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Question 1
ప్రశ్న 1
Which statement correctly describes the current Indian-code framework for the design of ordinary structural-steel compression members?
Explanation:
IS 800:2007 is the currently published general structural-steel design code used for ordinary compression-member design. Section 7 deals specifically with compression members and includes design strength, effective length, detailing and special compression-member provisions. The code accounts for practical column behaviour through buckling classes rather than relying solely on ideal Euler theory. IS 808 is primarily a dimensional and sectional-property standard for rolled sections and does not replace the strength-design provisions of IS 800. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which statement best distinguishes a column from the broader term compression member or strut?
Explanation:
A column is normally an upright structural member that supports floors, roofs or other structural components and is predominantly subjected to compression. The word strut is used more generally for a member whose principal action is axial compression. A strut may be vertical, horizontal or inclined, as in a roof truss or bracing system. Both types can be governed by buckling rather than by simple compressive yielding. Therefore Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
Why does IS 800 use buckling curves for ordinary hot-rolled and built-up steel compression members instead of treating them as perfectly straight ideal Euler columns?
Explanation:
Real steel compression members are not perfectly straight or perfectly stress-free before loading. Rolling, welding and fabrication can introduce residual stresses, while geometric imperfections produce initial bow. Small unintended eccentricities can also occur when loads and connections are introduced into the member. IS 800 incorporates these practical effects through different column buckling curves and associated imperfection factors. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
What is the effective slenderness ratio of a steel compression member?
Explanation:
The effective slenderness ratio is defined as the effective length KL divided by the appropriate radius of gyration r. The effective length incorporates the influence of end restraints and frame behaviour on buckling. The radius of gyration measures the distribution of cross-sectional area about the axis being considered. A larger value of KL/r generally reduces the compressive buckling strength of the member. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
A compression member has the same effective length about both principal axes, with radii of gyration ry = 70 mm and rz = 40 mm. If both directions are otherwise governed by the same buckling class, which direction is more critical for flexural buckling?
Explanation:
Slenderness is KL/r, so a smaller radius of gyration produces a larger slenderness ratio when effective length is unchanged. Here rz is 40 mm, which is substantially smaller than ry = 70 mm. The member is consequently more slender about the axis associated with rz. Under the stated assumption of the same buckling class in both directions, that direction develops the lower buckling resistance. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
According to IS 800, how is the actual length L of an ordinary compression member generally measured in the plane of the buckling deformation?
Explanation:
IS 800 defines the actual length of a compression member with reference to its supporting system. It is generally measured centre-to-centre of the intersections with the supporting members in the plane of the buckling deformation. Effective length is subsequently obtained from this actual length after considering rotational and translational restraints. For a member having a free end, the appropriate free-standing length from the supported intersection is used. Therefore Option D is correct.
వివరణ:
Question 7
ప్రశ్న 7
A prismatic steel column acts as a cantilever: translation and rotation are restrained at its base and the top is completely free. If its unsupported length is 3.0 m, what effective length is given by the IS 800 idealized end-condition table?
Explanation:
For a compression member fixed at one end and completely free at the other, IS 800 Table 11 gives an effective length of 2.0L. The unsupported length in this problem is 3.0 m. Hence KL = 2.0 × 3.0 = 6.0 m. This large effective length reflects the poor buckling restraint of a cantilever column. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
A prismatic compression member has both translation and rotation restrained at each end under the idealized IS 800 end conditions. If its unsupported length is 4.0 m, what effective length should be taken?
Explanation:
For the idealized condition in which translation and rotation are restrained at both ends, IS 800 Table 11 gives KL = 0.65L. The unsupported length is 4.0 m. Therefore KL = 0.65 × 4.0 = 2.60 m. The reduction below the actual length represents the beneficial effect of rotational restraint at both ends. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
A 4.0 m prismatic column has one end with translation and rotation restrained and the other end with translation restrained but rotation free. Using the IS 800 effective-length factor 0.8, and taking r = 40 mm about the governing axis, what is KL/r?
Explanation:
The stated fixed-pinned idealization has an effective length of 0.8L. Hence KL = 0.8 × 4000 = 3200 mm. The effective slenderness ratio is therefore 3200/40 = 80. This dimensionless slenderness is then used in assessing the design compressive stress for the relevant buckling class. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
A steel column has effective length KL = 3.2 m. Its radii of gyration are ry = 65 mm and rz = 40 mm. Which set correctly gives the two slenderness ratios and the governing value, assuming no other buckling mode controls?
Explanation:
The effective length is 3200 mm. About the axis with ry = 65 mm, the slenderness ratio is 3200/65 = approximately 49.2. About the axis with rz = 40 mm, it is 3200/40 = 80. The larger value is 80 and is the more critical slenderness under the assumptions stated. Therefore Option D is correct.
వివరణ:
Question 11
ప్రశ్న 11
For a steel compression member with effective slenderness ratio KL/r = 100 and E = 2.0 × 10^5 MPa, what is the Euler buckling stress fcc = pi^2 E/(KL/r)^2?
Explanation:
Euler buckling stress is fcc = pi²E/(KL/r)². Substituting E = 200000 MPa and KL/r = 100 gives pi² × 200000/10000. The result is approximately 197.39 MPa. IS 800 uses this elastic buckling stress in defining the non-dimensional effective slenderness used in the column-strength equation. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the member in the previous question, take fy = 250 MPa and fcc = 197.4 MPa. What is the non-dimensional effective slenderness lambda = sqrt(fy/fcc)?
Explanation:
The IS 800 non-dimensional effective slenderness is the square root of fy/fcc. Here fy/fcc = 250/197.4 = approximately 1.266. Taking the square root gives approximately 1.125. This non-dimensional quantity enters the buckling-reduction expression together with the imperfection factor. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
Which set correctly gives the IS 800 imperfection factors alpha for column buckling classes a, b, c and d respectively?
Explanation:
IS 800 Table 7 assigns a separate imperfection factor to each column buckling curve. The values are 0.21 for class a, 0.34 for class b, 0.49 for class c and 0.76 for class d. A larger imperfection factor generally leads to a lower buckling resistance for the same non-dimensional slenderness. These factors represent differing sensitivity to residual stresses and geometric imperfections for different section forms and buckling axes. Therefore Option D is correct.
వివరణ:
Question 14
ప్రశ్న 14
For otherwise identical material and slenderness, which IS 800 column buckling class generally provides the highest design compressive stress and which provides the lowest?
Explanation:
Buckling class a has the smallest imperfection factor, while class d has the largest. For the same yield stress and non-dimensional slenderness, the smaller imperfection factor gives a more favourable reduction factor. Consequently class a normally produces the highest design compressive stress of the four curves. Class d is the least favourable and normally gives the lowest design compressive stress. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
A rolled I-section has h/bf greater than 1.2 and flange thickness tf not exceeding 40 mm. According to IS 800 Table 10, which buckling classes apply about the z-z and y-y axes respectively?
Explanation:
IS 800 Table 10 assigns buckling curves according to section type, proportions, thickness and buckling axis. For the stated rolled I-section with h/bf greater than 1.2 and tf not exceeding 40 mm, buckling about z-z is assigned class a. Buckling about y-y is assigned class b. The difference recognizes the different residual-stress and stability behaviour associated with the two axes. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
What buckling class does IS 800 Table 10 generally assign to channel, angle, T and solid sections for flexural compression-member design?
Explanation:
IS 800 Table 10 groups channel, angle, T and solid sections together for the general flexural-buckling classification. These sections are assigned buckling class c for any axis under that table entry. The corresponding imperfection factor for class c is 0.49. A single angle loaded through only one leg additionally requires the special treatment of Clause 7.5 because flexural-torsional effects and connection eccentricity become important. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
Which statement about hollow-section buckling classes in IS 800 Table 10 is correct?
Explanation:
IS 800 Table 10 distinguishes hot-rolled hollow sections from cold-formed hollow sections. A hot-rolled hollow section is assigned buckling class a for any axis. A cold-formed hollow section is assigned buckling class b. The distinction reflects the different residual-stress and manufacturing characteristics associated with these product forms. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
A class-c compression member has KL/r = 80, fy = 250 MPa, E = 2.0 × 10^5 MPa, alpha = 0.49 and gamma_m0 = 1.10. Using the IS 800 column-strength equation, what design compressive stress fcd is obtained approximately?
Explanation:
For KL/r = 80, the Euler stress is approximately 308.4 MPa and the non-dimensional slenderness is about 0.900. For class c, alpha is 0.49, giving phi = 0.5[1 + 0.49(0.900 - 0.2) + 0.900²], which is approximately 1.077. The resulting stress-reduction factor chi is approximately 0.600. Hence fcd = chi fy/gamma_m0 is approximately 0.600 × 250/1.10 = 136 MPa, consistent with the IS 800 tabulated value. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
A compression member has effective sectional area Ae = 6000 mm^2 and design compressive stress fcd = 136 MPa. What is its design axial compressive strength Pd = Ae fcd?
Explanation:
IS 800 gives the design compressive strength as Pd = Ae fcd. Substitution gives Pd = 6000 × 136 = 816000 N. This equals 816 kN. The factored axial compression acting on the member must not exceed this design strength when this is the governing compression check. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
A factored axial compression of 900 kN is to be carried by a member for which the governing design compressive stress is fcd = 136 MPa. Ignoring other design constraints, what minimum effective sectional area is required?
Explanation:
The required effective area is obtained by rearranging Pd = Ae fcd. Thus Ae = 900000/136 = approximately 6617.6 mm². A section must therefore provide at least about 6620 mm² of effective area on this strength basis. Final selection must additionally satisfy slenderness, buckling class, local-element and connection requirements. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
Two trial steel columns have the same effective sectional area, yield stress, effective length and buckling class. Section X has minimum radius of gyration 40 mm, while Section Y has minimum radius of gyration 50 mm. Which is generally the better choice with respect to flexural buckling strength alone?
