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Question 1
ప్రశ్న 1
In slope-deflection analysis of a rigid portal frame, which quantities are normally treated as the primary unknowns?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The slope-deflection method is a displacement method of structural analysis. Joint rotations are therefore fundamental unknowns in rigid-frame analysis. If the frame can sway, one or more independent joint translations must also be determined. Member-end moments are then expressed in terms of these displacement coordinates and the fixed-end moments. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which statement correctly distinguishes a non-sway portal frame from a sway portal frame in elementary slope-deflection analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Sidesway refers to lateral translation of one or more joints of the frame. A non-sway frame may still have substantial joint rotations and member-end moments. A sway frame requires the translation terms to be included in the slope-deflection equations of affected members. Sway may result from lateral loading, unsymmetrical vertical loading, support movement or stiffness asymmetry when lateral translation is not restrained. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
A one-bay portal frame has identical columns, a symmetric beam, fixed bases, no support settlement and negligible axial deformation. If the loading is also symmetric about the vertical centreline of the frame, what is the usual sidesway displacement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Symmetric geometry, stiffness and loading produce a symmetric deformation pattern. Lateral translation of both upper joints in one direction would violate that symmetry in the absence of another asymmetric action. Consequently the frame does not undergo sidesway under the stated assumptions. The upper joints may nevertheless rotate because the members bend. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A geometrically symmetric unbraced portal frame with fixed bases is subjected to an unsymmetrical vertical load on its beam. Axial deformations are neglected. Which statement is generally correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Absence of an applied horizontal force does not by itself guarantee zero sidesway. Unsymmetrical vertical loading can produce unequal beam-end moments at the two beam-column joints. In an unbraced frame, the associated frame deformation may contain a lateral translation. The resulting column chord rotations must then be included in the slope-deflection equations. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
Throughout this test, counterclockwise member-end moments, joint rotations and chord rotations are positive. For a prismatic frame member AB of length L, which slope-deflection equation for end A is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The general slope-deflection equation contains contributions from member loading, end rotations and chord rotation. Under the stated convention, the fixed-end moment contribution is MABF. The rotational contribution is 2EI/L multiplied by 2θA + θB. The chord-rotation contribution enters as -3ψAB within the same bracket. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A vertical column AB is initially straight, with A at the fixed base and B directly above it at height h. Joint B translates a small distance Δ to the right while A does not translate. If ψAB is positive counterclockwise, what is the approximate chord rotation from A toward B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The original chord from A to B is vertical. When B moves to the right, the line from A to the displaced position of B rotates clockwise. Clockwise rotation is negative under the stated counterclockwise-positive convention. For small displacements, the rotation magnitude is approximately Δ/h. Hence ψAB = -Δ/h and Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
In a conventional one-storey rectangular portal frame, the two upper joints B and C translate by the same horizontal amount Δ while their relative vertical displacement is zero. Axial deformation of the horizontal beam BC is neglected. What is the chord rotation ψBC of the beam due solely to this sidesway?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Chord rotation depends on relative transverse displacement of the two ends of a member. Both ends of the horizontal beam translate horizontally by the same amount. They have no relative vertical displacement under the stated idealization. The chord therefore remains horizontal even though the complete frame undergoes sidesway. Hence ψBC = 0 and Option D is correct.
వివరణ:
Question 8
ప్రశ్న 8
For an unloaded prismatic frame member of length L, what are the standard near-end rotational stiffnesses when the far end is respectively fixed and hinged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
If the far end is fixed against rotation, a unit near-end rotation produces a moment corresponding to stiffness 4EI/L. If the far end is hinged, its final moment must be zero and it is allowed to rotate. This release reduces the effective near-end stiffness to 3EI/L. These stiffness relationships are frequently used in displacement methods for beams and rigid frames. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A prismatic beam member BC of span L in a portal frame carries a downward UDL w over its complete 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 UDL on a fixed-ended prismatic member produces hogging fixed-end moments of magnitude wL²/12. At the left end B, the physical hogging moment acts counterclockwise under the convention used here. At the right end C, the corresponding physical hogging moment acts 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 the centre of a prismatic beam member BC of span L whose ends are initially fixed. Using counterclockwise member-end moments as positive, what are the fixed-end moments?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A central concentrated load on a fixed-ended prismatic beam produces fixed-end moment magnitudes PL/8. The moments are hogging at the two beam ends. Under one common counterclockwise-positive convention, the left-end moment is positive and the right-end moment is negative. These fixed-end moments become the load terms in the beam slope-deflection equations. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
At rigid joint B of a portal frame, members BA and BC meet and no external couple is applied directly at B. Which equation expresses rotational equilibrium of the joint?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The end moments contributed by all members meeting at a rigid joint act on that joint. If no external couple is applied there, their algebraic sum must vanish. For a joint connecting only members BA and BC, this gives MBA + MBC = 0. Neither individual moment needs to be zero. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A one-bay fixed-base portal frame has two upper rigid joints B and C. Axial deformations are neglected, the frame is free to sidesway and no symmetry condition eliminates that sway. How many displacement unknowns are normally required in a slope-deflection analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each of the two upper rigid joints can undergo an independent rotation. Neglecting axial deformation makes the upper joints share a common horizontal sidesway coordinate in the usual rectangular one-storey portal. Thus θB, θC and Δ are the three independent displacement unknowns. Two joint moment-equilibrium equations and one horizontal force-equilibrium equation are normally used to determine them. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A rectangular portal frame has fixed bases A and D, columns AB and DC each 4 m high, and a horizontal beam BC 6 m long. All three members have EI = 24000 kN·m², axial deformations are neglected, and a UDL of 12 kN/m acts over the complete beam BC. Because geometry and loading are symmetric, sidesway is zero. Using counterclockwise rotations as positive, what are θB and θC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The fixed-end moment magnitude of the loaded beam is 12(6²)/12 = 36 kN·m. Symmetry requires the two top-joint rotations to have equal magnitudes and opposite directions. Applying the two joint-equilibrium equations gives EIθB = -27 kN·m and EIθC = +27 kN·m. Division by EI = 24000 kN·m² gives θB = -0.001125 rad and θC = +0.001125 rad. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
