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Question 1
ప్రశ్న 1
Which situation is most appropriately described as unsymmetrical bending of a homogeneous prismatic beam?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Unsymmetrical bending occurs when the applied bending moment is not confined to one centroidal principal bending axis. The moment can then be resolved into components about the two principal centroidal axes. Each component produces its corresponding linear bending-stress contribution. The total normal stress is obtained by algebraically superposing these contributions in linear-elastic bending. Therefore Option B correctly describes unsymmetrical bending.
వివరణ:
Question 2
ప్రశ్న 2
For centroidal x-y axes of a plane area, with product of inertia defined as Ixy = ∫xy dA, what condition identifies a pair of principal centroidal axes?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Principal centroidal axes are orientations for which the product of inertia vanishes. Thus Ixy equals zero when the coordinate axes coincide with principal axes. The corresponding second moments of area are called the principal moments of inertia. One principal moment is the maximum and the perpendicular principal moment is the minimum among rotated centroidal axes. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
A homogeneous cross-section has one axis of symmetry passing through its centroid. Which statement about centroidal principal axes is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reflection symmetry causes the positive and negative contributions to product of inertia about the symmetry axis and its perpendicular centroidal axis to cancel. Consequently the product of inertia for that orthogonal centroidal pair is zero. A zero product of inertia identifies principal axes. The two axes are mutually perpendicular and pass through the centroid. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
For centroidal axes x and y having second moments Ix and Iy and product of inertia Ixy = ∫xy dA, which expression gives the two principal moments of inertia?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The transformed second moment of area varies with twice the axis-rotation angle. Its maximum and minimum values occur when the transformed product of inertia becomes zero. Solving the transformation equations gives the mean value (Ix + Iy)/2 plus or minus a radius term. That radius is sqrt([ (Ix - Iy)/2 ]^2 + Ixy^2). Therefore Option A gives the correct principal moments.
వివరణ:
Question 5
ప్రశ్న 5
A cross-section has centroidal properties Ix = 8.0 × 10^6 mm^4, Iy = 2.0 × 10^6 mm^4 and Ixy = 1.5 × 10^6 mm^4, where Ixy = ∫xy dA. What are its principal moments of inertia approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The mean moment of inertia is (8 + 2)/2 = 5.0 × 10^6 mm^4. The radius term is sqrt(3^2 + 1.5^2) × 10^6 = approximately 3.354 × 10^6 mm^4. Adding this to the mean gives approximately 8.354 × 10^6 mm^4. Subtracting it gives approximately 1.646 × 10^6 mm^4. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For the section with Ix = 8.0 × 10^6 mm^4, Iy = 2.0 × 10^6 mm^4 and Ixy = 1.5 × 10^6 mm^4, let centroidal axes x'-y' be obtained by rotating x-y counterclockwise through θ, with Ixy = ∫xy dA and tan(2θ) = -2Ixy/(Ix - Iy). Which θ gives the x' axis associated with the maximum principal moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated transformation convention gives tan(2θ) = -3/6 = -0.5. The relevant solution is 2θ = approximately -26.565 degrees. Thus θ is approximately -13.28 degrees. Substitution into the transformed inertia equation gives Ix' approximately 8.354 × 10^6 mm^4, which is the larger principal moment. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
An L-shaped area is formed by a vertical rectangle occupying 0 <= x <= 20 mm, 0 <= y <= 100 mm and a horizontal extension occupying 20 <= x <= 80 mm, 0 <= y <= 20 mm. Coordinates are measured rightward and upward from the lower-left outer corner. What are the centroid coordinates and centroidal product of inertia Ixy = ∫(x-xbar)(y-ybar)dA?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The vertical rectangle has area 2000 mm^2 with centroid at (10,50) mm. The horizontal extension has area 1200 mm^2 with centroid at (50,10) mm. Area-weighted first moments give xbar = 25 mm and ybar = 35 mm. Because each component rectangle has zero local product of inertia about its own centroidal axes, the centroidal product is the sum of AΔxΔy terms, giving -1.20 × 10^6 mm^4. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For the L-shaped area described in the previous question, the centroidal properties are Ix = 2.9067 × 10^6 mm^4, Iy = 1.6267 × 10^6 mm^4 and Ixy = -1.2000 × 10^6 mm^4. Using tan(2θ) = -2Ixy/(Ix-Iy), what are the principal moments and the counterclockwise angle θ of the maximum-principal-moment x' axis from x?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The mean second moment is approximately 2.2667 × 10^6 mm^4. The transformation radius is sqrt(0.64^2 + 1.20^2) × 10^6 = 1.36 × 10^6 mm^4. Therefore the principal moments are approximately 3.6267 × 10^6 and 0.9067 × 10^6 mm^4. The stated angle equation gives tan(2θ) = 1.875, leading to θ approximately 30.96 degrees for the maximum principal moment. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
