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Question 1
ప్రశ్న 1
Which statement correctly describes the current Indian-code basis for ordinary structural-steel plate-girder design?
Explanation:
IS 800:2007 remains the published general code governing ordinary structural-steel design, including plate-girder provisions. Its applicable amendments must be read with the base standard. BIS has progressed a fourth revision, but a finalized draft is not itself the operative published replacement. IS 808:2021 mainly supplies dimensions and properties of standard rolled sections rather than the design rules for a fabricated plate girder. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
In a conventional welded I-shaped plate girder carrying major-axis bending and vertical shear, which statement best describes the structural roles of the flanges and web?
Explanation:
Major-axis bending produces the largest longitudinal normal stresses near the extreme fibres of an I-shaped girder. The flange plates are located far from the neutral axis and therefore provide a large share of the bending resistance. Vertical shear is concentrated mainly in the web, so the web is commonly proportioned principally for shear and stability. The web still contributes to the section's flexural properties and the flanges are not completely free of shear effects. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which situation most reasonably favours a fabricated plate girder over a single available rolled beam section?
Explanation:
Plate girders allow the designer to choose web depth and thickness and flange dimensions to suit the required actions. They become attractive when standard rolled beams are unavailable or inefficient for the required span, load or stiffness. Material can be concentrated in the flanges for bending resistance while a comparatively thin web carries shear. This efficiency must be balanced against fabrication, welding, stiffener and erection costs. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
In a simplified plate-girder flange-force estimate, the web contribution to bending is neglected and the compression and tension flange resultants are separated by 1.6 m. If the design bending moment is 2400 kN·m, what axial force magnitude is required in each flange?
Explanation:
Under the stated simplified flange-couple model, the resisting moment is the flange force multiplied by the distance between the flange resultants. Hence F = M/h. Substitution gives F = 2400/1.6 = 1500 kN. One flange is predominantly in compression and the other predominantly in tension under the corresponding bending sense. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
For the simplified flange force of 1500 kN in the previous question, take fy = 250 MPa and gamma_m0 = 1.10, and assume the required flange area may be estimated from the design stress fy/gamma_m0 with local buckling and other effects checked separately. What flange area is required?
Explanation:
The stated design stress is fy/gamma_m0 = 250/1.10 = approximately 227.27 N/mm^2. The flange force is 1500 kN, or 1,500,000 N. Dividing force by design stress gives 1,500,000/227.27 = approximately 6600 mm^2. This is only a preliminary flange-area estimate because section classification, lateral stability, connections and the actual web contribution must still be considered. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
A welded I-shaped plate girder has clear web depth d = 1500 mm and web thickness tw = 10 mm. What effective shear area Av is used for major-axis shear under the IS 800 welded-I-section rule?
Explanation:
For a welded I-section under major-axis shear, IS 800 takes the relevant shear area as d times tw. Here d = 1500 mm and tw = 10 mm. Thus Av = 1500 × 10 = 15000 mm^2. This web-based shear area reflects the dominant role of the web in carrying vertical shear. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
For the web in the previous question, take fyw = 250 MPa and gamma_m0 = 1.10. If shear yielding governs and web shear buckling does not, what design plastic shear strength Vd = Av fyw/(sqrt(3) gamma_m0) is obtained?
Explanation:
The web shear area is 15000 mm^2. The design plastic shear strength is 15000 × 250 divided by sqrt(3) × 1.10. This gives approximately 1,968,000 N. Thus the design plastic shear strength is about 1968 kN, provided shear buckling or another limit state does not reduce the usable resistance. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
For an unstiffened steel web, when does IS 800 require resistance to shear buckling to be explicitly verified?
