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THEORETICAL STRENGTH AND EFFICIENCY LIMITS
OF COMPOSITE
4.
(STRESS-STRAIN CALCULATION EXAMPLES RESULTING FROM STRUCTURAL LOADING)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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4. Theoretical Strength And Efficiency Limits of Composite
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4.1 Our Aims in This Section are:
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4.2 Load Distribution fraction Calculation in Composite Structure:
Tip for direction 1: Unit elongations are equal to:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(4.1)
4. Theoretical Strength And Efficiency Limits of Composite
Total Force on composite
Force carried by fibers
Force carried by matrix
;
Representative volume element :
How much of the tensile load in direction 1 do the fibers carry? First we are looking for an answer to this question :
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Stage 1: Elastic deformation occurs in the fibers and matrix.
Stage 2: While elastic deformation continues to occur in the fibers, the matrix undergoes plastic deformation.
Stage 3: Plastic deformation occurs in both fibers and matrix.
Stage 4: First the fibers and then the matrix are damaged.
(such as brittleness-ductility).
4.3 Deformation Stages of Composite Structure
When a unidirectionally reinforced composite layer with continuous fibers is subjected to tension in the 1-direction, it deforms in four stages as the load increases.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
(It is accepted that the composite is damaged when the fibers break.)
Fiber
damage
damage
matrix
composite
Stage 1
Stage 2
Stage 3
Stage 4
Stress
Strain
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(3.5)
koma
fiber
composite
matrix
rupture
rupture
a
b
c
d
rupture
Stages of deformation in a brittle structure
Figure 4.1
(4.5)
(4.2)
(4.3)
(4.4)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(It was explained in the calculation of E1.)
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Because the stresses in the fiber and matrix in the 2 direction will be equal, the matrix with lower strength will break first.
Under industrial working conditions, the direction of loading should coincide with the fiber direction (1 direction) in the composite. Because the 1st direction of the composite is more durable than the 2nd direction. Coinciding the load direction with the direction perpendicular to the fibers (direction 2) would be a wrong practice for unidirectional fiber reinforced composites. If there is loading in both 1st and 2nd directions, bidirectional fiber reinforced (cross-ply or woven fabric) composites must be used.
4.5 What is the Composite Strength in the (2) direction perpendicular to the fibers?�
(4.6)
4. Theoretical Strength And Efficiency Limits of Composite
For a unidirectional, continuous fiber reinforced layer,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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4.6 Theoratical Compressive Strengths �
It is assumed that when the fibers in a layer subjected to compressive loading in the 1st direction are damaged, the composite is also damaged. Accordingly;
kpma
fiber
composite
matrix
damage
a
c
damage
Compression
Tension Region
(4.7)
(4.8)
(4.11.a)
Equation 3.5, which gives the compressive stress in the composite at any moment in direction 1, is written for the moment when the fibers are subjected to compressive damage.
(4.9)
(4.10)
4. Theoretical Strength And Efficiency Limits of Composite
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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As stated in the theoretical calculation of G12 , which is the subject of 3.9, the shear stresses occurring in the composite, fiber and matrix are equal in case of shear loading.
Accordingly, the first component to be damaged due to shear loading will be the matrix with the lowest shear strength. (Because the matrix will reach its first strength limit.)
As a result, the shear strength of the composite is:
(4.11.c)
Any shear loading instant
(4.11.b)
Note: In some special cases, the shear strength of the fiber may be lower than that of the matrix. In this case, the theoretical shear strength of the composite should be taken as the shear strength of the fiber.
4. Theoretical Strength And Efficiency Limits of Composite
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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Example 4.1
A unidirectional and continuous fiber reinforced orthotropic composite layer will be produced from the matrix and fiber materials whose properties are given in the table. In this layer with dimensions of 400mm x 400mm x 4mm, 25% volumetric fiber will be used. According to this,
a-) When pulling in direction 1, find the stresses in the fibers and matrix at the moment when the composite will be damaged.
b-) If tensile forces of F1 = 80kN, F2 = 48kN are applied to this layer simultaneously in directions 1 and 2, calculate the stresses and total deformations (extensions) that will occur in the composite, fiber and matrix in directions 1 and 2.
c-) If we had not neglected the Poisson effect, how much would the total elongation in the 2 direction we found in option c change?
