Jump to content

shear and tensile force questions


koii123

Recommended Posts

suppose i have 4 solids like this:
and the solid have shear stress and tensile stress 12N/mm2 and 15N/mm2 respectively,
suppose i know the area of each surface,
how can i judge whether this solid is able to withstand 20N load?
what kind of formula should i use?

 

MFtfnlF.jpg

Link to comment
Share on other sites

Hello koii,

 

Is this homework of some sort?

 

You need to start by considering what you mean by 'gounded', which is an electricalengineering term, not a structural engineering term.

In particular your fig3. Which way do you think the forces are acting?

 

Then please say what you man by a 20N uniform force.

Do you mean a total of 20N uniformly distributed or 2oN/linear m of block or 20N/ sq metre?

Edited by studiot
Link to comment
Share on other sites

Hello koii,

 

Is this homework of some sort?

 

You need to start by considering what you mean by 'gounded', which is an electricalengineering term, not a structural engineering term.

In particular your fig3. Which way do you think the forces are acting?

 

Then please say what you man by a 20N uniform force.

Do you mean a total of 20N uniformly distributed or 2oN/linear m of block or 20N/ sq metre?

 

sorry, perhaps i should make it more clear....

this is not a homework, this is a real case,

because i have a component damaged, i would like to repair it with a glue,

and i would like to find out whether the glue is strong enough to withstand the 50N upward force, so i try to apply mathematical techniques,

actually, the following picture describes my current situation,

do you know how to do it? i really don't have any idea wacko.png

KRkczXi.jpg

Edited by koii123
Link to comment
Share on other sites

As I understand your intention you wish to repair the grey board by gluing the blue T shaped splint to it.

 

Is the grey board cracked or broken under the blue?

 

Can you not simply glue the large flat faces together?

 

The four brown wedge shaped pieces are a reasonable fixture, but why bother to cut the wedges, why not just use rectangular section?

 

The problem with edge fixing is differential movement between the grey board and the blue, particularly die to flexing. I see you attempting to address this with the by gluding on a flat edge strip. This is unlikely to hold for long.

 

Does the blue board carry circuitry and tracks or is it just a mechanical splint?

If you can drill through both boards a various places and provide through rivets, perhaps with soldered links, this would be a better solution.

Link to comment
Share on other sites

As I understand your intention you wish to repair the grey board by gluing the blue T shaped splint to it.

 

Is the grey board cracked or broken under the blue?

 

Can you not simply glue the large flat faces together?

 

The four brown wedge shaped pieces are a reasonable fixture, but why bother to cut the wedges, why not just use rectangular section?

 

The problem with edge fixing is differential movement between the grey board and the blue, particularly die to flexing. I see you attempting to address this with the by gluding on a flat edge strip. This is unlikely to hold for long.

 

Does the blue board carry circuitry and tracks or is it just a mechanical splint?

If you can drill through both boards a various places and provide through rivets, perhaps with soldered links, this would be a better solution.

 

thank you for help,

actually, the grey board is fixed and it is not cracked,

i cannot add glue under the blue board, this is a very special application, and it is complicated to explain it here, sorry for that,

due to some boundary conditions, i can only add glue in this way,

do you know how to do calculations in a situation like that ?

Link to comment
Share on other sites

Well yes, I can "do calculations" but I am still not clear what you are trying to achieve. I really would like to be sure that any calculations will be of some value.

 

Are you trying to make a mechanical repair or strengthening to the grey board

 

or

 

Are you trying to add a piggy back electrical board to repair the electrical circuitry or add further functionality?

 

These are two very different situations with very different mechanical requirements.

Link to comment
Share on other sites

Well yes, I can "do calculations" but I am still not clear what you are trying to achieve. I really would like to be sure that any calculations will be of some value.

 

Are you trying to make a mechanical repair or strengthening to the grey board

 

or

 

Are you trying to add a piggy back electrical board to repair the electrical circuitry or add further functionality?

 

These are two very different situations with very different mechanical requirements.

 

thank you,

actually, this is a mechanical repair (for pure mechanical loading), there is nothing related to electric circuit,

 

Here are the dimensions of the adhesives:

All the wedges have height 3mm (Z-direction) (the same as the PCBs), and width is 1mm (x-direction),

The long 80mm adhesive is 1mm thick (y-direction) and height is 6mm (z-direction).

 

Also, I read from the adhesive (Epoxy) specification sheet, the adhesive have the following properties:

tensile: 22N/mm2

shear: 18N/mm2

 

i would like to know whether the adhesive i selected is strong enough to hold the boards together under the 50N force. smile.png

and this is the only objective of this problem.

Edited by koii123
Link to comment
Share on other sites

OK

 

From the figures supplied.

 

The area resisting detachment of the blue board comprises the 4 blocks 4 (3x10 =30 sq mm) plus half the long strip attached ie (3 x 80 = 240 sq mm)

 

This makes a total resisting area of 360sqmm.

 

All of this is in single shear.

 

You say that the shear strength is 18 N/sqmm so total resisting capacity resisting detachment of the blue board is (18 x 360 = 6480) N.

 

Resisting the blocks pulling away is (4 x 30= 120 sqmm) plus in tension = 2640N

 

plus the other half of the area of the strip in shear = 18 x 240 = 4320N

 

making a total of 6060N

 

There are quite a few ifs and buts about this however, since the distribution of the 50N force is not known and no moments have been checked.

 

Further the strength of any glued system is dependent on the quality of workmanship and subject to considerable variation and can easily delaminate under vibration or flexing.

 

Are the strengths quoted the stength of the bond or the strength of the glue?

 

You need the bond strengths, and I have assumed these are quoted in the calculations.

Edited by studiot
Link to comment
Share on other sites

OK

 

From the figures supplied.

 

The area resisting detachment of the blue board comprises the 4 blocks 4 (3x10 =30 sq mm) plus half the long strip attached ie (3 x 80 = 240 sq mm)

 

This makes a total resisting area of 360sqmm.

 

All of this is in single shear.

 

You say that the shear strength is 18 N/sqmm so total resisting capacity resisting detachment of the blue board is (18 x 360 = 6480) N.

 

Resisting the blocks pulling away is (4 x 30= 120 sqmm) plus in tension = 2640N

 

plus the other half of the area of the strip in shear = 18 x 240 = 4320N

 

making a total of 6060N

 

There are quite a few ifs and buts about this however, since the distribution of the 50N force is not known and no moments have been checked.

 

Further the strength of any glued system is dependent on the quality of workmanship and subject to considerable variation and can easily delaminate under vibration or flexing.

 

Are the strengths quoted the stength of the bond or the strength of the glue?

 

You need the bond strengths, and I have assumed these are quoted in the calculations.

 

Thanks so much,

 

you are right, i think i need to find out the real bond strength in terms of N/mm2, the results looks too strong!

assume that the shear and the tensile are right,

so, at least 6060N pulling force is required to pull out the blue board, because it is smaller that 6480N,

am i right?

 

if the wedge change to other shape (for example, rectangular), how will the calculation be affected?

Edited by koii123
Link to comment
Share on other sites

  • 3 weeks later...

The stiffness of the materials, and glue, come into effect, as well as the distribution of the forces that make up 50N net force. If, for instance, the 50N was all in that location and the blue material flexible enough, it would peel off, especially if the glue and board were rigid as it would limit the spreading of the load and concentrate it where the peeling was taking place.

 

The higher numbers calculated would only be true if the whole area failed at once (In peel it could seem instantaneous, but it would not be)

Edited by J.C.MacSwell
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.