3.1 Dimensional analysis
3.1.2 Dimensional analysis
First, we will consider an example to illustrate a typical dimensional analy- sis process. Suppose a thin ring of radius Ris subjected to two equal, oppo- site inward point loads Pas shown in Fig. 3.1. The cross-section of the ring is rectangular with thickness hand width B. We want to find the initial col- lapse load Pofor a given ring material. This problem can be easily studied using the plasticity theory presented in Chapter 2 and indeed we will do this in Chapter 4. However, for our present purpose, we will pretend that the analytical solutions are not known; we only have experimental results for tubes of various dimensions and materials.
From our insight into the physics of this problem, we recognise that yield stress Ymay be chosen as the only material parameter; Young’s modulus is irrelevant as we are dealing with a plastic collapse problem. The geo- metrical parameters are radius R, thickness hand width B. Hence we may argue that the collapse load Pois a function of the above mentioned physi- cal variables which describe geometric and material properties of the struc- tures. We can write the following equation
[3.1]
or
Po =F Y R B h1( , , , )
P
h
R
P
B h
3.1 A circular ring under two opposite point loads. The cross-section of the ring is rectangular (h¥B).
[3.2]
where F1and F2are functions. There are a total of five physical variables in Eq. [3.2]. From the requirement for dimensional homogeneity, we deduce that Eq. [3.2] can be expressed in dimensionless form. To choose suitable dimensionless groups(usually called pgroups), we follow the simple pro- cedure below.
The two fundamental dimensions applicable here are force Fand length L. We obtain a particular dimensionless group by involving each physical variable raised to its own specific power. Hence, for this example a dimen- sionless group DGis of the form
[3.3]
The values of powers a,b,c,d andeare obtained from the fact that DGis dimensionless. Substituting the units for each physical variable in terms of the two fundamental dimensions, e.g.Poas Fand Yas F/L2, we have
[3.4]
or
[3.5]
Because DGis dimensionless, we have
[3.6]
The above two equations have five unknowns whose values cannot be determined uniquely. If, however, we assume the values for three unknowns, we can then solve the other two. Let a = -1, b = c = 0, then d = -2 and e =1. Hence, we could choose as our first dimensionless group
[3.7]
Similarly, if we let a =0,b =1,c = -1, then d =0,e =0, i.e.
[3.8]
Finally, if a=0,b=0,c=1, then e=0,d= -1, i.e.
[3.9]
p3=B h p2=R B p1= P2
Yh
o
a e a b c d
+ =
- + + + =
0
2 0
DG=( ) ( )F a e+ L -2a b c d+ + + DG F
L L L L F
a
b c d e
=Ê Ë
ˆ
¯ ( ) ( ) ( ) ( )
2
DG Y R B h Pa b c d oe
=
F P Y R B h2( o, , , , )=0
In principle, we may choose a number of other dimensionless groups, but we argue for the moment that there are only three independent dimen- sionless groups. Other groups can be obtained by various combinations of the above three groups. For example, if we choose a =0,b =1,c =0, then d = -1, and e =0. The corresponding dimensionless group is (R/h). But this is in effect (p2·p3), and is therefore dependent upon two of the three pre- vious groups, Eqs [3.7–3.9]. Equation [3.2] can now be written in dimen- sionless form
[3.10]
or
[3.11]
The dimensional analysis is now complete; we have successfully reduced the number of variables from five to three. The functional form of f1will need to be determined from experiments. We could, for example, plot p1
against p2for a given value of p3. For this particular problem, it turns out from experiments (which will be presented later) that p3has little effect on p1and may be discarded. A simple relationship between the remaining two dimensionless groups can be established as follows
[3.12]
Buckingham Pi theorem
The above example demonstrates two fundamental points. First, a dimen- sionally homogeneous equation of five physical variables, Eq. [3.2], can be written as an equation with a set of three dimensionless groups, Eq. [3.11].
Second, the number of independent dimensionless groups is only three, when five physical variables are involved with two fundamental dimensions.
In general, the Buckingham theorem (Buckingham, 1914) states that a dimensionally homogeneous equation with a number of physical variables can be reduced to an equation with several dimensionless groups. The number of independent dimensionless groups is equal to the difference between the number of physical variables and the number of the funda- mental dimensions. This is reflected in our example above where the number of independent dimensionless groups is three (= 5 -2).
Remarks on the dimensional analysis
At this stage, we may highlight several points in relation to the choice of dimensionless groups and the use of dimensional analysis.
P Yh f R
oµ ÊB
Ë ˆ
¯
2 2
f P Yh
R B
B h
o
1Ê 2, , 0
Ë
ˆ
¯= f1(p p p1, 2, 3)=0
(1) One needs to have a sound insight into the nature of the engineering problem under study. All the physical variables in question must be included; but irrelevant ones should be discarded. An intelligent choice of the physical variables will simplify the problem. The follow- ing example illustrates this point.
(2) The dimensionless groups must be independent.
(3) In our example above, we have chosen force and length as the two fundamental dimensions. This is convenient in static loadings. Alter- natively, mass, length and time could be chosen as the three funda- mental dimensions, especially in dynamic cases. The final result will be the same for both the methods, because force can be related to mass through Newton’s second law (which involves acceleration, or length/
time2).
(4) The exact form for each dimensionless group is not unique. It is more a matter of personal choice. Dimensionless groups can be obtained following the procedure explained in the above example, or indeed by inspection. Other, more rigorous approaches are available (Bridgman, 1922; Gibbings, 1982; Sedov, 1993; Harris and Sabnis, 1999).
(5) There are, nevertheless, some common practices used with regard to the dimensionless groups: (a) quite often, several different forms may be tried before those producing the simplest results are finally selected; (b) it is preferable to choose a dimensionless group that retains a certain physical meaning. Examples are the aspect ratio of a rectangular cross-section and the relative thickness of a circular tube section (h/R). (c) Mathematical formulation of a much simplified, though probably crude, model may give us a lot of clues to the way each physical variable should come into a dimensionless group.
(6) Dimensional analysis is particularly useful when an engineering problem is so complicated that it is not possible to obtain exact solu- tions (which could be due to a lack of information on the exact mechanics involved). Dimensional analysis enables us to reduce the total number of variables, and to design an experimental programme so as to cover a wide range of dimensionless parameters instead of merely certain physical variables. For example, to study the behaviour of a tube, we should vary, as widely as possible, the value of h/R, rather than just h or R. Two tests with the same value of h/R, though with different values of hand R, would not reveal much extra information.
Importance of physical insight – an example
In our earlier example for the collapse load of a circular ring under two point loads (Fig. 3.1), we argued that five physical variables are involved.
In the beginning, the only knowledge we had was that the collapse load was
independent of the elastic behaviour of the material and hence Young’s modulus was not relevant. Now suppose we have a better understanding of this problem: the collapse of this tube is largely governed by bending of the tube wall without any stretching. Therefore, the effect of thickness h and the material property Yshould be present in the form of bending resistance, which in this case is the fully plastic bending moment per unit width, Mo(= Yh2/4). Consequently, we can now argue that the initial collapse load Pois only a function of Mo, the tube width Band radius R, i.e.
[3.13]
Now there are only four variables and we may choose the following two dimensionless groups
[3.14]
It is straightforward to show that this equation is in effect the same as Eq.
[3.12]. An elegant choice of physical variables based on physical insight into the problem greatly simplifies the analysis of experimental results.
A further example: energy absorption in laterally loaded circular rings
The previous example deals with the initial collapse load and hence dis- placement is not involved. Now we are interested in energy absorption during the whole crushing process – the subject of this book. Take the same ring and loading as in the previous example Fig. 3.2. We need to select a parameter to describe the crushing. Let the total displacement between the two loading points be u. The work done by the two loads,W, is equal to the energy dissipated by plastic bending of the tube. Hence, considering all physical variables
[3.15]
The number of independent dimensionless groups involved is again three (= 5 -2) and we could choose the following dimensionless groups
F W M R B u4( , o, , , )=0 f P
M R B
o o
3Ê , 0
Ë
ˆ
¯= F P M R B3( o, o, , )=0
u 2 u
2 P
P
3.2 Collapse of an initially circular ring.
[3.16]
The above equation suggests that we should plot W/(MoB) against u/Rfor various values of B/R. Figure 3.3(b) is such a plot for a number of experi- ments of which some ‘raw’ test data are shown in Fig. 3.3(a). Again, the curves indicate that B/Rhas little effect and therefore this parameter can
f W M B
B R
u
o R
4Ê , , 0
Ë
ˆ
¯=
R = 28.91 mm, h = 1.79 mm R = 28.44 mm, h = 3.13 mm R =38.94 mm, h = 1.93 mm R =38.48 mm, h = 2.84 mm 300
200
100
0
0 20 40 60 80
W (N m)
u (mm)
R = 28.91 mm, h = 1.79 mm R = 28.44 mm, h = 3.13 mm R =38.94 mm, h = 1.93 mm R =38.48 mm, h = 2.84 mm
W/MoB
u/R
0 0.5 1.0 1.5 2.0
0 3 6 9 12 15
3.3 (a) Energy–deflection curves for tubes of different values of thickness; (b) non-dimensionalised energy–deflection curves.
(a)
(b)
be discarded as irrelevant. An analytical study of this problem to be pre- sented in Chapter 4 also demonstrates this.