Unformatted text preview: VII. Boundary Layer Flows
The previous chapter considered only viscous internal flows.
Viscous internal flows have the following major boundary layer characteristics:
* An entrance region where the boundary layer grows and dP/dx ≠ constant,
* A fully developed region where:
• The boundary layer fills the entire flow area.
• The velocity profiles, pressure gradient, and τw are constant;
i.e., they are not equal to f(x),
• The flow is either laminar or turbulent over the entire length of the flow ,
i.e., transition from laminar to turbulent is not considered.
However, viscous flow boundary layer characteristics for external flows are
significantly different as shown below for flow over a flat plate:
U∞ laminar to
turbulent
transition free stream y edge of boundary layer
δ(x) turbulent x laminar
xcr Fig. 7.1 Schematic of boundary layer flow over a flat plate
For these conditions, we note the following characteristics:
• The boundary layer thickness δ grows continuously from the start of the
fluidsurface contact, e.g., the leading edge. It is a function of x, not a
constant.
• Velocity profiles and shear stress τ are f(x,y).
• The flow will generally be laminar starting from x = 0.
• The flow will undergo laminartoturbulent transition if the streamwise
dimension is greater than a distance xcr corresponding to the location of
the transition Reynolds number Recr.
• Outside of the boundary layer region, free stream conditions exist where
velocity gradients and therefore viscous effects are typically negligible.  v v VII1  eText Main Menu  Textbook Table of Contents  Study Guide As it was for internal flows, the most important fluid flow parameter is the
local Reynolds number defined as
Re x = ρ U ∞ x U ∞x
=
µ
υ where
ρ = fluid density
µ = fluid dynamic viscosity
ν = fluid kinematic viscosity U∞ = characteristic flow velocity
x = characteristic flow dimension
It should be noted at this point that all external flow applications will not use a
distance from the leading edge x and the characteristic flow dimension. For
example, for flow over a cylinder, the diameter will be used as the characteristic
dimension for the Reynolds number.
Transition from laminar to turbulent flow typically occurs at the local transition
Reynolds number which for flat plate flows can be in the range of 500, 000 ≤ Re cr ≤ 3,000, 00
With xcr = the value of x where transition from laminar to turbulent flow occurs,
the typical value used for steady, incompressible flow over a flat plate is Re cr = ρ U∞ xcr
= 500, 000
µ Thus for flat plate flows for which: x < xcr the flow is laminar x ≥ xcr the flow is turbulent The solution to boundary layer flows is obtained from the reduced “Navier –
Stokes” equations, i.e., NavierStokes equations for which boundary layer
assumptions and approximations have been applied.  v v VII2  eText Main Menu  Textbook Table of Contents  Study Guide Flat Plate Boundary Layer Theory
Laminar Flow Analysis
For steady, incompressible flow over a flat plate, the laminar boundary layer
equations are:
Conservation of mass: ∂u + ∂v = 0
∂x ∂ y 'X' momentum: u 'Y' momentum: −∂p = 0 ∂u
∂u
1 dp 1 ∂ ∂u
µ +v
=−
+
∂x
∂y
ρ d x ρ ∂ y ∂ y
∂y The solution to these equations was obtained in 1908 by Blasius, a student of
Prandtl's. He showed that the solution to the velocity profile, shown in the table
below, could be obtained as a function of a single, nondimensional variable
defined as η= y U∞ υx η 1/ 2 with the resulting ordinary
differential equation: f ′′ ′ +
and 1
f f ′′ = 0
2
u
f ′(η ) =
U∞ Boundary conditions for the differential equation are expressed as follows:
at y = 0, v = 0 → f (0) = 0 ; y component of velocity is zero at y = 0
at y = 0 , u = 0 → f ′(0 ) = 0 ; x component of velocity is zero at y = 0  v v VII3  eText Main Menu  Textbook Table of Contents  Study Guide The key result of this solution is written as follows: ∂2 f τw = 0.332 =
∂η2 y= 0
µ U∞ U∞ / υ x
With this result and the definition of the boundary layer thickness, the following
key results are obtained for the laminar flat plate boundary layer:
Local boundary layer thickness δ (x) = Local skin friction coefficient: C fx = 0.664
Re x CD = 1.328
Re x CD = FD / A
1
ρ U2
∞
2 (defined below)
Total drag coefficient for length L ( integration
of τw dA over the length of the plate, per unit
2
area, divided by 0.5 ρ U∞ ) C fx = where by definition τ w (x )
1
ρ U2
∞
2 and 5x
Re x With these results, we can determine local boundary layer thickness, local wall
shear stress, and total drag force for laminar flow over a flat plate.
Example:
Air flows over a sharp edged flat plate with L = 1 m, a width of 3 m and
U∞ = 2 m/s . For one side of the plate, find: δ(L), Cf (L), τw(L), CD, and FD.
3 2 ρ = 1.23 kg/m Air:
First check Re: Re L = U∞ L υ = ν = 1.46 E5 m /s 2 m / s * 2.15 m
= 294,520 < 500,000
2
1.46 E − 5 m / s Key Point: Therefore, the flow is laminar over the entire length of the plate and
calculations made for any x position from 0  1 m must be made using laminar
flow equations.  v v VII4  eText Main Menu  Textbook Table of Contents  Study Guide Boundary layer thickness at x = L: δ (L ) = 5L
5* 2.15 m
=
= 0.0198 m = 1.98 cm
Re L
294, 520 Local skin friction coefficient at x = L: C f ( L) = 0.664
0.664
=
= 0.00122
Re L
294, 520 Surface shear stress at x = L:
2
τ w = 1 / 2 ρ U∞ C f = 0.5 *1.23 kg / m3 * 2 2 m 2 / s 2 * 0.00122 τ w = 0.0030 N / m2 (Pa)
Drag coefficient over total plate, 0 – L: CD (L ) = 1.328
1.328
=
= 0.00245
Re L
294, 520 Drag force over plate, 0 – L: FD = 1/ 2 ρ U∞ CD A = 0.5 *1.23 kg / m * 2 m / s * 0.00245 * 2 * 2.15 m
FD = 0.0259 N
3 2 2 2 2 2 Two key points regarding this analysis:
1. Each of these calculations can be made for any other location on the plate
by simply using the appropriate x location for any x ≤ L .
2. Be careful not to confuse the calculation for Cf and CD. Cf is a local
calculation at a particular x location (including x = L) and can only be
used to calculate local shear stress, not drag force. CD is an integrated
average over a specified length (including any x ≤ L ) and can only be
used to calculate the average shear stress and the integrated force over the
length.  v v VII5  eText Main Menu  Textbook Table of Contents  Study Guide Turbulent Flow Equations
While the previous analysis provides an excellent representation of laminar, flat
plate boundary layer flow, a similar analytical solution is not available for
turbulent flow due to the complex nature of the turbulent flow structure.
However, experimental results are available to provide equations for key flow
field parameters.
A summary of the results for boundary layer thickness and local and average skin
friction coefficient for a laminar flat plate and a comparison with experimental
results for a smooth, turbulent flat plate are shown below.
Laminar δ (x) =
C fx =
CD = Turbulent 5x
Re x δ (x ) = 0.664
Re x C fx = 1.328
Re L C fx =
where CD = τw
1
2
ρU∞
2 0.074
.2
Re L 0.37 x
.2
Re x 0.0592
.2
Re x
for turbulent flow over
entire plate, 0 – L, i.e.
assumes turbulent flow
in the laminar region. local drag coefficient based on local
wall shear stress (laminar or turbulent
flow region). and
CD = total drag coefficient
based on the integrated
force over the length 0 to L F/ A
CD = 1
2=
ρ U∞
2 ( 1
2
ρ U∞
2 A ) −1 L ∫ τ w ( x ) w dx
0 A careful study of these results will show that in general, boundary layer thickness
grows faster for turbulent flow and wall shear and total friction drag are greater for
turbulent flow than for laminar flow given the same Reynolds number.  v v VII6  eText Main Menu  Textbook Table of Contents  Study Guide It is noted that the expressions for turbulent flow are valid only for a flat plate with
a smooth surface. Expressions including the effects of surface roughness are
available in the text.
