) (“Cosine Eqn”) is
used to calculate the response, u t , at each time value. Then the column of u-values
is plotted against the column of t-values to get the plot.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 22
Using Matlab
Although MS Excel is very helpful since it provides direct access to the numbers in
each equation, as more concepts are introduced, we will need to use loops and create
regularly-used functions. Matlab is ideally suited to these tasks, and so we will begin
to use it also on the simple problems as a means to its introduction.
A script to directly generate Figure 1.3, and calculate the system properties is given
below:
% Script to plot the undamped response of a single degree of freedom system % and to calculate its properties
k = 100; % N/m - stiffness m = 10; % kg - mass delta_t = 0.1; % s - time step u0 = 0.025; % m - initial displacement v0 = 0; % m/s - initial velocity
w = sqrt(k/m); % rad/s - circular natural frequency f = w/(2*pi); % Hz - natural frequency T = 1/f; % s - natural period ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle
t = 0:delta_t:4; u = ro*cos(w*t-theta); plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)');
The results of this script are the system properties are displayed in the workspace
window, and the plot is generated, as shown below:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 23
Whilst this is quite useful, this script is limited to calculating the particular system of
Figure 1.3. Instead, if we create a function that we can pass particular system
properties to, then we can create this plot for any system we need to. The following
function does this.
Note that we do not calculate f or T since they are not needed to plot the response.
Also note that we have commented the code very well, so it is easier to follow and
understand when we come back to it at a later date.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 24
function [t u] = sdof_undamped(m,k,u0,v0,duration,plotflag) % This function returns the displacement of an undamped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response
Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1);
w = sqrt(k/m); % rad/s - circular natural frequency ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle
t = 0:delta_t:duration; u = ro*cos(w*t-theta);
if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end
To execute this function and replicate Figure 1.3, we call the following:
[t u] = sdof_undamped(10,100,0.025,0,4,1);
And get the same plot as before. Now though, we can really benefit from the
function. Let‟s see the effect of an initial velocity on the response, try +0.1 m/s:
[t u] = sdof_undamped(10,100,0.025,0.1,4,1);
Note the argument to the function in bold – this is the +0.1 m/s initial velocity. And
from this call we get the following plot:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 25
From which we can see that the maximum response is now about 40 mm, rather than
the original 25.
Download the function from the course website and try some other values.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 26
5.2.4 Free Vibration of Damped Structures
Figure 2.4: Response with critical or super-critical damping
When taking account of damping, we noted previously that there are 3, cases but only
when 1 does an oscillatory response ensue. We will not examine the critical or
super-critical cases. Examples are shown in Figure 2.4.
To begin, when 1 (5.2.17) becomes:
1,2 d
i (5.2.31)
where d
is the damped circular natural frequency given by:
21
d (5.2.32)
which has a corresponding damped period and frequency of:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 27
2
d
d
T
(5.2.33)
2
d
df
(5.2.34)
The general solution to equation (5.2.14), using Euler‟s formula again, becomes:
( ) cos sint
d du t e A t B t (5.2.35)
and again using the initial conditions we get:
0 0
0( ) cos sint d
d d
d
u uu t e u t t
(5.2.36)
Using the cosine addition rule again we also have:
( ) cost
du t e t (5.2.37)
In which
2
2 0 0
0
d
u uu
(5.2.38)
0 0
0
tand
u u
u
(5.2.39)
Equations (5.2.35) to (5.2.39) correspond to those of the undamped case looked at
previously when 0 .
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 28
Figure 2.5: SDOF free vibration response for:
(a) 0 ; (b) 0.05 ; (c) 0.1 ; and (d) 0.5 .
Figure 2.5 shows the dynamic response of the SDOF model shown. It may be clearly
seen that damping has a large effect on the dynamic response of the system – even for
small values of . We will discuss damping in structures later but damping ratios for
structures are usually in the range 0.5 to 5%. Thus, the damped and undamped
properties of the systems are very similar for these structures.
Figure 2.6 shows the general case of an under-critically damped system.
