Just a basic question on this not-so simple topic. Shouldn’t wind load have a sheer AND a normal force? Ok, that was the simple question.
Now…here’s what I’ve found so far: All the many texts I’ve found all seem to be addressing wind sheer. From there, these texts lead one to believe it IS the drag force on an object. As such, the formula applies a drag cofficient. Yet, other texts seem to derive almost the same formula - only without the use of a drag coefficient. Nonetheless, the body of text seems to be always describing sheer. BUT!!! No text (as of yet) addresses the fact that (in my little opinion) the wind also applies a normal load to any surface facing into the wind…doesn’t it?
Let’s say it’s a vertical cylinder in the wind. Heck, make it a rectangular smoke stack, if you prefer! One side will be windward and there would be a distributed load along the length of the cylinder, wouldn’t it? And, if so, how do I calculate this wind load if I know wind speed, the area of said surface, and the overall geometry of my object in the airstream?
Any links to websites and/or cites would be wonderful! Even if your link simply helps to serve to clarify some of my confusion!
First, it’s “shear”. if you mean the gradient in velocity along an axis perpendicular to the velocity. In the context of wind I think typically it’s a gradient in horizontal velocity along a vertical axis.
What are you trying to calculate? The load on a vertical structure due to wind? I think there are approximations for that but don’t know much about them. If it’s the force a moving fluid imposes on a solid object, well, that’s a major oblique slice through the entire field of fluid mechanics.
The drag force on a cannonball got studied heavily long ago - Newton, IIRC - and there’s this idea of a drag coefficient and the Reynolds number that help deal with it all, which has been extended in many directions, including no doubt the cylinder. Look up these topics.
About shouldn’t wind have a shear and a normal force, well, yes, if you mean shouldn’t it push against an object in both these ways. I think the normal force is way, way bigger for smokestacks.
Napier, thanks for the spelling correction. But, your reply has followed my footsteps bringing me around through the same circle I just followed. I got the drag part down, fine. It’s the normal force on a smokestack for which I cannot find any text discussion to reference. Hence, this brings me back to my original question. - Jinx
The force perpendicular to the freestream direction is usually referred to as “lift” when dealing with long, slender objects, even when they don’t resemble wings at all. Force parallel to the freestream is usually called “drag”, as you know.
A cylinder, just sitting there (not rotating or anything) will have zero lift in a time-averaged sense. However, assuming the flow is fast enough (i.e., the Reynolds number is high enough) it will shed alternating-sign vortices known as a Kármán vortex street (so named, I’m told, because the vortices somewhat resemble lightposts down either side of a street). This also produces lift forces of alternating sign - it’s a major issue when designing cylinders that will have to sit in a breeze. Books have been written on this topic, but here is a paper written in 1963 in which measurement and characterization of the forces is described.
If the cylinder isn’t rigidly fixed in space (as an example, think of a cable that has some ability to swing around laterally), things get even more complicated. Arguably the seminal work on this topic was a paper by Charles Williamson and Anatol Roshko in 1988 - Williamson was Roshko’s grad student at Caltech at the time. Now on the faculty at Cornell, Williamson is still at it on this topic - here is a meeting presentation he gave in 2005. That chart on page 9 (“Map of Wake Modes”) broadly characterizes the various patterns one can get if the cylinder is free to move transversely, depending on how several parameters relate to one another.
Mory Gharib at Caltech is another guy I know of who still does some work in this field - see the last entry on this page.
Assuming I’ve correctly guessed what you’re asking about, I hope this at least gives you some buzzwords to search under.
Brad_d, thanks for the details. Just to clarify, I wasn’t really picturing a force perpendicular to the airflow (i.e., lift), but say…the force I experience when walking into the wind that wants to push me backwards. If you are blocking the wind, I would presume it wouldn’t be drag per se…as I picture drag as the friction force created from air moving across a surface (i.e., brushing beside me in this example), but then again…
Re-reading the fluids text more carefully, I have come to understand that there is an overall Drag Coefficient (Cd) which is the sum of a friction drag coefficient (Cdf) from shear + pressure drag coefficient (Cdp). The latter, I presume, accounts for that “push” from a windload on a surface blocking the airflow. I say “presume” because the fluids text was not explicitly clear on this.
Also, I should add, the general equation for calculating Cd for common geometries (flat plates, spheres, and ellipses in an airstream) employes the product of width * length called the frontal area OR planform area…depending on the specific geometry.
Brad, would you say my summary is a fair assessment, in a nutshell?
Ah, OK - I see what you’re asking. “Drag” is sort of an all-encompassing term, referring to whatever net force a body experiences in the direction of the freestream flow.
You’re right - in most situations it’s a combination of two things - skin friction drag (shear stresses on the surface of the body) and pressure drag (caused by pressure differences). The main reasons that they’re usually lumped together are:[ol][li]Their combined effect is quite easy to measure with only a force balance; measuring them separately is a good deal more complicated; and[*]From an engineering perspective, you often don’t really care how they’re broken down - the total drag is what you’re after in the end.[/ol][/li]
I don’t have a lot of experience with this, but in a lot of situations the pressure term greatly outweighs the skin friction one. It’s only when the Reynolds number is pretty low (normally) that they skin friction component becomes significant. (Exceptions abound, I suspect.)
And, yes, the formulation for C[sub]D[/sub] involves the frontal area of the body. There are a number of ways to explain/justify doing it this way - here’s one that works well for me:
D = C[sub]D[/sub]*(1/2)rhoU[sup]2[/sup]*A[sub]f[/sub], where rho is the fluid density, U is the freestream speed, and A[sub]f[/sub] is the frontal area of the object. As a starting point for calculating drag, one simply hypothesizes that the entire area “blocked” by the body feels stagnation pressure plus ambient (while the backside feels only ambient), and that when integrated over the entire surface you’ll get most of that term above, in the proper units:
(1/2)rhoU[sup]2[/sup]*A[sub]f[/sub]
C[sub]D[/sub], basically, grabs the details that vary from body to body as a “fudge factor” to that simplified approach. The difference between and flat disk and a sphere of the same radius will be captured in the drag coefficient.
The drag coefficient for a given body need not be constant - but absent other complicating factors it will be a function strictly of the Reynolds number. Same Re <=> same C[sub]D[/sub], but note that that’s not the same thing as saying that the drag force is the same.