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AC
E
Example 2.79. A rigid member of length 2L is supported at ends A and C and at
its mid-point B by three vertical wires DA, EB and FC, as shown in Fig 2.97. The wires DA,
EB and FC extend ai, 22 and as respectively under unit load. Calculate the load in each wire
if the member AC weighs w per unit length.
Solution :
Total downward load = 2 w L
Let the loads in the three bars be P1, P2 and
P, respectively.
Pi + P2 + P3 = 2 w L... (1)
Taking moments about AD,
P2L + P32 L = 2 w L XL
P2 + 2 P = 2 w L ...(2a)
A,
A3
Also, by taking moments about FC,
A2
PiX 2 L + P2L = 2 w L.L
FIG. 2.97
or
2 Pi + P2 = 2 w L
.... (26)
*
... (2)
?? + ?;
From compatibility, ?2 :
2
1
2
A
B
or
-
or
From 2 (a) and 2 (b), we observe that Pa = Ps Doubt:-
> how this equation
us
attained a
= (+
Pz (Plant P3 a3 )
P, az = {(P, as + P, as) = (a1 + as
ai )
(antas)
P2a2 =
1
2
P1
1
2
=
or
Hence
Pi
P2=
2 az
....(3)
Substituting in (1), we get


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