Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which statement most accurately describes plastic analysis of a steel structure?
Explanation:
Plastic analysis is concerned with the ultimate load at which sufficient plastic hinges form to create a collapse mechanism. After first yielding, a ductile indeterminate steel structure can redistribute bending moments to other sections. Equilibrium must continue to be satisfied throughout this redistribution. The required sections and connections must possess sufficient strength and plastic rotation capacity without premature local or lateral instability. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which statement correctly distinguishes ordinary elastic analysis from plastic analysis of a ductile steel beam or frame?
Explanation:
In elastic analysis, stresses and deformations are evaluated while the material is treated essentially within its elastic range. First yield occurs when an extreme fibre reaches the yield stress at a critical section. In a sufficiently ductile section, further yielding can spread through the depth until the plastic moment is reached. In an indeterminate structure, additional loading can then produce redistribution and further plastic hinges before collapse. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a homogeneous steel section subjected to pure bending, which expressions correctly represent the first-yield moment My and the fully plastic moment Mp?
Explanation:
At first yield, the extreme-fibre elastic stress reaches fy. The corresponding moment is therefore My = fyZe, where Ze is the elastic section modulus. At full plasticity, the entire section has reached the idealized yield stress in tension or compression. The resulting plastic moment is Mp = fyZp, where Zp is the plastic section modulus. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a homogeneous cross-section subjected only to bending and having equal yield stress in tension and compression, where is the plastic neutral axis located?
Explanation:
At the fully plastic state, one part of the cross-section carries uniform compression at approximately fy while the other carries uniform tension of the same magnitude. Axial-force equilibrium requires the compressive resultant to equal the tensile resultant when no axial force is present. For a homogeneous material with equal yield strength in tension and compression, this requires equal areas on the two sides of the plastic neutral axis. In a symmetric section this axis coincides with the centroidal axis, but it need not do so for an unsymmetric section. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A solid rectangular steel section is 100 mm wide and 200 mm deep. If fy = 250 MPa, what is its first-yield moment My about the centroidal axis parallel to the width?
Explanation:
For a rectangle, the elastic section modulus about the stated axis is Ze = bd^2/6. Thus Ze = 100 × 200^2/6 = 666666.7 mm^3. The first-yield moment is My = fyZe = 250 × 666666.7 N·mm. This equals approximately 166.7 × 10^6 N·mm, or 166.7 kN·m. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
What is the plastic section modulus Zp of a solid rectangular section 100 mm wide and 200 mm deep about its centroidal axis parallel to the width?
Explanation:
For a symmetric rectangle, the plastic neutral axis is at mid-depth. Each half has area bd/2 and its centroid lies d/4 from the plastic neutral axis. The plastic section modulus is therefore 2(bd/2)(d/4), which simplifies to bd^2/4. Substituting b = 100 mm and d = 200 mm gives Zp = 1,000,000 mm^3. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
What is the plastic shape factor Mp/My = Zp/Ze for a solid rectangular cross-section?
Explanation:
The elastic section modulus of a rectangle is bd^2/6. Its plastic section modulus is bd^2/4. Their ratio is therefore (bd^2/4)/(bd^2/6) = 6/4 = 1.5. Thus the fully plastic moment is 50 percent greater than the first-yield moment for the idealized rectangular section. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
For the 100 mm × 200 mm rectangular section with fy = 250 MPa and Zp = 1.00 × 10^6 mm^3, what is its nominal plastic moment Mp?
Explanation:
The nominal plastic moment is Mp = fyZp. The yield stress is 250 N/mm^2 and the plastic section modulus is 1.00 × 10^6 mm^3. Thus Mp = 250 × 10^6 N·mm. Since 10^6 N·mm equals 1 kN·m, the plastic moment is 250 kN·m. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A steel section has elastic section modulus Ze = 800 × 10^3 mm^3 and plastic section modulus Zp = 920 × 10^3 mm^3. What are its shape factor and percentage increase of Mp over My?
