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Question 1
ప్రశ్న 1
Which failure behaviour is most characteristic of a genuinely short compression member under approximately concentric loading?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A short compression member has a comparatively small effective slenderness ratio. Its global flexural instability resistance is therefore relatively high. Material compressive strength becomes more important than Euler elastic buckling. Such a member generally fails by crushing, yielding, or another material-strength mechanism before a classical long-column instability develops. Therefore Option A correctly describes short-column behaviour.
వివరణ:
Question 2
ప్రశ్న 2
Which statement best describes a sufficiently long and slender ideal column that remains linearly elastic up to instability?
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Explanation:
A sufficiently slender column can become unstable while the average direct compressive stress is still below the material crushing or yield level. Classical Euler theory models this elastic flexural buckling behaviour. The critical load depends strongly on end restraint through the effective length. It decreases with the square of effective length rather than increasing with it. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Why are intermediate columns commonly treated differently from very short and very long ideal columns?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A very short member is commonly dominated by material-strength effects. A very long ideal member is commonly dominated by elastic buckling. Intermediate columns lie between these limiting behaviours. Their resistance can be influenced by both compression strength and instability. This motivates empirical or semi-empirical relations such as the Rankine-Gordon formula, so Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
Which use of Euler's formula is technically inappropriate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Classical Euler theory assumes elastic behaviour up to the buckling load. A stocky member may reach material yielding or crushing before the Euler instability load is reached. In that situation, using Euler load as the actual governing capacity is physically inappropriate. End conditions and the weak buckling axis remain essential when Euler theory is applicable. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
For a compression member having effective length Le and minimum radius of gyration k, which expression gives its effective slenderness ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Slenderness ratio compares an effective member length with a characteristic cross-sectional dimension. The relevant cross-sectional dimension for column buckling is the radius of gyration. Hence the effective slenderness ratio is Le/k. A greater effective length or a smaller radius of gyration increases the slenderness. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
Two otherwise identical columns have the same cross-section and material but different ideal end restraints. Why can they have different Rankine and Euler capacities?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
End restraint changes the buckled shape and is represented through the effective length. The slenderness ratio contains this effective length. Euler resistance therefore changes directly with the end condition. Rankine-type formulas also contain effective slenderness and consequently reflect the same restraint effect. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
A column section has principal radii of gyration kx = 60 mm and ky = 35 mm. It is free to buckle about either principal axis and has the same effective length for both directions. Which radius should normally be used for the governing column slenderness?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The governing buckling direction is normally associated with the smaller second moment of area. For a fixed area, the smaller second moment also corresponds to the smaller radius of gyration. Using ky = 35 mm produces the larger slenderness ratio. That direction therefore gives the lower Euler or Rankine-type resistance when other conditions are equal. Hence Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
Which expression is the commonly taught Rankine-Gordon form for the crippling load of a column, where σc is the crushing stress, A is area, a is the Rankine constant and λ = Le/k?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Rankine-Gordon expression combines a crushing-strength effect with a slenderness-dependent reduction. Its familiar form is PR = σcA/[1 + a(Le/k)²]. The denominator increases as slenderness increases, so predicted capacity decreases. For very small slenderness, the expression approaches the crushing load σcA. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
If Pc denotes the crushing load and Pe denotes the Euler elastic buckling load, which reciprocal relation represents the classical Rankine combination?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Rankine combination is commonly expressed as the sum of the reciprocals of the crushing and Euler loads. Thus 1/PR = 1/Pc + 1/Pe. Rearrangement gives PR = PcPe/(Pc + Pe). For positive Pc and Pe, this value is smaller than either individual limiting load. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
According to PR = σcA/(1 + aλ²), what does the Rankine load approach as the effective slenderness ratio λ tends toward zero?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
When λ approaches zero, the term aλ² also approaches zero. The Rankine denominator therefore approaches unity. The resulting load tends to σcA. This is consistent with the limiting behaviour of a very short compression member dominated by material compression. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
