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 : 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 : 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
Last Answer : (d) None
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
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 : In a body under hydrostatic pressure, the case exists (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 : (b) Secondly σ 1 = σ 2 =σ 3
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 : 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 : 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 : The principal strain due to σ1(tensile) and σ2 (Compressive ) stress is (a) Firstly (b)Secondly (c)Thirdly (d) None
Last Answer : (b)Secondly
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 : 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 : In a ductile material, the strength are (a)Firstly Ultimate >yield > elastic limit (b) Secondly Ultimate > yield =elastic limit (c) Thirdly Ultimate=yield=elastic limit (d) None
Last Answer : (a)Firstly Ultimate >yield > elastic limit
Description : In a brittle material, the strength are (a) Firstly Ultimate >yield > elastic limit (b) Secondly Ultimate > yield =elastic limit (c) Thirdly Ultimate=yield=elastic limit (d) None
Last Answer : (c) Thirdly Ultimate=yield=elastic limit
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 : 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 : Theories of elastic failure establishes the (a) Firstly Reasons of failure (b) Secondly Reasons of safety (c) Both (a) & (b) (d) None
Last Answer : (c) Both (a) & (b)
Description : Theories of elastic failure is the (a) Firstly analysis of the various failures (b) Secondly analysis of the strength of a material (c) Both (a) & (b) (d) None
Description : A ductile material may not meet a failure if it has been tested for the theories of failure (a) Firstly Maximum Shear Stress Theory (b) Secondly Maximum Shear Strain Energy Theory (c) Both (a) & (b) (d) None
Description : The angle of obliquity is the angle between the a. Firstly Resultant and the shear stress b. Secondly Resultant & the normal stress c. Both (a) & (b) d. None
Last Answer : b. Secondly Resultant & the normal stress
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 : The relationship between the alternating stress and mean stress is given by the following equation: σ a =σ e [1-(σ m /σ u )x], where σ e is the fatigue limit for completely reversed loading. The value of x for Gerber line is equal to _________ a) 1 b) 2 c) 0.5 d) -1
Last Answer : b) 2
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 : 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 : Under complex loading, theories of elastic failure establishes the (a) Margin of failure (b) Margin of safety (c) Both (a) & (b) (d) None
Last Answer : (b) Margin of safety
Description : Under complex loading, theories of elastic failures ensure (a) Stability (b) Instability (c) Both stability and instability (d) None
Last Answer : (a) Stability
Description : Under complex or simple loading, strain energy is (a) External work done (b) Internal work done (c) Both internal and external work (d) None
Last Answer : (b) Internal work done
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 : The maximum number of principal stresses is a. 1 b. 3 c. 5 d. None
Last Answer : b. 3
Description : The maximum number of principal stresses is a. 2 b. 4 c. 6 d. None
Last Answer : 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 : How many maximum shear stresses are there with three principal stresses? a. 1 b. 2 c. 3 d. None
Last Answer : c. 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 : 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 : 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 : Is principal a? a. Simple stress b. Complex stress c. Bending stress d. None
Last Answer : a. Simple stress
Description : The relationship between the alternating stress and mean stress is given by the following equation: σ a = σ e [1-(σ m /σ u ) x ]. The value of x for Goodman line is equal to _________ a) 1 b) 2 c) 0.5 d) -1
Last Answer : a) 1
Description : Which of the following equations is correct for Soderberg Criteria? a. (σ m / S ut ) + (σ a / S e ) = (1 / N f ) b. (σ m / S ut ) - (σ a / S e ) = (1 / N f ) c. (σ m / S yt ) + (σ a / S e ) = (1 / N f ) d. (σ m / S ut ) - (σ a / S e ) = (1 / N f )
Last Answer : c. (σ m / S yt ) + (σ a / S e ) = (1 / N f )
Description : Which of the following conditions is true for repeated stress? 1. σ m = 0 2. σ m = σ max / 2 3. σ m = σ a 4. σ min = 0 5. σ min = - σ max 6. σ a = σ max / 2 where σ m = mean ... amplitude a. condition 2 and 3 b. condition 1, 3 and 5 c. condition 2, 4, and 6 d. condition 3,4, 5 and 6
Last Answer : c. condition 2, 4, and 6
Description : The maximum tangential stress σ t = (σ x sin 2θ)/2 is maximum if, θ is equal to ________ a. 45 o b. 90 o c. 270 o d. all of the above
Last Answer : a. 45 o
Description : 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 - ... τ cos 2θ c. [(σ x - σ y )/2 ]sin θ - τ 2 cos θ d. None of the above
Last Answer : c. [(σ x – σ y )/2 ]sin θ – τ 2 cos θ