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Question 1
ప్రశ్న 1
Which set of assumptions is most consistent with the usual ideal analysis of a statically determinate plane truss?
Explanation:
An ideal plane truss is modelled as an assembly of straight members connected by frictionless pin joints. External loads are assumed to act at the joints so that individual members behave as two-force members. A two-force member carries axial tension or compression but no idealized bending moment or transverse shear between its ends. Member self-weight is commonly neglected or converted into equivalent joint loads when it must be included. Therefore Option A states the assumptions used in elementary determinate-truss analysis.
వివరణ:
Question 2
ప్రశ్న 2
For a plane pin-jointed truss having j joints, m members and r independent external reaction components, which counting relation is necessary for static determinacy when the truss is also geometrically stable?
Explanation:
Each joint of a plane truss provides two independent force-equilibrium equations. Therefore a truss with j joints provides 2j scalar joint-equilibrium equations. The unknowns consist of m axial member forces and r support-reaction components. Equality of the number of unknowns and equilibrium equations requires m + r = 2j. This counting equality is necessary but does not by itself guarantee geometric stability, so Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
A plane truss has 8 joints and 13 members and is supported by one pin and one roller, giving three independent reaction components. Provided its geometry is stable and properly triangulated, how is it classified by the usual counting criterion?
Explanation:
The pin and roller together provide three independent reaction components. Thus m + r = 13 + 3 = 16. For j = 8 joints, 2j is also 16. The truss therefore satisfies the plane-truss determinacy count exactly. Since the question additionally states that the geometry is stable and properly triangulated, Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A plane pin-jointed truss has 8 joints, 12 members, one pin support and one roller support. Assuming a generic geometry with no special stabilizing constraint, what does the member-and-reaction count indicate?
Explanation:
The pin and roller provide three reaction components, so the total number of unknown forces is 12 + 3 = 15. Eight plane joints provide 2 × 8 = 16 independent joint-equilibrium equations in the counting argument. The force-unknown count is therefore short by one relative to that required for a stable determinate arrangement. Such a truss is described as deficient by one member under the usual generic-geometry criterion. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A geometrically stable plane truss has 8 joints, 14 members, one pin support and one roller support. What is its degree of static indeterminacy according to the standard count?
Explanation:
The two supports provide three independent external reaction components. The total number of unknown member and reaction forces is therefore 14 + 3 = 17. The eight joints provide 2j = 16 scalar equilibrium equations. The excess number of unknown forces is 17 - 16 = 1, provided no mechanism is present. Hence the truss is statically indeterminate to degree one and Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
At an unloaded joint J of a plane truss, J is not a support and exactly two members JK and JL meet. The two members are not collinear. What can be concluded from joint equilibrium?
Explanation:
No external force or support reaction acts at joint J. Only the axial forces in members JK and JL can therefore act on the isolated joint. Because their lines of action are non-collinear, two nonzero forces cannot balance one another. Equilibrium is possible only if both axial member forces are zero. Therefore Option A is the standard two-member zero-force rule.
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Question 7
ప్రశ్న 7
At an unloaded, unsupported truss joint J, three members JK, JL and JM meet. Members JK and JL are collinear while JM is non-collinear with them. Which conclusion follows directly from equilibrium?
Explanation:
The forces in the collinear members JK and JL act along one common straight line. Any nonzero force in JM would have a component perpendicular to that line. There is no external load or support reaction at J to balance such a perpendicular component. Consequently the force in the non-collinear member JM must be zero. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
After the support reactions of a statically determinate plane truss have been found, which joint is usually the most convenient starting point for the method of joints?
Explanation:
An isolated joint in a plane truss is a concurrent-force system. Only the two equations ΣFx = 0 and ΣFy = 0 are independently available at that joint. A joint with no more than two unknown member forces can therefore normally be solved directly. Starting with more unknowns than available equations prevents an immediate solution unless additional information is already known. Hence Option A describes the usual method-of-joints strategy.
వివరణ:
Question 9
ప్రశ్న 9
Which statement best describes efficient use of the method of sections for a statically determinate plane truss?
