Composite Materials Questions and Answers – Elastic Properties of a Lamina

This set of Composite Materials Multiple Choice Questions & Answers (MCQs) focuses on “Elastic Properties of a Lamina”.

1. What is the equation of longitudinal modulus for a unidirectional continuous fiber 0° lamina?
a) E11 = Efvf + Emvm
b) E11 = Efvm + Emvf
c) E11 = Efvf – Emvm
d) E11 = Emvm – Efvf
View Answer

Answer: a
Explanation: The longitudinal modulus is the ratio of the longitudinal stress to the longitudinal strain. For a unidirectional continuous fiber 0° lamina, the longitudinal modulus can be calculated by the equation E = Efvf + Emvm. The longitudinal modulus is usually denoted as E11. The E indicates longitudinal strength, v for volume fraction and the subscripts f and m indicate fibers and matrix respectively.

2. Identify this fiber lamina and find the equation for the calculation of its transverse modulus from the following.

a) E22 = Efvf / Efvm + Emvf
b) E22 = Emvm / Efvm + Emvf
c) E22 = EfEm / Efvm + Emvf
d) E22 = Efvm / Efvm + Emvf
View Answer

Answer: c
Explanation: A unidirectional continuous fiber lamina is represented in the diagram. As the name suggests, the fibers are continuous in a single direction with a 0° angle. The transverse modulus is generally indicated by E22. The E indicates longitudinal strength, v for volume fraction and the subscripts f and m indicate fibers and matrix respectively.

3. The fibers contribute more to the development of the transverse modulus.
a) True
b) False
View Answer

Answer: b
Explanation: The fibers contribute more to the development of the longitudinal modulus. It is the matrix that contributes more to the development of the transverse modulus. The transverse modulus is the ratio of transverse stress to the strain.
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4. Identify this fiber lamina and find the equation for the calculation of its transverse modulus from the following.

a) E22 = Em(1+2ηTvf) / (1-ηTvf)
b) E22 = Ef(1+2ηTvf) / (1-ηTvf)
c) E22 = Ef(1-ηTvf) / (1+2ηTvf)
d) E22 = Em(1-ηTvf) / (1+2ηTvf)
View Answer

Answer: a
Explanation: A unidirectional discontinuous fiber is shown in the diagram. Unlike continuous fiber lamina, discontinuous fibers are arranged in a single direction with a 0° angle in this lamina. The transverse modulus is given by the equation E22 = Em(1+2ηTvf) / (1-ηTvf). Em indicates the tensile strength of the matrix, vf indicates volume fraction of the fiber and ηT = (Ef/Em)-1 / (Ef/Em)+2.

5. What is the equation of Major Poisson’s ratio for a unidirectional discontinuous fiber 0° lamina?
a) v12 = vfvf – vmvm
b) v12 = vfvm + vmvf
c) v12 = vfvf + vmvm
d) v12 = vfvf / vmvm
View Answer

Answer: c
Explanation: The equation of Major Poisson’s ratio for a unidirectional discontinuous fiber 0° lamina is v12 = vfvf + vmvm. The major Poisson’s ratio is generally indicated as v12. vf and vm indicates the Poisson ratio of fiber and matrix respectively whereas vf and vm indicate the volume fraction of the fiber and matrix respectively.

6. Fiber aspect ratio has a significant effect on longitudinal modulus.
a) True
b) False
View Answer

Answer: a
Explanation: Fiber aspect ratio is the ratio of average fiber length to the fiber diameter. Since the longitudinal stress and strain acts along the length of the fiber, the fiber aspect ratio has a significant effect on longitudinal modulus. On the other hand, the fiber aspect ratio does not affect the transverse modulus since it is acting normal to the fiber.

7. Which among these is the most difficult to test?
a) Longitudinal modulus
b) Transverse modulus
c) Major Poisson’s ratio
d) Shear modulus
View Answer

Answer: d
Explanation: The Longitudinal modulus(E11), Transverse modulus(E22) and Major Poisson’s ratio(E12) are relatively easy to test using unidirectional test specimens loaded in tension and biaxial strain gauges. While the Shear modulus(G12) is more difficult to test because it is highly sensitive to specimen effects.
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8. Identify this fiber lamina and find the equation for calculating its tensile modulus from the following.

a) E = ⅜ E11 + ⅝ E22
b) E = ¼ E11 + ¾ E22
c) E = ⅛ E11 + ⅞ E22
d) E = ⅝ E11 + ⅜ E22
View Answer

Answer: a
Explanation: A randomly oriented discontinuous fiber lamina is shown here. Such laminas exhibit planar isotropic behaviour. That is, the properties are ideally the same in all directions of the laminar plane. Its tensile modulus is obtained by the equation E = ⅜ E11 + ⅝ E22, where E11 is the longitudinal modulus and E22 is the transverse modulus.

9. What is the equation for the shear modulus for a randomly orientated discontinuous fiber lamina?
a) Grandom = ⅖ E11+ ¼ E22
b) Grandom = ⅜ E11+ ¼ E22
c) Grandom = ⅛ E11+ ¼ E22
d) Grandom = ¼ E11+ ⅛ E22
View Answer

Answer: c
Explanation: The shear modulus is one of the important factors that determines the stiffness of the composite. The equation for the shear modulus for a randomly orientated discontinuous fiber lamina is Grandom = ⅛ E11 + ¼ E22, where E11 is the longitudinal modulus and E22 is the transverse modulus.
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10. What is Poisson’s ratio?
a) -(Longitudinal strain/Transverse strain)
b) -(Transverse strain/Longitudinal strain)
c) (Longitudinal strain/Transverse strain)
d) (Transverse strain/Longitudinal strain)
View Answer

Answer: b
Explanation: The Poisson’s ratio is the ratio of the negative of the transverse strain to the longitudinal strain. It is used to measure the Poisson effect, which is a phenomenon where the materials tend to expand in a normal direction to the direction of compression.

Sanfoundry Global Education & Learning Series – Composite Materials.

To practice all areas of Composite Materials, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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