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The correct answer is (C) material to be used for fabrication

Easy explanation: In design of built up COLUMN, following conditions are considered : (i) buckling of column as WHOLE under axial load, (ii) buckling of COMPONENT column, (iii) failure of lattice member, (iv) distortion of cross section.

Right choice is (a) smaller than

Easy EXPLANATION: The buckling strength of latticed column is smaller than that of solid column having same AREA and same slenderness ratio provided that solid column does not buckle locally because of thinness of material. This is because SHEAR component of axial load produces deformation in lattice which tends to reduce overall stiffness of column and THEREFORE reduce buckling strength of column.

The correct option is (d) lacing bars and BATTEN PLATES are designed as load carrying elements

For explanation I would say: Lacing bars and batten plates are not designed as load carrying elements. Their function is to hold MAIN component members of built up column in relative position and equalize stress DISTRIBUTION in them. At ends and at intermediate points where it is necessary to interrupt the lacings, open sides are connected with TIE plates. Tie plates are also provided at top and bottom of column.

Right choice is (a) material used in COLUMN

Easy explanation: The shear in column may be due to (i) lateral loads from wind, earthquake, GRAVITY or other loads, (ii)slope of column with RESPECT to line of thrust due to both unintentional and initial curvature and increased curvature during bending, (iii)end of eccentricity of load due to EITHER end connection or FABRICATION imperfections.

The correct answer is (C) used when rolled section do not furnish required sectional area

Easiest explanation: Size and shape of rolled sections are limited because of limitation of rolling MILLS. When rolled section do not furnish required sectional area or when special shape or large RADIUS of gyration is required in two different direction, a BUILT up section is used.

Correct CHOICE is (B) critical stress is INVERSELY PROPORTIONAL toslenderness RATIO

The explanation: fcr= Pcr/Ag = π^2E/λ^2, critical stress is inversely proportional toslenderness ratio of column and very large values can be obtained by using L/r → 0 .

Right option is (c) PCR = 4π^2EI/L^2

The explanation is: The Euler critical load for column with both the ends HINGED is given by Pcr = 4π^2EI/L^2, where E is elastic MODULUS of material, I is MOMENT of inertia and L is length of column, effective length of column in this case = 2L.

The correct answer is (a) increasing member length causes reduction in stiffness

The best I can explain: Member with low stiffness will BUCKLE early than that the one with high stiffness. Increasing member length causes reduction in stiffness. Stiffness of member is influenced by AMOUNT and distribution of MATERIAL in cross section of column, the value of radius of gyration REFLECTS the way in which material is DISTRIBUTED.

The correct answer is (d) Pcr = π^2EI/L^2

For explanation: The EULER critical load for column with both the ends hinged is GIVEN by Pcr= π^2EI/L^2, where E is elastic MODULUS of material, I is moment of inertia and L is length of column.

The CORRECT option is (b) material is isotropic

To ELABORATE: According to Euler, the following are the attributes of ideal column: (i) Material isisotropic, homogenous and is assumed to be perfectly elastic, (ii) column is initially STRAIGHT and load acts along centroidal axis (no eccetricity of loads), (III) column has no imperfections, (IV) column ends are hinged.

The CORRECT answer is (C) by spreading material of section away from its AXIS

To explain: MAXIMUM RADIUS of gyration is obtained when material of section is farthest from centroid i.e. away from its axis.

The correct option is (a) ratio of EFFECTIVE LENGTH to radius of gyration

For explanation: The tendency of member to buckle is usually MEASURED by its slenderness ratio. Slenderness ratio of member is ratio of effective length to appropriate radius of gyration (λ = kL/r). This is valid only when COLUMN has EQUAL unbraced heights for both axes and end condition is same for particular section.

The correct choice is (B) 0.8L

The explanation is: The EFFECTIVE length of COMPRESSION member when ONE end is fixed and other end is hinged is 0.8L, where L is the length of the member.

Right choice is (d) 2L

Easiest explanation: The effective length of compression member when one END is fixed and other end is FREE is 2L, where L is the length of the member.

The correct answer is (c) L

For explanation: The EFFECTIVE length of COMPRESSION MEMBER when both ENDS of compression member are hinged is L, where L is the length of the member.

Correct CHOICE is (a) 0.65L

The BEST I can EXPLAIN: The effective length of COMPRESSION member when both ends of compression member are fixed is 0.65L (i.e. L/√2), where L is the length of the member.

The CORRECT choice is (d) smaller the effective LENGTH, smaller the danger of lateral buckling

Explanation: Smaller the effective length of particular compression MEMBER, smaller isthe danger of lateral buckling and GREATER is the LOAD carrying capacity.

The CORRECT answer is (b) rotational restraint SUPPLIED at end of compression member

The explanation: Magnitude of effective length depends UPON rotational restraint supplied at end of compression member and upon resistance to lateral movement provided.

The correct ANSWER is (c) distance between points of contraflexure

To explain: Effective LENGTH of compression member is distance between points of contraflexure. It should be DERIVED from actual length and END conditions.

The correct choice is (d) least radius of gyration of single angle section is small compared to channel and I-SECTIONS

The BEST explanation: Equal angle are desirable and ECONOMICAL over UNEQUAL angles because least radius of gyration of equal angle is greater than that of unequal angle for same area of steel. Single angle sections are not suitable for long LENGTHS. Least radius of gyration of single angle section is small compared to channel and I-sections.

Correct answer is (d) less than 3M

To ELABORATE: Rods and bars withstand very little COMPRESSION when length is more. Hence these are recommended for lengths less than 3m only.

Correct answer is (a) seismic load

To ELABORATE: The DECREASE in strength of slendermember is due to following parameter : imperfections- initial LACK of straightness, accidental eccentricities of loading, residual stress, and variation of material properties over the cross section.

Correct choice is (b) members will fail by both yielding and buckling

The BEST I can EXPLAIN: For INTERMEDIATE LENGTH COMPRESSION members, some fibres would have yielded and some fibres will still be elastic. They will fail by both yielding and buckling and their behaviour is said to be inelastic.

Right answer is (d) buckle elastically

Explanation: Long COMPRESSION MEMBERS will buckle elastically where AXIAL buckling stress REMAINS below proportional limit.

Right option is (a) L ≤ 88.5r

The best I can explain: For a MEMBER to be CLASSIFIED as short column, length of member should be L ≤ 88.5r , where r is RADIUS of gyration. The SLENDERNESS RATIO of column defines the column as short or long column.

The correct CHOICE is (c) FAILURE STRESS will equal yield stress

Explanation: For very short COMPRESSION MEMBERS, the failure stress will the equal yield stress and no buckling will occur.

Right option is (c) material being homogenous

To EXPLAIN I WOULD SAY: Imperfections in column INCLUDE material not being isotropic and homogenous, geometric variations of columns and eccentricity of load.

The correct answer is (c) WIND and crane load

Easy explanation: In industrial buildings, loads from crane and wind cause moments in COLUMNS. In such cases, wind load is applied to the COLUMN through sheeting rails and may be taken as UNIFORMLY DISTRIBUTED throughout the length of column.

Right option is (b) LOAD from foundation

Explanation: Axial loading on COLUMNS in buildings is due to LOADS from roofs, floors, and WALLS TRANSMITTED to the column through beams and also due to itsown self weight.

Correct choice is (c) STRUTS are compression members used in ROOF trusses

To elaborate: The compression members used in roof trusses and bracings are called as struts. They may be vertical or inclined and NORMALLY have small lengths. the top chord members of a roof truss are called principal RAFTER.

Right ANSWER is (a) vertical compression members in a building supporting floors or GIRDERS

The explanation: The vertical compression members in a building supporting floors or girders are normally CALLED as columns. They are SOMETIMES called as stanchions. They are subjected to heavy loads. Vertical compression members are sometimes called posts.

The correct answer is (C) member subjected to AXIAL force and bending moment

Explanation: If the NET END moments are not zero, the member will be subjected to axial force and bending moments along its length. Such members are CALLED beam-columns.

Right ANSWER is (d) net end moments are zero

To explain: if the net end moments are zero, the compression member is REQUIRED to RESIST LOAD acting concentric to original longitudinal axis of member and is CALLED axially loaded column or simply column.

Right choice is (b) structural member SUBJECTED to compressive force

The EXPLANATION: Structural member which is subjected to compressive forces along its axis is called compression member. Compression MEMBERS are subjected to LOADS that tend to DECREASE their lengths.

Right answer is (c) Perry-Robertson formula

Explanation: The design compressive stress of axially loaded COMPRESSION member in IS 800 is GIVEN by Perry-Robertson formula. IS 800:2007 PROPOSES multiple COLUMNS curves in nin-dimensional form based on Perry-Robertson approach.

Right option is (c) 0.21

To EXPLAIN: The value of imperfection factor, α for BUCKLING class a is 0.21. The imperfection factor CONSIDERS all the relevant defects in real structure when considering buckling, geometric IMPERFECTIONS, eccentricity of applied loads and RESIDUAL stresses.

Right answer is (a) (Vt / N) cotΘ

The explanation is: Strength of bolt /rivet should be GREATER than load COMING over rivet/bolt. The load on rivet/bolt when two lacing FLATS are connected at different POINT is (Vt/ N) cotΘ.

Correct choice is (d) (Vt /N) cosecΘ

The best I can EXPLAIN: Compressive strength in LACING BARS is equal to (Vt /N) cosecΘ for single lacing system and (Vt/2N) cosecΘ for double lacing system, where N = number of SHEAR resisting elements.

The CORRECT ANSWER is (a) t/√12

The best explanation: Minimum radius of gyration for lacing flats is t/√12 , where t is THICKNESS of FLAT.