Explanation:
Shape factor is Zp/Ze. Here the ratio is 920/800 = 1.15. Since both My and Mp contain the same yield stress fy, Mp/My is also 1.15. The plastic moment is therefore 15 percent greater than the first-yield moment. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
What is a plastic hinge in the idealized plastic analysis of a steel structure?
Explanation:
A plastic hinge forms when the section reaches its plastic moment and yielding has spread sufficiently through the cross-section. The idealized hinge can undergo significant rotation while the resisting moment remains approximately equal to Mp. It is not a literal mechanical pin because it continues to transmit the plastic bending moment. Plastic rotations at several critical locations permit an indeterminate structure to redistribute moment and eventually develop a mechanism. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
Under the IS 800 section-classification framework, what section class is required at a location where a plastic hinge is relied upon for a plastic collapse mechanism?
Explanation:
A plastic hinge must not only develop the plastic moment but also sustain the rotation required for mechanism formation. IS 800 therefore requires the proportions at plastic-hinge locations to satisfy the limits for Class 1 plastic sections. A Class 2 compact section can develop plastic moment but does not possess the same guaranteed plastic-hinge rotation capacity. Semi-compact and slender sections have still more restrictive local-buckling behaviour. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a structure having degree of static indeterminacy r, how many additional plastic hinges are generally required to convert it into a complete collapse mechanism, assuming a single independent mechanism is involved?
Explanation:
Each plastic hinge removes one relevant degree of static restraint from the structure. An r-times statically indeterminate system can sustain the formation of r suitably located plastic hinges while still remaining stable. One additional plastic hinge then converts the system into a mechanism. Thus the usual mechanism-count rule is r + 1 plastic hinges, subject to the geometry and possible local mechanisms of the particular structure. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A simply supported beam of span 8 m has a constant plastic moment capacity Mp = 200 kN·m and carries a central point load P. What is the collapse load?
Explanation:
A simply supported beam is statically determinate, so one plastic hinge at the point of maximum bending moment is sufficient to create a mechanism together with the two existing end hinges. For a central point load, the maximum moment is PL/4. At collapse, PL/4 = Mp. Hence P = 4Mp/L = 4 × 200/8 = 100 kN. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simply supported beam of span 10 m carries a point load P located 4 m from the left support and 6 m from the right support. If Mp = 240 kN·m, what is the collapse load?
Explanation:
For a point load at distances a and b from the supports of a simply supported beam, the bending moment beneath the load is Pab/L. The plastic hinge forms at this maximum-moment section. Setting Pab/L = Mp gives P = MpL/(ab). Substitution gives P = 240 × 10/(4 × 6) = 100 kN. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
A simply supported beam of span 6 m has constant plastic moment capacity Mp = 180 kN·m and carries a uniformly distributed load w over the full span. What is the collapse load intensity?
Explanation:
For a simply supported beam under a full-span UDL, the maximum bending moment is wL^2/8. A single midspan plastic hinge produces the collapse mechanism because the end supports are already hinges. Thus wL^2/8 = Mp at collapse. Therefore w = 8Mp/L^2 = 8 × 180/36 = 40 kN/m. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
A fixed-ended beam of span 8 m has the same plastic moment capacity Mp = 200 kN·m in positive and negative bending and carries a point load at midspan. What is the plastic collapse load?
Explanation:
A fixed-ended beam requires three plastic hinges to form the beam collapse mechanism: one at each fixed end and one at midspan. For a symmetric central-load mechanism, virtual work gives Pc = 8Mp/L. Substituting Mp = 200 kN·m and L = 8 m gives Pc = 200 kN. The additional restraint therefore gives twice the collapse load of a simply supported beam having the same Mp under the same central loading. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A fixed-ended beam of span 6 m has constant plastic moment capacity Mp = 180 kN·m and carries a uniformly distributed load over the full span. What is the collapse load intensity?
