146
5 Two-Dimensional Problems in Cartesian Co-ordinate System
= c 4 x
2
+ d 4 x y −
a 4
2
y
2
− 2c 4 y
2
σ y =
∂
2
φ
∂ x 2 =
a 4
2
x
2
+ b 4 x y + c 4 y
2
and
τ xy = −
∂
2
φ
∂ x∂ y
=
b 4
2
x
2
− 2c 4 x y −
d 4
2
y
2
For the given problem, the upper and lower edges ar free from load.
i.e. at
y = ±h, σ y = 0
and at
y = ±h, τ xy = 0
Further, the shearing forces, having a resultant P are distributed along the end x
= 0.
Now, taking the consultant c 4 = a 4 = b 4 = 0 and d 4 = 0,
then
σ x = d 4 x y
(5.43)
σ y = 0
(5.44)
and
τ xy = −
d 4
2
y
2
(5.45)
But the beam is subjected to a constant shear force ‘P’ resulting on constant shear
stress distribution. Hence, superimposing the pure shear component (from second
degree polynomial), i.e., τ xy = −b 2 in Eq. (5.45), the resulting final stresses are
given by
σ x = d 4 x y
σ y = 0
τ xy = −b 2 −
d 4
2
y
2
5 Two-Dimensional Problems in Cartesian Co-ordinate System
= c 4 x
2
+ d 4 x y −
a 4
2
y
2
− 2c 4 y
2
σ y =
∂
2
φ
∂ x 2 =
a 4
2
x
2
+ b 4 x y + c 4 y
2
and
τ xy = −
∂
2
φ
∂ x∂ y
=
b 4
2
x
2
− 2c 4 x y −
d 4
2
y
2
For the given problem, the upper and lower edges ar free from load.
i.e. at
y = ±h, σ y = 0
and at
y = ±h, τ xy = 0
Further, the shearing forces, having a resultant P are distributed along the end x
= 0.
Now, taking the consultant c 4 = a 4 = b 4 = 0 and d 4 = 0,
then
σ x = d 4 x y
(5.43)
σ y = 0
(5.44)
and
τ xy = −
d 4
2
y
2
(5.45)
But the beam is subjected to a constant shear force ‘P’ resulting on constant shear
stress distribution. Hence, superimposing the pure shear component (from second
degree polynomial), i.e., τ xy = −b 2 in Eq. (5.45), the resulting final stresses are
given by
σ x = d 4 x y
σ y = 0
τ xy = −b 2 −
d 4
2
y
2