Wind effect - Structural Design

A building in the path of wind causes wind pressure which in turn causes force, shear, and overturn moment at each level that must be resisted, following a load path to the foundations (wind wall pressure transfers to horizontal diaphragms, then to shear walls, finally to foundation).  Wind pressure times tributary area per level causes  lateral force per level.  Shear per level, the sum of wind forces above, defines required resistance.   Overturn moment per level is the sum of forces above times their height above the respective level.

1  Wind force, shear, and overturn moment per level

 Fx = wind force = wind pressure times tributary area per level exposed to wind
 VX = shear per level = sum of Fx  above
 Mx = overturn moment  = sum of all forces above times their distance above level x.

2 Overturn effect

3  Windward  pressure increase with height

4 Wind force

Fx = (windward pressure + leeward suction) times tributary area per level
(leeward wind suction is assumed constant for full height)
Fx = P A
P = wind pressure and suction in psf (Pa)
A = tributary area exposed to wind
(tributary area = building width times half the story height above and below)

5 Shear

Wind shear per level = sum of all wind forces above
Wind shear is the integration of wind forces above

6 Overturn moment

Overturn moment per level  = sum of all forces above times their distance
Overturn moment per level = integration of shear diagram above respective level

Wind Load - Structural Design

1  Wind load on gabled building
2  Wind load on dome or vault
3  Protected buildings inside a city
4  Exposed tall building inside a city
5  Wind flow around and above exposed building
6  Wind speed amplified by building configuration

Wind channeled between buildings causes a Venturi effect of increased wind speed.  Air movement through buildings causes internal pressure that effects curtain walls and cladding design.  Internal pressure has a balloon-like effect, acting outward if the wind enters primarily on the windward side.  Openings on leeward or side walls cause inward pressure.  In tall buildings with fixed curtain wall the difference between outside wind pressure and interior pressure causes air movement from high pressure to low pressure. 

This causes air infiltration on the windward side and outflow on the leeward side.  In high-rise buildings, warm air moving from lower to upper levels causes pressures at top levels on the leeward face and negative suction on lower levels.  Wind pressure is based on the equation developed by Daniel Bernoulli (1700-1782).   For steady air flow of velocity V, the velocity pressure, q, on a rigid body is

q = pV^2 / 2
p = air density   (air weight divided by the acceleration of gravity g = 32.2 ft/sec2)
Air of 15°C at sea level weighs 0.0765 Ib/ft
3, which yields:
q = 0.00256V^2   (q in psf)

The American National Standards Institute (ANSI) Minimum design loads for buildings and other structures  (ANSI A58.1 - 1982), converted dynamic pressure to velocity pressure qz (psf) at height z as

Wind Load


Deflection, shear, and bending diagrams are shown here for typical beams. The beam with deflection and load diagrams are drawn on top with shear and bending diagrams shown below.  With experience, these diagrams may be drawn by visual inspection prior to computing.  This is useful to verify computations and develop an intuitive sense and visualization regarding shear and bending on beams.  The deflection diagram is drawn, visualizing the deflection of a thin board, flexible ruler, or similar device.  It is drawn grossly exaggerated to be visible.  The shear diagram is drawn at a convenient force scale left to right, starting with zero shear to the left of the beam.  Downward uniform load yields downward sloping shear.  Downward point loads are drawn as downward offset, and upward reactions yield upward offset.  Bending diagrams are drawn, considering the area method; namely, bending at any point is equal to the area of the shear diagram up to that point.  Both, shear and bending must be zero to the right of the right beam end. To satisfy this, requires a certain amount of forward thinking and, in complex cases, even working backward from right to left as well as left to right.

