Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which statement best describes the Hardy Cross moment-distribution method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The moment-distribution method was developed by Hardy Cross for analyzing statically indeterminate structures. It is closely related to the slope-deflection method but avoids solving all joint-rotation equations simultaneously by hand. Initially, the joints are considered locked and fixed-end moments are calculated. Unbalanced joint moments are then distributed to connected members in proportion to their rotational stiffnesses and appropriate portions are carried over. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which sequence most accurately represents a standard moment-distribution analysis of a continuous beam with unyielding supports?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The process begins by calculating fixed-end moments produced by loads while joints are imagined locked. Member stiffnesses and distribution factors are then established at each joint. The algebraic unbalanced moment at a released joint is balanced and the balancing moment is distributed among the connected members. Appropriate carry-over moments are transferred to the far ends, and the process is repeated where further imbalance occurs. The final member-end moments are the algebraic sums of all fixed-end, distributed and carried-over contributions, so Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a prismatic member AB of length L, what rotational stiffness is used at end A in moment distribution when the far end B is fixed against rotation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Rotational stiffness is the moment required at a joint to produce unit rotation there under specified far-end conditions. For a prismatic member whose far end is fixed, the near-end stiffness is 4EI/L. This result follows directly from the slope-deflection equations. It is the standard stiffness used when all other joints are locked during moment distribution. Therefore Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a prismatic member AB whose far end B is an actual hinge and remains moment-free, what rotational stiffness is used at end A?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An actual hinge at the far end permits rotation and requires the final moment there to be zero. This reduces the restraint provided to rotation at the near end. Eliminating the far-end rotation from the slope-deflection equations gives a near-end stiffness of 3EI/L. The value is smaller than the 4EI/L stiffness obtained when the far end is fixed. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
At joint B, members BA, BC and BD have rotational stiffnesses KBA, KBC and KBD respectively. What is the distribution factor for member BA?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A distribution factor represents the fraction of a balancing joint moment assigned to a particular connected member. The fraction is proportional to that member's rotational stiffness at the joint. Consequently the stiffness of BA is divided by the sum of the stiffnesses of all members meeting at B. Distribution factors are dimensionless quantities. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For a rigid joint having no special release, what is the sum of the distribution factors of all members meeting at that joint?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each distribution factor is the corresponding member stiffness divided by the total joint stiffness. Adding the factors therefore reproduces the total stiffness divided by itself. Their sum must consequently equal unity. This ensures that the entire balancing moment is distributed among the connected members. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
For a prismatic member in moment distribution, what is the standard carry-over factor from a near end to a far end that is fixed, and what is it when the far end is an actual hinge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
If a moment increment is applied at one end of a prismatic member while the far end is fixed, one-half of that increment is induced at the far end. The carried-over moment has the same algebraic sign under the consistent member-end convention used in moment distribution. If the far end is an actual hinge, it cannot sustain a moment. The carry-over factor to such a hinged end is therefore zero. Hence Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
At a rigid joint, the algebraic sum of the currently accumulated member-end moments is +30 kN·m and no external joint couple acts there. What balancing moment must be distributed among the connected members?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The accumulated member-end moments produce an unbalanced joint moment of +30 kN·m. Joint equilibrium requires the final algebraic sum of moments at the joint to be zero when no external couple is applied. A balancing moment equal in magnitude and opposite in sign must therefore be introduced. This -30 kN·m balancing moment is then divided among the connected members according to their distribution factors. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
Throughout this test, counterclockwise member-end moments are 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 on a prismatic fixed-ended beam produces equal hogging fixed-end moment magnitudes of wL²/12. At the left end, the physical hogging moment is counterclockwise under the stated global convention. At the right end, the corresponding hogging moment is clockwise. Their algebraic values are therefore +wL²/12 and -wL²/12 respectively. Hence Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
