Transcript FRP vs. FIRE - An-Najah Staff - An

68402: Structural Design of Buildings II
61420: Design of Steel Structures
62323: Architectural Structures II
Tension Member Design
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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Slide # 1
Table of Contents

Typical Tension Members

Introductory Concepts

Design Strength

Effective and Net Areas

Staggered Bolted Connections

Block Shear

Design of Tension Members

Slenderness Requirements
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Slide # 2
Tension Members

Applications

In bridge, roof and floor trusses, bracing systems,
towers, and tie rods

Consist of angles, channels, tees, plates, W or S
shapes, or combinations
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Slide # 3
Typical Tension Members

Tension chord in a
truss
" Tension" Diagonal
Bottom " Tension" Chord
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Slide # 4
Typical Tension Members

Cables

Ties
" Tension" Tie
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Slide # 5
Tension Members

Commonly Used Sections:
• W/H shapes
• Square and Rectangular or round HSS
• Tees and Double Tees
• Angles and double angles
• Channel sections
• Cables
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Slide # 6
Introductory Concepts

Stress: The stress in the column cross-section can be
calculated as
f 
P
A
f - assumed to be uniform over the entire cross-section.
P - the magnitude of load
A - the cross-sectional area normal to the load

The stress in a tension member is uniform throughout the
cross-section except:
•
•
near the point of application of load
at the cross-section with holes for bolts or other discontinuities, etc.
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Slide # 7
Design Strength
Average Stress distribution
 
P
A
P
A A B  B
A A
Net Area
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BB
Gross Area
Slide # 8
Introductory Concepts

For example, consider an 200 x 10 mm. bar connected to a
gusset plate & loaded in tension as shown below in Fig. 2.1
Gusset plate
b
Section b-b
b
20 mm
hole diameter
7/8
in. diameter
hole
a
a
x 10
8 x200
½ in.
barmm plate
Section a-a
Fig. 2.1 Example of tension member
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Slide # 9
Introductory Concepts

Area of bar at section a – a = 200 x 10 = 2000 mm2

Area of bar at section b – b = (200 – 2 x 20 ) x 10 = 1600
mm2

Therefore, by definition the reduced area of section b – b
will be subjected to higher stresses

However, the reduced area & therefore the higher stresses
will be localized around section b – b.

The unreduced area of the member is called its gross area
= Ag

The reduced area of the member is called its net area = An
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Slide # 10
Steel Stress-strain Behavior
The stress-strain behavior of steel is shown below in
Fig. 2.2
E = 200 GPa
Fy = 248 MPa
Fu
Fy
Stress, f

E
y
Strain, 
u
Fig. 2.2 Stress-strain behavior of steel
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Slide # 11
Steel Stress-strain Behavior

In Fig. 2.2:
•
•
•
•
E - the elastic modulus = 200 GPa.
Fy the yield stress
Fu - the ultimate stress
y is the yield strain
u the ultimate strain
Deformations are caused by the strain . Fig. 2.2 indicates that the
structural deflections will be small as long as the material is elastic
(f < Fy)
•
Deformations due to the strain  will be large after the steel
reaches its yield stress Fy.
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Slide # 12
Design Strength

We usually determine the strength “capacity” of any
structural element based on possible scenarios of failure!

Possible failures of a tension member include
• Yield of the element
• Fracture of element



The stress of axially loaded elements can be determined
as f  P
A
The stress is therefore a function of the cross sectional
area thus the presence of holes will change the stress.
Bolted connections reduce the area of the cross section.
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Slide # 13
Design Strength

A tension member can fail by reaching one of two limit states:
•
•
excessive deformation
fracture

Excessive deformation can occur due to the yielding of the
gross section (for example section a-a from Fig. 2.1) along
the length of the member

Fracture of the net section can occur if the stress at the net
section (for example section b-b in Fig. 2.1) reaches the
ultimate stress Fu.

