6.13 Winkler–Bach Theory
191
Therefore, the tangential strain in the fibre gh = ε θ =
[ε c Rdθ+ydθ]
(R+y)dθ
.
Using Hooke’s law, the tangential stress acting on area dA is given by
σ θ =
ε c R + y((dθ/dθ)
(R + y)
E
(6.61)
Let angular strain
dθ
dθ
= λ.
Hence, Eq. (6.61) becomes
σ θ =
ε c R + yλ
(R + y)
E
(6.62)
Adding and subtracting ε c y in the numerator of Eq. (6.62), we get,
σ θ =
ε c R + yλ + ε c y − ε c y
(R + y)
E
Simplifying, we get
σ θ =
ε c + (λ − ε c )
y
(R + y)
E
(6.63a)
The beam section must satisfy the conditions of static equilibrium,
F z = 0 and M x = 0, respectively:
∴
σ θ dA = 0 and
σ θ ydA = M
(6.63b)
Substituting the above boundary conditions (6.63b) in (6.63a), we get
0 =
ε c + (λ − ε c )
y
(R + y)
dA
or
ε c dA = −(λ − ε c )
y
(R + y)
dA
or ε c
dA = −(λ − ε c )
y
(R + y)
dA
(6.63c)
Also,
M =
ε c
ydA + (λ − ε c )
y
2
(R + y)
dA
E
(6.63d)
Here
dA = A, and since y is measured from the centroidal axis,
ydA = 0.
Let
y
(R+y)
dA = −m A.
Or