Explanation:
IS 800 distinguishes stocky webs governed by plastic shear yielding from slenderer webs susceptible to shear buckling. For an unstiffened web, the shear-buckling verification becomes necessary when d/tw exceeds 67 epsilon_w. The parameter epsilon_w accounts for the web yield strength and is based on sqrt(250/fyw). Below the threshold, an explicit shear-buckling reduction is not required for the pure-shear check. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
A plate-girder web has d = 1500 mm, tw = 10 mm and fyw = 250 MPa. Taking epsilon_w = sqrt(250/fyw) = 1.0, what conclusion follows from the IS 800 shear-buckling trigger?
Explanation:
The web slenderness is d/tw = 1500/10 = 150. For fyw = 250 MPa, epsilon_w equals 1.0 and the unstiffened-web trigger is therefore 67. Since 150 is greater than 67, plastic shear yielding alone cannot automatically be assumed to govern. The web must be checked using the applicable shear-buckling provisions and any stiffening system provided. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
A web connected to flanges along both longitudinal edges has no transverse stiffeners, d = 1800 mm, tw = 8 mm and fyw = 250 MPa. Using the IS 800 serviceability limit d/tw not exceeding 200 epsilon_w, what is the correct conclusion?
Explanation:
The web slenderness is 1800/8 = 225. Since fyw is 250 MPa, epsilon_w is 1.0. The stated serviceability limit for a web connected along both longitudinal edges without transverse stiffeners is therefore 200. Because 225 exceeds 200, this unstiffened configuration does not satisfy that particular web-thickness requirement. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
For a transversely stiffened web panel with stiffener spacing c such that c/d = 0.75, what shear-buckling coefficient Kv is obtained from Kv = 4.0 + 5.35/(c/d)^2?
Explanation:
Because c/d is less than 1.0, the specified branch of the shear-buckling coefficient equation is used. The square of 0.75 is 0.5625. Thus Kv = 4.0 + 5.35/0.5625 = approximately 13.51. Closer transverse stiffener spacing generally raises the elastic shear-buckling resistance by increasing the relevant buckling coefficient. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
A stiffened web has c/d = 1.5, d/tw = 160, E = 2.0 × 10^5 MPa and Poisson's ratio mu = 0.30. Using Kv = 5.35 + 4.0/(c/d)^2 and tau_cr,e = Kv pi^2 E/[12(1 - mu^2)(d/tw)^2], what elastic critical shear stress is obtained approximately?
Explanation:
For c/d = 1.5, Kv = 5.35 + 4/1.5^2 = approximately 7.128. The web slenderness term is d/tw = 160 and mu is 0.30. Substitution in the elastic critical shear-stress equation gives approximately 50.33 MPa. This elastic critical stress is then used to obtain the non-dimensional web slenderness for the post-critical shear calculation. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
For fyw = 250 MPa and elastic critical shear stress tau_cr,e = 50.33 MPa, what non-dimensional web slenderness lambda_w = sqrt[fyw/(sqrt(3) tau_cr,e)] is obtained?
Explanation:
The denominator inside the ratio is sqrt(3) × 50.33, which is approximately 87.17 MPa. The ratio 250/87.17 is approximately 2.868. Taking the square root gives lambda_w of approximately 1.693. Since this exceeds 1.2, the high-slenderness branch of the IS 800 simple post-critical shear-stress expression would apply. Therefore Option D is correct.
వివరణ:
Question 14
ప్రశ్న 14
A web has d = 1200 mm, tw = 10 mm, fyw = 250 MPa, gamma_m0 = 1.10 and non-dimensional web slenderness lambda_w = 1.0. Using the simple post-critical relation tau_b = [1 - 0.8(lambda_w - 0.8)]fyw/sqrt(3) and Vd = d tw tau_b/gamma_m0, what design shear strength is obtained?
Explanation:
For lambda_w = 1.0, the multiplier in the middle branch is 1 - 0.8(1.0 - 0.8) = 0.84. The corresponding buckling shear stress is approximately 0.84 × 250/sqrt(3) = 121.24 MPa. The web area is 1200 × 10 = 12000 mm^2, so the nominal shear resistance is approximately 1454.9 kN. Dividing by gamma_m0 = 1.10 gives a design resistance of approximately 1322.7 kN. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
Which statement correctly describes the IS 800 simple post-critical method for plate-girder web shear buckling?