A downward point load P acts at midspan of a prismatic fixed-ended member AB of span L. With counterclockwise member-end moments positive, what are the fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A central point load produces equal hogging moment magnitudes PL/8 at the ends of a prismatic fixed-ended member. The physical rotations of these end moments are opposite when represented using a common counterclockwise-positive convention. Hence the left-end fixed-end moment is +PL/8. The right-end fixed-end moment is -PL/8. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
A 6 m prismatic fixed-ended member AB carries a downward 60 kN point load 2 m from A and 4 m from B. With counterclockwise member-end moments positive, what are the fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a point load P at distances a and b from the two ends, the left fixed-end moment magnitude is Pab²/L². Its value is 60 × 2 × 4²/6² = 53.33 kN·m. The right fixed-end moment magnitude is Pa²b/L² = 60 × 2² × 4/6² = 26.67 kN·m. Applying the stated counterclockwise-positive convention gives +53.33 kN·m at A and -26.67 kN·m at B. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
At joint B, member BA has length 4 m and flexural rigidity EI, while member BC has length 6 m and the same EI. Their far ends A and C are fixed during distribution. What are the distribution factors DFBA and DFBC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With fixed far ends, the stiffness at B of BA is 4EI/4 = EI. The stiffness at B of BC is 4EI/6 = 2EI/3. Their total stiffness is 5EI/3. Thus DFBA = 1/(5/3) = 0.60 and DFBC = (2/3)/(5/3) = 0.40. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
Three prismatic members BA, BC and BD meet at joint B and their far ends are fixed. BA has length 5 m and rigidity 2EI, BC has length 4 m and rigidity EI, and BD has length 6 m and rigidity 1.5EI. What are their distribution factors at B approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stiffness of BA is 4(2EI)/5 = 1.6EI. The stiffness of BC is 4EI/4 = 1.0EI, while that of BD is 4(1.5EI)/6 = 1.0EI. The total stiffness is therefore 3.6EI. Dividing each stiffness by this total gives 0.444, 0.278 and 0.278 approximately. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
A continuous beam ABC has A fixed, B as an intermediate vertical support with unrestricted rotation, and C fixed. Spans AB and BC are each 6 m long and have the same constant EI. Only span AB carries a UDL of 12 kN/m. Using counterclockwise member-end moments as positive, what are the final member-end moments after moment distribution?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The fixed-end moments of loaded span AB are +36 kN·m at A and -36 kN·m at B. At joint B the two members have equal stiffness, so their distribution factors are 0.5 each. The -36 kN·m unbalanced joint moment requires a +36 kN·m balancing moment, giving +18 kN·m to each member at B. One-half of each distributed moment carries over to the fixed far ends, producing +9 kN·m at A and C. Adding all contributions gives +45, -18, +18 and +9 kN·m, so Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
For the continuous beam in the previous question, what are the vertical reactions RA, RB and RC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For span AB, the final end moments are +45 kN·m at A and -18 kN·m at B. Member equilibrium with the 72 kN UDL gives the vertical reactions contributed by AB as 40.5 kN at A and 31.5 kN at B. For unloaded span BC, the +18 kN·m and +9 kN·m end moments require vertical forces of +4.5 kN at B and -4.5 kN at C. The total reaction at B is therefore 31.5 + 4.5 = 36 kN. Hence RA = 40.5 kN, RB = 36.0 kN and RC = -4.5 kN, making Option B correct.
వివరణ:
Question 16
ప్రశ్న 16
A two-span continuous beam ABC has simple supports at A and C and an intermediate support at B. Span AB is 4 m long with rigidity EI, span BC is 6 m long with rigidity 2EI, and only AB carries a UDL of 12 kN/m. What is the final bending moment at support B when sagging bending moment is positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because A and C are actual simple ends, the effective stiffnesses at B are 3EI/4 for BA and 3(2EI)/6 = EI for BC. Their distribution factors are therefore 3/7 and 4/7. Releasing the left simple end changes the effective joint-B load moment from -16 kN·m to -24 kN·m because -8 kN·m is carried to B during the release. Balancing the joint distributes +24 kN·m, leaving the BA end moment equal to -24 + 24(3/7) = -96/7 kN·m. Thus the support moment is approximately -13.71 kN·m and Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the continuous beam described in the previous question, what are the vertical reactions approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Span AB carries a total load of 48 kN and has end moments zero at A and -13.714 kN·m at B. Member equilibrium gives approximately 20.571 kN reaction at A and 27.429 kN at B from this span. Span BC is unloaded but has +13.714 kN·m at B and zero moment at C. These moments produce approximately +2.286 kN at B and -2.286 kN at C. Combining the contributions gives RA = 20.57 kN, RB = 29.71 kN and RC = -2.29 kN, so Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
