Maximizing Bending Moment in Simply Supported Beams: An Experimental Approach

[oxon88]

Homework Statement

The simply supported beam shown is 10 m long with
E = 200 × 109 Pa and I = 150 × 10–6 m4.

1. Determine the position and magnitude of the maximum bending
moment.2. Plot a graph of the deflection along the length of the beam
(calculate the deflection at 1 m intervals).3. What I value for the beam would be required to halve the maximum
deflection of the beam?4. The calculation for the maximum bending moment is to be verified
experimentally using a strain gauge bonded to the outer surface of the
beam, at the point where the maximum bending moment occurs.
Derive an equation which could be used to calculate the bending
moment from the measured strain value. State the meaning of all
symbols used in your equation.

Homework Equations

M/I = σ/y = E/R

d²y / dx² = M/EIcan anyone advise on where i should start?Thanks.

 

[SteamKing]

1. Set up a free body diagram for the beam.
2. Solve for the unknown support reactions.
3. Draw the shear diagram for the beam.
4. Use the shear diagram to construct the bending moment diagram.
5. Once the moment diagram is done, make the M/I diagram.
6. Integrate the M/I diagram twice, making sure you keep the constants of integration.
7. After doing the integrations, apply the boundary conditions for the beam to determine the constants of integration.
8. Adjust the last integration to calculate the deflections of the beam.

 

[oxon88]

Thanks for your response.

Is the free body diagram not the same as the attached diagram?

 

[SteamKing]

Yes.

 

[oxon88]

Ok I think i have calculated the unknown support reactions. does this look correct?

(3 * 20) + (7 * 30) + (5 * 10 *5) = r2 * 10

520 = r2 * 10

r2 = 52kN

upward forces=downward forces

r1 + 52 = 20 + 30 + (5 * 10)

r1 = 48kN

 

[SteamKing]

Looks good so far.

 

[pongo38]

oxon88, you didn't have to ask us if it was correct. You could check it yourself by taking moments about any axis other than the one you have already used going through R1.

 

[oxon88]

Thanks. I have drawn the shear force diagram and also calculated the bending moments (below). x=0 = 0kN-m
x=1 = (48*1)-(5*1*0.5) = 45.5kN-m
x=2 = (48*2)-(5*2*1) = 86kN-m
x=3 = (48*3)-(5*3*1.5) = 121.5kN-m
x=4 = (48*4)-(20*1)-(5*4*2) = 132kN-m
x=5 = (48*5)-(20*2)-(5*5*2.5) = 137.5kN-m
x=6 = (48*6)-(20*3)-(5*6*3) = 138kN-m
x=7 = (48*7)-(20*4)-(5*7*3.5) = 133.5kN-m
x=8 = (48*8)-(20*5)-(30*1)-(5*8*4) = 94kN-m
x=9 = (48*9)-(20*6)-(30*2)-(5*9*4.5) = 49.5kN-m
x=10 = (48*10)-(20*7)-(30*3)-(5*10*5) = 0kN-m

 

[SteamKing]

I'm not sure what you are doing in Post #8.

At x = 0, the shear force diagram has a value equal to R1, or 48 kN. The bending moment at x = 0 is equal to 0 kN-m (Note the units).

Can you clarify?

 

[oxon88]

post #8 are my calculations for the bending moments along the beam. I have amended the units.

 

[SteamKing]

The bending moments look good. Proceed with answering the questions in the OP.

 

[oxon88]

thanks. Please can you explain what the M/I diagram is?

 

[SteamKing]

Just what it says. Take the ordinates of your bending moment diagram and divide by the moment of inertia of the beam (in this case I = 150 * 10^-6 m^4 according to the OP). It's not really required for this problem, since I is a constant, but when I changes along the length of the beam, an M/I diagram is useful when calculating deflection.

 

[oxon88]

x=0 = 0kN-m
x=1 = 45.5kN-m / 150 * 10^-6 m^4 = 303333.3
x=2 = 86kN-m / 150 * 10^-6 m^4 = 573333.3
x=3 = 121.5kN-m / 150 * 10^-6 m^4 = 810000
x=4 = 132kN-m / 150 * 10^-6 m^4 = 880000
x=5 = 137.5kN-m / 150 * 10^-6 m^4 = 916666.7
x=6 = 138kN-m / 150 * 10^-6 m^4 = 920000
x=7 = 133.5kN-m / 150 * 10^-6 m^4 = 890000
x=8 = 94kN-m / 150 * 10^-6 m^4 = 626666.7
x=9 = 49.5kN-m / 150 * 10^-6 m^4 = 330000
x=10 = 0kN-m


so would the diagram be as pictured?

 

[SteamKing]

Looks OK. Make sure you use the proper units.

