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WBPSC AE Civil practice questions

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Question 1Engineering Mechanics and Strength of Materials

A simply supported beam of span 6 m carries a central point load of 40 kN. What is the maximum bending moment in the beam?

  1. 120 kN·m
  2. 240 kN·m
  3. 60 kN·m Correct answer.
  4. 30 kN·m

Explanation

For a simply supported beam with a central point load W over span L, the maximum bending moment occurs at midspan and equals WL/4. Substituting W = 40 kN and L = 6 m gives 40 × 6 / 4 = 60 kN·m.

Question 2Engineering Mechanics and Strength of Materials

The property of a material by which it returns to its original shape after the removal of load is called:

  1. Plasticity
  2. Malleability
  3. Ductility
  4. Elasticity Correct answer.

Explanation

Elasticity is defined as the ability of a material to regain its original dimensions on removal of the deforming load, provided the stress remains within the elastic limit. The other terms describe permanent deformation or forming behaviour, not recovery.

Question 3Engineering Mechanics and Strength of Materials

A bar of length 2 m and cross-sectional area 500 mm² is subjected to an axial tensile force of 50 kN. Taking E = 200 GPa, the elongation of the bar is:

  1. 0.5 mm
  2. 1.0 mm Correct answer.
  3. 0.25 mm
  4. 2.0 mm

Explanation

Axial elongation is given by PL/(AE). Here P = 50,000 N, L = 2000 mm, A = 500 mm² and E = 200,000 N/mm², so elongation = (50,000 × 2000)/(500 × 200,000) = 1.0 mm.

Question 4Engineering Mechanics and Strength of Materials

The point in a loaded beam where the shear force changes sign corresponds to:

  1. A point of contraflexure
  2. A point of zero slope
  3. A point of maximum shear force
  4. A point of maximum bending moment Correct answer.

Explanation

Since the shear force is the derivative of the bending moment along the beam, the bending moment reaches a local maximum or minimum where the shear force is zero or changes sign. This is why the maximum moment in a simply supported beam under transverse loading occurs where shear passes through zero.

Question 5Engineering Mechanics and Strength of Materials

For a circular shaft of diameter d subjected to a torque T, the maximum shear stress developed is:

  1. 32T/(πd³)
  2. T/(πd³)
  3. 8T/(πd³)
  4. 16T/(πd³) Correct answer.

Explanation

Torsion theory gives τ = T·r/J, with r = d/2 and polar moment J = πd⁴/32. Substituting these yields τ_max = 16T/(πd³), the standard result for a solid circular shaft.

Question 6Engineering Mechanics and Strength of Materials

The maximum shear stress in a beam of rectangular cross-section is related to the average shear stress by a factor of:

  1. 3/2 Correct answer.
  2. 2/3
  3. 1/2
  4. 4/3

Explanation

For a rectangular section the shear stress distribution is parabolic, being zero at the top and bottom fibres and maximum at the neutral axis. The maximum value equals 3/2 times the average shear stress V/(bd).

Question 7Engineering Mechanics and Strength of Materials

Two forces of 6 kN and 8 kN act at a point at right angles to each other. The magnitude of their resultant is:

  1. 10 kN Correct answer.
  2. 48 kN
  3. 14 kN
  4. 2 kN

Explanation

For two perpendicular forces, the resultant magnitude is the square root of the sum of squares: √(6² + 8²) = √100 = 10 kN. The direction makes an angle of about 53° with the 6 kN force.

Question 8Engineering Mechanics and Strength of Materials

A body of weight 100 N rests on a horizontal plane with coefficient of static friction 0.4. The minimum horizontal force required to just start motion is:

  1. 25 N
  2. 60 N
  3. 100 N
  4. 40 N Correct answer.

Explanation

Limiting friction equals μ times the normal reaction. On a horizontal plane the normal reaction equals the weight, 100 N, so the limiting friction force is 0.4 × 100 = 40 N, which is the minimum applied force to initiate sliding.

Question 9Engineering Mechanics and Strength of Materials

A simply supported beam of span L carries a uniformly distributed load w per unit length over its entire span. The maximum bending moment is:

  1. wL²/8 Correct answer.
  2. wL²/2
  3. wL²/12
  4. wL²/4

Explanation

For a uniformly distributed load over the full span, the reactions are wL/2 at each support and the maximum moment at midspan is wL²/8. This is obtained by integrating the shear force from the support to midspan.

