Civil Engineering Reference
In-Depth Information
M
L
2
P
L
EI
iy
iz
V
===− =−
q
∆
q
0
q
+
iconj
ireal
ij
iy
EI
2
y
y
=−
M
L
+
P
L
EI
2
L
2
3
L
iy
M
=
∆
=
0
iz
iconj
ireal
EI
22
y
y
Solving the second equation for
P
iz
in terms of
M
iy
and then substituting
into the first equation, the stiffness value can be found.
3
M
iy
P
=
iz
2
L
MMM
L
3
4
L
L
iy
iy
iy
−=−
q
+
=−
iy
EI
EI
4
EI
y
y
y
4
EI
L
y
M
=
q
y
(4.15)
iy
i
=
6
EI
L
y
P
q
iz
iy
2
It should be noted that the force
P
iz
was actually shown as negative in the
original free-body diagram so it does have a negative value for stiffness.
=−
6
EI
y
P
q
(4.16)
iz
iy
2
L
Example 4.16
∆
iz
stiffness
Derive the ∆
iz
stiffness using the area moment method for a linear member.
A free-body diagram is shown in Figure 4.29 with an imposed deflec-
tion of 1 unit on the
i
i-end of the member. The moments are assumed in the
negative
y
direction using the right-hand rule and the Cartesian right-hand
coordinate system. The forces are shown consistent with the deformation.
The moment diagram divided by
EI
is shown for the reaction forces on the
i
i-end of the member.
Since both ends of the beam are fixed for rotation, the change in rota-
tion from the
i
-end to the
j
i-end is zero. This is the area under the
M
/
EI
diagram between those points.
M
L
P
L
EI
2
iy
∆
q
==−=−
0
q
q
+
iz
ij
j
i
EI
2
y
y
2
M
iy
P
=
iz
L