Explanation:
The governing slenderness ratio is KL/r. With equal effective length, the section having the larger minimum radius of gyration has the lower governing slenderness ratio. A lower slenderness ordinarily produces a higher design compressive stress for the same steel grade and buckling curve. Thus Section Y is generally more favourable against flexural buckling under the assumptions stated. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
According to IS 800 Table 3, what is the maximum effective slenderness ratio for a member carrying compressive loads resulting from dead loads and imposed loads?
Explanation:
IS 800 places limits on effective slenderness in addition to the strength calculations. For a member carrying compression resulting from dead loads and imposed loads, the maximum KL/r is 180. This limit controls excessive slenderness even if a strength calculation might otherwise appear satisfactory. Different limits apply to members subjected to other loading situations or to members primarily acting in tension. Therefore Option D is correct.
వివరణ:
Question 23
ప్రశ్న 23
A member is subjected to compression only because of a load combination involving wind or earthquake action, and its deformation does not adversely affect stresses elsewhere in the structure. What maximum effective slenderness ratio is permitted by IS 800 Table 3 for this case?
Explanation:
IS 800 recognizes that compression arising only in combinations involving wind or earthquake may be temporary and conditional. For the specific case stated, Table 3 permits a maximum effective slenderness ratio of 250. The qualification that deformation must not adversely affect stresses in other parts of the structure is important. This provision should therefore not be applied indiscriminately to ordinary gravity-load columns. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement correctly describes the IS 800 effective-length treatment of compression members in ordinary bolted, riveted or welded trusses and braced frames?
Explanation:
IS 800 gives specific guidance for compression members in trusses and braced frames. In the plane of a bolted, riveted or welded truss, KL may be taken between 0.7 and 1.0 times the distance between centres of connections depending on the degree of end restraint. For buckling perpendicular to the plane of the truss, the effective length is generally taken as the distance between centres of intersection. Angle struts are additionally subject to the special provisions of Clause 7.5. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
How should a single-angle strut loaded in compression through only one leg generally be treated under the amended IS 800:2007 provisions?
Explanation:
Connecting only one leg of a compression angle introduces behaviour that is not represented adequately by a simple concentric KL/r check alone. IS 800 Clause 7.5.1.2 therefore provides an equivalent-slenderness procedure for flexural-torsional buckling of such an angle. Amendment No. 1 clarifies that this equivalent slenderness is used in place of the ordinary lambda in the compression equation. It also specifies use of buckling curve c with alpha = 0.49 for this calculation. Therefore Option C is correct.
వివరణ:
Question 26
ప్రశ్న 26
For a compression member whose cross-section is semi-compact or better, which statement correctly describes the effective sectional area Ae under IS 800?
Explanation:
IS 800 Clause 7.3.2 generally permits the gross sectional area to be used as the effective area for compression members that are semi-compact or better. This applies to members fabricated by welding, bolting or riveting under the stated conditions. Holes that are not filled with rivets, bolts or pins have to be deducted when determining the effective sectional area. Class 4 or slender sections require the modified effective-section provisions because local buckling prevents full use of the gross area. Therefore Option B is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. IS 800:2007 provides the currently published general steel-design provisions for compression members, including buckling-based Limit State design సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. A column is generally an upright member predominantly carrying compression, whereas a strut is a more general compression member that need not be vertical సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. Because residual stresses, initial bow and accidental load eccentricities influence practical buckling strength సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. KL/r సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. Buckling associated with rz = 40 mm సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: D. Centre-to-centre of its intersections with the supporting members సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: C. 6.00 m సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. 2.60 m సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: A. 80 సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: D. 49.2 and 80.0; governing 80.0 సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: B. 197.4 MPa సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 1.125 సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: D. 0.21, 0.34, 0.49, 0.76 సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. Class a highest and class d lowest సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: A. Class a about z-z and class b about y-y సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. Class c సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. Hot-rolled hollow sections are class a, while cold-formed hollow sections are class b సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 136 MPa సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. 816 kN సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Approximately 6620 mm^2 సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. Section Y, because its larger radius of gyration gives a smaller governing slenderness ratio సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 180 సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 250 సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. In-plane effective length may be taken as about 0.7 to 1.0 times the distance between connection centres depending on end restraint, while out-of-plane truss buckling is generally based on the distance between centres of intersection సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. Its flexural-torsional buckling is evaluated using the special equivalent-slenderness procedure of Clause 7.5.1.2 and buckling curve c సమాధానం: C.
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Question 26 ప్రశ్న 26Answer: B. The gross sectional area is generally taken as the effective area, except that holes not filled by rivets, bolts or pins are deducted; Class 4 sections require the separate effective-section treatment సమాధానం: B.
