How to calculate elastic modulus

[John54321]

Homework Statement

I am struggling with a part of the question. Please see diagram in the attached.

Calculate the relevant Elastic Modulus and select the lightest suitable beam section assuming a maximum allowable bending stress of 120 Mpa ?

Homework Equations

The Attempt at a Solution

To find the stress
the formula is stress = force / area
Force = 140KN
Area = 10m x 10m =100m2
Therefore 100 / 100 = 1 KN/m2

To find the strain
The formula is: strain change in length / original length
Change in length = 10.1m - 10.0 = 0.1m
Original length = 10m
Therefore strain = 0.1 / 10 = 0.01m

young modulus = strain / stress
Using the values from the stress and strain above
Elastic modulus = [/B] 1 / 0.01 =100Kn/m2

Im not sure if these calcs are correct would be greatful of any help thanks

 

[SteamKing]

You have provided only a portion of a much longer problem statement. PF Rules for the HW forums ask that you provide a complete statement of the problem, so that helpers have as much information to work with as possible.

The snippet of problem statement you provided along with the brief attempt at solution does not appear to make much sense, even with the attached diagram.

You state "Force = 140 KN". What force is this?

You have a calculation "Area = 10m x 10m = 100 m²" What area is this?

You state "Original length = 10m" Original length of what?

It doesn't appear that the attached diagram even goes with this problem.

 

[John54321]

You have provided only a portion of a much longer problem statement. PF Rules for the HW forums ask that you provide a complete statement of the problem, so that helpers have as much information to work with as possible.

The snippet of problem statement you provided along with the brief attempt at solution does not appear to make much sense, even with the attached diagram.

You state "Force = 140 KN". What force is this?

You have a calculation "Area = 10m x 10m = 100 m²" What area is this?

You state "Original length = 10m" Original length of what?

It doesn't appear that the attached diagram even goes with this problem.

10m being the total length of the beam in diagram and 140 KN being the total load I have worked out Ra and Rb if these are needed Ra = 95KN Rb = 165KN

The full question is below for some reason it wouldn't allow me to upload the 2 files

Calculate the relevant elastic modulus and select the lightest suitable beam from the attached table 3.1 universal beams (based on Bs5950) assuming a maximum allowable bending stress of 120Mpa

 

[SteamKing]

10m being the total length of the beam in diagram and 140 KN being the total load I have worked out Ra and Rb if these are needed Ra = 95KN Rb = 165KN

What happened to the UDL of 12 kN/m?

{Edit: I see now. The load on the beam is not 140 kN. That's the sum of the concentrated loads. The load on the beam should be the sum of the concentrated loads and the UDL.}

The full question is below for some reason it wouldn't allow me to upload the 2 files

Calculate the relevant elastic modulus and select the lightest suitable beam from the attached table 3.1 universal beams (based on Bs5950) assuming a maximum allowable bending stress of 120Mpa

The first part of this question doesn't make sense.

UB's are made from steel, so the elastic modulus of that material should already be known. What is unknown is the second moment of area I which will ensure that the maximum bending stress is below 120 MPa. That's what must be selected from the table of beam properties, with the further condition that the beam have the lowest total mass.

Is it possible that the question is asking you to determine the "elastic section modulus" of the beam?

You haven't provided any calculations for determining the maximum bending moment of this beam under the loading as shown.

 

[John54321]

Hi Steam King

In the attached is the maximum bending moment measured at Ra

 

[Chestermiller]

Hi SteamKing,

You are giving the OP more credit for knowledge on beams than he deserves. The only thing he has demonstrated so far is his ability to determine the reaction forces.

John54321: Have you drawn any shear and moment diagrams? Have you calculated the shear force and the bending moment distribution in the various sections of the beam. Do you know how the bending moment distribution is related to the shear force distribution? Once you know the bending moment distribution, you will be much closer to having a solution to your problem.

Chet

 

[John54321]

I have the shear force diagram in the attached I hope is correct? I have 4 parts left to answer for this question then the assignment is finished I am just struggling with these last few bits

 

[SteamKing]

Hi SteamKing,

You are giving the OP more credit for knowledge on beams than he deserves. The only thing he has demonstrated so far is his ability to determine the reaction forces.