-25
-20
-15
-10
-5
0
5
10
15
20
25
0 0.5 1 1.5 2 2.5 3 3.5 4
Dis
pla
cem
en
t (m
m)
Time (s)
(a)
(b)
(c)
(d)
m = 10
k = 100
varies
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 29
Figure 2.6: General case of an under-critically damped system.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 30
5.2.5 Computer Implementation & Examples
Using MS Excel
We can just modify our previous spreadsheet to take account of the revised equations
for the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response
(equation (5.2.37)), as well as the damped properties, to get:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 31
Using Matlab
Now can just alter our previous function and take account of the revised equations for
the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response
(equation (5.2.37)) to get the following function. This function will (of course) also
work for undamped systems where 0 .
function [t u] = sdof_damped(m,k,xi,u0,v0,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response
Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1);
w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency ro = sqrt(u0^2+((v0+xi*w*u0)/wd)^2); % m - amplitude of vibration theta = atan((v0+u0*xi*w)/(u0*w)); % rad - phase angle
t = 0:delta_t:duration; u = ro*exp(-xi*w.*t).*cos(w*t-theta);
if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end
Let‟s apply this to our simple example again, for 0.1 :
[t u] = sdof_damped(10,100,0.1,0.025,0,4,1);
To get:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 32
To plot Figure 2.5, we just call out function several times (without plotting it each
time), save the response results and then plot all together:
xi = [0,0.05,0.1,0.5]; for i = 1:length(xi) [t u(i,:)] = sdof_damped(10,100,xi(i),0.025,0,4,0); end plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Damping: 0%','Damping: 5%','Damping: 10%','Damping: 50%');
0 0.5 1 1.5 2 2.5 3 3.5 4-0.02
-0.01
0
0.01
0.02
0.03
Time (s)
Dis
pla
cem
ent
(m)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.03
-0.02
-0.01
0
0.01
0.02
0.03
Time (s)
Dis
pla
cem
ent
(m)
Damping: 0%
Damping: 5%
Damping: 10%
Damping: 50%
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 33
5.2.6 Estimating Damping in Structures
Examining Figure 2.6, we see that two successive peaks, n
u and n m
u
, m cycles apart,
occur at times nT and n m T respectively. Using equation (5.2.37) we can get the
ratio of these two peaks as:
2
expn
n m d
u m
u
(5.2.40)
where exp xx e . Taking the natural log of both sides we get the logarithmic
decrement of damping, , defined as:
ln 2n
n m d
um
u
(5.2.41)
for low values of damping, normal in structural engineering, we can approximate
this:
2m (5.2.42)
thus,
exp 2 1 2n
n m
ue m m
u
(5.2.43)
and so,
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 34
2
n n m
n m
u u
m u
(5.2.44)
This equation can be used to estimate damping in structures with light damping (
0.2 ) when the amplitudes of peaks m cycles apart is known. A quick way of
doing this, known as the Half-Amplitude Method, is to count the number of peaks it
takes to halve the amplitude, that is 0.5n m n
u u . Then, using (5.2.44) we get:
0.11
m when 0.5
n m nu u
(5.2.45)
Further, if we know the amplitudes of two successive cycles (and so 1m ), we can
find the amplitude after p cycles from two instances of equation (5.2.43):
1
p
n
n p n
n
uu u
u
(5.2.46)
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 35
5.2.7 Response of an SDOF System Subject to Harmonic Force
Figure 2.7: SDOF undamped system subjected to harmonic excitation
So far we have only considered free vibration; the structure has been set vibrating by
an initial displacement for example. We will now consider the case when a time
varying load is applied to the system. We will confine ourselves to the case of
harmonic or sinusoidal loading though there are obviously infinitely many forms that
a time-varying load may take – refer to the references (Appendix) for more.
To begin, we note that the forcing function F t has excitation amplitude of 0
F and
an excitation circular frequency of and so from the fundamental equation of
motion (5.2.3) we have:
0
( ) ( ) ( ) sinmu t cu t ku t F t (5.2.47)
The solution to equation (5.2.47) has two parts:
The complementary solution, similar to (5.2.35), which represents the transient
response of the system which damps out by exp t . The transient response
may be thought of as the vibrations caused by the initial application of the load.
