Design and Analysis of a Laminated Composite Tube Essay

Custom Student Mr. Teacher ENG 1001-04 7 August 2016

Design and Analysis of a Laminated Composite Tube

A composite material is a material made from two or more constituents with significantly different physical or chemical properties. There are two main categories of constituent materials, matrix and reinforcement. At least one material of each type is required. (Composite Material).

The phase distribution and geometry of the two materials have been ordered to optimise its properties, this has led to the increased use of composite materials due to their advantageous properties such as light weight, high strength to weight ratio and desirable stress-strain properties. Common types such as polymer matrix composites (PMC) and Ceramic matrix composites (CMC) are used in a wide use of applications.


The tube will be designed to sustain the following loading conditions provided:

1) An internal pressure of 3 MPa before burst when both ends are closed but free to deform (not to be tested but calculated design is required as a part of the report), assuming no leakage takes place before burst.

2) Axial compression of 25 kN.

3) Tube designed to achieve an angle of twist in a specified direction under axial compression as demonstrated by the diagram below. Maximum angle of twist should be achieved before the onset of identifiable mechanisms of failure.



Orthotropic lamina

The composite is made up of multiple thin laminate layers like the one in Figure 3 above. For an orthotropic material; one which has varying mechanical properties in each axis, a coordinate system is stated as demonstrated in Figure 3 above. The assumption of lamina in a composite laminated structure is that it is under plane stress state, therefore σ3=τ23=τ13=0.


From theory for the lamina, the strain-stress relationship in plane stress state is {ε}=[R]{σ}. This leads to the stress-strain relationship; {σ}=[Q]{ε}.


In most composites the coordinate system of a structure e.g. Figure 4 (x,y,z) is orientated differently to that of the materials principle axis e.g. Figure 3 (1,2,3). Therefore a coordinate transformation matrix [T] is required, for plane stress state this can be simplified to:

Classic Laminate Theory

Classic laminate theory describes a three dimensional laminate problem via a two dimensional representation for the simplification of analysis. This is done by assuming a deformation pattern through the thickness of the laminate, the simplest assumption is known as the ‘Love-Kirchoff’ assumption leading to the classic laminate theory (CLT).

The Love-Kirchhoff hypothesis generalises the plane section assumption in beam theory; assuming the normal to the laminate remains normal to the deformed laminate and the normal undergoes no extension of shortening. Leading to:

Resultant displacements:


u0 and v0 are in plane displacements.

w is the deflection.

z=0 as reference surface.


. (3)

Due to the assumptions that it demonstrates a linear distribution for plane strains throughout the laminate thickness and that out of plane strains can therefore be ignored.

{ε0} is the in plane strain and {k} is the curvature of the reference surface.

From equation (1) and appropriate coordinate transformations the following relationship is obtained.


Though integration and manipulation of the elasticity equations with respect to the “z” the membrane forces can be found as:


Thus the bending moments are as follows:


{N} and {M} are the generalised stresses can can be expressed as membrane strains and curvatures by using the laminar stress-strain relationship and Love Kirchhoff hypothesis.

, (7)

As {κ} = 0

Also as there is no bending, this can be assumed to be equal to {ε}.

Where [A], [B] and [D] are integrated over the layer thickness of the laminate, Figure 5:

Loading conditions

Axial loading case

Load acts over outer circumference of on end of the cylinder, while the other end remains against a rigid immoveable object. Due to the axial load only only x direction stresses will be present:


Internal pressure loading case

Pressure is applied through internal surfaces and outwards of the closed ends. In this case the cylinder experiences stress in the x and y directions and thus:

, (9)


Twist Angle

The twist angle of the cylinder can be calculated from the following formula:

rads (10)

Where Y​xy is the angle the generator deforms by, L the distance point A moves and R is the radius of the cylinder.

Maximum stress criterion

This failure mechanism was developed for brittle solids; the maximum stress criterion assumes that a material fails when the maximum principal stress in a material exceeds the uniaxial tensile strength of the material. (Material failure theory).

The criterion is the simplest and most widely used when dealing with composites, however it does suffer in accuracy compared to more laborious techniques. It is advantageous as it provides an indication of the method of failure compared to that of other methods.

