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

## Thursday, June 26, 2014

## Tuesday, June 17, 2014

### BEAMS MOMENT OF INERTIA

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.

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## Tuesday, June 10, 2014

### BEAMS SECTION MODULUS

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

The section modulus for such beams is derived here.

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:

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 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.

## Wednesday, June 4, 2014

### BEAMS 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.

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.

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

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