Explanation:
The method of sections isolates a finite portion of the truss as a rigid body. A plane rigid body provides the three equilibrium equations ΣFx = 0, ΣFy = 0 and ΣM = 0. A section passing through at most three unknown member forces is therefore especially convenient. Suitable moment centres can often eliminate two cut-member forces and give the third force directly. Thus Option A correctly describes the standard method-of-sections procedure.
వివరణ:
Question 10
ప్రశ్న 10
In a method-of-joints calculation, every unknown member force is initially assumed to act away from the isolated joint. After solving the equilibrium equations, one member force is obtained as -18 kN. How should it be interpreted?
Explanation:
A force drawn away from a joint corresponds to an assumed tensile member force. A positive result would therefore confirm tension under this convention. A negative result means that the actual member force acts opposite to the assumed tensile direction. The actual action is consequently compression with the magnitude of 18 kN. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A triangular truss has joints A(0,0), B(4,0) and C(2,3) in metres. Members are AB, AC and BC. A is a pin support, B is a roller providing only a vertical reaction, and a 12 kN downward load acts at C. What are the support reactions?
Explanation:
There is no external horizontal load, so the horizontal reaction at A is zero. The vertical load is applied at the midspan coordinate x = 2 m between supports separated by 4 m. Taking moments about A gives By × 4 = 12 × 2, so By = 6 kN upward. Vertical equilibrium then gives Ay = 12 - 6 = 6 kN upward. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A triangular truss has joints A(0,0), B(4,0) and C(2,3) in metres, with members AB, AC and BC. A is pinned, B is a vertical roller, and a 12 kN downward load acts at C. What are the forces in the two inclined members AC and BC?
Explanation:
The truss and loading are symmetric, so AC and BC carry equal axial forces. Each inclined member has length √13 m and a vertical direction cosine of 3/√13. At joint C, the two upward components required to balance 12 kN give 2F(3/√13) = 12 in magnitude. This gives F = 2√13 ≈ 7.21 kN, and the members must push on joint C rather than pull away from it. Therefore both inclined members are in compression and Option A is correct.
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Question 13
ప్రశ్న 13
A triangular truss has A(0,0), B(4,0) and C(2,3) in metres, with members AB, AC and BC. A is a pin, B is a vertical roller, and a 12 kN downward load acts at C. What is the force in member AB?
Explanation:
The vertical support reactions are 6 kN at A and 6 kN at B. Member AC has already been found to carry 7.21 kN compression. Its horizontal component at joint A has magnitude 7.21(2/√13) = 4 kN and acts toward the left. Horizontal equilibrium at A therefore requires member AB to pull 4 kN toward the right. Hence AB carries 4.00 kN tension and Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
A plane truss has joints A(0,0), B(2,0), C(4,0) and D(2,3) in metres. Its members are AB, BC, AD, CD and BD. A is pinned, C is a vertical roller, and a 10 kN downward load acts at D. Which member is a zero-force member?
Explanation:
At joint B, members AB and BC are collinear along the horizontal line AC. Member BD is the only non-collinear member meeting at B. No external load or support reaction acts at B. The standard three-member zero-force rule therefore requires the force in BD to be zero. Hence Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
A plane truss has A(0,0), B(2,0), C(4,0) and D(2,3) in metres, with members AB, BC, AD, CD and BD. A is pinned, C is a vertical roller, and a 10 kN downward load acts at D. Which set of member forces is correct?
Explanation:
Symmetry gives vertical reactions of 5 kN at A and C, while BD is zero by the unloaded-joint rule at B. Each inclined member has length √13 m and vertical direction cosine 3/√13. Vertical equilibrium gives an inclined-member compression magnitude of 5√13/3 ≈ 6.01 kN. Its horizontal component is 6.01(2/√13) = 3.33 kN, requiring equal tensile forces in AB and BC. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
A plane truss has joints A(0,0), B(3,0), C(6,0) and D(3,4) in metres. Members are AB, BC, AD, CD and BD. A is a pin support, C is a vertical roller, and joint D carries a 6 kN horizontal load to the right together with a 20 kN downward load. What are the support reactions?
