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Question 1
ప్రశ్న 1
In a two-dimensional state of stress, what is the defining characteristic of a principal plane?
Explanation:
A principal plane is a plane on which the shear-stress component is zero. The normal stress acting on such a plane is called a principal stress. In a general plane-stress state, the two principal stresses are the maximum and minimum normal stresses obtainable by rotating the plane orientation. The corresponding principal planes are mutually perpendicular in physical space. Therefore Option B correctly gives the defining property of a principal plane.
వివరణ:
Question 2
ప్రశ్న 2
For a plane-stress state with tensile normal stress taken as positive, which expression gives the two in-plane principal stresses?
Explanation:
The mean normal stress in plane stress is (σx + σy)/2. The radius of the corresponding stress-transformation circle is √[((σx - σy)/2)^2 + τxy^2]. The principal stresses are obtained by adding and subtracting this radius from the mean stress. This analytical expression follows directly from setting the transformed shear stress equal to zero. Therefore Option C is the correct principal-stress relationship.
వివరణ:
Question 3
ప్రశ్న 3
At a point, σx = 70 MPa tension, σy = 30 MPa tension and τxy = 0. What are the in-plane principal stresses and principal directions?
Explanation:
There is no shear stress on the original x- and y-faces. Therefore those faces already correspond to principal planes. The normal stresses acting on them are 70 MPa and 30 MPa, so these are the two in-plane principal stresses. No rotation is required to eliminate shear because it is already zero. Hence Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A point is subjected to a uniaxial tensile stress of 120 MPa, with all other in-plane stress components zero. What are the in-plane principal stresses and the maximum in-plane shear stress?
Explanation:
For uniaxial tension, the original stress components may be taken as σx = 120 MPa, σy = 0 and τxy = 0. These directions are already principal because the shear stress is zero. Thus the in-plane principal stresses are 120 MPa and 0 MPa. The maximum in-plane shear stress is one-half of their difference, giving (120 - 0)/2 = 60 MPa. Therefore Option D is correct.
వివరణ:
Question 5
ప్రశ్న 5
A plane-stress state has σx = 60 MPa, σy = 60 MPa and τxy = 25 MPa. What are the principal stresses and maximum in-plane shear stress?
Explanation:
The mean normal stress is (60 + 60)/2 = 60 MPa. Since the two normal stresses are equal, the radius of the in-plane stress-transformation circle is simply |τxy| = 25 MPa. The principal stresses are therefore 60 + 25 = 85 MPa and 60 - 25 = 35 MPa. The maximum in-plane shear stress equals the radius, which is 25 MPa. Hence Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
At a point, σx = 80 MPa, σy = 20 MPa and τxy = +30 MPa. What are the in-plane principal stresses approximately?
Explanation:
The mean stress is (80 + 20)/2 = 50 MPa. Half the normal-stress difference is (80 - 20)/2 = 30 MPa. The radius is √(30² + 30²) = 42.43 MPa. Therefore the principal stresses are 50 ± 42.43 MPa, giving approximately 92.43 MPa and 7.57 MPa. Hence Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A plane-stress state has σx = 100 MPa, σy = -20 MPa and τxy = 40 MPa. What are the principal stresses and maximum in-plane shear stress approximately?
Explanation:
The mean normal stress is (100 - 20)/2 = 40 MPa. Half the normal-stress difference is [100 - (-20)]/2 = 60 MPa. The radius is √(60² + 40²) = approximately 72.11 MPa. The principal stresses are therefore 40 ± 72.11 MPa, giving 112.11 MPa and -32.11 MPa, and the maximum in-plane shear stress is 72.11 MPa. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
Use the convention that positive τxy acts in the +y direction on the +x face, and let θ be the counterclockwise rotation of the x'-axis from the x-axis. For σx = 80 MPa, σy = 20 MPa and τxy = +30 MPa, what are the smallest positive orientations of the x'-axis for the major principal stress and for one maximum-magnitude in-plane shear stress, respectively?
