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1
TENSION MEMBER
- LRFD -
Ari Wibowo, Ph.D
Steel Structure
2
Cross-Sectional Shapes for Tension Members
The more common shapes used for tension members.
The solid circular bar is frequently used, either as
threaded rod or welded to other members. The
threaded end provides simple connection but mus take
into account the reduction of the cross sectional area.
3
Cross-Sectional Shapes for Tension Members
Some typical built-up cross section
Square, rectangular and circular tubes have attractive appearance and
ease of maintenance, but the end connections may become
complicated and expensive. The tubes is useful for longer tension
components, where slenderness and related serviceability
consideration may be important.
4
Behavior and Strength of Tension members
Load-Deformation Relationship
The differences between the curves in (a) and (b) may be attributed to a
number of factors:
1. The load P is not applied concentrically to the member
2. The full-size shape has certain built-in or residual stress
3. The tension member is not perfectly straight.
4. The cross-sectional shape and size may vary to some extent along the
member length.
5
Behavior and Strength of Tension members
Items 1 and 3 imply that there will be a certain amount of bending in the member,
in addition to the effect of the axial load.
Therefore, as the load
increased, the fiber that is
stressed the most in tension
due to bending plus axial
load will yield before any
other fiber.
It is clear that the
moments that are
produced create a
nonuniform stress
distribution in the cross-
section
That fiber reaches the yield plateau of the stress-strain curve before all of the
others, which are still strained in the elastic range.
6
Behavior and Strength of Tension members
item 2 : Effect of Residual Stress
7
Behavior and Strength of Tension members
The presence of residual stresses in
compression members results in a
significant lowering of the actual
strength.
For tension members, on the other
hand, the only effect is a shift in the
load-deformation curve that reflects
the larger deformations, but the
strength stays the same.
8
Behavior and Strength of Tension members
Item 4 : a member of cross-sectional area is not a constant along
the length.
That hole creates a stress concentration. As a result, the
application of an increasing tensile force will cause yielding at
the edges of the hole first.
On the other hand, while stress concentrations may not
reduce static yield strength of the member, they do have
significant effect on the dynamic or fatigue-related structural
performance.
9
Behavior and Strength of Tension members
Aspect of behavior of the load-deflection curve for the full-size
member
1. Range 0-1 : All fibers in the cross section are stressed to less than yield
stress. The member is therefore fully elastic, and all deformations are
recoverable upon unloading.
2. Range 1-2 : First yield occurs for P = Pprop, which is the proportional
limit.
3. Range 2-3 : Certain fibers that yielded early are bound to reach strain-
hardening.
4. Range 3-4 : a number fibers will already have reached the descending
portion of the stress-strain curve.
10
Behavior and Strength of Tension members
Eccentricity Load is less dangerous than lateral load case.
11
BATANG TARIK PADA STRUKTUR
nRQ
For tension members, this takes the form
untntu PP PP or
The central problem of all member design,
including tension member design, is to find a
cross section such that the sum of the factored
loads does not exceed the member strenght;
that is :
12
BATANG TARIK PADA STRUKTUR
75.0t
where Pu is the sum of the factored loads. To
avoid reaching the limits state of yielding,
y
ugugy
F.
PAPAF.
900or 900
u
ueueu
F.
P A PAF.
750or 750
To avoid fracture,
If Ae = UAn, then,
UF
PA
u
un
75.0
90.0t
13
BATANG TARIK PADA STRUKTUR
The slenderness ratio limitation will be satisfied if,
300
Lr
where r is the minimum radius of gyration of the cross
section and L is the member length.
14
Example 1
A bar 5 x ½ of A36 steel is used as a tension
member. It is connected to a gusset plate with four
5/8-in diameter bolts as shown here. Assume that
the effective net area Ae is equal to the actual net
area and compute the design strength.
15
• For the limit state of yielding,
Solution
2
21 .5.25 inAg
The nominal strength is
kipsAFP gyn 90)5.2(36
And the design strength is
kipsPnt 81)90(9.0
• For fracture of the net section,
example) for this(.75.1
.75.175.05.2
25.2
2
2
43
21
inAA
in
holes
AAA
ne
holesgn
16
The nominal strength is
kipsAFP eun 5.101)75.1(58
And the design strength is
kipsPnt 1.76)5.101(75.0
The smaller value control
17
Limit State
In the design of tension members, the following cross-sectional area definitions are important :
1. Gross cross-sectional area, Ag
2. Net Area, An
3. Effective net area, Ae
For s steel plate with width b, thickness t and single hole of diameter d :
and
btAg
tdbAn
The effective net area is a function of the cross section and the
method of end connection
18
Two modes of failure
1. Yielding on the gross area , = 0.9
If the size of the gross area is such that the condition
is reached before the ultimate stress Fu has been developed anywhere in
the member.
