PRINCIPAL STRAINS

MULTIPLE CHOICE QUESTIONS

(MCQ) WITH ANSWERS

MCQ are helpful in understanding Principal

Strains. It also increases the level of clarity.

Stresses are not measurable. Strains are

measured with mechanical and electrical

strain gauges. Principal stresses are found

from strains. Planes of principal strains are

same as the planes of principal stresses.

These planes are at right angles to each

other. These planes carry zero shear strain.

Mohr’ s strain circle is drawn from

complex strains and from principal strains

respectively.

 

1. Principal strain is

1. €_x

2. €_y

3. γ

4. None

ANS: (d)

2. Symbols for principal strains are

1. €₁, €₂ and €₃

2. €_x, €_y and €_z

3. Both (a) & (b)

4. None

ANS: (a)

3. Units of principal strains are

1. mm

2. m

3. Both (a) & (b)

4. None

ANS: (d)

4. Units of principal strains are

1. Radians

2. Degrees

3. No units

4. None

ANS: (c)

5. Principal strains causes

1. Deformation

2. Distortion

3. Both (a) & (b)

4. None

ANS: (a)

6. Principal strains can be found from

1. Principal stresses

2. Complex stresses

3. Both (a) &(b)

4. None

ANS: (c)

7. Principal strains are

1. Tensile strain

2. Compressive strain

3. Both (a) & (b)

4. None

ANS: (c)

8. Strains are measured with

1. Pressure gauge

2. Temperature gauge

3. Both (a) & (b)

4. None

ANS: (d)

9. Strains can be measured with

1. Mechanical strain gauges

2. Electrical strain gauges

3. Both (a) & (b)

4. None

ANS: (c)

10. Principal strains are found from

1. Mohr’s strain circle

2. Principal stresses

3. Both (a) & (b)

4. None

ANS: (c)

11. Principal strains are

1. Maximum normal strain

2. Minimum normal strain

3. Both (a) & (b)

4. None

ANS: (c)

12. The directions of principal strains are in the direction of

1. Principal stresses

2. Maximum shear stresses

3. Both (a) & (b)

4. None

ANS: (a)

13. Shear strain in the direction of principal strain is

1. Maximum

2. Minimum

3. Zero

4. None

ANS: (c)

14. The angles among the three principal strains are

1. 30⁰

2. 60⁰

3. 75⁰

4. None

ANS: (d)

15. The angles among the principal strains are

1. 90⁰

2. 120⁰

3. 180⁰

4. None

ANS: (a)

16. Principal strains from the three tensile principal stresses are

1. Firstly         €₁ = σ₁/E -µσ₂ /E – µσ₃ /E, €₂ = σ₁/E -µσ₂ /E – µσ₃ /E, €₃ = σ₃/E -µσ₁ /E – µσ₂ /E

2. Secondly     €₁ = σ₁/E -µσ₂ /E – µσ₃ /E, €₂ = σ₂ /E -µσ₃ /E – µσ₁ /E and €₃ = σ₃/E -µσ₁ /E – µσ₂ /E

3. Thirdly      €₁ = σ₁/E -µσ₂ /E – µσ₃ /E, €₂ = σ₂/E -µσ₃ /E – µσ₁ /E and €₃ = σ₃/E -µσ₁ /E – µσ₃ /E

4. None

ANS: (b)

17. Principal strains from a 2-dimensional strained element under complex strains €_x, €_y and γ are

1. Firstly          +[(€_x–€_y)² +(γ)²]⁵ and –[(€_x–€_y)² +(γ)²]^0.5

2. Secondly      +[(€_x–€_y)² +(γ)²]⁵ and –[(€_x–€_y)² +(γ/2)²]^0.5

3. Thirdly          +(1/2)[(€_x–€_y)² +(γ)²]⁵ and –(1/2)[(€_x–€_y)² +(γ/2)²]^0.5

4. None

ANS: (c)

18. As per maximum principal strain theory for a brittle material, say €₁ being the largest with σ1, σ2 and σ3 tensile and

       σ1> σ2 > σ3

1. Firstly          σ₁/E -µσ₂ /E – µσ₃ /E =σ _ult /(FOS x E)

2. Secondly      σ₁/E -µσ₂ /E – µσ₃ /E =σ _yp /(FOS x E)

3. Thirdly          σ₁/E -µσ₂ /E – µσ₃ /E =σ _{elastic limit} /(FOS x E)

4. None

ANS: (a)

19. As per maximum principal strain theory for a ductile material, say €₁ being the largest with σ1, σ2 and σ3 tensile and σ1> σ2 > σ3

1.  Firstly          σ₁/E -µσ₂ /E – µσ₃ /E =σ _ult /(FOS x E)

2. Secondly     σ₁/E -µσ₂ /E – µσ₃ /E =σ _yp /(FOS x E)

3. Thirdly         σ₁/E -µσ₂ /E – µσ₃ /E =σ _{elastic limit} /(FOS x E)

4. None

ANS: (b)

20. As per maximum principal strain energy theory for a brittle material, say €₁ being the largest with σ1, σ2 and σ3 tensile and σ₁> σ₂ > σ₃

1. Firstly        (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_ult /2E

2. Secondly    (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_yp /2E

3. Thirdly        (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_{elastic limit} /2E

4. None

ANS: (a)

21. As per maximum principal strain energy theory for a ductile material, say €₁ being the largest with σ1, σ2 and σ3 tensile and σ₁> σ₂ > σ₃

1. Firstly        (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_ult /2E

2. Secondly    (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_yp /2E

3. Thirdly        (1/2)(€₁σ₁+ €₂σ₂+ €₃σ₃) =σ²_{elastic limit} /2E

4. None

ANS: (b)

22. For a 3 dimensional complex strained body, the number of Mohr’s strain circles is

1. 3

2. 6

3. 9

4. None

ANS: (a)

23. For drawing the Mohr’s strain circle from complex strains, replace

1. Firstly              σ_x=€_x, σ_y =€_y and τ =γ

2. Secondly          σ_x=€_x, σ_y =€_y and τ =γ/2

3. Thirdly              σ_x=€_x, σ_y =€_y and τ =2γ

4. None

ANS: (b)

24. For drawing the Mohr’s strain circle from principal strains, replace

1. Firstly        σ₁=€₁, σ₂ =€₂ and τ=γ

2. Secondly    σ₁=€₁, σ₂ =€₂ and τ=γ/2

3. Thirdly        σ₁=€₁, σ₂ =€₂ and τ=0

4. None

ANS: (c)

25. Plane of maximum principal strain carries

1. Maximum shear strain

2. Zero shear strain

3. Can’t say

4. None

ANS: (b)

26. In a two dimensional strain system, the number of principal planes of strain is

1. 1

2. 2

3. 3

4. None

ANS: (b)

27. When principal strain are like and equal, the maximum shear strain is

1. γ

2. 2γ

3. γ/2

4. None

ANS: (d)

28. When principal strain are like and equal, the maximum shear strain is

1. Infinity

2. Zero

3. –infinity

4. None

ANS: (b)

29. When principal strains are like and equal, the Mohr’s strain circle is a point. Then Mohr’s point strain circle

1.  Coincides with pole

2.  Do not coincide with pole

3.   Passes though the point of maximum shear strain

4. None

ANS: (b)

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