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Question 1
ప్రశ్న 1
Which statement correctly describes the present design-method framework of IS 800:2007 for general construction in steel?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800:2007 is fundamentally organized around the Limit State Method for general structural-steel design. The code requires strength and serviceability limit states to be considered with the appropriate design actions and resistances. It nevertheless retains a separate Working Stress Design section for situations where the Limit State Method cannot conveniently be adopted. The presence of permissible-stress terminology in an examination syllabus therefore does not mean that the current code is exclusively a Working Stress code. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
In conventional Working Stress Design terminology for structural steel, what is meant by a permissible stress?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Working Stress Design is based primarily on elastic behaviour under service or working loads. The stress produced by those loads is calculated using elastic structural analysis and section properties. That actual working stress is then compared with a permissible or allowable stress specified with an appropriate margin against failure. This differs from Limit State Design, in which factored actions are compared with design strengths derived using partial safety factors. Therefore Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which is the current Indian Standard specification for the principal hot-rolled medium and high tensile structural-steel products used in structural work?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 2062 (Part 1):2025 is the current specification for structural steel covered by its scope, including hot-rolled medium and high tensile steel. It supersedes the older IS 2062:2011 acceptance basis for these products. The standard deals with such matters as grades, chemical requirements, tensile properties, bend tests and impact requirements where applicable. IS 800, by contrast, is principally a structural design code rather than the product specification for the steel itself. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
Which current Indian Standard primarily gives dimensions and sectional properties of hot-rolled steel beam, column, channel and angle sections?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 808:2021 is titled for hot-rolled steel beam, column, channel and angle sections and gives their dimensions and properties. Designers use such section data for properties including area, mass, moments of inertia and radii of gyration. IS 800:2007 provides the structural design rules in which many of those properties are subsequently used. A section standard and a structural design code therefore serve different but complementary purposes. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
What value of Young's modulus E may normally be used for structural steel in design according to IS 800:2007?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800 gives a modulus of elasticity of structural steel equal to 2.0 × 10^5 N/mm². This property describes the elastic relationship between normal stress and normal strain within the elastic range. It is essentially independent of the usual structural-steel strength grade for ordinary design calculations. The value is used in deflection, elastic buckling and stiffness calculations. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
Which pair correctly gives the approximate unit mass density and Poisson's ratio used for structural steel in IS 800:2007?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800 permits the unit mass of structural steel to be taken as approximately 7850 kg/m³. The corresponding Poisson's ratio used for elastic calculations is 0.30. These are physical properties rather than yield-strength parameters. They are used in calculations such as self-weight, elastic deformation and the relationship between elastic and shear moduli. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
Which statement correctly distinguishes the symbols fy and fu in structural-steel design?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The symbol fy represents the characteristic yield stress used in steel design. The symbol fu represents the ultimate tensile strength of the steel. Different limit states depend on these properties in different ways, so the two must not be interchanged. For example, yielding resistance is principally related to fy, whereas certain rupture and connection limit states involve fu. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
Under IS 2062 (Part 1):2025, an E250 product having thickness not exceeding 16 mm has which minimum yield stress and minimum tensile strength?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The current IS 2062 Part 1 mechanical-property table uses thickness-dependent yield-strength bands. For E250 steel with thickness up to and including 16 mm, the specified minimum yield stress is 250 MPa. Its specified minimum tensile strength is 410 MPa. The yield stress can reduce for greater product thickness, so the grade number should not be blindly treated as the minimum yield stress at every thickness. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
An E250 structural-steel plate conforming to IS 2062 (Part 1):2025 is 50 mm thick. What minimum yield stress applies to this thickness band?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 50 mm plate falls in the greater-than-40-mm and up-to-100-mm thickness band of the current E250 mechanical-property table. The minimum yield stress for E250 in that band is 230 MPa. The 250 MPa value applies to the thinner band up to 16 mm, while 240 MPa applies above 16 mm and up to 40 mm. This illustrates why structural design should use the yield strength corresponding to the actual product thickness. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
A flat steel plate has width 200 mm and thickness 12 mm. What is its gross cross-sectional area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The area of a rectangular plate is its width multiplied by its thickness. Here the dimensions are 200 mm and 12 mm. Thus A = 200 × 12 = 2400 mm². Gross area is a basic section property used before considering deductions such as bolt holes where net area is relevant. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
A rectangular steel section is 100 mm wide and 200 mm deep. What is its second moment of area about its centroidal horizontal axis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a rectangle about its centroidal axis parallel to the width, I = bd³/12. Here b = 100 mm and d = 200 mm. Therefore I = 100 × 200³/12 = 66.67 × 10^6 mm^4 approximately. The cubic dependence on depth explains why increasing section depth is highly effective in increasing bending stiffness. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
Which expression defines the radius of gyration r of a cross-section of area A and second moment of area I about the same axis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Radius of gyration is defined so that I = Ar². Rearranging gives r = √(I/A). It provides a convenient measure of how the area is distributed about a particular centroidal axis. In column design, the slenderness ratio is based on effective length divided by the relevant radius of gyration. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A steel section has area A = 8000 mm² and second moment of area I = 80 × 10^6 mm^4 about a specified centroidal axis. What is its radius of gyration about that axis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The radius of gyration is r = √(I/A). Substitution gives r = √[(80 × 10^6)/8000]. The quantity inside the square root is 10000 mm². Its square root is 100 mm. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
Which expression gives the elastic section modulus Ze about a bending axis when I is the second moment of area and ymax is the distance from the neutral axis to the extreme fibre?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The elastic flexure equation is sigma = My/I. At the extreme fibre, y equals ymax and the stress is therefore M ymax/I. Defining the elastic section modulus as Ze = I/ymax permits the equation to be written as sigma = M/Ze. A larger elastic section modulus consequently reduces the extreme-fibre elastic bending stress for a given moment. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
A steel section has I = 120 × 10^6 mm^4 about its bending axis, and the extreme fibre is 200 mm from the neutral axis. What is its elastic section modulus?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The elastic section modulus is Ze = I/ymax. Substituting gives Ze = 120 × 10^6/200. This equals 0.60 × 10^6 mm³, or 600 × 10^3 mm³. This property directly relates elastic bending moment to the maximum normal stress. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
For a steel section with elastic section modulus Ze = 600 × 10^3 mm³ and yield stress fy = 250 N/mm², what elastic yield moment My = fyZe is obtained?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The first-yield elastic moment is obtained from My = fyZe. Thus My = 250 × 600000 = 150000000 N·mm. Dividing by 10^6 converts N·mm to kN·m, giving 150 kN·m. This is the moment at which the extreme fibre first reaches the specified yield stress in simple elastic bending. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
What does the plastic section modulus Zp represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Plastic section modulus is a geometrical property associated with the fully yielded stress distribution of a cross-section. When an appropriate section can develop its full plastic capacity, the basic plastic moment is related to fyZp before application of the relevant design factors and code limits. It differs from the elastic section modulus Ze, which corresponds to the first-yield elastic stress distribution. The ability to exploit Zp also depends on section classification and stability requirements. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
For a solid rectangular cross-section, the elastic section modulus is bd²/6 and the plastic section modulus is bd²/4. What is its shape factor Zp/Ze?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Shape factor is defined as the ratio of plastic section modulus to elastic section modulus. For the rectangle, Zp/Ze = (bd²/4)/(bd²/6). The common terms cancel, leaving 6/4 = 1.5. This means the idealized fully plastic moment of the rectangular section is 1.5 times its first-yield elastic moment when other limit states do not intervene. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
Under the IS 800 section-classification concept, which description corresponds to a plastic section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A plastic section satisfies sufficiently restrictive width-to-thickness limits for its compression elements. It can develop the full plastic moment resistance without premature local buckling. It also possesses sufficient plastic-hinge rotation capacity for the redistribution required in a plastic mechanism. This additional rotation capacity distinguishes it from a merely compact section. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which statement correctly describes a compact steel section under IS 800 classification?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A compact section has compression elements sufficiently stocky to develop the plastic moment resistance. However, its available plastic rotation is not sufficient for the full hinge rotation demanded in plastic mechanism analysis. A plastic section has greater rotational capability. A semi-compact section, by contrast, can reach yield at the extreme compression fibre but cannot develop the full plastic moment. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
Which statement correctly describes a semi-compact section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A semi-compact section is less stocky than a compact section. Its extreme compression fibre can reach the yield stress under bending. However, local buckling prevents the stress distribution from developing into the fully plastic distribution required for the plastic moment. Elastic section properties are consequently important when determining its flexural resistance. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