Explanation:
The principal-plane orientation satisfies tan 2θp = 2τxy/(σx - σy). Substitution gives tan 2θp = 60/60 = 1, so the smallest positive angle associated with the major principal stress is θp = 22.5°. Planes of maximum in-plane shear are rotated by 45° from the principal directions. Therefore the corresponding smallest positive orientation is 22.5° + 45° = 67.5°. Hence Option D is correct.
వివరణ:
Question 9
ప్రశ్న 9
Use the stress-transformation equations with positive τxy acting in the +y direction on the +x face. For σx = 80 MPa, σy = 20 MPa and τxy = +30 MPa, the x'-axis is rotated 30° counterclockwise from the x-axis. What are σx' and τx'y' approximately?
Explanation:
The transformed normal stress is σx' = (σx + σy)/2 + [(σx - σy)/2]cos2θ + τxy sin2θ. With θ = 30°, cos60° = 0.5 and sin60° = 0.8660, giving σx' ≈ 90.98 MPa. The transformed shear is τx'y' = -[(σx - σy)/2]sin2θ + τxy cos2θ. Substitution gives approximately -10.98 MPa. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which quantity remains invariant when the coordinate axes are rotated in a two-dimensional stress state?
Explanation:
The sum of the two in-plane normal stresses is invariant under rotation of the coordinate axes. It is equal to the sum of the two in-plane principal stresses, σ1 + σ2. Individual transformed components such as σx' and τx'y' generally change with orientation. The numerical angle of a principal direction relative to an arbitrary coordinate axis also changes when the reference axes change. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
For plane stress, which expression equals the product of the two in-plane principal stresses?
Explanation:
The determinant of the two-dimensional stress tensor is invariant under coordinate rotation. For the stress matrix, this determinant is σxσy - τxy². In principal coordinates the shear terms vanish, so the determinant becomes σ1σ2. Therefore σ1σ2 = σxσy - τxy². Hence Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A material point is in a state of pure in-plane shear with τxy = 50 MPa and σx = σy = 0. What are the principal stresses and their orientation relative to the original axes?
Explanation:
For pure shear, the mean normal stress is zero. The radius of the stress-transformation circle is equal to the shear-stress magnitude of 50 MPa. Therefore the principal stresses are +50 MPa and -50 MPa. Setting the transformed shear equal to zero gives principal directions rotated 45° from the original axes. Hence Option D is correct.
వివరణ:
Question 13
ప్రశ్న 13
At a point, σx = σy = 60 MPa and τxy = 0. Which statement is correct regarding in-plane stress transformation?
Explanation:
The in-plane stress tensor is 60 MPa times the identity matrix. Rotating the coordinate axes therefore does not change its components. Every in-plane face has the same 60 MPa normal stress and zero shear stress. The two in-plane principal stresses are equal, so the principal directions are not unique. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a general plane-stress state, what normal stress acts on the planes of maximum in-plane shear stress?
Explanation:
The maximum in-plane shear planes correspond to the top and bottom points of the stress-transformation circle. Their horizontal coordinate is the centre of that circle. The centre is the mean normal stress, (σx + σy)/2. Equivalently, it is also (σ1 + σ2)/2. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
The mean in-plane normal stress at a point is 35 MPa and the maximum in-plane shear stress is 25 MPa. What are the two in-plane principal stresses?
Explanation:
The mean normal stress is the centre of the in-plane stress-transformation circle. The maximum in-plane shear stress is its radius. Therefore the principal stresses are obtained as 35 ± 25 MPa. This gives 60 MPa and 10 MPa. Hence Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
The in-plane principal stresses at a point are 120 MPa and 40 MPa. What are the mean normal stress and maximum in-plane shear stress?
Explanation:
The mean in-plane normal stress is the average of the two principal stresses. Thus the mean is (120 + 40)/2 = 80 MPa. The maximum in-plane shear stress is one-half of their difference. Hence τmax = (120 - 40)/2 = 40 MPa. Therefore Option D is correct.
వివరణ:
Question 17
ప్రశ్న 17
Use the convention that positive τxy acts in the +y direction on the +x face. A stress state has σx = 60 MPa, σy = -20 MPa and τxy = 0. If the x'-axis is rotated 45° counterclockwise from x, what are σx' and τx'y'?
