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

294
bENdINg
To satisfy equilibrium, the resultant of all bending stresses must reduce to a zero net axial
force:
Also, the following moment equations must be satisfied:
∫ σ x dA = 0
(b)
A
∫ zσ x dA = My
(c)
A
∫ yσ x dA =− Mz
(d)
A
Next, substitute the expression for σ x given by Equation (a) into Equation (b) to obtain
⎛ Ey Ez ⎞ y z 1 1
∫ dA
dA ydA z dA 0
A⎜
− −
ρ ρ
∫A
ρ ρ ρ
∫A
ρ
∫
⎝ ⎠
⎟ = ⎛
+
⎞
⎜ ⎟ = + = (e)
⎝ ⎠
A
z y z y z y
This equation can be satisfied only if the neutral axis n–n passes through the centroid of the
cross section.
Substitution of Equation (a) into Equation (c) then gives
⎛ Ey Ez ⎞ E
E
2
∫ z
dA yz dA z dA My
A
⎜ − −
ρ ρ
⎟ =−
ρ
∫ −
A ρ
∫ = (f )
⎝ ⎠
A
z y z y
But the integral terms are simply the moment of inertia about the z axis and the product of
inertia, respectively:
2
I = z dA I = yzdA
y
∫
A
yz
∫
A
Moments of inertia and the
product of inertia for areas are
reviewed in Appendix A.
Therefore, Equation (f) can be rewritten as
− EI yz EI y
M y
ρ
− ρ
= (g)
z
y
Similarly, Equation (a) can be substituted into Equation (d) to give
where
− EIz
EI yz
Mz
ρ
+ ρ
= (h)
I
z
z
=
∫
A
y
2
y dA
Equations (g) and (h) can be solved simultaneously to derive expressions for the curvatures
in the x–y and x–z planes, respectively, due to bending moments M y and M z :
1
ρ
z
=
MI
+ MI
EII ( − I )
1
ρ
MI y z+
MI z yz
=−
EII ( − I )
z y y yz
2 2
y z yz y
y z yz
(i)