A two-span continuous beam ABC has equal spans L, constant EI, simple supports at A and C and an intermediate support at B. Both spans carry the same full-span UDL w. What is the final support moment at B when sagging bending moment is positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The geometry, flexural rigidity and loading are symmetric about support B. Moment distribution gives equal stiffness participation from the two spans at B. After releasing the simple outer ends and balancing the central joint, the resulting interior support moment is hogging. Its magnitude is wL²/8. With sagging moment defined as positive, the support moment is -wL²/8 and Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
A symmetric non-sway portal frame has fixed bases A and D, columns AB and DC each 4 m high, and beam BC 6 m long. All members have the same EI. When joint B is released while the other joints are held fixed, what are the distribution factors for column BA and beam BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The column stiffness at B is 4EI/4 = EI because its fixed base A is the far end. The beam stiffness at B while joint C is locked is 4EI/6 = 2EI/3. Their total is 5EI/3. Thus the column distribution factor is 1/(5/3) = 0.60 and the beam distribution factor is 0.40. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
The symmetric fixed-base portal frame described in the previous question carries a UDL of 12 kN/m over the full 6 m beam BC. Sidesway is zero. After moment distribution has converged, what are the final beam-end moments MBC and MCB using counterclockwise member-end moments as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The beam fixed-end moments initially have magnitudes of 36 kN·m. Releasing joints B and C produces successive distributions between each beam end and its adjoining column. Carried-over moments from one joint create smaller imbalances at the other joint, so the procedure is repeated to convergence. The converged member-end moments are +27 kN·m at B and -27 kN·m at C. These agree with the independent slope-deflection solution, so Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
At a joint, two members have distribution factors 0.60 and 0.40. The algebraic unbalanced joint moment is +50 kN·m, and both far ends are fixed. What distributed moments and corresponding carry-over moments result from balancing this joint?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A +50 kN·m unbalanced moment requires a -50 kN·m balancing moment. Applying the distribution factors gives -30 kN·m to the first member and -20 kN·m to the second. Because both far ends are fixed, the carry-over factor is one-half. The corresponding carried-over moments are therefore -15 kN·m and -10 kN·m. Hence Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
Why are repeated distribution and carry-over cycles required in a continuous beam having more than one rotatable interior joint?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Balancing a joint changes the end moments of the members connected to it. When their far ends are restrained, fractions of these changes are carried over to adjacent joints. Those carried-over moments can disturb joints that had previously been balanced. The process is therefore repeated, producing progressively smaller corrections in a stable linear-elastic structure. Hence Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Which statement best describes the treatment of sidesway in moment-distribution analysis of an unbraced rigid frame?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A frame capable of lateral translation has an additional displacement mode compared with a non-sway frame. A conventional moment-distribution solution can separate the loading effect from an assumed sway effect. The sway correction is scaled so that horizontal or storey-force equilibrium is satisfied. This is analogous to including a translation unknown in the slope-deflection formulation. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about moment distribution:
1. Distribution factors at a joint are based on relative rotational stiffnesses of the connected members.
2. For a prismatic member with the far end fixed, the carry-over factor is one-half.
3. A member ending in an actual hinge has a near-end stiffness of 3EI/L and zero carry-over to that hinge.
4. The balancing moment applied at an unbalanced joint has the same algebraic sign as the unbalanced moment.
Which statements are correct?
1. Distribution factors at a joint are based on relative rotational stiffnesses of the connected members.
2. For a prismatic member with the far end fixed, the carry-over factor is one-half.
3. A member ending in an actual hinge has a near-end stiffness of 3EI/L and zero carry-over to that hinge.
4. The balancing moment applied at an unbalanced joint has the same algebraic sign as the unbalanced moment.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because a joint moment is distributed according to the relative stiffnesses of the connected members. Statement 2 is correct for a prismatic member whose far end is fixed. Statement 3 is also correct because a hinged far end reduces near-end stiffness to 3EI/L and cannot receive a carried-over moment. Statement 4 is incorrect because the balancing moment must oppose the algebraic unbalanced moment to restore equilibrium. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each moment-distribution concept with its correct value or definition:
1. Member stiffness when far end is fixed
2. Member stiffness when far end is hinged
3. Carry-over factor to a fixed far end of a prismatic member
4. Distribution factor of member BA at joint B
P. 3EI/L
Q. KBA divided by the sum of stiffnesses at B
R. 1/2
S. 4EI/L
Which matching is correct?
