05-Strength of Double Angle Bolted Tension Members (Steel Structural Design & Prof. Shehab Mourad)

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