Subject :- Design of Steel Structures / DOS-I(Steel)LSM / SE-I (Steel)
Unit-III
:- TENSION MEMBER
1. What do you mean by tension members?
a) Structural elements that are
subjected to direct tensile loads
b) Structural elements that are subjected to direct compressive loads
c) Structural elements that are subjected to indirect tensile loads
d) Structural elements that are subjected to indirect compressive loads
Answer: a
Explanation: Steel tension members are those structural elements that
are subjected to direct axial tensile loads, which tend to elongate the
members. A member in pure tension can be stressed up to and beyond the yield
limit and does not buckle locally or overall.
2. The strength of tensile members is not influenced
by:
a) length of connection
b) length of plate
c) net area of cross section
d) type of fabrication
Answer: b
Explanation: The strength of tensile members is influenced by factors
such as length of connection, size and spacing of fasteners, size and spacing
of fasteners, net area of cross section, type of fabrication, connection
eccentricity, and shear lag at the end connection.
3. Which of the following statement is correct?
a) single angle section with bolted
connection produce eccentricity about one plane only
b) single angle section with welded connection produce eccentricity about both
planes
c) single angle section with bolted connection produce eccentricity about both
planes
d) single angle section with welded connection does not produce eccentricity
about one plane
Answer: c
Explanation: Single angle section with bolted connection produce
eccentricity about both planes, whereas single angle section with welded
connection may produce eccentricity about one plane only.
4. Which of the following statement is correct?
a) Double angle members are used as
web members in trusses
b) Single angle members are used where members are subjected to reversal of
stresses
c) Double angle members are used in towers
d) Single angle members are used as web members in trusses
Answer: d
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.
5. The difference between strand and wire rope
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
b) Wire rope consists of individual wires wound straight around a central core,
strand is made of several wire ropes laid helically around a core
c) Wire rope consists of individual wires wound helically around a central
core, strand is made of several wire ropes laid helically around a core
d) Strand consists of individual wires wound straight around a central core,
wire rope is made of several strand laid helically around a core
Answer: a
Explanation: 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.
6. Which of the following statement is incorrect?
a) Cables in form of wires ropes and
strands are used in application where high strength is required
b) They are recommended in bracing systems
c) Cables are generally long and their flexural rigidity is negligible
d) They are flexible
Answer: b
Explanation: Cables used as floor suspenders in suspension bridges are
made from individual strands wound together in rope like fashion. Cables in
form of wires ropes and strands are used in application where high strength is
required and flexural rigidity is unimportant. Cables are generally long and
their flexural rigidity is negligible. They are flexible. They are not
recommended in bracing systems as they cannot resist compression.
7. Bars and rods are not used as :
a) tension members in bracing
systems
b) sag rods to support purlin
c) friction resistant members
d) to support girts in industrial buildings
Answer: c
Explanation: Bars and rods are used as tension members in bracing
systems, sag rods to support purlin between trusses, to support girts in
industrial buildings, where light structure is desirable. Rods are also used in
arches to resist thrust of arch.
8. Sagging of members by built up bars and rods
may be minimised by_____
a) fabricating rod/bar more than its
required theoretical length
b) increasing length diameter
c) increasing thickness ratio
d) fabricating rod/bar short of its required theoretical length
Answer: d
Explanation: Sagging of members by built up bars and rods may be
minimised by limiting length diameter or thickness ratio or by fabricating the
rod/bar short of its required theoretical length by some arbitrary amount and
drawing into place to provide an initial tension. The same effect may be
produced by providing turnbuckle in the rod.
9. ________ type of tension member is not mainly
used in modern practice?
a) circular section
b) open section such as angles
c) flat bars
d) double angles
Answer: c
Explanation: Tension members were generally made of flat bars earlier.
But modern practice is to use mainly the following sections for tension
members: (i)open sections such as angles, channels and I-sections, (ii)compound
and built up sections such as double angle and double channels with or without
additional plates, (iii)closed sections such as circular, square, rectangular
or hollow sections.
10. Which among the following comparison between
angle and flat bars is not true?
a) for light loads, angles are
preferred over flat bars
b) flat bars are used instead of angles in case of stress reversal
c) flat bar tension members tend to vibrate during passage of load in light
bridges
d) angles are used instead of flat bars in case of stress reversal
Answer: b
Explanation: For light loads, angles are preferred over flat bars. In
many light bridges, flat bar tension members tend to vibrate during passage of
load. In case of stress reversal angles are more suitable whereas flat bars are
unfit to carry compressive load on reversal due to their small radius of
gyration in one direction.
