Civil Engineering Reference
In-Depth Information
Observing that there are two new terms that vary with
x
,
we will substitute
S
5
and
S
6
as follows:
dx
EI
(
)
∫
S
=−
x
a
5
y
dx
EI
(
)
∫
S
=
x
2
−
ax
6
y
0
0
=
FEP
S
+
FEM
S
−
PS
iz
3
iy
2
6
=
FEP
S
FEM
FEP
FEM
+
S
−
PS
iz
2
iy
1
5
SS
SS
S
S
=
3
2
iz
6
P
2
1
iy
5
FEP
FEM
=
S S
SS
−
S
S
SS SS
−
1
P
D
iz
1
2
6
16 25
SSS SS
35 26
P
=
D
−
−
iy
2
3
5
Apply equilibrium on Figure 5.8 to find the
j-
end forces and moments.
FEP
=−
=− −
P
Pb
FEP
jz
iz
FEM
FEP
L
FEM
jy
iz
iy
The process for solving stiffness problems involving non-prismatic mem-
bers is the same as with prismatic members with the same four modifica-
tions for members with end releases. The following example shows this
process.
Example 5.7
Non-prismatic member stiffness
Determine the deformations at the free end of the non-prismatic beam
using only three degrees of freedom in the stiffness solution. Also deter-
mine the final end forces and the support reactions. The beam is shown in
Figure 5.9.
Z
A=10 in
2
I=100 in
4
A=20 in
2
I=200 in
4
W=600 lb/ft
X
100in
100in
E=10,000 ksi
Figure 5.9.
Example 5.7 Non-prismatic member stiffness.
