140
5 Two-Dimensional Problems in Cartesian Co-ordinate System
The couple due to the shearing stresses τ xy is
M =
d 4
2
h
2 L2h + L
+h
−h
d 4
2
y
2 dy
= −
2
3
d 4 Lh
3
Therefore, the net couple on the plate is zero.
From the above consideration, it is to be noted that for stress function in the
form of polynomials of second and third degree, all constants are arbitrary since
the Biharmonic equation is satisfied (i.e. ∇
4
φ = 0) regardless of the values of
the constants. However, for polynomials of higher order, the Biharmonic equation
is satisfied under certain conditions and all the constants are not arbitrary. As an
example, the fourth degree polynomial equation considered here is satisfied when
e 4 = −(a 4 + 4 c 4 ).
(e) Polynomial of the Fifth Degree
Let
φ 5 =
a 5
5(4)
x
5
+
b 5
4(3)
x
4 y +
c 5
3(2)
x
3 y
2
+
d 5
3(2)
x
2 y
3
+
e 5
4(3)
x y
4
+
f 5
5(4)
y
5
The corresponding stress components are given by
σ x =
∂
2
φ 5
∂ y 2 =
c 5
3
x
3
+ d 5 x
2 y − (2c 5 + 3a 5 )x y
2
−
1
3
(b 5 + 2d 5 )y
3
σ y =
∂
2
φ 5
∂ x 2 = a 5 x
3
+ b 5 x
2 y + c 5 x y
2
+
d 5
3
y
3
τ xy = −
∂
2
φ 5
∂ x∂ y
= −
1
3
b 5 x
3
− c 5 x
2 y − d 5 x y
2
+
1
3
(2c 5 + 3a 5 )y
3
Here the coefficients a 5 , b 5 , c 5 , d 5 are arbitrary, and in adjusting them we obtain
solutions for various loading conditions of the beam.
Now, if all coefficients, except d 5 , equal to zero, we find
σ x = d 5
x
2 y −
2
3
y
3
σ y =
1
3
d 5 y
3
τ xy = −d 5 x y
2
Case (i) The normal forces are uniformly distributed along the longitudinal sides of
the beam.