Principal stresses are found by

a. Analytical method

b. Graphical method

c. Analytical & graphical methods

d. None

1 Answer

Answer :

c. Analytical & graphical methods

Related questions

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Last Answer : c. Principal stress

Description : Symbols for principal stresses are a. Firstly σ, τ & γ b. Secondly σ 1 , σ 2 & σ 3

Last Answer : b. Secondly σ 1 , σ 2 & σ 3

Description : Why do we determine principal stresses? a. Failure is due to simple stress or strain b. Failure is due to complex stress or strain c. Both (a) & (b) d. None

Last Answer : a. Failure is due to simple stress or strain

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Last Answer : b. Secondly Maximum and minimum normal stresses

Description : Under complex loading, principal stresses exist as (a) Firstly σ 1 > σ 2 =σ 3 (b) Secondly σ 1 = σ 2 =σ 3 (c) Thirdly σ 1 > σ 2 < σ 3 (d) None

Last Answer : (d) None

Description : The total strain energy for a unit cube subjected to three principal stresses is given by? a) U= [(σέ) 1 + (σέ) 2+ (σέ) 3]/3 b) U= [(σ12+σ22+σ32)/2E] – (σ1σ2+σ2σ3+σ3σ1)2μ c) U= [(σέ) 1 + (σέ) 2+ (σέ) 3]/4 d) None of the mentioned

Last Answer : b) U= [(σ12+σ22+σ32)/2E] – (σ1σ2+σ2σ3+σ3σ1)2μ

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Last Answer : b. 3

Description : The maximum number of principal stresses is a. 2 b. 4 c. 6 d. None

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Description : Shear strain energy under principal tensile stresses σ1 and σ2 is (a) (1/12E) (σ1 — σ2) 2 + σ2 2 — σ1 2 ) (b) (1/12G) (σ1 — σ2) 2 + σ2 2 + σ1 2 ) (c) (1/12K) (σ1 — σ2) 2 + σ2 2 + σ1 2 ) (d) None

Last Answer : (b) (1/12G) (σ1 — σ2) 2 + σ2 2 + σ1 2 )

Description : Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ1 2 + σ2 2 –μ σ1 σ2) (b) (1/2E)( σ1 2 + σ2 2 –4μ σ1 σ2) (c) (1/2E)( σ1 2 + σ2 2 –2μ σ1 σ2) (d) None

Last Answer : (c) (1/2E)( σ1 2 + σ2 2 –2μ σ1 σ2)

Description : Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ1 2 + σ2 2 –3μ σ1 σ2) (b) (1/2E)( σ1 2 + σ2 2 –4μ σ1 σ2) (c) (1/2E)( σ1 2 + σ2 2 –5μ σ1 σ2) (d) None

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Description : Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ1 2 + σ2 2 –3μ σ1 σ2) (b) (1/2E)( σ1 2 + σ2 2 –4μ σ1 σ2) (c) (1/2E)( σ1 2 + σ2 2 –5μ σ1 σ2) (d) None

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Last Answer : c. Thirdly (σ 1 –σ 2 )/2

Description : In the analysis, all the principal stresses are assumed as a. Shear stresses b. Compressive stresses c. Tensile stresses d. None

Last Answer : c. Tensile stresses

Description : How many maximum shear stresses are there with three principal stresses? a. 1 b. 2 c. 3 d. None

Last Answer : c. 3

Description : Nature of the three principal stresses is a. Firstly All tensile b. Secondly All compressive c. Thirdly All shear d. None

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Description : The order of magnitude of the principal stresses is a. Firstly σ 1 >σ 2 >σ 3 b. Secondly σ 2 >σ 3 >σ 1 c. Thirdly σ 1 >σ 3 >σ 2 d. None

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Last Answer : (a)90 0

Description : All the principal stresses are at an angle of (a) 45 0 (b) 60 0 (c) 75 0 (d) None

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Description : The magnitude of principal stresses due to complex stresses is (a) (1/2)[ (σ x + σ y ) ± ((σ x –σ y ) 2 + 4 τ 2 )) 0.5 ] (b) (1/2)[ (σx + σy) ± (1/2)((σx –σy) 2 + 4 τ 2 )) 0.5 ] (c) (1/2)[ (σx + σy) ± ((1/2)(σx –σy) 2 + 4 τ 2 )) 0.5 ]

Last Answer : (a) (1/2)[ (σ x + σ y ) ± ((σ x –σ y ) 2 + 4 τ 2 )) 0.5 ]

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