Explore topic-wise InterviewSolutions in .

The correct ANSWER is (c) Short end plates are used for MODERATE moments

Easiest explanation: End plate SPLICES can be used in COLUMN to PROVIDE load reversals in columns. Short end plates are used for moderate moments and extended end plates are used for heavy moments.

Correct answer is (a) load is transferred through splices

Best EXPLANATION: When the columns carry predominantly axial forces, leaving a gap between the ENDS MAY be used. In this case, the whole load is transmitted through means of splice plates. HSFG bolts may be used in the CONNECTIONS and changes in size of column may be accommodated using packing plates.

The correct choice is (B) splices are DESIGNED only to resist accidental tension

Easiest explanation: End bearing arrangement MAY be used when the columns carry predominantly axial forces. In this load is transferred through CONTACT area and splices are designed only to resist accidental tension due to some uplift loading or internal EXPLOSION in the building.

Right option is (a) when change in size between TWO sections of beam OCCURS

Best explanation: When change in size between two sections of beam occurs at an end PLATE splice, it can be easily accommodated by introducing longitudinal stiffeners to the LARGER beam.

The CORRECT answer is (c) Flush end plates are USED when bending moments to be resisted are modest

Explanation: Flush end plates are used when bending moments to be resisted are modest. Singly or DOUBLY extended plates are used when RESISTING high moments of one SIGN or full reversal respectively.

Correct answer is (a) should not be LESS than 50%

Best EXPLANATION: As perIS CODE, strength of spliced PORTION should not be less than 50% of the effective strength of MATERIAL spliced.

Correct option is (d) plates on the flange should be DESIGNED to do the WORK of the flange and plates on the web should be designed to do the work of the web

Easy EXPLANATION: For beam splices, each element of the splice is designed to do the work the sections underlying the splice plates could do, if UNCUT. Plates on the flange should be designed to do the work of the flange and plates on the web should be designed to do the work of the web.

The correct CHOICE is (b) combined splice PLATE/end plate connection

Explanation: In a beam-beam connection, when top flanges are required to be at same level combined splice plate/end plate connection with coped ends is USED. HSFG bolts may be utilized in these connections for avoiding LOCAL rotation.

Correct option is (b) for easy transportation

The best I can EXPLAIN: Rolled beams, plate girders and columns are spliced due to following reasons : (i)FULL length of the MEMBER may not be available from the mill, (ii)size of section which can be TRANSPORTED depends on size of truck, so for easy transportation, (III)splice points may be used to camber the beam, (iv)when a change in section is required to fit variation in strength required along span of beam.

Correct choice is (d) SHEAR FORCE and bending moment

The explanation: When CANTILEVER beam is supported by girder, the end of cantilever will transfer shear force and bending moment and this bending moment will be TRANSFERRED to PRIMARY beam as torsion.

Right option is (b) cantilever BEAM supported by GIRDER

To elaborate: Rigid connections are necessary if a cantilever beam is supported by girder. Rigid connections may be PROVIDED for moment continuity between SECONDARY beams.

The correct answer is (c) SIMPLE Connection

To explain: Simple CONNECTIONS such as clip and seating ANGLE connection, WEB angle connection and flexible END plate connections, etc. used for connecting beam-to-columns, can be adopted for beam-beam connections.

The correct choice is (a) web cleats

Explanation: In PRACTICE, SECONDARY beams are connected to main beams by web cleats and bolts since web of the main beam may not be STRONG enough to support SEATING angles.

Correct option is (B) SHEAR effects caused by eccentric load on a bolt group

Easy explanation: When bolt groups are subjected to shear and MOMENT in shear plane, the load that is eccentric is eccentric with RESPECT to centroid of bolt group can be replaced with a force acting through the centroid of bolt group and a moment (Magnitude = Pe). Both the moment and the force RESULT in shear effects in the bolts of the group and is called as eccentric shear.

Correct answer is (C) tension and compression

Easiest explanation: In BOLTED moment end plate connection, bending moment, AXIAL force and SHEAR force are transferred by tension and compression or shear through flange WELDS and by shear through the web welds to the end plate.

The correct choice is (a) More rigid than any other connection

Explanation: When the lever arm is to be extended to accommodate more NUMBER of BOLTS, bracket connection is USED. The bracket type connection are more rigid than any other type of connection. But the fabrication cost is very high, so they are not ADOPTED in GENERAL practice.

Correct ANSWER is (a) transfer SMALL end moments in addition to large end shear

To explain I WOULD say: When additional pair of angles in angle seat connection is used to connect the web of BEAM to flange of column, the connection can be designed to transfer small end moments in addition to large end shear. Such connections are called clip angle connections or LIGHT moment connection.

Right answer is (c) Bolts should be designed to fail by bearing of CONNECTED plies

To elaborate: The following CONDITION must be considered for web side plate CONNECTION (i) only ordinary bolts should be USED, (ii) bolts should be designed to fail by bearing of connected plies AD not by shear of bolt, (iii) edge distances must be greater than two times the bolt diameter.

Correct option is (b) ELIMINATES eccentricity

To elaborate: Double cover butt joint preferred over single cover butt joint or lap joint because (i) eccentricity of load is eliminated, hence no BENDING in bolts, (II) total shear force to be transmitted is split into two parts, hence bolts are in double shear. Shear CAPACITY of double cover butt joint is double the shear capacity of single cover butt joint or lap joint.

Right OPTION is (a) lateral SUPPORT

The best explanation: Clip and seating angle connection transfer reaction from BEAM to COLUMN through angle seat.The cleat angle is PROVIDED for lateral or torsional support to the top flange of thebeam and bolted to the top flange.

