1. The document discusses the limit states and failure modes of bolted double angle tension members, including yielding of the gross section, fracture at the net section, and block shear failure. 2. It provides equations to calculate the effective net area considering shear lag effects, and the block shear strength considering both shear and tensile strengths. 3. An example calculation is shown to determine the tensile resistance of a double unequal angle member bolted at one leg, where fracture at the net section governs with a strength of 393.9 kN.

1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Strain, ε
εy
εuεy
εu
Stress,f
Fy
Fu
E
Strain, ε
εy
εuεy
εu
Stress,f
Fy
Fu
εy
εuεy
εu
Stress,f
Fy
Fu
E
Strength of Double Angle Bolted Tension Members
Limit States of a Tension Member
• A tension member can fail by reaching one of two limit states:
1. Excessive deformation: can occur due to the yielding of the gross section
along the length of the member, for example section a-a in Figure 2.
2. Fracture in the net section: can occur if the stress at the net section
(section b-b in Figure 2) reaches the ultimate stress Fu.
• The objective of design is to prevent these failures before reaching the
ultimate loads on the structure.
b b
aa
Gusset plate
b b
aa
200 x 12 mm bar
Gusset plate
22 mm diameter hole
Section a-a
Section b-b
Section a-a
Section b-b

2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Fig 1. Bolted tension member
1) Yielding of gross area
2) Fracture at net area
Fig 2 Dimension of cross section
Fu = Ultimate Tensile Strength of angles
Net area = Anet = Gross area – area of holes = {Ag – ∑ dh t }
dh = hole diameter = bolt diameter + 3mm (or 1/8 in)
Ag = Gross Area of angles
Fy = Yield Tensile Strength of angles
Ø Rn = 0.75* Ae * Fu
Ø Rn = 0.9* Fy * Ag
Effective Area , Ae = Anet * U
y*
y"
t
g2
g1
b
h
tg
section (1-1)
hg
Pu
Pu
SS Le2Le1
1
1
t
Le1 S S
Lc

3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Shear Lag effect
• Shear lag occurs when the tension force is not transferred uniformly
to all elements of the cross-section. This will occur when some
elements of the cross-section are not connected.
Strength reduction factor , U = (1 – x / Lc ) < 0.9
Lc = For bolted connections, l is the distance between the first and last
fasteners. For staggered bolts, the out-to-out dimension is used .

4. 4 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Strength reduction factor , U = (1 – x / Lc ) < 0.9
x' = c.g of angle along horizontal leg
y' = c.g of angle along vertical leg
y* = c.g of the shaded area of angle
y* = Ag (one angle) * y' – {(Ye)* tangle}* (hangle– [Ye /2] )
Ag – (Ye) * tangle
• SBC 306 gives values of U for some connection configurations
that can be used instead of using Equation . These values are
summarized in Table below.
1 For W, M, and S shapes
or Tee cut from these
shapes
With flange
connected with 3 or
more fasteners per
line in the direction of
loading
bf ≥ 2/3d …..
U=0.9
b f < 2/3d ….
U=0.85
2 With web connected
with 4 or more
fasteners per line in
the direction of
loading
U=0.7
3 For all other shapes
including built up
sections
with at least 3
fasteners per line in
the direction of
loading
U=0.85
4 For all members with only two
fasteners per line
U=0.75
5 For all tension members where tension load is
transmitted onlybytransverse welds to some but
not all of the cross-sectional elements: Ae=UA,
A=area of the directly connected elements.
U = 1.0
6 For plates where tension
load is transmitted by
longitudinal welds only.
For l ≥ 2w
For 2w>l ≥ 1.5w
For 1.5w>l ≥ w
U = 1.00
U = 0.87
U = 0.75
x'
h
g1
ye
y*
y"
dh
t
t
y'
l
w

5. 5 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
T
T
(a)
(b)
(c)
T
T
Tension failure plane
(a)
(b)
3) Block shear in angle
• For some connection configurations, the tension member can
fail due to ‘tear-out’ of material at the connected end. This is
called block shear.
case 1 case 2
Fig 3 Block shear failure in bolted connection
Lt
Lv 1
Lt
Lv 1
P
Lc