Question 10Engineering Mechanics and Strength of Materials

The polar moment of inertia of a solid circular area of diameter d about its centroidal axis is:

  1. πd⁴/32 Correct answer.
  2. πd⁴/128
  3. πd⁴/64
  4. πd⁴/16

Explanation

The polar moment of inertia of a solid circle about the centroid is J = πd⁴/32, which equals twice the second moment of area I = πd⁴/64 about a diameter. This quantity governs torsional stiffness of circular shafts.

Question 11Engineering Mechanics and Strength of Materials

In a two-dimensional stress state, the principal stresses are 80 MPa and 20 MPa. The maximum shear stress in the plane is:

  1. 30 MPa Correct answer.
  2. 60 MPa
  3. 100 MPa
  4. 50 MPa

Explanation

The in-plane maximum shear stress equals half the difference of the principal stresses: (80 − 20)/2 = 30 MPa. It acts on planes inclined at 45° to the principal planes.

Question 12Engineering Mechanics and Strength of Materials

A cantilever beam of length 3 m carries a point load of 10 kN at its free end. The bending moment at the fixed support is:

  1. 30 kN·m Correct answer.
  2. 90 kN·m
  3. 10 kN·m
  4. 15 kN·m

Explanation

For a cantilever with an end load, the maximum moment occurs at the fixed end and equals the load times the lever arm, PL = 10 × 3 = 30 kN·m. The bending moment diagram is triangular, zero at the free end.

Question 13Engineering Mechanics and Strength of Materials

Poisson's ratio for a perfectly incompressible material, and for most common metals respectively, are closest to:

  1. 1.0 and about 0.25
  2. 0.25 and about 1.0
  3. 0.5 and about 0.3 Correct answer.
  4. 0.3 and about 0.5

Explanation

A perfectly incompressible material has zero volume change under hydrostatic stress, which requires Poisson's ratio of 0.5. Most metals such as steel and aluminium have values near 0.25 to 0.33, commonly quoted as about 0.3.

Question 14Engineering Mechanics and Strength of Materials

The relationship between Young's modulus E, shear modulus G and Poisson's ratio ν for an isotropic material is:

  1. E = G(1 − 2ν)
  2. E = 2G(1 + ν) Correct answer.
  3. E = 2G(1 − ν)
  4. E = G(1 + ν)

Explanation

The standard isotropic elastic relation is E = 2G(1 + ν), which links tensile and shear stiffness through Poisson's ratio. It follows from combining the generalised Hooke's law with the definition of shear modulus.

Question 15Engineering Mechanics and Strength of Materials

A simply supported beam of span L has a rectangular section of width b and depth d. For a given maximum bending moment M, the maximum bending stress is:

  1. M/(bd²)
  2. 6M/(bd²) Correct answer.
  3. 12M/(bd³)
  4. 4M/(bd²)

Explanation

Bending stress is σ = M·y/I with y = d/2 and I = bd³/12 for a rectangle. Substituting gives σ_max = M(d/2)/(bd³/12) = 6M/(bd²), the standard flexure result.

Question 16Engineering Mechanics and Strength of Materials

A horizontal beam is loaded by a vertical downward force and a clockwise couple. For static equilibrium, the number of independent equations available from the general coplanar force system is:

  1. Three Correct answer.
  2. Two
  3. Six
  4. Four

Explanation

A general coplanar (two-dimensional) force system provides three independent equilibrium equations: summation of forces in two perpendicular directions and summation of moments about any point. Three unknown reactions can therefore be determined for a statically determinate beam.

Question 17Engineering Mechanics and Strength of Materials

A thin cylindrical shell of internal diameter 1 m and thickness 10 mm is subjected to an internal pressure of 2 MPa. The hoop stress developed is:

  1. 100 MPa Correct answer.
  2. 10 MPa
  3. 50 MPa
  4. 200 MPa

Explanation

For a thin cylinder the hoop (circumferential) stress is pd/(2t). With p = 2 N/mm², d = 1000 mm and t = 10 mm, the hoop stress is (2 × 1000)/(2 × 10) = 100 MPa.