Explanation:
The simple post-critical method is one of the IS 800 methods for evaluating shear resistance of slender I-section girder webs. It may be used whether or not intermediate transverse stiffeners are present. The web is required to have transverse stiffeners at the supports for the method as specified. Unlike the tension-field method, it does not rely on adding the post-buckling tension-field reserve to the web-buckling resistance. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
Which set of conditions is essential for using the IS 800 tension-field method for a plate-girder web panel?
Explanation:
Tension-field action is a post-buckling mechanism and therefore requires a stable surrounding frame for anchorage. IS 800 permits the method for webs having intermediate transverse stiffeners in addition to the stiffeners at supports. The adjacent panels or end-post arrangement must be capable of anchoring the diagonal tension field. The specified method also imposes the applicable c/d condition stated in the question. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
A plate girder has design shear strength Vd = 1200 kN. Above what factored shear does the usual IS 800 high-shear interaction with bending become relevant?
Explanation:
IS 800 treats shear as high for the bending-interaction provisions when the factored shear exceeds 0.6Vd. For Vd = 1200 kN, this threshold is 0.6 × 1200 = 720 kN. Below this level, the ordinary low-shear bending resistance can generally be used subject to the other beam requirements. Above it, the reduction or interaction prescribed by the code must be considered. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
What is the physical meaning of tension-field action in a slender plate-girder web after initial shear buckling?
Explanation:
Elastic shear buckling of a slender web does not necessarily represent complete structural failure. After buckling, diagonal tensile stresses can develop across the web panel while the surrounding boundary elements provide anchorage. This membrane action can supply additional post-buckling shear resistance. The flanges, transverse stiffeners and end-post arrangement must therefore be capable of supporting the forces associated with the assumed tension field. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
What is the primary purpose of an intermediate transverse web stiffener in a slender plate girder when it is not carrying an external concentrated load?
Explanation:
A thin deep web can buckle under the diagonal compression associated with shear. Intermediate transverse stiffeners divide the web into shorter panels and modify the panel aspect ratio. This increases the elastic shear-buckling resistance and may permit use of post-critical design methods. Such a stiffener is distinct from one specifically introduced to carry a concentrated external load or reaction. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which statement correctly distinguishes load-carrying stiffeners from bearing stiffeners in the IS 800 plate-girder provisions?
Explanation:
A concentrated load or support reaction can create more than one local limit state in the girder web. IS 800 identifies load-carrying stiffeners for cases in which the unstiffened web is inadequate against local buckling under the applied compression. Bearing stiffeners are associated with insufficient local bearing or crushing capacity at the web-to-flange load-introduction region. In practical design the same stiffener arrangement may need to satisfy several relevant checks, but the underlying limit states should not be confused. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
A support reaction applied through the flange is 900 kN, while the calculated local bearing capacity of the unstiffened web at that location is 620 kN. What is the appropriate design conclusion?
Explanation:
The factored reaction is greater than the stated local bearing resistance of the unstiffened web. The unsupported web therefore cannot safely be credited with transmitting the full concentrated force by local bearing alone. IS 800 requires bearing stiffening where the concentrated force through the flange exceeds the applicable local web capacity. The stiffener and its connections must then provide an adequate force path into the web and girder section. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
An intermediate transverse web stiffener has actual length L = 1600 mm between the relevant flange regions. What effective length is used for its IS 800 buckling-resistance calculation?
Explanation:
IS 800 specifies an effective length of 0.7L for an intermediate transverse stiffener when calculating its buckling resistance. With L = 1600 mm, the effective length is 0.7 × 1600. This equals 1120 mm. The stiffener must still satisfy the applicable cross-sectional, connection and force requirements in addition to this effective-length assumption. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
What is required when a flange plate in a plate girder is curtailed at a theoretical cut-off point?