6. 6 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
• Block shear strength is determined as the sum of the shear strength
on the shear path and the tensile strength on a tension path:
•
­ Block shear strength = net section fracture strength on shear path
+ gross yielding strength on the tension path
­ OR
­ Block shear strength = gross yielding strength of the shear path +
net section fracture strength of the tension path
Agt = Lt * ∑ tangle
Ant = ( Lt - ∑ dh ) * ∑ t angle
Agv = Lv * ∑ tangle
Anv = ( Lv - ∑ dh) * ∑ tangle
Where
Lv = 2*Lv1 (for given case 1)
Lv = Lv1 (for given case 2)
Agt = gross Area in tensile plane for 2 angle
Ant = net Area in tensile for 2 angle
Agv = gross Area in shear for 2 angle
Anv = net Area in shear for 2 angle
Fu = Ultimate Tensile Strength of angles
Fy = Yield Tensile Strength of angles
Effect of Staggered bolt holes on net area
If 0.6 * Fu * Anv > Fu * Ant , Ø Rn = 0.75* (0.6 * Fu * Anv + Fy * Agt)
If 0.6 * Fu * Anv < Fu * Ant , Ø Rn = 0.75 * (Fu * Ant + 0.6 * Fy * Agv)
S
g
1
1 2
2
For path 1-1
An = Ag – ∑ dh * t
For path 2-2
An = Ag + ∑ S2
t - ∑ dh *t
4 g
S
g
1
1 2
2
For angles bolted at one leg
For angles bolted at both legs

7. 7 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example 1.
Determine the factor tensile resistance of the given double unequal
angles, if are bolted at the long leg only.
Fig 4 Bolted tension member for example
given :
Fy = 250 MPa
Fu = 400 MPa
Le1 =Le2 = 51 mm
s = 76 mm
g = 51 mm
dbolt = 19 mm (for standard hole)
dhole = 19 + 3 = 22 mm
Solution :
1- Yielding of Ag.
Ag = 2*1020 = 2040 mm2
Ø Rn = 0.9 * Fy * Ag = 0.9 * 250 * 2040 * 10-3
= 459 kN
2- Fracture on Ae.
Ae = An * U
An = Ag – 2*dh*t = 2040 – 2* (22*6.4) = 1758.4 mm2
U = ( 1 – x/Lc) < 0.9
x the largest of
i) x'
ii) y" = g – y'
Fig 5 Dimension of cross section
g = 51 mm
ye = 38 mm
X' = 19.8 mm
y'
y"
g
b
t
h
tg
section (1-1)
hg
Pu
1
Pu
SS Le2Le1 1
Le1 S
Le2S
2L 89 x 76 x 6.4 mm
for single angle
Ag = 1020 mm2
x' = 19.8 mm
y' = 26.2 mm

8. 8 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
y'' = {1020 * 26.2 – 38 * [89 * (38/2)] * 6.4}/{1020 – 38*6.4} = 12.49 mm
x = 20 mm or x = 51 – 12.49 = 38.51 mm
x∴ = 38.51 mm
Lc = 2s = 2 * 76 = 152 mm
U = ( 1 – 38.51/152) = 0.747 < 0.90
Ae = An * U = 1758.4 * 0.747 = 1313.5 mm2
Ø Rn = 0.75 * Ae * Fu = 0.75 * 1313.5 * 400 * 10-3
= 393.9 kN
3- Block shear rupture.
Fig 6 Block shear failure of single angle
Lv = 2s + Le = 2 * 76 + 51 = 203 mm
Agv = Lv* t = 203 * 2 * 6.4 = 2598.4 mm2
Anv = Agv – 2.5 * dhole* 2 * t = 2598.4 – 2.5 * 22 * 2 * 6.4 = 1894.4 mm2
Lt = leg – g = 89 – 51 = 38 mm
Agt = 38 * 2 * 6.4 = 486.4 mm2
Ant = 486.4 – 0.5 * 22 * 6.4 * 2 = 345.6 mm2
Fu * Ant = 400 * 345.6 = 138240 N
0.6 * Fu * Anv = 0.6 * 400 * 1894.4 = 454656 N > Fu * Ant
Ø Rn = 0.75 * (0.6 * Fu * Anv + Fy * Agt)
= 0.75 * (0.6 * 400 * 1894.4 + 250 * 486.4) * 10-3
= 432.2 kN
∴the strength of the bolted angles
Ø Rn = 393.9 kN which is governing by fracture
Lt
Lv