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Homework Help: 2 point moment deflection calc

  1. Jun 17, 2010 #1
    I am cerious how to calculate a simple 2 point moment using the formula Mm dx/EI
    Lets say i have a span of 43 1/2" with 2 5000lb loads 10 3/4" from each side. Each side is simply supported. My inertia of my beam is 4

    I have drawn a moment diagram but i cant seem to translate it to diflection. I know that M represents moment. My first moment is 5000*10 3/4 = 500011 lb-in. So with the formula above i can calculate a deflection at any point. This is where i get lost. I read that m is the unit load where i want to calculate my deflection. I suspect that would be 5000lbs? What is dx in this equation? Obviously E for steel is 3.0x10^8 psi. I listed above as 4

    I would appreciate it if someone could enlighten me on this formula. This is a simple calc and can be done other ways, I just want to be familar with this formula so i can use it on more adavanced calcs.

    Thank you all!
  2. jcsd
  3. Jun 18, 2010 #2
    This is not a complete answer to your question but it is how I would do it, and you might be able to work out what m and dx are from my answer. (What is missing from your question is the integral sign in front of "Mm dx/EI", and the limits that go with it. My method attached is a practical way to solve this integral in simpler cases. Let's call it the An method. A is the area of the M/EI diagram, which in this case is partitioned into areas a1 a2 and a3. We want the deflection at (say) any point D. Put a unit load at D. Draw its M diagram, and identify in that diagram the ordinates corresponding to the centroids of the areas a1 a2 and a3 in the M/EI diagram. Then Deflection at D is a1n1+a2n2+a3n3. Of course this can be done with formal integrals, but, even though I am perfectly capable of integrating, this semi graphical approach appeals to me for its practicality, even when a UDL and parabolic diagrams are involved. I hope the figure comes out. I haven't attached one before...

    Attached Files:

  4. Jun 18, 2010 #3
    So if i placed D in the center of the beam I come up with a deflection of .0012" with an I value of 4. Is this correct?

    To find a1 I did (10.75*5000)/2 /EI = 2.240e-4 - a3 has the same value because its symmetrical where my loads are.

    a2 is 5000*22 /EI = 9.167e-4

    To find n1 I used methods from this site to find Cx - http://www.efunda.com/math/areas/triangle.cfm
    Cx = 7.167 from left. I then found my angle of triangle to be arctanØ = 1/10.75 = 5.315° - tan(5.315) = y/7.167 = .667. n1 is .667 - n3 is the same because my unit load is placed symmetrically.

    n2 is on D so n2 is 1.

    2.240e-4*.667 = 1.494e-4 * 2 (both sides) = 2.988e-4 + 9.167e-4 = 0.0012"

    Did I do this correctly? I do like your method. Its similar to what I was asking, For all I know, it seems to be the same - Yes you are right, i forgot the sum of in front of my equation.

    Thanks for the post. Let me know if I was interpreting your method properly.
  5. Jun 18, 2010 #4
    I noticed an error... my triangle is 1/21.75 making n1 = .334 Ironically I did that initially on paper and when I typed my answer here i thought i made an error. Answer still comes up pretty close to the same even with the error. If my process was correct.
  6. Jun 19, 2010 #5
    I think you may have misunderstood the n diagram. It is the bending moment diagram for a unit load at the point of interest. So the central ordinate is not 1 but 1xL/4 which is 43.5/4 = 11.75 lbin. I think this makes n1 = 5.22, but I usually work in SI units, so could have made a silly error.
  7. Jun 20, 2010 #6
    OHHH okay.. let me review the problem again.

    So using trig to find my point at n1 was correct tho? sure theres different ways to find it but that seemed to be the easiest for me

    Also, what if my equation involved different types of supporting? Like lets say the ends were fixed? or i had a cantilevered side? How would this effect my problem? I know it is a different coeficent on the M/EI formula right? How do i know what coeficent to use?


    Also, what makes the formula 1*L/4? Why divided by 4? like i said above is this effected by my type of supporting?
  8. Jun 21, 2010 #7
    Am I using the right method for finding Cx (centroid on the x axis)? On a1 I come up with an x value of 7.167 from the left. To find my intersection with my hypotenuse i used trig. With my new understanding (and I understand why its 1*L/4 because if i draw my shear diagram it shows 1/2 * 21 3/4 is my moment which is 43.5/4). I come up with 10 7/8 lb-in. That makes a right triangle of 21.75 x 10.875. With my centroid of a1 being at 7.167 I took arctanØ=10.875/21.75 = 26.57° which makes my Y value at n1 = tan26.57 = y/7.167 = 3.58. This conflicts with your value which is why I think I may be calculating my centroid incorrectly. I appreciate the help, could you enlighten me? I'm coming up with a value of 0.012" at D now.

