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The terms in brackets are the virtual work expressions
] 0
[
δ
V
w
V
δw δ
M
θ
M
δθ
on that portion of the boundary where the displacement boundary conditions occur, i.e.,
on S u .
N u
Introduce the trial solution
w(
x
) =
w , and note that
w T N u
δw =
N u δ
w
= δ
Then
w T EI L
0
N T
u
L
N u
N u
N T
u
N u
N u N u
on S u
N u +
N T
u
N u dx
= δ
δ
W
+
w
0
p z dx
L
N u
=
0
0
Let
EI N u
N u L
N u
N u
R
=
(7.70)
0
which corresponds to the term [
V
δw
M
δθ
] 0
on S u
.
Then
w T
EI L
0
L
!
" =
N T
u
N u dx
R T
= δ
N u p z dx
δ
W
+ (
R
+
)
w
0
(7.71a)
0
k u
p u
Thus, the free parameters
w can be found from
R T
+ (
+
)
=
[ k u
R
]
w
p u
0
(7.71b)
EXAMPLE 7.9 Beam with Linearly Varying Loading
Consider again the beam of Fig. 7.1. This time we will employ a trial function that does not
necessarily satisfy the displacement boundary conditions.
We begin with a polynomial trial solution in the form
w
1
w 2
w 3
w 4
2
3
4 ]
w(
x
) =
N u
w
=
[
ξξ
ξ
ξ
(1)
Although it is not necessary that any boundary conditions be satisfied, note that the con-
dition
w(
0
) =
0isfulfilled by (1). With d
ξ =
dx
/
L the needed derivatives become
1
L [1
N u =
2
3 ]
2
ξ
3
ξ
4
ξ
1
L 2 [026
N u =
2 ]
ξ
12
ξ
(2)
1
L 3 [00624
N u =
ξ
]
which are obtained using
d N u
dx
d N u
d
d
ξ
dx
N u =
=
ξ
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