Explanation:
The mean stress is (60 - 20)/2 = 20 MPa. The half-difference is [60 - (-20)]/2 = 40 MPa. At θ = 45°, cos2θ = cos90° = 0 and sin2θ = 1. Thus σx' = 20 MPa and τx'y' = -40 MPa under the stated sign convention. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Consider the following statements about principal stresses in plane stress:
1. Shear stress is zero on a principal plane.
2. Principal stresses are extreme values of normal stress obtainable by rotating the plane orientation.
3. Planes of maximum in-plane shear coincide with the principal planes.
4. The two principal directions in plane stress are perpendicular when the principal stresses are distinct.
Which statements are correct?
1. Shear stress is zero on a principal plane.
2. Principal stresses are extreme values of normal stress obtainable by rotating the plane orientation.
3. Planes of maximum in-plane shear coincide with the principal planes.
4. The two principal directions in plane stress are perpendicular when the principal stresses are distinct.
Which statements are correct?
Explanation:
Statement 1 is correct because principal planes are defined by zero shear stress. Statement 2 is correct because the principal stresses are the maximum and minimum transformed normal stresses. Statement 3 is incorrect because maximum in-plane shear planes are rotated 45° from the principal directions, not coincident with them. Statement 4 is correct for distinct principal stresses because the principal directions of the symmetric stress tensor are orthogonal. Therefore Statements 1, 2 and 4 are correct and Option C is the unique answer.
వివరణ:
Question 19
ప్రశ్న 19
Match the plane-stress quantity with its analytical expression:
1. Sum of principal stresses
2. Product of principal stresses
3. Mean normal stress
4. Maximum in-plane shear stress
P. σx + σy
Q. σxσy - τxy^2
R. (σx + σy)/2
S. √[((σx - σy)/2)^2 + τxy^2]
Which matching is correct?
1. Sum of principal stresses
2. Product of principal stresses
3. Mean normal stress
4. Maximum in-plane shear stress
P. σx + σy
Q. σxσy - τxy^2
R. (σx + σy)/2
S. √[((σx - σy)/2)^2 + τxy^2]
Which matching is correct?
Explanation:
The sum of the principal stresses equals the trace σx + σy, so item 1 matches P. Their product equals the determinant σxσy - τxy², so item 2 matches Q. The mean normal stress is (σx + σy)/2, making item 3 match R. The maximum in-plane shear stress is the radius √[((σx - σy)/2)^2 + τxy²], so item 4 matches S. Therefore Option A gives the correct matching.
వివరణ:
Question 20
ప్రశ్న 20
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. Which statement about the principal stresses is correct?
Explanation:
The mean stress is (80 + 20)/2 = 50 MPa. The stress-circle radius is √(30² + 40²) = 50 MPa. Therefore the two principal stresses are 50 + 50 = 100 MPa and 50 - 50 = 0 MPa. As an independent check, σxσy - τxy² = 80 × 20 - 40² = 0, so the product of the principal stresses is zero. Hence Option D is correct.
వివరణ:
Question 21
ప్రశ్న 21
At a point, σx = 50 MPa, σy = 10 MPa and the larger principal stress is 70 MPa. What is the magnitude of τxy?
Explanation:
The mean normal stress is (50 + 10)/2 = 30 MPa. Since the larger principal stress is 70 MPa, the radius of the stress-transformation circle must be 70 - 30 = 40 MPa. Half the difference of the original normal stresses is (50 - 10)/2 = 20 MPa. Thus 40² = 20² + τxy², giving |τxy| = √1200 = approximately 34.64 MPa. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
Use the convention that positive τxy acts in the +y direction on the +x face. If σx = σy = 40 MPa and τxy = +20 MPa, which orientation statement is correct?
Explanation:
When σx = σy, the denominator in tan2θp = 2τxy/(σx - σy) is zero. For positive τxy, the major principal orientation under the stated convention is θp = 45°. The original x- and y-faces carry shear of magnitude 20 MPa, which is already the maximum in-plane shear because the normal-stress difference is zero. The principal stresses in this case are 60 MPa and 20 MPa. Therefore Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
For a plane-stress state, rotate the coordinate axes by exactly 90° counterclockwise. Under the standard tensor transformation convention, how do the stress components change?