    Thanks again!
  9. Jun 21, 2010 #8
    I made a silly error. I now agree your value of n1. Also n2 is not what I said before but 10.875 So a1n1 + a2n2 + a1n1 again is .008 +.1 +.008 = 0.116 inches. So how did you get 0.012"? This may not help but the steel designers manual has a formula for this deflection and it gives .00973". So I conclude we are both struggling here. If the point loads were at the 1/3 points (14.5" from the ends) the formula is [23PL^3]/[648EI] and your deflection should be a bit less than that.
  10. Jun 21, 2010 #9
    You said a1 would be - area1/EI? So if i got 5000lbs x 10.75in = 53750lb-in then because its a triangle divide that by 2 = 26875lb-in/ E = (30,000,000 psi for steel) * I = 4 = 2.240x10^-4would this be correct? I take that value times my n1 value of 3.58 = 8.018x10^-4 * 2 (because my opposite side is the same because my load is symetrical) = 0.0016" + a2n2

    a2 = 5000lb * 22in = 110000lb-in/EI = 9.167x10^-4 * 10.875 = 0.00996 or 0.0100"

    0.0016" + 0.0100" = 0.0116"

    It seems correct in my mind. You having almost 1/8" seems like alot.

    My original formula was referenced from my Roarks book. In there they show an example of a different formula where they have a cantilevered situation and they multiply it by 1/3 (just like if you had a simple cantilever it would be pl^3/3EI) But sometimes they use coeficients of 1/24 and such, and I wonder when to use those values as well.

    Yes, I dont have a my mechinarys handbook near me but since this is a simple symetrical load they have a formula for it in there. But i want to learn how to use more advanced methods for not so simple situations. This why i can learn how to do it in a simple application and apply it later. I will have to reference that other formula and see if its getting values we are gathering or not.
  11. Jun 22, 2010 #10
    Yes, I did the calc out of my mechinery's handbook and got the same .0973" at center. So its in the middle of what we are both gathering. Can anyone else enlighten us with our formulas and what we might be doing wrong?
  12. Jun 24, 2010 #11
    I got this information from a different forum. I asked a question but i was banned because I told them i was still learning in school. Apparently you cant post there if your not "an engineer"

    Does anyone understand how to do this? Very simular to what we have talked about here.
    1) Draw the bending moment diagram for your loaded beam (M)

    2) Remove all of your loads and place a unit load (1.0 no units) at the point on your beam where you want to evaluate the deflection - say mid span

    3) Draw a second bending moment diagram for this loading (m)

    4) Multiply the values of the two diagrams together to obtain a third diagram

    5) Calculate the area of this third diagram (Mm dx)

    6) And lastly divide by EI

    I know this method as the `method of unit loads`. The formula is - deflection = Mm dx/EI

    It is easier than it sounds, give it a try and remember to keep your units consistent.

    Thanks guys
  13. Jun 26, 2010 #12
    This last message shows me how I slightly misled you. The sum of Ai*ni was right in principle, but a2, the central zone should have been taken as two parts, because the m diagram is different to the left of the centre line than the right. What i would like to know is: Can you do definite integrals? Can you take the origin as the left hand support and write down the three functions: (1) M diagram in the range 0 to 10.75 (2) M diagram in the range 10.75 to 21.75 (3) m diagram in the range 0 to 21.75, all as a function of x?. If you can do that, you should get an expression that you can integrate between the limits, add the two results together, divide by EI. Finally multiply by 2 because of symmetry (or develop new functions for the right hand side of the diagram, and show you get the same values, which you add to the first lot). You wanted a generalised method, which the integration method will do if you can write down the functions etc. You could use the sum of An method as a check. (An engineer will always look for independent checks on important calculations). Good luck.
  14. Jun 28, 2010 #13
    Okay, I investigated this more with your recent post and here is what I gathered.

    I know when I have a supported beam at both ends with a load in the center I have a formula of Pl^3/48EI. With that being said I have a coefficient of 1/48 right?

    I calculate my area of my M diagram.

    53750 * 10.75 / 2 = 288906.25 = a1
    53750 * 22 = 1182500 = a2
    a3 = a1.

