5.4Application to the Solution of Statically Indeterminate Systems 129
Weobservethatifthebeamwereonlycapableofsupportingdirectloads,thenthestructurewould
be a relatively simple statically determinate pin-jointed framework. Since the beam resists bending
moments(weareignoringsheareffects),thesystemisstaticallyindeterminatewithasingleredundancy,
thebendingmomentatanysectionofthebeam.Thetotalcomplementaryenergyoftheframeworkis
given,withthenotationpreviouslydeveloped,by
C=
∫
ABC
∫M
0
dθdM+
∑k
i= 1
∫Fi
0
λjdFi−P (i)
IfwesupposethatthetensileloadinthememberEDisR,then,forCtohaveastationaryvalue,
∂C
∂R
=
∫
ABC
dθ
∂M
∂R
+
∑k
i= 1
λi
∂Fi
∂R
=0(ii)
Atthispoint,weassumetheappropriateload–displacementrelationships;againweshalltakethesystem
tobelinearsothatEq.(ii)becomes
∫L
0
M
EI
∂M
∂R
dz+
∑k
i= 1
FiLi
AiE
∂Fi
∂R
=0(iii)
The two terms in Eq. (iii) may be evaluated separately, bearing in mind that only the beam ABC
contributes to the first term, while the complete structure contributes to the second. Evaluating the
summationtermbyatabularprocess,wehaveTable5.5.Summationofcolumn⑥inTable5.5gives
∑k
i= 1
FiLi
AiE
∂Fi
∂R
=
RL
4 E
(
1
AB
+
10
A
)
(iv)
ThebendingmomentatanysectionofthebeambetweenAandFis
M=
3
4
Pz−
√
3
2
Rz hence
∂M
∂R
=−
√
3
2
z
Table 5.5Tension positive
① ② ③ ④ ⑤ ⑥
Member Length Area F ∂F/∂R (F/A)∂F/∂R
AB L/ 2 AB −R/ 2 − 1 / 2 R/ 4 AB
BC L/ 2 AB −R/ 2 − 1 / 2 R/ 4 AB
CD L/ 2 AR 1 R/A
DE L/ 2 AR 1 R/A
BD L/ 2 A −R − 1 R/A
EB L/ 2 A −R − 1 R/A
AE L/ 2 AR 1 R/A