1  Cantilever beam with point load
2  Cantilever beam with uniform load
3  Cantilever beam with mixed load
4  Simple beam with point loads
5  Simple beam with uniform load
6  Simple beam with mixed load
7  Beam with one overhang and point load
8  Beam with one overhang and uniform load
9  Beam with one overhang and mixed load
10  Beam with two overhangs and point loads
11  Beam with two overhangs and uniform load
12  Beam with two overhangs and mixed load


This beam, supporting a column point load of 96 k over a door, is a composite beam consisting of a wide-flange base beam with 8x½ in plates welded to top and bottom flanges. The beam is analyzed with and without  plates.  As shown before, for steel beams shear stress is assumed to be resisted by the web only, computed as fv  = V/Av.  The base beam is a W10x49 [10 in (254 mm) nominal depth, 49 lbs/ft (6.77 kg/m) DL] with a moment of inertia Ixx= 272 in^4 (11322 cm^4).  Shear in the welds connecting the plates to the beam is found using the shear flow formula q = VQ/(I).

1  Beam of L= 6 ft (1.83 m) span with P = 96 k point load
2  Composite wide-flange beam W10x49 with 8x½ inch stiffener plates

Shear force V = P/2 = 96/2          V = 48 k
Bending moment M = 48(3)         M = 144 k’

Wide-flange beam 

Since the beam would fail in bending, a composite beam is used.

Composite beam

Since the shear force remains unchanged, the web shear stress is still ok.

Shear flow q in welded plate connection

Since there are two welds, each resists half the total shear flow

Note: in this steel beam, bending is stress is more critical than shear stress; this is typical for steel beams, except very short ones.



Since this is not a rectangular beam, shear stress must be computed by the general shear formula.  The maximum shear stress at the neutral axis as well as shear stress at the intersection between flange and web (shear plane As) will be computed.  The latter gives the shear stress in the glued connection.  To compare shear- and bending stress the latter is also computed.

Beam of L= 10 ft length, with uniform load w= 280plf (W = 2800 lbs)
Cross-section of wood I-beam

For the formula v= VQ/(Ib) we must find the moment of inertia of the entire cross-section. We could use the parallel axis theorem of Appendix A.  However, due to symmetry, a simplified formula is possible, finding the moment of inertia for the overall dimensions as rectangular beam minus that for two rectangles on both sides of the web.

Note c= 10/2 = 5 (half the beam depth due to symmetry)

Static moment Q of flange about the neutral axis:

Shear stress at flange/web intersection:

Static moment Q of flange plus upper half of web about the neutral axis

Maximum shear stress at neutral axis:

Note: Maximum shear stress reaches almost the allowable stress limit, but bending stress is well below allowable bending stress because the beam is very short.  We can try at what span the beam approaches allowable stress, assuming L= 30 ft, using the same total load W = 2800 lbs to keep shear stress constant:

At 30 ft span bending stress is just over the allowable stress of 1450 psi.  This shows that in short beams shear governs, but in long beams bending or deflection governs.



Based on the forgoing general derivation of shear stress, the formulas for shear stress in rectangular wood beams and flanged steel beams is derived here.  The maximum stress in those beams is customarily defined as fv instead of v in the general shear formula.

Shear at neutral axis of rectangular beam (maximum stress),
Note: this is the same formula derived for maximum shear stress before

Shear stress at the bottom of rectangular beam.  Note that y= 0 since the centroid of the area above the shear plane (bottom) coincides with the neutral axis of the entire section. Thus Q= Ay = (bd/2) 0 = 0, hence

v = V 0/(I b) = 0 = fv, thus
fv = 0 

Note: this confirms an intuitive interpretation that suggests zero stress since no fibers below the beam could resist shear

3  Shear stress at top of rectangular beam.  Note A = 0b = 0 since the depth of the shear area above the top of the beam is zero.  Thus  

Q = Ay = 0 d/2 = 0, hence v = V 0/(I b) = 0 = fv, thus
fv = 0

Note: this, too, confirms an intuitive interpretation that suggests zero stress since no fibers above the beam top could resist shear.

  Shear stress distribution over a rectangular section is parabolic as implied by the formula Q=b(d^2)/8 derived above.

5  Shear stress in a steel beam is minimal in the flanges and parabolic over the web.

The formula v = VQ/(I b) results in a small stress in the flanges since the width b of flanges is much greater than the web thickness.  However, for convenience, shear  stress in steel beams is computed as “average” by the simplified formula:


The distribution of shear stress over the cross-section of beams is derived, referring to a beam part of length x marked on diagrams.  Even though horizontal and vertical shear are equal at any part of a beam, horizontal  shear stress is derived here because it is much more critical in wood due to horizontal fiber direction.

1  Beam, shear and bending diagrams with marked part of length x
2  Beam part with bending stress pushing and pulling to cause shear
3  Beam part with bending stress above an arbitrary shear plane

Let M be the differential bending moment between m and n.  M is equal to the shear area between m and n (area method), thus M = V x.  Substituting V x for M in the flexure formula f= M c /I yields bending stress f= V x c/I in terms of shear.  The differential bending stress between m and n pushes top and bottom fibers in opposite directions,  causing shear stress.  At any shear plane y1 from the neutral axis of the beam the sum of  shear stress above this plane yields a force F that equals average stress fy times the  cross section area A above the shear plane, F = A fy.  The average stress fy is found from similar triangles; fy relates to y as f relates to c, i.e., fy/y = f/c; thus fy = f y/c.  Since f= V x c/I, substituting V x c/I for f yields fy = (V x c/I) y/c = V x y/I.  Since F = A fy, it follows that  F = A V x y/I.  The horizontal shear stress v equals the force F divided by the area of the  shear plane;

V = F/(x b)  =  A V x y/(I x b)  =  V A y/(I b)  

The term A y is defined as Q, the first static moment of the area above the shear plane times the lever arm from its centroid to the neutral axis of the entire cross-section. Substituting Q for A y yields the working formula

v = horizontal shear stress.
Q = static moment (area above shear plane times distance from centroid of that area to
      the neutral axis of the entire cross-section
I = moment of Inertia of entire cross section
b = width of shear plane

The formula for shear stress can also be stated as shear flow q, measured in force per unit length (pound per linear inch, kip per linear inch, or similar metric units); hence



The formula for the moment of inertia I=∫(y^2)da reveals that the resistance of any differential area da increases with its distance y from the neutral axis squared, forming a parabolic distribution.  For a beam of rectangular cross-section, the resistance of top and bottom fibers with distance y = d/2 from the neutral axis is (d/2)^2.  Thus, the moment of inertia, as geometric resistance, is the volume of all fibers under a parabolic surface,  which is 1/3 the volume of a cube of equal dimensions, or I= bd ((d/2)^2)/3, or
the moment of inertia of a rectangular beam of homogeneous material.  A formal calculus derivation of this formula is given in Appendix A.  The section modulus gives only the maximum bending stress, but the moment of inertia gives the stress at any distance c from the neutral axis as f= Mc/I.  This is useful, for example, for bending elements of  asymmetrical cross-section, such as T- and L-shapes.

1  Bending stress distribution over beam cross-section
2  Moment of inertia visualized as volume under parabolic surface
3  T-bar with asymmetrical stress: max. stress at c2 from the neutral axis
4  Angle bar with asymmetrical stress distribution about x, y, and z-axes:  maximum resistance about x-axis and minimum resistance about z-axis



Rectangular beams of homogeneous material, such as wood, are common in practice. 

The section modulus for such beams is derived here.

1  Stress block in rectangular beam under positive bending.
2  Large stress block and lever-arm of a joist in typical upright position.
3  Small, inefficient, stress block and lever-arm of a joist laid flat. 

Referring to 1, the section modulus for a rectangular beam of homogeneous material may be derived as follows.  The force couple C and T rotate about the neutral axis to provide the internal resisting moment.  C and T act at the center of mass of their  respective triangular stress block at d/3 from the neutral axis.  The magnitude of C and T  is the volume of the upper and lower stress block, respectively. 

C = T = (f/2) (bd/2) = f b d/4.

