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Question 1
ప్రశ్న 1
A stepped member consists of several prismatic segments connected in series and remains within the linear-elastic range. If segment i carries constant axial force Ni over length Li and has area Ai and Young's modulus Ei, which expression gives the algebraic total axial deformation?
Explanation:
The axial deformation of an individual prismatic segment is δi = NiLi/(AiEi) within the linear-elastic range. A stepped member is treated segment by segment whenever axial force, cross-sectional area or Young's modulus changes. The total end-to-end deformation is the algebraic sum of the individual segment deformations. Tensile segment deformation is taken as positive and compressive segment deformation as negative when that sign convention is used. Therefore the required expression is Σ(NiLi)/(AiEi), making Option B correct.
వివరణ:
Question 2
ప్రశ్న 2
A uniform steel bar 2 m long and 500 mm² in area carries a tensile load of 100 kN. If E = 200 GPa, what is its elastic elongation?
Explanation:
For a uniform prismatic bar under a constant axial load, δ = PL/(AE). Here P = 100000 N, L = 2000 mm, A = 500 mm² and E = 200000 N/mm². Substitution gives δ = (100000 × 2000)/(500 × 200000). The resulting elongation is 2.0 mm. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
A stepped steel bar carries a constant tensile force of 80 kN. One segment is 800 mm long with area 400 mm² and the second is 1200 mm long with area 800 mm². If E = 200 GPa throughout, what is the total elongation?
Explanation:
Because no intermediate axial load is specified, both segments carry the same tensile force of 80 kN. The first segment elongates by 80000 × 800/(400 × 200000) = 0.80 mm. The second segment elongates by 80000 × 1200/(800 × 200000) = 0.60 mm. The total elongation is the sum, 0.80 + 0.60 = 1.40 mm. Hence Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
A two-segment bar carries a constant tensile force of 60 kN. The first segment is steel with L = 1000 mm, A = 500 mm² and E = 200 GPa, while the second is aluminium with L = 800 mm, A = 800 mm² and E = 70 GPa. What is the total elongation?
Explanation:
The steel segment elongates by 60000 × 1000/(500 × 200000) = 0.60 mm. The aluminium segment elongates by 60000 × 800/(800 × 70000) = approximately 0.857 mm. Both segments are in tension, so their elongations have the same positive sign. The total elongation is approximately 0.60 + 0.857 = 1.457 mm. Rounded suitably, this is 1.46 mm and Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A stepped bar ABC is fixed at A. At B, a 30 kN axial load acts toward A, while at free end C a 90 kN load acts away from A. Segment AB has L = 1000 mm and A = 600 mm², while BC has L = 800 mm and A = 400 mm². If E = 200 GPa throughout, what is the displacement of C relative to A?
Explanation:
A cut through BC shows that segment BC carries 90 kN tension because the 90 kN end load lies to its right. A cut through AB includes both applied loads, giving a net tensile internal force of 90 - 30 = 60 kN. The elongation of AB is 60000 × 1000/(600 × 200000) = 0.50 mm, while that of BC is 90000 × 800/(400 × 200000) = 0.90 mm. Both contributions are tensile, so the total displacement of C is 0.50 + 0.90 = 1.40 mm away from A. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
A bar ABC is fixed at A. Axial loads of 40 kN and 70 kN act to the right at B and C respectively. The areas of segments AB and BC are 550 mm² and 350 mm² respectively. What are the average normal stresses in AB and BC?
Explanation:
Segment BC carries only the 70 kN end load, so its tensile stress is 70000/350 = 200 MPa. Segment AB must transmit both external loads to the support, so its internal tensile force is 40 + 70 = 110 kN. Its average stress is therefore 110000/550 = 200 MPa. Equal stress here results from the particular combination of forces and areas rather than from a general rule for stepped bars. Hence Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
A bar ABC is fixed at A. A load of 80 kN acts to the right at B and a load of 30 kN acts to the left at free end C. Segment AB has L = 1200 mm and A = 500 mm², while BC has L = 800 mm and A = 400 mm². For E = 200 GPa, what is the algebraic displacement of C, taking elongation as positive?
Explanation:
Equilibrium of segment BC shows that its internal force is 30 kN compression because the end load at C acts toward the bar. Segment AB carries a net force of 80 - 30 = 50 kN in tension. Its elongation is 50000 × 1200/(500 × 200000) = +0.60 mm, while BC shortens by 30000 × 800/(400 × 200000) = 0.30 mm. The algebraic displacement of C is therefore +0.60 - 0.30 = +0.30 mm. Hence Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
Two adjacent segments of an axially loaded member undergo opposite deformations. Segment AB carries 80 kN tension and has L = 1000 mm, A = 500 mm² and E = 200 GPa. Segment BC carries 40 kN compression, has L = 800 mm and the same E. What area must BC have so that the total end-to-end deformation is zero?
Explanation:
The elongation of AB is 80000 × 1000/(500 × 200000) = 0.80 mm. For zero total deformation, BC must shorten by exactly 0.80 mm. Using 0.80 = 40000 × 800/(A × 200000) gives A = 200 mm². The tensile and compressive deformations then have equal magnitudes and opposite signs. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
Two prismatic bars have the same length and carry the same tensile load. Bar 1 has A = 400 mm² and E = 200 GPa, while Bar 2 has A = 800 mm² and E = 100 GPa. Which statement concerning their elastic elongations is correct?
Explanation:
For equal load and length, axial deformation is inversely proportional to the axial rigidity AE. For Bar 1, AE is proportional to 400 × 200 = 80000 in consistent relative units. For Bar 2, AE is also proportional to 800 × 100 = 80000. Their axial rigidities are equal, so their elastic elongations are equal. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
A stepped bar ABC is fixed at A. A 20 kN load acts toward A at B, and an 80 kN load acts away from A at C. Segment AB has L = 1000 mm and A = 600 mm², while BC has L = 800 mm and A = 400 mm². If E = 200 GPa, what are the displacements of B and C relative to A?
Explanation:
Segment BC carries 80 kN tension because the 80 kN end load acts at C. Segment AB carries a net tensile force of 80 - 20 = 60 kN. Its elongation is 60000 × 1000/(600 × 200000) = 0.50 mm, which is the displacement of B. Segment BC elongates by 80000 × 800/(400 × 200000) = 0.80 mm, so C moves by 0.50 + 0.80 = 1.30 mm. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A two-segment member carries the same tensile load P throughout. Segment 1 has L = 1000 mm, A = 400 mm² and E = 200 GPa. Segment 2 has L = 800 mm, A = 600 mm² and E = 70 GPa. What value of P will produce a total elongation of 2.0 mm?
Explanation:
For the two segments in series, δ = P[L1/(A1E1) + L2/(A2E2)]. The combined compliance is 1000/(400 × 200000) + 800/(600 × 70000) mm/N. Setting the resulting deformation equal to 2.0 mm and solving gives P approximately equal to 63396 N. This is approximately 63.4 kN. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A two-segment bar carries a constant tensile force of 50 kN. Segment 1 has L = 800 mm, A = 400 mm² and E = 200 GPa. Segment 2 has L = 1000 mm, A = 500 mm² and unknown Young's modulus E2. If the total elongation is 1.50 mm, what is E2?
Explanation:
The elongation of Segment 1 is 50000 × 800/(400 × 200000) = 0.50 mm. Since the total elongation is 1.50 mm, Segment 2 must elongate by 1.00 mm. Using 1.00 = 50000 × 1000/(500 × E2) gives E2 = 100000 N/mm². This is equal to 100 GPa. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
Two steel segments are connected in series and carry the same axial load. Segment 1 has L = 1000 mm and A = 400 mm², while Segment 2 has L = 500 mm and A = 800 mm². If E = 200 GPa, what is the equivalent axial stiffness P/δ of the complete member?
Explanation:
For segments in series, the total axial compliance is the sum L1/(A1E) + L2/(A2E). Substitution gives 1000/(400 × 200000) + 500/(800 × 200000) = 1.5625 × 10^-5 mm/N. The equivalent axial stiffness is the reciprocal of this compliance. Therefore P/δ = 64000 N/mm = 64 kN/mm. Hence Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
Every cross-sectional area of a stepped linear-elastic bar is doubled, while its segment lengths, Young's moduli and internal axial forces remain unchanged. What happens to its total axial deformation?
Explanation:
Each segment deformation is NiLi/(AiEi). Doubling every Ai while keeping Ni, Li and Ei unchanged halves the deformation of every segment. Since the total deformation is the algebraic sum of those segment deformations, the complete member deformation is also halved. This result assumes the internal force distribution itself is not altered by the stated change in areas. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
A straight bar is subjected only to concentrated axial loads and has no distributed axial loading between them. Which statement about the internal axial force is correct?
Explanation:
Internal axial force is determined from equilibrium of a free body obtained by cutting the member. If no axial load is distributed along a segment, the internal axial force remains constant throughout that segment. Crossing a concentrated axial load changes the force resultant by the magnitude and sign required by equilibrium. A change in cross-sectional area changes stress and deformation but does not by itself create an axial-force jump. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
At an abrupt change in cross-section of an axially loaded stepped bar, how should the elementary stress expression P/A be interpreted?
Explanation:
The elementary expression P/A gives the nominal average normal stress over a cross-section carrying an axial resultant P. An abrupt shoulder or change in geometry creates a local non-uniform stress field and a stress concentration. Consequently P/A should not be interpreted as the exact maximum pointwise stress immediately at that discontinuity. Away from the localized disturbance, the simple axial-stress model is appropriate for the prismatic portions under the usual assumptions. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
Match the axial-member quantity with its appropriate expression for a prismatic elastic segment:
1. Average normal stress
2. Axial rigidity
3. Axial compliance of a segment
4. Segment deformation
P. AE
Q. N/A
R. L/(AE)
S. NL/(AE)
Which matching is correct?
1. Average normal stress
2. Axial rigidity
3. Axial compliance of a segment
4. Segment deformation
P. AE
Q. N/A
R. L/(AE)
S. NL/(AE)
Which matching is correct?
Explanation:
Average normal stress is the axial force divided by area, so item 1 matches Q. Axial rigidity is the product AE, so item 2 matches P. The elastic axial compliance of a prismatic segment is L/(AE), so item 3 matches R. Multiplying that compliance by the internal axial force gives deformation NL/(AE), so item 4 matches S. Therefore Option A gives the correct complete matching.
వివరణ:
Question 18
ప్రశ్న 18
Consider the following statements about stepped axially loaded members:
1. With tension taken positive, the sign of NL/(AE) follows the sign of internal axial force N when A, E and L are positive.
2. Adjacent segments necessarily carry the same internal axial force even when a concentrated axial load acts at their junction.
3. For fixed N and L, increasing AE reduces the magnitude of elastic axial deformation.
4. Total end-to-end deformation is obtained by the algebraic sum of segment deformations.
Which statements are correct?
1. With tension taken positive, the sign of NL/(AE) follows the sign of internal axial force N when A, E and L are positive.
2. Adjacent segments necessarily carry the same internal axial force even when a concentrated axial load acts at their junction.
3. For fixed N and L, increasing AE reduces the magnitude of elastic axial deformation.
4. Total end-to-end deformation is obtained by the algebraic sum of segment deformations.
Which statements are correct?
Explanation:
Statement 1 is correct because positive tensile force produces positive elongation under the stated convention. Statement 2 is incorrect because a concentrated axial load at a junction generally causes a jump in the internal axial-force diagram. Statement 3 is correct because deformation magnitude varies inversely with axial rigidity AE for fixed N and L. Statement 4 is correct because compatible end-to-end deformation is obtained from the algebraic sum of all segment deformations. Therefore Statements 1, 3 and 4 are correct and Option B is the unique answer.
వివరణ:
Question 19
ప్రశ్న 19
A stepped circular steel bar carries a constant tensile load of 60 kN. The first segment is 500 mm long and 20 mm in diameter, while the second is 1000 mm long and 30 mm in diameter. If E = 200 GPa, what is the total elongation approximately?
Explanation:
The area of the 20 mm diameter segment is π(20²)/4 = approximately 314.16 mm². The area of the 30 mm diameter segment is π(30²)/4 = approximately 706.86 mm². Applying δ = PL/(AE) to the two segments gives approximately 0.4775 mm and 0.4244 mm respectively. Their total elongation is therefore approximately 0.9019 mm. Rounded suitably, this is 0.902 mm and Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
Two segments connected in series carry the same axial tensile force P. Segment 1 has area 300 mm² and allowable tensile stress 150 MPa, while Segment 2 has area 500 mm² and allowable tensile stress 120 MPa. What is the maximum permissible value of P based on these stress limits?
Explanation:
Segment 1 permits a maximum force of 150 × 300 = 45000 N, or 45 kN. Segment 2 permits a maximum force of 120 × 500 = 60000 N, or 60 kN. Because the segments are in series and carry the same axial force under the stated loading, both stress restrictions must be satisfied simultaneously. The smaller allowable force therefore governs the complete member. Hence the maximum permissible load is 45 kN and Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
A steel bar and an aluminium bar have equal lengths and equal cross-sectional areas and carry the same tensile load. Take Esteel = 200 GPa and Ealuminium = 70 GPa. What is the ratio of aluminium elongation to steel elongation?
Explanation:
Axial deformation is δ = PL/(AE). Since P, L and A are equal for the two bars, deformation is inversely proportional to Young's modulus. Thus δaluminium/δsteel = Esteel/Ealuminium = 200/70. This ratio is approximately 2.857. Therefore the aluminium bar elongates about 2.86 times as much and Option D is correct.
వివరణ:
Question 22
ప్రశ్న 22
A three-segment bar ABCD is fixed at A. Loads of 20 kN to the right at B, 50 kN to the left at C and 100 kN to the right at D are applied. Segment AB has L = 600 mm, A = 350 mm² and E = 200 GPa; BC has L = 700 mm, A = 500 mm² and E = 70 GPa; CD has L = 800 mm, A = 400 mm² and E = 200 GPa. What is the displacement of D relative to A?
Explanation:
Starting from the free end, segment CD carries 100 kN tension. Segment BC carries the net load 100 - 50 = 50 kN tension, while AB carries 100 - 50 + 20 = 70 kN tension. Their respective elongations are 70000 × 600/(350 × 200000) = 0.60 mm, 50000 × 700/(500 × 70000) = 1.00 mm, and 100000 × 800/(400 × 200000) = 1.00 mm. All three are tensile, so the displacement of D is 0.60 + 1.00 + 1.00 = 2.60 mm. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
Two segments connected in series carry the same tensile force. A steel segment has L = 1000 mm, A = 500 mm² and E = 200 GPa. An aluminium segment has L = 600 mm and E = 70 GPa. What aluminium area will make the elongations of the two segments equal?
Explanation:
For equal axial force, the two segment deformations are equal when Ls/(AsEs) = La/(AaEa). Substitution gives 1000/(500 × 200000) = 600/(Aa × 70000). Solving gives Aa = approximately 857.14 mm². The larger aluminium area compensates for its lower Young's modulus and the different segment length. Therefore Option D is correct.
వివరణ:
Question 24
ప్రశ్న 24
A stepped axial member experiences both tensile and compressive segment forces because of multiple applied axial loads. If the distance between its two end points is required to remain unchanged, which compatibility condition must be satisfied within the small-deformation linear-elastic model?
Explanation:
Compatibility concerns the actual relative displacement of connected points rather than merely the magnitudes of individual deformations. Tensile segments contribute positive elongation and compressive segments contribute negative shortening under the stated sign convention. If the overall end-to-end length is unchanged, these signed contributions must cancel. Thus the required condition is Σ(NiLi/AiEi) = 0 for the linear-elastic segmented model. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements for a linearly elastic stepped bar subjected to concentrated axial loads:
1. The internal axial force in any segment can be determined by equilibrium of either portion obtained by cutting that segment.
2. Total axial deformation is the algebraic sum of the individual segment deformations.
3. A compressive deformation in one segment can partially or completely offset tensile elongation in another segment.
4. The internal axial force in every segment must equal the load applied at the free end even when additional concentrated loads act at intermediate points.
Which statements are correct?
1. The internal axial force in any segment can be determined by equilibrium of either portion obtained by cutting that segment.
2. Total axial deformation is the algebraic sum of the individual segment deformations.
3. A compressive deformation in one segment can partially or completely offset tensile elongation in another segment.
4. The internal axial force in every segment must equal the load applied at the free end even when additional concentrated loads act at intermediate points.
Which statements are correct?
Explanation:
Statement 1 is correct because segment force follows directly from axial equilibrium of a suitable free body. Statement 2 is correct because connected segment deformations combine algebraically to give the relative displacement of the member ends. Statement 3 is correct because tension and compression contribute deformations of opposite signs under a consistent sign convention. Statement 4 is incorrect because intermediate concentrated axial loads cause the internal axial force to change from one segment to another. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Σ(NiLi)/(AiEi) సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. 2.0 mm సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: C. 1.40 mm సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: B. 1.46 mm సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 1.40 mm elongation సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. AB = 200 MPa tension and BC = 200 MPa tension సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. +0.30 mm సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. 200 mm² సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. Both bars have the same elongation. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. uB = 0.50 mm and uC = 1.30 mm సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. 63.4 kN సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 100 GPa సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 64 kN/mm సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: B. It becomes one-half. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: B. It is constant between adjacent concentrated-load points and changes when an axial concentrated load is crossed. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: B. It gives a nominal average normal stress away from the local stress-concentration region. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. 1-Q, 2-P, 3-R, 4-S సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. 1, 3 and 4 only సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 0.902 mm సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. 45 kN సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: D. 2.86 సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: C. 2.6 mm elongation సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: D. 857 mm² సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. The algebraic sum Σ(NiLi/AiEi) must be zero. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 1, 2 and 3 only సమాధానం: B.
