Why don't light bulbs implode/ break in it contains a vacuum?

[Elite Engineer]

Obviously light bulbs don't shatter, but my questions is, why DON'T they? What specifically prevents the higher external pressure from

caving in the bulb?

 

~ee

[StringJunky]

i don't think they they are vacuumed but have an inert gas in them which has the same effect in preventing oxidation of the filament.

[Janus]

You don't need them to contain an inert gas to keep them from caving in.(after all, vacuum tubes don't collapse in on themselves.  The trick is that the pressure is equally distributed over its surface. In this case the force is a compressive force and the compressive strength of glass is actually in the 1000's of psi.  A glass bulb will shatter when hit because the force is not evenly distributed over its surface.

[StringJunky]

24 minutes ago, Janus said:

You don't need them to contain an inert gas to keep them from caving in.(after all, vacuum tubes don't collapse in on themselves.  The trick is that the pressure is equally distributed over its surface. In this case the force is a compressive force and the compressive strength of glass is actually in the 1000's of psi.  A glass bulb will shatter when hit because the force is not evenly distributed over its surface.

That's interesting. cheers.

[studiot]

5 hours ago, Elite Engineer said:

Obviously light bulbs don't shatter, but my questions is, why DON'T they? What specifically prevents the higher external pressure from

caving in the bulb?

I think Janus' post was a bit short and may lead to false impressions so here is a longer version.

I was the Romans who are credited with being the first to apply the principle of the arch, although they didn't properly understand it. They believed the arch had to be circular in shape to work successfully.

To emphasise that this is not true I have used quite pointy ellipses in my diagrams.

Fig1 shows the principle that direct normal (perpendicular) loads can be transformed into tangential stresses within a body.

The figure shows a complete ellipse, as with an egg, but again this is not necessary, although it explains how radial forces can be directed into a self sustaining loop of circumferential forces.

But a light bulb is not a complete loop of glass.

So Fig 2 shows that we must suppl;y stout abutments or springings at the palces where the circumference terminates.
These abutments must be able to supply the reaction forces to the outward pointing circumferential forces at each termination point of the circumference.
There must be at least two such points and they are labelled A and B in the figure.
Note the abutments are shown as faces perpendicular to the circumference.

In the case of the light bulb the abutments are provided by the sides/faces of the metal base of the bulb.
 

Obviously we can't block off the space under an arch bridge so bridge arches require separate abutments.

Equally we know that bridges support quite heavy loads travelling over them so the system is capable of supporting unevenly distributed loads.

So it cnnot be the differnce in pressure alone that breaks the bulb.

So how does it break?

Well Fig3 shows how the situation changes when the load is unevenly distributed.
So far all the forces have been direct forces.
Just two forces F1 and F2 are shown with one greater than the other, and they are still direct forces.
Since the forces act in the same direction they do not form a couple.

But because one is greater than the other their moments about the midpoint between them are not equal.

So uneven loading leads to bending moments within the supporting circumference shell.

Glass is particularly weak in bending.

So it breaks.

 

[John Cuthber]

Glass is actually very strong, especially in compression (most things are better in compression than tension).

Bulbs are typically filled with a low pressure of some inert gas- often argon (because it is cheap). Krypton and xenon are also used.

[studiot]

53 minutes ago, John Cuthber said:

Glass is actually very strong, especially in compression (most things are better in compression than tension).

Bulbs are typically filled with a low pressure of some inert gas- often argon (because it is cheap). Krypton and xenon are also used.

Strong maybe, but it is not tough.

 

The glass shatters not because it fails in compression - it does not.

It fails because it is an exceedingly brittle material with a very low fracture toughness.

The glass fails suddenly in bending because bending introduces tension and tension introduces cracks.

And cracks propagate at an alarming rate in materials of low K_1c.

 

Edited October 9, 2017 by studiot

[John Cuthber]

True, it has a "work of fracture" of about sod all.

Steel light bulbs would be much less likely to shatter if dropped. (And they would be about as much use as a photochromic light bulb)

 

[studiot]

I also didn't mention this, to avoid information overload, but the thinness of a light bulb glass is also significant.