Explanation:
Horizontal equilibrium requires the pin reaction at A to oppose the 6 kN rightward load, so Ax is 6 kN to the left. Taking moments about A includes the clockwise moment 6 × 4 from the horizontal load acting 4 m above A. Thus 6Cy - 20 × 3 - 6 × 4 = 0, giving Cy = 14 kN upward. Vertical equilibrium then gives Ay + 14 - 20 = 0, so Ay = 6 kN upward. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
A plane truss has A(0,0), B(3,0), C(6,0) and D(3,4) in metres, with members AB, BC, AD, CD and BD. A is pinned, C is a vertical roller, and D carries 6 kN to the right and 20 kN downward. What are the forces in AD, CD and BD?
Explanation:
Joint B is unloaded and contains collinear members AB and BC together with non-collinear member BD, so BD is a zero-force member. The support reactions are Ax = 6 kN left, Ay = 6 kN upward and Cy = 14 kN upward. Members AD and CD are 3-4-5 diagonals, so their vertical direction cosine is 4/5. Joint A equilibrium gives an AD compression magnitude of 6/(4/5) = 7.5 kN, while joint C gives a CD compression magnitude of 14/(4/5) = 17.5 kN. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
A plane truss has A(0,0), B(3,0), C(6,0) and D(3,4) in metres. Members are AB, BC, AD, CD and BD. A is pinned, C is a vertical roller, and D carries 6 kN rightward and 20 kN downward. What are the axial forces in AB and BC?
Explanation:
Member AD carries 7.5 kN compression and has a horizontal component of 7.5(3/5) = 4.5 kN at joint A. Together with the 6 kN leftward horizontal support reaction, joint A requires AB to provide 10.5 kN toward the right. Thus AB is in tension with magnitude 10.5 kN. The same equilibrium check at joint C with CD = 17.5 kN compression gives a 10.5 kN horizontal component that requires BC to be in tension. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
A determinate truss has joints A(0,0), B(4,0), C(8,0), D(2,3) and E(6,3) in metres. Its members are AB, BC, AD, DB, BE, EC and DE. A is a pin, C is a vertical roller, and downward loads of 10 kN and 20 kN act at D and E respectively. What are the vertical support reactions?
Explanation:
There is no horizontal external loading, so the horizontal reaction at A is zero. Taking moments about A gives 8Cy = 10 × 2 + 20 × 6. Therefore Cy = 140/8 = 17.5 kN upward. Vertical equilibrium then gives Ay = 30 - 17.5 = 12.5 kN upward. Hence Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
A determinate truss has A(0,0), B(4,0), C(8,0), D(2,3) and E(6,3) in metres, with members AB, BC, AD, DB, BE, EC and DE. A is pinned, C is a vertical roller, and 10 kN and 20 kN downward loads act at D and E. A section cuts members AB, DB and DE, and the left portion containing A and D is considered. What is the force in member DE?
Explanation:
The support reaction at A is 12.5 kN upward. For the cut left portion, the lines of action of cut-member forces AB and DB intersect at joint B. Taking moments about B therefore eliminates both of those unknown cut forces. Moment equilibrium gives -12.5(4) + 10(2) - 3FDE = 0 when FDE is initially assumed tensile on the left section. This gives FDE = -10 kN, so member DE carries 10 kN compression and Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
A determinate truss has A(0,0), B(4,0), C(8,0), D(2,3) and E(6,3) in metres. Members are AB, BC, AD, DB, BE, EC and DE. A is a pin, C is a vertical roller, and downward loads of 10 kN at D and 20 kN at E are applied. What are the forces in bottom-chord members AB and BC?
Explanation:
The vertical reactions are 12.5 kN at A and 17.5 kN at C. At joint A, member AD has direction cosines 2/√13 and 3/√13, and vertical equilibrium gives AD = 15.02 kN compression. Its horizontal component is 15.02(2/√13) = 8.33 kN, so AB must carry 8.33 kN tension. A corresponding calculation at joint C gives EC = 21.03 kN compression and a horizontal component of 11.67 kN, requiring BC to carry 11.67 kN tension. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A determinate truss has A(0,0), B(4,0), C(8,0), D(2,3) and E(6,3) in metres. Its members are AB, BC, AD, DB, BE, EC and DE. A is pinned, C is a vertical roller, and downward loads of 10 kN at D and 20 kN at E act on the truss. Which pair correctly gives the forces in diagonal members DB and BE?