What characterizes a slender section in the IS 800 section-classification framework?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The term slender in cross-section classification refers to local slenderness of compression plate elements. If their width-to-thickness ratios are excessive, local buckling can occur before the full yield stress is developed. This is different from overall member slenderness, which involves member length and radius of gyration. Appropriate effective-section or other code provisions are therefore required when slender compression elements are present. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
An I-section has a plastic flange but a semi-compact web for the loading condition being considered. How is the overall cross-section classified?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A rolled or built-up steel cross-section contains several compression elements whose local buckling behaviour must be classified. The overall section cannot be assigned a more favourable class than its governing compression element. Here the flange satisfies the plastic limits but the web satisfies only the semi-compact limits. Consequently the section as a whole must be treated as semi-compact for the relevant loading condition. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Why are width-to-thickness ratios central to the classification of steel cross-sections?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Flanges, webs, angle legs and similar parts of steel sections behave as plate elements. Thin wide plate elements can buckle locally while their stresses are still comparatively low. Restricting width-to-thickness ratios delays this local buckling and permits greater utilization of the steel yield strength. The required degree of compactness depends on whether elastic yield, plastic resistance or plastic-hinge rotation is needed. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
For a typical rolled I-section bent about its major and minor centroidal axes, which statement is generally correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A conventional I-section places much of its flange area far from the major centroidal axis. This produces a large second moment of area about that axis. The corresponding major-axis radius of gyration is therefore normally larger than the minor-axis value. The smaller minor-axis radius is especially important when assessing column buckling because it often produces the larger slenderness ratio. Therefore Option A is correct.
వివరణ:
Question 26
ప్రశ్న 26
A rolled steel section has a gross area of 5000 mm². Using a steel mass density of 7850 kg/m³, what is its theoretical mass per metre length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An area of 5000 mm² is equal to 0.005 m². For one metre length, the corresponding steel volume is therefore 0.005 m³. Multiplying by the mass density gives 0.005 × 7850 = 39.25 kg. Standard rolled-section tables provide mass per metre directly, but the same value follows from area multiplied by steel density. Therefore Option C is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Limit State Design is generally adopted, while Working Stress Design may be used where Limit State Design cannot conveniently be adopted సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. An allowable stress against which elastic stresses calculated under working or service loading are checked సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. IS 2062 (Part 1):2025 సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: A. IS 808:2021 సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 2.0 × 10^5 N/mm² సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: B. 7850 kg/m³ and 0.30 సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. fy denotes yield stress, while fu denotes ultimate tensile stress సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: B. 250 MPa and 410 MPa సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: B. 230 MPa సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 2400 mm² సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 66.67 × 10^6 mm^4 సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. r = √(I/A) సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. 100 mm సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: B. Ze = I/ymax సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 600 × 10^3 mm³ సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: C. 150 kN·m సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. A property used to calculate the fully plastic moment resistance of a cross-section సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: C. 1.50 సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. A section that can develop the plastic moment and has sufficient rotation capacity for formation of a plastic mechanism సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. It can develop the plastic moment resistance but has insufficient rotation capacity for a full plastic mechanism సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: B. It can reach yield stress at the extreme compression fibre but local buckling prevents development of the full plastic moment సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. Its compression elements can locally buckle before reaching the yield stress సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: C. Semi-compact, because the most adverse compression-element classification governs సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. They control whether compression plate elements are susceptible to local buckling before the desired stress distribution and rotation capacity develop సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: A. Its radius of gyration is usually larger about the major axis than about the minor axis సమాధానం: A.
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Question 26 ప్రశ్న 26Answer: C. 39.25 kg/m సమాధానం: C.