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Question 1
ప్రశ్న 1
Which statement most accurately describes load transfer through a plain jointed cement-concrete pavement resting on a prepared foundation?
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Explanation:
Rigid pavement action involves bending of a comparatively stiff concrete slab over an underlying foundation. Wheel loading creates tensile as well as compressive stresses in the slab. The supporting foundation supplies an upward reaction and affects slab deflections. Slab thickness, elastic properties and the state of support influence the resulting stress distribution. Concrete strength in compression does not eliminate the need to examine flexural tensile fatigue.
వివరణ:
Question 2
ప్రశ్న 2
Which statement best represents the structural design philosophy of IRC:58-2015 for plain jointed rigid highway pavements?
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Explanation:
IRC:58-2015 uses a fatigue-based approach for plain jointed concrete pavement design. Both bottom-up and top-down cracking mechanisms are addressed. Traffic loading and temperature gradients affect critical flexural tensile stresses. Repeated applications of different axle-load categories contribute to cumulative damage. Foundation support, joint load transfer and practical construction provisions are also important to the design.
వివరణ:
Question 3
ప్రశ్న 3
A plate-load assessment gives foundation reaction pressure p = 0.075 MPa for a deflection of 1.5 mm. Using the simplified Winkler relation k = p/δ, what is the modulus of subgrade reaction?
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Explanation:
The modulus of subgrade reaction represents reaction pressure per unit deflection under the adopted foundation model. The stated pressure is 0.075 MPa. The deflection of 1.5 mm equals 0.0015 m. Hence k = 0.075/0.0015 = 50 MPa/m. This parameter represents an effective support response rather than a fundamental elastic modulus of soil.
వివరణ:
Question 4
ప్రశ్న 4
Two trial rigid pavements have the same concrete slab properties and axle loading, but one rests on a foundation with a smaller effective modulus of subgrade reaction k. Which conclusion is appropriate?
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Explanation:
The modulus k expresses the support reaction developed for a given foundation deflection. A smaller effective k indicates more compliant support in the simplified foundation model. This can affect slab deflection and the distribution of bending stresses. The direction and magnitude of a particular critical stress change require the relevant structural analysis. Foundation improvement and drainage may therefore be important parts of a rigid pavement design.
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Question 5
ప్రశ్న 5
For a simplified Westergaard relative-stiffness calculation, take E = 30000 MPa, slab thickness h = 0.25 m, Poisson's ratio μ = 0.15 and foundation k = 50 MPa/m. Using l = [Eh³/{12k(1-μ²)}]^(1/4), what is l approximately?
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Explanation:
The radius of relative stiffness measures the relationship between concrete slab stiffness and foundation support. The stated expression uses a slab thickness of 0.25 m and the given elastic constants. Its numerator is 30000 times 0.25 cubed, or 468.75 in consistent units. Its denominator is 12 times 50 times (1 - 0.15 squared), or 586.5. Taking the fourth root of 468.75/586.5 gives approximately 0.946 m.
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Question 6
ప్రశ్న 6
Why are interior, edge and corner wheel-loading positions distinguished in classical rigid-pavement stress analysis?
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Explanation:
Interior, edge and corner are distinct classical load positions in rigid pavement analysis. At an edge or corner, slab continuity and load distribution differ from those at an interior location. The resulting deflection and flexural-stress fields are therefore not identical. The governing design condition also depends on joint load transfer, slab configuration and temperature effects. Modern fatigue checks should not assume one position controls every pavement arrangement.
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Question 7
ప్రశ్న 7
A rigid pavement tyre transmits a wheel load of 40 kN at an assumed uniform contact pressure of 0.80 MPa. What is the equivalent contact area?
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Explanation:
Uniform contact pressure is the wheel force divided by the loaded contact area. The wheel force is 40000 N. The pressure is 0.80 MPa, equivalent to 800000 N/m². Therefore the equivalent contact area is 40000/800000 = 0.050 m². This is an idealized loading area and not necessarily the exact observed tyre footprint.
వివరణ:
Question 8
ప్రశ్న 8
A simply supported concrete test beam is loaded at midspan by a single load P = 22.5 kN over a span L = 450 mm. Its cross-section is b = 150 mm and d = 150 mm. Using the explicitly stated flexural formula fr = 3PL/(2bd²), what is fr?