John54321: Have you drawn any shear and moment diagrams? Have you calculated the shear force and the bending moment distribution in the various sections of the beam. Do you know how the bending moment distribution is related to the shear force distribution? Once you know the bending moment distribution, you will be much closer to having a solution to your problem.

Chet

I appreciate the comment Chet. I'm just trying to make sense out of a confusing and incomplete problem statement which was originally posted.

 

[SteamKing]

I have the shear force diagram in the attached I hope is correct? I have 4 parts left to answer for this question then the assignment is finished I am just struggling with these last few bits

If you have calculated the shear force diagram for this beam, then you should be able to calculate the bending moment diagram as well.

If you can't post the diagrams, then post the values for SF and BM at different locations along the beam.

 

[John54321]

Shear force diagram in attached

 

[John54321]

Bending moments in attached again not sure if I am correct !

 

[SteamKing]

Bending moments in attached again not sure if I am correct !

The shear forces and the bending moments appear to be correct.

Now, use this information to solve the rest of the problem.

 

[John54321]

That's good then. That's the problem I am struggling on how to work out that question and 3 other parts ?

 

[SteamKing]

That's good then. That's the problem I am struggling on how to work out that question and 3 other parts ?

Well,start with the basics.

Given the bending moment, how do you calculate the bending stress in the beam?

 

[John54321]

would this be the flexure formula ?

 

[SteamKing]

would this be the flexure formula ?

Yes.

 

[John54321]

So the formula I need to use is 0 = m y / l

or is this wrong ?

bending moment maximum = 3.75m

to find modulus = strain/ stress

As this is a self study course its hard going thanks for helping.

 

[SteamKing]

So the formula I need to use is 0 = m y / l

or is this wrong ?

bending moment maximum = 3.75m

to find modulus = strain/ stress

As this is a self study course its hard going thanks for helping.

Ah, another piece of the puzzle falls into place.

The bending stress σ = M y / I. It is also written as σ = M / SM, where SM is called the elastic section modulus of the beam, which I think you are confusing with the elastic modulus of the material, which is denoted by E. E does not figure into the calculation of the bending stress of the beam.

Note, SM = I / y, where I is the moment of inertia (or second moment of area) of the beam cross section, and y is the distance to the outer fiber of the beam, measured from the neutral axis.

You have determined the maximum bending moment from your calculations of the beam.

You are given a maximum bending stress σ = 120 MPa.

What you want to do is to calculate the SM of the beam such that the bending stress is less than this maximum stress.

Once you have determined the minimum SM of the beam which produces bending stresses less than 120 MPa, then you can use the table of UB properties to find which section can be used to make the beam which has the least total mass.

 

[John54321]

Sorry what do you mean by the outer fibre of the beam ? this has to be explained in laymans terms I do apologise.

 

[SteamKing]

Sorry what do you mean by the outer fibre of the beam ? this has to be explained in laymans terms I do apologise.

That's what you should have learned when studying about calculating the bending stress in a beam.

The outer fiber is that part of the beam which is furthest from the neutral axis. The bending stress is greater at locations further away from the neutral axis, as seen here:

For the UB sections, y = d/2.

 

[John54321]

Thanks for explaining showing the diagram.

So please can you explain what I need to put into the formula to work out the question as I am still not understanding what you mean ? Thanks

 

[SteamKing]

Thanks for explaining showing the diagram.

So please can you explain what I need to put into the formula to work out the question as I am still not understanding what you mean ? Thanks

The bending stress σ = M y / I

M - maximum bending moment, which you have already calculated for this beam and loading in Post #11.

y - the distance at the point where σ is calculated, measured from the neutral axis of the beam.
For UBs, y = d/2, as explained above.

I - the second moment of area of the beam cross section. This is supplied by the Table of Properties of UBs, attached in Post #3.

