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The correct choice is (a) Strand CONSISTS of individual wires wound helically around a CENTRAL core, wire rope is made of several strand laid helically around a core

The explanation is: Strand consists of individual wires wound helically around a central core, wire rope is made of several strand laid helically around a core. Wire ropes are exclusively USED for hoisting purposes and as guy wires in steel STACKS and TOWERS.

The CORRECT choice is (c) SINGLE angle members are used as web members in TRUSSES

Explanation: Single angle members are used in TOWERS and as web members in trusses. DOUBLE angle sections are used as chord members in light roof trusses or in situations where some rigidity is required and where members are subjected to reversal of stresses.

The CORRECT OPTION is (d) 416.62 kN

The best explanation: dh=18 MM, fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25

Anc = (100 – 10/2 – 18) x 10 = 770 mm^2, Ag0 = (75 – 10/2) x 10 = 700 mm^2

β = 1.4 – 0.076(w/t)(fy/fu)(bs/Lc)

= 1.4 – 0.076 [(75-5)/10] [250/410] [{(75-5)+(100-40)}/{40×5}]

= 1.19 > 0.7 and < 1.44[(410/250)(1.1/1.25)] (=2.07)

Tdn = 0.9fuAnc /γm1 + βAg0fy /γm0

= [0.9x410x770/1.25 + 1.19x700x250/1.1] x 10-3 = 416.62 kN.

Correct answer is (C) 432.27 KN

The explanation is: FU = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25,

For ISA 125 x 75 x 10mm : gross AREA Ag = 1902 mm^2

Tensile strength due to gross section yielding, Tdg = Agfy/γm0 = 1902 x 250 x 10-3 / 1.1 = 432.27 kN.

The correct answer is (b) 216.49 kN

Easiest explanation: fu = 410 MPA, fy = 250 MPa, γm0 = 1.1, γm1 = 1.25

Avg = ( 1×100 + 50 ) x 8 = 1200 mm^2

Avn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm^2

Atg = 35 x 8 = 280 mm^2

Atn = (35 – 1/2 x 20) x 8 = 200 mm^2

Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN

Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1) = [ (0.9x1440x410 / 1.25x√3) + (280×250/1.1) ] x 10-3 = 309.06 kN

Block shear strength of tension MEMBER is 216.49 kN.

The correct CHOICE is (c) 1956 mm2

For EXPLANATION I would SAY: Net area of CONNECTED leg, Anc = (100 – 12/2) x 12 = 1128 mm^2

Net area of OUTSTANDING leg, Ago = (75 – 12/2) x 12 = 828 mm^2

Total net area = 1128 + 828 = 1956 mm^2.

The CORRECT OPTION is (b) 2681mm^2

For explanation I would say: B = 300MM, t = 10mm, dh = 18+2 =20MM, n = 3, n’ = 1, p = 75mm, g = 50mm

Effective NET area = (B-ndh+n’p^2/4g)t = (300 – 3×20 + 1×75^2/4×50)x10 = 2681.25 mm^2.

The CORRECT choice is (d) 1-5-7-6-3

Easiest EXPLANATION: The section giving minimum area of PLATE is CONSIDERED for DESIGN. So, section 1-5-7-6-3 is used for net area of flat.

Right OPTION is (a) 38.2 cm^2

For explanation I would SAY: B = 40cm = 400mm, t = 10MM, dh = 18mm

Net section AREA = 400×10 – 16×10 = 3820mm^2 = 38.2 cm^2.

Right option is (c) ndt – n’p^2t/4g

The EXPLANATION is: The net area of flat with staggered hole is given by : A = (B – ndh + n’p^2/4g)t, where b = width of plate, n = number of HOLES in zig-zag line, n’ = number of staggered pitches, p = pitch distance, g= gauge distance, t = THICKNESS of flat.

The correct option is (c) LINES of action of TRUSS members meeting at a joint shouldnot coincide

Explanation: A gusset plate is plate provided at ends of tension members through which forces are transferred to MAIN MEMBER. Gusset plates are used to join more than one member at a joint. The lines of action of truss members meeting at a joint should coincide. The size and shape of gusset plates are usually decided from direction of members meeting at joint. The plate outlines are fixed to meet minimum edge distances for bolts used for connection.

Correct answer is (d) 12MM

The EXPLANATION is: The thickness of gusset plate in any case should not be less than 12mm. Structurally a gusset plate is SUBJECTED to SHEAR stresses, direct stresses and BENDING stresses and therefore it should be of ample thickness to resist all these at the critical section.

