Maximum shear stress is (a) Average sum of principal stresses (b) Average difference of principal stresses (c) Average sum as well as difference of principal stresses (d) None

Maximum shear stress is

(a) Average sum of principal stresses

(b) Average difference of principal stresses

(c) Average sum as well as difference of principal stresses

(d) None
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1 Answer

Answer :

(b) Average difference of principal stresses
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Related questions

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

There are in all (a) Two principal stresses (b) Three principal stresses (c) Four principal stresses (d) None

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Torsional shear stresses are induced in the spring wire when (A) spring is under compression (B) spring is under tension (C) both (A) and (B) (D) none of the above

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The helical spring ad wire of helical torsion spring, both are subjected to torsional shear stresses. a) True b) False

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Transverse fillet welds are under (i) Bending and shear stresses (ii)Compressive and torsion shear stresses (iii)Tensile and compressive stresses (iv)None

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Parallel fillet welds are under  Shear and bending stresses  Compressive and torsion shear stresses  Tensile and compressive stresses  None

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Under complex loading, if elastic limit reaches in tension, then failure occurs due to (a) Firstly Maximum principal strain theory (b) Secondly Maximum principal theory of strain energy (c) Thirdly Maximum Principal stress theory (d) None

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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