The internal resisting moment is the sum of C and T times their respective lever arm, d/3, to the neutral axis.  Hence

M = C d/3 + T d/3.  Substituting C = T = f bd/4 yields
M = 2 (f bd/4) d/3 = f bd2/6, or M = f S, 

where S = bd2/6, defined as the section modulus for rectangular beams of homogeneous material.
Solving M = f S for f yields the maximum bending stress as defined before: 

f = M/S

This formula is valid for homogeneous beams of any shape; but the formula S = b(d^2)6 is valid for rectangular beams only.  For other shapes S can be computed as S = I  /c as  defined before for the flexure formula.

Comparing a joist of 2”x12” in upright and flat position as illustrated in 2 and 3 yields an interesting observation:

The upright joist is six times stronger than the flat joist of equal cross-section. This demonstrates the importance of correct orientation of bending members, such as beams  or moment frames.



The flexure formula gives the internal bending stress caused by an external moment on a beam or other bending member of homogeneous material.  It is derived here for a rectangular beam but is valid for any shape. 

1  Unloaded beam with hatched square
2  Beam subject to bending with hatched square deformed
3  Stress diagram of deformed beam subject to bending

Referring to the diagram, a beam subject to positive bending assumes a concave curvature (circular under pure bending).  As  illustrated by the hatched square, the top shortens and the bottom elongates, causing compressive stress on the top and tensile stress on the bottom.  Assuming stress varies linearly with strain, stress distribution over the beam depth is proportional to strain deformation.  Thus stress varies linearly over the depth of the beam and is zero at the neutral axis (NA).  The bending stress fy at any  distance y from the neutral axis is found, considering similar triangles, namely fy relates  to y as f relates to c; f is the maximum bending stress at top or bottom and c the distance  from the Neutral Axis, namely fy / y = f / c. Solving for fy yields

fy = y f / c

To satisfy equilibrium, the beam requires an internal resisting moment that is equal and  opposite to the external bending moment.  The internal resisting moment is the sum of all partial forces F rotating around the neutral axis with a lever arm of length y to balance the external moment.  Each partial force F is the productof stress fy and the partial area a on which it acts, F = a fy.  Substituting fy = y f / c, defined above, yields F = a y f / c.  Since the internal resisting moment M is the sum of all forces F times their lever arm y to the  neutral axis, M = F y = (f/c) Σ y y a = (f/c) Σ y^2 a, or M = I f/c, where the term Σy^2 a is defined as  moment of inertia (I =  Σy^2 a) for convenience.  In formal calculus the summation of area  a is replaced by integration of the differential area  da, an infinitely small area:
The internal resisting moment equation M = I f/c solved for stress f yields

which gives the bending stress f at any distance c from the neutral axis.  A simpler form  is used to compute the  maximum fiber stress as derived before.  Assuming c as  maximum fiber distance from the neutral axis yields: 
Both the moment of inertia I and section modulus S define the strength of a cross-section regarding its geometric form.



Indeterminate beams include beams with fixed-end (moment resistant) supports and beams of more than two supports, referred to as continuous beams. The design of  statically indeterminate beams cannot be done by static equations alone. However, bending coefficients, derived by mechanics, may be used for analysis of typical beams. 

The bending moment is computed, multiplying the bending coefficients by the total load W and span L between supports.  For continuous beams, the method is limited to beams  of equal spans for all bays.  The coefficients here assume all bays are loaded. 

Coefficients for alternating live load on some bays and combined dead load plus live load  on others, which may result in greater bending moments, are in Appendix A.  Appendix A  also has coefficients for other load conditions, such as various point loads. The equation  for bending moments by bending coefficients is:

M = C L W

M = bending moment
C = bending coefficient
L = span between supports
W = w L (total load per bay)
w = uniform load in plf (pounds / linear foot

1 Simple beam
2  Fixed-end beam (combined positive plus negative moments equal the simple-beam moment)
3 Two-span beam
4 Three-span beam





Find x, where shear = 0 and bending = maximum:

Vbr-w2 x = 0;  x = Vbr/w2 = 2000/200      x = 10 ft


Section modulus 

Bending stress

Shear stress
Note: stress is figured, using absolute maximum bending and shear, regardless if positive  or negative.  Lumber sizes are nominal, yet  actual sizes are used for computation.   Actual sizes are ½ in. less for lumber up to 6 in. nominal and ¾ in. less for larger sizes:  4x8 nominal is 3½x7¼ in. actual. 