For the portal frame described in the previous question, which set of final member-end moments is correct using counterclockwise member-end moments as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For each 4 m column with its base fixed, the top-end rotational moment is 4EIθ/L and the base-end moment is 2EIθ/L. Using the rotations from the previous problem gives -27 kN·m at B from the left column and +27 kN·m at C from the right column. Joint equilibrium therefore requires beam moments +27 kN·m at B and -27 kN·m at C. The corresponding column base moments are -13.5 kN·m at A and +13.5 kN·m at D. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
A symmetric fixed-base portal frame has 4 m high columns and a 6 m beam. The beam has flexural rigidity EI = 24000 kN·m², each column has flexural rigidity 2EI, axial deformation is neglected, and the beam carries a full-span UDL of 12 kN/m. Sidesway is zero by symmetry. What is the magnitude of the final beam-end moment at each end of BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The beam fixed-end moment magnitude is 36 kN·m. The columns are twice as stiff in EI as the beam and therefore provide greater rotational restraint at B and C. Solving the symmetric joint-equilibrium equations gives EIθB = -108/7 kN·m and EIθC = +108/7 kN·m, where EI refers to the beam rigidity. Substitution into the beam slope-deflection equation gives an end-moment magnitude of 216/7 = approximately 30.86 kN·m. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
A rectangular portal frame has fixed bases A and D, 4 m high columns AB and DC, a 6 m beam BC, and EI = 24000 kN·m² for every member. An ideal brace prevents sidesway. A 60 kN downward point load acts on beam BC at 2 m from B and 4 m from C. Using counterclockwise rotations as positive, what are θB and θC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The beam fixed-end moments are +53.33 kN·m at B and -26.67 kN·m at C. Because the brace prevents sidesway, the column chord-rotation terms are zero. Joint equilibrium at B and C gives EIθB = -36.667 kN·m and EIθC = +23.333 kN·m. Dividing these values by EI = 24000 kN·m² gives approximately -0.001528 rad and +0.000972 rad. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the braced portal frame described in the previous question, 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 rotations of B and C modify the fixed-end moments of the loaded beam. Substitution into MBC = MBCF + (2EI/L)(2θB + θC) gives approximately +36.67 kN·m. The corresponding equation at C gives MCB approximately -23.33 kN·m. These moments balance the top-column moments at their respective rigid joints. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
When a one-storey portal frame is free to sidesway and its slope-deflection unknowns are θB, θC and Δ, the two rigid-joint moment equations provide only two equations. What type of additional equation is normally required to determine Δ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Moment equilibrium at joints B and C supplies two independent equations. A third equation is required because the sway displacement Δ is also unknown. Column shears can be expressed in terms of their end moments and column heights. Horizontal equilibrium of the storey then relates these shears to the externally applied lateral load. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A rectangular 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 EI = 20000 kN·m², axial deformations are neglected, and there are no transverse member loads. Horizontal loads of 15 kN to the right act at each of the upper joints B and C. What is the lateral sidesway Δ of the beam level?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Symmetry gives equal rotations at B and C and equal horizontal shears in the two columns. Joint moment equilibrium relates the common rotation to the lateral translation. Story equilibrium requires each identical column to resist 15 kN of the total 30 kN lateral load. Solving the slope-deflection and shear equations gives Δ = 0.0064 m. This is 6.4 mm to the right, so Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
For the sway portal frame described in the previous question, using counterclockwise rotations as positive, what are θB and θC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The frame sways to the right by 0.0064 m. The column chord rotations are therefore clockwise and equal to -Δ/4 under the adopted convention. Joint equilibrium between each column and the beam yields a common top-joint rotation equal to -0.1875Δ when Δ is measured in metres. Substitution of Δ = 0.0064 m gives θB = θC = -0.0012 rad. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
For the same sway portal frame, which set of member-end moments is correct using counterclockwise member-end moments as positive?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
There are no member loads, so all fixed-end moments are zero. The moments arise entirely from joint rotation and column chord rotation associated with sidesway. Substitution of Δ = 0.0064 m and θB = θC = -0.0012 rad gives +36 kN·m at each column base and +24 kN·m at each column top. Joint equilibrium then requires a beam-end moment of -24 kN·m at both B and C. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
For the same 30 kN total lateral load case, what horizontal shear is carried by each identical 4 m column?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The frame geometry, member stiffnesses and lateral loading are symmetric. The two columns therefore carry equal horizontal shears. Horizontal equilibrium requires their sum to equal the total applied load of 30 kN. Each column consequently carries 15 kN. This is also obtained from the column end moments as (36 + 24)/4 = 15 kN, so Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A vertical prismatic column AB is 4 m high with EI = 20000 kN·m². Both end rotations are restrained to zero, the base A does not translate, and the top B translates 5 mm to the right. There are no member loads. Using counterclockwise member-end moments as positive and taking the chord from A to B, what end moments are produced?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The rightward top displacement produces chord rotation ψAB = -0.005/4 = -0.00125 rad. Both joint rotations are zero, so only the translation term contributes to the slope-deflection equations. Each end moment is (2EI/L)(-3ψAB). Substitution gives 37.5 kN·m at both ends with the positive counterclockwise sign. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about slope-deflection analysis of portal frames:
1. A sway frame may require joint rotations and one or more joint translations as displacement unknowns.
2. A horizontal load applied directly at a frame joint does not by itself create a fixed-end moment in an otherwise unloaded member; it enters the appropriate joint or storey equilibrium equations.
3. In a non-sway rectangular frame with no support settlement, the column chord-rotation terms associated with lateral translation are zero.
4. Symmetric frame geometry alone guarantees zero sidesway regardless of how asymmetrically the frame is loaded.
Which statements are correct?
1. A sway frame may require joint rotations and one or more joint translations as displacement unknowns.
2. A horizontal load applied directly at a frame joint does not by itself create a fixed-end moment in an otherwise unloaded member; it enters the appropriate joint or storey equilibrium equations.
3. In a non-sway rectangular frame with no support settlement, the column chord-rotation terms associated with lateral translation are zero.
4. Symmetric frame geometry alone guarantees zero sidesway regardless of how asymmetrically the frame is loaded.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because sidesway introduces translational displacement coordinates in addition to joint rotations. Statement 2 is correct because fixed-end moments represent the effect of loads acting on the member span while its ends are restrained. Statement 3 is correct because zero lateral translation and zero settlement give zero column chord rotation in the standard rectangular-frame idealization. Statement 4 is false because unsymmetrical loading of an otherwise symmetric unbraced frame can produce sidesway. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each portal-frame 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. Chord rotation of a vertical column whose top sways right by Δ while its base is fixed, with counterclockwise rotation positive
4. Additional equilibrium relation required for a one-storey sway frame
P. 3EI/L
Q. Horizontal story-force equilibrium
R. 4EI/L
S. -Δ/h
Which matching is correct?
1. Near-end rotational stiffness with far end fixed
2. Near-end rotational stiffness with far end hinged
3. Chord rotation of a vertical column whose top sways right by Δ while its base is fixed, with counterclockwise rotation positive
4. Additional equilibrium relation required for a one-storey sway frame
P. 3EI/L
Q. Horizontal story-force equilibrium
R. 4EI/L
S. -Δ/h
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A member with its far end fixed has near-end rotational stiffness 4EI/L, so item 1 matches R. A hinged far end reduces the effective stiffness to 3EI/L, making item 2 match P. Rightward translation of the top of a vertical column rotates its chord clockwise, which is -Δ/h under the stated convention, so item 3 matches S. A sway displacement requires an additional horizontal force-equilibrium relation, 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. Joint rotations and any independent joint translations associated with sidesway సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. A non-sway frame has no lateral joint translation, whereas a sway frame has one or more independent lateral joint translations. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. Zero సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. The frame can undergo sidesway because the unsymmetrical loading can produce unequal end actions and lateral translation. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. MAB = MABF + (2EI/L)(2θA + θB - 3ψAB) సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. -Δ/h సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: D. 0 సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. Fixed far end: 4EI/L; hinged far end: 3EI/L సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. MBCF = +wL²/12 and MCBF = -wL²/12 సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. MBCF = +PL/8 and MCBF = -PL/8 సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. MBA + MBC = 0 సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. Three: θB, θC and one common lateral translation Δ సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. θB = -0.001125 rad and θC = +0.001125 rad సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. MAB = -13.5, MBA = -27, MBC = +27, MCB = -27, MCD = +27 and MDC = +13.5 kN·m సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: C. 30.86 kN·m సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. θB = -0.001528 rad and θC = +0.000972 rad సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. MBC = +36.67 kN·m and MCB = -23.33 kN·m సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: B. Horizontal force equilibrium involving the column shears and applied lateral load సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. 6.4 mm to the right సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. θB = θC = -0.0012 rad సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. Each column base moment = +36 kN·m, each column top moment = +24 kN·m, and each beam-end moment = -24 kN·m సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 15 kN సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. MAB = MBA = +37.5 kN·m 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.