11. Which of the following statement is correct?
a) angles placed on same side of
gusset plate produce eccentricity about one plane only
b) angles placed on opposite side of gusset plate produce eccentricity about
two planes
c) angles placed on same side of gusset plate produce eccentricity about two
planes
d) angles placed on opposite side of gusset plate produce eccentricity about
one plane only
Answer: a
Explanation: Two angle sections can either be placed back-to-back on the
same side of gusset plate, or back-to-back on the opposite side of gusset
plate. When angles are connected on the same side of gusset plate, the
eccentricity is about one plane only, which can be almost eliminated when the
same angles are connected on opposite side of gusset plate.
12. Which of the following statement is true about
built up section?
a) Single rolled section are formed
to meet required area which cannot be provided by built up members
b) Built up members can be made sufficiently stiff
c) Built up sections are not desirable when stress reversal occurs
d) Built up members are less rigid than single rolled section
Answer: b
Explanation: Built-up members, made up of two or more plates or shapes
and connected to act as single member, are formed primarily to meet required
area which cannot be provided by single rolled section. Built up members are
more rigid because for same area much greater moment of inertia can be obtained
than single rolled section. Built up members can be made sufficiently stiff to
carry compression and tension thus desirable when stress reversal occurs.
13. What is slenderness ratio of a tension member?
a) ratio of its unsupported length
to its maximum radius of gyration
b) ratio of its least radius of gyration to its unsupported length
c) ratio of its unsupported length to its least radius of gyration
d) ratio of its maximum radius of gyration to its unsupported length
Answer: c
Explanation: Slenderness ratio of tension member is ratio of its
unsupported length to its least radius of gyration. This limiting slenderness
ratio is required in order to prevent undesirable lateral movement or excessive
vibration.
14. What is the maximum effective slenderness
ratio for a tension member in which stress reversal occurs?
a) 200
b) 280
c) 300
d) 180
Answer: d
Explanation: The maximum effective slenderness ratio for a tension
member in which stress reversal occurs due to loads other than wind or seismic
forces is 180.
15. What is the maximum effective slenderness
ratio for a member subjected to compressive forces resulting only from
combination of wind/earthquake actions?
a) 200
b) 340
c) 250
d) 180
Answer: c
Explanation: The maximum effective slenderness ratio for a member
subjected to compressive forces resulting only from combination of wind or
earthquake actions, such that the deformation of such member does not adversely
affect stresses in any part of structure is 250.
16. What is the maximum effective slenderness ratio
for a member normally acting as a tie in roof truss or a bracing member?
a) 180
b) 350
c) 400
d) 200
Answer: b
Explanation: The maximum effective slenderness ratio for a member
normally acting as a tie in roof truss or a bracing member, which is not considered
when subject to stress reversal resulting from action of wind or earthquake
forces is 350.
17. What is the maximum effective slenderness
ratio for members always in tension?
a) 400
b) 200
c) 350
d) 150
Answer: a
Explanation: The maximum effective slenderness ratio for members always
in tension other than pre-tensioned members is 400.
18. The limits specified by IS code for
slenderness ratio are not
a) applicable to circular sections
b) applicable to cables
c) applicable to angle sections
d) applicable to built-up sections
Answer: b
Explanation: The limits specified for slenderness ratio in the IS code
are not applicable to cables. They are applicable to angle sections, built-up
sections, circular sections.
19. The displacement of tension member under
service load is given by
a) P/LEAg
b) PLEAg
c) PL/EAg
d) PLE/Ag
Answer: c
Explanation: The displacement, that is increase in length of tension
member, under service load is given by Δ = PL/EAg, where Δ =
Elongation of member in mm, P= unfactored axial load in N, L = length of member
in mm, E = elastic modulus = 2×105MPa, Ag = gross cross sectional area of member in mm2.
20. What do you mean by gross section yielding?
a) it is considerable deformation of
the member in longitudinal direction may take place before it fractures, making
the structure serviceable
b) it is considerable deformation of the member in lateral direction may take
place before it fractures, making the structure unserviceable
c) it is considerable deformation of the member in lateral direction may take
place before it fractures, making the structure serviceable
d) it is considerable deformation of the member in longitudinal direction may
take place before it fractures, making the structure unserviceable
Answer: d
Explanation: Tension member without bolt holes can resist loads up to
ultimate load without failure. But such a member will deform in longitudinal
direction considerably (10-15% of its original length) before fracture and the
structure becomes unserviceable.
21. What is net section rupture failure?
a) rupture of member when the cross
section is reaches very less value than ultimate stress
b) rupture of member when the cross section reaches yield stress
c) rupture of member when the cross section reaches ultimate stress
d) rupture of member when the cross section reaches less value than yield
stress
Answer: c
Explanation: The point adjacent to hole reaches yield stress first when
tension member with hole is loaded statically. The stress at that point remains
constant and each fibre away from hole progressively reaches yield stress on
further loading. With increasing load, deformations continue until finally
rupture of member occurs when entire net cross section of member reaches
ultimate stress.
22. The tensile stress adjacent to hole will be
____________
a) about five times the average
stress on the net area
b) about two to three times the average stress on the net area
c) about half the average stress on the net area
d) equal to average stress on the net area
Answer: b
Explanation: From the theory of elasticity, the tensile stress adjacent
to hole will be about two to three times the average stress on the net area,
depending upon the ratio of diameter of hole to the width of plate normal to
direction of stress.
23. What is block shear failure?
a) failure of member occurs along
path involving tension on one plane and shear on perpendicular plane along
fasteners
b) failure of member occurs along path involving tension on one plane and shear
on parallel plane along fasteners
c) failure of fasteners occurs along path involving tension on one plane and
shear on parallel plane along fasteners
d) failure of fasteners occurs along path involving tension on one plane and
shear on perpendicular plane along fasteners
Answer: a
Explanation: Failure of member occurs along path that involves (i)
tension on one plane and (ii) shear on perpendicular plane along fasteners in
block shear failure mode.
24. The possibility of block shear failure
increases by
a) larger connection length
b) with use of high bearing strength material
c) increasing the number of bolts per connection
d) with use of low strength bolts
Answer: b
Explanation: The block shear failure becomes a possible mode of failure
when material’s bearing strength and bolt shear strength are higher. When high
bearing strength of material and high strength bolts are used, only few bolts
are required in connection. This Decreasing number of bolts per connection
results in smaller connection length, but the possibilities of block shear
failure increases.
25. Which of the following statement is correct?
a) stress and strain calculated
using initial cross section area and initial gauge length are referred to as
true stress and true strain
b) stress and strain calculated using current cross section area and initial
gauge length are referred to as true stress and engineering strain
c) stress and strain calculated using initial cross section area and initial
gauge length are referred to as engineering stress and engineering strain
d) stress and strain calculated using current cross section area and gauge
length are referred to as engineering stress and engineering strain
Answer: c
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.
26. Arrange the regions of engineering
stress-strain curve in order from right to left as in graph
a) strain hardening region, strain
softening region, linear elastic region, yielding region
b) strain softening region, yield plateau, linear elastic region, strain
hardening region
c) strain hardening region, linear elastic region, yielding region, strain
softening region
d) strain softening region, strain hardening region, yielding region, linear
elastic region
Answer: d
Explanation: The engineering stress-strain curve is typically
represented by four regions : linear elastic region, yielding points, strain
hardening region, strain softening (unloading)region.
27. What is the yield point for high strength
steel?
a) 0.2% of offset load
b) 0.1% of offset load
c) 1.5% of offset load
d) 0.5% of offset load
Answer: a
Explanation: High-strength steel tension members do not exhibit well
defined yield point and yield plateau. Hence, 0.2% of offset load is usually
taken as yield point for such high strength steel.
28. The design strength of tension member
corresponding to gross section yielding is given by :
a) γm0fy/Ag
b) fy/Ag γm0
c) fyAg/ γm0
d) γm0 fyAg
Answer: c
Explanation: The design strength of tension member corresponding to
gross section yielding is given by Tdg = fyAg/ γm0, where fy = yield
strength of material in MPa, Ag = gross
cross-sectional area in mm2, γm0 = partial safety factor for failure in
tension by yielding = 1.10.
29. Which of the following relation is correct?
a) Net area = Gross area –
deductions for holes
b) Net area = Gross area x deductions for holes
c) Net area = Gross area + deductions for holes
d) Net area = Gross area / deductions for holes
Answer: a
Explanation: Net area = Gross area – deductions, that is net area of
tensile members is calculated by deducting areal of holes from the gross area.
30. The design strength of tension member
corresponding to net section rupture is given by :
a) 0.9Anfyγm1
b) 0.9An/fyγm1
c) 0.9Anfy/γm1
d) Anfyγm1
Answer: c
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
mm2, fy = ultimate
strength of material in MPa, γm1 = partial
safety factor for failure due to rupture of cross section = 1.25.
