Deflection of a beam - Two point loads simply supported

[Jack. C]

Homework Statement

Derive an equation for the deflection using Macaulay's method.

Hello all,

New member here, I come across this forum for advice regularly and have decided to register :).

I have been given a problem that is highlighted in the attached picture. I am competent in calculating beam deflections with a simple two support system, with a point load somewhere between, but in this scenario I am thrown off by the 6.5 kN.

I am undecided on where to 'cut' the beam, and more to the point, how to include the 6.5kN load into the equation.

To simplify the derivation I have substituted

R1=2.5kN
P1=11.5kN
R2=15.5kN
P2=6.5kN
a=0.4m

Homework Equations

EI d^2y/dx^2 = M

The Attempt at a Solution

I'm really not sure about the 6.5kN. I have searched countless resources for a similar question format to no avail.

My trouble is with with the initial set up of the question, as I feel comfortable with the integration to get y.

Attempt 1) I have attempted to take the moment 'cut' just before P2 which I come to the solution of:
M=R1x - P1<x-a> + R2<x-2a>
I have a feeling this is definitely not right..

Attempt 2) Treat the equation as a simple two support deflection, assuming the bending moment of P2 is absorbed by the reaction at R2. Taking the 'cut' just before R2:
M=R1x - p1<x-a>
Again, probably not..

Attempt 3) Reverse the beam layout as to have P2 at the left hand side. This means that both of the point loads will be included in the equation. 'Cut' just before R1:
M=-P2x + R2<x-a> - P1<x-2a>

This module is self taught so forgive me if there is something blindingly obvious that I have missed.

Am I on the right track with any of these attempts?

Thank you.

 

[SteamKing]

Homework Statement

Derive an equation for the deflection using Macaulay's method.

Hello all,

New member here, I come across this forum for advice regularly and have decided to register :).

I have been given a problem that is highlighted in the attached picture. I am competent in calculating beam deflections with a simple two support system, with a point load somewhere between, but in this scenario I am thrown off by the 6.5 kN. View attachment 83591
I am undecided on where to 'cut' the beam, and more to the point, how to include the 6.5kN load into the equation.

To simplify the derivation I have substituted

R1=2.5kN
P1=11.5kN
R2=15.5kN
P2=6.5kN
a=0.4m

Homework Equations

EI d^2y/dx^2 = M

The Attempt at a Solution

I'm really not sure about the 6.5kN. I have searched countless resources for a similar question format to no avail.

My trouble is with with the initial set up of the question, as I feel comfortable with the integration to get y.

Attempt 1) I have attempted to take the moment 'cut' just before P2 which I come to the solution of:
M=R1x - P1<x-a> + R2<x-2a>
I have a feeling this is definitely not right..

Attempt 2) Treat the equation as a simple two support deflection, assuming the bending moment of P2 is absorbed by the reaction at R2. Taking the 'cut' just before R2:
M=R1x - p1<x-a>
Again, probably not..

Attempt 3) Reverse the beam layout as to have P2 at the left hand side. This means that both of the point loads will be included in the equation. 'Cut' just before R1:
M=-P2x + R2<x-a> - P1<x-2a>

This module is self taught so forgive me if there is something blindingly obvious that I have missed.

Am I on the right track with any of these attempts?

Thank you.

While this simply supported beam is loaded in a somewhat unusual fashion, with P₂ being applied so far beyond the support, don't panic! Treat this beam as you would any other simply supported beam.

If you haven't done so, make a quick sketch of the shear force curve for this beam, and while you're at it, roughly sketch in what the bending moment curve might look like.

Don't worry about making a super-accurate drawing, something approximately right is all you're after.

If you are self-teaching this method, the document I'm uploading is a pretty good guide at explaining how to apply it to construct functions for the SF and BM for a beam, from which you can integrate and obtain the slope and deflection functions.

Since you are dealing with actual loads and known locations here, when you construct the shear force function or the bending moment function, it is OK to substitute the actual location of the load for the parameter "a" in the expression for each load. For example, the load P₁ would be 11.5[x-0.4]⁰ kN when expressed as a Macauley function, and this expression can be used directly in the SF function.

The great thing about either the Macauley method or the singularity function method is you don't have to worry about where to cut the beam in order to calculate the shear force, bending moment, etc. All you have to do is construct the functions for each of the applied loads and reactions and add them together. All the data you need is shown on the beam diagram. The only thing you need to do is make sure the beam is in static equilibrium.

 

[SteamKing]

Since you are dealing with actual loads and known locations here, when you construct the shear force function or the bending moment function, it is OK to substitute the actual location of the load for the parameter "a" in the expression for each load. For example, the load P₁ would be 11.5[x-0.4]⁰ kN when expressed as a Macauley function, and this expression can be used directly in the SF function.

I made a slight mistake. According to the sign convention, the load P₁ should be -11.5[x-0.4]⁰ kN.

 

[Jack. C]

SteamKing,

Fantastic! Some really great information here thank you very much!

The attachment is very helpful, there is also a small example of overhangs which agrees with the advice you have given: treat this as an ordinary beam.

Quote from attachment: 'Though the 6.5 kN load is to the right of the cut, and so not in the diagram, its effect is accounted for in the reactions which are included' - Would this statement be correct in this application?

I understand that you may simply construct the functions for each component, and when applied it will sort itself out.