Right option is (b) available length is less than required length of a tension member

The best explanation: Splices are provided when the available length is less than required length of a tension member.Splices in tension members are used to JOIN sections when a joint is to be provided that is these REPLACE the members at the joint where it is cut. If the sections to spliced are not of same THICKNESS, then packing plates are INTRODUCED.

Correct answer is (d) lug angles and their connection to gusset should be capable of developing a strength of not LESS than 20% of excess of force of outstanding leg of angle

The BEST I can explain: In CASE of angle MEMBERS, lug angles and their connection to gusset should be capable of developing a strength of not less than 20% of excess of force of outstanding leg of angle, and the connection of lug angle to angle member should be capable of developing a strength of 40% of excess of force.

The correct option is (d) lug angles and their connection to gusset should be capable of developing a STRENGTH of less than 5% of excess of force in flange of channel

Easiest EXPLANATION: In case of channel members, lug angles and their connection to gusset should be capable of developing a strength of not less than 10% of excess of force in flange of channel, and the attachment of lug angle to angle member should have a strength not less than 20% of excess of that force.

Right answer is (b) by providing unequal angle sections with wider leg as connected leg

To explain I WOULD SAY: Lug angles can be eliminated by providing unequal angle sections with wider leg as connected leg and USING two ROWS of staggered bolts.

Correct choice is (c) BEGINNING of CONNECTION

Easy explanation: Lug angles are found to be more effective at beginning of connection rather than the end DUE to non-uniform distribution of load AMONG connecting BOLTS.

Right answer is (a) additional angles USED to reduce joint length

The BEST I can explain: When tension MEMBER is SUBJECTED to heavy load, the number of bolts or length of welds required for making connection becomes large, it results in uneconomical size of gusset plates. In such situations, additional SHORT angles called lug angles may be used to reduce joint length and shear lag.

Right choice is (a) residual STRESS result in local early STRAIN hardening

The EXPLANATION is: Residual stress result in local early strain hardening and reduce plastic range of MEMBER. Residual stresses have no consequences with respect to static strength of member, they can be important if FATIGUE is involved.

Right choice is (d) reduction in ductility REDUCES strength of member

To EXPLAIN: Reduction in ductility tends to reduce strength of member. An increase in ductility tends to increase net section strength by allowing BETTER plastic redistribution of stress concentration over CROSS section.

Correct choice is (b) reduces

The EXPLANATION is: The shear lag effect INCREASES with increase in connection LENGTH. The shear lag reduces the EFFECTIVENESS of component plates of tension member that are not connected directly to GUSSET plate.

Right ANSWER is (d) BOLT diameter

The explanation is: The actual FAILURE mode in bearing DEPENDS on end distance, bolt diameter and thickness of the CONNECTED material.

The correct option is (b) improves

For explanation I would SAY: Staggering of holes improves the load carrying capacity of tension member for given number of bolts. The failure paths may occur along sections NORMAL to axis of member, or they may include zigzag sections when more than one bolt hole is presentand staggering of holes may help to make the net AREA MINIMUM.

Correct choice is (a) strength of members with PUNCHED HOLES is less than members with drilled holes

The explanation: Strength of members with punched holes may be 10-15% less than the members with drilled holes. This is due to strain hardening EFFECT of MATERIAL around punched holes and consequent loss of DUCTILITY.

The correct ANSWER is (c) reduces

Best explanation: The BOLT holes REDUCE the area of CROSS section available to carry TENSION and hence reduce the strength of tension member.

The CORRECT answer is (b) 0.7

Best explanation: As PER IS code, the equation for PRELIMINARY design of angle tension memberwith partial factor of safety for material is given by Tdn = αAnfy/γm1, where α = 0.6 for one or TWO bolts, 0.7 for two bolts, 0.8 for four or more bolts in the end connection or equivalent WELD length.

Correct choice is (a) area of unconnected leg

To elaborate: β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] , where fu and fy are ultimate and yield stress of material, w and t are size and thickness of outstanding leg respectively, bs is the shear DISTANCE from edge of outstanding leg to nearest line of fasteners, Lc is the LENGTH of end connection measured from centre of FIRST bolt hole to centre of LAST bolt hole in the end connection.

Right answer is (b) βAg0fy/γm0

The explanation is: The design strength of angle section GOVERNED by tearing at net section is given byTdn = (0.9Ancfu/γm1) + βAg0fy/γm0 , where Anc = net area of connected leg, Ag0 = gross area of outstanding leg, fu = ultimate strength of material, γm1 = PARTIAL safety factor for FAILURE DUE to rupture of cross section = 1.25, γm0 = partial safety factor for failure in tension by YIELDING = 1.10.

