Airy Example

7/31/2019 Airy Example

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EXAMPLE: Airy’s stress function — Cantilever

There are known forms of Airy’s stress function that are solutions to thepartial differential stress equation. One is that for a cantilever:

φ(x, y) = Dxy3 + Bxy. (1)

We will consider a cantilever of unit depth (rectangular cross-section) with ashear loading on the free end, such that the average shear stress is equivalentto a vertical load, P . Given this information, the task is firstly to find theconstants D and B from the boundary conditions. The solution is that tothe problem where the load is applied as a plate on the end being pulleddown with force P (diagram given in lectures). Note that x = 0 is at the freeend of the beam, x = L is the fixed end and the top and bottom surfaces of the beam are at y = ±c respectively (saves me drawing a diagram — see inclass diagram if you need one).First determine stresses:

σx = 6Dxyσy = 0

σxy = −3Dy2−B.

Boundary Conditions:

1. σxy = 0 on top and bottom (at y = ±c, where 2c is the thickness of the beam). This leads to

0 = −3Dc2 −B → B = −3Dc2. (2)

2. When using this technique, it is not possible to precisely specify apoint load (it is a singularity). So, some clever people came up wiht

equation 1 to very effectively approximate a point load at the end of the cantilever through a complicated end shear stress distribution. If that makes no sense, it will suffice to know that at the end of thebeam, the loading is such that

(Area)× (Avg. shear stress on free end) = (Effective point load at end)

(2c)×

1

2c

c

−c

σxydy

=

c

−c

σxydy = P

c

−c

(−3Dy2−B)dy = P

.

.

.

D =P

4c3

∴ B = −3P

4c

1



Therefore, we have Airy’s stress function and stresses:

φ(x, y) = −P x

4

3

y

c−

y3

c3

σx =3

2c3

P xy

σxy = −3

4c3P y2 +

3P

4c

Now, the second moment of area for a unit depth beam with thickness 2ccan be shown to be

I =

c

−c

y2dy

=2

3c3.

Inserting this into the stress equations:

σx = P I xy

σxy =P

2I

c2 − y2

So we have stress, what is the strain? Use Hooke’s Law to get:

εx =σxE −

νσyE

=P

EI xy

εy =σyE −

νσx

E = −

νP

EI xy

εxy =(1 + ν )

E σxy =

P

4IG

c2 − y2

,

where 2G = E/(1 + ν ) (I’m sure about this now!).From the strains, we can get deflections, but this is hard work. Here we go...

u1 = x-displacement =

εxdx =

P x2y

2EI + f 1(y)

u2 = y-displacement =

εydy = −

νPxy2

2EI + f 2(x)

Need to find f 1 and f 2. Use compatability equation:

εxy =1

2

∂u1

∂y+

∂u2

∂x

P 2IG

c2 − y2

= P

2EI x2 + df 1

dy− νP y

2

2EI + df 2

dx

−P

2EI x2−

df 2dx

= −P

2IG

c2 − y2

+

df 1dy−

νP y2

2EI

(3)

2



Now we have a separation of variables problem since the left hand side is afunction of x only and the right hand side is a function of y only. The onlysolution is that both x and y are constant and the same constant. Let’s callthat constant −C 1.So,

df 2dx

= C 1 −P x2

2EI

f 2 = C 1x−P x3

6EI + C 4

∴ u2 = −νP

2EI xy + C 1x−

P x3

6EI + C 4

We will only consider this component of displacement, because we are inter-ested in the deflection of the cantilever. The task is now to find the constantsC 1 and C 4 through boundary conditions. We will need two boundary con-ditions (obviously).

1. u2 = 0 at y = 0, x = L.

0 = −P L3

6EI + C 1L + C 4

C 1 =P L2

6EI −

C 4L

2. ∂u2/∂x = 0 at x = L.

∂u2

∂x= −

νP y

2EI −

P x2

2EI + C 1

0 = −

P L2

2EI + C 1

C 1 =P L2

2EI

∴ C 4 = −P L3

3EI

So in the end we get the deflection of the cantilever to be:

u2 = −νP

2EI xy −

P

6EI x3 +

P L2

2EI x−

P L3

3EI . (4)

The maximum deflection (which we will see plenty of during this

course) is

u2max = −P L3

3EI , (5)

which occurs at x = 0, of course.

3

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Airy Example