The objective of design is to prevent these failure before
reaching the ultimate loads on the structure (Obvious).

This is also the load & resistance factor design approach for
designing steel structures
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Slide # 14
Load & Resistance Factor Design

The load & resistance factor design approach is
recommended by AISC for designing steel
structures. It can be understood as follows:
• Step
I. Determine the ultimate loads acting on the
structure
• The values of D, L, W, etc. are nominal loads (not maximum or
ultimate)
• During
its design life, a structure can be subjected to some
maximum or ultimate loads caused by combinations of D, L, or
W loading.
• The
ultimate load on the structure can be calculated using
factored load combinations. The most relevant of these load
combinations are given below:
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Slide # 15
Load & Resistance Factor Design
• 1.4 D
• 1.2 D + 1.6 L + 0.5 (Lr or S)
• 1.2 D + 1.6 (Lr or S) + (0.5 L or 0.8 W)
• 1.2 D + 1.6 W + 0.5 L + 0.5 (Lr or S)
• 0.9 D + 1.6 W
•
Step II. Conduct linear elastic structural analysis
•
Determine the design forces (Pu, Vu, & Mu) for each structural member
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Slide # 16
Load & Resistance Factor Design
• Step III. Design the members
• The failure (design) strength of the designed member must be
greater than the corresponding design forces calculated in Step
II:
 Rn 
 i Qi
Rn - the calculated failure strength of the member
 - the resistance factor used to account for the reliability of the
material behavior & equations for Rn
Qi - the nominal load
i - the load factor used to account for the variability in loading & to
estimate the ultimate loading
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Slide # 17
Design Strength of Tension Members

Yielding of the gross section will occur when the stress f
reaches Fy.
P
f 
 Fy
Ag

Therefore, nominal yield strength = Pn = Ag Fy
Factored yield strength = t Pn
t = 0.9 for tension yielding limit state

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Slide # 18
Design Strength of Tension Members

Facture of the net section will occur after the stress on the
net section area reaches the ultimate stress Fu
P
f 
 Fu
Ae

Therefore, nominal fracture strength = Pn = Ae Fu

Where, Ae is the effective net area, which may be equal to
the net area or smaller.

The topic of Ae will be addressed later.

Factored fracture strength = t Ae Fu
where: t = 0.75 for tension fracture limit state
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Slide # 19
Net Area

We calculate the net area by deducting the width of the
“bolts + some tolerance around the bolt”

Use a tolerance of 1.6 mm above the diameter hole
which is typically 1.6 mm larger than the bolt diameter

Rule
b
t
dbolt
d Hole
An 
 dbolt  3.2 mm
b  (nholes d hole)
t
dhole
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Slide # 20
Design Strength


Tensile strength of a section is governed by two limit states:
•
•
Yield of gross area (excessive deformation)
Fracture of net area
Thus the design strength is one of the following
Load Effect
Pu 

t Pn  t Fy Ag
t Pn  t Fu An
t 
0.9
YIELD
t  0.75 FRACTURE
The difference in the  factor for the two limit states represent the
•
•
Seriousness of the fracture limit state
The reliability index (probability of failure) assumed with each limit state
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Slide # 21
Important Notes

Why is fracture (& not yielding) the relevant limit state at
the net section?
Yielding will occur first in the net section. However, the
deformations induced by yielding will be localized around the net
section. These localized deformations will not cause excessive
deformations in the complete tension member. Hence, yielding at
the net section will not be a failure limit state.