Combined Laminar and Turbulent Flow
U laminar to
turbulent
transition free stream y edge of boundary layer δ(x) turbulent x laminar
xcr Flat plate with both laminar and turbulent flow sections
For conditions (as shown above) where the length of the plate is sufficiently long
that we have both laminar and turbulent sections:
* Local values for boundary layer thickness and wall shear stress for either
the laminar or turbulent sections are obtained from the expressions for δ(x)
and Cfx for laminar or turbulent flow as appropriate for the given region. * The result for average drag coefficient CD and thus total frictional force
over the laminar and turbulent portions of the plate is given by
(assuming a transition Re of 500,000) CD =
* 0.074
.2
Re L − 1742
Re L Calculations assuming only turbulent flow can be made typically for two cases
when some physical situation (a trip wire) has caused the flow to be
leading from the leading edge or
if the total length L of the plate is much greater than the length xcr of
the laminar section such that the total flow can be considered turbulent
from x = 0 to L. Note that this will overpredict the friction drag force
since turbulent drag is greater than laminar. 1.
2. With these results, a detailed analysis can be obtained for laminar and/or turbulent
flow over flat plates and surfaces that can be approximated as a flat plate.  v v VII7  eText Main Menu  Textbook Table of Contents  Study Guide Example:
Water flows over a sharp flat plate 2.55 m long, 1 m wide, with U∞ = 2 m/s.
Estimate the error in FD if it is assumed that the entire plate is turbulent.
3 2 Water: ρ = 1000 kg/m ν = 1.02 E m /s Re L = Reynolds number: U∞ L υ = 2 m / s * 2.55 m
= 5E 6 > 500, 000
1.02 E − 6 m 2 / s with Re cr = 500,000 ⇒ x cr = 0.255 m ( or 10% laminar)
a. Assume that the entire plate is turbulent CD = 0.074
0.074
.2 = 0.00338
.2 =
Re L
5 E 6)
(
2 kg
m
FD = 0.5 ρ U CD A = 0.5 *1000 3 * 2 2 2 * 0.00338 * 2.55 m 2
s
m
2
∞ FD = 17.26 N This should be high since we have assumed that
the entire plate is turbulent and the first 10% is
actually laminar. b. Consider the actual combined laminar and turbulent flow: 0.074
.2
Re L CD = − 1742
1742
= 0.00338 −
= 0.00303
Re L
5 E6 Note that the CD has decreased when both the laminar and turbulent sections are
considered.
2 kg
m
FD = 0.5 ρ U CD A = 0.5 *1000 3 * 2 2 2 * 0.00303* 2.55 m 2
s
m
2
∞  v v VII8  eText Main Menu  Textbook Table of Contents  Study Guide FD = 15.46 N {Lower than the fully turbulent value}
Error = 17.26 − 15.46
* 100 = 11.6% high
15.46 Question: Since xcr = 0.255 m, what would your answers represent if you had
calculated the Re, CD, and FD using x = xcr = 0.255 m?
Answer: You would have the value of the transition Reynolds number and the
drag coefficient and drag force over the laminar portion of the plate
(assuming you used laminar equations). If you used turbulent
equations, you would have red marks on your paper.  v v VII9  eText Main Menu  Textbook Table of Contents  Study Guide Von Karman Integral Momentum Analysis
While the previous results provide an excellent basis for the analysis of flat plate
flows, complex geometries and boundary conditions make analytical solutions to
most problems difficult.
