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ElasticityRectangular_02_StressFunction.pdf
Section 3.2
3.2 The Stress Function Method
An effective way of dealing with many two dimensional problems is to introduce a new
“unknown”, the Airy stress function φ , an idea brought to us by George Airy in 1862.
The stresses are written in terms of this new function and a new differential equation is
obtained, one which can be solved more easily than Navier’s equations.
3.2.1
The Airy Stress Function
The stress components are written in the form
σ xx =
∂ 2φ
∂y 2
∂ 2φ
∂x 2
∂ 2φ
=−
∂x∂y
σ yy =
σ xy
(3.2.1)
Note that, unlike stress and displacement, the Airy stress function has no obvious physical
meaning.
The reason for writing the stresses in the form 3.2.1 is that, provided the body forces are
zero, the equilibrium equations are automatically satisfied, which can be seen by
substituting Eqns. 3.2.1 into Eqns. 2.2.3 {▲Problem 1}. On this point, the body forces,
for example gravitational forces, are generally very small compared to the effect of
typical surface forces in elastic materials and may be safely ignored (see Problem 2 of
§2.1). When body forces are significant, Eqns. 3.2.1 can be amended and a solution
obtained using the Airy stress function, but this approach will not be followed here. A
number of examples including non-zero body forces are examined later on, using a
different solution method.
3.2.2
The Biharmonic Equation
The Compatability Condition and Stress-Strain Law
In the previous section, it was shown how one needs to solve the equilibrium equations,
the stress-strain constitutive law, and the strain-displacement relations, resulting in the
differential equation for displacements, Eqn. 3.1.4. An alternative approach is to ignore
the displacements and attempt to solve for the stresses and strains only. In other words,
the strain-displacement equations 3.1.2 are ignored. However, if one is solving for the
strains but not the displacements, one must ensure that the compatibility equation 1.3.1 is
satisfied.
Eqns. 3.2.1 already ensures that the equilibrium equations are satisfied, so combine now
the two dimensional compatibility relation and the stress-strain relations 3.1.1 to get
{▲Problem 2}
Solid Mechanics Part II
46
Kelly
Section 3.2
⎛ ∂ 4φ
∂ 4φ
∂ 4φ ⎞
plane stress : ⎜⎜ 4 + 2 2 2 + 4 ⎟⎟ = 0
∂x ∂y
∂y ⎠
⎝ ∂x
⎛ ∂ 4φ
∂ 4φ
∂ 4φ ⎞
plane strain : ⎜⎜ 4 + 2 2 2 + 4 ⎟⎟(1 − ν ) = 0
∂x ∂y
∂y ⎠
⎝ ∂x
(3.2.2)
Thus one has what is known as the biharmonic equation:
∂ 4φ
∂ 4φ
∂ 4φ
+ 2 2 2 + 4 = 0 The biharmonic equation
∂x 4
∂x ∂y
∂y
(3.2.3)
The biharmonic equation is often written using the short-hand notation ∇ 4φ = 0 .
By using the Airy stress function representation, the problem of determining the stresses
in an elastic body is reduced to that of finding a solution to the biharmonic partial
differential equation 3.2.3 whose derivatives satisfy certain boundary conditions.
Note that the biharmonic equation is independent of elastic constants, Young’s modulus E
and Poisson’s ratioν . Thus for bodies in a state of plane stress or plane strain, the stress
field is independent of the material properties, provided the boundary conditions are
expressed in terms of tractions (stress) 1; boundary conditions on displacement will bring
the elastic constants in through the stress-strain law. Further, the plane stress and plane
strain stress fields are identical.
3.2.3
Some Simple Solutions
Clearly, any polynomial of degree 3 or less will satisfy the biharmonic equation. Here
follow some elementary examples.
(i)
φ = Ay 2
one has σ xx =
∂ 2φ
= 2 A, σ yy = σ xy = 0 , a state if uniaxial tension
∂y 2
(ii)
φ = Bxy
here, σ xx = σ yy = 0, σ xy = − B , a state of pure shear
(iii)
φ = Ay 2 + Bxy
here, σ xx = 2 A, σ yy = 0, σ xy = − B , a superposition of (i) and (ii)
1
technically speaking, this is true only in simply connected bodies, i.e. ones without any “holes”, since
problems involving bodies with holes have an implied displacement condition (see, for example, Barber
(1992), §2.2).
Solid Mechanics Part II
47
Kelly
Section 3.2
3.2.4
Pure Bending of a Beam
Consider the bending of a rectangular beam by a moment M 0 , as shown in Fig. 3.2.1.
