# Composite Materials Questions and Answers – Stiffness, Compliance and Engineering Constant for Orthotropic Materials

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

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

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

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 (Sij) and the compliance matrix (Cij)?
a) [Sij] = [Cij]
b) [Sij] = [Cij]-1
c) [Sij] + [Cij] = [Kij]
d) [Sij] α [Cij]

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 S12, S11, S22 are known, then the formula for finding the compliance coefficient C11 is __________
a) C11 = S22 / (S11 * S12 – S222)
b) C11 = S11/ (S11 * S12 – S222)
c) C11 = S22 / (S11 * S22 – S122)
d) C11 = S12/ (S11 * S12 * S222)

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 υ21 if the values of E11, E22 and υ12 are given by E11 = 138GPa, E22 = 10GPa and υ12 = 0.21?
a) 0.818
b) 1.225
c) 0.9898
d) 0.0152

Explanation: Given that, E11 = 138GPa, E22 = 10GPa and υ12 = 0.21
We know that,
υ21 / υ12 = E22/E11
⇨ υ21 = (E22/E11) * υ12
⇨ υ21 = (10/138) * 0.21
⇨ υ21 = 0.0152.

7. When the lamina is reinforced with fibers in 1 (principal axis frame reference coordinate system) direction, if E1 > E2, then what is the relationship between 1Δ2 and 2Δ1?
a) 1Δ2 > 2Δ1
b) 1Δ2 < 2Δ1
c) 1Δ2 = 2Δ1
d) 1Δ2 = 2 * 2Δ1

Explanation: Because of the reciprocal relations, irrespective of the values E1 and E2, 1Δ2 = 2Δ1 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 υ23 be related to υ12 and υ21?
a) υ23 = υ32
b) υ12 = υ21
c) υ23 = υ12 * ((1 – υ21) / (1 – υ12))
d) υ32 = υ12 * ((1 – υ21) / (1 – υ12))

Explanation: Christensen, a researcher, has shown that in the case of unidirectional fiber – reinforced composites with fibers oriented in the 1-direction, υ23 can be related to υ12 and υ21 using the following equation. υ23 = υ32 = υ12 * ((1 – υ21)/ (1 – υ12)). 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 (ε11) 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 E11 as 138GPa.
a) 72.5 * 10-4
b) 725.6 *10-6
c) 7.25 * 10-4
d) 0.725 * 10-4

Explanation: Tensile load is applied parallel to the fiber direction, that is, in the 1-direction.
Therefore,
σ11 = F/A = (1000 N)/ ((100 mm) * (1 mm))
= 10 MPa and σ22 = 0.
Now, we calculate the normal strain ε11
ε11 = σ11/E11
= (10MPa)/(138GPa)
= 0.725 * 10-4.

10. What is the value of ε22, when the values of ε11 and υ12 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

Explanation: Given, ε11 = 0.725*10-4 and υ12 = 0.21
We know that,
ε22 = – υ12 * ε11
= -(0.21*0.725*10-4)
= -0.152*10-4.

Sanfoundry Global Education & Learning Series – Composite Materials.

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