According to the maximum stress failure criterion for a given stress state the material will fail if the following conditions are void:

2D case:

This demonstrates that a ratio greater than 1 for either case will result in a failure in a given direction.


The following data was input into the MathCAD sofware, where P is the axial compression, q the internal pressure, E the Young’s moduli, ν the poisons ratio, G the shear modulus and L the length of the tube:

P:= -25000 N, q:=3×106 Pa, R:=0.0255 m , L:=0.300 m,

E1:=236×109 Pa, E2:=5×109 Pa, G:=2.6×109 Pa, ν:=0.25.

Calculate the stiffness matrix [Q], equation (1), for the material coordinate system. Calculated by substituting above values into equation 1.

Input the winding angles in vector form for lay-ups in the tube in the specific [α/β/α/β] layup desired. The process to determine the optimum angles is iterative, optimum winding angles need to be input to achieve the greatest angle of twist without resulting in failure.

Calculate the stiffness matrix [A]. To obtain [A] we initially determine the coordinate transformation matrix for the laminate; using estimated values of α and β. Substituting the obtained matrix into equation 7 provides a value for [A].

Calculate membrane forces {N} for the axial compression load case. Due to the load only being applied in the x-direction {N} is simply obtained by substituting equation 8 into equation 5.

The mid-plane strains ({ε0}) are obtained by using the generalised stress-strain relationship stated above in equation 7. Substituting the previously calculated values for {N} and [A] into equation 7 will provide {ε0}.

Calculate the twist angle of the tube. The angle of twist ‘ρ’ is calculated using equation 10 above.

Calculate the layer strains in the material principal direction. Achieved by multiplying the mid-plane strains {ε0} by the coordinate transformation matrix (equation 2), using the estimated value of α for the first lamina and β for the second lamina.

Calculate the layer stresses in their material principal direction. The stiffness matrix (equation 1), with an input of a 0 degree angle, is multiplied by the above calculated strain in their material principal direction.

Calculate the Maximum Stress Criterion as explained in equation 5. The values calculated are of a magnitude less than “one”, indicating that no failure is occurring with these parameters.

Calculate the load {N} for the internal pressure case. Here the relationships discussed in equation 9 are used. {Nx} is obtained by multiplying half of the axial load by the radius of the cylinder. {Ny} is obtained by multiplying the total load by the cylinder radius.

Mid-plain strains are again calculated by inverting the extensional stiffness matrix [A] and multiplying it by the newly obtained load matrix {N} above.

Steps 8 and 9 are now repeated to find the layer stresses and strains in the other principal directions.

Step 10 was repeated to determine the maximum stress failure criterion of the piece; its magnitude was checked to ensure the avoidance of any failures in the lamina (value less than one).

In order to determine the optimum winding angles it is now necessary to alter the winding angle values of α and β. The optimum being one that gives you the highest angle of twist while preventing failures according to the maximum stress failure criterion. From here steps 4-14 are repeated to check
the suitability of the values chosen. The angle iteration process is explained in more detail below.


From The figure 9 above we see that MathCad was used to produce a graph for the twist angles for combinations between [-90o, -90o] and [+90o, +90o]. The highest angle of twist in the positive and negative directions is depicted by the red and purple areas on the figure. The purple areas in figures 9 and 10 demonstrate the winding angles that yield the maximum amount of twist and provide the range of angles for which to test, additionally the angles are to be between [-75o, -45o] and [+75o, +45o] as set in the brief. The sections under observation can therefore be constricted.

From figure 2 we know the direction of twist is in the anti-clockwise direction; which according to the right hand rule, is opposite to the positive angle of twist due to compression, the maximum amount of twist according to the diagram is therefore in the negative direction. Therefore the area of interest is the bottom left portion of figure 9 (yellow box) as this is the only section that produces negative twist angles in the winding angle range.

Below are the tabulated results for the various angle combinations tested in the MathCad analyses, as previously discussed the angle must lie within the range [-75o, -45o] and [+75o, +45o]. Due to the symmetry of the graph either axis can represent both winding angles α and β.

Free Design and Analysis of a Laminated Composite Tube Essay Sample


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  • University/College: University of Arkansas System

  • Type of paper: Thesis/Dissertation Chapter

  • Date: 7 August 2016

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