The resisting moment is generated by a couple comprising the tension and compression separated by a lever arm slightly smaller than the thickness of the shell.

Since the lever arm is so small, the forces (and thus the tension)  must be correspondingly larger to create the necessary moment.

[Elite Engineer]

17 hours ago, studiot said:

I think Janus' post was a bit short and may lead to false impressions so here is a longer version.

I was the Romans who are credited with being the first to apply the principle of the arch, although they didn't properly understand it. They believed the arch had to be circular in shape to work successfully.

To emphasise that this is not true I have used quite pointy ellipses in my diagrams.

Fig1 shows the principle that direct normal (perpendicular) loads can be transformed into tangential stresses within a body.

The figure shows a complete ellipse, as with an egg, but again this is not necessary, although it explains how radial forces can be directed into a self sustaining loop of circumferential forces.

But a light bulb is not a complete loop of glass.

So Fig 2 shows that we must suppl;y stout abutments or springings at the palces where the circumference terminates.
These abutments must be able to supply the reaction forces to the outward pointing circumferential forces at each termination point of the circumference.
There must be at least two such points and they are labelled A and B in the figure.
Note the abutments are shown as faces perpendicular to the circumference.

In the case of the light bulb the abutments are provided by the sides/faces of the metal base of the bulb.
 

Obviously we can't block off the space under an arch bridge so bridge arches require separate abutments.

Equally we know that bridges support quite heavy loads travelling over them so the system is capable of supporting unevenly distributed loads.

So it cnnot be the differnce in pressure alone that breaks the bulb.

So how does it break?

Well Fig3 shows how the situation changes when the load is unevenly distributed.
So far all the forces have been direct forces.
Just two forces F1 and F2 are shown with one greater than the other, and they are still direct forces.
Since the forces act in the same direction they do not form a couple.

But because one is greater than the other their moments about the midpoint between them are not equal.

So uneven loading leads to bending moments within the supporting circumference shell.

Glass is particularly weak in bending.

So it breaks.

 

Superb explanation!

[Area54]

20 hours ago, studiot said:

The glass shatters not because it fails in compression - it does not.

It fails because it is an exceedingly brittle material with a very low fracture toughness.

The glass fails suddenly in bending because bending introduces tension and tension introduces cracks.

And cracks propagate at an alarming rate in materials of low K_1c.

 

I am not sure if this is universally true, but my understanding is that rocks fail only in tension. That tension may be a local effect caused by compression, but the point of failure is in a zone of tension. Is my understanding correct?

[studiot]

On 10/10/2017 at 9:10 AM, Area54 said:

I am not sure if this is universally true, but my understanding is that rocks fail only in tension. That tension may be a local effect caused by compression, but the point of failure is in a zone of tension. Is my understanding correct?

Hello Area54 this is a big question. I'm not sure what your understanding is but this is often the case.
But, as ever, life is actually more complicated.

I admit to being a bit glib with my Fig3, but it seems to have got the main point across for light bulbs.

Details of rock and soil mechanics really shoulf be in another thread but here is an outline.

We haven't really discussed the spatial distribtion of the stress systems, but the work so far has been about what are called uniaxial stresses.
That is stress applied in one dimension.

Uniaxial compression is always accompanied by induced tension in the other two spatial axes - these are often called bursting forces or stresses.
This occurs in all materials, not just rocks.

This is amply demonstrated in the well known cylinder splitting test or the 'Brazilian test'.
Here a cylinder of rock or soil is subject to a uniaxial compression applied across a diameter, untill it splits across the diameter.
Cylinders are good because that is what core drills cut out.

The test is a measure of the tensile strength of the material, although a compressive load is applied.

This is an exmaple of what you are talking about.

 

If a triaxial stress compressive is applied (simultaneous compression on all three axes) then the rock cannot fail structurally in that it will continue to carry the load as it has nowhere to go.
It does however undergo plastic flow and possibly heating in these circumstance.

Compressive failure may occur in bending where the material in the compression zone is crushed to a powder before any cracks in the tensile zone develop.
This is the dangerous situation that reinforced concrete design codes seek to avoid by forbidding what is known as 'over reinforcement'.
Failure in this mode can be sudden and without warning, catastrophic, even explosive.