Explanation:
The collapse mechanism of the symmetric fixed-ended beam contains plastic hinges at both ends and at midspan. Applying the virtual-work equation to this mechanism gives wc = 16Mp/L^2. With Mp = 180 kN·m and L = 6 m, wc = 16 × 180/36. The result is 80 kN/m. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
A propped cantilever of span 8 m has constant plastic moment capacity Mp = 200 kN·m and carries a central point load. What is the collapse load?
Explanation:
A propped cantilever is once statically indeterminate and therefore requires two plastic hinges to form a collapse mechanism. For a central point load, the relevant hinges form at the fixed end and at the load point. Virtual work gives the standard result Pc = 6Mp/L. Substitution gives Pc = 6 × 200/8 = 150 kN. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A cantilever beam of span 3 m has constant plastic moment capacity Mp = 180 kN·m and carries a point load at its free end. What is the collapse load?
Explanation:
The maximum bending moment in a cantilever carrying an end load occurs at the fixed support. Once that section reaches Mp, a plastic hinge at the fixed end together with the free end produces a mechanism. The collapse condition is therefore PL = Mp. Hence P = Mp/L = 180/3 = 60 kN. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A single-bay, single-storey portal frame has rigid beam-column joints, both column bases fixed, column height h = 4 m and equal plastic moment capacity Mp = 180 kN·m at all four potential sway-hinge locations. Neglect gravity loads and second-order effects. What horizontal load H at beam level produces the simple sway collapse mechanism?
Explanation:
The simple sway mechanism contains plastic hinges at the two column bases and at the two beam-column joints. If each column rotates through a small angle theta, the roof translates by approximately h theta. The total internal work is 4Mp theta, while the external work is Hh theta. Equating the two gives Hc = 4Mp/h = 4 × 180/4 = 180 kN. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
What principle is used in the virtual-work method for determining the collapse load of an assumed plastic mechanism?
Explanation:
At the idealized collapse mechanism, the structure can undergo an incremental compatible displacement without an increase in the plastic moments at the hinges. Each plastic hinge dissipates work equal to Mp multiplied by its corresponding hinge rotation. The applied loads perform external work through their associated displacements. Equating total external and internal work gives the load corresponding to the assumed mechanism. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Which statement correctly describes the lower-bound or static theorem of plastic collapse?
Explanation:
The static theorem is based on finding an admissible equilibrium distribution of internal forces or moments. The distribution must satisfy the applied loading and structural boundary conditions. It must also respect the yield criterion, so the plastic moment cannot be exceeded at any section. A load for which such a distribution exists cannot be greater than the actual collapse load. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Which statement correctly describes the upper-bound or kinematic theorem of plastic collapse?
Explanation:
The kinematic theorem begins with an assumed geometrically admissible collapse mechanism. The associated plastic-hinge rotations and load displacements must be compatible with that mechanism. Virtual work then provides the corresponding collapse-load estimate. Among competing admissible mechanisms, the lowest calculated upper-bound load is the critical kinematic estimate. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
A lower-bound equilibrium solution proves that a frame can safely carry a load factor of 1.42, while the lowest kinematically admissible mechanism gives a collapse factor of 1.46. What can be concluded about the true plastic collapse factor?
Explanation:
A valid lower-bound solution cannot exceed the actual collapse load. Thus the true collapse factor is at least 1.42. A valid kinematic solution is an upper bound, so the true collapse factor cannot exceed 1.46. Without an exact coincident static and kinematic solution, its precise value is not established. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
Which set of material requirements is consistent with the IS 800 provisions for use of plastic analysis, unless adequate ductility is demonstrated by other permitted means?
Explanation:
Plastic analysis depends on substantial inelastic deformation and redistribution before the structure forms a mechanism. IS 800 therefore imposes ductility-related material requirements for plastic analysis. These include a yield stress not exceeding 450 MPa, a tensile-to-yield strength ratio of at least 1.2, and adequate yield-plateau and elongation characteristics. The steel must also possess suitable strain-hardening capability, with steels satisfying the prescribed material provisions deemed suitable. Therefore Option A is correct.
