Architectural Structures

Flexure Formula

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.

1Unloaded beam with hatched square

2Beam subject to bending with hatched square deformed

3Stress 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 product of 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 y2a, or M = I f/c, where the term Σy2a is defined as moment of inertia (I = Σy2a) 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.

6-14 FUNDAMENTALS: Bending

Moment of inertia

The formula for the moment of inertia I=∫y2da 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

I = bd3 / 12 (for rectangular beams only)

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.

1Bending stress distribution over beam cross-section

2Moment of inertia visualized as volume under parabolic surface

3T-bar with asymmetrical stress: max. stress at c2 from the neutral axis

4Angle bar with asymmetrical stress distribution about x, y, and z-axes: maximum resistance about x-axis and minimum resistance about z-axis

6-16 FUNDAMENTALS: Bending

Shear stress in wood and steel beams

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.

v = V Q / I b = V (bd2/8) / [(bd3/12)b] = fv, or fv = 1.5 V / (bd)

Note: this is the same formula derived for maximum shear stress before

2Shear 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

3Shear 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.

4Shear stress distribution over a rectangular section is parabolic as implied by the formula Q=bd2/8 derived above.

5Shear 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:

fv = V / Av

fv = shear stress in steel beam

V = shear force at section investigated

Av = shear area, defined as web thickness times beam depth

6-19 FUNDAMENTALS: Bending

Shear stress in steel beam

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 in4 (11322 cm4) (see Appendix). Shear in the welds connecting the plates to the beam is found using the shear flow formula q = VQ/(I).

1Beam of L= 6 ft (1.83 m) span with P = 96 k point load

2Composite wide-flange beam W10x49 with 8x½ inch stiffener plates

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

6-21 FUNDAMENTALS: Bending

Deflection formulas

Based on the moment-area method, the following formulas for slope and deflection are derived for beams of common load and support conditions. Additional formulas are provided in Appendix A. Although downward deflection would theoretically be negative, it is customary to ignore the sign convention and define upor downward deflection by inspection. The angle θ is the slope of the tangent to the elastic curve at the free end for cantilever beams and at supports for simple beams; ∆ is the maximum deflection for all cases. As derived before, θ is the area of the bending moment diagram divided by EI, the elastic modulus and moment of inertia, respectively; ∆ equals θ times the lever-arm from the centroid of the bending moment diagram between zero and maximum deflection to the point where θ is maximum.

6-24 FUNDAMENTALS: Bending

Typical beam diagrams

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.

1Cantilever beam with point load

2Cantilever beam with uniform load

3Cantilever beam with mixed load

4Simple beam with point loads

5Simple beam with uniform load

6Simple beam with mixed load

7Beam with one overhang and point load

8Beam with one overhang and uniform load

9Beam with one overhang and mixed load

10Beam with two overhangs and point loads

11Beam with two overhangs and uniform load

12Beam with two overhangs and mixed load

6-27 FUNDAMENTALS: Bending