An alternative procedure provides the basis for an approximate solution which in
many cases can provide excellent results.
The key to practical results is to use a reasonable approximation to the boundary
layer profile, u(x,y). This is used to obtain the following:
&
m= a. Boundary layer mass flow: δ ∫ ρ u bdy
0 where b is the width of the area for which the flow rate is being obtained. τw = µ b. Wall shear stress: d u d y y =0 You will also need the streamwise pressure gradient dP
dx for many problems. The Von Karman integral momentum theory provides the basis for such an
approximate analysis. The following summarizes this theory.
Displacement thickness:
Consider the problem
indicated in the adjacent
figure:
A uniform flow field with
velocity U∞ approaches a
solid surface. As a result
of viscous shear, a
boundary layer velocity
profile develops. y=h +δ * y U∞ Streamline U∞ U∞
h h
u
Simulated
effect 0
x  v v VII10  eText Main Menu  Textbook Table of Contents  Study Guide δ* A viscous boundary layer is created when the flow comes in contact with the solid
surface.
Key point: Compared to the uniform velocity profile approaching the solid
surface, the effect of the viscous boundary layer is to displace
streamlines of the flow outside the boundary layer away from the
wall.
* With this concept, we define δ = displacement thickness
* δ = distance the solid surface would have to be displaced to maintain the
same mass flow rate as for nonviscous flow.
From the development in the text, we obtain
δ * δ = ∫ 1 −
0 u dy
U∞ * Therefore, with an expression for the local velocity profile we can obtain δ = f(δ)
Example:
2
u y − y
=2
δ δ
U∞ Given: * determine an expression for δ = f(δ) Note that for this assumed form for the velocity profile:
1. At y = 0, u = 0 correct for no slip condition 2. At y = δ, u = U∞ correct for edge of boundary layer
3. The form is quadratic
To simplify the mathematics,
let η = y/δ, at y = 0, η = 0 ; u
= 2η − η 2
U∞ Therefore: at y = δ , η = 1 ; d y = δ d η  v v VII11  eText Main Menu  Textbook Table of Contents  Study Guide 1 2η2 η3 δ = ∫ 1 − 2 η + η δ d η = δ η −
+
2
3 0
0 * Substituting: 1 ( 2 ) 1 δ* = 3δ which yields Therefore, for flows for which the assumed quadratic equation approximates the
velocity profile, streamlines outside of the boundary layer are displaced
approximately according to the equation
1 δ* = 3δ
This closely approximates flow for a flat plate.
Key Point: When assuming a form for a velocity profile to use in the Von
Karman analysis, make sure that the resulting equation satisfies both surface and
free stream boundary conditions as well as has a form that approximates u(y).
Momentum Thickness:
The second concept used in the Von Karman momentum analysis is that of
momentum thickness  θ
The concept is similar to that of displacement thickness in that θ is related to the
loss of momentum due to viscous effects in the boundary layer.
Consider the viscous flow
regions shown in the adjacent
figure. Define a control
volume as shown and
integrate around the control
volume to obtain the net
change in momentum for the
control volume.  v v VII12  eText Main Menu  Textbook Table of Contents  Study Guide If D = drag force on the plate due to viscous flow, we can write
 D = ∑ ( momentum leaving c.v. )  ∑ ( momentum entering c.v. )
Completing an analysis shown in the text, we obtain D= u
u
1− dy
U ∞ U∞ 0 δ 2
ρ U∞ θ θ=∫ CD = Using a drag coefficient defined as CD = We can also show that D/A
1
ρ U2
∞
2 2 θ (L)
L where: θ(L) is the momentum thickness evaluated over the length L.
Thus, knowledge of the boundary layer velocity distribution u = f(y) allows the
drag coefficient to be determined. Momentum integral:
The final step in the Von Karman theory applies the previous control volume
analysis to a differential length of surface. Performing an analysis similar to the
previous analysis for drag D we obtain d
τw
d U∞
*
2
= δ U∞
+
(U∞ θ )
ρ
dx
dx This is the momentum integral
for 2D, incompressible flow
and is valid for laminar or
turbulent flow.  v v VII13  eText Main Menu  Textbook Table of Contents  Study Guide where d U∞
δ* d P
δ U∞
=−
dx
ρ dx Therefore, this analysis also accounts for
the effect of freestream pressure
gradient. * For a flat plate with nonaccelerating
flow, we can show that P = const ., U∞ = const., d U∞
=0
dx Therefore, for a flat plate, nonaccelerating flow, the Von Karman momentum
integral becomes τw d
2
2
=
(U∞ θ )= U∞ d θ
ρ dx
dx
From the previous analysis and the assumed velocity distribution of
2
u y − y = 2η − η 2
=2
δ δ
U∞ The wall shear stress can be expressed as τw = µ d u
2 µ U∞ 2 2 y = 2 U∞ − 2 =
δ
d y w
δ δ y = 0 (A) Also, with the assumed velocity profile, the momentum thickness θ can be
evaluated as u
u
1− dy
U ∞ U∞ 0 δ θ=∫
or  v v VII14  eText Main Menu  Textbook Table of Contents  Study Guide 2δ
15 δ θ = ∫ (2η − η 2 )( − 2η + η 2 )δ d η =
1
0 We can now write from the previous equation for τw
2
τ w = ρ U∞ dθ 2
2 dδ
= ρ U∞
d x 15
dx Equating this result to Eqn. A we obtain τw =
or δ dδ = 15 µ
dx
ρ U∞ 2
2 µ U∞
2 dδ
ρ U∞
=
dx
δ
15
which after integration yields 1/ 2 30 µ x δ= ρU∞ δ= or 5.48
Re x Note that the this result is within 10% of the exact result from Blasius flat plate
theory.
Since for a flat plate, we only need to consider friction drag (not pressure drag),
we can write C fx = τ w (x ) 2 µ U∞ 1
=
1
δ 1 ρ U2
ρ U2
∞
∞
2
2 Substitute for δ to obtain C fx = 2 µ U∞ Re
0.73
=
Re x
5.48 1 ρ U 2
∞
2  v v VII15  eText Main Menu  Textbook Table of Contents  Study Guide Exact theory has a numerical constant of 0.664 compared with 0.73 for the
previous result.
It is seen that the von Karman integral theory provides the means to determine
approximate expressions for
δ, τw, and Cf
using only an assumed velocity profile.
Solution summary: 1. Assume an analytical expression for the velocity profile for
the problem.
2. Use the assumed velocity profile to determine the solution
for the displacement thickness for the problem.
3. Use the assumed velocity profile to determine the solution
for the momentum thickness for the problem.
4. Use the previous results and the von Karman integral
momentum equation to determine the solution for the
drag/wall shear for the problem.  v v VII16  eText Main Menu  Textbook Table of Contents  Study Guide Bluff Body, Viscous Flow Characteristics
( Immersed Bodies)
In general, a body immersed in a flow will experience both externally applied
forces and moments as a result of the flow about its external surfaces. The typical
terminology and designation of these forces and moments are given in the diagram
shown below.
The orientation of the axis for the drag force is typically along the principal body
axis, although in certain applications, this axis is aligned with the principal axis of
the free stream, approach velocity U. Since in many cases the drag force is aligned with the principal axis of the body
shape and not necessarily aligned with the approaching wind vector. Review all
data carefully to determine which coordinate system is being used: body axis
coordinate system or a wind axis coordinate system.
