I-beam Shear Force calculation

[SacredTear]

So I'm taking a Strengths of Materials class right now (aka Mechanics of Materials) and my teacher absolutely sucks. We ask him questions but he doesn't have answers, so in turn I have no idea what is going on and why the book does what it does.

Right now we're working with I-beams and some of the problems deal with Shear Force (not to be confused with Shear stress). The question says "If the wide flanged beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam." Then it gives the dimensions and all that. I've asked my teacher to do this problem but he just copies from the solutions manual and doesn't explain anything.

So what the solution did was find the things needed for the equation T(shear) = VQ/It. It first finds I, which I can figure out, but I don't get how it calculates Q. It takes the height from the Neutral Axis to the top (.155 in this case), then adds an arbitrary y value. It divides this value by 2, then multiplies by (.155 - y)*(.2) ...where .2 is the length of the top of the I-beam. And that's it. Usually when Q is calculated (that I've seen) it's the distance of the centroid of the part to the NA, multiplied by that parts area. (Part being...that you can break I-beams into the horizontal part and the vertical part). But I don't understand why in this problem it seems to be taking the whole area (of a box instead of the top half of the I-beam) and multiplying it by (.155 + y)/2. Where does this value come from?

Then in a DIFFERENT problem reading, "If the wide flange beam is subjected to a shear of V = 25 kip, determine the shear force resisted by the web of the beam." I is calculated the same way...but Q is totally different than the previous problem. It calculates the horizontal part like I thought, but then it adds what I presume is the vertical part, making it look like:

(.5y + 2.5)*(5-y)*(1)

Where the total height of the vertical part is 10 inches (thus the 2.5 is 1/4 of 10 [half away from the NA]) and 5-y I guess is the arbitrary height of the "point". I have no idea where the .5 comes from (except maybe half of the thickness of the horizontal part, which is 1 in), but why is it multiplied by y? And why is it added to the horizontal part whereas in the last problem it was just the Q of what seemed to be the whole box?

Sorry if this is terribly confusing, but I don't know how else to explain it without using pictures. I'm just confused what their method of calculating Q is and why it's different between the two seemingly same problems. Thank you for any help in advance and let me know if you need me to clear some things up! ^_^

 

[jasonbot]

I know this is really old but I have exactly the same issue! I'll be asking my lecturer within the week though. I think it has to do with the thickness value used though...

lol

 

[Studiot]

Hello Jason, you would do better to post mechanics of materials questions in the mech eng forum.

Here is a good presentation.

The key phrase is 'shear flow'.

http://www.google.co.uk/#sclient=ps...&aq=f&aqi=&aql=&oq=&pbx=1&fp=e8605a0cb6e399f5

go well

 

[jasonbot]

Thanks. Its just that I had exactly the same question as the OP which I thought was interesting. Anywyas I figured it was just a different method of calculation.

Hello Jason, you would do better to post mechanics of materials questions in the mech eng forum.

Here is a good presentation.

The key phrase is 'shear flow'.

http://www.google.co.uk/#sclient=ps...&aq=f&aqi=&aql=&oq=&pbx=1&fp=e8605a0cb6e399f5

go well

 

[alphy]

I understand your frustration with your teacher's lack of explanation and the confusion surrounding the calculations for shear force on I-beams. Let me try to break down the process for you and hopefully provide some clarity.

First, let's define what we mean by shear force. Shear force is the force that acts parallel to the cross-sectional area of a material, causing it to slide or shear. In the case of an I-beam, the shear force acts along the web (the vertical part) of the beam.

To calculate the shear force resisted by the web of the I-beam, we use the equation T(shear) = VQ/It. This equation tells us that the shear force is equal to the product of the shear stress (V) and the first moment of area (Q) divided by the moment of inertia (I) and the distance from the neutral axis (t).

The first thing we need to determine is the moment of inertia, which is a measure of the beam's resistance to bending. This can be calculated using the dimensions and properties of the I-beam.

Next, we need to find the first moment of area, which is the product of the area and its distance from the neutral axis. In the case of an I-beam, we can break it down into two parts - the top flange and the web. The top flange has a larger area, so its contribution to the first moment of area (Q) will be greater.

Now, to answer your question about the different methods of calculating Q in the two problems. In the first problem, the Q value is calculated by taking the distance from the neutral axis to the top of the beam (.155) and adding an arbitrary distance (y) to account for the contribution of the web. This is then multiplied by the length of the top flange (.2) to get the first moment of area for the top part of the beam. This is because the web only contributes to the first moment of area for the top part of the beam, not the entire cross-sectional area.

In the second problem, the Q value is calculated by breaking down the I-beam into two parts - the top flange and the web. The first moment of area for the top flange is calculated using its area and distance from the neutral axis. The first moment of area for the web is calculated using its area and the distance from the neutral axis to the point at which we