The characteristics are : ductile, large deflection/deformation
phenomenon.
2. Fracture on the effective area , = 0.75
If the size of the effective area is such that the condition
is reached before the yielding stress Fy has been developed on the gross
area.
The characteristics are : brittle, small deflection/deformation
phenomenon, occurs suddenly.
y
g
n FA
P
u
e
n FA
P
19
Stress Definition
Stress are defined as
Then can be rewritten as :
and
Two limit states occur simultaneously if
or
u
e
neffy
g
ngross F
A
PF
A
P and
eueeffn
gyggrossn
AFAP
AFAP
gyeu AFAF
u
y
g
e
F
F
A
A
20
The limit state of yielding on the gross section governs if :
or
Conversely, fracture on the effective net section governs
when
or
gyeu
u
y
g
eg
AFAF
F
F
A
AAonYielding
:
gyeu
u
y
g
ee
AFAF
F
F
A
AAonFracture
:
21
Computation of Areas1. Gross Area
No holes or other area reductions can be present where the
section is made
+ plates and bars, Ag = bt
+ solid circular, Ag = d2/4
or in general :
+ angles, can be treated as an equivalent plate
width : we = l1 + l2 – t , thickness : te = t
so : Gross area : Ag = wete = (l1 + l2 –t)t
iig twA
22
Computation of Areas
2. Net Area
Net area is a gross area that reduced by holes resulting from
fasteners (bolts, rivets) and certain type of welds (plug and
slots).
Such welds (fillet, full and partially groove welds) do not
reduce or increase the cross-sectional area, so :
Welded joints : An = Ag
23
Size of Standard Holes
Recognizing the needs for fabrication and erection tolerances, standard
bolt holes are made 1/16 in. larger in diameter than that of the bolt.
The hole making may damage the material, as in case punch holes, the
edges around the hole are deformed. This region discounted and the
effective holes increased by another 1/16 in.
161
161: dddiameterholeEffective e
24
Gross area: Ag = 6(½) = 3.0 in.2
Net Area : Ag = ( b – de ) t
For a ¾-in.-diameter bolt, the effective hole size is
Consequently,
Example 2
.87
161
161
43 inde
1. A single line of standard holes for ¾-in bolts are placed
in a 6½ plate. Determine the gross and net areas
Solution
2
21
87 in.56.20.6 nA
25
Gross area: Ag = 10( ¾ ) = 7.5 in.2
Net Area : An = ( b – de ) t
For a 7/8-in.-diameter bolt, the effective hole size is
Consequently,
.in0.1161
161
87 ed
2. A double line of standard holes for 7/8-in bolts are placed
in a 10¾ plate. Determine the gross and net areas
Solution
2
43 in.00.60.120.10 nA
26
Oversized and Slotted Holes
Section J3.7 LRFD Specs. gives the
measurement for oversized holes, short
slotted and long slotted holes.
Basically, this types of holes are used in
many instances to facilitate the erection of
the structure, or to allow certain rotations or
deformations for specific loading case.
When the tension member is short
connecting element, such as stiffener or a
gusset plate,
An < 0.85 Ag
27
Influence of Hole Placement
The location of the governing net section is not obvious for
either of the two cases shown above. (a) is a plate and (b) is an
angle with pitch s and gages g between the holes.
28
• For the plate, perpendicular section 1-1 and 2-2 give identical An values,
where only a deduction for one hole has been taken.
• For the angle, it is clear that section 2-2, which incorporates one hole
deduction for each of the legs, for a total of two is more critical than
section 1-1.
But the elastic stress distribution on plate sec. 1-1 and 2-2 is different.
If s is small enough (section 1-1 and 2-2 is close enough), the stress will
influence each other and elastic distribution will changed. On this basis, it is
possible that a section or chain such 1-2 on plate and 2-3-2 on angle may be
the governing one.
29
• The influence of the staggered hole could be accounted for by
using the following expression for the net width of the section :
wn = wg – d – xd
where : wn = the net width, wg is the gross width, d is the
diameter of the hole, and xd represents the strength-reducing
influence of the staggered hole.