These externally applied forces and moments are generally a function of
a. Body geometry
b. Body orientation
c. Flow conditions  v v VII17  eText Main Menu  Textbook Table of Contents  Study Guide These forces and moments are also generally
expressed in the form of a nondimensional
force/moment coefficient, e.g. the drag
coefficient: CD = FD /A
1
2
2 ρ U∞ It is noted that it is common to see one of three reference areas used depending on
the application:
1. Frontal (projected) area: Used for thick, stubby, nonaerodynamic
shapes, e.g., buildings, cars, etc.
2. Planform (top view, projected) area: Used for flat, thin shapes, e.g.,
wings, hydrofoils, etc.
3. Wetted area: The total area in contact with the fluid. Used for surface
ships, barges, etc.
The previous, flat plate boundary layer results considered only the contribution of
viscous surface friction to drag forces on a body. However, a second major (and
usually dominant) factor is pressure or form drag.
Pressure drag is drag due to the integrated surface pressure distribution over the
body. Therefore, in general , the total drag coefficient of a body can be expressed
as C D = CD,press + C D,friction
or CD = FD, total /A
1
2 ρ U2
∞ FD,press. /A = 1
2 ρ U2
∞ + FD,friction /A
1
2 ρ U2
∞ Which factor, pressure or friction drag, dominates depends largely on the
aerodynamics (streamlining) of the shape and to a lesser extent on the flow
conditions.  v v VII18  eText Main Menu  Textbook Table of Contents  Study Guide Typically the most important factor in the magnitude and significance of pressure
or form drag is the boundary layer separation and resulting low pressure wake
region associated with flow around non  aerodynamic shapes.
Consider the two shapes shown below:
large pressure drag
boundary layer separation
Low
Pressure
Wake High Pressure no separated flow region Low pressure drag The flow around the streamlined airfoil remains attached, producing no boundary
layer separation and comparatively small pressure drag. However, the flow
around the less aerodynamic circular cylinder separates, resulting in an area of
high surface pressure on the front side and low surface pressure on the back side
and thus significant pressure drag.
This effect is shown very graphically in the following figures from the text.  v v VII19  eText Main Menu  Textbook Table of Contents  Study Guide Fig. 7.12 Drag of a 2D, streamlined cylinder
The previous figure shows the effect of streamlining and aerodynamics on the
relative importance of friction and pressure drag. While for a thin flat plate (t/c =
0), all the drag is due to friction with no pressure drag, for a circular cylinder (t/c =
1), only 3% of the drag is due to friction with 97% due to pressure. Likewise for
most bluff, nonaerodynamic bodies, pressure (also referred to as form drag) is the
dominant contributor to the total drag.
However, the magnitude of the pressure (and therefore the total) drag can also be
changed by reducing the size of the low pressure wake region. One way to do this
is to change the flow conditions from laminar to turbulent. This is illustrated in
the following figures from the text for a circular cylinder. Fig. 7.13 Circular cylinder with (a) laminar separation and (b) turbulent separation
Note that for the cylinder on the left, the flow is laminar, boundary layer
separation occurs at 82o and the CD is 1.2, whereas for the cylinder on the right,
the flow is turbulent and separation is delayed and occurs at 120o. The drag
coefficient CD is 0.3, a factor of 4 reduction due to a smaller wake region and
reduced pressure drag.  v v VII20  eText Main Menu  Textbook Table of Contents  Study Guide It should also be pointed out that the
friction drag for the cylinder on the right
is probably greater (turbulent flow
conditions) than for the cylinder on the
left (laminar flow conditions).
However, since pressure drag
dominates, the net result is a significant
reduction in the total drag.
The pressure distribution for laminar
and turbulent flow over a cylinder is
shown in Fig. 7.13c to the right. The
fronttorear pressure difference is
greater for laminar flow, thus greater
drag.
Finally, the effect of streamlining on total drag is shown very graphically with the
sequence of modifications in Fig. 7.15.
Two observations can
be made: (1) As body
shape changes from a
bluff body with fixed
points of separation to
a more aerodynamic
shape, the effect of
pressure drag and the
drag coefficient will
decrease.