 

[oxon88]

ok Thankyou. What units should be used? Can you provide some guidance on how to integrate the M/I diagram twice?

 

[SteamKing]

ok Thankyou. What units should be used? Can you provide some guidance on how to integrate the M/I diagram twice?

Obviously, the units which result from dividing bending moment M by moment of inertia I.

The integration of the M/I diagram proceeds just like any other numerical integration: divide the x-axis into regular spacings and determine the M/I ordinates. You can use a trapezoidal rule to calculate the actual values of the integrals.

Remember, you must include the constants of integration in your calculation. There will be two unknown constants which are determined by applying the support conditions for your beam. You can start the integration by assuming the value of the integrals is zero at x = 0.

Since I is constant, the slope \vartheta = (1/EI) ∫ M dx + C1
The deflection δ = (1/EI)∫∫ M dx + C1*x + C2

The final values for slope and deflection are determined by applying these constants of integration to your calculations of the integral values at the various x-locations.

 

[oxon88]

am i not able to use this formula to solve y?

 

[SteamKing]

am i not able to use this formula to solve y?

View attachment 63703

Where did you get this formula? It looks like only a point load P is present. What about deflection due to the distributed load?

 

[oxon88]

Our learning material gives it as an example. Yes it states that the formula is used for simply supported beams at ends with concentrated load P at any point.

Would it not be posssible to calculate the deflection for each of the loads and then sum the answers?

For the UDL, could this formula not be used?

 

[SteamKing]

Yes, you can use these formulas as long as you understand how to apply them properly.

 

[oxon88]

ok thanks. So i can work out the deflection for each of the loads, then sum them all together?Do you think Macaulay’s Method would be a better method to use?

 

[oxon88]

I have got the following answers for total deflection. do these look feasible?

0m = 0mm
1m = 15.20mm
2m = 28.9mm
3m = 39.75mm
4m = 46.67mm
5m = 49.20mm
6m = 47.17mm
7m = 40.55mm
8m = 29.66mm
9m = 15.65mm
10m = 0mm

 

[SteamKing]

ok thanks. So i can work out the deflection for each of the loads, then sum them all together?

Yes. That's what superposition is all about w.r.t. beams.

Do you think Macaulay’s Method would be a better method to use?

It's up to you. You already have your deflection formulas. I don't think you would see any savings in terms of calculating the deflections as asked in the OP.

 

[oxon88]

one thing i can't understand...

when i work out δ_max for the 20kN load. i get δ_max=10.9033mm

however when working out the deflection in terms of x, i can see that at 4.1m δ = 11.051 mm

have i done something wrong?


δ_max = =20000*7*((100-49)^1.5)/(9*(SQRT(3)*10*30000000))*1000 = 10.9033mm

δ_x4.1 = (20000*3*(4.1-10)/(6*10*30000000))*(9-2*10*4.1+4.1*4.1)*1000 = 11.0507mm

 

[SteamKing]

one thing i can't understand...

when i work out δ_max for the 20kN load. i get δ_max=10.9033mm

however when working out the deflection in terms of x, i can see that at 4.1m δ = 11.051 mm

have i done something wrong?


δ_max = =20000*7*((100-49)^1.5)/(9*(SQRT(3)*10*30000000))*1000 = 10.9033mm

δ_x4.1 = (20000*3*(4.1-10)/(6*10*30000000))*(9-2*10*4.1+4.1*4.1)*1000 = 11.0507mm

It doesn't look like you have used the proper formula to calculate δ_x4.1

It doesn't match what you showed in an earlier post and it doesn't match the formula on this table:

http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf

 

[pongo38]

Note that because the beam is unsymetrically loaded, the maximum bending moment does not coincide with the geometric centre of the beam. The same is true for the maximum deflection. The point of maximum deflection is not necessarily the same as the point of maximum bending moment. If your formula is correct, you can presumably differentiate it to find the point of maximum deflection. It is worth pointing out that your calculations show a precision which no engineer would believe. This is because, in addition to deflection due to bending - given by the formula quoted-, there is also a further deflection (perhaps 2 or 3% more, depending on the material used) due to shear.

 

[oxon88]

It doesn't look like you have used the proper formula to calculate δ_x4.1

It doesn't match what you showed in an earlier post and it doesn't match the formula on this table:

http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf

im a little confused as to which formula to use. could you provide some guidance?

for x=4.1m,Should i be using this formula?

or this one?

 

[SteamKing]

For the 20 kN load, a = 3 m. You want to calculate the deflection at x = 4.1 m. This should make it easy to select the correct formula.

 

[oxon88]

ok got it. here are my answers for the deflection.

and the graph looks like this

does this look reasonable? Would the 5kN UDL really cause more of a deflection that the 30kN point load?

 

[SteamKing]

All you have to do is compare the expressions for deflection for a point load versus the distributed load.

 

[oxon88]

ok Thank you. so my figures for deflection look acceptable?

 

[SteamKing]

Yes.

 

[oxon88]

Q3. What I value for the beam would be required to halve the maximum
deflection of the beam?

A. In order to halve the maximum deflection of the beam, the value of I would need to be doubled to 300x10^-6m⁴, this is because the deflection δ is inversely proportional to I

 

[oxon88]

does the above answer look acceptable?

 

[SteamKing]

Yes.

 

[oxon88]

ok thanks.

Are you able to provide some guidance on the last question?

4. The calculation for the maximum bending moment is to be verified
experimentally using a strain gauge bonded to the outer surface of the
beam, at the point where the maximum bending moment occurs.
Derive an equation which could be used to calculate the bending
moment from the measured strain value. State the meaning of all
symbols used in your equation.

 

[SteamKing]

Can't help with this part. Don't know anything about strain gauges and have never worked with them.

 

[oxon88]

Im not sure the strain gauge part is all that relevant. Just looking to Derive an equation which could be used to calculate the bending moment from the measured strain value.


the strain is the ratio of the deflection to the radius of curvature


e = δ / R = Mδ / EI


e=strain
δ=Deflection
E=young's modulus
I=the second moment of area of section about neutral axis
M=the bending moment at the section concerned
R=radius of curvature of bent beam

 

[Big Jock]

Which formulas did you use for the various deflections? I am having a real nightmare trying to work this out

 

[SteamKing]

Which formulas did you use for the various deflections? I am having a real nightmare trying to work this out

Re-read the earlier posts in this thread. There is a reference to the deflection formulas.

 

[Big Jock]

Still a little lost with this. The question asks to Plot a graph of the deflection along the length of the beam
(calculate the deflection at 1 m intervals). So unsure which formulas to use for the 5kN udl the 20 and 30 kN point loads. So if that deflection or maximum deflection and why would you need the slope for this isn't that something completely different from the asked question?

 

[SteamKing]

It's very simple. The problem asks that you calculate and plot the values of the beam deflection at 1 m intervals along the length of the beam. Sure, the beam will have a maximum deflection associated with a particular loading, but the beam deflection is not zero at other points on the beam. The formulas in the beam table allow one to select a distance from one end of the beam at which deflection is to be calculated. The beam table includes formulas for the deflection due to a UDL and point loads. Remember, the total deflection due to several different loads is the sum of the deflection due to each load by itself. This part of the question is basically plug and chug and plot the results.

 

[Big Jock]

I know this but unsure which formulas to use for the 5kN udl and the 20 and 30kN point loads for the distances of x=0-10 this is where my problem lies. I think the udl formula should be w/EI(Lx^2/12-x^4/24-L^3/24) and the point load formula w/EI(x^3/12-L^2x/16) is this correct? I so I then have all the various deflection point for all the info then sum them together?...

 

[SteamKing]

I don't recognize your formulas for the deflections. Did you read the earlier posts in this thread? If you go to Post #26, the reference there gives deflection formulas for cantilever beams on p. 1 and simply supported beams on p. 2.

 

[Big Jock]

Yeah seen them so reference to the attachment beam 7 formula 2 and beam 8 formula 2 would that be correct and give me all the info I need for my graph? The other thing there are two separate formulas for beam 7 formula 2 which one do you use?

 

[SteamKing]

Yeah seen them so reference to the attachment beam 7 formula 2 and beam 8 formula 2 would that be correct and give me all the info I need for my graph? The other thing there are two separate formulas for beam 7 formula 2 which one do you use?

It shouldn't be this hard. For the point load, the range of x is given for which each formula is applicable (e.g., formula #1; 0 < x < a). You use the formula which is appropriate for the value of x at which you are calculating deflection. Since you are calculating deflection along the entire length of the beam, you will have used both formulas by the time you complete your calculations.

 

[Big Jock]

Think I have it. Been working on the udl first, for the first metre of the udl y=6.8125x10^-5(micromillimters) would this be the correct measurement for the y-axis or should it be converted to mm?

 

[SteamKing]

Think I have it. Been working on the udl first, for the first metre of the udl y=6.8125x10^-5(micromillimters) would this be the correct measurement for the y-axis or should it be converted to mm?

Be careful with your units. I checked you calculation and I come up with a deflection of 6.8125*10-3 m, which is equal to 6.8 mm. If you are not sure, post your work here where I can check it in detail.

 

[Big Jock]

5x1/24x200x10^9x150x10^-6(10^3-2x10x1^2+1^3) was what I did. Which I have re done instead of 5 the udl should have been 5000 which gave the figure you quoted. Many, many thanks SteamKing for you patience over this. Glad I asked though as I have learned how to do this now correctly with your help