Civil Engineering Reference
In-Depth Information
F
AB
A
60
°
F
AE
R
A
2.75 kN
F
IGURE
4.15
Equilibrium of forces at joint A
Hence
F
AB
sin 60
◦
+
2
.
75
=
0
(i)
so that
F
AB
=−
3
.
18 kN
the negative sign indicating that AB is in compression as expected.
Referring again to Fig. 4.15 and resolving forces horizontally
F
AB
cos 60
◦
=
F
AE
+
0
(ii)
Substituting the
negative
value of
F
AB
in Eq. (ii) we obtain
3
.
18 cos 60
◦
=
F
AE
−
0
which gives
F
AE
=+
1
.
59 kN
the positive sign indicating that
F
AB
is a tensile force.
We now inspect the truss to determine the next joint at which there are no more than
two unknown forces. At joint E there remain three unknowns since only
F
EA
(
=
F
AE
)
has yet been determined. At joint B there are now two unknowns since
F
BA
(
F
AB
)
has been determined; we can therefore proceed to joint B. The forces acting at B
are shown in Fig. 4.16. Since
F
BA
is now known we can resolve forces vertically and
therefore obtain
F
BE
directly. Thus
=
F
BE
cos 30
◦
+
F
BA
cos 30
◦
+
2
=
0
(iii)
Substituting the negative value of
F
BA
in Eq. (iii) gives
F
BE
=+
0
.
87 kN
which is positive and therefore tensile.