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Explanation:
The formula given in the question is for a rectangular beam under central point loading. Use the total load of 22500 N and all dimensions in millimetres. Substitution gives fr = 3 times 22500 times 450 divided by (2 times 150 times 150 squared). The result is 4.5 N/mm², equal to 4.5 MPa. This is the flexural strength under the stated loading arrangement and must not be confused with concrete compressive strength.
వివరణ:
Question 9
ప్రశ్న 9
A rigid pavement slab has calculated critical tensile stress of 2.4 MPa and specified design flexural strength of 4.8 MPa. What is its stress ratio, defined as tensile stress divided by flexural strength?
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Explanation:
The stress ratio is a dimensionless measure of applied flexural tensile demand relative to flexural strength. The calculated tensile stress is 2.4 MPa. The design flexural strength is 4.8 MPa. Division gives 2.4/4.8 = 0.50. An allowable fatigue life cannot be inferred from this ratio alone without the appropriate fatigue model.
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Question 10
ప్రశ్న 10
Which interpretation correctly distinguishes a positive daytime temperature gradient from a negative nighttime gradient in concrete slabs?
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Explanation:
Temperature differences through slab depth produce differential thermal strains. A hotter upper surface and a cooler lower surface create one curvature tendency. A cooler upper surface relative to the bottom produces the opposite tendency. These temperature states interact with wheel loading and support conditions in fatigue analysis. IRC rigid-pavement practice therefore considers temperature-related contributions to both bottom-up and top-down distress.
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Question 11
ప్రశ్న 11
A freely expanding 6.0 m long concrete slab undergoes a uniform temperature increase of 25°C. If the coefficient of linear expansion is 10 × 10^-6 per °C, what is the change in length?
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Explanation:
For unrestrained uniform thermal movement, the length change is ΔL = αLΔT. The coefficient is 10 × 10^-6 per degree Celsius. The slab length and uniform temperature change are 6.0 m and 25°C. Hence ΔL = 10 × 10^-6 × 6.0 × 25 = 0.0015 m, or 1.50 mm. Actual joint movement can differ if friction, restraint or nonuniform temperatures influence the slab.
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Question 12
ప్రశ్న 12
Which statement most accurately distinguishes slab warping from friction-related temperature stress?
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Explanation:
A through-thickness temperature gradient causes different thermal strains near the slab top and bottom. This difference creates a curling or warping tendency. A nearly uniform temperature change tends instead to expand or contract the entire slab. Resistance to that overall movement at the slab-support interface can generate frictional stresses. The two effects have different causes even though both may influence concrete pavement performance.
వివరణ:
Question 13
ప్రశ్న 13
In an idealized, fully restrained, linear-elastic concrete member undergoing a uniform temperature increase of 20°C, take E = 30000 MPa and α = 10 × 10^-6/°C. Ignoring creep, cracking and relaxation, what is the magnitude of the thermal stress σ = EαΔT?
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Explanation:
Full restraint prevents the free thermal strain of the idealized member. The magnitude of the prevented strain is αΔT. Here that strain is 10 × 10^-6 times 20 = 0.0002. Multiplication by E = 30000 MPa gives 6.0 MPa. A real pavement is not generally fully restrained, so this specified idealization should not be used as a universal pavement warping-stress calculation.
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Question 14
ప్రశ్న 14
Why is the slab–subbase interface friction relevant when selecting joint spacing and longitudinal restraint details?
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Explanation:
Concrete slabs undergo volume changes caused by temperature variation and shrinkage. Contact with the supporting layer can restrain part of the associated movement. This restraint creates in-plane forces and may contribute to cracking if poorly accommodated. Joint layout and tie-bar arrangements must therefore account for the anticipated movement and restraint conditions. Interface friction is different from both tyre skid resistance and vertical foundation stiffness.
వివరణ:
Question 15
ప్రశ్న 15
A rectangular rigid-pavement panel is 4 m long, 3.5 m wide and 0.25 m thick. Its concrete unit weight is 24 kN/m³. If the entire panel were sliding on a horizontal base with an assumed friction coefficient of 1.25, what is the limiting sliding-friction force F = μW?