Why is the resistance factor (t) smaller for fracture than
for yielding?
The smaller resistance factor for fracture (t = 0.75 as compared to
t = 0.90 for yielding) reflects the more serious nature &
consequences of reaching the fracture limit state.
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Slide # 22
Important Notes

What is the design strength of the tension member?
The design strength of the tension member will be the lesser value of
the strength for the two limit states (gross section yielding & net
section fracture).
yp

Where are the Fy & Fu values for different steel materials?
The yield & ultimate stress values for different steel materials are
dependent on type of steel.
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Slide # 23
Ex. 2.1 – Tensile Strength

A 125 x 10 mm bar of A572 (Fy = 344 MPa) steel is used as a tension
member. It is connected to a gusset plate with six 20 mm. diameter
bolts as shown below. Assume that the effective net area Ae equals
the actual net area An & compute the tensile design strength of the
member.
Gusset plate
b
b
20 in.
mmdiameter
hole diameter
7/8
bolt
a
a
5 200
x ½ in.
barmm plate
x 10
A572(Fy
Gr. =50344 MPa)
A572
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Slide # 24
Ex. 2.1 – Tensile Strength

Gross section area = Ag = 125 x 10 = 1250 mm2

Net section area (An)

•
•
•
•
Bolt diameter = db = 20 mm.
Nominal hole diameter = dh = 20 + 1.6 = 21.6 mm
Hole diameter for calculating net area = 21.6 + 1.6 = 23.2 mm
Net section area = An = (125 – 2 x (23.2)) x 10 = 786 mm2
Gross yielding design strength = t Pn = t Fy Ag
•
Gross yielding design strength = 0.9 x 344 x 1250/1000 = 387 kN
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Slide # 25
Ex. 2.1 – Tensile Strength


Fracture design strength = t Pn = t Fu Ae
•
•
Assume Ae = An (only for this problem)
Fracture design strength = 0.75 x 448 x 786/1000 = 264 kN
Design strength of the member in tension = smaller of
264 kN & 387 kN
•
Therefore, design strength = 264 kN (net section fracture
controls).
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Slide # 26
Effective Net Area

The connection has a significant influence on the
performance of a tension member. A connection almost
always weakens the member & a measure of its influence
is called joint efficiency.

Joint efficiency is a function of:

•
•
•
•
•
material ductility
fastener spacing
stress concentration at holes
fabrication procedure
shear lag.
All factors contribute to reducing the effectiveness but
shear lag is the most important.
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Slide # 27
Effective Net Area


Shear lag occurs when the tension force is not transferred
simultaneously to all elements of the cross-section. This
will occur when some elements of the cross-section are
not connected.
For example, see the figure below, where only one leg of
an angle is bolted to the gusset plate.
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Slide # 28
Effective Net Area

A consequence of this partial connection is that the
connected element becomes overloaded & the
unconnected part is not fully stressed.

Lengthening the connection region will reduce this effect

Research indicates that shear lag can be accounted for by
using a reduced or effective net area Ae

Shear lag affects both bolted & welded connections.
Therefore, the effective net area concept applied to both
types of connections.
•
•
For bolted connection, the effective net area is Ae = U An
For welded connection, the effective net area is Ae = U Ag
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Slide # 29
Effective Net Area



The way the tension member is connected affects its efficiency
because of the “Shear Lag” phenomenon
Shear lag occurs when the force is transmitted to the section
through part of the section (not the whole section)
To account for this stress concentration in stress, an area
reduction factor “U” is used
x
Bolted Connections
Ae  U An
Welded Connections
Ae  U Ag
Over stressed
Under stressed
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Slide # 30
Effective Net Area

Where, the reduction factor U is given by:
x
U = 1≤ 0.9
L
(4.7)
x- the distance from the centroid of the connected area to the plane of
the connection
L - the length of the connection.
•
If the member has two symmetrically located planes of connection,
is measured from the centroid of the nearest one – half of the area.
x
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Slide # 31
Effective Net Area
L
L
Bolted Connections


Welded Connections
Increasing the connection length reduces the shear lag effect
Some special cases govern bolted and welded connections
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Slide # 32
Effective Net Area