Note: in the above two beams shear stress is more critical (closer to the allowable stress)  than bending stress, because negative cantilever moments partly reduces positive  moments. 



V   Shear diagram.
M  Bending diagram.
∆   Deflection diagram. 
I    Inflection point (change from + to - bending).


  R= 400plf (24)/2      R = 4800 lbs



Try 4x10 beam  

Bending stress  

f b = Mmax/S= 5000 (12)/50      f b = 1200 psi
                                                  1200 < 1450, ok

Shear stress  

f v = 1.5V/bd=1.5(2800)/[3.5(9.25)]       f v = 130 psi
                                                               130 > 95, not ok

Try 6x10 beam  
f v = 1.5V/bd=1.5 (2800)/[5.5 (9.25)]      f v= 83 psi
                                                                83 < 95, ok

Note: increased beam width is most effective to reduce shear stress; but increased depth  is most effective to reduce bending stress.



The area method for beam design simplifies computation of shear forces and bending moments and is derived, referring to the following diagrams:
1  Load diagram on beam
2 Beam diagram
3 Shear diagram
4 Bending diagram

The area method may be stated:

•  The shear at any point n is equal to the shear at point m plus  the area of the load diagram between m and n. 

•  The bending moment at any point n is equal to the moment at point m plus the area of the shear diagram between m and n.

The shear force is derived using vertical equilibrium:

∑ V = 0;   Vm - w x - Vn = 0;  solving for Vn

Vn = Vm-wx

where w x is the load area between m and n (downward load w = negative).

The bending moment is derived using moment equilibrium: 

∑ M = 0;    Mm + Vmx – w x x/2 - Mn = 0;  solving for Mn

where Vmx – wx^2/2 is the shear area between m and n, namely, the rectangle    Vm x less the triangle w x^2/2.  This relationship may also be stated as Mn = Mm + Vx, where V is the average shear between m and n.

By the area method moments are usually equal to the area of one or more rectangles and/or triangles.  It is best to first compute and draw the shear diagram and then compute the moments as the area of the shear diagram.

From the diagrams and derivation we may conclude:

•  Positive shear implies increasing bending moment. 
•  Zero shear (change from + to -) implies peak bending moment (useful to locate maximum bending moment). 
•  Negative shear implies decreasing bending moment. Even though the forgoing is for uniform load, it applies to concentrated load and non-uniform load as well.  The derivation for such loads is similar.



1  Beam of L= 20 ft span, with uniformly distributed load w = 100 plf
2  Free-body diagram of partial beam x units long
3 Shear diagram
4  Bending moment diagram

To find the distribution of shear and bending along the beam, we investigate the beam at intervals of 5’, from left to right.  This is not normally required. 

Reactions R are half the load on each support due to symmetry 

Shear force Vx at any distance x from left is found using ∑ V = 0 

V  = R - w x

Bending moment Mx at any distance x from left is found by ∑ M = 0. 

∑ M = 0;    R x – w x (x/2) - Mx = 0;   solving for Mx 

Bending is zero at both supports since pins and rollers have no moment resistance. Since the bending formula Mx= Rx-wx2/2 is quadratic, bending increase is quadratic (parabolic curve) toward maximum at center, and decreases to zero at the right support. For simple beams with uniform load the maximum shear force is at the supports and the maximum bending moment at mid-span (x= L/2) are:

Vmax = R = w L / 2 

Mmax = wL^2 / 8

This formula is only for simple beams with uniform load.  Verifying example: