Civil Engineering Reference
In-Depth Information
B
C
D
2m
A
F
G
100 kN
2m
2m
F
IGURE
P.16.7
Ans.
AB
=
=−
=
=−
=
=
=
=
FD
29.2 kN BC
CD
29.2 kN AG
GF
20.8 kN BG
DG
41.3 kN AC
=
FC
=−
29.4 kN CG
=
41.6 kN.
P.16.8
Calculate the forces in the members of the truss shown in Fig. P.16.8 and the
vertical and horizontal components of the reactions at the supports; all members of
the truss have the same cross-sectional properties.
100 kN
E
C
B
2.5 m
F
5m
D
A
5m
5m
F
IGURE
P.16.8
Ans
.
R
A,V
=
67
.
52 kN
R
A,H
=
70
.
06 kN
=
R
F,H
R
F,V
=
32
.
48 kN
=−
=−
=−
=
AB
32
.
48 kN AD
78
.
31 kN BC
64
.
98 kN BD
72
.
65 kN
=−
=−
=
=−
CD
100
.
0kN CE
64
.
98 kN DE
72
.
65 kN DF
70
.
06 kN
EF
=−
32
.
49 kN
.
P.16.9
The plane truss shown in Fig. P.16.9(a) has one member (24) which is loosely
attached at joint 2 so that relative movement between the end of the member and
the joint may occur when the framework is loaded. This movement is a maximum
of 0.25mm and takes place only in the direction 24. Figure P.16.9(b) shows joint 2
in detail when the framework is unloaded. Find the value of
P
at which the member
24 just becomes an effective part of the truss and also the loads in all the members