Question 18Engineering Mechanics and Strength of Materials

The strain energy stored in an axially loaded bar of length L, area A and Young's modulus E under an axial force P is:

  1. P²L/(AE)
  2. PL²/(2AE)
  3. P²L/(2AE) Correct answer.
  4. PL/(AE)

Explanation

Strain energy for gradual axial loading equals half the product of load and extension, U = ½P·δ. Since δ = PL/(AE), this gives U = P²L/(2AE), showing energy is proportional to the square of the load.

Question 19Engineering Mechanics and Strength of Materials

A rigid body is in equilibrium under the action of three non-parallel coplanar forces. The lines of action of these forces must:

  1. Be concurrent at a single point Correct answer.
  2. Form a closed triangle of forces
  3. Pass through the centre of gravity
  4. Be parallel to each other

Explanation

Lami's theorem and the three-force principle state that three non-parallel coplanar forces in equilibrium must be concurrent, that is, their lines of action meet at one point. The force triangle being closed is a vector condition, not a statement about the lines of action.

Question 20Engineering Mechanics and Strength of Materials

The deflection at the free end of a cantilever of length L carrying a point load W at the free end is:

  1. WL²/(2EI)
  2. WL³/(8EI)
  3. WL⁴/(8EI)
  4. WL³/(3EI) Correct answer.

Explanation

For a cantilever with an end point load, integrating the moment-curvature relation twice with zero slope and deflection at the fixed end gives δ = WL³/(3EI). The cube of the length shows deflection is highly sensitive to span.

Question 21Structural Analysis

A simply supported beam of span L carries a central point load W. The maximum bending moment developed in the beam is

  1. WL/8
  2. WL/4 Correct answer.
  3. WL²/8
  4. WL/2

Explanation

For a simply supported beam with a central point load, each reaction equals W/2 and the moment diagram is triangular with its peak at midspan. The peak value is (W/2)(L/2) = WL/4. The value WL/8 applies to a uniformly distributed load, not a central point load.

Question 22Structural Analysis

In a cantilever beam of length L fixed at one end and free at the other, a point load W acts at the free end. The deflection at the free end is

  1. WL⁴/(8EI)
  2. WL²/(2EI)
  3. WL³/(8EI)
  4. WL³/(3EI) Correct answer.

Explanation

The bending moment at a section distance x from the free end is Wx, and integrating M/EI twice with the fixed-end conditions gives a tip deflection of WL³/(3EI). This is the standard result for a cantilever with an end point load.

Question 23Structural Analysis

The degree of static indeterminacy of a continuous beam with four spans supported on five simple supports and having no internal hinges is

  1. 1
  2. 2 Correct answer.
  3. 3
  4. 4

Explanation

For a continuous beam, the degree of static indeterminacy equals the number of internal supports, since each interior support supplies one redundant vertical reaction. Four spans on five supports gives three interior supports, and the simple end supports provide only two reactions for a statically determinate base. Evaluating with the general formula for a single rigid body gives 5 reactions minus 3 equilibrium equations, which equals 2.

Question 24Structural Analysis

A truss is classified as statically determinate when, for a plane truss with j joints and m members supported on r reaction components, the relation satisfied is

  1. m + r = 2j Correct answer.
  2. m + r > 2j
  3. m + r < 2j
  4. m = 2j + r

Explanation

Each joint of a plane truss provides two equilibrium equations, giving 2j equations in total. The unknowns are the m member forces plus the r support reactions, so determinacy requires m + r = 2j exactly.

Question 25Structural Analysis

The influence line for the vertical reaction at the left support A of a simply supported beam of span L, due to a unit load moving across the span, is

  1. A straight line equal to 1.0 at A and 0 at the right support B Correct answer.
  2. A triangular diagram peaking at midspan
  3. A rectangle of constant ordinate 0.5 throughout
  4. A parabola with maximum at midspan

Explanation

Taking moments about B for a unit load at distance x from A gives the reaction at A as (L − x)/L, a linear function. Its ordinate is 1.0 when the unit load is at A and falls to zero when the load reaches B.

Question 26Structural Analysis

In the slope-deflection method, the fixed-end moment at each end of a prismatic member of length L carrying a central point load W is

  1. WL/4
  2. WL/12
  3. WL/6
  4. WL/8 Correct answer.

Explanation

For a fixed-ended beam with a central point load, the fixed-end moments are each WL/8, with hogging moment at both ends. The total static moment WL/4 is shared equally between the two ends.