The particular solution, pu t , representing the steady-state harmonic response of
the system to the applied load. This is the response we will be interested in as it
will account for any resonance between the forcing function and the system.
m
k u(t)
c
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 36
The complementary solution to equation (5.2.47) is simply that of the damped free
vibration case studied previously. The particular solution to equation (5.2.47) is
developed in the Appendix and shown to be:
sinp
u t t (5.2.48)
In which
1 2
2 220 1 2F
k
(5.2.49)
2
2tan
1
(5.2.50)
where the phase angle is limited to 0 and the ratio of the applied load
frequency to the natural undamped frequency is:
(5.2.51)
the maximum response of the system will come at sin 1t and dividing
(5.2.48) by the static deflection 0
F k we can get the dynamic amplification factor
(DAF) of the system as:
1 2
2 22DAF 1 2D
(5.2.52)
At resonance, when , we then have:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 37
1
1
2D
(5.2.53)
Figure 2.8 shows the effect of the frequency ratio on the DAF. Resonance is the
phenomenon that occurs when the forcing frequency coincides with that of the
natural frequency, 1 . It can also be seen that for low values of damping, normal
in structures, very high DAFs occur; for example if 0.02 then the dynamic
amplification factor will be 25. For the case of no damping, the DAF goes to infinity
- theoretically at least; equation (5.2.53).
Figure 2.8: Variation of DAF with damping and frequency ratios.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 38
The phase angle also helps us understand what is occurring. Plotting equation
(5.2.50) against for a range of damping ratios shows:
Figure 2.9: Variation of phase angle with damping and frequency ratios.
Looking at this then we can see three regions:
1 : the force is slowly varying and is close to zero. This means that the
response (i.e. displacement) is in phase with the force: for example, when the
force acts to the right, the system displaces to the right.
1 : the force is rapidly varying and is close to 180°. This means that the
force is out of phase with the system: for example, when the force acts to the
right, the system is displacing to the left.
1 : the forcing frequency is equal to the natural frequency, we have
resonance and 90 . Thus the displacement attains its peak as the force is
zero.
0 0.5 1 1.5 2 2.5 30
45
90
135
180
Frequency Ratio
Phase A
ngle
(degre
es)
Damping: 0%
Damping: 10%
Damping: 20%
Damping: 50%
Damping: 100%
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 39
We can see these phenomena by plotting the response and forcing fun(5.2.54)ction
together (though with normalized displacements for ease of interpretation), for
different values of . In this example we have used 0.2 . Also, the three phase
angles are 2 0.04, 0.25, 0.46 respectively.
Figure 2.10: Steady-state responses to illustrate phase angle.
Note how the force and response are firstly “in sync” ( ~ 0 ), then “halfway out of
sync” ( 90 ) at resonance; and finally, “fully out of sync” ( ~180 ) at high
frequency ratio.
0 0.5 1 1.5 2 2.5 3
-2
0
2
Dis
p.
Ratio
0 0.5 1 1.5 2 2.5 3
-2
0
2
Dis
p.
Ratio
0 0.5 1 1.5 2 2.5 3
-2
0
2
Dis
p.
Ratio
Time Ratio (t/T)
Dynamic Response
Static Response = 0.5; DAF = 1.29
= 0.5; DAF = 2.5
= 2.0; DAF = 0.32
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 40
Maximum Steady-State Displacement
The maximum steady-state displacement occurs when the DAF is a maximum. This
occurs when the denominator of equation (5.2.52) is a minimum:
1 22 22
2 2
2 22
2 2
0
1 2 0
4 1 4 210
2 1 2
1 2 0
d D
d
d
d
The trivial solution to this equation of 0 corresponds to an applied forcing
function that has zero frequency –the static loading effect of the forcing function. The
other solution is:
21 2 (5.2.54)
Which for low values of damping, 0.1 approximately, is very close to unity. The
corresponding maximum DAF is then given by substituting (5.2.54) into equation
(5.2.52) to get:
max 2
1
2 1D
(5.2.55)
Which reduces to equation (5.2.53) for 1 , as it should.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 41
Measurement of Natural Frequencies
It may be seen from equation (5.2.50) that when 1 , 2 ; this phase
relationship allows the accurate measurements of the natural frequencies of
structures. That is, we change the input frequency in small increments until we can
identify a peak response: the value of at the peak response is then the natural
frequency of the system. Example 2.1 gave the natural frequency based on this type
of test.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 42
5.2.8 Computer Implementation & Examples
Using MS Excel
Again we modify our previous spreadsheet and include the extra parameters related
to forced response. We‟ve also used some of the equations from the Appendix to
show the transient, steady-sate and total response. Normally however, we are only
interested in the steady-state response, which the total response approaches over time.
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 43
Using Matlab
First let‟s write a little function to return the DAF, since we will use it often:
function D = DAF(beta,xi) % This function returns the DAF, D, associated with the parameters: % beta - the frequency ratio % xi - the damping ratio
D = 1./sqrt((1-beta.^2).^2+(2*xi.*beta).^2);
And another to return the phase angle (always in the region 0 ):
function theta = phase(beta,xi) % This function returns the pahse angle, theta, associated with the % parameters: % beta - the frequency ratio % xi - the damping ratio
theta = atan2((2*xi.*beta),(1-beta.^2)); % refers to complex plane
With these functions, and modifying our previous damped response script, we have:
function [t u] = sdof_forced(m,k,xi,u0,v0,F,Omega,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % F - amplitude of forcing function, N % Omega - frequency of forcing function, rad/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response
Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1);
w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency
beta = Omega/w; % frequency ratio D = DAF(beta,xi); % dynamic amplification factor ro = F/k*D; % m - amplitude of vibration theta = phase(beta,xi); % rad - phase angle
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 44
% Constants for the transient response Aconst = u0+ro*sin(theta); Bconst = (v0+u0*xi*w-ro*(Omega*cos(theta)-xi*w*sin(theta)))/wd;
t = 0:delta_t:duration; u_transient = exp(-xi*w.*t).*(Aconst*cos(wd*t)+Bconst*sin(wd*t)); u_steady = ro*sin(Omega*t-theta); u = u_transient + u_steady;
if(plotflag == 1) plot(t,u,'k'); hold on; plot(t,u_transient,'k:'); plot(t,u_steady,'k--'); hold off; xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Total Response','Transient','Steady-State'); end
Running this for the same problem as before with 0
As can be seen, the total response quickly approaches the steady-state response.
0 1 2 3 4 5 6-0.06
-0.04
-0.02
0
0.02
0.04
0.06
Time (s)
Dis
pla
cem
ent
(m)
Total Response
Transient
Steady-State
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 45
Next let‟s use our little DAF function to plot something similar to Figure 2.8, but this
time showing the frequency ratio and maximum response from equation (5.2.54):
% Script to plot DAF against Beta for different damping ratios xi = [0.0001,0.1,0.15,0.2,0.3,0.4,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) D(i,:) = DAF(beta,xi(i)); end % A new xi vector for the maxima line xi = 0:0.01:1.0; xi(end) = 0.99999; % very close to unity xi(1) = 0.00001; % very close to zero for i = 1:length(xi) betamax(i) = sqrt(1-2*xi(i)^2); Dmax(i) = DAF(betamax(i),xi(i)); end plot(beta,D); hold on; plot(betamax,Dmax,'k--'); xlabel('Frequency Ratio'); ylabel('Dynamic Amplification'); ylim([0 6]); % set y-axis limits since DAF at xi = 0 is enormous legend( 'Damping: 0%','Damping: 10%','Damping: 15%',... 'Damping: 20%','Damping: 30%','Damping: 40%',... 'Damping: 50%', 'Damping: 100%', 'Maxima');
This gives:
0 0.5 1 1.5 2 2.5 30
1
2
3
4
5
6
Frequency Ratio
Dynam
ic A
mplif
ication
Damping: 0%
Damping: 10%
Damping: 15%
Damping: 20%
Damping: 30%
Damping: 40%
Damping: 50%
Damping: 100%
Maxima
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 46
Lastly then, using the phase function we wrote, we can generate Figure 2.9:
% Script to plot phase against Beta for different damping ratios xi = [0.0001,0.1,0.2,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) T(i,:) = phase(beta,xi(i))*(180/pi); % in degrees end plot(beta,T); xlabel('Frequency Ratio'); ylabel('Phase Angle (degrees)'); ylim([0 180]); set(gca,'ytick',[0 45 90 135 180]); grid on; legend('Damping: 0%','Damping: 10%','Damping: 20%','Damping: 50%',... 'Damping: 100%','Location','SE');
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 47
5.2.9 Numerical Integration – Newmark’s Method
Introduction
The loading that can be applied to a structure is infinitely variable and closed-form
mathematical solutions can only be achieved for a small number of cases. For
arbitrary excitation we must resort to computational methods, which aim to solve the
basic structural dynamics equation, at the next time-step:
1 1 1 1i i i i
mu cu ku F (5.2.56)
There are three basic time-stepping approaches to the solution of the structural
dynamics equations:
