Statically Indeterminant Truss
B
P
C
L
A
D
L
Overall Equilibrium
P
P
P
P
We can find all the external
reactions but the truss is still
statically indeterminant
since there are too many
members to solve for all the
forces
Take out, for example, BD
B
C P
R
B
D
A
P
D
R
P
P
Then for the remaining truss we can solve for all the
forces in terms of P and R . R is our redundant force
B
C P
R
R
A
P
D
P
P
∑F
For example, at B
x
=0
R 2
2
∑ Fy = 0
FBC = −
FBC
R
FAB = −
FAB
R 2
2
Now, for the entire truss
Fi 2 Li
U =∑
i =1 2 Ai Ei
6
where we will take
From the Principle of Least Work
6
F L ∂F
∂U
=∑ i i i =0
∂R i =1 AE ∂R
Ai = A
Ei = E
6
F L ∂F
∂U
=∑ i i i =0
∂R i =1 AE ∂R
member
∂Fi / ∂R
Fi Li ∂Fi / ∂R
−R 2 / 2
− 2/2
RL / 2
L
−R 2 / 2
− 2/2
RL / 2
L
− P+R 2/2
− 2/2
L P 2 /2+ R/2
L
−R 2 / 2
Li
AB
Fi
L
BC
CD
DA
AC
2L
BD
2L
(
)
− 2/2
P 2+R
1
R
1
(
RL / 2
(
L 2 R+P 2
RL 2
Fi Li ∂Fi ( 4.83RL + 2.707 PL )
=
=0
∑
AE
i =1 AE ∂R
R = −0.56 P
6
)
)
Now, suppose that in assembling the truss, member BD was
slightly short so that there was a lack of fit
∆R
B
C
R
R
A
P
D
P
P
In this case R does complimentary virtual work and we have
6
F L ∂F
∂U
δU =
δ R = ∑ i i i δ R = δ R∆ R
∂R
i =1 AE ∂R
which must be satrisfied for all δR so
Fi Li ∂Fi
= ∆R
∑
i =1 AE ∂R
6
( 4.83RL + 2.707 PL ) = ∆
AE
R = −0.56 P +
R
AE
∆R
4.83L
so the force in BD is less compressive,
which is what we expect since we have
to pre-tension BD to get it to fit.