With this information I have had another attempt. If you could give it a quick look over I would be very grateful indeed:

EI (d^2y/dx^2) = M = R1x - P1<x-a> + R2<x-2a>

EI (dy/dx) = (R1^2/2) - (P1<x-a>^2/2) + (R2<x-2a>^2/2) + C1

EI y = (R1x^3/6) - (P1<x-a>^3/6) + (R2<x-2a>^3/6) +C1x +C2

When x = 0, y = 0

0 = 0 - 0 + 0 + 0 + C2 Therefore C2 = 0

When x = 0.8, y = 0

0 = (R1*0.8^3/6) - (P1<0.8-a>^3/6) + 0 + C1*0.8

0 = (2.5*0.8^3/6) - (11.5<0.8-0.4>^3/6) + 0 + C1*0.8

0= 0.21333 - 0.12267 + C1*0.8

C1 = (0.12267 - 0.21333)/0.8 = -0.11333

Therefore solution is:

y = 1/EI ((R1x^3/6) - (P1<x-a>^3/6) + (R2<x-2a>^3/6) - 0.11333x

Sorry if you have trouble with the format, I'm not sure how to create an equation on the forum.

The question goes on to ask to calculate the deflection at every 0.1m interval, so determining the maximum deflection will not be necessary

Thanks again, I feel I'm getting close now! Or maybe not :)

 

[SteamKing]

SteamKing,

Fantastic! Some really great information here thank you very much!

The attachment is very helpful, there is also a small example of overhangs which agrees with the advice you have given: treat this as an ordinary beam.

Quote from attachment: 'Though the 6.5 kN load is to the right of the cut, and so not in the diagram, its effect is accounted for in the reactions which are included' - Would this statement be correct in this application?

I understand that you may simply construct the functions for each component, and when applied it will sort itself out.

With this information I have had another attempt. If you could give it a quick look over I would be very grateful indeed:

EI (d^2y/dx^2) = M = R1x - P1<x-a> + R2<x-2a>

1. You have omitted showing P2. I know it may not have an influence on the moment, but I like to include it for completeness.
2. Since the values of R1, R2, P1, and P2 are known, they can be inserted into the BM equation instead of the force names.
3. Since the locations of R1, R2, P1, and P2 are also known, use them instead of "a" in the BM equation.
4. The singularity function method uses <> brackets to represent the singularity function. Since it is slightly
different, some authors chose to use [] brackets to indicate the Macauley function.

EI (dy/dx) = (R1^2/2) - (P1<x-a>^2/2) + (R2<x-2a>^2/2) + C1

To avoid confusion, write out the complete expression for each load, so that instead of R1x,
you write R1*[x - 0]¹.
On the terms for P1 and R2, write (P1/2)*[x - 0.4]², for example, to avoid confusion in the exponent.
Also note, the R1 term in the slope equation above is incorrect.

EI y = (R1x^3/6) - (P1<x-a>^3/6) + (R2<x-2a>^3/6) +C1x +C2

When x = 0, y = 0

0 = 0 - 0 + 0 + 0 + C2 Therefore C2 = 0

When x = 0.8, y = 0

0 = (R1*0.8^3/6) - (P1<0.8-a>^3/6) + 0 + C1*0.8

This equation for the deflection is not correct for the reasons outlined above. You need to correct it and recheck the calculation of C1 and C2.

0 = (2.5*0.8^3/6) - (11.5<0.8-0.4>^3/6) + 0 + C1*0.8

0= 0.21333 - 0.12267 + C1*0.8

C1 = (0.12267 - 0.21333)/0.8 = -0.11333

Therefore solution is:

y = 1/EI ((R1x^3/6) - (P1<x-a>^3/6) + (R2<x-2a>^3/6) - 0.11333x

Sorry if you have trouble with the format, I'm not sure how to create an equation on the forum.

You can create simple formulas using the items on the toolbar above the editing box.

1. You can create subscripts by hitting the "x₂" button.
2. You can create superscripts by hitting the x² button.
3. Greek letters and certain math symbols can be accessed by hitting the ∑ button on the right end of the toolbar. The symbols will along the bottom of the editing box. You put your cursor in the box where the symbol is to be inserted, the select the symbol with your mouse.

The question goes on to ask to calculate the deflection at every 0.1m interval, so determining the maximum deflection will not be necessary

Thanks again, I feel I'm getting close now! Or maybe not :)

Make the changes indicated and repost.

 

[Jack. C]

Again, thank you for helping me with this.

Ok, all taken on board. When typing the slope equation out I appear to have missed out the x, but have correctly typed my workings for the deflection equation; with the correct formatting:

EI y = (2.5/6)*[x-0]³ - (11.5/6)*[x-0.4]³ + (15.5/6)*[x-0.8]³ - (6.5/6)*[x-1.2]³ + C1x + C2

I have cleared up the equation and included the P2 term and the x for R1, but If this is correct then I will again get the same values for C1 and C2.

Is it that these terms are incorrect?

Thanks,

Jack

 

[SteamKing]

Again, thank you for helping me with this.

Ok, all taken on board. When typing the slope equation out I appear to have missed out the x, but have correctly typed my workings for the deflection equation; with the correct formatting:

EI y = (2.5/6)*[x-0]³ - (11.5/6)*[x-0.4]³ + (15.5/6)*[x-0.8]³ - (6.5/6)*[x-1.2]³ + C1x + C2

I have cleared up the equation and included the P2 term and the x for R1, but If this is correct then I will again get the same values for C1 and C2.

Is it that these terms are incorrect?

Thanks,

Jack

This result looks good, but you must not forget that R1, R2, P1, and P2 were originally in kN. If you have to calculate a deflection value for some value of x, you must not forget that factor of 1000 on those quantities in the deflection formula.

 

[Jack. C]

SteamKing,

Brilliant, seems to all check out when compared to experimental results!

Yes must remember to use the correct figures i.e. *10³.

Have a good day and once again thank you for your help,

Jack