The distance L is defined as the length of the connection in
the direction of load.
•
For bolted connections, L is measured from the center of the bolt at
one end to the center of the bolt at the other end.
•
For welded connections, it is measured from one end of the
connection to other.
•
If there are weld segments of different length in the direction of
load, L is the length of the longest segment.
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Slide # 33
U for Bolted Connections
U  1

x
 0.9
L
OR
Two major groups of bolted connections
•
Connections with at least three bolts per line
•
•
•
W,M and S shapes and T cut from them connected in flange with
U  0.9
All other shapes
U  0.85
Connections with only two bolts per line
U  0.75
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Slide # 34
bf
2

d
3
Ex. 2.2 – Design Strength

Determine the effective net area & the corresponding
design strength for the single angle tension member in the
figure below. The tension member is an L 4 x 4 x 3/8 made
from A36 steel. It is connected to a gusset plate with 15
mm diameter bolts, as shown in Figure below. The spacing
between the bolts is 75 mm center-to-center.
a
x
L 4L x44xx43/x83/8
d bd= 5/8
= 15in.mm
b
Section a-a
L 4 x 4 x 3/ 8
a
Gusset plate
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Slide # 35
Ex. 2.2 – Design Strength
•
•
Gross area of angle = Ag = 1850 mm2
T = 9.5 mm
Net section area = An
• Bolt diameter = 15 mm.
• Hole diameter for calculating net area = 15 +3.2 = 18.2 mm.
• Net section area = Ag – 18.2 x 9.5 = 1850 –172.9 = 1677.1 mm2
• x
is the distance from the centroid of the area connected to
the plane of connection
• For this case is equal to the distance of centroid of the angle
from the edge.
• This value is given in the section property table.
•x
= 28.7 mm.
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Slide # 36
Ex. 2.2 – Design Strength
•
•
L is the length of the connection, which for this case will be
equal to 2 x 75 = 150 mm.
x
28.7
U  1  1
 0.809
L
150
•
Effective net area = Ae = 0.809 x 1677.1 in2 = 1357 mm2
•
Gross yielding design strength = t Ag Fy = 0.9 x 1850 x
248/1000 = 412.9 kN
•
Net section fracture = t Ae Fu = 0.75 x 1357 x 400/1000 = 407.1
kN
•
Design strength = 407.1 kN
•
(Lower of the two values)
(net section fracture governs)
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Slide # 37
Ex. 2.3 – Design Strength

Determine the design strength of an ASTM A992 W8 x 24 with
four lines if 20 mm diameter bolts in standard holes, two per
flange, as shown in the Figure below. Assume the holes are
located at the member end & the connection length is 225 mm.
Also calculate at what length this tension member would cease
to satisfy the slenderness limitation in LRFD specification.
dbdiameter
= 20 mm
¾ in.
bolts
W 8 x 24
753mm
in. 753 mm
in. 753 mm
in.
Holes in beam flange
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Slide # 38
Ex. 2.3 – Design Strength
•
•
For ASTM A992 material: Fy = 344 MPa; & Fu = 448 MPa
For the W8 x 24 section:
• Ag = 4570 mm2
d = 201 mm.
• tw = 6.2 mm.
bf = 165 mm.
• tf = 10.2 mm.
ry = 40.9 mm.
•
Gross yielding design strength = t Pn = t Ag Fy = 0.90 x 4570 x
344/1000 = 1414.9 kN
•
Net section fracture strength = t Pn = t Ae Fu = 0.75 x Ae x 448
•
•
•
•
Ae = U An
- for bolted connection
An = Ag – (no. of holes) x (diameter of hole) x (thickness of flange)
An = 4570 – 4 x (20+3.2) x 10.2.
An = 3623 mm2
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Slide # 39
Ex. 2.3 – Design Strength
•
•
•
•
•
U  1
x
 0.9
L
What is x for this situation?
x is the distance from the edge of the flange to the centroid of
the half (T) section
x
(b f  t f ) 
tf
2
(
d  2t f
bf  t f 
 tw )  (
2
d  2t f
2
d  2 tf
4
)
 tw
 201  2 10.2 
 201  2 10.2 
165 10.2  5.1  

