Civil Engineering Reference
In-Depth Information
fibres to expand more than the lower ones so that bending strains, without bending
stresses, are induced as shown in Fig. 15.19(a). Note that the undersurface of the
member is unstrained since the change in temperature in this region is zero.
Consider an element,
δ
x
, of the member. The upper surface will increase in length
to
x
as shown in
Fig. 15.19(c);
α
is the coefficient of linear expansion of the material of the member.
Thus, from Fig. 15.19(c)
δ
x
(1
+
α
t
), while the length of the lower surface remains equal to
δ
R
δ
R
+
h
x
=
δ
x
(1
+
α
t
)
so that
h
α
t
R
=
Also
=
δ
x
R
δ
θ
whence
α
t
δ
x
δ
θ
=
(15.41)
h
If we require the deflection,
Te,B
, of the free end of the member due to the tempera-
ture rise, we can employ the unit load method as in Ex. 15.9. Thus, by comparison with
Eq. (ii) in Ex. 15.9
L
d
θ
∂
M
∂
P
f
Te,B
=
(15.42)
0
in which, as we have seen,
∂
M
/∂
P
f
=
M
1
, the bending moment at any section of the
member produced by a unit load acting vertically downwards at B. Now substituting
for
δ
θ
in Eq. (15.42) from Eq. (15.41)
L
M
1
α
t
Te,B
=−
h
d
x
(15.43)
0
In the case of a beamcarrying actual external loads the total deflection is, fromthe prin-
ciple of superposition (Section 3.7), the sum of the bending, shear (unless neglected)
and temperature deflections. Note that in Eq. (15.43)
t
can vary arbitrarily along the
length of the beam but only linearly with depth. Note also that the temperature gradi-
ent shown in Fig. 15.19(b) produces a hogging deflected shape for the member. Thus,
strictly speaking, the radius of curvature,
R
, in the derivation of Eq. (15.41) is negative
(compare with Fig. 9.4) so that we must insert a minus sign in Eq. (15.43) as shown.
E
XAMPLE
15.10
Determine the deflection of the free end of the cantilever beam
in Fig. 15.20 when subjected to the temperature gradients shown.
