Example of Composite
(Laminated) Section Analysis
"Equivalent Area Method"

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 10⁶ 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 10⁶ 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 in²

M1 = (r1)(A1) = (0.125")(3.5 in²) = 0.4375 in³

I1 = (r1)²(A1) + Ic1 =(0.125")²(3.5 in²)+(1/12)(14")(0.25")³= 0.729 in⁴

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 in²

M2 = (r2)(A2) = (0.25" + 2")(11.6 in²) = 26.1 in³

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 in²) + (1/12)(2")(4")³(1.45) = 74.192 in⁴

For the entire transformed area section, the following calculations are made:

A = A1 + A2 = 3.5 + 11.6 = 15.1 in²   this is the total area of the transformed section

M = M1 + M2 = 0.4375 + 26.1 = 26.538 in³   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 in⁴    this is the moment of inertia about the assumed neutral axis

I = IT - R²A = 74.921 - (1.757)² 15.1 = 28.31 in⁴    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 in³

for the lowest fibers of this material

y1bot = R  = 1.757"

SM1bot = I / y1bot = 28.31 / 1.757 = 16.11 in³

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 in³

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 in³

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 in³

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 in³

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 in³ = 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 in³ = 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 in³ = 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 in³ = 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 in³, ABS required section modulus for Material 1, this is assumed value only for this illustrative example

SM2R = SM1R / N2 = 4.0 in³ / 1.45 = 2.76 in³, 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 8-87, 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 5-6), 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, Glen-L Marine Designs, Bellflower, California.

Applications: The concepts described in this article are utilized in the following templates:

• Equivalent Area Method, Composite (Laminated) Section Analysis, English Units

• Equivalent Area Method, Composite (Laminated) Section Analysis, Metric Units