# Determining the first moment of area (Q) when calculating the shear flow of a built up member

Consider the two scenarios in this picture: On the left, calculating Q=A'y[bar]' is easy. A is the area of the top board since the cross sectional area of that board corresponds to the area on the opposite side of the shear plane to the centroid of the member.

I am confused about how to choose Q on the right hand side scenario. The two shear planes are perpendicular to the plane that separates A' from the rest of the member, meaning the definition I used for the first case on the left simply doesn't apply. Based on other sources I have consulted, the area I should choose is the area of the board in the center of the two upright members. I have no good justification for this choice.

Edit: More information: Before I was introduced to the concept of shear flow, I was taught how to make shear stress calculations at any given height above or below the centroid of a beam cross section. To calculate this shear stress, a first moment of area (Q) needed to be calculated. This Q value was explained to me to be the area of the cross section of the beam on the opposite side of the desired horizontal shear plane.

Take the example above. In that diagram, the shear plane is a plane running along the longitudinal axis of the beam at a height $y$ above the neutral axis. The $Q$ value = [Area of the hatched region (which is all of the area on the opposite side of the shear plane from the neutral axis)]$\times$[$\bar{y}'$(height of the centroid of this area above the neutral axis)].

That is how I understand Q. Now let's apply this understanding of Q to figure b in the first picture. The two shear planes are vertical planes that run along the longitudinal axis of the beam. Thus, a horizontal plane cannot be defined co-planer to the 2 vertical shear planes to denote that I want to calculate the shear flow along this horizontal plane. In short, my definition of Q has broken down spectacularly. How is Q defined for case (b)?

• engineering.stackexchange.com/questions/12232/… this question may help. My answer and the comments explains some of the reasoning (and also supports what you found at other resources).
– JMac
Dec 15, 2016 at 11:36
• Thank you for your response, JMac. Unfortunately, that question thread did not answer my question. I have edited my original post to include more information. about my confusion. Dec 15, 2016 at 17:20
• Can I check that the scenarios in (a) and (b) are intended to show sections through two long members extending into the page, which are subject to bending (and consequently shear)? Dec 19, 2016 at 12:27
• That is correct. Dec 20, 2016 at 19:19

As preamble, I note that the equation for this shear stress / force / flow is something that seems to suffer from almost everyone adopting a different nomenclature. I learnt it as longitudinal shear = Q A y[bar] / I. In that form, the longitudinal shear is a force per unit length along the section, and Q is the shear force in the section.

S = longitudinal shear force per unit length across a defined boundary of interest, that boundary being perpendicular to a plane section. This boundary is normally a single plane on which you could slice off a part of the section, but it doesn't have to be.

V = vertical shear force in the plane section.

A = an area (see below).

y = a distance (see below).

I = second moment of area of the whole plane section.

The equation in question is then S = V A y / I.

(The question has an intermediate value Q = A y, but I'm going to ignore that - you can substitute it back in if you like, having calculated A and y, at any point in this discussion.)

The value of A is the area of a 'chunk' of the section. The value of y is the distance from the neutral axis of the whole section to the centroid of the sliced off part. It doesn't actually matter if the chunk is the result of a single planar slice, or an arbitrary portion of the section.

In case b there's a chunk that's the top plank. It's a rectangle d wide and b high. The centroid of that part is exactly central, b/2 from the top and bottom faces of that plank. The y value will be the distance from the centre of that plank to the neutral axis of the whole section.

For sake of discussion, suppose the neutral axis of the whole section is 0.75xc from the bottom edge. Suppose also that a=d+2xb and e=b+c (ie, as it appears to be drawn).

In case a):

A = a * b

y = 0.25 * c + 0.5 * b

in case b):

A = b * d

y = exactly the same as case a), because the centroid of area A in case a) is at the same place as the centroid of area A in case b)

The second half of the question contains an error about what y is. The value y in the equation is not the distance from the centroid of the whole section to the cut plane. That is, y is not what is labelled as y in the diagram, it should be the distance from the whole section neutral axis at NA to the centroid of the hatched area.

It's worth noting that the S = V A y / I equation is only valid in the case of a prismatic section where the chunk is arranged such that A y and I do not vary along the length of the member. Some thinking about the derivation of the equation is relevant (and explains the assertions made above):

The axial bending fibre stress at any point on a section is calculated from stress = M y / I, where M is bending moment, y is now the distance from the neutral axis to the point where the stress is calculated and I is the second moment of area of the whole section.

If you consider any arbitrary chunk of the cross section, the axial force (F) on that whole chunk is the integration across the area of the chunk of (M y / I). That is, F = M A y / I, if A is the area of the chunk and y is back to being the distance from neutral axis of the whole section to centroid of the chunk.

Now, consider repeating that an increment along the member. There's now a different axial force on the chunk, and the way that force got there was by longitudinal shear across the perimeter of the chunk. Or, to put it another way, if x is position along the beam, the longitudinal shear flow across the chunk boundary = dF/dx = d(M A y / I)/dx. However in that differentiation, A y and I are all constant with varying x, so dF/dx = dM/dx A y / I. However, we know dM/dx (rate of change of bending moment along the beam) as something else - it's the vertical shear force in the section: dM/dx = V.

Hence longitudinal shear across the chunk boundary S = dF/dx = V A y / I

The important thing to note is that this derivation depends absolutely upon A y and I not varying with x - if they do vary, you can't simply recalculate S = V A y / I with different A values along the section, because with an A that varies with x (for example), d(M A y / I)/dx does NOT equal dM/dx A y / I. In that case you need to apply the differentiation product rule, so d(M A y / I)/dx = (dM/dx A + M dA/dx) y / I. Actually, of course, if A is varying with x it's almost certain that y and I are likewise varying with x, and you'll need to multiply out all four terms as varying-with-x.

• Thank you very much for your response. My main confusion still exists however. Why did you choose in case b that $A=b\times d$? In the calculation of shear flow (I know it as $q$, and I think you called it S), it seems that the way to calculate the area $A$ in $Q = A\times \bar{y}'$ is different than when you want to calculate $Q$ for the shear stress $\tau=\frac{VQ}{It}$ at a given height of a beam cross section. The rationale for choosing $Q$ in the former case is wholly opaque to me. Can you help me with that problem? Dec 22, 2016 at 23:38
• I'm sorry, I don't understand why you have difficulty. In the equation, the area used is the area the 'other side' of some boundary. In case b, the chunk is the top plank, the boundary is its perimeter, so the area the other side of that boundary is A = b d. It is exactly the same in your second case. There, the chunk is the top part of the beam, the A is the area of the top part (that part the other side of some boundary). It isn't different. It's the same. Why do you say it is different? Dec 24, 2016 at 8:46
• I think what I have not explicitly asked yet is why should any part of the area on the other side of the horizontal shear plane be excluded from the area calculation? (i.e. why is $A \neq b\times (d+2b)?$ (the whole horizontal section above the shear plane in case b, nothing excluded)). I see just as much reason for $A$ to equal $2b^2$ (the two little squares on the top left and top right) as $b\times d$ (the area of the separate plank in the middle of case b). Dec 26, 2016 at 6:45
• I thought you were asking how much shear force is conducted by the nails in your diagram. If so, all that matters is the amount of area held onto the section by the nails, which is d x b. If you want to know the shear force conducted to the whole of the top, all that area beyond the horizontal plane at the bottom of the top plank, you do use all the area in the equation. The equation calculates the shear applied to an arbitrary chunk of cross-section. Whatever chunk you want to examine, the area of that chunk is what goes into the equation. Dec 26, 2016 at 22:18