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
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Solution:
a)
From equation (4.3):
To solve another options of this problem, we must first calculate the orthotropic properties:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
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Stresses in the composites:
F1=80kN
F1
A1
F2
A2
F2=48kN
b-)
Unit elongation (strain) in the composite in direction 1:
Total elongation in the composite in direction 1:
From equation (2.17.a)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Total elongation in the composite in direction 2:
From equation (2.17.b)
Strain in the composite in direction 2:
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0
0
0
0
That is, from equation 3.3, strains in the 1 direction in the fiber and matrix:
Strains in direction 1 are equal (from tip-1)
Stress in direction 2 are equal (from tip-2)
That is, from equation 3.6, stresses in the 2 direction in the fiber and matrix:
We can apply Hooke's equations, which are valid for isotropic materials, separately for fiber and matrix. (Because fiber and matrix materials are isotropic.)
From equation (2.15.a):
From equation (2.15.b):
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
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c-)
Poisson effect factor :
Strain in direction 2 in the composite :
When we take the Poisson effect into consideration, the changing values are E2 and ν21.
If the Poisson effect is not neglected, we put (*) above the affected values.
Total elongation in composite in direction 2:
If the poisson effect is not neglected :
If the poisson effect is neglected :
From equation (3.10b):
From equation (3.13):
From equation (3.10.a):
From equation (2.17.b):
Difference between elongations:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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Example 4.2
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
The layer in Example 4.1 is placed in a fixed cavity suitable for its dimensions, as shown in the figure, and is subjected to a compression force of F = 60kN in the 1 direction. According to this; Calculate the changes in the side lengths of this layer in directions 1 and 2. (The bottom, back and side surfaces of the layer are in contact with the cavity.)
Çözüm:
stress in direction 1 :
Due to the constraints, the unit and total strains in the 2 direction are zero:
Total shortening in direction 1 :
From equation (2.13.b):
From equation (2.13.a):
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An aluminum plate with an initial temperature of 23 C is placed between two fixed walls as shown in the figure. According to this,
a-) Can the aluminum plate be used at 120 oC under these boundary conditions? Calculate.
Example 4.3* (video 3)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Material | E �Modulus of Elasticity | ν�Poisson ratio | α�Coefficient of Thermal Expansion (CTE) | Tensile/Yield Strength |
Steel - fiber (ductule) | 210GPa | 0,28 | 10x10-6 1/ oC | 800 / 400 MPa |
Aluminum - matrix (ductule) | 70 GPa | 0,27 | 23x10-6 1/ oC | 200/150 MPa |
(* Attention: This example also includes formula inferences regarding thermal loads.)
b-)Can this composite structure be used at 120 oC?
c-) When the right wall is removed, to what temperature can the composite structure be heated within its strength limits?
d-) When the right wall is removed, what will be the stress value that will occur in the matrix component of the composite structure at the maximum allowable (safe) temperature?
The same aluminum plate will be unidirectionally reinforced with 25% steel fibers to create a composite structure and this structure will be placed between two fixed walls. Accordingly, for this composite structure, answer the following questions with calculations.
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Aluminum
Aluminum
Aluminum
a-) Can the aluminum plate be used at 120 oC ?
P
As a result, when the temperature exceeds 116.1 °C, the aluminum material will flow. This means that Aluminum plate cannot be used at 120 oC. (We can make this calculation with our strength information)
= 0
Aluminum can be heated to its yield limit. In the limit case,σ = σakma.
According to the superposition principle, we first lift the right wall and increase the temperature, then we apply the reaction force P coming from the wall. We start from the fact that the total extension (δ) is zero.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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P
b-) Can the plate be used at the same operating temperature (120 oC) if it is unidirectionally reinforced with 25% steel fibers?
(4.12)
In an orthotropic layer constrained in direction 1, the temperature difference at any moment is:
This time we will use the same solution as in part a for the orthotropic composite structure. Note that the only thing that changes are the material constants in Hooke's relations.
From the last equality …>>
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Steel (fiber)
Aluminum (matrix)
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Then from the Equation 4.12:
:
The maximum temperature difference that can be applied to an orthotropic layer limited (constrained) in the 1 direction, within the strength limits:
(4.13)
The composite can be used up to this temperature in a constrained condition.
Note: In the constrained state, deformation is completely prevented. For this, it is necessary to have both walls.
Kompozit Malzeme Mekaniği-Ders Notları-Prof.Dr.Mehmet Zor
From equ.4.4..>>
From equ. 4.3..>>
(See: chapter 4)
From eq. 3.4..>>
From equ. 4.13...>>
From equ. 3.34..>>
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c-) When the right wall is removed, to what temperature can the composite structure be heated within its strength limits?
As explained in article 3.14, even if the composite is allowed to expand freely, stresses in the 1 direction will occur in the fiber and matrix. In addition to thermal elongation, the fibers also lengthen a little more due to the pull of the matrix itself.
