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All 100 questions from the WBPSC AE 1998 (P.W. Deptt., common to all categories) paper (1998 — P.W. Deptt., Common to All Categories). Answers and explanations for the first 10 are free below; the rest are in the full previous-papers pack, along with timed practice and scoring.
| Paper | 1998 — P.W. Deptt., Common to All Categories |
| Questions | 100 |
| Marking (current pattern) | +2 for a right answer, −2/3 for a wrong one |
| Answers on this page | Q1–Q10 (rest in the full pack) |
Q1.A bar is subjected to an uni-axial loading. The material of the bar will fall in shear if the ratio of the shear strength to the tensile strength is not more than
Answer: (a) 1/2
Under a uniaxial pull σ the direct stress on a transverse plane is σ, but the shear stress varies with the inclination of the plane and reaches its maximum on the 45° plane: τmax = (σ/2) sin 2θ |θ=45° = σ/2 The bar therefore carries a shear intensity equal to half the applied tensile intensity. Whichever limit is reached first governs the mode of failure.
Q2.The maximum shear stres in a bar of rectangular corssection subjected to beinding bears the following ratio to the average shear stress
Answer: (d) 1.5
For a rectangular section the shear stress from τ = VAȳ/(Ib) is parabolic — zero at the extreme fibres, maximum at the neutral axis: τmax = 3V/(2bd) = 1.5 × V/(bd) = 1.5 τavg The average value V/(bd) is what a naive "load ÷ area" calculation gives; the true peak at the neutral axis is 50% higher. This 1.5 factor is why the shear check in a rectangular RCC beam is written on the nominal stress τv = Vu/(bd) with permissible values that already allow for the real distribution.
Q3.A beam AB of length L and of flexural figidly EI is hinged at its ends and is subjected to two clockwise couples of value M each at A and B. The central deflection is
Answer: (d) Zero
Take A at x = 0, B at x = L, both couples M clockwise. Vertical equilibrium requires an equal and opposite reaction couple, so RB = 2M/L up and RA = 2M/L down. The bending moment is then M(x) = M(1 − 2x/L) i.e. +M at A, zero at mid-span and −M at B. The M/EI diagram is antisymmetric about the centre. Load a conjugate beam with that antisymmetric diagram: the conjugate bending moment (= real deflection) is itself antisymmetric, so it must vanish at mid-span.
Q4.In a simply supported beam of uniform rectangular crossection carrying an isolated force at the centre, the depth is halved, the deflection of the beam at the centre will be increased by a factor
Answer: (c) 8
For a simply supported beam with a central point load, δ = WL³/(48EI), with I = bd³/12 for a rectangle. Deflection is inversely proportional to I, and I goes as the cube of the depth. Halving d divides I by 8, so the deflection is multiplied by 8. This cube law is the reason depth, not width, is the economical variable in beam design: doubling the depth is eight times as effective against deflection as doubling the width.
Q5.The bending moment at the central section of a beam of span equal to L loaded by a uniformly distributed load of w/unit run covering a distance equal to L/4 form either ends of the support is given by
Answer: (c) WL/32
The load is a UDL of intensity w over a length L/4 at each end, so the total load is wL/2 and, by symmetry, each reaction is wL/4. Cut at mid-span and take the left half.
Q6.A simply supported beam of span = L is loaded by a concentrated load W at a distance 'a' (a<L/2) from the right-hand support. The bending moment at the centre of span is given by
Answer: (c) Wa/2
The load W sits at a from the right support, hence at L − a from the left. Taking moments about B: RA = Wa/L Since a < L/2, the load lies in the right half and the mid-span section is to the left of the load. Cutting there and taking the left segment, only RA acts: MC = RA × L/2 = (Wa/L)(L/2) = Wa/2 The trap is to reach for the standard Wab/L, which is the moment under the load, not at mid- span.
Q7.The bending moment at the ends of a fixed ended beam of span = L, loaded by two loads each equal to W acting at L/3 from each support is
Answer: (a) 2WL/9
Use the standard fixed-end moment MA = Wab²/L² for each load and superpose. Load 1 at a = L/3, b = 2L/3: MA1 = W(L/3)(2L/3)²/L² = 4WL/27. Load 2 at a = 2L/3, b = L/3: MA2 = W(2L/3)(L/3)²/L² = 2WL/27. MA = 4WL/27 + 2WL/27 = 6WL/27 = 2WL/9 By symmetry MB = 2WL/9 as well. The free bending moment under each load is WL/3, so the net sagging moment in the middle third is WL/3 − 2WL/9 = WL/9 — the fixed ends have absorbed two- thirds of the simple-beam moment.
Q8.The slope of the free end of a cantilver AB, fixed at A, free at B, of lenght = L and of constant 'EI' loaded by a uniformity distributed load of w/unit length run over half the length from A is given by
Answer: (c) WL³/48EI
The outer half (L/2 to L) carries no load, so it is subjected to no bending moment at all: it stays straight and simply rotates with the section at x = L/2. Hence θB = θL/2 For a cantilever of length a with a UDL w over its whole length, the tip slope is wa³/(6EI). Putting a = L/2: θB = w(L/2)³/(6EI) = wL³/(48EI) The deflection at B, by contrast, is not that of the half-cantilever: it is δL/2 + θL/2(L/2) = wL⁴/(128EI) + wL⁴/(96EI) = 7wL⁴/(384EI).
Q9.In a closed circular ring the number of statical indeterminacies are
Answer: (b) three
A closed plane ring is a structure with no ends: it is statically determinate externally (three simple supports would do) but internally indeterminate. To find the internal forces you must cut it, and at any cut a plane frame member transmits three actions — axial force, shear force and bending moment. Cutting the ring once turns it into a curved determinate member; the three released actions are the three redundants.
Q10.A cross-section of a body is acted upon by a stress, intensity 'p'. The tangential stress 'pt' acting one a plane inclined at an angle θ to the direction of applied stress is given by
Answer: (b) pt = p.(sin²θ)/2
On a plane inclined at θ in a body carrying a single direct stress p, resolving the force on the plane into components normal and tangential to it gives pn = p cos²θ and pt = p sinθ cosθ = (p/2) sin 2θ The tangential (shear) component is therefore half of p times sin 2θ. It is zero on the transverse plane and on the longitudinal plane, and maximum = p/2 at θ = 45° — the result that governs slip- plane failure in ductile metals.
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Unlock all 12 papers — Rs. 349Q11.The principal plane of a stressed body is the plane where
Q12.A cantilever beam of leangth 'l' acted upon by an external moment 'M' at the free end. Shear force (S.F) at a distance 'x' from free end shall be
Q13.Position of maximum bending moment in an arbitrarily loaded beam is were
Q14.For calculating the maximum bending moment at any section of a simply support beam for a UDL of length smaller than the span-length of the beam, the load is to be placed in such a way that
Q15.To calculate by 'conjugte beam method' the slope and deflection at any section of a beam for any loading the conjugate beam is assumed to be loaded by
Q16.A single bay single storied symetrical portal with indentical stiffness of al the there members and fixed at supports carries un UDL covering the horizontal member. Number of points of contraflecture will be
Q17.A square element id subjected to biaxial tensile stresses of equal magnitudes. The shear strees on an inclined plane is equal to
Q18.In limit state method of design the partial safety factor for concrete is
Q19.The minimum reinforcement of steel of grade Fe 415 in either direction of a slab is given by
Q20.The development length is
Q21.In case of a single angle strut the 'l/r' ratio should not be exceeded
Q22.The permissible stress for Fe 415 steel in tension used near water face for water tanks is
Q23.The sulphate content as SO4 in water to be used for concreting should not exceed
Q24.Corner steel is provided in slab
Q25.The minimum percentage of steel in case of an axially loaded R.C.C. colum is given by
Q26.An R.C.C. column may be considered as short when the ratio of effective length of column to the least lateral dimension does not exceed
Q27.In under-reinforced section
Q28.Permitted maximum slenderness ratio of a compression steel member under dead load and live load only in
Q29.In pre-stressed concrete members, percentage concrete sectional area utilised
Q30.The top reinforcement at the support section of a continuous beam shall be extended form eithre face of the column up to a distance equal to
Q31.The maximum spacing for vertical stirrups of shear reinforcements shall not exceed
Q32.Vicant apparatus is used to determine
Q33.For purpose of determining the seismic force in structure, India has been classfied into several zones an calcutta falls in
Q34.Which of the following materials has the highest carbon percentage?
Q35.Long term losses in a post-tensioned girded are due to
Q36.The degree of seturation of a soil at its shrinkage limit is
Q37.Phreatic line indicates the surface of seepage wate within the body of
Q38.Maximum deflection in a steel beam should not generally exceed
Q39.The maximum deflection in a steel beam should not generally exceed
Q40.The vertical stiffener of a plate girder is provided is d/tv exceeds-
Q41.The critical section for bending in a footing should be considered at
Q42.Indicate the correct response in respect of analysis of a symmetrical building frame for lateral loads by portal method.
Q43.The most active clay mineral is
Q44.If the plasticity index of a soil mass is zero the soil is
Q45.The pressure transmitted through grain to train at the contact points through a soil mass is termed as
Q46.Liquid and plastic limits exist in
Q47.Colluvial soils are those which are
Q48.For estimating bearing capacity of a saturated clay deposit the shear strength perometers should be determined from
Q49.Loam is a mixture of
Q50.The voiled ratio of a soil smaple decreased from 1.59 to 1.25 when pressure was increassed from 25 to 50 tonne/m². The coefficient at compressibility in m²/tonne is
Q51.If Kx and Kz are horizontal and verbical permeability respectively of a straified deposit, effective permeability is given by
Q52.The total active earth pressure on a retaining wall due to dry sandy back fill acts at H/3 above the base of the wall and is directly proportional to
Q53.If water is pumped out, the effective stress within a soil mass is
Q54.For a fully saturated soil of mositure content 22% P.L. and L.L. are found to be 18% and 45% respectively. P.I. of the soil is
Q55.Whole circle bearing of S45°W is
Q56.R being radius of earth, curvature correction in levelling between two stations depart is expressed as (with usual notation)
Q57.Log sine formula is used in triangulation to check
Q58.A well-conditioned triangle is one which
Q59.Three-point problem is most suitable in
Q60.To minimum errors in fly levelling
Q61.Contour gradient is
Q62.The 'alidate' is used in
Q63.The process of fixing intermediate points is
Q64.Error due to 'Sag' in chain survey is
Q65.The imaginary line joining the inter-section of crosswise of diaphragm to opticaloentre of object glass and its continuation is
Q66.If h is the difference in height between the end points of a chain of length L the slope correction is approximately equal to
Q67.A prismatic compass has 0° graduation at the
Q68.The horizontality of trunnion axis in theodolite is checked by
Q69.Local attraction in a prismatic compass is due to
Q70.Seasoning of timber is done to
Q71.Strength of cement is mainly due to
Q72.The purpose of adding pozzolana in cement is
Q73.Vicant's apparatus is used for determination of
Q74.Critical path of a network is
Q75.Which of the following trees yields soft wood?
Q76.Which cement is expected to have the highest compressive strength after 72 hours?
Q77.Heat of hydration in cement is mainly due to
Q78.Initial setting time of ordinary protland cement
Q79.In steel as the perdentage of carbon increase which of the following
Q80.The percentage of carbon in low carbon steel is
Q81.Membrane curing of concreate is adopted
Q82.Purpose of slumb test in concrete is to ensure
Q83.Suction lift of a water supply pump is equal is
Q84.House connections are given from the water mains through
Q85.For laminar flow in a circular pipe, momentum correction factor is
Q86.In the computation of force exerted by a steady jet striking a fixed falt plate held perpendicular to it exis the assumption is
Q87.In a steady pipe flow, hydraulic grade line
Q88.For complete turbulent flow through pipes, friction factor
Q89.For a primematic channel of small alope relation between flow depths imain diately before and after a hydraulic jump are derived by using
Q90.In laminar flow through a pipe
Q91.In rainfall studies, a Thiessen polygon
Q92.Base flow of a stream is
Q93.A venturimeter is used for measuring
Q94.The rating curve applicable to a section of a stream, depends on
Q95.In uniform open channel flow of small slope and curvature
Q96.The maximum velocity in open channels occurs
Q97.Alternate depths of flow in a sloping channel are the depth
Q98.For the same discharge of a given liquid carried by a circular pipe having laminar flow, reduction of diameter to half will increase the wall shear stress by a factor of
Q99.Ryve's formula for flood estimate in cumecs is
Q100.The most essential condition for a hydraulic jump to form is
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