Fig. 7.15 The effect of streamlining on total drag
(2) The addition of surface area from (a) to (b) and (b) to (c) increases the
friction drag, however, since pressure drag dominates, the net result is a reduction
in the drag force and the CD .  v v VII21  eText Main Menu  Textbook Table of Contents  Study Guide The final two figures show results for the drag coefficient for two and three
dimensional shapes with various geometries.
Table 7.2 CD for TwoDimensional Bodies at Re ≥ 10 4 First note that all
values in Table 7.2 are
for 2D geometries,
that is, the bodies are
very long (compared to
the crosssection
dimensions) in the
dimension perpendicular to the page.
Key Point: Non –
aerodynamic shapes
with fixed points of
separations (sharp
corners) have a single
value of CD,
irrespective of the
value of the Reynolds
number, e.g. square
cylinder, halftube,
etc.
Aerodynamic shapes
generally have a
reduction in CD for a
change from laminar
to turbulent flow as a
result of the shift in
the point of boundary
layer separation, e.g.
elliptical cylinder.  v v VII22  eText Main Menu  Textbook Table of Contents  Study Guide Table 7.3 Drag of threedimensional bodies at Re ≥ 10 4 The geometries in Table 7.3 are all 3D and thus are finite perpendicular to the
page. Similar to the results from the previous table, bluff body geometries with
fixed points of separation have a single CD, whereas aerodynamic shapes such as
slender bodies of revolution have individual values of CD for laminar and
turbulent flow.  v v VII23  eText Main Menu  Textbook Table of Contents  Study Guide In summary, one must remember that broad generalizations such as saying that
turbulent flow always increases drag, drag coefficients always depend on
Reynolds number, or increasing surface area increases drag are not always valid.
One must consider carefully all effects (viscous and pressure drag) due to
changing flow conditions and geometry.
Example:
A square 6in piling is acted on by
a water flow of 5 ft/s that is 20 ft
deep. Estimate the maximum
bending stress exerted by the flow
on the bottom of the piling.
Water: ρ = 1.99 slugs/ft 3 2 ν = 1.1 E – 5 ft /s Assume that the piling can be treated as 2D and thus end effects are negligible.
Thus for a width of 0.5 ft, we obtain: Re = 5 ft / s .5 ft
= 2.3 E 5
1.1 E − 5 ft 2 / s In this range, Table 7.2 applies for 2D bodies and we read CD = 2 .1. The
2
frontal area is A = 20*0.5 = 10 ft
2 slug
ft
FD = 0.5 ρ U CD A = 0.5 *1.99 3 *52 2 * 2.1*10 ft 2 = 522 lbf
s
ft
2
∞ For uniform flow, the drag should be uniformly distributed over the total length
with the net drag located at the midpoint of the piling.
Thus, relative to the bottom of the piling, the bending moment is given by
Mo = F * 0.5 L = 522 lbf* 10 ft = 5220 ftlbf  v v VII24  eText Main Menu  Textbook Table of Contents  Study Guide From strength of materials, we can write σ= M o c 5220 ft − lbf * 0.25 ft
=1
= 251,000 psf = 1740 psi
3
3
0.5 ft * 0.5 ft
I
12
3 where c = distance to the neutral axis, I = moment of inertia = b h /12
Question: Since pressure acts on the piling and increases with increasing depth,
why wasn’t a pressure load considered?
Answer: Static pressure does act on the piling, but it acts uniformly around the
piling at every depth and thus cancels. Dynamic pressure is considered in the
drag coefficients of Tables 7.2 and 7.3 and does not have to be accounted for
separately.  v v VII25  eText Main Menu  Textbook Table of Contents  Study Guide ...
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 Spring '08
 Sakar
 Fluid Dynamics, Aerodynamics, eText Main Menu, Laminar Flow Analysis

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