• xd could be expressed as
• This ratio reduces to s2/4g, if it is assumed that 2d is significantly
less than 4g.
dg
sdxd
24
2
g
sdxd
4
2
30
• So the net width of a two-hole section, with one-hole offset with
a pitch of s and a gage of g, becomes :
or
Equation above states that the net width is found by deducting fully
for all holes in the chain, but since this overestimates the effect
of the staggered holes, a certain width contribution is added.
The generalized form of the net width equation for m holes with
(m-1) staggers for a chain :
g
sddww gn
4
2
g
sdww gn
42
2
i
iegn
g
smdww
4
2
31
Picture (a).
Given :
The hole pattern for an 18-in.-wide plate is loaded in tension.
Determine the net width that governs the design.
Example 3
32
Solution
Step 1. Chain A B C E F
Deduct for 3 holes @ ( ¾ + ⅛ ) = - 2.63 in.
For BC, add s2/4g = 2.02/(4)(4) = + 0.25 in.
For CE, add s2/4g = 2.52/(4)(10) = + 0.16 in.
Total deduction = - 2.22 in.
Step 2. Chain A B C D E F
Deduct for 4 holes @ ( ¾ + ⅛ ) = - 3.50 in.
For BC, add s2/4g = 2.02/(4)(4) = + 0.25 in.
For CD, add s2/4g = 4.52/(4)(6) = + 0.85 in.
For DE, add s2/4g = 2.02/(4)(4) = + 0.25 in.
Total deduction = - 2.15 in.
33
Step 3. Net width = 18 – 2.22 = 15.78 in.
Step 4. Check maximum allowable net area :
Max allowable net width = 18.0(0.85) = 15.3 in.
Since 15.3 is less than 15.78, the maximum allowable
value of step 4 governs : 15.3 in.
34
Picture (b).
Given :
A 6 4 ½ angle is used as a tension member. It has
holes for ⅞-in. bolts that are placed as shown below.
Example 4
35
Step 1. Determine the width of the folded-out angle “plate” :
we = l1 + l2 – t = 9.5 in
Step 2. Gross area of plate:
Ag = we t = 9.50 (0.5) = 4.75 in.2
Step 3. The gages for the holes are shown above. The gage
between the holes closest to the heel of the angle in the two legs is
found as ( g + g1 – t ) = 2.50 + 2.25 – 0.50 = 4.25 in.
Step 4. Computation of net area: The governing net section will be
section 2-2 or 2-1-2.
Section 2-2: An2 = [9.50 – 2 (⅞ + ⅛)] 0.50 = 3.75 in.2
Section 2-1-2:
Solution
222
3 in.5.350.0)25.4(4
50.2
)50.2(4
50.2)1(350.9
nA
36
Step 5. Governing net area :
Since An3 = 3.50 in.2 is less then An2 as well as An,max =
0.85 Ag = 4.04 in.2, it governs the net area of the given
angle and hole configuration.
37
Effective Net Area• A connection always weakens the member, and the
measure of its influence is called the joint efficiency
• This factor is a function of :
– The ductility of the material
– Fastener spacing
– Stress concentration at holes
– Fabrication procedure
– Shear lag phenomenon
Shear Lag occurs when all elements of the cross section are not
connected. This happens, for example, when only one leg of an
angle is bolted to a gusset plate (figure below)
38
Shear Lag
9.01
L
xU
The consequence of this partial connection is that the
connected element becomes overloaded and the unconnected
part is not fully stressed. Lengthening the connected region
relative to the member length will reduce this effect.
Shear Lag is accounted for by using a reduced, or effective, net
area obtained by multiplying the actual net area by the reduction
factor, U :
x
Where :
= the distance from the centroid of the connected area to
the shear plane of the connection.
L = the length of the connection
39
For bolted connection, the effective net area is :
Ae = U An
For welded connection,
Ae = U Ag
If all element of the cross section are connected, then there
is no reduction, and Ae = An
40
Shear Lag – bolted connection
L
x
L
xL
L
LU
1
'
The reduction can be approximated by multiplying the net area
by the factor U :
A typical example calculation of the
shear lag factor is shown for both a
bolted and a welded case.
For the bolted case, N is the
total number of bolts, and s is
the pitch.
41
In case where there are
transverse and
longitudinal weld :
a. Transverse :
• Longitudinal
L
xUL 1
Ag
tdU
ftransverse
T
.
