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Homework Help: A tower against wind

  1. Jun 6, 2008 #1
    1. Suppose we have a tower at seaside (with length L) under a constant speed of wind.
    Let's assume the force is constant. If we took the drag force equation, only serious variable here is the speed of wind and for a specific time frame let's say for hour or few hours, wind speed and direction do not change so we surmise it's constant.

    How can we calculate the tension created all over the tower due to wind? Albeit it's constant it is applied from bottom to the top so there must be a cumulative effect.
    How can I formulate this using an integral?

    {INTEGRAL -from 0 to L} F dL yields to FL which gives the torque at the top, NOT accruing one.

    2. Torque = Force x Length
  2. jcsd
  3. Jun 8, 2008 #2
    No reply? I'm suprised. Is it that much simple or that much diffucult?
    or perhaps that much insignificant? :)
  4. Jun 8, 2008 #3


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    Hi yasar1967! :smile:

    I don't know about the tension, but if you want the torque (measured about a horizontal line through the base), it's the integral of pressure x area x height, = pressure x width x d(height) x height, = 1/2 x pressure x width x height-squared. :smile:
  5. Jun 8, 2008 #4
    Thank you but why "area"? Torque formula consist no area just force multipled by level arm.
    Force x distance = Torque
  6. Jun 8, 2008 #5
    If I've understood your question right this should be something like part 1 in the figure I've attached. That case can be transformed to a free body diagram, see part 2. To find the unknowns we can apply three equations of equilibrium:
    Sum of vertical forces: This easily concludes that Ay = 0
    Sum of horizontal forces: qL - Ax = 0 => Ax = qL
    Sum of torque around point A (positive counterclockwise): Ma - qL*L/2 = 0 => Ma = qL²/2

    Then take a look at part 3 of the figure. This is a cut placed a distance x from point A. If we apply the same equations of equilibrium here we get:
    Sum of vertical forces: N=0
    Sum of horizontal forces: V + q*x - qL = 0 => V = qL - qx
    Sum of torque around the cut (positive counterclockwise): qL²/2 - qL*x + qx*x/2 - M = 0 => M = qL²/2 + qx²/2 - qLx

    In other words, the shear force is given by the function V(x) = qL - qx. The torque is given by the function M(x) = qL²/2 + qx²/2 - qLx, with positive rotation counterclockwise. In both cases x ranges from 0 to L. This gives a maximum torque of qL²/2 at x = 0. At the top (x = L), zero torque will occur.

    I hope I understood your problem correctly.
    If I went a bit to fast ahead somewhere, please feel free to ask. This was a rather quick discussion of the task and all of it might not be clear.
    Also, if you want to calculate the stress/tension distribution in a cross section this gets a bit more complicated, but I can try to explain that too if you wish. The stress distribution across a cross section depends on the cross section's moment of inertia so you can't calculate that without knowing how the tower looks like.

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    Last edited: Jun 8, 2008
  7. Jun 8, 2008 #6


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    'cos wind has pressure, and pressure depends on area.

    To put it another way, if the tower is much wider on the north/south side than on the east/west side, then a North wind will exert much more force on it than an equally strong East wind, won't it? :smile:

    That's because there's more area exposed to a North wind!
  8. Jun 9, 2008 #7
    Tiny-tim is right. The distributed load because of the wind depends on the exposed area. I forgot to take that into account when I explained my way of doing it. However, if the tower has the same exposed area all over it's easy to add this. Say that the area has a width, b and (of course) a height L. If we call the wind pressure p (dimension [N/m²]). The distributed load q over the tower (dimension [N/m]) will the be: q = p*b

    That gives a distribution of torque by the function M(x) = M(x) = pbL²/2 + pbx²/2 - pbLx and a maximum torque of pbL²/2, the same as tiny-tim suggested.

    Remember that not all real towers have a constant exposed area. Many towers tend to be larger at the bottom and shrinking upwards. This will cause the distributed load to shrink the further up you get and the distribution of torque will be different. It's quite logical that they're built that way. Since the torque is biggest at the bottom, a stronger construction is required to hold back for this than it is required further up.
    Also, as you said, the wind is seldom distributed evenly. This makes this problem even more complicated if you want it to be more realistic.
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