Civil Engineering Reference
In-Depth Information
W
M
3
1
2
L
2
L
F
IGURE
17.11
Beam of Ex. 17.2
The beam may be idealized into two beam-elements, 1-2 and 2-3. From Fig. 17.11 we
see that
v
1
=
M
. Therefore, eliminating rows and columns
corresponding to zero displacements from Eq. (17.40), we obtain
v
3
=
0,
F
y
,2
=−
W
,
M
2
=+
.
/
!
.
/
!
27/2
L
3
9/2
L
2
6/
L
2
3/2
L
2
F
y
,2
=−
W
−
v
2
θ
2
θ
1
θ
3
9/2
L
2
M
2
=
M
6/
L
2/
L
1/
L
=
EI
(i)
6/
L
2
0
"
0
"
M
1
=
0
2/
L
4/
L
0
3/2
L
2
M
3
=
−
1/
L
2/
L
0
0
Equation (i) may be written such that the elements of [
K
] are pure numbers
.
/
!
.
/
!
F
y
,2
=−
W
3
912 4 2
12
27
9
12
−
v
2
θ
2
L
θ
1
L
θ
3
L
M
2
/
L
=
M
/
L
EI
2
L
3
=
(ii)
0
M
1
/
L
=
0
"
4
8
0
0
"
M
3
/
L
=
0
−
3204
Expanding Eq. (ii) by matrix multiplication we have
27 9
912
v
2
θ
2
L
12
θ
1
L
θ
3
L
−
W
M
/
L
EI
2
L
3
3
42
−
=
+
(iii)
and
0
0
12
v
2
θ
2
L
80
04
θ
1
L
θ
3
L
EI
2
L
3
4
=
+
(iv)
−
32
Equation (iv) gives
θ
1
L
θ
3
L
−
v
2
θ
2
L
3
2
1
2
−
=
(v)
3
4
1
2
−
−
Substituting Eq. (v) in Eq. (iii) we obtain
v
2
θ
2
L
−
L
3
9
EI
−
4
−
2
W
M
/
L
=
(vi)
−
23