Total elongation in composite
Total elongation in fiber:
Thermal elongation
Elongation caused by the matrix pulling the fiber
=
=
Temperature difference at any instant in free state: :
: Stress caused by the matrix pulling the fibers:
(4.14)
Maximum allowable temperature difference in free state:
(4.15)
From the above equation
For option c of the example we are examining, there is an unconstrained situation since the right wall is removed. Maximum temperature difference from equation 4.15:
(Maximum operating temperature that can be reached within endurance limits when the right wall is removed)
Note: If one or both of the right or left walls are removed, extension is allowed and the Free state is obtained.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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=
Thermal elongation
Shortening caused by fibers preventing the matrix from elongating
: Stress caused by the fibers working to prevent the matrix from elongating
(4.16)
From the above equation, the stress in the matrix for any ∆𝑇 :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Total elongation in composite
Total elongation in matrix
=
d-) We will calculate the stress in the matrix at the maximum allowable temperature in the unconstrained case:
The maximum temperature difference was found in option c. At this instant the stress in the matrix:
(4.17)
or 2nd way
For free-state thermal loading:
From eq. 3.5:
If you pay attention, at maximum temperature, the yield strength of the matrix (150 MPa) is not exceeded and no damage occurs to the matrix.
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Sample Question 4.4
A cylindrical reinforcement sample with a diameter of 12cm and a length of 24cm, obtained by placing iron rods in concrete, is placed between two cylindrical rigid plates at the bottom and top and is subjected to a compression test. During the test, a 0.4mm collapse (shortening) in the length of the sample was measured when the compressive force P1 = 700kN. According to this,
P1
φ D = 12cm
24cm
| E(GPa) | ν | σmax (MPa) |
Iron | 200 | 0,3 | 400 |
Concrete | 32 | 0,2 | 65 |
a-) Which special type of anisotropic material does this material fall into? (Answer: Transversely Isotropic Material)
c-) Calculate the stresses in the concrete and iron bars for a compressive force of 700kN in this sample.(Answer: stress in concrete: -53.34MPa, stress in iron: -333.4MPa)
d-) Calculate theoretically the maximum compression force that the sample can withstand. (Answer: 839.66kN)
e-) How many 24mm diameter iron rods should be used in a 1m x 0.5m rectangular cross-section column with the same material properties? (Answer: 34)
b-) Find the narrowing in the sample diameter.(Answer: 40,56x10-3mm)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
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4.8 Limits of Efficiency from Composite
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4. Theoretical Strength And Efficiency Limits of Composite
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0
Strength of composite at break:
(4.18)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
kompozit
matris
kopma
kopma
d
e
(Remember: Equation 4.18 is valid for the case where the fibers do not carry any load and break immediately.)
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or
(4.18)
=
(4.4)
the fibers are said to carry load.
Considering equations 4.2, 4.3 and 4.5,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
koma
fiber
composite
matrix
rupture
rupture
rupture
a
b
c
d
When
in terms of strains :
(4.19a)
(4.19b)
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or
in terms of strains
It is always valid.
From the above equaiton,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
Equation (4.4)
(4.20a)
(4.20b)
4.8.3 Diagram: Fiber Volume fraction – Composite Strength
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A graphical summary of what is explained in this 4th topic can be seen on the side.
Important points :
The fibers break immediately and the matrix carries all the load. Composite strength is considered when the matrix is damaged.
2-) In the region Vfmin< Vf < Vfcr :
fibers carry the load, but the composite is not efficient.
fibers carry load and efficiency is obtained from the composite.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
Matrix dominant
Fiber dominant
Try to interpret the diagram for different situations.
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Example 4.5
It was produced in a unidirectionally reinforced orthotropic layer from matrix and fiber materials, both of which exhibit brittle character. The dimensions of the layer are 400mm x 400mm x 8mm and the material properties are given in the table above. 25% fiber was used in the structure. According to this;
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
Matrix Fiber
Modulus of Elasticity: Em=16GPa, Ef=82 GPa,
Maximum Strain( at breaking) εm-max=2,5x10-3 εf-max=1,71 x 10-3
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Solution:
b) Critical fiber volume fraction:
a) Minimum fiber fraction:
From equation(4.19b):
From equ. (4.20b):
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
fibers carry load.
efficiency is obtained from the composite.
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Fiber strength:
c-)
F1max
F1max
A1
= 55,575 x 400 x 8
As explained in article 4.5, the strength of the composite in the 2 direction is equal to the strength of the matrix:
Matris strength:
Stress in the matrix when fibers break :
Composite Strength in direction 1:
Maximum tensile force that can be applied in direction 1:
Maximum tensile force that can be applied in direction 2:
F2max
F2max
A2
Composite Strength in direction 2 :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite