Civil Engineering Reference
In-Depth Information
d
d
x
kd s p wx
EI
/
0
c
(A.13)
=
a
Differentiating Equation A.11 and eliminating
φ
from Equation A.13,
F
from Equation A.4, and
v
L
from Equation A.1:
d
d
s
kd s p wd x
EI
/
−
ks
EAp
ks
pE I
⎛
⎜
I
A
⎞
⎟
−
wd x
EI
2
2
c
c
0
c
=
+
=
d
2
+
c
x
2
a
0
a
0
a
0
0
a
0
Introducing
A
′
from Equation 2.21,
α
2
from Equation 2.23 and
β
from
Equation 2.24 gives Result 2.25, which is in a standard form:
d
d
s
2
−=−
α
2
s
α β
2
wx
(2.25)
x
2
Solving for
s
,
s
=
K
1
sinh
α
x
+
K
2
cosh
α
x
+
β
wx
(A.14)
The boundary conditions, Equations A.2 and A.3, give
K
2
=
0
ε
c
=
−
K
1
α
cosh(
α
L
/2)
−
β
w
and substitution in Equation A.14 gives
s
in terms of
x
:
s x
w
βε
α
+
α
L
⎛
⎜
⎞
⎟
⎛
⎝
⎞
⎠
c
=
β
−
sec
h
sinh
α
x
(2.27)
2
Other results can now be found as required. For example, the slip strain at
mid-span is
⎛
⎜
d
d
s
x
⎞
⎟
=− +
ββε
ww L
(
) sec (
h
α
/ )
2
(A.15)
c
x
=
0
and the slip at
x
=
L
/2 due to
ε
c
alone (i.e. with
w
=
0), is
ε
α
α
L
⎛
⎝
⎞
⎠
c
( )
s
=−
tanh
(A.16)
x L
=
/
2
2
A.2
Example: partial interaction
These calculations are introduced in Section 2.7. They relate to a beam
shown in section in Fig. 2.16, which carries a distributed load
w
per unit
