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Question 1
ప్రశ్న 1
A proposed foundation has an irregular L-shaped plan and applies approximately uniform vertical pressure to the ground. Which statement best describes the purpose of Newmark's influence chart in evaluating stresses beneath the foundation?
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Explanation:
Newmark's influence chart provides a graphical method for estimating vertical stress increments beneath loaded areas. It is particularly useful when the foundation plan has an irregular shape. The method uses the influence of different surface loading elements on a selected subsurface observation point. The contributions of the loaded elements are added to obtain the total vertical stress increment. It does not directly determine soil shear strength, groundwater conditions or ultimate bearing capacity.
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Question 2
ప్రశ్న 2
Which combination of soil behaviour assumptions forms the theoretical basis of the conventional Newmark influence chart derived from Boussinesq's elastic stress distribution?
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Explanation:
The conventional Newmark chart is developed from Boussinesq's elastic stress solution. The underlying soil is idealized as homogeneous and isotropic. Its mechanical response is assumed to be linearly elastic. The ground is represented as a semi-infinite continuum extending beneath the loaded surface. Strong stratification, anisotropy and nearby rigid boundaries can make these assumptions unsuitable without further analysis.
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Question 3
ప్రశ్న 3
A conventional Newmark chart is designed with 20 equal angular sectors and ten radial influence increments of 0.10 each. Which statement about the chart construction is correct?
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Explanation:
The chart divides the complete theoretical influence into ten equal radial increments. Each radial increment is further divided into 20 angular sectors. This produces 10 times 20, or 200 theoretical influence units. The cumulative influence factors from 0.10 to 0.90 correspond to nine finite circular boundaries. A cumulative influence factor of 1.00 is approached only as the circular loaded radius tends to infinity.
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Question 4
ప్రశ్న 4
Why are the concentric circular boundaries in a conventional Newmark chart unequally spaced?
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Explanation:
The influence of a uniformly loaded surface element depends on its distance from the observation point. Elements farther from the point generally contribute less vertical stress for a given area. The circular boundaries are therefore selected using the Boussinesq stress distribution. Their spacing makes the individual influence units contribute equal stress increments under the same uniform pressure. Consequently the chart units are equal in influence rather than equal in physical plan area.
వివరణ:
Question 5
ప్రశ్న 5
In using a Newmark chart to determine vertical stress at a point P situated 5 m below an irregular foundation, which plotting procedure is correct?
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Explanation:
The stress increment is required at one specified point within the soil mass. The loaded area must first be drawn using the scale corresponding to that point's depth. The surface projection of the observation point is then positioned at the centre of the Newmark chart. The influence units covered by the loaded plan are counted in this position. Placing the foundation centroid at the chart centre instead would generally give the stress at a different location.
వివరణ:
Question 6
ప్రశ్న 6
The reference depth line AB on a Newmark chart measures 5 cm. Vertical stress is required at an actual depth of 8 m, and the foundation plan must be drawn to the corresponding chart scale. What is the required representative fraction?
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Explanation:
The chart reference line AB must represent the actual depth at which stress is required. In this case, 5 cm on the drawing represents 8 m on the ground. Converting 8 m to centimetres gives 800 cm. The representative fraction is therefore 5/800 = 1/160. Thus the required drawing scale is 1:160.
వివరణ:
Question 7
ప్రశ్న 7
For construction of the first circular boundary of a conventional 200-unit Newmark chart, the Boussinesq solution gives R1/z = 0.270 approximately. If the stress observation depth is z = 6 m, what actual radial distance R1 does this boundary represent?
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Explanation:
The first circle corresponds to the initial cumulative influence increment of 0.10. Its radius-to-depth ratio is supplied as R1/z = 0.270. Therefore the actual radius is R1 = 0.270z. Substitution gives R1 = 0.270(6) = 1.62 m. This is the ground distance represented by the first circular boundary at the specified observation depth.
వివరణ:
Question 8
ప్రశ్న 8
A stress observation point is situated outside the plan boundary of a uniformly loaded rectangular foundation. Which procedure is correct when applying a Newmark influence chart?
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Explanation:
A surface foundation can produce vertical stress increments at points outside its plan boundary. The elastic stress influence extends laterally as well as downward. The observation point must remain the reference location for the chart. Its surface projection is placed at the chart centre, even when this position lies outside the loaded area. The offset loaded area is then counted to estimate its vertical stress contribution.
వివరణ:
Question 9
ప్రశ్న 9
An irregular foundation applies a uniform net pressure of 160 kPa. When its correctly scaled plan is placed on a Newmark chart having influence value 0.005, the loaded area covers 75 equivalent influence units. What is the vertical stress increment at the selected point?
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Explanation:
The Newmark stress expression for uniform pressure is delta_sigma_z = qNI. The applied net pressure is 160 kPa. The counted number of equivalent influence units is 75, and each has an influence value of 0.005. Therefore delta_sigma_z = 160(75)(0.005) = 60 kPa. The result represents the additional vertical stress at the selected subsurface point.
వివరణ:
Question 10
ప్రశ్న 10
A Newmark chart has influence value 0.005. An irregular area carrying uniform pressure q = 150 kPa produces a calculated vertical stress increment of 45 kPa at a selected point. How many equivalent influence units must the loaded area cover?
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Explanation:
The relation between stress increment and chart count is delta_sigma_z = qNI. Rearranging gives N = delta_sigma_z/(qI). The denominator is 150(0.005) = 0.75 kPa per influence unit. Dividing 45 kPa by 0.75 kPa gives N = 60. Therefore the plotted loaded area must cover 60 equivalent influence units.
వివరణ:
Question 11
ప్రశ్న 11
A nonstandard Newmark influence chart is constructed with 100 equal-influence units, giving influence value 0.010 per unit. A uniformly loaded area carrying 200 kPa covers 34 equivalent units. What is the resulting vertical stress increment?
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Explanation:
Different chart constructions can have different influence values. The question specifies 100 equal units, so each has an influence value of 0.010. The uniformly loaded area covers 34 equivalent units. Therefore the vertical stress increment is 200(34)(0.010). This gives 68 kPa, showing that the actual chart influence value must be used rather than assuming 0.005 for every chart.
వివరణ:
Question 12
ప్రశ్న 12
An irregular loaded area covers 32 complete Newmark influence units and approximately half of one additional unit. The chart influence value is 0.005, and the uniform applied pressure is 120 kPa. What approximate vertical stress increment should be reported?
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Explanation:
A partially covered influence unit may be counted fractionally in an approximate graphical calculation. The 32 complete units and one half-unit give 32.5 equivalent units. The total dimensionless influence factor is 32.5(0.005) = 0.1625. Multiplication by the applied pressure gives 120(0.1625) = 19.5 kPa. The result is an approximate stress increment consistent with the stated fractional-unit counting procedure.
వివరణ:
Question 13
ప్రశ్న 13
Two separate, nonoverlapping irregular loaded areas influence the same subsurface point. Area A carries 120 kPa and covers 40 Newmark units, while Area B carries 180 kPa and covers 30 units on the same chart. If each unit has influence value 0.005, what is the combined vertical stress increment?
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Explanation:
The vertical stress contributions from separate loaded areas can be combined by elastic superposition. Area A contributes 120(40)(0.005) = 24 kPa. Area B contributes 180(30)(0.005) = 27 kPa. Adding the individual stress increments gives 24 + 27 = 51 kPa. The different loading intensities must be applied separately rather than using an unweighted total count.
వివరణ:
Question 14
ప్రశ్న 14
An irregular foundation carries a uniform base pressure of 100 kPa over a region covering 70 influence units. An additional uniform surcharge of 50 kPa acts only on a smaller portion of this region covering 20 of those units. Using Newmark influence value 0.005, what is the combined vertical stress increment at the observation point?
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Explanation:
The original pressure acts over the entire region represented by 70 influence units. Its contribution is 100(70)(0.005) = 35 kPa. The additional surcharge acts only over the specified 20 units. Its separate contribution is 50(20)(0.005) = 5 kPa. Adding the original and incremental contributions gives a total vertical stress increment of 40 kPa.
వివరణ:
Question 15
ప్రశ్న 15
Two overlapping plan regions A and B together form one irregular foundation area carrying the same uniform pressure of 200 kPa. Region A covers 80 influence units, Region B covers 60 units, and their common overlap covers 20 units. If each unit has influence value 0.005, what is the stress increment due to the uniformly loaded union of the two regions?
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Explanation:
The two regions form a single uniformly loaded union. Influence units in the overlap must be counted only once because the pressure there is not doubled. The total covered count is 80 + 60 - 20 = 120 units. The corresponding influence factor is 120(0.005) = 0.60. Therefore the vertical stress increment is 200(0.60) = 120 kPa.
వివరణ:
Question 16
ప్రశ్న 16
An L-shaped foundation area is represented by subtracting an unloaded rectangular cutout from a larger uniformly loaded rectangle. At the observation point, the large rectangle covers 95 equivalent Newmark units and the cutout covers 25 units. If the foundation pressure is 150 kPa and the chart influence value is 0.005, what is the vertical stress increment?
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Explanation:
An L-shaped loaded area can be represented as a larger rectangle minus an unloaded cutout. The cutout contribution must be subtracted because it does not carry the specified pressure. The effective influence-unit count is 95 - 25 = 70. The vertical stress increment is 150(70)(0.005). Therefore the correct result is 52.5 kPa.
వివరణ:
Question 17
ప్రశ్న 17
A construction stage applies an additional uniform pressure of 180 kPa over a loaded region covering 65 Newmark influence units. At the same time, excavation removes an existing uniform pressure of 80 kPa from a separate region covering 15 units. Assuming elastic superposition and influence value 0.005, what is the net change in vertical stress at the observation point?
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Explanation:
Newly applied loading creates a positive stress increment in the stated calculation. The added loading contributes 180(65)(0.005) = 58.5 kPa. Excavation unloading is represented as a negative pressure increment and contributes -80(15)(0.005) = -6 kPa. The net change is 58.5 - 6 = 52.5 kPa. Thus the combined construction stage produces a vertical stress increase of 52.5 kPa at the selected point.
వివరణ:
Question 18
ప్రశ్న 18
A uniformly loaded irregular foundation applies q = 150 kPa. Using the correct drawing scale separately at two observation depths, the loaded area covers 90 equivalent Newmark units at the shallower point and 54 units at the deeper point. For influence value 0.005, what are the respective vertical stress increments?
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Explanation:
The loaded area must be drawn to the appropriate scale for each observation depth. At the shallower point, the stress increment is 150(90)(0.005) = 67.5 kPa. At the deeper point, it is 150(54)(0.005) = 40.5 kPa. The specified counts represent different plan overlays using their corresponding depth scales. The calculated vertical stress increment therefore decreases from 67.5 kPa to 40.5 kPa.
వివరణ:
Question 19
ప్రశ్న 19
A fixed Newmark chart has a reference depth line AB measuring 6 cm. For a stress calculation at depth 6 m, a 4 m actual foundation dimension is plotted as 4 cm. If the same foundation is analysed at depth 12 m using the unchanged chart, what should the plotted length of that 4 m dimension become?
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Explanation:
The chart reference line must represent the observation depth in every calculation. Initially, 6 cm represents 6 m, giving a drawing scale of 1 cm to 1 m. At the new depth of 12 m, the same 6 cm reference line represents twice the actual depth. The corresponding scale becomes 1 cm to 2 m. Therefore a 4 m foundation dimension must be plotted as 4/2 = 2 cm.
వివరణ:
Question 20
ప్రశ్న 20
For a conventional Newmark chart with 20 angular sectors and influence value 0.005, the second circular boundary corresponds to R2/z = 0.40 and encloses two complete radial influence increments. A uniformly loaded circular area coincides with this boundary at depth z = 6 m and carries q = 150 kPa. What are its actual radius and calculated stress increment directly beneath its centre?
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Explanation:
The actual radius represented by the second boundary is R2 = 0.40z. At z = 6 m, the radius is 0.40(6) = 2.4 m. Two complete radial increments with 20 sectors each contain 40 influence units. Their combined influence factor is 40(0.005) = 0.20. Therefore the stress increment is 150(0.20) = 30 kPa.
వివరణ:
Question 21
ప్రశ్న 21
An observation point lies outside the plan of a uniformly loaded waste-storage platform. With the surface projection of the observation point correctly centred on a Newmark chart, the platform covers 16 equivalent influence units. If the uniform pressure is 180 kPa and the influence value is 0.005, what is the vertical stress increment at the observation point?
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Explanation:
A loaded area can produce stresses in the soil outside its vertical plan projection. The Newmark chart accounts for this effect through its radial influence distribution. The offset platform covers 16 equivalent influence units at the specified observation point. The vertical stress increment is therefore 180(16)(0.005) = 14.4 kPa. Taking the increment as zero merely because the observation point is outside the platform would be incorrect.
వివరణ:
Question 22
ప్రశ్న 22
A foundation is constructed after excavation of soil producing an initial vertical overburden pressure of 45 kPa at the founding level. The completed foundation applies a gross uniform contact pressure of 250 kPa. If the project analysis uses the net pressure increment and the appropriately scaled Newmark chart gives 64 influence units with value 0.005, what is the vertical stress increment at the selected underlying point?
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Explanation:
The net foundation pressure is the gross contact pressure less the stress removed by excavation. Under the stated assumptions, qnet = 250 - 45 = 205 kPa. The chart influence factor for the observation point is 64(0.005) = 0.32. The vertical stress increment is therefore 205(0.32) = 65.6 kPa. Using the gross pressure directly would neglect the specified removal of the original overburden.
వివరణ:
Question 23
ప్రశ్న 23
An irregular raft foundation applies a uniform net pressure of 200 kPa above two compressible clay layers. For Layer 1, H1 = 2 m, mv1 = 4.0 × 10^-4 kPa^-1 and the representative average Newmark count is 90 units. For Layer 2, H2 = 3 m, mv2 = 3.0 × 10^-4 kPa^-1 and the representative average count is 40 units. Using influence value 0.005 and assuming one-dimensional primary compression with constant mv in each layer, what is the combined eventual settlement?
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Explanation:
The representative average vertical stress increment in Layer 1 is 200(90)(0.005) = 90 kPa. Its settlement is mv1 DeltaSigma1 H1 = (4.0 × 10^-4)(90)(2) = 0.072 m. The average stress increment in Layer 2 is 200(40)(0.005) = 40 kPa. Its settlement is (3.0 × 10^-4)(40)(3) = 0.036 m. Adding the two contributions gives a combined eventual settlement of 0.108 m, or 108 mm.
వివరణ:
Question 24
ప్రశ్న 24
The same uniformly loaded irregular area is analysed using two Newmark charts at the same observation point and depth. Chart A has influence value 0.005 and gives 44 equivalent covered units, while a finer Chart B has influence value 0.0025 and gives 88 equivalent units. If the applied uniform pressure is 200 kPa, what are the calculated stress increments from Charts A and B?
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Explanation:
Different influence charts may use different numbers of elements and corresponding influence values. Chart A has a total influence factor of 44(0.005) = 0.22. Chart B has a total influence factor of 88(0.0025) = 0.22. Both therefore give a vertical stress increment of 200(0.22) = 44 kPa. A larger unit count in a finer chart does not necessarily represent a larger stress when its individual influence value is smaller.
వివరణ:
Question 25
ప్రశ్న 25
A major raft foundation is proposed over a deposit containing alternating stiff and very soft soil layers, with shallow bedrock and substantial differences in stiffness. A conventional Newmark chart gives a vertical stress increment under the homogeneous elastic half-space assumption. Which engineering interpretation is most appropriate?
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Explanation:
The conventional Newmark influence chart is based on an ideal homogeneous elastic half-space. A strongly layered deposit can respond differently because its stiffness and stress distribution vary with depth. Shallow bedrock also conflicts with the assumed semi-infinite soil medium. Foundation rigidity and nonuniform contact pressure may further influence the actual behaviour. Therefore the chart result should be interpreted alongside representative soil properties and, where needed, more suitable layered or numerical analyses.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. It estimates the vertical stress increment at a selected subsurface point by summing the influences of the loaded portions of the plan area. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. A homogeneous, isotropic, linearly elastic and semi-infinite soil continuum subjected to surface loading. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. There are 200 theoretical influence units, nine finite circular boundaries and an outer influence zone extending theoretically to infinity. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. To make each chart unit contribute the same vertical stress increment under identical uniform surface pressure. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Draw the loaded area to the required depth scale and place the surface projection of P at the chart centre. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 1:160. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 1.62 m. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. The surface projection of the external observation point must be placed at the chart centre, and the influence units covered by the offset foundation area must be counted. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 60.0 kPa. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 60. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. 68 kPa. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: C. 19.5 kPa. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: A. 51 kPa. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. 40 kPa. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: D. 120 kPa. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 52.5 kPa. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 52.5 kPa increase. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. 67.5 kPa and 40.5 kPa. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 2 cm. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. Radius 2.4 m and stress increment 30 kPa. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 14.4 kPa. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 65.6 kPa. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 108 mm. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. 44 kPa and 44 kPa. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. The chart provides an idealized estimate, and the actual stratification, foundation rigidity and ground deformation characteristics should be considered before using the result in detailed settlement design. సమాధానం: B.