Question 27Structural Analysis

A propped cantilever of span L is fixed at A and simply supported at B, and carries a uniformly distributed load w over the whole span. The vertical reaction at the prop B is

  1. 5wL/8
  2. wL/4
  3. wL/2
  4. 3wL/8 Correct answer.

Explanation

The prop reaction is the redundant in this one-degree indeterminate beam. Compatibility of deflection at B gives a prop reaction of 3wL/8, with the remaining 5wL/8 carried at the fixed end.

Question 28Structural Analysis

A three-hinged arch of span L has its two supports at the same level and carries a point load W at the crown. The horizontal thrust developed at each support is

  1. WL/2h
  2. WL/8h
  3. WL/4h Correct answer.
  4. WL/h

Explanation

Taking moments about one hinge for the half-arch gives the thrust H as the moment of the applied load about that hinge divided by the rise h. With the crown load, the moment about the springing hinge is WL/4, so H = WL/4h.

Question 29Structural Analysis

In a two-hinged parabolic arch of span L and rise h carrying a uniformly distributed load w over the entire span, the bending moment at any section is

  1. Zero throughout the arch Correct answer.
  2. Maximum at the crown
  3. Parabolic and always sagging
  4. Constant along the arch

Explanation

For a parabolic arch under a uniform load over the whole span, the funicular shape of the arch coincides with the parabolic profile. The arch then carries the load purely by axial thrust and the bending moment vanishes at every section.

Question 30Structural Analysis

The stiffness matrix of a plane frame element with three degrees of freedom per node has an order of

  1. 6 × 6 Correct answer.
  2. 3 × 3
  3. 12 × 12
  4. 4 × 4

Explanation

A plane frame element has two nodes, and each node has three degrees of freedom: axial translation, transverse translation and rotation. The element stiffness matrix therefore has order 2 × 3 = 6.

Question 31Structural Analysis

A simply supported beam of span L carries a uniformly distributed load w over its entire span. The maximum shear force occurs

  1. At the supports and equals wL
  2. At midspan
  3. At quarter span and equals wL/4
  4. At the supports and equals wL/2 Correct answer.

Explanation

The shear force diagram for a simply supported beam under a full UDL is linear, varying from +wL/2 at one support to −wL/2 at the other. Its magnitude is therefore greatest at the supports, where it equals half the total load.

Question 32Structural Analysis

A fixed beam of span L carries a central point load W. The ratio of the fixed-end moment to the maximum free bending moment (WL/4) is

  1. 1/3
  2. 1/4
  3. 2/3
  4. 1/2 Correct answer.

Explanation

The fixed-end moments are each WL/8 while the free bending moment is WL/4. Their ratio is therefore one half, meaning fixity halves the design moment at midspan compared with a simply supported beam.

Question 33Structural Analysis

In the moment-distribution method, the distribution factor for a member meeting a joint is computed as

  1. The product of its stiffness and the sum of joint stiffnesses
  2. The reciprocal of the number of members meeting at the joint
  3. The ratio of its carry-over factor to the joint stiffness sum
  4. The ratio of its stiffness to the sum of stiffnesses of all members meeting at that joint Correct answer.

Explanation

The distribution factor expresses the share of an unbalanced moment taken by each member and equals the member stiffness divided by the total stiffness at the joint. The sum of distribution factors at any joint is therefore unity.

Question 34Structural Analysis

A simply supported beam of span L carries a point load W at a distance a from the left support, with b = L − a. The deflection under the load is

  1. Wab(L + a)/(3EI)
  2. Wa²b²/(3EIL) Correct answer.
  3. Wa²b²/(8EIL)
  4. Wab/(3EIL)

Explanation

Using Macaulay's method or unit-load integration, the deflection beneath an eccentric point load on a simply supported beam is Wa²b²/(3EIL). The expression is symmetric in a and b, as expected when the load position is interchanged.

Question 35Structural Analysis

A simply supported beam of span L carries a uniformly distributed load w. The maximum deflection occurs at midspan and equals

  1. 5wL⁴/(384EI) Correct answer.
  2. wL⁴/(384EI)
  3. wL³/(48EI)
  4. wL⁴/(185EI)

Explanation

Double integration of M/EI with the simply supported boundary conditions gives a maximum midspan deflection of 5wL⁴/(384EI) for a full UDL. The coefficient 5 arises from the parabolic moment diagram.

