BEAMS SECTION MODULUS

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.