Use centroidal principal axes x and y, with x positive rightward, y positive upward and tensile longitudinal stress positive. Define positive bending moments so that the normal bending stress is σ = (Mx/Ix)y - (My/Iy)x. Which expression is therefore correct for biaxial bending about these principal axes?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because x and y are principal axes, their product of inertia is zero. The bending effect of each moment component can therefore be evaluated independently about its own principal axis. Under the sign convention explicitly stated in the question, Mx contributes Mx y/Ix. The My component contributes -My x/Iy, and linear superposition gives the total stress. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
A section has principal centroidal properties Ix = 8 × 10^6 mm^4 and Iy = 4 × 10^6 mm^4. With the convention σ = (Mx/Ix)y - (My/Iy)x, moments Mx = 16 kN·m and My = 8 kN·m act simultaneously. What is the stress at the point x = +40 mm, y = +50 mm?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Convert the moments to N·mm before using the section properties. The Mx contribution is (16 × 10^6/8 × 10^6)(50) = 100 MPa tension. The My contribution is -(8 × 10^6/4 × 10^6)(40) = -80 MPa. Their algebraic sum is +20 MPa. Therefore the point carries 20 MPa tension and Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
For the same principal-axis section with Ix = 8 × 10^6 mm^4, Iy = 4 × 10^6 mm^4, Mx = 16 kN·m and My = 8 kN·m, what is the stress at x = -40 mm, y = +50 mm using σ = (Mx/Ix)y - (My/Iy)x?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The contribution from Mx remains +100 MPa because y is still +50 mm. The My term becomes -(8/4)(-40) = +80 MPa when moments are expressed consistently in N·mm and inertias in mm^4. Both bending components therefore produce tension at this point. The total stress is 100 + 80 = 180 MPa tension. Hence Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the principal-axis section with Ix = 8 × 10^6 mm^4, Iy = 4 × 10^6 mm^4, Mx = 16 kN·m and My = 8 kN·m, what is the neutral-axis equation under the convention σ = (Mx/Ix)y - (My/Iy)x?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The neutral axis is the locus where longitudinal bending stress is zero. Here the stress equation becomes 2y - 2x = 0 when x and y are measured in the same length units. Rearranging gives y = x. The neutral axis passes through the centroid because no direct axial force is included. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
A cross-section has principal moments Ix = 12 × 10^6 mm^4 and Iy = 6 × 10^6 mm^4. A positive moment Mx = 18 kN·m acts, and a positive My is to be applied so that the stress at x = +30 mm, y = +40 mm is zero. Using σ = (Mx/Ix)y - (My/Iy)x, what My is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Mx stress contribution at the specified point is (18/12)(40) = 60 MPa. The My contribution has magnitude (My/6)(30) when My is expressed in kN·m with the common factor of 10^6 accounted for. Setting the total stress equal to zero gives 60 - 5My = 0. Thus My = 12 kN·m. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
A solid rectangular section is 100 mm wide in the x direction and 200 mm deep in the y direction, with centroidal principal axes x and y. Thus Ix = 66.667 × 10^6 mm^4 and Iy = 16.667 × 10^6 mm^4, and its four corners have coordinates x = ±50 mm, y = ±100 mm. Moments Mx = 12 kN·m and My = 4 kN·m act under σ = (Mx/Ix)y - (My/Iy)x. What is the maximum absolute bending stress among the four corners?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated dimensions give the consistent rectangular-section values Ix = bh^3/12 = 66.667 × 10^6 mm^4 and Iy = hb^3/12 = 16.667 × 10^6 mm^4. The Mx contribution at y = ±100 mm has a magnitude of 18 MPa. The My contribution at x = ±50 mm has a magnitude of 12 MPa. At the two corners where these contributions reinforce, the absolute stress is 18 + 12 = 30 MPa, while at the other two corners it is 6 MPa. Therefore the maximum absolute bending stress is 30 MPa and Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For non-principal centroidal axes x-y, define Ixy = ∫xy dA, tensile longitudinal stress as positive, and internal moments by Mx = ∫yσ dA and My = -∫xσ dA. If D = IxIy - Ixy^2, which expression gives the bending stress?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For non-principal axes, the stress distribution is still linear in the centroidal coordinates x and y. However, the nonzero product of inertia couples the effects of the two moment components. Enforcing the two resultant-moment equations gives coefficients containing the determinant D = IxIy - Ixy^2. Under the exact moment and product-of-inertia conventions stated in the question, the resulting expression is the one given in Option A. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