Explanation:
A flange plate carries longitudinal force as part of the bending-resisting flange. That force cannot disappear abruptly at the theoretical point where the plate first becomes unnecessary in a sectional-strength calculation. IS 800 therefore requires the plate to continue beyond its theoretical cut-off. The connection in the extension must be capable of developing the load in that plate corresponding to the bending moment at the theoretical cut-off section. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
A symmetric welded plate girder has a 1200 mm × 10 mm web and two 300 mm × 20 mm flange plates placed directly above and below the web, giving an overall depth of 1240 mm. What is its elastic major-axis section modulus approximately?
Explanation:
The web contributes I = 10 × 1200^3/12 = 1.44 × 10^9 mm^4. Each flange has area 6000 mm^2 with its centroid 610 mm from the neutral axis, so the parallel-axis contribution of the two flanges is dominant. Including each flange's own centroidal inertia gives a total I of approximately 5.9056 × 10^9 mm^4. The extreme-fibre distance is 1240/2 = 620 mm, so Ze = I/620 = approximately 9.53 × 10^6 mm^3. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
At a section of a welded plate girder, the vertical shear is V = 600 kN, the first moment of one flange area about the neutral axis is Q = 4.0 × 10^6 mm^3, and the girder moment of inertia is I = 6.0 × 10^9 mm^4. Using q = VQ/I, what total longitudinal shear flow must the web-to-flange connection transfer, and what is the demand per weld line if two identical weld lines share it equally?
Explanation:
The longitudinal interface shear flow is q = VQ/I. Substitution gives 600000 × 4.0 × 10^6 divided by 6.0 × 10^9, which equals 400 N/mm. If two identical web-to-flange weld lines are arranged to share this interface transfer equally, each carries 200 N/mm under the stated idealization. These welds are essential because they transfer the longitudinal shear required for the web and flange plates to act together as one girder section. Therefore Option B is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. IS 800:2007, with its applicable amendments, remains the published general steel-design code; the fourth revision is still a finalized draft rather than the operative replacement సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. The flanges resist a major part of the longitudinal bending forces, while the web carries a major part of the vertical shear, although both remain parts of the complete section సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. When the required bending strength, shear capacity, stiffness or depth cannot be provided economically by a standard rolled section సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 1500 kN సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: C. 6600 mm^2 సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: C. 15000 mm^2 సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. Approximately 1968 kN సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. When d/tw exceeds 67 epsilon_w సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: C. d/tw = 150, which exceeds 67, so shear-buckling resistance must be evaluated సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: C. d/tw = 225 and the unstiffened-web serviceability limit is not satisfied సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: C. Approximately 13.51 సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: B. Approximately 50.3 MPa సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: D. Approximately 1.69 సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: C. Approximately 1323 kN సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. It may be used for I-section girder webs with or without intermediate transverse stiffeners, provided transverse stiffeners are present at the supports సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. Intermediate transverse stiffeners and support stiffeners are provided, the required anchorage is available from adjacent panels or end posts, and the applicable panel satisfies c/d not less than 1.0 సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. 720 kN సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. The buckled web can develop a diagonal tensile membrane field anchored by the surrounding flanges and stiffeners, providing post-buckling shear resistance సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: A. To subdivide the web into panels and improve its resistance to shear buckling సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: A. Load-carrying stiffeners address local web buckling under concentrated force, while bearing stiffeners address local crushing or bearing capacity of the web at a concentrated load or reaction సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: B. A bearing stiffener or another adequate load-transfer arrangement is required because the applied force exceeds the local web bearing capacity సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 1120 mm సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. It must extend beyond the theoretical cut-off point, and the bolts, rivets or welds in that extension must be sufficient to develop the plate force corresponding to the girder moment at the theoretical cut-off section సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: C. Approximately 9.53 × 10^6 mm^3 సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 400 N/mm total and 200 N/mm per weld line సమాధానం: B.