1. Interpolation of the excitation function;
2. Use of finite differences of velocity and acceleration;
3. An assumed variation of acceleration.
We will examine one method from the third category only. However, it is an
important method and is extensible to non-linear systems, as well as multi degree-of-
freedom systems (MDOF).
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 48
Development of Newmark’s Method
In 1959 Newmark proposed a general assumed variation of acceleration method:
1 11
i i i iu u t u t u
(5.2.57)
2 2
1 10.5
i i i i iu u t u t u t u
(5.2.58)
The parameters and define how the acceleration is assumed over the time step,
t . Usual values are 1
2 and
1 1
6 4 . For example:
Constant (average) acceleration is given by: 1
2 and
1
4 ;
Linear variation of acceleration is given by: 1
2 and
1
6 .
The three equations presented thus far (equations (5.2.56), (5.2.57) and (5.2.58)) are
sufficient to solve for the three unknown responses at each time step. However to
avoid iteration, we introduce the incremental form of the equations:
1i i i
u u u
(5.2.59)
1i i i
u u u
(5.2.60)
1i i i
u u u
(5.2.61)
1i i i
F F F
(5.2.62)
Thus, Newmark‟s equations can now be written as:
i i iu t u t u (5.2.63)
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 49
2
2
2i i i i
tu t u u t u
(5.2.64)
Solving equation (5.2.64) for the unknown change in acceleration gives:
2
1 1 1
2i i i i
u u u utt
(5.2.65)
Substituting this into equation (5.2.63) and solving for the unknown increment in
velocity gives:
12
i i i iu u u t u
t
(5.2.66)
Next we use the incremental equation of motion, derived from equation (5.2.56):
i i i i
m u c u k u F (5.2.67)
And introduce equations (5.2.65) and (5.2.66) to get:
2
1 1 1
2
12
i i i
i i i i i
m u u utt
c u u t u k u Ft
(5.2.68)
Collecting terms gives:
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 50
2
1
1 11
2 2
i
i i i
m c k utt
F m c u m t c ut
(5.2.69)
Let‟s introduce the following for ease of representation:
2
1k̂ m c k
tt
(5.2.70)
1 1ˆ 1
2 2i i i i
F F m c u m t c ut
(5.2.71)
Which are an effective stiffness and effective force at time i. Thus equation (5.2.69)
becomes:
ˆ ˆi i
k u F (5.2.72)
Since k̂ and ˆi
F are known from the system properties (m, c, k); the algorithm
properties ( , , t ); and the previous time-step (i
u , i
u ), we can solve equation
(5.2.72) for the displacement increment:
ˆ
ˆi
i
Fu
k
(5.2.73)
Once the displacement increment is known, we can solve for the velocity and
acceleration increments from equations (5.2.66) and (5.2.65) respectively. And once
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 51
all the increments are known we can compute the properties at the current time-step
by just adding to the values at the previous time-step, equations (5.2.59) to (5.2.61).
Newmark‟s method is stable if the time-steps is about 0.1t T of the system.
The coefficients in equation (5.2.71) are constant (once t is), so we can calculate
these at the start as:
1
A m ct
(5.2.74)
1
12 2
B m t c
(5.2.75)
Making equation (5.2.71) become:
ˆi i i i
F F Au Bu (5.2.76)
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 52
Newmark’s Algorithm
1. Select algorithm parameters, , and t ;
2. Initial calculations:
a. Find the initial acceleration:
0 0 0 0
1u F cu ku
m (5.2.77)
b. Calculate the effective stiffness, k̂ from equation (5.2.70);
c. Calculate the coefficients for equation (5.2.71) from equations (5.2.74) and
(5.2.75).
3. For each time step, i, calculate:
ˆi i i i
F F Au Bu (5.2.78)
ˆ
ˆi
i
Fu
k
(5.2.79)
12
i i i iu u u t u
t
(5.2.80)
2
1 1 1
2i i i i
u u u utt
(5.2.81)
1i i i
u u u
(5.2.82)
1i i i
u u u
(5.2.83)
1i i i
u u u
(5.2.84)
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 53
5.2.10 Computer Implementation & Examples
Using MS Excel
Based on our previous spreadsheet, we implement Newmark Integration. Download it
from the course website, and see how the equations and algorithm are implemented.
In the example shown, we‟ve applied a sinusoidal load of 10 N for 0.6 secs to the
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 54
Using Matlab
There are no shortcuts to this one. We must write a completely new function that
implements the Newmark Integration algorithm as we‟ve described it:
function [u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, plotflag) % This function computes the response of a linear damped SDOF system % subject to an arbitrary excitation. The input parameters are: % m - scalar, mass, kg % k - scalar, stiffness, N/m % xi - scalar, damping ratio % t - vector of length N, in equal time steps, s % F - vector of length N, force at each time step, N % u0 - scalar, initial displacement, m % v0 - scalar, initial velocity, m/s % plotflag - 1 or 0: whether or not to plot the response % The output is: % u - vector of length N, displacement response, m % ud - vector of length N, velocity response, m/s % udd - vector of length N, acceleration response, m/s2
% Set the Newmark Integration parameters % gamma = 1/2 always % beta = 1/6 linear acceleration % beta = 1/4 average acceleration gamma = 1/2; beta = 1/6;
N = length(t); % the number of integration steps dt = t(2)-t(1); % the time step w = sqrt(k/m); % rad/s - circular natural frequency c = 2*xi*k/w; % the damping coefficient
% Calulate the effective stiffness keff = k + (gamma/(beta*dt))*c+(1/(beta*dt^2))*m; % Calulate the coefficients A and B Acoeff = (1/(beta*dt))*m+(gamma/beta)*c; Bcoeff = (1/(2*beta))*m + dt*(gamma/(2*beta)-1)*c;
% calulate the change in force at each time step dF = diff(F);
% Set initial state u(1) = u0; ud(1) = ud0; udd(1) = (F(1)-c*ud0-k*u0)/m; % the initial acceleration
for i = 1:(N-1) % N-1 since we already know solution at i = 1 dFeff = dF(i) + Acoeff*ud(i) + Bcoeff*udd(i); dui = dFeff/keff; dudi = (gamma/(beta*dt))*dui-(gamma/beta)*ud(i)+dt*(1-
Bear in mind that most of this script is either comments or plotting commands –
Newmark Integration is a fast and small algorithm, with a huge range of applications.
In order to use this function, we must write a small script that sets the problem up and
then calls the newmark_sdof function. The main difficulty is in generating the
forcing function, but it is not that hard:
% script that calls Newmark Integration for sample problem m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; t = 0:0.1:4.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set sinusoidal force of 10 over 0.6 s Famp = 10; Tend = 0.6; i = 1; while t(i) < Tend F(i) = Famp*sin(pi*t(i)/Tend); i = i+1; end
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 56
Explosions are often modelled as triangular loadings. Let‟s implement this for our
system:
0 0.5 1 1.5 2 2.5 3 3.5 4-5
0
5
10
Time (s)
Forc
e (
N)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.1
0
0.1
Time (s)
Dis
pla
cem
ent
(m)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.5
0
0.5
Time (s)
Velo
city (
m/s
)
0 0.5 1 1.5 2 2.5 3 3.5 4-1
0
1
Time (s)
Accele
ration (
m/s
2)
Structural Analysis IV Chapter 5 – Structural Dynamics
Dr. C. Caprani 57
% script that finds explosion response m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; Fmax = 50; % N Tend = 0.2; % s t = 0:0.01:2.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set reducing triangular force i = 1; while t(i) < Tend F(i) = Fmax*(1-t(i)/Tend); i = i+1; end