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Question 1
ప్రశ్న 1
In the slope-deflection method, what are normally treated as the primary unknowns in the analysis of a continuous beam?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The slope-deflection method is a displacement method of structural analysis. Its primary unknowns are the independent joint displacement coordinates rather than redundant forces. For an ordinary continuous beam with unyielding supports, these unknowns are mainly joint rotations. Support settlement or frame sidesway may introduce translational or chord-rotation terms where applicable. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Throughout this test, counterclockwise member-end moments and joint rotations are positive. For a prismatic member AB of length L with no support settlement, which pair of slope-deflection equations is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a prismatic member with unyielding supports, the member-end moments consist of fixed-end moments plus the moments caused by end rotations. With the stated counterclockwise-positive convention, MAB equals MABF plus 2EI/L times 2θA + θB. Similarly, MBA equals MBAF plus 2EI/L times 2θB + θA. These equations follow from compatibility of the elastic curve and member force-deformation relations. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For member AB, let ψAB denote the small counterclockwise rotation of the member chord from A toward B. Using the counterclockwise-positive convention, which slope-deflection equations include the effect of support settlement correctly?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Support settlement causes the chord joining the member ends to rotate. Joint rotations must therefore be measured relative to this rotated chord when member deformation is evaluated. Under the stated convention, the chord-rotation contribution enters both member-end equations as -3ψAB inside the bracket. Multiplication by 2EI/L gives the corresponding settlement-induced moment contribution. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A prismatic fixed-ended member AB of length L carries a downward UDL w over its entire span. Using counterclockwise member-end moments as positive, what are its fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A full-span downward UDL on a fixed-ended prismatic member produces equal hogging moment magnitudes wL²/12 at its two ends. Because the test uses a common global counterclockwise-positive convention, the physical hogging end moments have opposite algebraic directions at A and B. The left-end member moment is therefore +wL²/12. The right-end member moment is -wL²/12. Hence Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A downward point load P acts at the midspan of a prismatic fixed-ended member AB of length L. Using counterclockwise member-end moments as positive, what are the fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A central point load on a fixed-ended prismatic beam produces hogging fixed-end moments of magnitude PL/8 at both supports. Their physical senses are opposite when expressed using one common counterclockwise-positive direction. Thus the end moment acting at A is +PL/8 under the convention used here. The end moment at B is -PL/8. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A 6 m fixed-ended prismatic member AB carries a 60 kN downward point load 2 m from A and 4 m from B. Using counterclockwise member-end moments as positive, what are the fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a point load P located a from A and b from B, the fixed-end moment magnitudes are Pab²/L² at A and Pa²b/L² at B. Here P = 60 kN, a = 2 m, b = 4 m and L = 6 m. The magnitudes are 53.33 kN·m at A and 26.67 kN·m at B. Under the stated global convention their signs are positive at A and negative at B. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
For an unloaded prismatic member AB with no chord rotation, end B is fixed against rotation while end A undergoes a rotation θA. What are the near-end rotational stiffness at A and the moment carried over to B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With θB = 0, no member load and no chord rotation, the slope-deflection equation gives MAB = 4EIθA/L. Thus the rotational stiffness at the near end when the far end is fixed is 4EI/L. The corresponding far-end moment is MBA = 2EIθA/L. It is therefore one-half of the near-end moment under this convention. Hence Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For a prismatic member whose far end is hinged and therefore carries zero final moment, what is the effective rotational stiffness at the near end when support translation is absent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Let the near-end rotation be θA while the far-end moment is constrained to zero. The far-end slope-deflection equation then determines the accompanying far-end rotation as θB = -θA/2. Substitution into the near-end equation gives MAB = 3EIθA/L. Therefore the near-end rotational stiffness with the far end hinged is 3EI/L. Hence Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
At an interior joint B of a continuous beam, no external couple is applied directly at the joint. What joint-equilibrium condition must be satisfied after the member-end moments are expressed by slope-deflection equations?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Slope-deflection equations express member-end moments in terms of displacement unknowns and loads. The unknown joint rotations are then determined from joint equilibrium. If no external couple acts at an interior joint, all member-end moments applied to that joint must balance algebraically. For two members BA and BC, this condition is MBA + MBC = 0 under a consistent convention. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
An unloaded prismatic member AB has L = 6 m and EI = 24000 kN·m². There is no chord rotation. If θA = +0.002 rad and θB = -0.001 rad, what are MAB and MBA using counterclockwise end moments as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The coefficient 2EI/L is 2(24000)/6 = 8000 kN·m per radian. Therefore MAB = 8000[2(0.002) - 0.001] = 24 kN·m. Similarly, MBA = 8000[2(-0.001) + 0.002] = 0. No fixed-end moments are present because the member carries no external transverse load. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A 6 m fixed-fixed beam carries a UDL of 10 kN/m over its entire span. There is no support settlement. Using counterclockwise member-end moments as positive, what are the final member-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At fixed supports both joint rotations are zero. With no support settlement, the rotational terms in the slope-deflection equations therefore vanish. The final member-end moments are consequently equal to the fixed-end moments. Their magnitude is wL²/12 = 10(6²)/12 = 30 kN·m. Under the stated convention the pair is +30 kN·m and -30 kN·m, so Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A propped cantilever AB has A fixed and B supported by a vertical roller. Its span is 6 m and it carries a UDL of 12 kN/m over the entire span. Neglect support settlement. What is the magnitude of the hogging bending moment at the fixed end A?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At A the rotation is zero, while the roller at B permits rotation and requires the final end moment at B to be zero. Applying the B-end slope-deflection equation gives the required rotation θB. Substitution into the A-end equation gives a fixed-end support moment magnitude wL²/8. With w = 12 kN/m and L = 6 m, this magnitude is 54 kN·m. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
For the propped cantilever in the previous question, what are the vertical reactions at A and B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a propped cantilever carrying a full-span UDL, the roller reaction is 3wL/8. Here this gives RB = 3(12)(6)/8 = 27 kN. The total applied vertical load is 72 kN. Vertical equilibrium therefore gives RA = 72 - 27 = 45 kN. Hence Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
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 UDL w over their entire lengths. What is the bending moment at support B when sagging bending moment is taken as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The beam and loading are symmetric about support B. The joint rotation at B is therefore zero, although the outer simple-support rotations are not zero. Applying the zero-moment conditions at A and C together with joint equilibrium at B gives the interior support moment. Its magnitude is wL²/8 and it is hogging. With sagging moment defined as positive, MB = -wL²/8, so Option D is correct.
