Finding Shear Centre of Lipped Channel

In summary, the conversation is about finding the shear centre of a lipped channel, with the person sharing their attempt at a solution by locating the centroid, calculating the second moment of area, and determining the Q-distribution for each section of the cross-section. They also mention finding the area under the Q-distribution diagram and getting stuck on determining the bounds of integration for the first section. They end by questioning the accuracy of their method.
  • #1
Seven7
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Homework Statement



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

http://img149.imageshack.us/img149/6324/88754244yh7.th.png [Broken]

I need to locate the shear centre based this.

Homework Equations



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);

http://img144.imageshack.us/img144/6072/10584649lt5.th.png [Broken]

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.
 
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  • #2
Since the links are now broken, it is impossible to comment.
 

1. What is the shear centre of a lipped channel?

The shear centre of a lipped channel is the point on the cross-section where the applied shear force causes no twisting or rotation. It is an important parameter in structural engineering as it helps determine the bending and shear stresses in the channel.

2. How is the shear centre of a lipped channel calculated?

The shear centre of a lipped channel can be calculated using the geometric properties of the cross-section, such as the area, moment of inertia, and centroid. It can also be determined experimentally through physical testing or through finite element analysis.

3. Why is it important to find the shear centre of a lipped channel?

Knowing the shear centre of a lipped channel is crucial in designing and analyzing the structural integrity of a component. It helps determine the distribution of stresses and deformations, which can affect the overall stability and strength of the structure.

4. Can the shear centre of a lipped channel be located outside the cross-section?

Yes, the shear centre of a lipped channel can be located outside the cross-section. This is often the case when the channel has an asymmetrical shape or when there are additional elements attached to the cross-section, such as flanges or stiffeners.

5. How does the location of the shear centre affect the design of a lipped channel?

The location of the shear centre can significantly impact the design of a lipped channel. If the shear centre is not located at the centroid of the cross-section, it can result in additional bending and torsional stresses, which may require additional reinforcements or modifications to the design to ensure structural stability.

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