# Stress calculations in a perforated paper

You have a sheet of paper (torn out of a good quality foolscap notebook) as shown above, and you start pulling it apart with both your hands (forces indicating by the blue arrows). Its difficult to tear the paper apart this way --- it has a very high tensile strength, I assume. If you try this at home, you are bound to fail.

Now if you make tiny perforations in the paper (indicated by red circles), pull it apart the same way, you'll notice that its very easy to tear apart the paper. And the line of tearing will definitely have a few circles on them.

Is this happening because of the high stresses that are formed around the circle? Can someone give a good mathematical as well as verbal explanation of this phenomenon.

• See en.wikipedia.org/wiki/Stress_concentration. The problem is getting a tear started. The perforations concentrate the stress and make it easier to initiate a tear. Good luck with getting a mathematical treatment - I suspect you'll need a finite element approach. – John Rennie Feb 12 '14 at 18:16
• You would need to include the fact that paper is not isotropic, because of the way the fibers of cellulose line up in the direction of motion of the paper through the paper-making machine... – DJohnM Feb 12 '14 at 18:41
• @JohnRennie To the rescue as always. Any good books on FEM you have on mind? – Black Dagger Feb 13 '14 at 6:47
• @vardhanamdaga: I'm afraid I know absolutely nothing about finite element analysis software. It might be worth asking on scicomp.stackexchange.com. – John Rennie Feb 13 '14 at 7:54

Yes. The tear is initiated at stress concentrations around the holes, where stress is highest. After initiation, the tear continues to propagate along the line of highest stress.

Stress is a function of force and geometry ($\sigma_{n} = \frac {F}{A_{n}}$). In a piece of paper without holes, the stress is uniform, and the paper tears when stress exceeds the ultimate tensile strength of the paper ($\sigma_{n} \gt \sigma_{UT, paper}$).

## Average Stress- the most basic explanation

When holes are present, they effectively reduce the cross sectional area ($A_{eff} = A_{o} - n_{holes}A_{hole}$)* that transmits force. Static equilibrium requires that the stress increases proportional to the reduction in area, visualized below.

$$\therefore \sigma_{eff} = \frac {F}{A_{eff}}$$

Because $\sigma_{eff} \gt \sigma_{n}$, it follows that the paper will tear along cross sections where holes are present. While this correctly calculates the average stress, it assumes the stress between holes is uniform (and equal to the average stress). In reality, the stress profile between holes is not uniform, as discussed in the following.

* Note that 'area' refers to cross sectional area

## Stress Concentrations- approximate stress state

Stress concentrations describe the stress state at abrupt changes in geometry, where the stress profile is non-uniform. Analogous to lines of pressure in (laminar) fluid flow around an immersed body, lines of force 'flow' through geometry, becoming concentrated around the holes (where no material exists to transmit force).

A stress concentration factor ($K_{s}$) is applied to the nominal stress to calculate the maximum stress, where $\sigma_{max} = K_{s}\sigma_{n}$. Stress concentration factors depend on geometry and are determined analytically, or by experimental data. From the analytical solution of an infinite plate with a single hole, loaded uniaxially, $K_{s} = 3$. More applicable, from Peterson's Stress Concentration Factors, an infinite plate with a linear hole pattern: $$\therefore \sigma_{SC} \approx 3 \sigma_{n}$$

## Finite Element Method- complete stress state

Complete, accurate solutions are readily obtained by Finite Element Methods (FEM), where analytic solutions are not possible with complex geometry. With assumed dimensions (similar to the posed problem), $K_{s} = 3.75$, determined from the converged solution shown below (where 'brighter' colors indicate higher stress, consistent with the solution given by stress concentration).

$$\therefore \sigma_{FEM} = 3.75 \sigma_{n}$$

All solution methods demonstrate that stress increases in cross sections where holes are present: When stress at any location in the paper exceeds the paper's strength ($\sigma_{max} \gt \sigma_{UT, paper}$), a tear is initiated and follows the line of highest stress- this validates your insight.

There are several other considerations, not discussed, but listed under 'References':

• Paper is not a ductile material- it does not plastically deform
• Paper is not (usually) isotropic- it is orthotropic, where its strength depends on orientation
• Fracture mechanics (founded on the assumption that all materials have defects)- material irregularities act as micro stress concentrations

References:

Mechanical Properties of Paper- Basic