Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which statement best describes Kani's method of structural analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Kani's method is commonly called the rotation-contribution method. It is a displacement-based iterative procedure for statically indeterminate beams and rigid frames. Joint rotations are accounted for indirectly through successive rotation contributions rather than by solving all slope-deflection equations simultaneously. Member stiffness and fixed-end moments are essential parts of the procedure. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which comparison among slope-deflection, moment-distribution and Kani's methods is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Slope-deflection expresses member-end moments directly in terms of joint displacements and usually leads to simultaneous equations. Moment distribution avoids solving those equations directly by distributing unbalanced joint moments and carrying part of each correction to adjacent ends. Kani's method is another displacement-based iterative formulation that works with rotation contributions. All three methods depend on member stiffness and compatibility. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a non-sway structure, if member ij has rotational stiffness Kij at joint i, what is its Kani rotation factor μij?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Kani rotation factor allocates the effect of joint rotation among the members meeting at that joint. It is proportional to the relative rotational stiffness of the member. The standard factor is minus one-half times the member stiffness divided by the sum of joint stiffnesses. Thus μij = -(1/2)(Kij/ΣKi). Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a rigid joint whose connected members all participate in rotation, what is the algebraic sum of their Kani rotation factors?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each rotation factor equals minus one-half times the corresponding stiffness fraction. The stiffness fractions of all participating members at a joint sum to one. Their rotation factors must therefore sum to minus one-half. This property provides a useful numerical check when rotation factors are calculated. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
If DFij is the ordinary moment-distribution factor of member ij at a joint, what is the corresponding Kani rotation factor μij for the same stiffness model?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A moment-distribution factor is Kij divided by the sum of stiffnesses at the joint. The Kani rotation factor uses the same stiffness ratio. It additionally contains the multiplier minus one-half. Hence μij is equal to minus one-half of DFij. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
For a prismatic member, what near-end rotational stiffnesses are used when the far end is respectively fixed and actually hinged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A unit rotation at the near end of a prismatic member with its far end fixed requires a moment corresponding to 4EI/L. If the far end is an actual hinge, it is free to rotate and must remain moment-free. This release reduces the effective near-end rotational stiffness to 3EI/L. Equivalently, when relative stiffness I/L is used in Kani tables, the hinged-far-end value is three-fourths of the fixed-far-end value. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
At joint B, members BA and BC have the same EI, lengths 4 m and 6 m respectively, and their far ends are fixed for stiffness evaluation. What are the Kani rotation factors μBA and μBC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The rotational stiffnesses are proportional to EI/L because the common factor 4 cancels in their ratio. Thus the relative stiffnesses are 1/4 and 1/6. Their normalized stiffness fractions are 0.60 and 0.40 respectively. Multiplying these fractions by minus one-half gives -0.30 and -0.20. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
Three members BA, BC and BD meet at joint B. Their respective EI/L values are 2EI/5, EI/4 and 1.5EI/6. What are their Kani rotation factors approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The relative stiffness quantities are 0.40EI, 0.25EI and 0.25EI. Their sum is 0.90EI. The corresponding stiffness fractions are 0.4444, 0.2778 and 0.2778. Multiplying each by minus one-half gives approximately -0.2222, -0.1389 and -0.1389. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
For this question, take clockwise member-end moments as positive. A prismatic fixed-ended member AB of length L carries a downward UDL w over its full span. What are its fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A full-span UDL produces hogging fixed-end moments of magnitude wL²/12. Under a common clockwise-positive member-end convention, the left-end hogging moment is negative. The right-end hogging moment is positive. Their algebraic values are therefore -wL²/12 and +wL²/12. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
Using clockwise member-end moments as positive, what are the fixed-end moments of a prismatic fixed-ended member AB carrying a downward point load P at midspan?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A central point load on a fixed-ended prismatic member produces hogging fixed-end moments of magnitude PL/8. The left-end moment is counterclockwise physically and is therefore negative under the stated clockwise-positive convention. The right-end hogging moment is clockwise and is positive. Hence the algebraic pair is -PL/8 and +PL/8. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
In a non-sway Kani analysis, what is meant by the restrained moment at a rotatable joint before rotation contributions are applied?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The initial member-load effects are represented by fixed-end moments while the joints are regarded as restrained. At a particular joint, the fixed-end moments of all connected members are added algebraically. This joint sum is commonly called the restrained moment in the Kani procedure. It forms the starting quantity used to evaluate the first rotation contributions. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
At a joint, two connected members have rotation factors -0.30 and -0.20. The restrained joint moment is +50 kN·m, and all far-end rotation contributions are taken as zero in the first cycle. What are the first-cycle near-end rotation contributions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In the first approximation, the unknown far-end rotation contributions may be taken as zero. The quantity multiplied by each rotation factor is therefore the restrained moment of +50 kN·m. The two contributions are -0.30(50) = -15 kN·m and -0.20(50) = -10 kN·m. Their sum is -25 kN·m, which equals minus one-half of the restrained moment. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
At the same joint as in the previous question, suppose that in a later cycle the algebraic sum of the far-end rotation contributions entering the joint is -6 kN·m. What are the updated near-end rotation contributions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The updated quantity inside the Kani iteration is the restrained joint moment plus the relevant far-end rotation contributions. Here it equals 50 - 6 = 44 kN·m. Multiplication by -0.30 gives -13.2 kN·m. Multiplication by -0.20 gives -8.8 kN·m. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a non-sway member ij after Kani rotation contributions have converged, which expression gives its final member-end moment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Kani's rotation contribution M'ij represents one-half of the end-moment change associated with rotation of joint i. The final near-end effect therefore contains twice the near-end rotation contribution. Rotation at the opposite joint also influences end i and enters through the far-end contribution M'ji. The fixed-end moment caused by member loading must also be retained. Thus Mij = MijF + 2M'ij + M'ji, making Option B correct.
