 In this example, we are going to examine the shear flow distribution in a thin walled I-beam section shown on the left. The problem asks us to determine this shear flow distribution for an arbitrary internal shear force V that can occur in the section, and the problem also specifies that it is safe to assume that the beam section is thin walled. We are going to look at trying to determine the distribution of this shear flow through this section, and because the problem is thin walled, it is safe to assume, although we do not have the magnitudes of the dimensions, that T is much, much smaller than the height or width of the cross section. It is also important to recognize that the problem is double symmetric. This will come in handy because we only need to solve for the shear flow distribution in one half of one flange and one half of the web, and then we know the distribution in the other parts due to symmetry. Let's first start off by examining shear flow in the flange of the I-beam section. To help us remember that it is a thin walled section, and to simplify some of the calculations, let's consider the median line of the I-beam section as the actual section, therefore the height to the middle of the flange is approximately equal to d over 2. This is the thin walled approximation saying that the thickness of this flange is much smaller compared to the dimensions of the section. Now you could actually put the real height in here, which would be d over 2 minus thickness over 2, but for the simplicity of the expressions written here, we'll leave it here, we'll say that that thickness is negligible compared to this value. Now we have to apply our shear formula that the shear flow, little q, is equal to vq divided by I. Now for this problem we don't actually have magnitudes of the dimensions, we're going to keep everything in variable form, therefore I'm going to leave I for the cross section just as I, I won't calculate it in terms of the different dimensions of the cross section, I recognize that it is a constant for the cross section and for the simplicity of the expressions, I'll just leave it at I. What we need to do is figure out what q is in the flange. Now if you recall what q is when we developed the shear formula, we looked at sectioning a portion of the beam away and finding the shear flow or shear stress that maintained equilibrium of that small component sectioned away from the beam. So we're going to do this process to develop q in our flange here, but instead of doing it exactly as shown here, I'm going to do it in the positive quadrant, so on the left here just so that I don't have to play around with negative signs in my x variable. So I will specify a section point, a distance x away from the center of the web and that will define this green area which is equivalent to this small area sectioned away. So the question becomes what is q for this green sectioned area? Well q is y bar times a prime. Now if you forget what those two things are, y bar is the distance from the neutral access to the centroid of our area of this sectioned portion and in this case that will be the height d over 2. A prime is the cross sectional area of that section and if we calculate that it will be b over 2 minus x times the thickness and it gets us this expression. Therefore I can take this expression, substitute it into our original value for q, v is constant given by the problem and i will be left as i for this problem and we will get that q or little q shear flow is v times t times d divided by 2i multiplied by bracket b over 2 minus x closed bracket. What we can recognize from this is that shear flow linearly decreases with x. Now if we move on to the shear flow in the web we have to repeat the same procedure however now we need to make sections in the web that create a little bit more complicated area to deal with. Now again I'm sectioning down below the neutral access to stay in the positive quadrant of y but we know due to symmetry of this cross section the distribution we have on this side will be equal to the distribution we have on the top side. So this is just to keep the math simple for now. So we have to go through the same procedure and determine what q is in the web because we know that i is a constant and v is a constant for this section. So again q of this sort of green cross sectional area is y bar times a prime. What we can recognize that this area defined by this green block is actually a compound area and make our lives a little bit easier by dividing that into two rectangular areas and I'll label them 1 and 2. So q for all of this area then becomes y bar 1 times a prime 1 plus y bar 2 times a prime 2. Now looking at the first area 1 this value will actually be a constant. If I'm sectioning through the web this area will always be the same and it will be equal to the distance to the centroid is just d over 2 as it was in the previous analysis for the flange and now the area is constant it's not changing because we're not sectioning at different various points along x. It is just the area of the flange which is b times t. Moving on to the second area y bar 2 will be defined as a distance from the centroid of that area to the neutral axis and in this case that distance will be y the distance to our sectioning plane plus one half the height of this rectangle. So it's plus one half d over 2 is this distance minus y. Then we have to look at the area of that second compound area and that will be the thickness times this height which we already established was d over 2 minus y. If we combine that and simplify it we can factor out the thickness and multiply through and get this following expression. Now what we need to do is substitute this expression that we've established for q of this compound area into our shear formula. So if we do that we again get that v and i are constant we can factor out t from this expression and get vt over i times open bracket db over 2 plus one half opened circular bracket d squared over 4 minus y squared closed circular bracket closed square bracket. Now what we can recognize from this is this expression is parabolic. It varies with y squared and it varies from maximum at the neutral axis and it decreases as we go away from the neutral axis. So now that we've solved for the shear flow distribution in both the positive quadrant of the flange and the positive quadrant of the web we need to put everything together to take a look at what our overall shear flow distribution is. We saw that in the flange we had a linear distribution for shear flow which is maximum at the center line of the web and it decreases to zero at the outer edges of the flange. We can see that the maximum value occurs at that central portion and is given by the value here vt times db divided by 4i. Now if we look at the web we found that the shear flow varied due to this parabolic distribution and it is maximum at the neutral axis so it has a maximum value here and it decreases as we go away from the neutral axis. It doesn't quite go to zero which makes sense because there's still shear at this intersection point due to the stresses carried by the flange. Now we can calculate what the minimum shear flow in the web is by substituting in y is equal to d over 2 and we find that the minimum shear flow in the web is vt times db over 2i. Now it's interesting to note that this value is twice the minimum q in the flange and hopefully that will make sense in a second when we again overlay these graphs are showing the distribution, the linear increase and the parabolic variation but shear is acting on the section according to these arrows and it actually flows through the section so it makes sense that the shear flow in the web here is equal to twice the shear flow in the flange because we have two streams of shear flowing into that web and same at the bottom the stream splits and so we go from twice the shear flow here to half the left and right in the flange. And there you have it, this is a simple example looking at shear flow in an I-beam section. Now the interesting thing to keep in mind in future problems is sort of that property of the shear flow combining so that you can double check your work to see that when you have two shear flows combined into a single one that you should be able to do the addition of those magnitudes to check your work. The other interesting thing that we see is that in any portion of the section that is horizontal or perpendicular to the internal shear force our distribution of shear flow is linear and that will hold in other sections as well and in vertical or portions of the section that are aligned with the internal shear force our distribution is parabolic and that will also hold in other sections as you'll see in sample problems that you do. And then finally our maximum shear flow is always at the neutral axis and if you recall how we derive the shear formula that should make sense because if you do the section at the neutral axis it's right at the point where the normal stress due to bending switches from compression to tension and so sectioning it there will get you the maximum amount of normal stress in the same direction to be balanced by the shear flow or shear stress.