A section has centroidal properties Ix = 10 × 10^6 mm^4, Iy = 6 × 10^6 mm^4 and Ixy = 2 × 10^6 mm^4. Under the conventions Mx = ∫yσ dA, My = -∫xσ dA and Ixy = ∫xy dA, a moment Mx = 12 kN·m acts while My = 0. What is the stress at x = +30 mm, y = +40 mm?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The determinant is D = (10)(6) - 2^2 = 56 in units of 10^12 mm^8. With My = 0, the stress coefficients are -MxIxy/D for x and +MxIy/D for y. These evaluate to approximately -0.42857 MPa/mm for x and +1.28571 MPa/mm for y. At x = 30 mm and y = 40 mm, σ = -12.857 + 51.429 = +38.57 MPa. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the non-principal-axis section with Ix = 10 × 10^6 mm^4, Iy = 6 × 10^6 mm^4 and Ixy = 2 × 10^6 mm^4, only Mx = 12 kN·m acts under the stated convention. What is the neutral-axis equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For this loading the stress distribution is σ = -0.42857x + 1.28571y in MPa when coordinates are in millimetres. The neutral axis is obtained by setting this stress equal to zero. Thus 1.28571y = 0.42857x. Dividing gives y = x/3. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
A section has Ix = 10 × 10^6 mm^4, Iy = 6 × 10^6 mm^4 and Ixy = 2 × 10^6 mm^4. Using the convention Mx = ∫yσ dA and My = -∫xσ dA, only My = +10 kN·m acts. What is the stress at x = +20 mm, y = -30 mm?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The determinant is again D = 56 × 10^12 mm^8. With Mx = 0, the coefficient of x is -MyIx/D = approximately -1.78571 MPa/mm. The coefficient of y is +MyIxy/D = approximately +0.35714 MPa/mm. At x = 20 mm and y = -30 mm, the stress is -35.714 - 10.714 = -46.43 MPa. The negative value denotes compression, so Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A section has principal centroidal axes x-y with Ix = 9 × 10^6 mm^4 and Iy = 3 × 10^6 mm^4. A resultant bending moment of 12 kN·m lies in the x-y moment plane at 30 degrees counterclockwise from +x, so Mx = 12 cos30 degrees and My = 12 sin30 degrees. Using σ = (Mx/Ix)y - (My/Iy)x, what is the stress at x = +30 mm, y = +40 mm?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The moment components are Mx approximately 10.392 kN·m and My = 6 kN·m. The Mx contribution to stress is (10.392/9)(40) = approximately 46.19 MPa. The My contribution is -(6/3)(30) = -60 MPa. Their sum is approximately -13.81 MPa. Therefore the point is in compression and Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
For any pair of mutually perpendicular centroidal axes obtained by rotating another centroidal x-y pair through an arbitrary angle, which quantity remains invariant?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The sum Ix + Iy is the polar second moment of area about the centroidal axis perpendicular to the cross-sectional plane. Rotating the in-plane coordinate axes does not alter the radial distance of any area element from the centroid. Consequently the sum of the two perpendicular second moments remains unchanged. Individual moments and the product of inertia generally change with axis orientation. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
At a centroid of a plane area, suppose Ix = Iy and Ixy = 0 for one orthogonal pair of axes. What follows for axes obtained by any further in-plane rotation about the same centroid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
If Ix and Iy are equal and Ixy is zero, the in-plane inertia matrix is proportional to the identity matrix. Rotation therefore leaves that matrix unchanged. The transformed product of inertia remains zero for every orientation. The two transformed second moments also remain equal to their original common value. Therefore every orthogonal centroidal pair is principal and Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A homogeneous beam section is subjected to pure bending with no resultant axial force. Which statement about the neutral axis is correct within linear-elastic unsymmetrical-bending theory?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pure bending without an axial resultant produces a linear normal-stress field containing no constant uniform-stress term. Centroidal coordinates therefore satisfy zero stress at x = 0 and y = 0. Hence the neutral-axis line passes through the centroid. When the moment is not applied about a principal axis, the neutral-axis orientation generally differs from the moment-axis orientation. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
A bending moment acts about a centroidal x axis for which the product of inertia Ixy is nonzero. Which statement is generally correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A nonzero product of inertia means that the chosen centroidal axes are not principal axes. A moment applied about one such axis generally couples the two curvature components. The resulting normal-stress expression contains both x and y coordinate terms. Consequently the zero-stress line is generally inclined relative to the applied moment axis. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about unsymmetrical bending:
1. Principal centroidal axes have zero product of inertia.
2. A centroidal axis of symmetry is a principal axis.
3. In pure bending without axial force, the neutral axis passes through the centroid.
4. For bending about non-principal centroidal axes, the neutral axis must always be parallel to the resultant bending moment.
Which statements are correct?
1. Principal centroidal axes have zero product of inertia.
2. A centroidal axis of symmetry is a principal axis.
3. In pure bending without axial force, the neutral axis passes through the centroid.
4. For bending about non-principal centroidal axes, the neutral axis must always be parallel to the resultant bending moment.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because vanishing product of inertia is the defining property of principal centroidal axes. Statement 2 is correct because symmetry causes the relevant product-of-inertia contributions to cancel. Statement 3 is correct for homogeneous linear-elastic pure bending without a direct axial resultant. Statement 4 is incorrect because a non-principal bending moment generally produces a neutral axis whose orientation differs from that of the applied moment. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each unsymmetrical-bending concept with its correct description:
1. Principal centroidal axes
2. Product of inertia Ixy
3. Principal moments of inertia
4. Neutral axis in pure bending without axial force
P. Passes through the centroid but may be inclined to the applied moment axis
Q. ∫xy dA for the stated sign convention
R. Orthogonal centroidal axes for which the product of inertia is zero
S. Maximum and minimum second moments obtainable by rotation of centroidal axes
Which matching is correct?
1. Principal centroidal axes
2. Product of inertia Ixy
3. Principal moments of inertia
4. Neutral axis in pure bending without axial force
P. Passes through the centroid but may be inclined to the applied moment axis
Q. ∫xy dA for the stated sign convention
R. Orthogonal centroidal axes for which the product of inertia is zero
S. Maximum and minimum second moments obtainable by rotation of centroidal axes
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Principal centroidal axes are the orthogonal centroidal directions for which Ixy vanishes, so item 1 matches R. Under the convention stated throughout the test, product of inertia is ∫xy dA, so item 2 matches Q. The corresponding extreme second moments are the maximum and minimum principal moments, so item 3 matches S. In pure bending without axial force, the neutral axis passes through the centroid but is not generally aligned with the applied moment, so item 4 matches P. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The bending moment has components about both centroidal principal axes, so bending occurs simultaneously in two principal planes. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. Ixy = 0 సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. The symmetry axis and the centroidal axis perpendicular to it form a pair of principal axes. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. (Ix + Iy)/2 ± sqrt([ (Ix - Iy)/2 ]^2 + Ixy^2) సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. 8.354 × 10^6 mm^4 and 1.646 × 10^6 mm^4 సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. -13.28 degrees సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. xbar = 25 mm, ybar = 35 mm and Ixy = -1.20 × 10^6 mm^4 సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. Imax = 3.6267 × 10^6 mm^4, Imin = 0.9067 × 10^6 mm^4 and θ = 30.96 degrees సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. σ = (Mx/Ix)y - (My/Iy)x సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. 20 MPa tension సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 180 MPa tension సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. y = x సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: C. 12 kN·m సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 30 MPa సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. σ = [(MxIy + MyIxy)y - (MyIx + MxIxy)x]/D సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. 38.57 MPa tension సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: A. y = x/3 సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. 46.43 MPa compression సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 13.81 MPa compression సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. Ix + Iy సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. Every rotated orthogonal pair is also a principal pair and the two second moments remain equal. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. The neutral axis passes through the centroid, although its orientation need not coincide with the applied moment axis. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. The stress distribution usually contains both x and y terms, and the neutral axis need not coincide with the x axis. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. 1, 2 and 3 only సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1-R, 2-Q, 3-S, 4-P సమాధానం: A.