MecMovies 4.0 : M15.2: Principal Stresses in a Tee Beam

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At $D$ , which is located $2.96 i n .$ below the neutral axis, the bending stress is:

σ_{x} = - \frac{M y}{I} = - \frac{(- 25 k i p - f t) (- 2.96 i n .) (12 i n . / f t)}{70.55 i n .^{4}} = - 12.59 k s i = 12.59 k s i (c o m p r e s si o n)

The normal stresses produced by a bending moment is computed from the flexure formula:

$σ_{x} = - \frac{M y}{I}$

where $y =$ distance measured from neutral surface, $M =$ bending moment, and $I =$ moment of inertia about the centroidal axis.