Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

These curvature expressions can now be substituted into Equation (a) to give a general relationship
for the bending stresses produced in a prismatic beam of arbitrary cross section
subjected to bending moments M y and M z :
295
uNSyMMETRIC bENdINg
( MI z y + MI y yz ) y ( MI y z+
MI z yz ) z
σ x =−
+
2 2
II − I
II − I
y z
yz
y z
yz
(8.21)
or
Neutral-Axis orientation
⎛ Iz z − Iyz
y⎞
Iyy
Iyzz
σ x = ⎜ M y
M
2 2 z
⎝ II − I ⎠
⎟ + ⎛ − + ⎞
⎜
⎝ II − I
⎟ (8.22)
⎠
y z
yz
The orientation of the neutral axis must be determined in order to locate points in the cross
section where the normal stress has a maximum or minimum value. Since σ is zero on the
neutral surface, the orientation of the neutral axis can be determined by setting Equation (8.21)
equal to zero:
Solving for y then gives
y z
− ( MI + MI ) y + ( M I + M I ) z = 0
z y y yz y z z yz
yz
y =
MI
MI
+ MI
y z z yz
+ MI
z y y yz
z
which is the equation of the neutral axis in the y–z plane. If the slope of the neutral axis is
expressed as dy/dz = tan β, then the orientation of the neutral axis is given by
tan β =
MI
MI
+ MI
y z z yz
+ MI
z y y yz
(8.23)
Beams with Symmetric cross Sections
If a beam cross section has at least one axis of symmetry, then the product of inertia for the
cross section is I yz = 0. In this case, Equations (8.21) and (8.22) each reduce to
and the neutral-axis orientation can be expressed by
Mz y My z
σ x = − (8.24)
I I
y
z
MI y z
tan β = (8.25)
MI
Notice that if the loading acts entirely in the x–y plane of the beam, then M y = 0 and Equation
(8.24) reduces to
My z
σ x =−
I
which is identical to the elastic flexure formula [Equation (8.7)] developed in Section 8.3.
z
z
y