Bending and Shear Stresses in Beams
The flexure formula M/I = σ/y = E/R, section modulus, beams of uniform strength, shear stress distribution in rectangular and circular sections, and the shear centre.
A bending moment stretches one side of a beam and compresses the other. The flexure formula gives these stresses; a separate formula gives the shear stress caused by the shear force.
The flexure formula
For a beam in pure bending (plane sections remain plane, linear elastic material):
M/I = σ/y = E/R
- σ varies linearly with distance y from the neutral axis, which passes through the centroid.
- The largest stress is at the outer fibre: σmax = M/Z, where Z = I/ymax is the section modulus.
| Section | I (about centroidal axis) | Z |
|---|---|---|
| Rectangle b × d | bd³/12 | bd²/6 |
| Solid circle, diameter d | πd⁴/64 | πd³/32 |
| Hollow circle D, d | π(D⁴ − d⁴)/64 | π(D⁴ − d⁴)/(32D) |
For a given area, an I-section has a far larger Z than a rectangle, which is why it is used for beams.
Shear stress in beams
τ = VAȳ/(Ib), where Aȳ is the first moment of the area above the level considered and b is the width at that level.
- Rectangle: parabolic distribution, zero at top and bottom, τmax = 1.5 τavg at the neutral axis.
- Solid circle: τmax = (4/3) τavg.
- I-section: most of the shear is carried by the web.
Shear centre
For sections that are not symmetric about the load axis (a channel, an angle), the load must pass through the shear centre to bend the beam without twisting it. For sections with two axes of symmetry, the shear centre is the centroid.
Worked example
A rectangular beam 100 mm wide and 200 mm deep carries a bending moment of 10 kN·m and a shear force of 20 kN.
- Z = 100 × 200²/6 = 666,667 mm³; σmax = 10 × 10⁶ / 666,667 = 15 MPa.
- τavg = 20000/(100 × 200) = 1 MPa; τmax = 1.5 MPa at the neutral axis.
Practice
A solid circular beam of diameter 100 mm carries a bending moment of 2 kN·m. What is the maximum bending stress in MPa? (Give 2 decimal places.)
Show answer and explanation
Answer: 20.37Z = πd³/32 = 98,175 mm³, so σ = 2 × 10⁶/98,175 = 20.37 MPa.
For a rectangular section, the maximum shear stress is how many times the average shear stress?
Show answer and explanation
Answer: 1.5The parabolic distribution gives τmax = 1.5 V/A at the neutral axis.
If the depth of a rectangular beam is doubled (width unchanged), its section modulus becomes:
Show answer and explanation
Answer: CZ = bd²/6 grows with d², so doubling d multiplies Z by 4 (I grows by 8).
Which statements about pure bending are true? (Select all that apply.)
Show answer and explanation
Answer: A, BBending stress grows linearly from zero at the neutral axis; shear stress (from V) is largest at the neutral axis.