1. Member stiffness when far end is fixed
2. Member stiffness when far end is hinged
3. Carry-over factor to a fixed far end of a prismatic member
4. Distribution factor of member BA at joint B
P. 3EI/L
Q. KBA divided by the sum of stiffnesses at B
R. 1/2
S. 4EI/L
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A prismatic member with its far end fixed has a near-end stiffness of 4EI/L, so item 1 matches S. If the far end is hinged, the near-end stiffness becomes 3EI/L, so item 2 matches P. The carry-over factor to a fixed far end is one-half, so item 3 matches R. A distribution factor is the member stiffness divided by the total stiffness at the joint, so item 4 matches Q. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: B. It is an iterative displacement-based procedure in which unbalanced joint moments are successively distributed according to member stiffnesses. సమాధానం: B.
-
Question 2 ప్రశ్న 2Answer: A. Calculate fixed-end moments, determine stiffnesses and distribution factors, balance joints, carry over moments, repeat until convergence, and sum all contributions. సమాధానం: A.
-
Question 3 ప్రశ్న 3Answer: C. 4EI/L సమాధానం: C.
-
Question 4 ప్రశ్న 4Answer: B. 3EI/L సమాధానం: B.
-
Question 5 ప్రశ్న 5Answer: A. KBA/(KBA + KBC + KBD) సమాధానం: A.
-
Question 6 ప్రశ్న 6Answer: C. 1.0 సమాధానం: C.
-
Question 7 ప్రశ్న 7Answer: B. Fixed far end: 1/2; hinged far end: 0 సమాధానం: B.
-
Question 8 ప్రశ్న 8Answer: B. -30 kN·m సమాధానం: B.
-
Question 9 ప్రశ్న 9Answer: A. MABF = +wL²/12 and MBAF = -wL²/12 సమాధానం: A.
-
Question 10 ప్రశ్న 10Answer: B. MABF = +PL/8 and MBAF = -PL/8 సమాధానం: B.
-
Question 11 ప్రశ్న 11Answer: B. MABF = +53.33 kN·m and MBAF = -26.67 kN·m సమాధానం: B.
-
Question 12 ప్రశ్న 12Answer: C. DFBA = 0.60 and DFBC = 0.40 సమాధానం: C.
-
Question 13 ప్రశ్న 13Answer: A. BA = 0.444, BC = 0.278, BD = 0.278 సమాధానం: A.
-
Question 14 ప్రశ్న 14Answer: B. MAB = +45, MBA = -18, MBC = +18, MCB = +9 kN·m సమాధానం: B.
-
Question 15 ప్రశ్న 15Answer: B. RA = 40.5 kN, RB = 36.0 kN, RC = -4.5 kN సమాధానం: B.
-
Question 16 ప్రశ్న 16Answer: B. -13.71 kN·m సమాధానం: B.
-
Question 17 ప్రశ్న 17Answer: B. RA = 20.57 kN, RB = 29.71 kN, RC = -2.29 kN సమాధానం: B.
-
Question 18 ప్రశ్న 18Answer: A. -wL²/8 సమాధానం: A.
-
Question 19 ప్రశ్న 19Answer: C. DFBA = 0.60 and DFBC = 0.40 సమాధానం: C.
-
Question 20 ప్రశ్న 20Answer: C. MBC = +27 kN·m and MCB = -27 kN·m సమాధానం: C.
-
Question 21 ప్రశ్న 21Answer: B. Distributed: -30 and -20 kN·m; carry-over: -15 and -10 kN·m సమాధానం: B.
-
Question 22 ప్రశ్న 22Answer: A. Because balancing one joint can create new moments at adjacent joints through carry-over. సమాధానం: A.
-
Question 23 ప్రశ్న 23Answer: C. A sway frame requires the translational effect to be accounted for, commonly by combining a no-sway analysis with a sway correction satisfying lateral equilibrium. సమాధానం: C.
-
Question 24 ప్రశ్న 24Answer: B. 1, 2 and 3 only సమాధానం: B.
-
Question 25 ప్రశ్న 25Answer: A. 1-S, 2-P, 3-R, 4-Q సమాధానం: A.