The elementary beam theory predicts that the stress σ xx varies linearly with y, Fig. 3.2.1,
with the y = 0 axis along the beam-centre, so a good place to start would be to choose, or
guess, as a stress function φ = Cy 3 , where C is some constant to be determined. Then
σ xx = 6Cy, σ yy = 0, σ xy = 0 , and the boundary conditions along the top and bottom of
the beam are clearly satisfied.
b
M0
M0
Figure 3.2.1: a beam in pure bending
The moment and stress distribution are related through
+b
+b
−b
−b
M 0 = ∫ σ xx ydy = 6C ∫ y 2 dy = 4Cb 3
(3.2.4)
and so C = M 0 / 4b 3 and σ xx = 3M 0 y / 2b 3 . The fact that this last expression agrees with
the elementary beam theory ( σ = − My / I with I = 2b 3 h / 3 , where h is the depth “into
the page”) shows that that the beam theory is exact in this simple loading case.
Assume now plane strain conditions. In that case, there is another non-zero stress
component, acting “perpendicular to the page”, σ zz = ν (σ xx + σ yy ) = 3Mνy / 2b 3 . Using
Eqns. 3.1.1b,
⎛ 1 − ν 2 3M ⎞
1 +ν
⎟ y = αy, say
(1 − ν )σ xx − νσ yy = ⎜⎜
3 ⎟
E
E
2
b
⎝
⎠
1 +ν
⎛ v(1 + ν ) 3M ⎞
=
− νσ xx + (1 − ν )σ yy = ⎜ −
⎟ y = β y, say
E
E 2b 3 ⎠
⎝
ε xx =
ε yy
[
[
]
]
(3.2.5)
and the other four strains are zero.
As in §1.2.4, once the strains have been found, the displacements can be found by
integrating the strain-displacement relations. Thus
Solid Mechanics Part II
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Kelly
Section 3.2
∂u x
= αy
∂x
→ u x = αxy + f ( y )
ε xx =
ε yy =
∂u y
∂y
= βy
(3.2.6)
→ u y = β y + g ( x)
1
2
2
∂u y ⎞
1 ⎛ ∂u
1
⎟ = (αx + f ′( y ) + g ′( x) ) ≡ 0
ε xy = ⎜⎜ x +
∂x ⎟⎠ 2
2 ⎝ ∂y
→ g ′( x) + αx = − f ′( y )
Therefore f ′( y ) must be some constant, − C say, so f ( y ) = −Cy + A , and
g ( x) = Cx − 12 αx 2 + B . Finally,
u x = αxy − Cy + A
u y = − 12 αx 2 + 12 βy 2 + Cx + B
(3.2.7)
which are of the form 1.2.18. For the case when the mid-point of the beam is fixed, so
has no translation, u x (0,0) = u y (0,0) = 0 , and if it has no rotation there, ω z (0,0) = 0 , then
the three arbitrary constants are zero, and
u x = αxy
u y = − 12 αx 2 + 12 β y 2
3.2.5
(3.2.8)
A Cantilevered Beam
Consider now the cantilevered beam shown in Fig. 3.2.2. The beam is subjected to a
uniform shear stress σ xy = τ over its free end, Fig. 3.2.2a. The boundary conditions are
σ xx (0, y ) = 0, σ xy (0, y ) = τ , σ yy ( x,±b) = σ xy ( x,±b) = 0
(3.2.9)
It is difficult, if not impossible, to obtain concise expressions for stress and strain for
problems even as simple as this2. However, a concise solution can be obtained by
relaxing one of the above conditions. To this end, consider the similar problem of Fig.
3.2.2b – this beam is subjected to a shear force F, the resultant of the shear stresses. The
applied force of Fig. 3.2.2b is equivalent to that in Fig. 3.2.2a if
+b
∫σ
xy
(0, y )dy = F
(3.2.10)
−b
2
an exact solution will usually require an infinite series of terms for the stress and strain
Solid Mechanics Part II
49
Kelly
Section 3.2
This is known as a weak boundary condition, since the stress is not specified in a pointwise sense along the boundary – only the resultant is. However, from Saint-Venant’s
principle (Part I, §3.3.2), the stress field in both beams will be the same except for in a
region close to the applied load.
y
τ
y
b
x
F
(a )
x
( b)
Figure 3.2.2: A cantilevered beam subjected to; (a) a uniform distribution of shear
stresses along its free end, (b) a shear force along its free end
The elementary beam theory predicts a stress σ xx = − My / I = Fxy / I . Thus a good place
to start is to choose the stress function φ = αxy 3 , where α is a constant to be determined.
The stresses are then
σ xx = 6αxy, σ yy = 0, σ xy = −3αy 2
(3.2.11)
However, it can be seen that σ xy ( x,±b) = −3αb 2 ≠ 0 . To offset this, one can superimpose
a constant shear stress 3αb 2 , in other words amend the stress function to
φ = αxy 3 − 3αb 2 xy
(3.2.12)
The boundary conditions are now satisfied and, from Eqn. 3.2.10,
α=
F
4b 3
(3.2.13)
and so
σ xx =
3.2.6
(
3F
3F
xy, σ yy = 0, σ xy = 3 b 2 − y 2
3
2b
4b
)
(3.2.14)
Problems
1. Verify that the relations 3.2.1 satisfy the equilibrium equations 2.2.3.
2. Derive Eqn. 3.2.2.
3. A large thin plate is subjected to certain boundary conditions on its thin edges (with
its large faces free of stress), leading to the stress function
Solid Mechanics Part II
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Kelly
Section 3.2
φ = Ax 3 y 2 − Bx 5
(i)
(ii)
(iii)
(iv)
(v)
(vi)
use the biharmonic equation to express A in terms of B
calculate all stress components
calculate all strain components (in terms of B, E, ν )
derive an expression for the volumetric strain, in terms of B, E, ν , x and y.
check that the compatibility equation is satisfied
check that the equilibrium equations are satisfied
4. A very thick component has the same boundary conditions on any given crosssection, leading to the following stress function:
φ = x 4 y + 4x 2 y 3 − y 5
(i)
is this a valid stress function, i.e. does it satisfy the biharmonic equation?
calculate all stress components (with ν = 1 / 4 )
(ii)
(iii) calculate all strain components
find the displacements
(iv)
specify any three displacement components which will render the arbitrary
(v)
constant displacements of (iv) zero
5. For the cantilevered beam discussed in §3.2.5, evaluate the resultant shear force and
moment on an arbitrary cross-section x = x . Are they as you expect? (You will find
that the beam is in equilibrium, as expected, since the equilibrium equations have
been satisfied.)
6. For the cantilevered beam discussed in §3.2.5, evaluate the strains and displacements,
assuming plane stress conditions.
Note: to evaluate the three arbitrary constants of integration, one would be tempted to
apply the obvious u x = u y = 0 all along the built-in end. However, since only weak
boundary conditions were imposed, one cannot enforce these strong conditions (try
it). Instead, apply the following weaker conditions: (i) the displacement at the built-in
end at y = 0 is zero ( u x = u y = 0 ), (ii) the slope there, ∂u y / ∂x , is zero.
7. Show that the stress function
p
φ=
− 20 y 3 (L2 − x 2 ) − 4 y 5 − 15h 2 x 2 y + 2h 2 y 3 − 5h 3 x 2
3
20h
satisfies the boundary conditions for the simply supported beam subjected to a
uniform pressure p shown below. Check the boundary conditions in the weak (SaintVenant) sense on the shorter left and right hand sides (for both normal and shear
stress). Since the normal stress σ xx is not zero at the ends, but only its resultant,
check also that the moment is zero at each end.
[
]
p
y
x
Lp
L
Solid Mechanics Part II
Lp
h
L
51
Kelly
Section 3.2
Note that the elementary beam theory predicts an approximate flexural stress but an
exact shear stress:
⎞
6p
6 p ⎛ h2
σ xx = − 3 y (L2 − x 2 ), σ xy = 3 x⎜⎜ − y 2 ⎟⎟
h
h ⎝ 4
⎠
8. Consider the dam shown in the figure below. Assume first a general cubic stress
function
1
1
1
1
φ = C1 x 3 + C 2 x 2 y + C3 xy 2 + C 4 y 3
6
2
2
6
Apply the boundary conditions to determine the constants and hence the stresses in
the dam, in terms of ρ , the density of water. (Use the stress transformation equations
for the sloped boundary and ignore the weight of the dam.)
[Just consider the effect of the water; to these must be added the stresses resulting
from the weight of the dam itself, which are given by
⎡ 1
⎤
σ xx = 0, σ yy = ρ s g ⎢
x − y ⎥, σ xy = 0
⎣ tan β
⎦
where ρ s is the density of the dam material.]
x
β
ρ
y
Solid Mechanics Part II
52
Kelly
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