Correct option is (C) lap joint results in eccentricity of applied load, butt joint eliminates eccentricity at connection

Explanation: Lap joints and butt joints are USED to connect plates or members composed of PLATE elements. Lap joint results in eccentricity of applied load, butt joint eliminates eccentricity at connection.

The correct answer is (c) 2

For EXPLANATION I would SAY: In lap joint, MEMBERS to be connected are simply overlapped and connected together by means of bolts and WELDS. To minimize the effect of bending due to eccentricity in a lap joint, at least two bolts in a line should be PROVIDED.

The correct answer is (a) forces

Explanation: Simple CONNECTION is required to transmit force only and there MAY not be any MOMENT acting on the group of connectors. This connection may be capable of transmitting some amount of moment. Simple connections are also called FLEXIBLE connections.

Correct option is (c) flexure

To elaborate: Large bending moments are generated SINCE members joined by PIN CONNECTIONS are SEPARATED some distance. So the pin diameter is generally governed by flexure.

The CORRECT option is (c) 1.5fypZ

To explain I would say: MOMENT capacity of pin GIVEN by (i) 1.5fypZ, when ROTATION is not required, (ii) 1.0fypdt, when rotation is not required, where fyp=design strength of pin, Z=section modulus of the pin.

The correct answer is (d) 1.5fypdt

The BEST explanation: Bearing capacity of PIN given by (i) 1.5fypA, when ROTATION is not REQUIRED, (ii) 0.8fypdt, when rotation is not required, where fyp=design strength of pin.

Correct answer is (a) 0.5fypA

To ELABORATE: Shear capacity of PIN is given by (i) 0.5fypA, when ROTATION is REQUIRED, (II) 0.6fypA, when rotation is not required, where fyp=design strength of pin, A = cross sectional area of pin.

Right answer is (c) tall buildings

Explanation: PIN connections are USED in following cases : (i) truss BRIDGE girders, (ii) HINGED arches, (iii)tie rod connection in water tanks, (iv)as diagonal bracing connections in beams and columns, (v)chain-link cables suspension bridges.

The correct CHOICE is (a) hinge joint

Best explanation: Pin connections are provided when HINGED joints are required , where zero MOMENTS or free rotation is desired and horizontal and vertical movement are not desired.

The correct ANSWER is (c) reduce secondary stresses

For explanation: PINS USED for the connection reduce secondary stresses. It serves the same PURPOSE as SHANK of bolt.

Right ANSWER is (B) structural members connected by CYLINDRICAL pins

Explanation: When two structural members are connected by means of cylindrical-shaped pins, the CONNECTION is known as pinned connection. It resists horizontal and vertical movement, but not MOMENT.

Correct answer is (c) tensile forces

Easiest explanation: In connections subjected to tensile STRESSES, the flexibility of connected parts can lead to deformations that INCREASES tension applied to BOLTS. This additional tension is CALLED prying FORCE.

Right answer is (a) MINIMUM of shear strength and bearing capacity of BOLT

Easy explanation: Strength of bolt is minimum of shear strength and bearing capacity of bolt. DESIGN shear strength = nominal shear capacity/1.25, Design bearing strength = nominal bearing capacity/1.25.

The CORRECT option is (d) 40%

Explanation: EFFICIENCY = (strength of bolt PER pitch length/ strength of PLATE per pitch length)X100 = 60×100/150 = 40%.

The correct answer is (b) 4

For explanation I would say: Fo=0.7fubAnb=0.7x800x0.78x(π/4)x20^2x10^-3=137.22 kN

Assume μf=0.5, ne=1, KH=1

Slip RESISTANCE of bolt = μf ne kh Fo/1.25 = 0.5x1x1x137.22/1.25 =54.88 kN

Number of bolts required = 200/54.88 = 3.64 = 4(approximately).

Correct answer is (C) 50.18 kN

Easy explanation: diameter of bolt = 16mm, diameter of hole =16+2 =18mm

e=1.5×18=27mm, p=2.5×16=40mm

e/3d0 = 27/(3×18) = 0.5, p/3d0 -0.25 = 40/(3×18) -0.25=0.49, fub /FB = 400/410=0.975

kb = minimum of (e/3d0, p/3d0 -0.25, fub /fb,1) = 0.49

bearing strength = 2.5kbdtfu/1.25 = 2.5×0.49x16x8x400x10-3/1.25 = 50.18 kN.

The CORRECT option is (a) 0.5

To explain I would say: DIAMETER of bolt = 20MM, diameter of hole = 20+2 =22mm

e=1.5×22=33mm, p=2.5×20=50mm

e/3d0 = 33/(3×22) = 0.5, p/3d0-0.25 = 50/(3×22) -0.25=0.5, fub /fb= 400/410=0.975

kb= MINIMUM of (e/3d0 , p/3d0-0.25, fub /fb, 1) = 0.5.

The correct ANSWER is (c) 29 kN

To explain I WOULD say: BOLTS will be in single shear. Diameter of bolt = 16mm. NET area = 0.78x(π/4)x16^2=156.83mm^2.

Strength of bolt in shear = Anbfub/(√3 x 1.25) = 156.83x400x10^-3/1.25x√3 = 28.97kN.

Right ANSWER is (a) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1

To elaborate: Bolt required to SATISFY both shear and tension at the same time should satisfy (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1 , where Vsb= FACTORED shear force, Vdb = design shear CAPACITY, Tsb = factored tensile force, Tdb= design tensile capacity.

Right OPTION is (a) 0.7fub

Easy explanation: Proof STRESS for minimum bolt tension is 0.7fub, where FUB= ultimate TENSILE stress of bolt. Proof stress is an approximation for yield point for materials which do not have a definite one because of their structure.