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

308
bENdINg
The circumferential normal stress acting on area dA can now be obtained from Hooke’s
law as
(
E r n − r ) d θ
σ x =−
(b)
rθ
The location of the neutral axis (N.A.) follows from the condition that the summation
of the forces acting perpendicular to the section must equal zero; that is,
(
F dA E r n r )
0
d
x ∫ σ
− θ
Σ = x =−
dA 0
A ∫ =
A rθ
However, since r n , E, θ, and dθ are constant at any one section of a stressed bar, they may
be taken outside the integral to obtain
Ed rn
r Ed dA
∫ σ θ −
θ ⎛
xdA
=−
dA rn
A ∫ =−
θ A r
θ
∫ −
⎝
⎜
A r
To satisfy equilibrium, the value of r n must be
r
n
∫
A
⎞
dA 0
⎠
⎟ = (c)
= A
dA
(8.27)
∫ A r
where A is the cross-sectional area of the bar and r n locates the neutral surface (and thus the
neutral axis) of the curved bar relative to the center of curvature. (Note that the location of
the neutral axis does not coincide with that of the centroidal axis.)
Formulas for the areas A and the radial distances r n from the center of curvature, O, to the
neutral axis are given in Table 8.2 for several typical cross sections. These formulas can be
combined as necessary for a shape made up of several shapes. For example, a tee shape can
be calculated as the sum of two rectangles, where Equation (8.27) is expanded by the addition
of a second term. Shapes can also be treated as negative areas. For example, the radial distance
to the neutral axis of a box-shaped cross section can be calculated by the combination of (a) an
outer rectangle whose area A is treated as a positive quantity in Equation (8.27), and (b) an
inner rectangle whose area is treated as a negative quantity. Clearly, the sum of the positive
area of the outer rectangle and the negative area of the inner rectangle gives the area of the
box shape.
After the location of the neutral axis is established, the equation for the circumferential
normal stress distribution is obtained by equating the external moment M to the internal
resisting moment developed by the stresses expressed in Equation (b). The summation of
moments is made about the z axis, which is placed at the neutral-axis location found in
Equation (8.27):
Σ Mz
= 0 M + ∫ yσ
x dA = 0
The distance y can be expressed as y = r n − r. The normal stress σ x was defined in Equation (b).
We substitute these expressions into the integral term to obtain
M r r dA E r r 2
( n )
∫ σ
−
( )
d θ
=− x n − =
dA
A ∫A
rθ
Since the variables E, θ, and dθ are constant at any one section of a stressed bar, Equation
(d) can be expressed as
M Ed θ ( r n − r )
=
θ
∫A
r
A
2
dA
(d)
(e)