31. The block shear strength at an end connection
for shear yield and tension fracture is given by :
a) (0.9Avgfy/√3 γm0)+(Atnfu/γm1)
b) (Avgfy/√3 γm0)+(0.9Atnfu/γm1)
c) (Atgfy/√3 γm0)+(0.9Avnfu/γm1)
d) (0.9Atgfy/√3 γm0)+( Avnfu/γm1)
Answer: b
Explanation: 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.
32. The block shear strength at an end connection
for shear fracture and tension yield is given by :
a) (0.9Atgfy/√3 γm0)+( Avnfu/γm1)
b) (Avgfy/√3 γm0)+(0.9Atnfu/γm1)
c) (Atgfy/ γm0)+(0.9Avnfu/√3 γm1)
d) (0.9Avgfy/√3 γm0)+(Atnfu/γm1)
Answer: c
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.
33. The block shear strength of connection is
________
a) larger of block shear strength at
an end connection for (shear fracture, tension yield) and (shear yield, tension
fracture)
b) block shear strength at an end connection for shear fracture and tension
yield
c) block shear strength at an end connection for shear yield and tension
fracture
d) smaller of block shear strength at an end connection for (shear fracture,
tension yield) and (shear yield, tension fracture)
Answer: d
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).
34. The design tensile strength of tensile member
is
a) minimum of strength due to gross
yielding, net section rupture, block shear
b) strength due to gross yielding
c) strength due to block shear
d) maximum of strength due to gross yielding, net section rupture, block shear
Answer: a
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)).
35. Which of the following is not true for angles
as tension members?
a) Angles if axially loaded through
centroid can be designed as plates
b) Angles connected to gusset plates by welding or bolting only through one of
the two legs results in eccentric loading
c) When load is applied by connecting only one leg of member, there is shear
lag at the end connection
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.
Answer: d
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.
36. Which of the following is true statement?
a) effect of gusset plate thickness
on ultimate tensile strength is significant
b) thickness of angle has no significant influence on member strength
c) net section efficiency is lower when long leg of angle is connected rather
than short leg
d) when length of connection decreases, the tensile strength increases
Answer: b
Explanation: (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.
37. The additional factor to be added for angles
for design strength of tension member in net section rupture is given by :
a) βAg0/fyγm0
b) βAg0fyγm0
c) βAg0fy/γm0
d) βAg0γm0
Answer: c
Explanation: The design strength of angle section governed by tearing at
net section is given by Tdn = (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.
38. The constant β in βAg0fy/γm0 for tensile strength of angle section does
not depend on :
a) area of unconnected leg
b) thickness of outstanding leg
c) size of outstanding leg
d) ultimate stress of material
Answer: a
Explanation: 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.
39. Which of the following is correct?
a) β ≥ fuγm0/fyγm1
b) β ≥ fuγm1/fy γm0
c) β ≤ fuγm0/fyγm1
d) β ≤ fuγm1/fy γm0
Answer: c
Explanation: As design tensile strength of angle section in net section
rupture is, Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0, where β = 1.4
– 0.076[(bs/Lc)(w/t)(fy/fu)] and β ≤ fuγm0/fyγm1, β ≥ 0.7.
40. What is the maximum value of β in βAg0fy/γm0 for tensile strength of angle section?
a) 1.2
b) 0.7
c) 0.9
d) 1.4
Answer: d
Explanation: as we know design tensile strength, Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0, where β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] and β ≥ 0.7. That's why minimum value is 0.7 and maximum 1.4.
41. What is the value of partial factor of safety for
material α for preliminary design for angle section as per IS code for three
bolts in connection?
a) 0.6
b) 0.7
c) 0.8
d) 1.0
Answer: b
Explanation: As per IS code, the equation for preliminary design of
angle tension member with 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.
42. Which of the following statement is correct?
a) strength of members with punched
holes is equal to members with drilled holes
b) strength of members with punched holes is less than members with drilled
holes
c) strength of members with drilled holes is less than members with punched
holes
d) strength of members with punched holes is greater than members with drilled
holes
Answer: b
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.
43. The presence of holes _____ the strength of
tension member
a) doubles
b) does not affect
c) improves
d) reduces
Answer: d
Explanation: The bolt holes reduce the area of cross section available
to carry tension and hence reduce the strength of tension member.
44. Staggering of holes __________ the load carrying
capacity of tension member
a) halves
b) improves
c) reduces
d) does not affect
Answer: b
Explanation: 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 present and staggering of holes may help to make the
net area minimum.