Of course cracks can't develop in compression zones as the compression closes them.

These are a few modes of failure, if you would like to discuss these and others in rocks we will start a new thread.

If you would like to expand on my explanation for glass light bulbs we can do so here.

 

 

[Mordred]

lol side note just read and article involving triaxial stress involvement in DM. Just found it amusing to see the term being applied here at the same time roughly lol. Very few papers will write triaxial stress.

 Typically they just write stress tensor. Same basic physics apply.

Edited October 12, 2017 by Mordred

[Elite Engineer]

Been thinking about this for a bit since I posted the topic. If you were drop to a glass oval and a glass sphere, from say 50 cm onto a wooden floor,

would the oval not shatter but the sphere would? My understanding is that the oval will distribute the force around evenly while the sphere will not

[studiot]

9 hours ago, Elite Engineer said:

Been thinking about this for a bit since I posted the topic. If you were drop to a glass oval and a glass sphere, from say 50 cm onto a wooden floor,

would the oval not shatter but the sphere would? My understanding is that the oval will distribute the force around evenly while the sphere will not

I am assuming solid glass objects.

I am not sure about this prognosis. Glass marbles are spherical and pretty tough cookies.

I can't remember ever seeing one shatter when dropped from twice that height onto a concrete floor. Chip maybe, but not shatter.

Anyway this is all about contact forces/contact stresses, not stress distribution.

Contact stresses are generally much higher than ordinary stresses in a body, but they are concentrated over the contact area.

Since this is an impact loading the impact force will also be much higher than the resting force due to the weight of the glass object.

Let P be the resting weight force
a be the contact area
C_E be  a constant due to the poisson's ratio and elasticities of the glass and the floor
D is the diameter of the glass ball at the contact point.

Then Roark gives the following formulae for [math]{{\sigma _t}}[/math], the maximum tensile stress due to contact.

[math]\max \left( {{\sigma _t}} \right) = 0.2\frac{P}{{\pi {a^2}}}[/math]

Where

[math]{C_E} = \frac{{1 - {\nu _1}^2}}{{{E_1}}} + \frac{{1 - {\nu _2}^2}}{{{E_2}}}[/math]

and

[math]a = 0.721\sqrt[3]{{PD{C_E}}}[/math]

You can see from these equations that the max stress is inversely proportional to the cube root of the radius of curvature of the glass ball.

A spherical ball will have a constant radius of curvature, but an oval one, of the same weight,  will have a pointy end of greater curvature and a flatter side of lesser curvature.
Further, most of the weight  volume and surface area, will be concentrated in the equatorial, flatter regions of the oval ball.

So presumabably the oval ball is more likely to land on a point of lower curvature than the equivalent spherical one, thereby suffering a lower impact stress.

 

 

 

11 hours ago, Mordred said:

lol side note just read and article involving triaxial stress involvement in DM. Just found it amusing to see the term being applied here at the same time roughly lol. Very few papers will write triaxial stress.

 Typically they just write stress tensor. Same basic physics apply.

The use of triaxial is common in earth scineces as is the splitting of the stress tensor into two tensors.

Engineers in particular know that whilst it is convenient for theoreticians to collect everything together into a nice compact package, as soon as to want to work the numbers in a real world situation you have to break them all apart.

In the words of a wise old engineer

"You cannot avoid the arithmetic"

Not long after I joined here there was a long discussion about just this, with a Russian who was doing a Phd in ice Rheology.

[math]T = {T_m} + {T_D}[/math]

 

[math]{T_m} = \left[ {\begin{array}{*{20}{c}}{{\sigma _m}} & 0 & 0  \\0 & {{\sigma _m}} & 0  \\0 & 0 & {{\sigma _m}}  \\\end{array}} \right][/math]

 

[math]{T_D} = \left[ {\begin{array}{*{20}{c}}
   {\frac{{\left( {2{\sigma _{xx}} - {\sigma _{yy}} - {\sigma _{zz}}} \right)}}{3}} & {{\sigma _{xy}}} & {{\sigma _{xz}}}  \\
   {{\sigma _{xy}}} & {\frac{{\left( {2{\sigma _{yy}} - {\sigma _{xx}} - {\sigma _{zz}}} \right)}}{3}} & {{\sigma _{yz}}}  \\  {{\sigma _{xz}}} & {{\sigma _{yz}}} &{\frac{{\left( {2{\sigma _{zz}} - {\sigma _{yy}} - {\sigma _{xx}}} \right)}}{3}}  \\\end{array}} \right][/math]

Tm is known as the mean stress or hydrostatic stress tensor and is a summary of the pure direct stresses.