A pin-ended rectangular column has cross-section 100 mm × 150 mm and length 3.0 m. Use the least radius of gyration, crushing stress σc = 320 MPa and Rankine constant a = 1/7500. What is the Rankine crippling load approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The area is 15000 mm² and the least radius of gyration is 100/sqrt(12) = 28.87 mm. For pin-ended conditions, Le = 3000 mm, giving λ approximately 103.92. The crushing load is 320 × 15000 = 4.80 MN. Substitution in PR = σcA/(1 + λ²/7500) gives approximately 1.967 MN. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
The same 100 mm × 150 mm column is 3.0 m long, but both ends are ideally fixed. Use σc = 320 MPa, a = 1/7500 and the least radius of gyration. What is its Rankine load approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For ideal fixed-fixed end conditions, the effective length is L/2 = 1500 mm. The least radius of gyration remains 28.87 mm. Hence the effective slenderness ratio is approximately 51.96. The Rankine expression gives 4.80/[1 + 51.96²/7500] MN, which is approximately 3.529 MN. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
The same 100 mm × 150 mm column is 3.0 m long but is fixed at one end and free at the other. Using σc = 320 MPa, a = 1/7500 and the least radius of gyration, what is its Rankine load approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A fixed-free column has effective length 2L, so Le = 6000 mm. With k = 28.87 mm, the effective slenderness ratio is approximately 207.85. The crushing load is still 4.80 MN because the area and material are unchanged. The Rankine denominator becomes 1 + 207.85²/7500, giving a load of approximately 0.710 MN. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a particular column, the crushing load Pc is 3000 kN and the Euler load Pe is 1500 kN. Using 1/PR = 1/Pc + 1/Pe, what is the Rankine load?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The reciprocal Rankine relation gives 1/PR = 1/3000 + 1/1500. The right-hand side equals 1/3000 + 2/3000 = 3/3000. Therefore PR = 1000 kN. This value is less than both the crushing and Euler loads, as expected from the reciprocal combination. Hence Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
A column has crushing load Pc = 5000 kN and Euler load Pe = 2000 kN. What is the Rankine load according to the reciprocal formula?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Rankine load is PcPe/(Pc + Pe). Substitution gives 5000 × 2000 divided by 7000. This equals approximately 1428.6 kN. The result is below the smaller individual limiting load of 2000 kN. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A column has Rankine load PR = 1200 kN and crushing load Pc = 3000 kN. If 1/PR = 1/Pc + 1/Pe, what is the corresponding Euler load Pe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The required reciprocal Euler term is 1/Pe = 1/1200 - 1/3000. This difference equals 1/2000. Therefore Pe = 2000 kN. Substituting 3000 kN and 2000 kN back into the Rankine expression returns 1200 kN. Hence Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A pin-ended rectangular column has cross-section 100 mm × 200 mm, length 2.4 m, crushing stress 300 MPa and Rankine constant a = 1/7500. It is free to buckle about either principal axis. What is its governing Rankine load approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The area is 20000 mm² and the smaller radius of gyration is 100/sqrt(12) = 28.87 mm. Since the member can buckle about either axis, this weaker direction governs. The effective slenderness ratio is 2400/28.87 = approximately 83.14. Using PR = 300 × 20000/[1 + 83.14²/7500] gives approximately 3.122 MN. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
For the same 100 mm × 200 mm, 2.4 m pin-ended column, suppose ideal lateral bracing completely prevents buckling about the weak axis so that buckling can occur only about the strong axis. Using σc = 300 MPa and a = 1/7500, what Rankine load is predicted for the permitted strong-axis mode?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For buckling about the strong axis, the relevant radius of gyration is 200/sqrt(12) = 57.74 mm. The slenderness ratio is therefore approximately 41.57. The crushing load remains 300 × 20000 = 6.0 MN. Substitution in the Rankine equation gives approximately 4.876 MN. Thus suppression of the weak-axis mode raises the predicted capacity, and Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
A column has Rankine constant a = 1/8000 and initial effective slenderness ratio λ = 80. If its effective length increases by 20 percent while the cross-section and material remain unchanged, what is the approximate ratio of the new Rankine load to the original Rankine load?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Increasing effective length by 20 percent increases the slenderness ratio from 80 to 96. The original Rankine denominator is 1 + 80²/8000 = 1.8. The new denominator is 1 + 96²/8000 = 2.152. Since the crushing-load numerator is unchanged, the load ratio is 1.8/2.152 = approximately 0.836. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
For a column obeying PR = Pc/(1 + aλ²), the Rankine load is measured as 60 percent of its crushing load. If a = 1/7500, what is the effective slenderness ratio approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ratio PR/Pc equals 1/(1 + aλ²). Setting this equal to 0.60 gives 1 + aλ² = 1/0.60 = 1.6667. Hence aλ² = 0.6667. With a = 1/7500, λ² = 5000 and λ is approximately 70.7. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