45. The actual failure mode in bearing depends on
a) hole diameter
b) length of metal plate
c) length of bolt
d) bolt diameter
Answer: d
Explanation: The actual failure mode in bearing depends on end distance,
bolt diameter and thickness of the connected material.
46. The shear lag effect _____ with increase in
connection length
a) increases
b) reduces
c) doubles
d) does not change
Answer: b
Explanation: 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.
47. For the calculation of net area of flat with
staggered bolts, the area to be deducted from gross area is :
a) nd + n’p2t/4g
b) nd
c) n’p2t/8g
d) ndt - n’p2t/4g
Answer: d
Explanation: The net area of flat with staggered hole is given by : A =
(b – ndh + n’p2/4g)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.
48. What is the net section area of steel plate
40cm wide and 10mm thick with one bolt if diameter of bolt hole is 18mm?
a) 480 mm2
b) 38.2 cm2
c) 20 cm2
d) 240 mm2
Answer: b
Explanation: b = 40cm = 400mm, t = 10mm, dh = 18mm
Net section area = 400×10 – 16×10 = 3820mm2 = 38.2 cm2.
49. Which section to be considered in the design
for the net area of flat?
a) 2-7-4
b) 1-5-7-4
c) 1-5-7-6-3
d) 1-5-6-3
Answer: c
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.
50. What is the net area for the plate 100 x 8 mm
bolted with a single bolt of 20mm diameter in case of drilled hole ?
a) 640 mm2
b) 624 mm2
c) 756 mm2
d) 800 mm2
Answer: a
Explanation: In case of drilled hole, dh = 20mm
Net Area An = Ag – dht = 100 x 8 – 20 x 8 = 640mm2.
51. Determine the effective net area for angle
section ISA 100 x 75 x 12 mm, when 100mm leg is connected to a gusset plate
using weld of length 140mm.
a) 1795 mm2
b) 1812 mm2
c) 1956 mm2
d) 2100 mm2
Answer: c
Explanation: Net area of connected leg, Anc = (100 –
12/2) x 12 = 1128 mm2
Net area of outstanding leg, Ago = (75 –
12/2) x 12 = 828 mm2
Total net area = 1128 + 828 = 1956 mm2.
52. Calculate the value of β for the given angle
section ISA 150x115x8mm of Fe410 grade of steel connected with gusset plate :
Length of weld = 150mm
a) 0.89
b) 1
c) 0.75
d) 0.5
Answer: a
Explanation: w=115mm, t=8mm, b=115mm, Lc=150mm, fy=250MPa, fu=410MPa
β = 1.4 – [0.076 (w/t)(fy/fu)(bs/Lc)] = 1.4 – [ 0.076 x (115/8) x (250/410) x
(115/150)] = 0.89 (>0.7) .
53. Calculate the tensile strength due to gross
section yielding of an angle section 125 x 75 x 10mm of Fe410 grade of steel
connected with a gusset plate.
a) 586.95 kN
b) 225.36 kN
c) 432.27 kN
d) 780 kN
Answer: c
Explanation: fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25,
For ISA 125 x 75 x 10mm : gross area Ag = 1902 mm2
Tensile strength due to gross section yielding, Tdg = Agfy/γm0 = 1902 x 250 x 10-3 / 1.1 = 432.27 kN.
54. A single unequal angle 100 x 75 x 10 of Fe410
grade of steel is connected to a 10mm thick gusset plate at the ends with six
16mm diameter bolts with pitch of 40mm to transfer tension. Find the tensile
strength due to net section rupture if gusset is connected to 100mm leg.
a) 416.62 kN
b) 526.83 kN
c) 385.74 kN
d) 450.98 kN
Answer: a
Explanation: dh=18 mm, fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25
Anc = (100 – 10/2 – 18) x 10 = 770 mm2, Ag0 = (75 – 10/2)
x 10 = 700 mm2
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.
55. Determine block shear strength of tension
member shown in figure if grade of steel is Fe410.
a) 326.54 kN
b) 309.06 kN
c) 216.49 kN
d) 258.49 kN
Answer: c
Explanation: fu = 410 MPa,
fy = 250 MPa, γm0 = 1.1, γm1 = 1.25
Avg = ( 1×100 + 50 ) x 8 = 1200 mm2
Avn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm2
Atg = 35 x 8 = 280 mm2
Atn = (35 – 1/2 x 20) x 8 = 200 mm2
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
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