TL
TTLL
dd
dUdUU
..
It is recommended that weighted average U value should
be used for such cases.
Shear Lag – welded connection
42
Simplified method: 3 condition of rules
for bolted conection
1. For W-, M-, and S-shapes that have a flange width to web depth ratio of
at least ⅔ (and tee shapes cut from these) and are connected through
flanges with at least three fasteners per line in the direction of applied
load,
U = 0.9
2. For all other shapes (including built-up shapes) with at least three
fasteners per line,
U = 0.85
3. For all members with only two fasteners per line,
U = 0.75
Based on average values of x/L for various types of bolted tension
member connections, the Commentary to AISC B3 gives values of
reduction factor U :
43
Example : Application of the rules
44
1. For any W-, M-, and S-shapes or tee shapes connected by transverse
weld alone,
Ae = area of connected element
2. For plates or bars connected by longitudinal welds at their ends,
U = 1.0 for l ≥ 2w
= 0.87 for 1.5w ≤ l < 2w
= 0.75 for w ≤ l < 1.5w
The Commentary to AISC B3 gives values of reduction factor U
for welded connection :
Where :
l = length of the pair welds ≥ w
w = distance between the welds
Simplified method: 2 condition of rules
for welded conection
45
Determine the effective net area for the tension member shown
below
Example 5
Solution
An = Ag – Aholes
= 5.75 – ½ (⅝ + ⅛)(2) = 5.00 in.2
Since only one element (one leg) of the cross section is
connected, the net area must be reduced. From the
properties tables, the distance from the centroid to the
outside face of the leg of an L 6x6x½ is
2in.68.1x
46
The length of the connection is
L = 3 + 3 = 6 in.
Ae = U An = 0.720 (5.00) 3.60 in.2
)(9.0720.06
68.111 OK
L
xU
The average value of U from Commentary can also be used.
Since this shape is not a W, M, S, or tee and has more than
two bolts in the direction of the load, the reduction factor U
can be taken as 0.85, and
Ae = U An = 0.85 (5.00) = 4.25 in.2
Either U value is acceptable, but the value obtained from
AISC equation is more accurate. However, the average values
of U are useful during preliminary design, when actual
section properties are not known.
47
Example 6
Solution
If the tension member 66½ is welded as shown, determine
the effective net area
Only part of the cross section is connected, and reduced
effective net area must be used. The connection is made with
a combination of longitudinal and transverse welds, so it is
not one of the special case for welded members.
9.0695.05.5
68.111
L
xU L
522.075.5
6 21
g
ftransv
TA
tdU
48
2mm381.375.5588.0. ge AUA
Therefore, the effective area of welded members is 3.381
mm2.
588.0
65.5
6522.05.5659.0
..
TL
TTLL
dd
dUdUU
49
Block Shear
For certain configuration, a segment or “block” of material at
the end of the member can tear out. For example as shown
below, is called “block shear”.
The shaded block would tend to fail :
By shear along the longitudinal section ab
By tension on the transverse section bc
• The design strength of tension members are not always
controlled by factor of safety or by the strength of the bolts or
welds with which they are connected. They may instead be
controlled by block shear strength.
• In block shear mode, the failure of the member occurs along a
path involving tension on one plane and shear on a
perpendicular plane along the fasteners. When a tensile load
applied to a particular connection is increased, the fracture
strength of the weaker plane approaches.
• This plane does not fail instantly, because it is restrained by the
stronger plane. The load can be increased until the fracture
strength of the stronger plane is reached and during this time,
the weaker plane yields.
• The total strength of the connection equals the fracture strength
of the stronger plane plus the yield strength of the weaker plane.
50
51
The procedure is based on the assumption that one of the two
failure surface fractures and the other yields :
Fracture on the shear surface is accompanied by yielding on
the tension surface, or
Fracture on the tension surface with yielding on the shear
surface
There are two possible failure modes :
• For shear yield and tension fracture
Rn = [0.6Fy Agv + Fu Ant] (AISC equation J4-3a)
• For shear fracture and tension yield
Rn = [0.6Fu Anv + Fu Agt] (AISC equation J4-3b)
In both cases, = 0.75. Since the limit state is fracture, the controlling
equation will be the one which has the larger fracture term.
52
Example 7
Check the block shear design strength of the tension member
shown below. The holes are for ⅞-in. diameter bolts, and A36
steel is used.