6.2




2
4



  17.6 mm.
x
 201  2 10.2 
165 10.2  
  6.2
2


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Slide # 40
Special Cases for Welded Connections

If some elements of the cross-section are not connected,
then Ae will be less than An

For a rectangular bar or plate Ae will be equal to An

However, if the connection is by longitudinal welds at the
ends as shown in the figure below, then Ae = UAg

Where,
U = 1.0
for L ≥ 2w
U = 0.87
for 1.5 w ≤ L < 2 w
U = 0.75
for w ≤ L < 1.5 w

L = length of the pair of welds ≥ w

w = distance between the welds or width of plate/bar
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Slide # 41
Ex. 2.3 – Design Strength
• The calculated value is not accurate due to the deviations
in the geometry
•
U  1
x
17.6
 1
 0.922
L
225
• But, U ≤ 0.90. Therefore, assume U = 0.90
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Slide # 42
Ex. 2.3 – Design Strength
•
•
•
Net section fracture strength = t Ae Fu = 0.75 x 0.9 x 3623 x
448/1000 = 1095.6 kN
The design strength of the member is controlled by net section
fracture = 1095.6 kN
According to LRFD specification, the maximum unsupported
length of the member is limited to 300 ry = 300 x 40.9 = 12270
mm = 12.27 m.
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Slide # 43
Special Cases for Welded Connections
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Slide # 44
Special Cases for Welded Connections

For any member connected by transverse welds alone,
Ae = area of the connected element of the cross-section
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Slide # 45
U for Welded Connections
x
 0.9
L
U  1

OR
Two major groups of welded connections
•
General case
•
•
•
W,M and S shapes and T cut from them connected in flange with
U  0.9
All other shapes
U  0.85
Special case for plates welded at their ends
L  2W  U  1.0
W
1.5W  L  2W U  0.87
•
W  L  1.5W  U  0.75
Any member with transverse welds all around ONLY
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L
U  1.0
Slide # 46
bf
d

2
3
Ex. 2.4 – Tension Design Strength

Consider the welded single angle L 6x 6 x ½ tension
member made from A36 steel shown below. Calculate
the tension design strength.
42.4 mm
140 mm
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Slide # 47
Ex. 2.4 – Tension Design Strength
•
•
•
Ag = 3720 mm2
An = 3720 mm2
- because it is a welded connection
Ae = U An
• x = 42.4 mm for this welded connection
• L = 152 mm for this welded connection
x
42.4
U

1


1

 0.72
•
L
152
•
Gross yielding design strength = t Fy Ag = 0.9 x 248 x
3720/1000 = 830 kN
•
Net section fracture strength = t Fu Ae = 0.75 x 400 x 0.72 x
3720/1000 = 803 kN
•
Design strength = 803 kN (net section fracture governs)
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Slide # 48
Design of Tension Members

The design of a tension member involves finding the
lightest steel section (angle, wide-flange, or channel
section) with design strength (Pn) greater than or equal
to the maximum factored design tension load (Pu) acting
on it.
•
•
 Pn ≥ Pu
•
 Pn is the design strength based on the gross section yielding,
net section fracture & block shear rupture limit states.
Pu is determined by structural analysis for factored load
combinations
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Slide # 49
Design of Tension Members

For net section fracture limit state, Pn = 0.75 x Ae x Fu
•
•
•
•

Therefore, 0.75 x Ae x Fu ≥ Pu
Pu
Therefore, Ae ≥
0.75 Fu
But, Ae = U An
U & An - depend on the end connection.
Thus, designing the tension member goes hand-in-hand
with designing the end connection, which we have not
covered so far.
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Slide # 50
Design of Tension Members

Therefore, for this chapter of the course, the end
connection details will be given in the examples &
problems.

