The representation (5.10) is ideal for a straight forward implementation for calculating the stress-resultants of a specific theory. However, we will renumber the summation indices again, so that both sums start form 0, to get a deeper insight into the structure of the stress-resultants.
For anNth-order theory we get
Mk(m−k)ij = hb
l2d2dk2ec2(dm2e−dk2e)
N−dm2e X
n=0
bn+dm2e−m2+k2c−dk2e X
q=0
3n+dm2e
(2q+ 2lk2m+ 1)(2n+ 2m2−2q−2lk2m+ 1) d2qc2(n−q)ω(2q+2dk2e−k)(2n+2dm2e−m−2q−2dk2e+k)
ij
!
+O(e2(N+1)). (5.11) The sum of the indices of the stress resultants m gives us basically the magnitude (2m2) of an excludable factor in c2 and d2, whereas k indicates how the common power is distributed betweend2 andc2. For example,M20ij has an excludable factor of magnitude 2, because the sum of the upper indices is 2. Since the first index is 2, it isd2. Analogously,M01ij has an excludable factorc2 andM12ij has an excludable factorc2d2. Furthermore,m basically defines how many summands the double sum actually has, since the complex-looking upper bound of the second sum simply yields
n+
m 2
−m 2 +k
2
− k
2
=
(n−1 ifmeven and kodd
n otherwise .
Basically the double sum “fills up” the stress resultants with summands containing all possible logical combinations of powers of the factorsc2 andd2, so that the common power ofcand d of each summand is less or equal 2N after the multiplication with the excludable factor. The greater the common power of the excludable factor 2m2 is, the fewer possibilities there are to “fill up” the stress-resultant. 2n is the common power of the summand (disregarding the excludable factor) and the sum in q basically arranges the distribution of the common power betweend2 and c2 through all possible combinations.
We used the word “basically” in the last paragraph, because there is one exception. Ifm is even andkis odd, or formulated differently, if both upper indices of Mare odd, then the double sum does not contain all possible logical combinations of powers of the factorsc2 and d2, so that the maximal common power is 2N. In this case, the upper bound forq results in an extra excludablec2. For example, M11ij has an excludable factorc2d2.
If we turn our attention to the question whether a stress resultant is to be neglected in an approximative theory, we derive
m 2
> N or (m= 2N and k odd) =⇒ Mk(m−k)ij = 0 +O(e2(N+1)), (5.12) directly from the upper bounds of the double sum of equation (5.11). For example we have
Mk(m−k)ij = 0 +O(e6) ifm≥5 or (m= 4 and kodd),
for a second-order approximation. In general, equation (5.12) implies that there are only finitely many stress-resultants for anyNth-order theory and their number increases with N.
Another remarkable fact is that there are four classes of the linear combinationsωij(g)(h) that can appear in a stress resultant, identified by the divisibility of gand h by 2 (i.e., the parities of gand h). A stress resultant only contains linear combinations of elements of one of these classes.
This class is determined by the divisibility ofm andk by 2, since the terms in the indices of the linear combinations that depend onm andk yield
2 k
2
−k=
(1 ifk odd 0 ifk even.
(Also note that the indicesg and h of theωij(g)(h) appearing in (5.11) are always non-negative: g is obviously non-negative. For the second index we haveh= 2n+ 2m2−m−2q−2lk2m+k.
2n−2q is non-negative, because of the summation bounds for q, whereas, 2m2−m−2lk2m+k is−1, if and only if, mis even andk is odd and non-negative otherwise. Especially in this case the upper bound forq is n−1 and we have 2n−2q ≥2, so that h stays non-negative.) If m andk have different parities for two different stress resultants, the resultants do not have an intersection in the sets of their linear combinationsω(g)(h)ij . If the parities of m andk are the same, there is an intersection, if m differs in the two stress resultants, and the sets are even equal, ifm is the same, while the specific value of khas no influence on the sets. However, this generally does not imply proportionality in anNth-order theory between two stress resultants with the same parities of mand k. By proportionality in an Nth-order theory we understand:
a∝b+O(e2(N+1)) :⇐⇒ There existsα=O(e0) :αa=b+O(e2(N+1)).
The reason for the non-proportionality is the denominator of the numerical factor in round brackets, which can not be factorized such that one factor only depends on k andm and the other one only depends onn and q. For example, we have
M02ij = hb l2
c2ω00ij + 9
5c4ω02ij +d2c2ω20ij
+O(e6), M20ij = hb
l2
d2ωij00+d2c2ωij02+9 5d4ω20ij
+O(e6),
so that M02ij and M20ij are not proportional, although the m’s are equal and the k’s are both even. As a trivial fact, we have proportionality, if the double sum has only one summand. For a second-order theory this is the case if we havem= 4 orm= 3, which gives us
m= 4 : 5 9
d2
c2M04ij O(e
6)
= M22ij O(e
6)
= 5 9
c2
d2M40ij, M13ij O(e
6)
= M31ij O(e
6)
= 0 +O(e6), (5.13) m= 3 : 5
9 d2
c2M03ij O(e=6)M21ij, M12ij O(e=6) 5 9
c2
d2M30ij, (5.14) where we used the notation
aO(e
6)
= b :⇐⇒ a=b+O(e6),
for convenience. (As a matter of fact, we also have only one summand for m = 2 and k= 1, but this generates no proportionalities like the ones above, since there is no other pair (m, k)
with one summand that has the same parities. Therefore, the list of proportionalities one can find without the use of the reduction equations, cf. 8.3, is complete.) On the other hand, for multiplications of stress resultants by c2 and d2, we have to retruncate the power series, i.e., decrement the upper bound of nby 1
1− m
2
= 2− m
2
+ 1
= 2−
m+ 2 2
.
Therefore, e2Mk(m−k)ij and Mk(m+2−ij˜ k)˜ have representations that use the same set of linear combinations ωijgh, when k and ˜k have the same parity and e is c or d. Again, this leads to relations between these stress-resultants, if the double sum has only one summand. For a second-order theory this gives us
m= 2 :c2M02ij O(e=6) c4
d2M20ij O(e=6) 5
9M04ij, c2M11ij O(e= 0 +6) O(e6), (5.15) m= 1 :c2M01ij O(e=6) 5
9M03ij, c2M10ij O(e=6)M12ij. (5.16) However, these are no proportionality relations. In particular the relations donot imply that the series expansions of the two connected stress-resultants are basically equal. Analogously, for the factore4, we get another relation, which is
m= 0 :c4M00ij O(e=6)c2M02ij. (5.17)
This relation finally completes the list of relations, if one does not use any information from the equilibrium equations, which will imply further relations, cf. section 8.3 and 8.4 .
Some additional Notes:
In the special casem even andk odd, we have an extrac2 as excludable factor and the stress resultant actually depends on fewer linear combinationsωijgh, because of the upper bound for q.
For deriving an explicit formula for the number of linear combinations and the actual excludable scaling prefactor incanddwe renumber the sum again. Formeven andkodd, we can substitute nwithn+ 1, which leads to a case-independent upper bound forq
2−dm2e X
n=0 n−1
X
q=0
f(n, q) =
2−dm2e X
n=1 n−1
X
q=0
f(n, q)
=
2−dm2e−1
X
n=0 n
X
q=0
f(n+ 1, q).
Therefore, we get
2−dm2e X
n=0
bn+dm2e−m2+k2c−dk2e X
q=0
f(n, q) =
2−dm2e−γhmiηhki X
n=0
n
X
q=0
f(n+γhmiηhki, q), where we introduced the symbol ηhni for any integern by
ηhni..=
(1 forn odd
0 forn even = 1−γhni. (5.18)
Inserting this into formula (5.11), while replacing all occurrences of the floor and ceiling function with the η-symbol, and by the use of the identity
ηhmi= (ηhmi)2 ⇒
2γhmiηhki+ηhmi −ηhki=ηhmi −2ηhmiηhki+ηhki= (ηhmi −ηhki)2
=
1 for (m even andk odd) 1 for (k even andm odd) 0 otherwise
=ηhm−ki,
we get
Mk(m−k)ij = hb
l2dk+ηhkicm−k+ηhm−ki
N−m
2−1
2ηhki−1
2ηhm−ki
X
n=0
n
X
q=0
3n+m2+12ηhki+12ηhm−ki
(2q+k+ηhki+ 1)(2(n−q) +m−k+ηhm−ki+ 1) d2qc2(n−q)ω(2q+ηhki)(2(n−q)+ηhm−ki)
ij
!
+O(e2(N+1)). (5.19)
The actual excludable factor of a stress resultant Mk(m−k)ij has, therefore, the representation dk+ηhkicm−k+ηhm−ki, independent of the order of the theory. In anNth-order theory the upper bound for the sum overn isι..=N−m2 −12ηhki − 12ηhm−ki and the actual number of linear combinations is computable via
ι
X
n=0
n+ 1 = 1 2ι2+ 3
2ι+ 1, for non-negativeι.