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 ]

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 ]
by

1 Answer

Answer :

(a) (1/2)[ (σ x + σ y ) ± ((σ x –σ y ) 2 + 4 τ 2 )) 0.5 ]
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Related questions

The magnitude of maximum shear stress is (a) ± (1/2)[ ((σx –σy) 2 + 4 τ 2 )) 0.5 ] (b) ± (1/2)[ (1/2)((σx –σy) 2 + 4 τ 2 )) 0.5 ] (c) ± (1/2)[ ((1/2)(σx –σy) 2 + 4 τ 2 )) 0.5 ] (d) None

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Last Answer : (a) ± (1/2)[ ((σx –σy) 2 + 4 τ 2 )) 0.5 ]

Maximum principal stress is equal to (a) (σx + σy)/2 + [ (σx –σy) 2 + τ 2 ] 0.5 (b) (σx + σy)/2 + 0.5 [ (σx –σy) 2 + τ 2 ] 0.5 (c) (σx + σy)/2 + 0.5 [ (σx –σy) 2 + 4τ 2 ] 0.5 (d) None

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Last Answer : (c) (σx + σy)/2 + 0.5 [ (σx –σy) 2 + 4τ 2 ] 0.5

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

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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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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

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The equations for principal stresses are valid only when (a)σ x and σ y are both tensile (b) σ x is compressive and σ y is tensile (c) σ x is tensile and σ y is compressive (d) None

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Which of the following formulae is used to calculate tangential stress, when a member is subjected to stress in mutually perpendicular axis and accompanied by a shear stress? a. [(σ x – σ y )/2 ]sin θ – τ cos 2θ b. [(σ x – σ y )/2 ]– τ cos 2θ c. [(σ x – σ y )/2 ]sin θ – τ 2 cos θ d. None of the above

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