వివరణ:
Question 26
ప్రశ్న 26
Which statement correctly describes the member and connection requirements at plastic-hinge regions in an IS 800 plastic analysis?
Explanation:
Plastic analysis is valid only if the assumed collapse mechanism can physically develop. The connections must therefore possess sufficient strength and deformation capacity to transmit the actions while the required hinges rotate. Local buckling must not prevent the hinge section from sustaining its plastic moment through the necessary rotation. Lateral-torsional instability must also be controlled by suitable restraint at or near plastic-hinge regions in accordance with the code provisions. Therefore Option A is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: A. It determines collapse by allowing redistribution of bending moments as plastic hinges form, provided equilibrium, ductility and stability requirements are satisfied సమాధానం: A.
-
Question 2 ప్రశ్న 2Answer: A. Elastic analysis generally limits behaviour by an elastic stress distribution, whereas plastic analysis can utilize moment redistribution up to formation of a collapse mechanism సమాధానం: A.
-
Question 3 ప్రశ్న 3Answer: A. My = fyZe and Mp = fyZp సమాధానం: A.
-
Question 4 ప్రశ్న 4Answer: A. At an axis that divides the cross-sectional area into equal areas in tension and compression సమాధానం: A.
-
Question 5 ప్రశ్న 5Answer: C. 166.7 kN·m సమాధానం: C.
-
Question 6 ప్రశ్న 6Answer: C. 1.00 × 10^6 mm^3 సమాధానం: C.
-
Question 7 ప్రశ్న 7Answer: C. 1.50 సమాధానం: C.
-
Question 8 ప్రశ్న 8Answer: C. 250.0 kN·m సమాధానం: C.
-
Question 9 ప్రశ్న 9Answer: B. 1.15 and 15 percent సమాధానం: B.
-
Question 10 ప్రశ్న 10Answer: A. A yielded zone capable of significant inelastic rotation while sustaining approximately the plastic moment సమాధానం: A.
-
Question 11 ప్రశ్న 11Answer: A. Class 1 plastic section సమాధానం: A.
-
Question 12 ప్రశ్న 12Answer: B. r + 1 సమాధానం: B.
-
Question 13 ప్రశ్న 13Answer: B. 100 kN సమాధానం: B.
-
Question 14 ప్రశ్న 14Answer: B. 100 kN సమాధానం: B.
-
Question 15 ప్రశ్న 15Answer: C. 40 kN/m సమాధానం: C.
-
Question 16 ప్రశ్న 16Answer: C. 200 kN సమాధానం: C.
-
Question 17 ప్రశ్న 17Answer: C. 80 kN/m సమాధానం: C.
-
Question 18 ప్రశ్న 18Answer: B. 150 kN సమాధానం: B.
-
Question 19 ప్రశ్న 19Answer: B. 60 kN సమాధానం: B.
-
Question 20 ప్రశ్న 20Answer: C. 180 kN సమాధానం: C.
-
Question 21 ప్రశ్న 21Answer: A. External work done by the collapse loads during a compatible virtual displacement equals the internal work dissipated at the plastic hinges సమాధానం: A.
-
Question 22 ప్రశ్న 22Answer: A. Any bending-moment distribution that satisfies equilibrium and boundary conditions and nowhere exceeds the plastic moment provides a safe lower bound to the true collapse load సమాధానం: A.
-
Question 23 ప్రశ్న 23Answer: A. A kinematically admissible collapse mechanism evaluated by virtual work gives a collapse-load estimate that is not less than the true collapse load సమాధానం: A.
-
Question 24 ప్రశ్న 24Answer: B. It lies between 1.42 and 1.46 inclusive సమాధానం: B.
-
Question 25 ప్రశ్న 25Answer: A. Yield stress not exceeding 450 MPa, tensile-to-yield strength ratio at least 1.2, adequate yield plateau and elongation, and strain-hardening capability సమాధానం: A.
-
Question 26 ప్రశ్న 26Answer: A. Members and connections must permit the required plastic hinges and rotations to develop without premature local or lateral instability or connection failure సమాధానం: A.
