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

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bENdINg
as those in the actual cross section. The width dimension (i.e., the dimension parallel to the
neutral axis), however, can be modified. Let the equivalent area dA′ of Material 1 be given by
the height dy and a modified width n dz, where n is a factor to be determined (Figure 8.11c).
Then the force transmitted by this area of Material 1 can be expressed as
dF′ = ( E1ε
)( ndz)
dy
If the transformed section is to be equivalent to the actual cross section, then the forces dF′
and dF must be equal:
( E ε)( ndz) dy = ( E ε)
dzdy
Therefore,
1 2
E2
n = (8.12)
E
1
The ratio n is called the modular ratio.
This analysis shows that the actual cross section consisting of two materials can be transformed
by use of the modular ratio into an equivalent cross section consisting of a single material.
The actual cross section is transformed in the following manner: The area of Material 1 is
unmodified, meaning that its original dimensions remain unchanged. The area of Material 2 is
transformed into an equivalent area of Material 1 by multiplication of the actual width (i.e., the
dimension that is parallel to the neutral axis) by the modular ratio n. The height of Material 2
(i.e., the dimension perpendicular to the neutral axis) is kept the same. This procedure produces
a transformed section, made entirely of Material 1, that transmits the same force (for
any given strain ε) as the actual cross section, which is composed of two materials, transmits.
Does the transformed section have the same neutral axis as the actual cross section?
If the transformed cross section is equivalent to the actual cross section, then it must produce
the same strain distribution. Therefore, it is essential that both cross sections have the
same neutral axis location. For a homogeneous beam, the neutral axis was determined by
summing forces in the x direction in Equation (8.4). Application of this same procedure to
a beam made up of two materials gives
∫
Σ F = σ dA+ σ dA = 0
x
A
∫
x1 A
x2
1 2
in which σ x1 is the stress in Material 1 and σ x2 is the stress in Material 2. In this equation,
the first integral is evaluated over the cross-sectional area of Material 1 and the second integral
is evaluated over the cross-sectional area of Material 2. From Equation (8.3), the
normal stresses at y (measured from the neutral axis) for the two materials can be expressed
in terms of the radius of curvature ρ as
σ
x1
E1
E y and
2
=− σ x2
=− y
(8.13)
ρ
ρ
Substituting these expressions for σ x1 and σ x2 gives
Σ E E
Fx
= − 1 ydA
2
∫
ydA 0
A ∫
1 ρ
− A2
ρ
=
The radius of curvature can be cancelled out so that this equation reduces to
∫
E ydA + E ydA = 0
∫
1 2
A
A
1 2
In this equation, the integrals represent the first moments of the two portions of the cross
section with respect to the neutral axis. At this point, the modular ratio will be introduced
so that the previous equation can be rewritten in terms of n:
∫
E ydA + E yndA = 0
∫
1 1
A
A
1 2