Explanation:
The previously determined member forces include AD = 15.02 kN compression and DE = 10 kN compression. Applying equilibrium at joint D leaves DB as the remaining unknown member force. Its resolved components give DB = 3.0046 kN in tension. Applying equilibrium at joint E similarly gives BE = -3.0046 kN when tension is assumed positive. Thus BE carries 3.00 kN compression while DB carries 3.00 kN tension, making Option A correct.
వివరణ:
Question 23
ప్రశ్న 23
A determinate truss has joints A(0,0), B(4,0), C(8,0), D(0,3), E(4,3) and F(8,3) in metres. Members are AB, BC, DE, EF, AD, BE, CF, DB and BF. A is a pin, C is a vertical roller, and downward loads of 12 kN at E and 8 kN at F are applied. What is the force in diagonal member DB?
Explanation:
Whole-truss equilibrium gives Ay = 6 kN upward and Cy = 14 kN upward. At joint A, vertical equilibrium requires member AD to carry 6 kN compression, while AB is zero. At joint D, member DE is horizontal and member DB has a 3-4-5 geometry from D(0,3) to B(4,0). Vertical equilibrium requires the downward component of DB to equal 6 kN, so FDB(3/5) = 6 and FDB = 10 kN. The force acts in the initially assumed tensile direction, so Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
A triangular truss has joints A(0,0), B(6,0) and C(2,3) in metres. Members are AB, AC and BC. A is a pin support, B is a vertical roller, and a 15 kN downward load acts at C. What are the support reactions?
Explanation:
No horizontal external force acts on the truss, so Ax is zero. Taking moments about A gives By × 6 = 15 × 2. Therefore By = 5 kN upward. Vertical equilibrium then requires Ay = 15 - 5 = 10 kN upward. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
A triangular truss has A(0,0), B(6,0) and C(2,3) in metres, with members AB, AC and BC. A is pinned, B is a vertical roller, and a 15 kN downward load acts at C. Which set of member forces is correct?
Explanation:
The support reactions are Ay = 10 kN upward and By = 5 kN upward. Member AC has length √13 m, while BC has length 5 m because its horizontal and vertical projections are 4 m and 3 m. Solving joint C equilibrium gives FAC = -12.0185 kN and FBC = -8.3333 kN when tension is assumed positive. Joint A horizontal equilibrium then gives FAB = +6.6667 kN. Therefore AC and BC are in compression, AB is in tension, and Option A is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Members are straight two-force members connected by ideal pins, and external loads and reactions act at the joints. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. m + r = 2j సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. Statically determinate or perfect by count సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. It is deficient by one member and is kinematically unstable by the count criterion. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. One సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. Both JK and JL are zero-force members. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. JM is a zero-force member. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. A joint having no more than two unknown member forces సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. Cut through the desired members and preferably through no more than three unknown member forces, then apply planar rigid-body equilibrium to one side of the cut. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. 18 kN compression సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. Ax = 0, Ay = 6 kN upward and By = 6 kN upward సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. 7.21 kN compression in each సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: A. 4.00 kN tension సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: A. BD సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: A. AD = CD = 6.01 kN compression, AB = BC = 3.33 kN tension, BD = 0 సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. Ax = 6 kN left, Ay = 6 kN upward and Cy = 14 kN upward సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: A. AD = 7.5 kN compression, CD = 17.5 kN compression and BD = 0 సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: A. AB = BC = 10.5 kN tension సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: A. Ay = 12.5 kN and Cy = 17.5 kN సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: A. 10 kN compression సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. AB = 8.33 kN tension and BC = 11.67 kN tension సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. DB = 3.00 kN tension and BE = 3.00 kN compression సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. 10 kN tension సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. Ax = 0, Ay = 10 kN upward and By = 5 kN upward సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: A. AC = 12.02 kN compression, BC = 8.33 kN compression and AB = 6.67 kN tension సమాధానం: A.
