Civil Engineering Reference
In-Depth Information
T
ABLE
P.17.5(i)
Node
x
y
z
2
0
0
0
L
3
0
0
7
L
0
.
8
L
0
9
L
0
L
T
ABLE
P.17.5(ii)
Effect
Member
Axial
Bending
Torsional
23
-
EI
-
37
-
-
GJ
=
0
.
8
EI
6
√
2
EI
L
2
EA
=
29
-
-
Use the
direct stiffness
method to find all the displacements and hence calculate the
forces in all the members. For member 123 plot the shear force and bending moment
diagrams.
Briefly outline the sequence of operations in a typical computer program suitable for
linear frame analysis.
F
28
=
√
2
P
/6 (tension)
M
3
=−
Ans
.
F
29
=
M
1
=
PL
/9 (hogging)
M
2
=
2
PL
/9 (sagging)
F
y
,3
=−
F
y
,2
=
P
/3
.
Twisting moment in 37,
PL
/
18 (anticlockwise).
P.17.6
Given that the force-displacement (stiffness) relationship for the beamelement
shown in Fig. P.17.6(a) may be expressed in the following form:
.
/
!
.
/
!
F
y
,1
M
1
/
L
F
y
,2
M
2
/
L
12
−
6
−
12
−
6
v
1
θ
1
L
v
2
θ
2
L
EI
L
3
−
6462
=
0
"
0
"
−
12
6
12
6
−
6264
obtain the force-displacement (stiffness) relationship for the variable section beam
(Fig. P.17.6(b)), composed of elements 12, 23 and 34.
Such a beam is loaded and supported symmetrically as shown in Fig. P.17.6(c). Both
ends are rigidly fixed and the ties FB, CH have a cross-sectional area
a
1
and the ties
EB, CG a cross-sectional area
a
2
. Calculate the deflections under the loads, the forces
in the ties and all other information necessary for sketching the bending moment and
shear force diagrams for the beam.
Neglect axial effects in the beam. The ties are made from the same material as the
beam.
