E NERGY AND M ATRIX M ETHODS
6.3 Stiffness matrix for two elastic springs in line
Fx,1=ku1−ku2 Fx,2= −ku1+ku2
(6.5) Writing Eq. (6.5) in matrix form we have
Fx,1 Fx,2
=
k −k
−k k u1 u2
(6.6) and by comparison with Eq. (6.1) we see that the stiffness matrix for this spring element is
[K]=
k −k
−k k
(6.7) which is a symmetric matrix of order 2×2.
6.3 Stiffness matrix for two elastic springs in line
Bearing in mind the results of the previous section we shall now proceed, initially by a similar process, to obtain the stiffness matrix of the composite two-spring system shown in Fig. 6.2. The notation and sign convention for the forces and nodal displacements are identical to those specified in Section 6.1.
First let us suppose that u1=u1 and u2=u3=0. By comparison with the single spring case we have
Fx,1=kau1= −Fx,2 (6.8)
but, in addition, Fx,3=0 since u2=u3=0.
Secondly, we put u1=u3=0 and u2=u2. Clearly, in this case, the movement of node 2 takes place against the combined spring stiffnesses kaand kb. Hence
Fx,2=(ka+kb)u2
Fx,1= −kau2, Fx,3= −kbu2
(6.9) Hence the reactive force Fx,1(=−kau2) is not directly affected by the fact that node 2 is connected to node 3, but is determined solely by the displacement of node 2. Similar conclusions are drawn for the reactive force Fx,3.
Finally, we set u1=u2=0, u3=u3and obtain Fx,3=kbu3= −Fx,2 Fx,1=0
(6.10)
Fig. 6.2Stiffness matrix for a two-spring system.
Superimposing these three displacement states we have, for the condition u1=u1, u2=u2, u3=u3
Fx,1=kau1−kau2
Fx,2= −kau1+(ka+kb)u2−kbu3 Fx,3= −kbu2+kbu3
⎫⎬
⎭ (6.11)
Writing Eqs (6.11) in matrix form gives
⎧⎨
⎩ Fx,1 Fx,2 Fx,3
⎫⎬
⎭=
⎡
⎣ ka −ka 0
−ka ka+kb −kb
0 −kb kb
⎤
⎦
⎧⎨
⎩ u1 u2 u3
⎫⎬
⎭ (6.12)
Comparison of Eqs (6.12) with Eq. (6.1) shows that the stiffness matrix [K] of this two-spring system is
[K]=
⎡
⎣ ka −ka 0
−ka ka+kb −kb
0 −kb kb
⎤
⎦ (6.13)
Equation (6.13) is a symmetric matrix of order 3×3.
It is important to note that the order of a stiffness matrix may be predicted from a knowledge of the number of nodal forces and displacements. For example, Eq. (6.7) is a 2×2 matrix connecting two nodal forces with two nodal displacements; Eq. (6.13) is a 3×3 matrix relating three nodal forces to three nodal displacements. We deduce that a stiffness matrix for a structure in which n nodal forces relate to n nodal displacements will be of order n×n. The order of the stiffness matrix does not, however, bear a direct relation to the number of nodes in a structure since it is possible for more than one force to be acting at any one node.
So far we have built up the stiffness matrices for the single- and two-spring assemblies by considering various states of displacement in each case. Such a process would clearly become tedious for more complex assemblies involving a large number of springs so that a shorter, alternative, procedure is desirable. From our remarks in the preceding paragraph and by reference to Eq. (6.2) we could have deduced at the outset of the analysis that the stiffness matrix for the two-spring assembly would be of the form
[K]=
⎡
⎣k11 k12 k13 k21 k22 k23 k31 k32 k33
⎤
⎦ (6.14)
The element k11of this matrix relates the force at node 1 to the displacement at node 1 and so on. Hence, remembering the stiffness matrix for the single spring (Eq. (6.7)) we may write down the stiffness matrix for an elastic element connecting nodes 1 and 2 in a structure as
[K12]=
k11 k12 k21 k22
(6.15) and for the element connecting nodes 2 and 3 as
[K23]=
k22 k23 k32 k33
(6.16)
6.3 Stiffness matrix for two elastic springs in line 173 In our two-spring system the stiffness of the spring joining nodes 1 and 2 is kaand that of the spring joining nodes 2 and 3 is kb. Therefore, by comparison with Eq. (6.7), we may rewrite Eqs (6.15) and (6.16) as
[K12]=
ka −ka
−ka ka
[K23]=
kb −kb
−kb kb
(6.17) Substituting in Eq. (6.14) gives
[K]=
⎡
⎣ ka −ka 0
−ka ka+kb −kb
0 −kb kb
⎤
⎦
which is identical to Eq. (6.13). We see that only the k22term (linking the force at node 2 to the displacement at node 2) receives contributions from both springs. This results from the fact that node 2 is directly connected to both nodes 1 and 3 while nodes 1 and 3 are each joined directly only to node 2. Also, the elements k13and k31of [K] are zero since nodes 1 and 3 are not directly connected and are therefore not affected by each other’s displacement.
