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!

Structural mechanics; stress distribution on a cross section; locating shear centre

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data

    The figure shows the cross-section of a lipped channel;

    [​IMG]

    I need to locate the shear centre based this.

    2. Relevant equations

    Second moment of area / inertia; Ixx = (1/12)bh^3 + Ay(bar)^2

    First moment of area; Q = A*y(bar)

    3. The attempt at a solution

    Firstly, I located the centroid of the shape. I found it to be at x(bar) = 40.28mm and y(bar) = 150mm.

    Secondly, I found Ixx (the second moment of area / inertia) for the shape. I found Ixx = 9.04x10^7 mm^4.

    Next I worked out the Q-distribution for the cross-section (refer to image below);

    [​IMG]

    For the first section labelled 1;

    Q = 15s(150 - s/2) = 2250s - 15/2*s^2

    Intuitively, (or by taking the derivatives) we know the maximum value for s will be 150;

    Q(150) = 1.688x10^5 mm^3

    For the section labelled 2;

    Q = 1.688x10^5 + 15s x 150 = 1.688x10^5 + 2250s

    Same as before, s = 100 is the maximum;

    Q(100) = 1.688x10^5 + 2250(100) = 3.938x10^5 mm^3

    Finally for section 3;

    Q = 3.938x10^5 + 10s(150-s/2) = 3.938x10^5 + 1500s - 5s^2

    Q(150) = 3.938x10^5 + 1500(150) - 5x(150)^2 = 5.063x10^5

    Next I assumed that the cross-section was loaded through the shear centre with a force of 100KN. The idea being that I can resolve the Q-distribution on the webs of the cross-section and then take moments about a certain point and find the shear centre.

    In order to do this I tried to find the area under the Q-distribution diagram. For the purpose of this question we can ignore the section labelled 2 (because the only applied shear is vertically through the shear centre;

    For section 1;

    T1 = VQ/It = 100000 N / 9.04x10^7 mm^3 x 15 mm x (2250s - 15/2*s^2)

    T1 = 0.1659s - 5.5313x10^-4*s^2

    Now taking the integral;

    integral T1*15ds = 15 x ( (0.1659/2*s^2) - (5.5313*10^-4/3*s^3) )

    ****** Here is where I get stuck ******

    I'm not sure what the bounds of integration should be on the first section. I'm not sure whether it's 0 to 30; 270-300; or even 130-150. Any advice / direction would be greatly appreciated.

    In addition to this;

    T3 = VQ/It = 100000 N / 9.04x10^7 mm^3 x 10 mm x (3.938x10^5 +1500s - 5s^2)

    T3 = 43.562 + 0.16593s - 5.531x10^-4*s^2

    Now taking the integral;

    integral T3*10ds = 10 x (43.562s + (0.16593/2*s^2) - (5.531x10^-4/3*s^3) )

    This is bounded between 0 and 300 so we get;

    integral T3*10ds = 15756.75 N = 157.6 KN

    If this is correct then we can expect that section 1 will be taking roughly 50KN of shear stress across the two small lips; i know this is the case because by doing a simple for balance, we know that the sum of forces in the y-direction for the 3 sections of the web (large body and 2 small lips) will sum to 100KN (the applied force); 50KN on the small lips sounds a bit large, so I'm thinking there is something wrong in the method up until this point.

    Thanks for any help guys.
     
    Last edited: Mar 29, 2008
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Structural mechanics; stress distribution on a cross section; locating shear centre
Loading...