The figure shows the cross-section of a lipped channel;
I need to locate the shear centre based this.
Second moment of area / inertia; Ixx = (1/12)bh^3 + Ay(bar)^2
First moment of area; Q = A*y(bar)
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);
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 by a moderator: