This set of Composite Materials Multiple Choice Questions & Answers (MCQs) focuses on “Stiffness, Compliance and Engineering Constant for Orthotropic Materials”.

1. How is the Young’s modulus of an orthotropic material is measured?

a) Uniaxial tensile testing

b) Failure testing

c) Rigidity testing

d) Flexibility testing

View Answer

Explanation: The Young’s modulus of an orthotropic material is found through the experiments in which a sample is subjected to a controlled tension until failure. For a given load, when the specimen fails, the ratio between the load applied to the elongation in the direction of applied load gives the Young’s modulus.

2. Which of the following coupling is not exhibited by orthotropic materials when they are stressed in the principle material direction?

a) Extension – extension coupling

b) Shear – Shear coupling

c) Extension coupling

d) Shear coupling

View Answer

Explanation: Orthotropic materials when they are stressed in the principal material coordinates direction, the effect of shear on the material is zero. Hence, they do not exhibit either shear- extension coupling or shear – shear coupling.

3. The relation between the Shear stress (τ), Shear strain (γ) for any isotropic or orthotropic solid is given by _________

a) τ = Gγ

b) τ = 1/(Gγ)

c) τ = G/γ

d) τ = γ/G

View Answer

Explanation: Rigidity modulus (G) of a material is defined as the ratio of the Shear stress to the shear strain.

4. For a given 2D orthotropic lamina, what is the relationship between stiffness matrix (S_{ij}) and the compliance matrix (C_{ij})?

a) [S_{ij}] = [C_{ij}]

b) [S_{ij}] = [C_{ij}]^{-1}

c) [S_{ij}] + [C_{ij}] = [K_{ij}]

d) [S_{ij}] α [C_{ij}]

View Answer

Explanation: When the strains are known, the stresses of the lamina can be found out with the help of stiffness matrix according to the Hooke’s law. But, when the stresses are known, and if we are to find strains, we can simply multiply the compliance matrix from which we can get directly the strains.

5. If the stiffness coefficients S_{12}, S_{11}, S_{22} are known, then the formula for finding the compliance coefficient C_{11} is __________

a) C_{11} = S_{22} / (S_{11} * S_{12} – S_{22}^{2})

b) C_{11} = S_{11}/ (S_{11} * S_{12} – S_{22}^{2})

c) C_{11} = S_{22} / (S_{11} * S_{22} – S_{12}^{2})

d) C_{11} = S_{12}/ (S_{11} * S_{12} * S_{22}^{2})

View Answer

Explanation: The stiffness matrix and the compliance matrix are always inverse to each other. The compliance coefficients are found by the mathematical derivations involving the matrix algebra that their components are related to the stiffness co-efficient.

6. What is the value of υ_{21} if the values of E_{11}, E_{22} and υ_{12} are given by E_{11} = 138GPa, E_{22} = 10GPa and υ_{12} = 0.21?

a) 0.818

b) 1.225

c) 0.9898

d) 0.0152

View Answer

Explanation: Given that, E

_{11}= 138GPa, E

_{22}= 10GPa and υ

_{12}= 0.21

We know that,

υ

_{21}/ υ

_{12}= E

_{22}/E

_{11}

⇨ υ

_{21}= (E22/E11) * υ

_{12}

⇨ υ

_{21}= (10/138) * 0.21

⇨ υ

_{21}= 0.0152.

7. When the lamina is reinforced with fibers in 1 (principal axis frame reference coordinate system) direction, if E_{1} > E_{2}, then what is the relationship between ^{1}Δ_{2} and ^{2}Δ1?

a) ^{1}Δ_{2} > ^{2}Δ_{1}

b) ^{1}Δ_{2} < ^{2}Δ_{1}

c) ^{1}Δ_{2} = ^{2}Δ_{1}

d) ^{1}Δ_{2} = 2 * ^{2}Δ_{1}

View Answer

Explanation: Because of the reciprocal relations, irrespective of the values E

_{1}and E

_{2},

^{1}Δ

_{2}=

^{2}Δ

_{1}which is a generalization of Bettie’s law to the treatment of isotropic bodies. That is, the transverse strain is same when the stress is applied is 2-direction as when applied in 1-direction.

8. In the case of unidirectional fiber reinforced composites with fibers oriented in the 1-direction, how cannot υ_{23} be related to υ_{12} and υ_{21}?

a) υ_{23} = υ_{32}

b) υ_{12} = υ_{21}

c) υ_{23} = υ_{12} * ((1 – υ_{21}) / (1 – υ_{12}))

d) υ_{32} = υ_{12} * ((1 – υ_{21}) / (1 – υ_{12}))

View Answer

Explanation: Christensen, a researcher, has shown that in the case of unidirectional fiber – reinforced composites with fibers oriented in the 1-direction, υ

_{23}can be related to υ

_{12}and υ

_{21}using the following equation. υ

_{23}= υ

_{32}= υ

_{12}* ((1 – υ

_{21})/ (1 – υ

_{12})). Thus, fits the experimental data within the range of experimental accuracy. Thus, for a unidirectional fiber-reinforced composite, the number of independent elastic constants is reduced from 5 to 4.

9. What is the value of normal strain (ε_{11}) for a rectangular plate whose length and thickness are given by 100mm*1mm respectively, when a tensile load of 1000N is applied parallel to the fiber direction? Take the value of E_{11} as 138GPa.

a) 72.5 * 10^{-4}

b) 725.6 *10^{-6}

c) 7.25 * 10^{-4}

d) 0.725 * 10^{-4}

View Answer

Explanation: Tensile load is applied parallel to the fiber direction, that is, in the 1-direction.

Therefore,

σ

_{11}= F/A = (1000 N)/ ((100 mm) * (1 mm))

= 10 MPa and σ

_{22}= 0.

Now, we calculate the normal strain ε

_{11}

ε

_{11}= σ

_{11}/E

_{11}

= (10MPa)/(138GPa)

= 0.725 * 10

^{-4}.

10. What is the value of ε_{22}, when the values of ε_{11} and υ_{12} are given by 0.725*10^{-4} and 0.21 respectively?

a) 0.152*10^{-4}

b) 0.58*10^{-4}

c) -0.58*10^{-4}

d) -0.152*10^{-4}

View Answer

Explanation: Given, ε

_{11}= 0.725*10

^{-4}and υ

_{12}= 0.21

We know that,

ε

_{22}= – υ

_{12}* ε

_{11}

= -(0.21*0.725*10

^{-4})

= -0.152*10

^{-4}.

**Sanfoundry Global Education & Learning Series – Composite Materials**.

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