Composite Materials Questions and Answers – Stress-Strain Relations for a Anisotropic Materials – String

This set of Composite Materials Multiple Choice Questions & Answers (MCQs) focuses on “Stress-Strain Relations for a Anisotropic Materials – String”.

1. What does the term “Anisotropic material” refers to?
a) A material that possess same properties in different directions
b) A material that possess different properties along different directions
c) A material that possess many axes of symmetry
d) A material that possess different properties in same direction
View Answer

Answer: b
Explanation: For a general anisotropic body, there won’t be any planes of symmetry. So, it has different properties along different directions when measured at a particular point. These material properties are directionally dependent. They always depend upon the crystal orientations in a body.

2. Which of the following is true regarding the anisotropic materials?
a) σij & εkl are symmetric
b) σij & εkl are not symmetric
c) σij & εkl cannot be same at all
d) σij & εkl do not dependent upon direction
View Answer

Answer: a
Explanation: In an anisotropic material, generally σij & εkl are symmetric. This is the usual case when the body forces are absent. By the moment equilibrium concept, we can say that the shear stresses and the shear strains are same.

3. What is the best definition of Engineering Shear Strain (γij)?
a) It is the half of the angle of shearing under pure shear stress
b) It is half of the angle of shearing under pure shear strain
c) It is the total angle of shearing under a state of simple shear
d) They don’t have an accompanying rotation
View Answer

Answer: c
Explanation: The Engineering Shear Strain is defined as the total angle of the shearing under the state of simple shear. It implies that it is always accompanied with a rotation of an element. It differs from the tensor shear strain in the way that tensor shear strain doesn’t accompany any form of rotation.
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4. The relation between Engineering Shear Strain (γ12) and the Tensor Shear Strain (ε12) is given by __________
a) 4γ12 = ε12
b) γ12 = 4ε12
c) 2γ12 = ε12
d) γ12 = 2ε12
View Answer

Answer: d
Explanation: The Engineering Shear Strain is given by the total angle of the shearing under the state of simple shear. Whereas, the Tensor Shear Strain is given by the half of the angle of shearing under the action of the simple shear. Hence, γ12 = 2ε12.

5.The stiffness matrix (Cij) of an anisotropic material will have 36 elastic constants.
a) True
b) False
View Answer

Answer: a
Explanation: Yes, According to the moment equilibrium concept, ij = ji => Τxy = Τyx. Also, as kl = lk, γxy = γyx Since σij & εkl are symmetric for a material, the stiffness matrix is given by Cijkl = Cjilk. Hence, 9*9 matrix converges to 6*6 stiffness matrix with 36 elastic constants.

6. The alternate name for characterizing anisotropic materials are ___________
a) monoclinic materials
b) symmetric materials
c) triclinic materials
d) orthotropic materials
View Answer

Answer: c
Explanation: For an anisotropic material, there won’t be any plane of symmetry for the material properties. Similarly, a triclinic material is a material in which the three axes of the material are all oblique to each other as in anisotropic materials. Hence, they are also called as triclinic materials.

7. Which of the following is not true about the monoclinic materials?
a) Monoclinic materials possess one plane of material property symmetry
b) Monoclinic material has 13 independent elastic constants
c) There will be 2 axes of symmetry
d) The plane of symmetry is about XY plane (Z=0)
View Answer

Answer: c
Explanation: If there are 2 orthotropic planes of material property symmetry for a material, symmetry will exist relative to the third mutually orthogonal plane. Then, they are said to be Orthotropic materials.
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8. Which of the following is a wrong representation of compliance in terms of engineering constants?
a) S11 = 1/E1
b) S22 = 1/E2
c) S12 = -(γ21/E2)
d) S66 = 1/G12
View Answer

Answer: c
Explanation: The compliance term, S12 in terms of Engineering constant is given by S12 = -(γ21/E2) = -(γ12/E1). S12 cannot be given in terms of E2 .when Shear Strain is expressed as γ21.

9. What is the major objective in studying the stress – strain relations?
a) In order to conclude what deformation occurs because of a specific applied stress
b) In order to see the effect of only the anisotropic behavior of the body
c) In order to detect the properties of the material along only one direction
d) In order to know the behavior of materials when no load is applied
View Answer

Answer: a
Explanation: Stress–Strain relations gives us the idea about how the material is behaving towards the load that we are applying. It explains graphically when does the material goes into the plastic region from elastic zone. It elucidates when does the materials fail under the action of applied stresses and gives the corresponding elongations.
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10. Which of the following is not true with respect to the orthotropic lamina?
a) It follows plane stress condition
b) Stress components in all out of plane conditions are zero
c) It has the reduced stiffness matrix from 3D to 2D
d) It doesn’t possess material symmetry at all
View Answer

Answer: d
Explanation: Orthotropic materials will have orthogonal symmetry to both XY and XZ plane. Hence, these materials where 3 mutually orthogonal symmetry exists are called as orthotropic elements. They will have 9 independent elastic constants.

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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