    Sum of areas = 1760312.5 = area of M

    I calculate my area of my m diagram

    21.75 * 10.875 = 236.53125 = area of m

    Multiply the areas together.

    1760312.5 * 236.53125 = 416368916.016

    I now divide by EI

    416368916.016/(30,000,000)(4) = 3.46974" which is MAX deflection in any supporting application.

    Now multiply by 1/48 given how my beam is being supported

    Final answer = 0.0723" - This is closer to the answer we achieved with the basic formula.


    I've done some more research on deflection coefficients and depending on the application and amount of point loads, this coefficient that I used varies. I think that is what we are lacking. Based off the formula - Sum of Mm dx/EI, I feel like the above adding the areas of both moments, multiplying them together and dividing them by EI is correct. We are just missing the coefficient. Is this what dx represents?

    It appears they call this superposition. Appears to be a calculus method I will be learning this fall.

    So with that are we inching closer to this calculation?
  15. Jun 28, 2010 #14
    With further investigation with another beam I tried using this method and compared it to the equation I have and it is not working.

    Trying to figure it out but im burnt out for the day. Have to investigate with a fresh mind.
  16. Jun 29, 2010 #15
    Pl^3/48EI is correct for a centrally loaded simply supported beam. You shouldn't multiply the areas together, as you have done. The reason I asked if you could write down the functions of M and m is so that you could multiply those functions together, not their areas. If you are having to ask about dx then you haven't done calculus. This is not the time or place to teach it to you, but you can try it next year when you have done integration. Meanwhile I suggest you use approximate formulas or the An method. Best wishes.
  17. Jun 29, 2010 #16
    Im pretty determined and i am pretty good with numbers so im researching dx and am going to figure out how to use it. However, could u help me by writing a formula for my M moment so i can work with it and determine how you created it? Or, if not could you lead me to a site or book that might help explain it? I researched making equations from moment diagrams and I cant find a damn thing.

    I appreciate the help. If anything ive learned alot about some of this and im hoping to finish it and beable to calculate deflections with this formula.
  18. Jul 1, 2010 #17
    I admire your perseverance. So here is some extra work for you!
    First of all, dx is not d multiplied by x. It is a single concept called “dx” which is vanishingly small but not zero. You need to study calculus to get it. However, if you want to calculate the area enclosed by a function of x, you can divide it into a number of vertical strips, approximating to thin rectangles. Then add the areas of all the thin rectangles, each with a width delta x. Under a curved function there will be a small error, but the error is minimized by reducing the width of the rectangles; dx is the name given to delta x as delta x approaches zero. (This web site doesn't allow the lower case greek d we call delta) End of calculus lesson. Now for the functions.
    If you take the left hand support as the origin, then I hope you agree the m diagram is a triangle with point at the centre. In the region 21.75<x<43.5 the m diagram is a straight line that has to satify two points (43.5, 0) and (21.75, 10.875) where 10.875 is an interpretation of “PL/4” with P=1. If the equation of this straight line is y=ax+c, where a and c are coefficients, then you can substitute the two known points into that equation and obtain two simultaneous equations, which, when solved, should give you the equation y= -0.5x +21.75 As the signs depend on your sign convention for moment, you might get the negative of this: y= +0.5x-21.75 If you understood that, you should be able to work out what all the other functions are...
  19. Jul 2, 2010 #18
    Thanks for the post, I havnt gotten the time to investigate it to much farther yet. But that night I had my last post I recearched dx and i understand the concept i just need to put it together. Which i need my equation to determine. What i read and i will do more research for a better understanding, is its really the change in slope at the given point of where i want to calculate and it is a VERY small number. Taking 2 points very very close to gether to collect it. I had calculus in HS and i ended up having to drop it from a personal problem but we got to talking about limits and such, and unfortunitely i missed many days so i never got a full grasp of the concept then. Taking trig this last semester really helped me alot in the understanding of approaching zero. Example, sin to a very small angle times 180 gives you the value of pie. I found this very interesting.

    I will investigate the formula more when i get time. I think i can formulate them. Once i solve the 2 i suspect i need to set equal to x and fill the function in the 2nd equation correct?

    Thanks for your help! :-)
  20. Jul 2, 2010 #19
    I understand dx now watching youtube videos. They are very helpful. Now if i can make my funtion Mm I can then find the derivative of my Mm funtion and plug in x=21.75 then divide by EI. This would be correct right?
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