The formation of a stiffness matrix for a complete structure thus becomes a relatively simple matter of the superposition of individual or element stiffness matrices. The procedure may be summarized as follows: terms of the form kiion the main diagonal consist of the sum of the stiffnesses of all the structural elements meeting at node i while off-diagonal terms of the form kijconsist of the sum of the stiffnesses of all the elements connecting node i to node j.
An examination of the stiffness matrix reveals that it possesses certain properties. For example, the sum of the elements in any column is zero, indicating that the conditions of equilibrium are satisfied. Also, the non-zero terms are concentrated near the leading diagonal while all the terms in the leading diagonal are positive; the latter property derives from the physical behaviour of any actual structure in which positive nodal forces produce positive nodal displacements.
Further inspection of Eq. (6.13) shows that its determinant vanishes. As a result the stiffness matrix [K] is singular and its inverse does not exist. We shall see that this means that the associated set of simultaneous equations for the unknown nodal displacements cannot be solved for the simple reason that we have placed no limitation on any of the displacements u1, u2or u3. Thus the application of external loads results in the system moving as a rigid body. Sufficient boundary conditions must therefore be specified to enable the system to remain stable under load. In this particular problem we shall demonstrate the solution procedure by assuming that node 1 is fixed, i.e. u1=0.
The first step is to rewrite Eq. (6.13) in partitioned form as
⎧⎨
⎩ Fx,1 Fx,2 Fx,3
⎫⎬
⎭=
⎡
⎢⎢
⎢⎢
⎣
ka ... −ka 0
ã ã ã ã
−ka ... ka+kb −kb 0 ... −kb kb
⎤
⎥⎥
⎥⎥
⎦
⎧⎨
⎩ u1=0
u2 u3
⎫⎬
⎭ (6.18)
In Eq. (6.18) Fx,1 is the unknown reaction at node 1, u1 and u2 are unknown nodal displacements, while Fx,2and Fx,3are known applied loads. Expanding Eq. (6.18) by
matrix multiplication we obtain {Fx,1} =[−ka 0]
u2 u3
Fx,2 Fx,3
=
ka+kb −kb
−kb kb u2 u3
(6.19) Inversion of the second of Eqs (6.19) gives u2 and u3 in terms of Fx,2 and Fx,3. Substitution of these values in the first equation then yields Fx,1.
Thus
u2 u3
=
ka+kb −kb
−kb kb −1
Fx,2 Fx,3
or
u2 u3
=
1/ka 1/ka 1/ka 1/kb+1/ka
Fx,2 Fx,3
Hence
{Fx,1} =[−ka 0]
1/ka 1/ka 1/ka 1/kb+1/ka
Fx,2 Fx,3
which gives
Fx,1= −Fx,2−Fx,3
as would be expected from equilibrium considerations. In problems where reactions are not required, equations relating known applied forces to unknown nodal displacements may be obtained by deleting the rows and columns of [K] corresponding to zero dis- placements. This procedure eliminates the necessity of rearranging rows and columns in the original stiffness matrix when the fixed nodes are not conveniently grouped together.
Finally, the internal forces in the springs may be determined from the force–
displacement relationship of each spring. Thus, if Sais the force in the spring joining nodes 1 and 2 then
Sa =ka(u2−u1) Similarly for the spring between nodes 2 and 3
Sb=kb(u3−u2)