You have a maximum bending stress of 120 MPa which you cannot exceed. Since M is fixed, you must look at the table of UB properties to find a particular beam or beams which have a lower bending stress than 120 MPa. Once you have found possible candidate beams, the beam section which gives the lowest total mass for the 10 m beam will be the one to select.

This is not a matter of plugging a few numbers into a formula. It's a bit of trial and error, but you can make some simple calculations with which to help you eliminate beams which are either 1.) not strong enough, or 2.) too heavy.

 

[John54321]

Hi SteamKing

Thanks for taken the time and having the patience. The first parts to the question were done with guidance from you tube and I'm glad they were correct bit of a confidence boost. Its just these last 4 parts that I'm struggling with, well last 3 now. Thanks again much appreciated.

 

[John54321]

Morning SteamKing

I have looked at the tables which there are 3 of, the other 2 are in the attached. Obviously the Mass per unit is the total weight of the beam and not the max bending moment, which column covers what I am meant to be looking at please as there are no beams 10m long or have any max bending moments?

 

[SteamKing]

Morning SteamKing

I have looked at the tables which there are 3 of, the other 2 are in the attached. Obviously the Mass per unit is the total weight of the beam and not the max bending moment, which column covers what I am meant to be looking at please as there are no beams 10m long or have any max bending moments?

The tables are intended to compile design information about the different UB cross sections.

Beams of various lengths can be fabricated using such sections, which is why total lengths are not given. Some will be 5 m long, some will be 6-7/8 m long, etc.

The "mass/unit length" is just that: it tells you that a beam 1 m long fabricated from that particular section will have a mass of X kg. If your particular beam is 10 m long, then you'll obviously have to multiply the mass/unit length by 10 m in order to find the total mass of the beam.

As I mentioned previously, several times, knowing the maximum bending moment the beam must with withstand, coupled with the maximum allowable bending stress, will allow you to calculate the minimum value of the elastic section modulus, or I / y. These tables shorten this name to just elastic modulus, and you'll notice that two values are given: one for bending about the X-X axis, and another for bending about the Y-Y axis. Refer to the sketch on the first page of the tables to see which axis is which.

Remember: allowable bending stress, σ ≥ (max. bending moment) / (elastic modulus)
Here, σ = 120 MPa, and you have already calculated the max. bending moment as shown in Post #11 above.
All that's left to calculate is the minimum elastic modulus of the beam.

In this situation, we want to use the larger elastic modulus of the beam from the tables to compare with the value of the minimum elastic modulus calculated above, so that we can have a beam with the least mass while ensuring that the bending stresses are less than allowable.

Pay particular attention to the units of the elastic modulus, cm³. You want to make sure that the calculation of the minimum elastic modulus for the maximum bending moment in this particular beam has the same units.

 

[John54321]

Thanks so this is what your explaining to do BMmax 3.75m / 120 Mpa stress = 0.03125 this is the minimum elastic modulus correct ?

 

[SteamKing]

Thanks so this is what your explaining to doBMmax 3.75m / 120 Mpa stress = 0.03125 this is the minimum elastic modulus correct ?

It's not the location where the max. BM occurs, its the value of the BM itself which is used to calculate bending stress.

You're not even working through with the units on these quantities to keep from making silly mistakes in your calculations.

 

[John54321]

Please can you show me what figures I need to use then as I am not getting this thanks.

 

[SteamKing]

Please can you show me what figures I need to use then as I am not getting this thanks.

You attached the calculation of your bending moment values, plotted along the length of the beam, in Post #11.

Are you saying you can't read your own work?

 

[John54321]

So its got to be 184.375 (Knm) / 120 Mpa = 1.5365

 

[SteamKing]

So its got to be 184.375 (Knm) / 120 Mpa = 1.5365

What are the units of 1.5365? These are just as important as the number.

 

[John54321]

Kn/m^2

 

[SteamKing]

Kn/m^2

Don't guess. Write out the units for the numerator and the denominator and cancel as required.

Remember that a pascal is a derived unit. Be sure to write it in terms of basic units.

 

[John54321]

Correct that was a guess. But to be honest I'm not sure how the units would work out.

 

[SteamKing]

Correct that was a guess. But to be honest I'm not sure how the units would work out.

