## Search in the document preview

**Figure 8.2 Representation of uniaxial test specimen and resultant
stress state
**

x

y

These grips allow no displacement in the y-direction

--> v = 0 both @ x = 0, l The solution gives:

−νσ v = y ≠ 0 in general @ x = 0, l

E
**But**, “far” from the effects of load introduction, the solution φ holds.

Near the grips, biaxial stresses arise. Often failure occurs here.
(**Note**: not in a pure uniaxial field. This is a common problem with test
specimens.)

Fall, 2002

Example 2

Let’s now get a bit more involved and consider the…

Stress Distribution Around a Hole

**Figure 8.3 Configuration of uniaxially loaded plate with a hole
**

Large (“infinite”) plate subjected to uniform far - field tension

Since the “local specific” of interest is a circle, it makes sense to use polar coordinates.

By using the transformations to polar coordinates of Unit #7, we find:

1 ∂φ 1 ∂2φ σrr = r ∂r

+ r2 ∂θ2

+ V

∂2φ + Vσθθ =

∂r2

1 ∂2φ 1 ∂φ +

r σrθ = − r ∂ ∂θ r2 ∂θ

These are, again, defined such that equilibrium equations are automatically satisfied.

where: fr = − ∂V

fθ = − 1 ∂V

∂r r ∂θ

∂fθ + fθ 1 ∂frand V exists if: =

∂r r r ∂θ

The governing equation is again:

∇4φ = − Eα∇2 (∆T) − (1 − ν) ∇2 V for plane stress, isotropic

but the Laplace operator in polar coordinates is:

∇2 = ∂2

+ 1 ∂

+ 1 ∂2

∂r2 r ∂r r2 ∂θ2

Let’s go through the steps…

Step 1: Assume a φ(r, θ) “We know” the correct function is:

φ = [A0 + B0 lnr + C0r2 + D0r2 lnr] + A2r

2 + B

2 2 + C r4 + D2

cos2θ2 r

Does this satisfy equilibrium? **It must.
**

One can show it satisfies ∇4φ

However, it can also be shown that non-zero values of D0 result in mulitvalued displacements in the y-direction (v), so we must get D0 = 0

[don’t worry about this, just “accept” it…I do!]

Step 2: Determine stresses Performing the derivatives from the φ-σ relations in polar coordinates results in:

σrr = B

2 0 + 2C0 +

−2A2 −

6B 4

2 − 4D2 cos2θ

r r r2

σθθ = − B

2 0 + 2C0 +

2A2 +

6B 4

2 + 12C2r 2 cos2θ r r

2 inσrθ = 2A2 −

6B 4

2 + 6C2r 2 −

2D2 s 2θ r r

Note that we have a term involving r2. As r gets larger, the stresses would become infinite. This is not possible. Thus, the coefficient C 2 must be zero:

C2 = 0

So we have five constants remaining:

A2, B0, B2, C0, D2 we find these by…

Fall, 2002

Step 3: Satisfy the boundary conditions

What are the boundary conditions here?

• at the edge of the hole there are no stresses (stress-free edge)

⇒ @ r = a: σrr = 0, σθr = 0

• at the y-edge, there are no stresses

⇒ @ y = ± ∞: σyy = 0, σyx = 0

• at the x-edges, the stress is equal to the applied stress and there is no strain

⇒ @ x = ± ∞: σxx = σ0 , σxy = 0

Since we are dealing with polar coordinates, we need to change the last two sets of B. C.’s to polar coordinates. We use: σ̃ αρ = l2̃ σ l ̃βγ lσγ

and look at r = ∞

σ̃ xx = σrr = cos 2 θ σo

σ̃ yy = σθθ = sin 2θ σo

σ̃ xy = σrθ = − sinθ cos θ σo

**Figure 8.4 Representation of local rotation of stresses from polar to
Cartesian system
**

We use the double angle trigonometric identities to put this in a more convenient form:

**Unit 8 - p. 23 **

cos2 θ =� 1 (1 + cos2θ) 2

1 sinθ cos θ = sin2θ

2

sin2 θ =� 1 (1 − cos2θ) 2

Thus at r = ∞

σ σoσ = o + cos2θrr 2 2

σ σrθ = −

o sin2θ 2

Why don’t we include σθθ? At the boundary in polar coordinates, this stress does not act on the edge/boundary.

**Figure 8.5 Stress condition at boundary of hole
**

So we (summarizing) appear to have 4 B. C.’s: σ σ

σrr = o + o cos2θ @ r = ∞

2 2

σ σrθ = −

o sin2θ @ r = ∞ 2

σrr = 0 @ r = a

σrθ = 0 @ r = a

And we have 5 constants to determine.

What happened?

The condition of σrr at r = ∞ is really two B. C.’s

− a constant part − a part multiplying cos 2θ

Going through (and skipping) the math, we end up with:

σ σ

θ

σ σ

θ

σ σ

θθ

θ

rr o

2

2 o

2

2

4

4

o 2

2 o

4

4

o 2

2 1

a

r 2 1 4

a

r 3

a

r

2 1

a

r 2 1 3

a

r

2 1 2

a

r

= −

+ − +

=

− +

= − +

cos

cos

2

2

r 2

4

4 3 a

r in −

s 2θ

σ

σ +

for: …

**Figure 8.6 Polar coordinate configuration for uniaxially loaded plate
with center hole
**

So we have the solution to find the stress field around a hole. Let’s consider one important point. Where’s the largest stress?

At the edge of the hole. Think of “flow” around the hole:

**Figure 8.7 Representation of stress “flow” around a hole **

@ θ = 90°, r = a

σ a2 σ a4 oσθθ = σxx = 2 o

1 + a2

− 2

1 + 3 a4

(−1)

= 3 σo

Define the:

Stress Concentration Factor (SCF) = local stress far - field stress

SCF = 3 at hole in isotropic plate

The SCF is a more general concept. Generally the “sharper” the discontinuity, the higher the SCF.

The SCF will also depend on the material. For orthotropic materials, it depends on Ex and Ey getting higher as Ex/Ey increases. In a uni-directional composite, can have SCF = 7.

Can do stress functions for orthotropic materials, but need to go to complex variable mapping

--> (See Lekhnitskii as noted earlier)
**Unit 8 - p. 28 **

Example 3 -- Beam in Bending (we’ll save for a problem set or a recitation)

There are many other cases (use earlier references) − circular disks − rotary disks

.

. “classic” cases

But…what’s really the point? Are these still used?

Yes!

This is a very powerful technique which is especially well-suited for preliminary design and exploratory development

− parametric study − know assumptions and resulting limitations and then

interpret results accordingly
− linear solutions --> can use **principle of superposition
**− can find many solutions in books

(See attached page)

1.

2.

3.

Fall, 2002

Roark, Raymond J., “Roark’s Formulas for Stresses and Strain, 6th Ed.,” New York, McGraw-Hill, 1989.

Peterson, Rudolph E., “Stress Concentration Factors: Charts and Relations Useful in Making Strength Calculations for Machine Parts and Structural Elements,” New York, Wiley, 1974.

Pilkey, Walter D., “Formulas for Stress, Strain, and Structural Matrices,” New York, John Wiley, 1994.