Question 36Structural Analysis

A beam is hinged at A, roller-supported at B, and has an internal hinge at C between A and B. The degree of static indeterminacy of this beam is

  1. 0 Correct answer.
  2. 2
  3. 1
  4. 3

Explanation

The three reaction components (two at A, one at B) are just sufficient for equilibrium, and the internal hinge supplies an additional condition equation. Counting unknowns against equations shows the structure is statically determinate, with the hinge providing the extra relation needed.

Question 37Structural Analysis

A cantilever of length L is loaded by a uniformly distributed load w over its entire length. The bending moment at the fixed end is

  1. wL²/8
  2. wL²/6
  3. wL²/4
  4. wL²/2 Correct answer.

Explanation

The total load wL acts at its centroid, a distance L/2 from the fixed end, so the fixed-end moment is wL × L/2 = wL²/2. This is a hogging moment.

Question 38Structural Analysis

In the unit-load method, the deflection at a point in a linear elastic structure is obtained by integrating the product of

  1. The actual moment and the actual axial force, divided by EA
  2. The actual shear force and the actual bending moment, divided by EI
  3. The actual bending moment and the bending moment due to a unit load applied at that point, divided by EI Correct answer.
  4. The unit-load moment and the unit-load shear, divided by GA

Explanation

The unit-load or virtual work theorem states that the displacement equals the integral of M·m/EI over the member, where m is the moment caused by a unit load at the point and in the direction of the required displacement. Only the actual and virtual moments enter for bending deformation.

Question 39Structural Analysis

A two-hinged arch has span L and rise h, and the horizontal thrust is H. If the arch rib is treated as rigid in bending compared with the tie, the horizontal thrust due to a temperature rise is best described as

  1. Independent of the temperature rise but dependent on the rise-to-span ratio
  2. Zero, because no external load acts on the arch Correct answer.
  3. Equal to the product of the coefficient of thermal expansion and the span
  4. Directly proportional to the temperature rise and independent of the arch flexibility

Explanation

In a two-hinged arch the horizontal thrust arises from restrained deformation. When the rib is treated as infinitely stiff in bending relative to the tie, the arch can accommodate thermal movement without developing thrust, so no redundant force is generated by temperature alone.

Question 40Structural Analysis

A plane truss has j joints, m members and r reaction components, and satisfies m + r = 2j. Which additional condition is required before the truss can be declared stable and determinate?

  1. All members must have equal cross-sectional area
  2. The members must be arranged so that the equations are independent and the truss is not a mechanism Correct answer.
  3. The truss must be symmetrical about a vertical axis
  4. The truss must contain at least one redundant member

Explanation

The counting equation m + r = 2j is necessary but not sufficient. The equilibrium equations must also be linearly independent, which requires a proper triangular arrangement of members so that no part of the truss can deform as a mechanism.

Question 41Concrete and Steel Design (RCC and Steel Structures)

As per IS 456:2000, the minimum clear cover to the main reinforcement of a beam for mild exposure conditions is:

  1. 15 mm
  2. 20 mm Correct answer.
  3. 40 mm
  4. 25 mm

Explanation

IS 456:2000 Table 16 specifies nominal cover requirements related to exposure. For mild exposure the minimum clear cover for beams is 20 mm, which also satisfies the durability requirement for a typical reinforced concrete beam. Slabs get 15 mm and columns 40 mm under the same exposure class.

Question 42Concrete and Steel Design (RCC and Steel Structures)

In limit state design of a singly reinforced rectangular beam, the limiting value of the neutral axis depth ratio xu,lim/d for Fe 415 steel is:

  1. 0.53
  2. 0.46
  3. 0.48 Correct answer.
  4. 0.42

Explanation

The limiting neutral axis depth depends on the yield strain of the steel and the ultimate concrete strain of 0.0035. For Fe 415 steel, IS 456 gives xu,lim/d = 0.48, which corresponds to a limiting moment coefficient of about 0.138. Higher grade steel gives a smaller ratio because it needs more rotation to yield.

Question 43Concrete and Steel Design (RCC and Steel Structures)

The design bond stress in limit state method for deformed bars in tension, for M25 concrete, is taken as:

  1. 1.4 MPa Correct answer.
  2. 1.2 MPa
  3. 1.5 MPa
  4. 1.9 MPa

Explanation

IS 456 gives the design bond stress in limit state method as 1.2 MPa for M20 concrete, increased by 25 percent for M25, giving 1.5 MPa...

Question 44Concrete and Steel Design (RCC and Steel Structures)

A simply supported rectangular beam of effective span 6 m carries a factored uniformly distributed load of 30 kN/m. The maximum factored bending moment is:

  1. 135 kNm Correct answer.
  2. 270 kNm
  3. 180 kNm
  4. 90 kNm

Explanation

For a simply supported beam under a uniformly distributed load, the maximum bending moment is wl^2/8. Substituting w = 30 kN/m and l = 6 m gives 30 x 36 / 8 = 135 kNm. This is the factored design moment used for reinforcement design.

Question 45Concrete and Steel Design (RCC and Steel Structures)

For a reinforced concrete column subjected to axial load plus uniaxial bending, the interaction diagram is plotted between:

  1. axial load and deflection
  2. axial load and bending moment Correct answer.
  3. bending moment and shear force
  4. axial load and shear force

Explanation

A column interaction diagram plots the axial load capacity P against the moment capacity M for a given reinforcement and section. It defines the safe combinations of P and M, with the balanced point separating compression-controlled and tension-controlled behaviour. Design checks ensure the factored (P, M) point lies inside the envelope.

Question 46Concrete and Steel Design (RCC and Steel Structures)

In a two-way slab with corners held down and no torsion reinforcement provided, the bending moment in the short span direction at the support is:

  1. kept equal to the midspan value
  2. increased by 50 percent
  3. reduced by 50 percent Correct answer.
  4. taken as zero

Explanation

IS 456 permits a reduction of support moments in slabs where torsion is not accounted for by reinforcement. When corners are held down and torsion reinforcement is omitted, the support moment in each direction may be reduced by 50 percent, with the midspan moment increased correspondingly. This redistribution reflects the torsional stiffness of the slab.

Question 47Concrete and Steel Design (RCC and Steel Structures)

The purpose of providing a shear key or dowel bar at a construction joint in a reinforced concrete wall is to:

  1. transfer shear across the joint Correct answer.
  2. increase the concrete grade locally
  3. reduce shrinkage cracking in the wall
  4. improve the fire resistance of the joint

Explanation

Construction joints interrupt the continuity of concrete, so shear transfer across the interface must be ensured. A shear key or dowel bar provides mechanical interlock that transfers shear forces between the two pours. It does not change concrete grade, shrinkage behaviour, or fire resistance.

Question 48Concrete and Steel Design (RCC and Steel Structures)

As per IS 800:2007, the design strength of a bolt in bearing is governed by the minimum of the bearing capacity of the bolt and:

  1. the tensile capacity of the bolt
  2. the tensile strength of the connected plate
  3. the shear capacity of the bolt
  4. the bearing capacity of the connected plate Correct answer.

Explanation

Bearing-type bolted connections fail either by crushing of the bolt or by bearing failure of the plate around the hole. IS 800 therefore takes the design bearing strength as the smaller of the bolt bearing capacity and the plate bearing capacity. The plate capacity depends on plate thickness, ultimate tensile strength, and edge distance.

Question 49Concrete and Steel Design (RCC and Steel Structures)

In a fillet weld, the effective throat thickness is measured as:

  1. the root gap between plates
  2. the leg length of the weld
  3. the actual throat plus 2 mm
  4. 0.7 times the size of the weld Correct answer.

Explanation

For a fillet weld with equal legs, the throat thickness is the perpendicular distance from the root to the weld face. Geometry gives this as 0.7 times the weld size (leg length), and IS 800 uses this effective throat for strength calculations. The size of a fillet weld is measured as the leg length.

Question 50Concrete and Steel Design (RCC and Steel Structures)

The maximum effective slenderness ratio KL/r permitted for a compression member carrying dead and superimposed loads, as per IS 800:2007, is:

  1. 120
  2. 180 Correct answer.
  3. 250
  4. 200

Explanation

IS 800 limits slenderness to control buckling and second-order effects. For compression members carrying dead and superimposed loads, the maximum effective slenderness ratio is 180. The value 250 applies to members carrying only wind or seismic loads, and 120 applies to certain other categories.