Solution
Shear Areas :
Agv = ⅜(7.5) = 2.812 in.2
and since there are 2.5 hole diameters,
Anv = ⅜[7.5-2.5(1)] = 1.875 in.2
53
Tension areas :
Agt = ⅜(1.5) = 0.5626 in.2
Ant = ⅜[1.5 - 0.5 ] = 0.375 in.2
AISC Equation J4-3a :
Rn = [0.6Fy Agv + Fu Ant]
= 0.75[0.6(36)(2.812) + 58(0.375)]
= 0.75 [ 60.74 + 21.75 ] = 61.9 kips.
AISC Equation J4-3a :
Rn = [0.6Fu Anv + Fy Agt]
= 0.75[0.6(58)(1.875) + 36(0.5625)]
= 0.75 [ 65.25 + 20.25 ] = 64.1 kips.
The second equation has the larger fracture term, therefore, the
second equation governs.
54
Block Shear
Case (a)
The resistance to block shear is
primarily afforded by shear. Since
the shear fracture load is larger than
the shear yield load, failure will be
governed by shear fracture in
combination with tensile yield.
Case (b)
A case where tension is the primary
stress resultant; in this case the limit
state is that of combined tensile
fracture and shear yield.
55
Example 8 : Design
A tension member with the length of 5 feet 9 inches
must resist a service dead load of 18 kips and a
service live load of 52 kips. Select a member with a
rectangular cross section. Use A36 steel and
assume a connection with one line of ⅞ -in diameter
bolts. Solution
Pu = 1.2D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips
2
2
in 409.20.75(58)
104.8
)(0.75
in 3.2350.90(36)
104.8
)(0.90
y
ue
y
ug
F
PA
F
PA
56
Since Ae = An for this member, the gross area
corresponding to the required net area is
t
t
AAA holeng
409.2
8
1
8
7409.2
Try t = 1 in
2in 4093114092 .)(.Ag
57
Rounding to the nearest ⅛ in., try a 13½ cross
section. Check the slenderness ratio :
2in409.31
409.3
t
Aw
g
g
4
3
min .2917.012
15.3inI
Since 3.409 > 3/235, the required gross area is
3.409 in.2, and
A = 1(3.5) = 3.5 in.2
58
Since I = Ar2,
)(3002392887.0
)12(75.5 maximum
in.2887.05.3
2917.0minmin
OKr
L
A
Ir
Answer : Use a bar 3½ 1 in.
59
Example 9
For the gusset plate in the heavy bracing connection that is
shown below, check whether the plate thickness of ½ in. is
adequate to resist block shear. The steel grade is A36 (Fy = 36
ksi, Fu = 58 ksi), and the holes are drilled, for ⅞-in.-diameter
A325 high-strength bolts.
Solution
It is assumed that the total
factored load of 225 kips is
distributed evenly between
all of the bolts.
The loads listed are those
that are in the gusset plate at
each plane.
60
Assume that planes B and D
have already been evaluated
and found to be acceptable.
Plane A
Anv has two shear plane
Agt has one tension plane
Solution
2
21
161
87 in.53.250.000.32 nvA
2
21 in.00.800.2200.62 gtA
Block Shear Capacity
Pbs = 0.75(0.6FuAnv + FyAgt)
= 0.75(0.6(58)(2.53) + 36(8.00)]
= 261 kips > 80.4 kips (OK)
61
Plane C
Anv = 2[9.00 – 2.50(0.938)] ½ = 6.66 in.2
Agt = 8.00 in.2
Block shear capacity :
Pbs = 0.75[0.6(58)(6.66) + 36(8.00)]
= 390 kips > 192.9 kips (OK)
Since all planes are adequate, it is not necessary to
analyze the other limit state, shear yield plus
tension fracture.
62
Example 10
A W1443 wide-flange shape is connected by flange plates, as
shown below. The bolts are ⅞-in.-diameter A490-X, high
strength bolts. For a single flange the design strength of the
bolts in shear is 211 kips; and the design strength of the bolts in
bearing is 548 kips. Determine the design strength of the
member.
63
Solution
Yield Limit State of Shape :
Gross area of W1443 = Ag = 12.60 in.2
Design yield strength of the member = FyAg
= 0.9(36)(12.6) = 408 kips
Fracture Limit State of Member
• Step 1 Hole area to be deducted in the flange at each connection
L = (N/2 – 1)s = (6/2 -1)(3.00) = 6.00 in.