If the crushing load Pc and Euler load Pe of a column happen to be equal to the same value P0, what does the Rankine reciprocal formula predict?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The reciprocal equation becomes 1/PR = 1/P0 + 1/P0. Thus 1/PR = 2/P0. Inverting gives PR = P0/2. This illustrates that the Rankine interaction value is below either separate limiting load when both are finite and positive. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A column has area A = 12000 mm², minimum radius of gyration k = 35 mm, actual length L = 2.8 m, crushing stress σc = 300 MPa and Rankine constant a = 1/8000. One end is fixed and the other pinned; use effective-length factor K = 0.70. What is the Rankine load approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The effective length is 0.70 × 2800 = 1960 mm. The corresponding slenderness ratio is 1960/35 = 56. The crushing load is 300 × 12000 = 3.60 MN. The Rankine denominator is 1 + 56²/8000 = 1.392, giving PR approximately 2.586 MN. Therefore Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
A column has A = 10000 mm², k = 30 mm, crushing stress σc = 300 MPa and Rankine constant a = 1/7500. What is the maximum effective length approximately if its Rankine load must be at least 1500 kN?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The crushing load is 300 × 10000 = 3000 kN. Requiring PR = 1500 kN at the limiting case means the Rankine denominator is 2. Thus aλ² = 1 and λ = sqrt(7500) = approximately 86.60. Since Le = λk, the effective length is 86.60 × 30 = approximately 2598 mm. Therefore the limiting effective length is about 2.60 m and Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
A Rankine calculation gives an ultimate or crippling load of 2.4 MN for a compression member. If an overall factor of safety of 3 is specified for this simplified calculation, what working load corresponds to it?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A factor of safety relates the limiting load to the chosen working or allowable load in this simplified calculation. The working load is obtained by dividing the crippling load by the factor of safety. Thus 2.4 MN divided by 3 equals 0.8 MN. This calculation does not replace any separate code-specific design requirements that may apply in practice. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about column behaviour and Rankine-type formulas:
1. Short columns are generally more strongly influenced by compressive material strength than by elastic Euler buckling.
2. For sufficiently long elastic columns, Euler buckling becomes the dominant limiting behaviour.
3. For positive finite Pc and Pe, the Rankine load PR = PcPe/(Pc + Pe) is greater than the smaller of Pc and Pe.
4. For unchanged effective length and material, reducing the governing radius of gyration increases slenderness and reduces the Rankine load.
Which statements are correct?
1. Short columns are generally more strongly influenced by compressive material strength than by elastic Euler buckling.
2. For sufficiently long elastic columns, Euler buckling becomes the dominant limiting behaviour.
3. For positive finite Pc and Pe, the Rankine load PR = PcPe/(Pc + Pe) is greater than the smaller of Pc and Pe.
4. For unchanged effective length and material, reducing the governing radius of gyration increases slenderness and reduces the Rankine load.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because stocky members are commonly controlled primarily by material compression rather than Euler instability. Statement 2 is correct for sufficiently slender members that remain elastic up to buckling. Statement 3 is incorrect because the reciprocal Rankine combination is smaller than either positive finite Pc or Pe. Statement 4 is correct because a smaller radius of gyration increases Le/k and therefore increases the Rankine denominator. Hence 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: A. Material crushing or yielding with relatively little overall lateral buckling సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. It is expected to fail primarily by elastic buckling before reaching the material crushing stress. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. Both material compression and buckling effects can influence their load capacity. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. Applying it directly as the governing failure formula to a stocky short column whose predicted Euler stress is above the elastic material range సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. Le/k సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. Their end restraints produce different effective lengths and hence different effective slenderness ratios. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 35 mm సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. PR = σc A/(1 + aλ²) సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. 1/PR = 1/Pc + 1/Pe సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. The crushing load σcA సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. 1.97 MN సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 3.529 MN సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: A. 0.710 MN సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. 1000 kN సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: B. 1428.6 kN సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. 2000 kN సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 3.12 MN సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 4.88 MN సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. 0.836 సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. 70.7 సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. PR = P0/2 సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. 2.59 MN సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. 2.60 m సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. 0.8 MN సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: B. 1, 2 and 4 only సమాధానం: B.