The correct option is (a) THICKNESS of angle has no SIGNIFICANT INFLUENCE on member strength

The explanation is: (i) The effect of GUSSET plate thickness on ultimate TENSILE strength is not significant, (ii) the thickness of angle has no significant influence on member strength, (iii) the net section efficiency is higher(7-10%) when long leg of angle is connected rather than short leg, (iv) when length of connection increases, the tensile strength increases upto four bolts and effect of any further increase in number of bolts on tensile strength of member is not significant.

Correct OPTION is (a) MINIMUM of STRENGTH due to gross yielding, net section RUPTURE, block shear

The BEST explanation: The design tensile strength of tensile member is taken as the minimum of strength due to gross yielding (Tdg=fyAg/1.1), net section rupture(Tdn=0.9Anfy/γm1), block shear (Tdb1=(Avgfy/√3 γm0)+(0.9Atnfu/γm1),Tdb2=(Atgfy/ γm0)+(0.9Avnfu/√3 γm1)).

Correct answer is (d) When angles are connected to gusset plates by welding or bolting only through one of the two legs RESULTING in eccentric loading, there is a UNIFORM stress distribution over cross SECTION.

Easiest explanation: Angles if axially loaded through centroid can be designed as plates. Angles connected to gusset plates by welding or bolting only through one of the two legs results in eccentric loading, causing non-uniform stress distribution over cross section. When LOAD is applied by connecting only one LEG of member, there is shear lag at the end connection.

Right choice is (d) smaller of BLOCK shear STRENGTH at an END connection for (shear fracture, tension yield) and (shear yield, tension fracture)

EXPLANATION: The block shear strength of connection is smaller of block shear strength at an end connection for shear yield, tension fracture Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1) and block shear strength at an end connection for shear fracture, tension yield Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γvm1).

Right answer is (a) (Avgfy/√3 γm0)+(0.9Atnfu/γm1)

To explain: The block shear strength at an end connection for shear yield and tension fracture is given by Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1), where Avg = minimum gross area in shear ALONG line of action of force, Atn = minimum net area of CROSS SECTION in tension from hole to toe of angle or last ROW of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.

Right option is (b) (Atgfy/ γm0)+(0.9Avnfu/√3 γm1)

The best explanation: The block shear strength at an end connection for shear fracture and TENSION YIELD is GIVEN by Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1), where Avn = minimum net area in shear along line of action of force, Atg = minimum gross area in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, fy and fu are yield and ultimate stress of material respectively, γm1 = 1.25, γm0 = 1.10.

Right choice is (d) 0.9Anfy/γm1

The EXPLANATION: The design STRENGTH of tension member corresponding to NET section rupture is given by Tdn = 0.9Anfy/γm1,where An = net effective area of cross section in mm^2, fy = ultimate strength of material in MPa, γm1 = PARTIAL safety factor for failure DUE to rupture of cross section = 1.25.

Right choice is (c) Net area = Gross area – deductions

Easy EXPLANATION: Net area = Gross area – deductions, that is net area of TENSILE MEMBERS is calculated by DEDUCTING AREAL of holes from the gross area.

Correct CHOICE is (a) flow CURVE

The best I can explain: True stress strain curve is also known as flow curve since it REPRESENTS BASIC plastic flow characteristics if the material. Any POINT on the flow curve can be considered as local stress for metal strained in tension by magnitude shown on the curve.

Right choice is (d) post ultimate strain SOFTENING in engineering stress strain curve is ABSENT in true stress strain curve

To EXPLAIN I would say: The post ultimate strain softening in engineering stress strain curve CAUSED by necking of cross section is absent in true stress strain curve as engineering stress strain curve are based on true DIMENSIONS of specimen and true stress strain curve are based on actual cross sectional area of specimen.

Right CHOICE is (a) strain softening REGION, strain hardening region, YIELD plateau, linear elastic region

Best explanation: The ENGINEERING stress-strain curve is typically represented by four regions : linear elastic region, yield plateau, strain hardening region, strain softening (UNLOADING)region.

Right OPTION is (b) it does not gives TRUE INDICATION of DEFORMATION characteristics of metal because it is entirely based on true dimensions of specimen

Easiest explanation: The engineering stress-strain curve does not provide true indication of deformation characteristics of metal. It is entirely based on true dimensions of specimen and these dimensions change continuously as the load increases.

Correct choice is (c) stress and strain calculated using initial CROSS SECTION area and initial GAUGE LENGTH are referred to as engineering stress and engineering strain

Easiest explanation: Stress and strain calculated using initial cross section area and initial gauge length are referred to as engineering stress and engineering strain. Stress and strain calculated using current cross section area and gauge length are referred to as true stress and true strain.