Explanation:
Substituting θ = 90° into the plane-stress transformation equations gives cos180° = -1 and sin180° = 0. The two normal stresses therefore exchange positions, so σx' = σy and σy' = σx. The transformed shear becomes τx'y' = τxy cos180° = -τxy. This sign reversal is consistent with changing the orientation of the coordinate faces by 90°. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
A plane-stress state has in-plane principal stresses of 100 MPa and 40 MPa. Since plane stress also has the third principal stress σ3 = 0, which statement correctly distinguishes maximum in-plane shear stress from the absolute maximum shear stress at the point?
Explanation:
The maximum in-plane shear stress is one-half of the difference between the two in-plane principal stresses. Thus it equals (100 - 40)/2 = 30 MPa. In plane stress, the out-of-plane principal stress is zero, so the complete principal set is 100 MPa, 40 MPa and 0 MPa. The absolute maximum shear stress is one-half of the difference between the largest and smallest of all three principal stresses, giving (100 - 0)/2 = 50 MPa. Therefore Option D is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about analytical plane-stress transformation:
1. The mean in-plane normal stress is invariant under rotation of axes.
2. The maximum in-plane shear stress equals one-half of the difference between the two in-plane principal stresses.
3. A plane of maximum in-plane shear must always be inclined exactly 45° to the original x-axis.
4. Principal-plane orientation equations involve 2θ because the transformed stress components contain sin2θ and cos2θ terms.
Which statements are correct?
1. The mean in-plane normal stress is invariant under rotation of axes.
2. The maximum in-plane shear stress equals one-half of the difference between the two in-plane principal stresses.
3. A plane of maximum in-plane shear must always be inclined exactly 45° to the original x-axis.
4. Principal-plane orientation equations involve 2θ because the transformed stress components contain sin2θ and cos2θ terms.
Which statements are correct?
Explanation:
Statement 1 is correct because the average of the two transformed normal stresses is invariant. Statement 2 is correct because the radius of the in-plane stress-transformation circle is (σ1 - σ2)/2. Statement 3 is incorrect because maximum-shear planes are 45° from the principal planes, not necessarily 45° from the original x-axis. Statement 4 is correct because the plane-stress transformation equations depend on double-angle trigonometric terms. Therefore Statements 1, 2 and 4 are correct and Option C is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The shear stress is zero on the plane, and the normal stress is a principal stress. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. σ1,2 = (σx + σy)/2 ± √[((σx - σy)/2)^2 + τxy^2] సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: A. 70 MPa and 30 MPa, with the x- and y-directions already principal సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. σ1 = 120 MPa, σ2 = 0, τmax = 60 MPa సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: B. σ1 = 85 MPa, σ2 = 35 MPa, τmax = 25 MPa సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 92.43 MPa and 7.57 MPa సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. σ1 = 112.11 MPa, σ2 = -32.11 MPa, τmax = 72.11 MPa సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: D. 22.5° and 67.5° సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: B. σx' = 90.98 MPa and τx'y' = -10.98 MPa సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. The sum σx + σy సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. σxσy - τxy^2 సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. Principal stresses are +50 MPa and -50 MPa, on directions rotated 45° from the original axes. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. Every in-plane orientation has normal stress 60 MPa and zero shear stress. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. The average normal stress (σx + σy)/2 సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. 60 MPa and 10 MPa సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. Mean = 80 MPa, τmax = 40 MPa సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. σx' = 20 MPa and τx'y' = -40 MPa సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 1, 2 and 4 only సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. 1-P, 2-Q, 3-R, 4-S సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: D. The principal stresses are 100 MPa and 0 MPa. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: C. 34.64 MPa సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: B. The major principal direction is at 45° counterclockwise from x, while the original x- and y-faces are planes of maximum-magnitude in-plane shear. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: A. σx' = σy, σy' = σx and τx'y' = -τxy సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: D. Maximum in-plane shear is 30 MPa, while the absolute three-dimensional maximum shear stress is 50 MPa. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: C. 1, 2 and 4 only సమాధానం: C.
