Description : The maximum number of principal stresses is a. 1 b. 3 c. 5 d. None
Last Answer : b. 3
Description : Maximum shear stress in terms of principal stresses is a. Firstly (σ 1 +σ 2 )/2 b. Secondly (σ 1 /σ 2 ) c. Thirdly (σ 1 –σ 2 )/2 d. None
Last Answer : c. Thirdly (σ 1 –σ 2 )/2
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 : Principal stresses are a. Firstly Maximum and minimum shear stresses b. Secondly Maximum and minimum normal stresses c. Both (a) & (b) d. None
Last Answer : b. Secondly Maximum and minimum normal stresses
Description : 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
Last Answer : (b) Average difference of principal 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
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
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
Description : Principal stresses are found by a. Analytical method b. Graphical method c. Analytical & graphical methods d. None
Last Answer : c. Analytical & graphical methods
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 : In a body under pure shear, the magnitude and nature of the two principal stresses are a. Firstly Equals shear stress, opposite nature b. Secondly Equals shear stress, same nature c. Both (a) & (b) d. None
Last Answer : a. Firstly Equals shear stress, opposite nature
Description : Nature of the three principal stresses is a. Firstly All tensile b. Secondly All compressive c. Thirdly All shear d. None
Last Answer : a. Firstly All tensile
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
Last Answer : a. Firstly σ 1 >σ 2 >σ 3
Description : All the principal stresses are at an angle of (a)90 0 (b) 45 0 (c) 135 0 (d) None
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
Description : 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
Last Answer : (a)σ x and σ y are both tensile
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 ]
Description : There are in all (a) Two principal stresses (b) Three principal stresses (c) Four principal stresses (d) None
Last Answer : (b) Three principal stresses
Description : Which of the following stresses can be determined using Mohr's circle method? a. Torsional stress b. Bending stress c. Principal stress d. All of the above
Last Answer : c. Principal stress
Description : Total number of maximum shear stresses is (a) One (b) Three (c) Five (d) None
Last Answer : (b) Three
Description : The region of safety for biaxial stresses is of which shape in the case of maximum distortion energy theorem. a) Ellipse b) Circle c) Rectangle d) Square
Last Answer : a) Ellipse
Description : All the maximum shear stresses are at an angle of (a)45 0 (b) 90 0 (c) 135 0 (d) None
Last Answer : (b) 90 0
Description : On the planes of maximum shear, there are (a) Normal stresses (b) Bending stresses (c) Bucking stresses (d) None
Last Answer : (a) Normal stresses
Description : How many angles of obliquity are there for a cuboidal body under complex stresses? a. 6 b. 8 c. 4 d. None
Last Answer : a. 6
Description : A transmission shaft subjected to pure bending moment should be designed on the basis of (A) Maximum principal stress theory (B) Maximum shear stress theory (C) Distortion energy theory (D) Goodman or Soderberg diagrams
Last Answer : (A) Maximum principal stress theory
Description : For a homogeneous & isotropic body under hydrostatic pressure, which theory of elastic failure does not fail (a) Firstly Maximum Principal Theory (b) Secondly Maximum Shear Stress Theory (c) Thirdly Maximum Principal Energy Theory (d) None
Last Answer : (a) Firstly Maximum Principal Theory
Description : For a homogeneous & isotropic body under hydrostatic pressure, which theory of elastic failure fails (a) Firstly Maximum Principal Theory (b) Secondly Maximum Principal strain Theory (c) Thirdly Maximum Principal Energy Theory (d) None
Last Answer : (c) Thirdly Maximum Principal Energy Theory
Description : 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
Last Answer : (c) Thirdly Maximum Principal stress theory
Description : 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 shear stress theory (d) None
Description : A ductile material may not meet a failure if it has been tested for the theories of failure (a) Firstly Maximum Principal Theory (b) Secondly Maximum Principal Strain Theory (c) Thirdly Maximum principal strain energy theory (d) None
Description : Maximum principal strain is equal to when σ1 and σ2 are tensile (a) (σ1 –μσ2)/E (b) (σ1 + μσ2)/E (c) (–σ1 –μσ2)/E (d) None
Last Answer : (a) (σ1 –μσ2)/E
Description : Maximum principal strain theory is also called as (a) Guest’s theory (b) Haigh theory (c) St.Venant’s theory (d) None
Last Answer : (c) St.Venant’s theory
Description : Maximum principal strain theory is applicable to (a) Ductile materials (b) Brittle materials (c) Composite materials (d) None
Last Answer : (b) Brittle materials
Description : 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
Last Answer : (c) (σx + σy)/2 + 0.5 [ (σx –σy) 2 + 4τ 2 ] 0.5
Description : Maximum principal theory is also known as (a) Beltrami Theory (b) Maximum normal stress theory (c) Saint Venant’s theory (d) None
Last Answer : (b) Maximum normal stress theory
Description : Maximum principal theory is also known as (a) Guest Theory (b) Beltrami Theory (c) Rankine Theory (d) None
Last Answer : (c) Rankine Theory
Description : Under maximum principal stress theory, maximum principal stress is equal to (a) Allowable stress in tension (b) Allowable stress in compression (c) Allowable stress in shear (d) None
Last Answer : (a) Allowable stress in tension
Description : Maximum principal stress theory is applicable to (a) Ductile materials (b) Brittle materials (c) Composite materials (d) None
Description : In a general two dimensional stress system, planes of maximum shear stress are inclined at ___ with principal planes. a. 90 degree b. 180 degree c. 45 degree d. 60 degree
Last Answer : c. 45 degree
Description : Principal planes are those planes on which a. Normal stress is maximum b. Normal stress is minimum c. Normal stress is either maximum or minimum d. Shear stress is maximum
Last Answer : c. Normal stress is either maximum or minimum
Description : Maximum Principal Stress Theory is not good for brittle materials. a) True b) False
Last Answer : b) False
Description : If compressive yield stress and tensile yield stress are equivalent, then region of safety from maximum principal stress theory is of which shape? a) Rectangle b) Square c) Circle d) Ellipse
Last Answer : b) Square
Description : The magnitude of maximum principal stress is a. Firstly (σ x +σ y )/2+ (1/2)( σ x +σ y ) +4τ 2 ) 5 b. Secondly (σ x +σ y )/2+ (1/2)( σ x -σ y ) 2 +4τ 2 ) 5 c. Thirdly (σ x +σ y )/2+ (1/2)( σ x +σ y ) 2 +4τ 2 ) 5 d. None
Last Answer : b. Secondly (σ x +σ y )/2+ (1/2)( σ x -σ y ) 2 +4τ 2 ) 5
Description : Which is the maximum principal stress? a. Firstly σ 2 b. Secondly σ 3 c. Thirdly σ 1 d. None
Last Answer : c. Thirdly σ 1