TD is known as the devaitor stress tensor and is a summary of the pure shear stresses.

Another term is the for triaxial the confined compression test, which can be seen in the diagram below, which also adds to my discussion with Area54.

 

Edited October 12, 2017 by studiot

[DrP]

Regarding the oval - I would have thought that it might give better strength in the long direction but actually less across the wider side...  I think it would depend on where the impact was located on the shape.

[studiot]

5 minutes ago, DrP said:

Regarding the oval - I would have thought that it might give better strength in the long direction but actually less across the wider side...  I think it would depend on where the impact was located on the shape.

"I think it would depend on where the impact was located on the shape."

That's what I said.

The pointy end dropped onto a wooden floor might indent and receive some cushioning from the softer material.

I would expect that to come out when the different Elasticity moduli and poisson's ratios are taken into account.

[DrP]

Just now, studiot said:

That's what I said.

Yes I know (+1)...  it's just that all those stress tensors are a bit over my head. lol.

[Mordred]

2 hours ago, studiot said:

I am assuming solid glass objects.

I am not sure about this prognosis. Glass marbles are spherical and pretty tough cookies.

I can't remember ever seeing one shatter when dropped from twice that height onto a concrete floor. Chip maybe, but not shatter.

Anyway this is all about contact forces/contact stresses, not stress distribution.

Contact stresses are generally much higher than ordinary stresses in a body, but they are concentrated over the contact area.

Since this is an impact loading the impact force will also be much higher than the resting force due to the weight of the glass object.

Let P be the resting weight force
a be the contact area
C_E be  a constant due to the poisson's ratio and elasticities of the glass and the floor
D is the diameter of the glass ball at the contact point.

Then Roark gives the following formulae for σt , the maximum tensile stress due to contact.

max(σt)=0.2Pπa2

Where

CE=1−ν12E1+1−ν22E2

and

a=0.721PDCE−−−−−−√3

You can see from these equations that the max stress is inversely proportional to the cube root of the radius of curvature of the glass ball.

A spherical ball will have a constant radius of curvature, but an oval one, of the same weight,  will have a pointy end of greater curvature and a flatter side of lesser curvature.
Further, most of the weight  volume and surface area, will be concentrated in the equatorial, flatter regions of the oval ball.

So presumabably the oval ball is more likely to land on a point of lower curvature than the equivalent spherical one, thereby suffering a lower impact stress.

 

 

 

The use of triaxial is common in earth scineces as is the splitting of the stress tensor into two tensors.

Engineers in particular know that whilst it is convenient for theoreticians to collect everything together into a nice compact package, as soon as to want to work the numbers in a real world situation you have to break them all apart.

In the words of a wise old engineer

"You cannot avoid the arithmetic"

Not long after I joined here there was a long discussion about just this, with a Russian who was doing a Phd in ice Rheology.

T=Tm+TD

 

Tm=⎡⎣⎢σm000σm000σm⎤⎦⎥

 

TD=⎡⎣⎢⎢⎢⎢⎢(2σxx−σyy−σzz)3σxyσxzσxy(2σyy−σxx−σzz)3σyzσxzσyz(2σzz−σyy−σxx)3⎤⎦⎥⎥⎥⎥⎥

Tm is known as the mean stress or hydrostatic stress tensor and is a summary of the pure direct stresses.

TD is known as the devaitor stress tensor and is a summary of the pure shear stresses.

Another term is the for triaxial the confined compression test, which can be seen in the diagram below, which also adds to my discussion with Area54.

 

 

Precisely, though the above stress tensor is also modified to include the additional dimension for the time component. (Under 4 velocity treatments).

 Other than that is identical, you more commonly see this format with the Raychaudhuri equations.