Civil Engineering Reference
In-Depth Information
which simplifies to
L
0
T
α
C
A
C
E
C
+
α
S
A
S
E
S
δ =
(7.51)
A
C
E
C
+
A
S
E
S
Clearly when
α
C
=
α
S
=
α
,
say,
P
C
=
P
S
=
0,
σ
C
=
σ
S
=
0 and
δ =
L
0
α
T
as for
unrestrained expansion.
The above analysis also applies to the case,
α
C
>α
S
, when, as can be seen from Eqs
(7.48) and (7.49) the signs of
P
C
,
P
S
,
σ
C
and
σ
S
are reversed. Thus the load and stress
in the concrete become compressive, while those in the steel become tensile. A similar
argument applies when
T
specifies a temperature reduction.
Equation (7.44) is an expression of the compatibility of displacement of the concrete
and steel. Also note that the stresses could have been obtained directly by writing Eqs
(7.43) and (7.44) as
σ
C
A
C
=
σ
S
A
S
and
σ
C
L
0
E
C
+
σ
S
L
0
E
S
=
L
0
T
(
α
S
−
α
C
)
respectively.
E
XAMPLE
7.6
A rigid slab of weight 100 kN is supported on three columns each of
height 4m and cross-sectional area 300mm
2
arranged in line. The two outer columns
are fabricated from material having a Young's modulus of 80 000N/mm
2
and a coef-
ficient of linear expansion of 1
.
85
10
−
5
/
◦
C; the corresponding values for the inner
×
column are 200 000N/mm
2
and 1
.
2
10
−
5
/
◦
C. If the slab remains firmly attached to
each column, determine the stress in each column and the displacement of the slab if
the temperature is increased by 100
◦
C.
×
The problem may be solved by determining separately the stresses and displacements
produced by the applied load and the temperature rise; the two sets of results are then
superimposed. Let subscripts o and i refer to the outer and inner columns, respectively.
Using Eq. (7.38) we have
E
i
A
o
E
o
+
E
o
A
o
E
o
+
σ
i
(load)
=
P
σ
o
(load)
=
P
(i)
A
i
E
i
A
i
E
i
In Eq. (i)
10
6
A
o
E
o
+
A
i
E
i
=
×
×
+
×
=
×
2
300
80 000
300
200 000
108
.
0
Then
10
3
200 000
×
100
×
185
.
2N
/
mm
2
(compression)
σ
i
(load)
=
=
×
10
6
108
.
0
10
3
80 000
×
100
×
74
.
1N
/
mm
2
(compression)
σ
o
(load)
=
=
108
.
0
×
10
6
