This example is to
illustrate how a composite equivalent area calculations are evaluated and
how they apply to first principles and American Bureau of Shipping
requirements. For theoretical background for this example, refer to
the article entitled "Composite
(Laminated) Section Analysis, Equivalent Area Method."
This example is
only for illustrative purposes. Among other things, it leaves out
the required bonding angles between a stiffener and it's associated
plating. In this example the wood is not encased, in real life it
is normally encased in fiberglass. Details like these are
purposely left out of this example, in order to focus on understanding
the calculative process with respect to composite sections.
Example Section Dimensions and Material Properties
The example cross
consists of a 2" x 4" piece of Douglas Fir plywood that is standing
upright on a 1/4" thick 14" fiberglass plating element that is laying
flat. The fiberglass is standard laminate as per ABS requirements.
The material properties are:
For Standard
Laminate: E1 = flexural modulus = 1.1 x 10^{6} psi,
Stress1a = allowable stress = 25,000 psi, per ABS Ref. 5, Section 4.4.2
For Douglas Fir
Plywood: E2 = flexural modulus = 1.6 x 10^{6} psi,
Stress2a = allowable stress = 1,000 psi, per ABS Ref. 5, Section 4.6
Calculation of Section's Equivalent Area Properties
This will be done
for each material and then combined to get the results for the entire
section. For this example the assumed neutral axis is located at
the bottom of the plate (the lowest edge of the section).
For the fiberglass ABS Standard Laminate, the first material, the following
calculations are made.
A1 = (0.25")(14") = 3.5 in^{2}
M1 = (r1)(A1) = (0.125")(3.5 in^{2}) = 0.4375 in^{3}
I1 = (r1)^{2}(A1) + Ic1 =(0.125")^{2}(3.5 in^{2})+(1/12)(14")(0.25")^{3}= 0.729 in^{4}
For Douglas Fir, the second material, the following calculations are made.
N2 =
E2 / E1 = 1,600,000 psi / 1,100,000 psi = 1.45, this is the "transformation
factor" or "modulus ratio"
A2 = (2")(4")N2 = 11.6 in^{2}
M2 = (r2)(A2) = (0.25" + 2")(11.6 in^{2}) = 26.1 in^{3}
Note:
Ic2 (moment of inertia, about its own centroidal axis, for
Material 2) can be calculated in one of two ways; either with
the equivalent area width or multiplying the original section Ic
by the transformation ratio N2. The second method is
required when using tabular values or if the element is a non
rectangular geometric shape (like a triangle where the moments
of inertia values are calculated directly by formulas).
Let's say the element is a I beam or an angle, then the second
method would be required because only the moment of inertia for
the original section is readily available in tables. Both
these methods are illustrated as follows:
Ic2 = (1/12)(2"x N2)(4")^{3} <Method 1 using equivalent
area breadth
Ic2 = (1/12)(2")(4")^{3} x N2 < Method 2 using the original
section's moment of inertia
I2 = (r2)^{2}(A2)
+ Ic2 = (0.25" + 2")^{2}(11.6 in^{2}) +
(1/12)(2")(4")^{3}(1.45) = 74.192 in^{4}
For the entire
transformed area section, the following calculations are made:
A = A1 + A2 = 3.5 + 11.6 = 15.1 in^{2 }this is the total
area of the transformed section
M = M1 + M2 = 0.4375 + 26.1 = 26.538 in^{3 }this is the
total moment of the transformed section
R = M / A = 26.538 / 15.1 = 1.757 in this is the location of the
neutral axis
IT = I1 + I2 = 0.729 + 74.192 = 74.921 in^{4 }this is
the moment of inertia about the assumed neutral axis
I = IT  R^{2}A = 74.921  (1.757)^{2 }15.1 = 28.31 in^{4
}this is the moment of inertia about the neutral axis for the
entire transformed section
Section Modulus Calculations
For the fiberglass
for the highest fibers of this material
y1top = R  0.25" = 1.757"  0.25" = 1.507"
SM1top = I / y1top = 28.31 / 1.507 = 18.79 in^{3}
for the lowest fibers of this material
y1bot = R = 1.757"
SM1bot = I / y1bot = 28.31 / 1.757 = 16.11 in^{3}
for the governing section modulus of this material
y1 = the larger of y1top or y1bot = 1.757" for this material's fibers
that are furthest from the neutral axis
SM1 = I / y1 = 28.31 / 1.757 = 16.11 in^{3}
For the plywood
For this
second material, notice that the section modulus must be divided by
the transformation factor. For details refer to Formulas 10
and 10a of the
derivation article.
For the highest fibers of this material.
y2top = 4.25"  R = 4.25"  1.757" = 2.493"
SM2top = (I / y2top) / N2 = 28.31 / (2.493 x 1.45) = 7.831 in^{3}
for the lowest fibers of this material
y2bot = R  0.25" = 1.757"  0.25" = 1.507"
SM2bot = (I / y2bot) / N2 = 28.31 / (1.507 x 1.45) = 11.11 in^{3}
for the governing section modulus of this material
y2 = the larger of y2top or y2bot = 2.493" for this material's
fibers that are furthest from the neutral axis
SM2 = (I / y2) / N2 = 28.31 / (2.493 x 1.45) = 7.831 in^{3}
Application First Principles Requirements
Based on the conditions present a maximum moment is calculated. These
conditions include the beam's span, the orientation of supports, the
types of supports, the magnitude and the nature of loading or loadings.
Suppose the first principle requirements yielded the following moment.
M = 5,000 inch pounds, first principles applied moment for this example
Starting from the top down, the following stresses and factors of safety are present
within the cross section.
For the top of the section of plywood
Stress2top = M / SM2top = 5,000 inch pounds / 7.831 in^{3} = 638.5 psi,
this is the governing stress for the wood.
FS2 = Stress2a / Stress2top = 1,000 psi / 638.5 psi = 1.57, this is the
governing factor of safety for the wood
For the bottom of the plywood, at the top of the plating
Stress2bot = M / SM2bot = 5,000 inch pounds / 11.11 in^{3} = 450.0 psi
FS2 = Stress2a / Stress2bot = 1,000 psi / 450.0 psi = 2.22
For the top of the fiberglass plating
Stress1top = M / SM1top = 5,000 inchpounds / 18.79 in^{3} = 266.1 psi
FS1top = Stress1a / Stress1top = 25,000 psi / 266.1 psi = 93.9
For the bottom of the fiberglass plating
Stress1bot = M / SM1bot = 5,000 inchpounds / 16.11 in^{3} = 310.4 psi,
this is the governing stress for the fiberglass
FS1bot = Stress1a / Stress1bot = 25,000 psi / 310.4 psi = 80.5, this is the
governing FS for the fiberglass
Application ABS Section Modulus Requirements
Based on the conditions present a required section is calculated using Reference 5.
These input conditions include the beam's span, the orientation of
supports, the types of supports, the magnitude and the nature of loading
or loadings. Suppose that the ABS requirements contained in
Reference 5 yielded the following required section modulus.
SM1R = 4.0 in^{3}, ABS required section modulus for Material 1, this is
assumed value only for this illustrative example
SM2R = SM1R / N2 = 4.0 in^{3} / 1.45 = 2.76 in^{3}, ABS required
section modulus for Material 2. Note that the required section
modulus must be divided by the transformation factor for this second
material. For details refer to Formula 18 of the
derivation article.
The factors of
safety on required section modulus values are calculated below for each
material in the cross section.
For the fiberglass plating, the minimum available section modulus is
SM1, this is the governing section modulus for this material.
FS1SM = SM1 / SM1R = 16.11 / 4.0 = 4.03, since this value is greater
that 1 this material meets the requirements of ABS.
For the Wood stiffener, the minimum available section modulus is SM2
(this is the governing section modulus for this material).
FS2SM = SM2 / SM2R = 7.831 / 2.76 = 2.84, since this value is greater
that 1 this material meets the requirements of ABS.
Summary
In this simplified
example both materials meet first principles stress requirements and
also the ABS section modulus requirements. The stress at to top of
the plywood is 638.5 psi, then the stress linearly goes to zero at
1.757" above the bottom of the member, then the stress changes linearly
and at the bottom of the plywood it is 450 psi, then it jumps down to
266.1 psi at the top of the plating and then it linearly varies to 310.4
psi at the bottom of the plating.
Primary References
Composite beam
analysis references for this example are listed as follows:

EN358 Ship Structures, Course Notes, U. S. Naval Academy,
Annapolis, MD. This is a large document, see the section
entitled "Composite Beam Approach (Dissimilar Materials)."

High Strength Composites Course, University of Utah. The
section entitled "Composite (laminated) beam" applies in this
discussion.

Mechanics eBook: Composite Beams, University of Oklahoma.
See last section entitled "Alternative Method  Equivalent Area."

Rules for Building and Classing Reinforced Plastic Vessels,
1978, American Bureau of Shipping.

NVIC 887,
Notes on Design, Construction, Inspection and Repair of
Fiber Reinforced Plastic (FRP) Vessels, USCG, Washington, D. C..
This is a very large document, for pertinent details go to Enclosure
1, Part B Determining Section Modulus and Moment of Inertia.

Buckling
of Transversely Framed Panels, Chapter 3, Marine
Composites, Eric Greene Associates, Annapolis, MD. See pages
172 and 173 for an example of composite section calculations.

Rules for Materials and Welding, Part 2 Aluminum & Fiber
Reinforced Plastic (FRP) (Chapters 56), 2006 American Bureau of Shipping.

Fiberglass Boat Design and Construction,
by Robert J. Scott, 1996 Second Edition, Society of Naval Architects
& Marine Engineers.

Fiberglass Boatbuilding for Amateurs,
by Ken Hankinson, 1982, GlenL Marine Designs, Bellflower,
California.
Applications: The concepts described in this article are utilized in
the following templates:
