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Question 1
ప్రశ్న 1
A compound bar consists of two different materials of equal length, perfectly connected so that they deform together under an axial load. Which pair of conditions forms the basic basis for determining the load shared by the two materials?
Explanation:
A compound bar must first satisfy equilibrium, so the sum of the axial forces carried by its constituents equals the externally applied axial load. The connection between the constituent materials also imposes a deformation compatibility condition. For equal-length constituents that deform together without slip, their axial deformations and hence their longitudinal strains are equal. Their stresses need not be equal because stress depends on Young's modulus through Hooke's law. Therefore equilibrium together with compatibility provides the basic equations required to determine load sharing.
వివరణ:
Question 2
ప్రశ్న 2
A steel bar and an aluminium bar of equal length are rigidly connected in parallel between common end plates and remain elastic. If their Young's moduli are 200 GPa and 80 GPa respectively, what is the ratio of steel stress to aluminium stress?
Explanation:
Equal-length members attached to common rigid end plates undergo the same axial deformation. Their longitudinal strains are therefore equal. Within the linear-elastic range, stress equals Young's modulus multiplied by strain. Hence the stress ratio is σs/σa = Es/Ea = 200/80 = 2.50. The cross-sectional areas affect the forces carried by the members but not this stress ratio under equal-strain compatibility.
వివరణ:
Question 3
ప్రశ్న 3
A steel member of area 600 mm² and E = 200 GPa and a brass member of area 900 mm² and E = 100 GPa are of equal length and rigidly connected in parallel. If they jointly carry an axial tensile load of 180 kN, what loads are carried approximately by the steel and brass members?
Explanation:
Because the two members have equal length and compatible deformation, they have the same axial strain. Their loads are therefore proportional to their axial rigidities AE. The steel rigidity is 600 × 200000 = 120 × 10^6 N, while the brass rigidity is 900 × 100000 = 90 × 10^6 N. Thus the steel carries 180 × 120/(120 + 90) = approximately 102.86 kN and the brass carries approximately 77.14 kN. The two component loads add to the applied load of 180 kN, satisfying equilibrium.
వివరణ:
Question 4
ప్రశ్న 4
A compound member 1500 mm long consists of a steel component of area 500 mm² and E = 200 GPa together with a brass component of area 700 mm² and E = 100 GPa. The equal-length components are rigidly connected and jointly carry 170 kN tension. What is the elongation of the compound member?
Explanation:
For equal-length components with common deformation, the combined axial rigidity is ΣAE. Here ΣAE = 500 × 200000 + 700 × 100000 = 170 × 10^6 N. The common strain is therefore P/ΣAE = 170000/(170 × 10^6) = 0.001. The elongation is strain multiplied by length, giving 0.001 × 1500 = 1.50 mm. Hence Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
A steel component of area 400 mm² and E = 200 GPa and an aluminium component of area 600 mm² and E = 70 GPa are perfectly bonded over the same length. If the compound member carries 122 kN tension, what are the stresses in steel and aluminium?
Explanation:
Perfect bonding and equal length impose the same longitudinal strain on the steel and aluminium. The combined rigidity is 400 × 200000 + 600 × 70000 = 122 × 10^6 N. Therefore the common strain is 122000/(122 × 10^6) = 0.001. The stresses are Esε = 200000 × 0.001 = 200 MPa and Eaε = 70000 × 0.001 = 70 MPa. The corresponding forces are 80 kN and 42 kN, which sum to the applied 122 kN.
వివరణ:
Question 6
ప్రశ్న 6
A compound member consists of a steel element having A = 500 mm² and E = 200 GPa and a brass element having A = 800 mm² and E = 100 GPa, both of equal length and connected to deform together. What is the equivalent axial rigidity ΣAE of the compound section?
Explanation:
The axial rigidity of a constituent is the product AE. The steel contribution is 500 × 200000 = 100 × 10^6 N. The brass contribution is 800 × 100000 = 80 × 10^6 N. Because the members act in parallel with compatible deformation, their axial rigidities add, giving 180 × 10^6 N. This quantity is axial rigidity and should be distinguished from member stiffness, which would additionally involve division by length.
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Question 7
ప్రశ్న 7
Two materials of equal length are perfectly bonded to form a compound bar and remain linearly elastic. Their Young's moduli are E1 = 200 GPa and E2 = 80 GPa. If the common longitudinal strain is ε, which stress relationship is correct?
Explanation:
Perfect bonding of equal-length constituents produces equal axial strain. Under linear elasticity, σ1 = E1ε and σ2 = E2ε. Dividing the two expressions gives σ1/σ2 = E1/E2. Substitution gives 200/80 = 2.5. Therefore Material 1 carries 2.5 times the stress of Material 2 even though their strains are identical.
వివరణ:
Question 8
ప్రశ్న 8
A steel member of area 300 mm² and E = 200 GPa is to be connected in parallel with an aluminium member of E = 75 GPa and the same length. What aluminium area is required if the two members are to carry equal axial loads?
Explanation:
Equal-length parallel members undergo equal strain when attached to common rigid ends. The force in each member is P = AEε. Equal force sharing therefore requires AsEs = AaEa. Thus Aa = 300 × 200/75 = 800 mm². The lower Young's modulus of aluminium requires the larger cross-sectional area to provide the same axial rigidity as the steel member.
వివరణ:
Question 9
ప్రశ్న 9
Three equal-length members are rigidly connected in parallel: steel with A = 400 mm² and E = 200 GPa, brass with A = 600 mm² and E = 100 GPa, and aluminium with A = 800 mm² and E = 75 GPa. If the total tensile load is 200 kN, what load is carried by the steel member?
Explanation:
The axial rigidities AE of the steel, brass and aluminium members are 80, 60 and 60 million newtons respectively. Their total axial rigidity is therefore 200 million newtons. The steel fraction of the applied load is 80/200. Hence the steel load is 200 × 80/200 = 80 kN. The remaining 120 kN is shared equally by the brass and aluminium members because their axial rigidities are equal.
వివరణ:
Question 10
ప్రశ్న 10
Two axial rods are attached to the same rigid loading plate, while their opposite ends are fixed to separate supports so that their original lengths may differ. The steel rod has L = 1000 mm, A = 500 mm² and E = 200 GPa, while the aluminium rod has L = 500 mm, A = 500 mm² and E = 70 GPa. If a 170 kN tensile load causes the rigid plate to translate axially without rotation, what are the plate displacement and respective rod forces?
Explanation:
The translating rigid plate imposes the same axial end displacement on both rods even though their original lengths differ. Their individual stiffnesses are AE/L. The steel stiffness is 500 × 200000/1000 = 100000 N/mm and the aluminium stiffness is 500 × 70000/500 = 70000 N/mm. The combined stiffness is 170000 N/mm, so the displacement under 170 kN is 170000/170000 = 1.0 mm. The corresponding forces are 100000 × 1 = 100 kN in steel and 70000 × 1 = 70 kN in aluminium, so Option A is correct.
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Question 11
ప్రశ్న 11
Two elastic rods of different original lengths have one end attached to the same rigid loading plate and their opposite ends fixed to separate supports. If the plate translates axially without rotation and the rods remain aligned with the loading direction, which compatibility statement is generally correct?
Explanation:
The common rigid loading plate undergoes one axial translation, while the opposite end of each rod is fixed. Consequently each rod undergoes the same change in length equal to the plate displacement. Strain is deformation divided by original length, so rods having different original lengths generally develop different strains. Their stresses and forces then depend on their individual lengths, areas and elastic moduli. Therefore equal deformation rather than equal strain is the correct compatibility condition, making Option C correct.
వివరణ:
Question 12
ప్రశ్న 12
A steel rod and an aluminium rod are attached to the same rigid loading plate, with their opposite ends fixed to separate supports. The steel rod has L = 1200 mm, A = 400 mm² and E = 200 GPa, while the aluminium rod has L = 800 mm, A = 600 mm² and E = 70 GPa. If the rigid plate translates axially by 0.60 mm without rotation, what total axial force is required?
Explanation:
The rigid plate imposes the same 0.60 mm axial deformation on each rod. The steel force is AEδ/L = 400 × 200000 × 0.60/1200 = 40000 N, or 40 kN. The aluminium force is 600 × 70000 × 0.60/800 = 31500 N, or 31.5 kN. Equilibrium requires the applied plate force to equal the sum of the rod forces, giving 40 + 31.5 = 71.5 kN. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
Consider the following statements about an equal-length, perfectly bonded compound bar under concentric axial loading:
1. The constituent materials undergo the same longitudinal strain.
2. The total load equals the sum of the loads carried by the constituents.
3. The stresses in all constituent materials must be equal.
4. The load carried by a constituent is proportional to AE when all constituents have the same strain.
Which statements are correct?
1. The constituent materials undergo the same longitudinal strain.
2. The total load equals the sum of the loads carried by the constituents.
3. The stresses in all constituent materials must be equal.
4. The load carried by a constituent is proportional to AE when all constituents have the same strain.
Which statements are correct?
Explanation:
Statement 1 is correct because perfect bonding and equal length impose equal longitudinal strain. Statement 2 follows directly from axial-force equilibrium. Statement 3 is incorrect because equal strain gives σ = Eε, so different elastic moduli generally produce different stresses. Statement 4 is correct because Pi = AiEiε for each constituent. Therefore Statements 1, 2 and 4 are correct and Option B is the unique answer.
