Elastic Flexure Formula

σ_{x} = \frac{M y}{I}, τ = \frac{V Q}{I t}

Bending stress varies linearly with distance y from the neutral axis. Maximum tension/compression is at the extreme fiber, σ = Mc/I. Transverse shear uses the shear formula with first moment Q.

$M$ (N·m): Internal bending moment
$I$ (m⁴): Moment of inertia about NA
$c$ (m): Distance to extreme fiber

Step-by-step solved example

A 50×100 mm wood beam (I = 4.167×10⁶ mm⁴) has M = 2 kN·m. Extreme-fiber stress?

Bending: σ = My/I. Extreme fiber at c; compression vs tension.

1. c and units
c = 50 mm. M = 2e6 N·mm. I = 4.167e6 mm⁴.
2. Flexure
σ = Mc/I = 2e6×50 / 4.167e6 = 24.0 MPa.
$σ = 24.0 MPa$

Answer: 24.0 MPa

10 practice questions

0/10 correct

1.Rectangle b×h: I about centroidal axis is

2.Doubling c (same I, M) will

3.Neutral axis of a homogeneous beam is at the

4.Simple span L, midspan point load P: M_max =

5.Uniform load w on simple span: M_max =

6.Section modulus S is

7.Shear stress τ = VQ/(It) is typically maximum at the

8.For a rectangle, τ_max / τ_avg =

9.Sign convention (usual): sagging midspan moment is

10.I of 20×60 mm rectangle about its strong centroidal axis