• Step 2 Determine x and L to be used in the shear lag calculation
The W1443 is
considered as two tee
section WT721.5
64
The x of each WT is found in the LRFD Manual as 1.31 in.
The value of L is :
L = (N/2 – 1)s = (6/2 – 1)(3.00) = 6.00 in.
• Step 3 Calculate the shear lag factor U :
U = 1 – x/L = 1 – 1.31/6.00 – 0.78
The LRFD specification, section B3, notes that for this case, a value of
U = 0.85 may be used.
• Step 4 The design strength of the member is twice the capacity of
each flange connection. The gross area of tee is 6.31 in.2
Design fracture strength = 2AnFuU
= 2(0.75)(6.31 – 1.06)(58)(0.85) = 388 kips
65
Block Shear of Section Flanges :
The design strength is provided
by the larger of the two basic
block shear combination :
(1) Shear yield + Tension fracture
(2) Tension yield + Shear fracture
Step 1 Fracture on shear plane :
Pbs = 2(0.6)FuAnv = kips20953.050.200.8586.021615
Step 2 Yield on tension plane :
Pbs = 2FyAg = 2(36)(2)(0.53) = 76.3 kips
Step 3 Design Strength for condition 1:
(Pbs)total = (Shear + Tension) = 0.75(209+76.3) = 214 kips
66
Step 4 Fracture on tension plane :
Pbs = 2FuAn = 2(58)[2-(½)(15/16)](0.53) = 94 kips
Step 5 Yield on shear plane:
Pbs = 2(0.6)FyAg = 2(0.6)(36)(8.00)(0.53) = 183 kips
Step 6 Design Strength for condition 2:
(Pbs)total = (Tension + Shear) = 0.75(94+183) = 208 kips
Step 7 Governing block shear strength:
The largest capacity from steps 3 and 6 is condition 1; thus
Block shear design strength = 214 kips.
67
Summary For Connection
Bolt design shear strength (given) 211 kips
Connection bolt bearing capacity (given) 548 kips
Gross section yielding of member 408 kips
Effective net section fracture of member 388 kips
Block shear capacity of flanges 214 kips
Conclusion :
The bolt design shear strength of the connection of 211 kips governs
68
Example 11
A WT5 11 in A36 steel is
connected to a gusset plate
with the ends completely
welded and each side
welded along 6 in.
Determine the effective net
area of the member. From
the LRFD Manual, Ag = 3.24
in.2,
in.36.0 and in.07.1 ftx
69
Solution
For the 6-in. welds, the value of U = 1 – 1.07/6.00 = 0.822. For the
end weld, the value of U = 5.75(0.36)/3.24 = 0.639.
The effective reduction coefficient is now found as the weighted
average of the above.
76.075.5)00.6(2
)75.5(639.0)12(822.0
effU
The effective net area is then
Ae = UeffAn = 0.76(3.24) = 2.46 in.2
70
Example 12
The truss diagonal
member consist of a pair
of angles L4 3 ⅜ that
are loaded in tension.
Determine the design
strength T of one angle.
The bolts that will be
used are ¾-in. A325-N,
and the steel is A36.
The bolt design strengths
for the connection in one
angle are 46.5 kips in
shear and 88.1 kips in
bearing
71
Solution
Step 1 Determine the angle design strength for the limit state of
yielding on the gross section :
Ag = 2.48 in.2
T = FyAg = 0.9(36)(2.48) = 80.4 kips
Step 2 Determine the angle design strength in tension fracture
Ag = 2.48 - ⅜(¾+⅛) = 1.23 in.2
Reduction coefficient :
U = 1 – x/L = 1 – 1.28/6.00 = 0.787
T = UFuAn = 0.787(0.75)(58)(1.23) = 42.1 kips
Step 3 Check angle design strength in block shear, for tension
yield and shear fracture combination
72
Tension yield load :
Pbs = FyAgt = 0.75(36)(⅜)(1.5) = 15.2 kips
Shear fracture load :
Pbs = (0.6)FuAnv
= 0.75(0.6)(58)(⅜)[7.25-2.5(⅞+1/16)]
= 48.0 kips
Total block shear capacity :
(Pbs)total = 15.2 + 48.0 = 63.2 kips
Since the lowest tensile design strength is that of angle in tension
fracture, the load 42.1 kips controls. Further, since the larger of the
two block shear mode controls, additional computations are not
needed.
Conclusion : The design tension strength of the angle is 42.1 kips.