వివరణ:
Question 15
ప్రశ్న 15
The two-span beam in the previous question has L = 6 m and w = 12 kN/m on both spans. What are the vertical support reactions RA, RB and RC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The interior support moment is -wL²/8 = -54 kN·m. For each identical span, the exterior support reaction is 3wL/8. This gives RA = RC = 3(12)(6)/8 = 27 kN. Each span contributes 5wL/8 = 45 kN to the reaction at B, so RB = 90 kN. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A continuous beam ABC has AB = BC = 6 m and constant EI = 30000 kN·m². Supports A and C are fixed, while B is an intermediate support whose rotation is free. Only span AB carries a UDL of 10 kN/m; BC is unloaded and there is no settlement. Using counterclockwise rotations as positive, what is θB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The UDL fixed-end moment magnitude on AB is wL²/12 = 30 kN·m. Since A and C are fixed, θA = θC = 0 and θB is the only rotational unknown. Joint equilibrium requires MBA + MBC = 0. Solving the two member slope-deflection expressions gives θB = wL³/(96EI) = 0.000750 rad. The sign is counterclockwise under the convention used, so Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the continuous beam described in the previous question, what are the member-end moments MAB, MBA, MBC and MCB using counterclockwise member-end moments as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For span AB, the fixed-end moments are +30 and -30 kN·m. Substitution of θB = 0.000750 rad gives MAB = +37.5 kN·m and MBA = -15 kN·m. Span BC is unloaded, so only rotation θB generates its end moments, giving MBC = +15 kN·m and MCB = +7.5 kN·m. At joint B, MBA + MBC equals zero as required by equilibrium. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
Member AB is initially horizontal. Support A does not move, while support B settles vertically downward by a small amount Δ. With counterclockwise chord rotation ψAB taken as positive, what is ψAB approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The chord is considered from end A toward end B. A downward settlement of B causes this chord to rotate clockwise from its original horizontal position. Clockwise rotation is negative under the stated counterclockwise-positive convention. For a small displacement, the magnitude of chord rotation is approximately Δ/L. Thus ψAB = -Δ/L and Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A 5 m prismatic beam AB is fixed against rotation at both ends and carries no external load. Its flexural rigidity is EI = 40000 kN·m². Support B settles 10 mm downward relative to A. Using counterclockwise member-end moments as positive, what settlement-induced end moments result?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The downward settlement is Δ = 0.01 m, so ψAB = -0.01/5 = -0.002 rad. Both end rotations are zero because the ends remain fixed against rotation. The settlement term in each slope-deflection equation is (2EI/L)(-3ψAB). Substitution gives 6EIΔ/L² = 96 kN·m at each member end in the positive counterclockwise direction. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
For the settlement-only fixed-fixed beam in the previous question, what vertical reactions are induced at A and B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
There is no external vertical load, so the two vertical reactions must be equal and opposite. The two member-end moments are both +96 kN·m under the stated global convention. Taking moments of the member free body gives a shear magnitude of 2(96)/5 = 38.4 kN. Equilibrium requires the reaction at A to act upward and that at B to act downward. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
An 8 m fixed-fixed prismatic beam carries an 80 kN downward point load at midspan. There is no support settlement. Using counterclockwise member-end moments as positive, which set of final end moments and vertical reactions is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Both support rotations are zero, so the final member-end moments equal the fixed-end moments. For a central point load, their magnitude is PL/8 = 80(8)/8 = 80 kN·m. Under the common counterclockwise-positive convention the pair is +80 kN·m at A and -80 kN·m at B. Symmetry gives equal vertical reactions of P/2 = 40 kN. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A two-span continuous beam ABC has equal spans of 6 m, constant EI, simple supports at A and C and an intermediate support at B. Only span AB carries a UDL of 12 kN/m; span BC is unloaded. What is the support bending moment at B when sagging moment is positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The outer supports impose zero bending moment at the exterior member ends A and C. The rotations θA, θB and θC are related through the two span slope-deflection equations and the equilibrium condition at the continuous joint B. Solving these equations gives the interior support moment magnitude wL²/16. With w = 12 kN/m and L = 6 m, the magnitude is 27 kN·m. It is hogging, so with sagging positive MB = -27 kN·m and Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
For a continuous beam with unyielding supports subjected only to prescribed external loads, every member flexural rigidity EI is multiplied by the same positive factor while all relative stiffness ratios remain unchanged. What happens to the final member-end moments and joint rotations in linear slope-deflection analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With unyielding supports, the fixed-end moments depend on loading and geometry rather than on the common magnitude of EI. Multiplying every EI by the same factor multiplies all rotational stiffness coefficients by that factor without changing their ratios. The joint-equilibrium equations are then satisfied by rotations reduced by the reciprocal of that factor. Products such as EIθ consequently remain unchanged. Thus the final member-end moments remain unchanged while rotations vary inversely with the common EI multiplier, making Option A correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about the slope-deflection method:
1. It is a displacement method in which joint rotations are important primary unknowns for continuous beams.
2. Fixed-end moments caused by member loading must be included in the member-end moment equations.
3. At a member end that is hinged or moment-released, the final member-end moment is zero even though the fixed-end moment calculated for the corresponding fully restrained member under its load may be nonzero.
4. Support settlement cannot generate end moments if no external transverse load acts.
Which statements are correct?
1. It is a displacement method in which joint rotations are important primary unknowns for continuous beams.
2. Fixed-end moments caused by member loading must be included in the member-end moment equations.
3. At a member end that is hinged or moment-released, the final member-end moment is zero even though the fixed-end moment calculated for the corresponding fully restrained member under its load may be nonzero.
4. Support settlement cannot generate end moments if no external transverse load acts.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because slope-deflection analysis is formulated in terms of displacement coordinates such as joint rotations. Statement 2 is correct because fixed-end moments represent the member-load contribution before joint displacements are introduced. Statement 3 is correct because an actual hinge or moment release imposes zero final moment at that member end, with the end rotation adjusting accordingly. Statement 4 is incorrect because support settlement can rotate a member chord and generate end moments even when no external transverse load acts. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each slope-deflection concept with its correct result:
1. Near-end rotational stiffness with far end fixed
2. Near-end rotational stiffness with far end hinged
3. Fixed-end moment magnitude at either end for a full-span UDL
4. Fixed-end moment magnitude at either end for a central point load
P. 3EI/L
Q. PL/8
R. 4EI/L
S. wL²/12
Which matching is correct?
1. Near-end rotational stiffness with far end fixed
2. Near-end rotational stiffness with far end hinged
3. Fixed-end moment magnitude at either end for a full-span UDL
4. Fixed-end moment magnitude at either end for a central point load
P. 3EI/L
Q. PL/8
R. 4EI/L
S. wL²/12
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With the far end fixed, a near-end rotation produces a rotational stiffness of 4EI/L, so item 1 matches R. Releasing the far end to a hinge reduces the near-end stiffness to 3EI/L, so item 2 matches P. A full-span UDL on a fixed-ended prismatic member produces fixed-end moment magnitude wL²/12, making item 3 match S. A central point load produces fixed-end moment magnitude PL/8, so item 4 matches Q. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Joint rotations and, where applicable, joint translations or chord rotations సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. MAB = MABF + (2EI/L)(2θA + θB) and MBA = MBAF + (2EI/L)(2θB + θA) సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. MAB = MABF + (2EI/L)(2θA + θB - 3ψAB) and MBA = MBAF + (2EI/L)(2θB + θA - 3ψAB) సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. MABF = +wL²/12 and MBAF = -wL²/12 సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. MABF = +PL/8 and MBAF = -PL/8 సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. MABF = +53.33 kN·m and MBAF = -26.67 kN·m సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. Near-end stiffness = 4EI/L and MBA = (1/2)MAB సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 3EI/L సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. The algebraic sum of the member-end moments meeting at B must be zero. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. MAB = +24 kN·m and MBA = 0 సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. MAB = +30 kN·m and MBA = -30 kN·m సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 54 kN·m సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. RA = 45 kN and RB = 27 kN సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. -wL²/8 సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: B. RA = 27 kN, RB = 90 kN, RC = 27 kN సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: B. 0.000750 rad counterclockwise సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. MAB = +37.5, MBA = -15, MBC = +15, MCB = +7.5 kN·m సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. -Δ/L సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: A. MAB = MBA = +96 kN·m సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: A. 38.4 kN upward at A and 38.4 kN downward at B సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. MAB = +80 kN·m, MBA = -80 kN·m, RA = RB = 40 kN సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. -27 kN·m సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: A. End moments remain unchanged, while joint rotations vary inversely with the common EI factor. 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-P, 3-S, 4-Q సమాధానం: A.