వివరణ:
Question 15
ప్రశ్న 15
For a member ij, the fixed-end moment MijF is -30 kN·m, the converged near-end rotation contribution M'ij is +8 kN·m, and the far-end rotation contribution M'ji is +3 kN·m. What is the final end moment Mij?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The non-sway Kani final-moment expression is Mij = MijF + 2M'ij + M'ji. Substitution gives Mij = -30 + 2(8) + 3. This equals -30 + 16 + 3 = -11 kN·m. The doubling of the near-end contribution is essential and must not be omitted. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Which statement concerning fixed or hinged ends in Kani's method is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A fixed support has zero joint rotation, so no rotation contribution is generated at that fixed joint. It can nevertheless develop a nonzero final support moment. An actual hinge is different because the bending moment at that member end must be zero. The member stiffness used at the opposite end must also reflect the hinged far-end condition. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
For Questions 17 to 23, clockwise member-end moments are positive. A two-span continuous beam ABC has fixed supports at A and C and an intermediate vertical support at B, where vertical translation is restrained but rotation is unrestricted. Spans AB and BC are each 6 m long and have the same EI. Only AB carries a UDL of 12 kN/m. What are the Kani rotation factors at joint B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Both members have the same EI and the same length. Their rotational stiffnesses at B are therefore equal because their far ends A and C are fixed. Each member consequently has one-half of the total joint stiffness. The Kani factor is minus one-half of that stiffness fraction. Hence each rotation factor is -0.25 and Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
For the continuous beam in the previous question, the fixed-end moments on AB are MABF = -36 kN·m and MBAF = +36 kN·m. What are the converged rotation contributions at B and the final member-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The restrained moment at joint B is +36 kN·m. Both rotation factors are -0.25, so each B-end rotation contribution is -9 kN·m. Ends A and C are fixed, so their rotation contributions are zero. Applying Mij = MijF + 2M'ij + M'ji gives -45, +18, -18 and -9 kN·m in the stated order. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
For the same two-span continuous beam with fixed supports at A and C and an intermediate vertical support at B, what are the vertical support reactions RA, RB and RC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Span AB carries a total UDL of 72 kN and has final end moments -45 and +18 kN·m under the stated clockwise-positive convention. Equilibrium of span AB gives vertical contributions of 40.5 kN at A and 31.5 kN at B. The unloaded span BC has end moments -18 and -9 kN·m, producing +4.5 kN at B and -4.5 kN at C. Combining the two contributions at B gives 36.0 kN. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A three-span continuous beam ABCD has equal 4 m spans and constant EI. Supports A and D are fixed, while B and C are intermediate vertical supports that restrain vertical translation but permit joint rotation. Only span BC carries a UDL of 12 kN/m, giving fixed-end moments MBCF = -16 and MCBF = +16 kN·m under the clockwise-positive convention. If a Kani cycle is started at B and then proceeds to C, what are the first contributions at B and the subsequently calculated first contributions at C?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each rotatable intermediate support joint connects two members of equal rotational stiffness, so every rotation factor is -0.25. At B the restrained moment is -16 kN·m, giving first contributions of +4 kN·m to BA and BC. When C is then treated, its restrained moment is +16 kN·m and it already receives the +4 kN·m far-end contribution from BC. The quantity used at C is therefore +20 kN·m, giving -5 kN·m to each connected member. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
For the three-span continuous beam in the previous question, the rotation contributions converge to +16/3 kN·m at each B end and -16/3 kN·m at each C end. What are the final member-end moments MAB, MBA, MBC, MCB, MCD and MDC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The fixed supports A and D have zero rotation contributions. At B, both near-end contributions are +16/3 kN·m, while at C both are -16/3 kN·m. The final-moment relation is applied separately to each member end using the appropriate fixed-end moment and the rotation contribution from the opposite end of that member. This gives approximately +5.33, +10.67, -10.67, +10.67, -10.67 and -5.33 kN·m. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A symmetric non-sway portal frame has fixed bases A and D, columns AB and DC each 4 m high, beam BC 6 m long, and the same EI for every member. What are the Kani rotation factors at joint B for column BA and beam BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The relative stiffness of column BA is proportional to I/4. The relative stiffness of beam BC is proportional to I/6. Their stiffness fractions at B are therefore 0.60 and 0.40. Multiplication by minus one-half gives rotation factors -0.30 and -0.20. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
The portal frame in the previous question carries a UDL of 12 kN/m over the full 6 m beam, and sidesway is zero. Using clockwise member-end moments as positive, which set of converged rotation contributions and final member-end moments is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The beam fixed-end moments are -36 kN·m at B and +36 kN·m at C under the stated convention. Successive Kani cycles converge to +13.5 and +9 kN·m at joint B and -9 and -13.5 kN·m at joint C. Substitution into the final-moment expression produces the listed column and beam end moments. The beam moments are -27 and +27 kN·m, while the column top moments balance them at the rigid joints. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about sway-frame analysis by Kani's method:
1. Rotation contributions alone are sufficient for every sway frame without any additional translation treatment.
2. Linear displacement contributions are introduced for columns when storey sway is present.
3. In the conventional equal-storey-height formulation, a linear displacement factor for a column is -(3/2)(k/Σk), where k is that column's relative sway stiffness quantity.
4. For two equal columns in the same storey under that formulation, each linear displacement factor is -0.75.
Which statements are correct?
1. Rotation contributions alone are sufficient for every sway frame without any additional translation treatment.
2. Linear displacement contributions are introduced for columns when storey sway is present.
3. In the conventional equal-storey-height formulation, a linear displacement factor for a column is -(3/2)(k/Σk), where k is that column's relative sway stiffness quantity.
4. For two equal columns in the same storey under that formulation, each linear displacement factor is -0.75.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is incorrect because sidesway introduces a translational effect that cannot generally be represented by joint-rotation contributions alone. Statement 2 is correct because the sway form of Kani's method includes linear displacement contributions for the columns. Statement 3 correctly states the conventional linear displacement factor for columns of a common storey under the stated formulation. For two equal columns, k/Σk equals one-half, so the factor becomes -(3/2)(1/2) = -0.75, making Statement 4 correct. Therefore Statements 2, 3 and 4 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each Kani-method concept with its correct expression or description:
1. Rotation factor of member ij
2. Sum of rotation factors at a rigid joint
3. Final non-sway member-end moment Mij
4. Main iterative quantity emphasized by Kani's method
P. -1/2
Q. Rotation contribution
R. -(1/2)(Kij/ΣKi)
S. MijF + 2M'ij + M'ji
Which matching is correct?
1. Rotation factor of member ij
2. Sum of rotation factors at a rigid joint
3. Final non-sway member-end moment Mij
4. Main iterative quantity emphasized by Kani's method
P. -1/2
Q. Rotation contribution
R. -(1/2)(Kij/ΣKi)
S. MijF + 2M'ij + M'ji
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The rotation factor is minus one-half of the member's stiffness fraction, so item 1 matches R. The sum of all rotation factors at one rigid joint is minus one-half, so item 2 matches P. The final non-sway member-end moment contains the fixed-end moment, twice the near-end rotation contribution and the far-end rotation contribution, so item 3 matches S. Kani's procedure is specifically known as the rotation-contribution method, making item 4 match Q. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. It is an iterative displacement method in which joint-rotation effects are evaluated through rotation contributions. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. Slope-deflection directly formulates displacement equations, moment distribution iteratively balances joint moments, and Kani's method iteratively evaluates rotation contributions. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. μij = -(1/2)(Kij/ΣKi) సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: B. -1/2 సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. μij = -(1/2)DFij సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. Fixed far end: 4EI/L; hinged far end: 3EI/L సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. μBA = -0.30 and μBC = -0.20 సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. -0.222, -0.139 and -0.139 సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. MABF = -wL²/12 and MBAF = +wL²/12 సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. MABF = -PL/8 and MBAF = +PL/8 సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. The algebraic sum of the fixed-end moments of the members meeting at that joint సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: B. -15 kN·m and -10 kN·m సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. -13.2 kN·m and -8.8 kN·m సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: B. Mij = MijF + 2M'ij + M'ji సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. -11 kN·m సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. A fixed joint has zero rotation contribution, while an actual hinged member end must have zero final end moment. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: A. μBA = μBC = -0.25 సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: A. M'BA = M'BC = -9; final moments MAB = -45, MBA = +18, MBC = -18, MCB = -9 kN·m సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. RA = 40.5 kN, RB = 36.0 kN, RC = -4.5 kN సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. At B: +4 and +4 kN·m; at C: -5 and -5 kN·m సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. +5.33, +10.67, -10.67, +10.67, -10.67, -5.33 kN·m సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. μBA = -0.30 and μBC = -0.20 సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. M'BA = +13.5, M'BC = +9, M'CB = -9, M'CD = -13.5 kN·m; final moments MAB = +13.5, MBA = +27, MBC = -27, MCB = +27, MCD = -27, MDC = -13.5 kN·m సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. 2, 3 and 4 only సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1-R, 2-P, 3-S, 4-Q సమాధానం: A.