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Explanation:
The panel volume is 4 times 3.5 times 0.25 = 3.5 m³. With a concrete unit weight of 24 kN/m³, its self-weight is 84 kN. Under the explicitly assumed whole-panel sliding model, friction force is coefficient times normal weight. Thus F = 1.25 times 84 = 105 kN. This total limiting force is not the same as a local tie-bar force distribution or the actual thermal stress in a partially restrained pavement.
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Question 16
ప్రశ్న 16
Which statement correctly describes the principal function of regularly spaced transverse contraction joints in jointed plain concrete pavement?
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Explanation:
Concrete undergoes shrinkage and temperature-related contraction after construction. Without controlled cracking locations, the slab may crack irregularly. Transverse contraction joints create planes at which the slab can accommodate these effects in a planned pattern. Joint sealing and suitable load-transfer provisions may also be needed for satisfactory service. These joints do not mean the slab is completely free of thermal stress or future cracking.
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Question 17
ప్రశ్న 17
Which joint arrangement most directly restrains separation between adjacent concrete lanes while permitting the necessary transverse-joint movement?
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Explanation:
Tie bars maintain contact between adjacent lanes or a tied concrete shoulder across a longitudinal joint. They are commonly deformed bars that develop anchorage in the adjacent concrete. Transverse joints have a different principal function because they accommodate slab length changes and control transverse cracking. Smooth dowels across those transverse joints permit movement while helping transfer vertical loads. The two types of steel reinforcement should therefore not be interchanged.
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Question 18
ప్రశ్న 18
Which explanation most accurately describes the function of smooth dowel bars crossing a transverse contraction joint in a jointed rigid pavement?
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Explanation:
Dowel bars transfer part of a wheel load from one slab across the transverse joint to the adjacent slab. This reduces differential deflection when the bars and supporting concrete perform properly. Smooth bars are aligned and detailed to accommodate longitudinal opening and closing of the joint. Tie bars, by contrast, are intended mainly to hold slabs together across longitudinal joints. Dowel design also requires consideration of spacing, anchorage, bearing and corrosion protection.
వివరణ:
Question 19
ప్రశ్న 19
At a transverse pavement joint, the measured deflection on the unloaded slab is 0.24 mm and that on the loaded slab is 0.40 mm. If deflection load-transfer efficiency is defined as LTE = 100(δ_unloaded/δ_loaded), what is LTE?
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Explanation:
The definition provided compares the deflection of the unloaded slab to the deflection of the loaded slab. The respective measured deflections are 0.24 mm and 0.40 mm. Their ratio is 0.24/0.40 = 0.60. Multiplying by 100 gives a load-transfer efficiency of 60%. This is a deflection-based measure for the specified testing arrangement and does not independently establish dowel bearing safety.
వివరణ:
Question 20
ప్రశ్న 20
A jointed concrete pavement repeatedly ejects water and fine foundation material from transverse joints under heavy vehicle passes. Which explanation best describes the primary distress mechanism and its consequence?
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Explanation:
Pumping is the ejection of water and fine particles through joints or cracks under repeated load action. It requires moisture and a foundation susceptible to erosion or displacement. The repeated slab deflection contributes to the movement of water and fines. Progressive material loss can create voids and reduce joint support. Drainage, erosion-resistant support and effective joint load transfer help mitigate this mechanism.
వివరణ:
Question 21
ప్రశ్న 21
Under a stated linear cumulative-fatigue rule D = Σ(ni/Ni), a concrete pavement receives 150000 repetitions from a group with allowable life 600000 and 100000 repetitions from another group with allowable life 500000. What is the total damage D?
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Explanation:
The problem explicitly specifies a linear summation of fatigue damage fractions. The first load group contributes 150000/600000 = 0.25. The second group contributes 100000/500000 = 0.20. Adding the two fractions gives D = 0.45. The calculation does not determine the allowable repetitions, which have been supplied, and it does not replace checking other damage mechanisms.
వివరణ:
Question 22
ప్రశ్న 22
A rigid pavement trial design has computed bottom-up cumulative fatigue damage 0.65 and top-down cumulative fatigue damage 0.47. If the specified combined criterion is D_bottom + D_top ≤ 1.0, what is the correct design judgement?