వివరణ:
Question 14
ప్రశ్న 14
Match the compound-member concept with the appropriate relationship:
1. Axial equilibrium
2. Equal-length compatibility
3. Load carried by constituent i
4. Equivalent axial rigidity
P. Pi = AiEiε
Q. ΣPi = P
R. ΣAiEi
S. ε1 = ε2 = ...
Which matching is correct?
1. Axial equilibrium
2. Equal-length compatibility
3. Load carried by constituent i
4. Equivalent axial rigidity
P. Pi = AiEiε
Q. ΣPi = P
R. ΣAiEi
S. ε1 = ε2 = ...
Which matching is correct?
Explanation:
Axial equilibrium requires the sum of constituent forces to equal the applied load, so item 1 matches Q. Equal-length compatible constituents have equal longitudinal strain, so item 2 matches S. Under linear elasticity, a constituent force is AiEiε, so item 3 matches P. The combined axial rigidity of parallel equal-length constituents is ΣAiEi, so item 4 matches R. Therefore Option A contains the correct complete matching.
వివరణ:
Question 15
ప్రశ్న 15
A fixed total cross-sectional area is available for an equal-length compound member made from two linearly elastic materials. If some area is transferred from the material having the higher Young's modulus to the material having the lower Young's modulus, with total area unchanged, what happens to the equivalent axial rigidity ΣAE?
Explanation:
Equivalent axial rigidity is obtained by summing AiEi over all constituents. An area assigned to a material with higher E contributes more axial rigidity than the same area assigned to a material with lower E. Transferring area from the stiffer material to the less stiff material therefore reduces ΣAE. The total geometric area remaining constant does not preserve elastic rigidity when the elastic moduli differ. Hence Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A steel member of area 300 mm² and E = 200 GPa is connected in parallel with an aluminium member of E = 80 GPa and the same length. If the steel member is required to carry exactly 60 percent of the total axial load, what aluminium area is required?
Explanation:
For equal-length compatible members, load is shared in proportion to AE. The steel axial rigidity is 300 × 200 = 60000 in consistent relative units. If steel carries 60 percent, the aluminium member must carry 40 percent, so the rigidity ratio steel to aluminium must be 60:40 or 3:2. The required aluminium rigidity is therefore 40000, giving Aa = 40000/80 = 500 mm². Thus Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A steel member with A = 500 mm², E = 200 GPa and allowable tensile stress 160 MPa is rigidly connected in parallel with an equal-length aluminium member having A = 800 mm², E = 70 GPa and allowable tensile stress 90 MPa. What is the maximum total tensile load the compound member can carry while both stress limits are satisfied?
Explanation:
Equal-length compatible constituents experience the same strain. The maximum strain allowed by steel is 160/200000 = 0.0008, while aluminium would permit 90/70000 = approximately 0.001286. Steel therefore governs and the common strain cannot exceed 0.0008. The total load at that strain is 0.0008[(500 × 200000) + (800 × 70000)] = 124800 N. Hence the maximum permissible total load is 124.8 kN and Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
An equal-length steel-brass compound bar has steel area 400 mm², Es = 200 GPa and allowable steel stress 150 MPa, while the brass area is 600 mm², Eb = 100 GPa and allowable brass stress 100 MPa. What maximum total tensile load is permitted?
Explanation:
Equal strain requires the stress ratio to equal the modulus ratio, so steel stress is twice the brass stress. The allowable strain for steel is 150/200000 = 0.00075, whereas brass permits 100/100000 = 0.001. Steel therefore reaches its allowable stress first. At ε = 0.00075, steel carries 400 × 150 = 60 kN and brass stress is 75 MPa, giving a brass force of 45 kN. The maximum total load is consequently 60 + 45 = 105 kN, so Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
An equal-length composite member consists of two parallel components whose axial rigidities AE are 80 × 10^6 N and 40 × 10^6 N. If the second component is removed and the same total axial load is then carried only by the first component, by what factor does the elastic deformation increase?
Explanation:
Initially the total axial rigidity is 80 + 40 = 120 × 10^6 N. For a given load and common length, deformation is inversely proportional to total axial rigidity. After removing the second component, only 80 × 10^6 N of axial rigidity remains. The deformation ratio is therefore 120/80 = 1.50. Thus the deformation becomes one and one-half times the original value, making Option B correct.
వివరణ:
Question 20
ప్రశ్న 20
A perfectly bonded equal-length steel-brass compound member experiences a measured tensile strain of 0.0006. Steel has A = 300 mm² and E = 200 GPa, while brass has A = 500 mm² and E = 100 GPa. What total tensile load is acting on the compound member?
Explanation:
The measured strain is common to both perfectly bonded equal-length constituents. The steel stress is 200000 × 0.0006 = 120 MPa, giving a steel force of 120 × 300 = 36 kN. The brass stress is 100000 × 0.0006 = 60 MPa, giving a brass force of 60 × 500 = 30 kN. Equilibrium gives a total load of 36 + 30 = 66 kN. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
In an equal-length perfectly bonded two-material compound bar under axial loading, can the stresses in the two materials generally be made equal merely by changing their cross-sectional areas while retaining different Young's moduli?
Explanation:
Equal-length perfectly bonded constituents undergo the same longitudinal strain. Their stresses are therefore σ1 = E1ε and σ2 = E2ε. Consequently the stress ratio is E1/E2 and does not contain either cross-sectional area. Changing area changes the amount of force carried by each constituent but not their stress ratio for a specified common strain. Thus unequal elastic moduli generally prevent equal stresses under this compatibility condition, making Option C correct.
వివరణ:
Question 22
ప్రశ్న 22
For an equal-length, perfectly bonded compound bar having total area At = ΣAi and common axial strain, which expression gives the equivalent Young's modulus Eeq of a homogeneous bar having the same total area and the same axial rigidity?
Explanation:
The actual compound section has equivalent axial rigidity ΣAiEi. A homogeneous replacement section of the same total area At would have axial rigidity EeqAt. Equating these two rigidities gives EeqAt = ΣAiEi. Therefore Eeq = Σ(AiEi)/At. This is an area-weighted elastic modulus for the specific equal-strain axial response, so Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A perfectly bonded compound bar contains 400 mm² of steel with E = 200 GPa and 600 mm² of aluminium with E = 70 GPa. What is the equivalent Young's modulus of a homogeneous bar having the same total cross-sectional area and axial rigidity?
Explanation:
The equivalent modulus for an equal-strain composite is ΣAiEi divided by the total area. The steel contribution is 400 × 200 = 80000 GPa·mm² in consistent relative units, while the aluminium contribution is 600 × 70 = 42000. Their sum is 122000 and the total area is 1000 mm². Therefore Eeq = 122000/1000 = 122 GPa. Hence Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
An equal-length three-material compound member 1200 mm long consists of steel with A = 300 mm² and E = 200 GPa, brass with A = 500 mm² and E = 100 GPa, and aluminium with A = 1200 mm² and E = 75 GPa. If the member carries 200 kN tension, what is its total elongation?
Explanation:
The respective axial-rigidity contributions are 60, 50 and 90 million newtons. Their total is therefore 200 × 10^6 N. The common strain is P/ΣAE = 200000/(200 × 10^6) = 0.001. Multiplying the strain by the common length of 1200 mm gives an elongation of 1.20 mm. Thus Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about axially loaded compound members:
1. Equal-length constituents connected so that no relative axial slip occurs have equal longitudinal strain.
2. For such constituents in the linear-elastic range, their stresses are proportional to their Young's moduli.
3. Their loads are necessarily equal regardless of cross-sectional area and modulus.
4. Parallel members of unequal original lengths attached to a common rigid moving plate and fixed at their other ends can undergo the same end displacement while developing different axial strains.
Which statements are correct?
1. Equal-length constituents connected so that no relative axial slip occurs have equal longitudinal strain.
2. For such constituents in the linear-elastic range, their stresses are proportional to their Young's moduli.
3. Their loads are necessarily equal regardless of cross-sectional area and modulus.
4. Parallel members of unequal original lengths attached to a common rigid moving plate and fixed at their other ends can undergo the same end displacement while developing different axial strains.
Which statements are correct?
Explanation:
Statement 1 is correct because equal-length constituents that deform together undergo the same change in length and therefore the same strain. Statement 2 is correct because Hooke's law gives σ = Eε under the linear-elastic assumption. Statement 3 is incorrect because load sharing depends on axial rigidity AE and is generally unequal when the constituent properties or areas differ. Statement 4 is correct because a common rigid plate can impose the same displacement on rods of different original lengths, while strain equals deformation divided by each rod's own original length. Therefore Statements 1, 2 and 4 are correct and Option B is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Equilibrium of the total axial load and compatibility of axial deformation సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. 2.50 సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. Steel 102.86 kN and brass 77.14 kN సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. 1.50 mm సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Steel 200 MPa and aluminium 70 MPa సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 180 × 10^6 N సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. σ1 = 2.5σ2 సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: D. 800 mm² సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: B. 80 kN సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. Displacement 1.0 mm; steel 100 kN and aluminium 70 kN సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. Their changes in length are equal, while their strains may differ. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: C. 71.5 kN సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 1, 2 and 4 only సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. 1-Q, 2-S, 3-P, 4-R సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. It decreases. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. 500 mm² సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 124.8 kN సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: B. 105 kN సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 1.50 సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 66 kN సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: C. No; under equal-strain compatibility their stress ratio is determined by the ratio of their Young's moduli. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: C. Eeq = Σ(AiEi)/At సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 122 GPa సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: C. 1.20 mm సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 1, 2 and 4 only సమాధానం: B.
