1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal and shear stress distribution due to shear force and bending

  1. Aug 11, 2013 #1
    Hi guys, I have a repeat exam in 10 days and I have a couple of things which I would appreciate if you would help me with.

    I have attached the mock exam and my attempt at question 2. I will describe where I have issues below.

    1. The problem statement, all variables and given/known data

    For the FIRST question, I have the equation: Sigmaz= N/A + MxY/Ix - MyX/Iy.

    I understand how to proceed with the question if the bending moment were to point at one direction (x or y axis), but when it is at an angle I get confused and I'm unable to proceed. Can anyone lead me to the right directions with this?

    For the SECOND question, I have attached a file of my attempt at it. I understand that the units are messy, but I believe everything is correct, until the I get stuck at calculating the shear stress distribution below the Neutral Axis. I believe the diagram should curve back down and go back to zero at ''GH'', but when sum all the areas ABOVE 'GH' I do not go back to zero and the diagram does not look right. Can anyone help me proceed after this?

    I am yet to attempt the THIRD question.


    3. The attempt at a solution

    Attempt of question 2 is attached.



    Any help will be greatly appreciated, thanks.
     

    Attached Files:

  2. jcsd
  3. Aug 11, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    1. You are given a bending moment magnitude and told that it acts at angle β to the horizontal axis. Decompose the bending moment into its horizontal and vertical components, just like it was a force. You will have a bending moment acting about the horizontal axis and another moment acting about the vertical axis. For the stress at a particular point, use the appropriate bending stress formula for each bending moment component. Don't forget to add the two bending stresses together after these calculations (both stresses will be normal to the cross section, but may act in different directions depending on location w.r.t. the N.A. of the beam and the direction of the moment.
     
  4. Aug 11, 2013 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    2. This problem is tricky because the diagram shows Vy = 400 kN, but the question text asks to calculate the shear stress distribution if Vy = 300 kN.
     
  5. Aug 11, 2013 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    2. Your calculation of the Q values is incorrect. Q is the FIRST moment of area of the cross-section above the shear plane, w.r.t. the N.A. of the whole cross-section of the beam. I don't know where you got Q = A * x^2 from, and the units (cm^2 * cm^2) certainly are not cm^3.
     
  6. Aug 12, 2013 #5
    Thank you StreamKing. I will will split the moment into vertical and bending moments and attempt the question and will post it hear if I get stuck.

    In question 2, I just recalled that my lecturer made a mistake when writing out the problem. The Vy is 400Kn not 300. Also, I realized I made a mistake when calculating Q. Q= Sum of AY above the point, not AY^2... I guess I got a little confused when solving the problem.
     
  7. Aug 12, 2013 #6
    So I've attached my attempts at questions 1 and 2.

    At the first question, I've found the vertical and horizontal components (Mx, My) and put them into the equation and the answer is what I got on the bottom of the attached page, in terms of X and Y. If I put x=0, y will also be zero and vice-versa.
    I've tried using values of x=1, then I get y... and then y=1, then I get x. Then I draw the x,y reference where the neutral axis is the origin. I then get a slope, draw a line across the two points and I guess that is where the normal will be equal to zero, but I'm not sure how to finish the question and I have no clue if what I'm doing is right.

    At the second question, I've redone it and you will see where my issue is at the end.
    I believe that the shear should be equal to zero at the top and bottom of the beam, but after the neutral axis, the Q value will increase and so and so will the shear stress along the beam, rather than go back down.
    As far as I know, Q is the sum of the areas x distance from center of the individual part to the whole beam, above a certain point.

    Any help please?

    Thanks
     

    Attached Files:

  8. Aug 12, 2013 #7

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    On p. 3 of your calculations, the bottom shear stress distribution is the correct one. The shear stress is zero at the top an bottom flanges of the beam. The shear stress jumps where the flanges connect to the web, and the shear stress is a maximum at the N.A. In calculating Q, you always start at the outermost fiber and work back toward the N.A. and stop there.
     
  9. Aug 12, 2013 #8

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    On p. 1 you have made a bonehead mistake: 500 kNm = 500,000 Nm, not 50,000 Nm

    Also, at the bottom of the page, your moment units are indicated as 'kNm' when clearly they should be 'Nm'
     
  10. Aug 12, 2013 #9
    500kNm= 50,000 kNcm (kilonewton CENTImeters). Not the most used units but I guess they are still correct.
     
  11. Aug 12, 2013 #10

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It would help you to use a tabular form to calculate section properties:

    Code (Text):

    Section properties for Problem 2:

    Note:  measurements in cm

    Item     A   y-bar   A*y    A*y^2        I

    U-Flg   20   9.00   180.0   1620        6.667
    Web      7   4.50    31.5    141.75    28.583
    L-Flg   10   0.50     5.0      2.50     1.667
    ----------------------------------------------
    Total   37   5.85   216.5   1764.25    36.917

    INA = 1764.25 + 36.917 - 216.5^2/37 = 534.35 cm^4

    [Note:  A*y^2 = (ƩA*y)^2 / ƩA]

    y-bar = 5.85 cm abv. bottom of btm. flg.


    Qcd = 20*(9-5.85) = 63 cm^3
     
     
  12. Aug 12, 2013 #11

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    In problem 1, shouldn't you be looking for points where the combined stresses are a maximum and not zero?
     
  13. Aug 12, 2013 #12
    Thanks steamking, the tabular form is a helpful technique!

    I have paid a person on fiverr do to the questions for me in detail, he will hopefully do it tomorrow, so I guess I'm done here.

    Thanks for the support :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Normal and shear stress distribution due to shear force and bending
Loading...