That's why I asked you to write them out, just like you were making a calculation with numbers, only using units in place of the numbers.

This is how you train yourself to figure out if the units of the answers to your calculation make sense.

 

[John54321]

I don't know that is why I guessed ? I'm not sure what the units would be this is why I'm using this forum for help.

 

[SteamKing]

I don't know that is why I guessed ? I'm not sure what the units would be this is why I'm using this forum for help.

The Rules in the HW Forums at PF allow helpers to guide those with questions. We are not supposed to do your work for you.
You must do some work in order to learn the material.

I asked you to write out the units in the calculation quoted below:

So its got to be 184.375 (Knm) / 120 Mpa = 1.5365

You have kN-m / MPa. I also told you that pascals (Pa) are derived units. Do you know the definition of a pascal?

 

[John54321]

I understand that this is why I have showed all of my workings for the first 4 parts of the question.

Mpa is a unit measure of pressure.

 

[SteamKing]

I understand that this is why I have showed all of my workings for the first 4 parts of the question.

Mpa is a unit measure of pressure.

Yes, that's what it measures, but the pascal is a derived unit, which means it is made up of a combination of other units. What are those units?

 

[John54321]

Yes internal stress, young modulus and ulltimate tensile strength.

 

[SteamKing]

Yes internal stress, young modulus and ulltimate tensile strength.

No. You keep describing things which Pascals are used to measure.

A pascal is made up of other units, things like meters, kilograms, and whatnot. Which other units are used to make pascals?

Hint: it's OK to google "pascal" if you don't know the answer off the top of your head.

 

[John54321]

It's made up of N for Newton M for metre kg kilogram s second

 

[SteamKing]

It's made up of N for Newton M for metre kg kilogram s second

Yes, but how? What is the particular combination of these units which make up a Pascal but not something else?

Complete, this sentence: 1 pascal = 1 ?

 

[John54321]

Hi SteamKing

1 pascal = 1 m ^-1 kg s^-2

 

[SteamKing]

Hi SteamKing

1 pascal = 1 m ^-1 kg s^-2

That's better.

A pascal is also expressed as a force / unit area. For our purposes, 1 Pa = 1 N / m²

Getting back to your last calculation:

So its got to be 184.375 (Knm) / 120 Mpa = 1.5365

if you divided 10³ N-m by (10⁶ N/m²), what's left over?

 

[John54321]

10^3N-m / 10^6N/^2 = 0.001n-m 0r 0.000001n-m^2 after its sqt

Not sure this is correct not done nothing like this before

 

[SteamKing]

10^3N-m / 10^6N/^2 = 0.001n-m 0r 0.000001n-m^2 after its sqt

Not sure this is correct not done nothing like this before

Sure you have, if you've studied algebra. Except instead of dividing units like N and m², you were working with things like x and y. It's the same principle.

Now, in your calculation above, you are close, but there's a small error. I'm going to switch to Latex and make it easier to see:

$\frac{10^3 N⋅m}{10^6\frac{N}{m^2}}$

Can you reduce the expression above, cancelling out the units which can cancel?

 

[John54321]

Morning SteamKing

Are you talking about seeing it like this 1/1000 n-m^2 ?

If not I am not sure what to do if this isn't correct.
I

 

[SteamKing]

Morning SteamKing

Are you talking about seeing it like this 1/1000 n-m^2 ?

If not I am not sure what to do if this isn't correct.

I

Yes, but you're still not cancelling the units properly.

Remember, if you have a fraction like this: $\frac{a}{\frac{a}{b}}$

you invert the denominator and multiply by the numerator, like so: $\frac{a}{\frac{a}{b}} = a ⋅ \frac{b}{a}=b$

Do this with your expression for the units: $\frac{10^3 N⋅m}{10^6\frac{N}{m^2}}$ and see what's left.

 

[John54321]

So N-m being the denominator makes it be multiplied to N-M^2 end up as the numerator.

Sorry I don't get this ? maybe I need to sit down with someone and get this covered as I am running out of time.

Thanks for your help and patience.