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Explanation:
The trial must satisfy the combined cumulative-fatigue criterion stated in the question. Bottom-up damage is 0.65 and top-down damage is 0.47. Their sum is 1.12. This exceeds the allowed combined value of 1.0, so the trial is not acceptable under the specified criterion. A revised slab or support arrangement must be analysed again rather than comparing the two damage components only in isolation.
వివరణ:
Question 23
ప్రశ్న 23
Using the simplified IRC tie-bar steel-area relation As = b f h γ / Sst per metre of longitudinal joint, take lane width b = 3.5 m, interface friction coefficient f = 1.5, slab thickness h = 0.25 m, concrete unit weight γ = 24000 N/m³, and allowable steel tensile stress Sst = 200 N/mm². What is As?
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Explanation:
The problem specifies the formula and all design parameters for the simplified tie-bar area calculation. The force-related numerator is 3.5 times 1.5 times 0.25 times 24000 = 31500 N per metre of joint. The allowable steel stress is 200 N/mm². Division gives As = 31500/200 = 157.5 mm² per metre of joint. Bar diameter, spacing, development length and construction provisions require separate checks before final detailing.
వివరణ:
Question 24
ప్రశ్న 24
Two rigid-pavement alternatives have the same axle loading and slab thickness, but one includes a tied concrete shoulder and sound transverse-joint load transfer. Which design implication is most reasonable?
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Explanation:
A tied concrete shoulder can provide additional lateral support to the outer traffic lane. Proper joint load transfer distributes part of a wheel load to adjoining slabs. Both features may influence deflections and critical tensile stresses. IRC rigid-pavement design distinguishes relevant shoulder and joint configurations rather than treating them as identical structures. Their benefits must be assessed alongside traffic, temperature and foundation characteristics.
వివరణ:
Question 25
ప్రశ్న 25
An engineer finds that a trial concrete slab passes a single wheel-load flexural check but fails the specified combined fatigue assessment and shows a high risk of pumping because of poor foundation drainage. Which next action is most defensible?
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Explanation:
Rigid pavement service life depends on repeated wheel loads and temperature effects, not one instantaneous stress check alone. An unsatisfactory fatigue result shows that the trial does not meet its specified structural criterion. Poor foundation drainage can additionally promote erosion, pumping and loss of support. The designer should consider revisions to slab thickness, support layers, drainage and joint provisions as necessary. Revised alternatives must then be checked again for all relevant damage and performance requirements.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. The relatively stiff slab distributes wheel loads by plate bending while the foundation provides supporting reaction. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. Evaluate cumulative fatigue associated with bottom-up and top-down cracking under axle loading and temperature effects, along with relevant joint and foundation requirements. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 50 MPa/m. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. The weaker support generally permits greater slab deflection, so stresses and fatigue performance must be reassessed. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 0.946 m. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: D. Because the available slab continuity and support geometry differ with load position, altering bending and critical tensile stresses. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 0.050 m². సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 4.5 MPa. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 0.50. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. A hotter slab top can produce one curling tendency and a cooler top the opposite tendency, with potentially different critical fatigue locations. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. 1.50 mm. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. A through-depth temperature difference produces curling or warping tendencies, whereas restrained overall temperature-related movement can generate frictional stresses along the slab-support interface. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. 6.0 MPa. సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: D. Resistance to thermal and shrinkage movements can generate in-plane forces and influence joint and tie-bar requirements. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: B. 105 kN. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: A. They control where shrinkage and thermal-contraction cracks form by creating predetermined weakened planes. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. Deformed tie bars across longitudinal joints, with appropriately detailed transverse joints for controlled lengthwise movements. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. They transfer shear-related wheel load between adjacent slabs while allowing joint opening and closing along the dowel axis. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 60%. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. Pumping of erodible fines caused by water and repeated slab deflections, potentially producing loss of support and joint faulting. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. 0.45. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: D. Fail, because the combined fatigue damage is 1.12, exceeding the stated limit. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 157.5 mm² per metre of joint. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. The improved lateral restraint and joint load sharing can change critical slab stresses and should be reflected in the structural analysis. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. Revise the slab and support or drainage design, verify joint load transfer, and repeat the critical fatigue and serviceability checks. సమాధానం: C.