The AISC manual tabulates the tension design strength of
standard steel sections
•
•
•
Include: wide flange shapes, angles, tee sections, & double angle
sections.
The gross yielding design strength & the net section fracture
strength of each section is tabulated.
This provides a great starting point for selecting a section.
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Slide # 51
Design of Tension Members

There is one serious limitation
•
•
•
The net section fracture strength is tabulated for an assumed value
of U = 0.75, obviously because the precise connection details are
not known
For all W, Tee, angle & double-angle sections, Ae is assumed to be
= 0.75 Ag
The engineer can first select the tension member based on the
tabulated gross yielding & net section fracture strengths, & then
check the net section fracture strength & the block shear strength
using the actual connection details.
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Slide # 52
Design of Tension Members

Additionally for each shape, the code tells the
value of Ae below which net section fracture will
control:
• Thus, for Grade 50 steel sections, net section fracture
will control if Ae < 0.923 Ag
• For
Grade 36 steel sections, net section fracture will
control if Ae < 0.745 Ag

Slenderness limits
• Tension
member slenderness l/r must preferably be
limited to 300 as per LRFD specifications.
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Slide # 53
Slenderness Requirements

Although tension elements are not likely to buckle, it is
recommended to limit their slenderness ratio to 300
L
max 
 300
rmin
rmin 
I min
A

The slenderness limitation of tension members is not for
structural integrity as for compression members.

The reason for the code limitation is to assure that the member
has enough stiffness to prevent lateral movement or vibration.

This limitation does not apply to tension rods and cables.
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Slide # 54
Steps for Design of Tension Members

Steps for design
•
•
•
•
•
Calculate the load
Decide whether your connection will be welded or bolted
Assume U factor of 0.75
Determine the gross area of the element
Assume An = 0.75 Ag
 Pu
Ag  
 0.9 Fy
 Pu
Ag  
for bolted
 0.45 Fu
Pu
for welded
0.45 Fu
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Slide # 55
Steps for Design of Tension Members

Steps for design
•
•
Choose the lightest section with area little larger than Ag
Calculate, Ag, An, U and Ae for the chosen section
0.9 Ag Fy

0.75 Ae FU
•
Check
•
Check slenderness ratio
Pu 
max 
L
 300
rmin
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Slide # 56
Ex. 2.7 – Design of Tension Members


Design a member to carry a factored maximum tension load of 350 kN.
Assume that the member is a wide flange connected through the
flanges using eight 20 mm diameter bolts in two rows of four each as
shown in the figure below. The center-to-center distance of the bolts in
the direction of loading is 100 mm. The edge distances are 40 & 50
mm as shown in the figure below. Steel material is A992
20¾
mm
in. d iameter bolts
50 mm
2 in.
100
mm
4 in.
in.
50 2
mm
4 in.
100
mm
401.5
mm
in.
W
40
1.5mm
in.
Holes in beam flange
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Fy = 344 MPa
Fu = 448 MPa
Slide # 57
Ex. 2.7 – Design of Tension Members
•
Select a section from the Tables
• Ag ≥ 350*1000/(0.9*344) = 1130 mm2.
• Assume U=0.75
• Ag  350*1000/(0.45*448)=1736 mm2
• Try W8x10 with Ag = 1910 mm2.
• An = 1910 – 4*(23.2*5.2) = 1427 mm2.
• x = 24.6 mm (students have to compute it)
• U = 1 – 24.6/100 = 0.754
• Gross yielding strength = 591 kN, & net section fracture
strength = 362 kN
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Slide # 58
Extra Slides
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Slide # 59
Block Shear

For some connection configurations, the tension member
can fail due to ‘tear-out’ of material at the connected end.
This is called block shear.

For example, the single angle tension member connected
as shown in the Fig. 2.3 below is susceptible to the
phenomenon of block shear.

For the case shown above, shear failure will occur along
the longitudinal section a-b & tension failure will occur
along the transverse section b-c.

AISC Specification on tension members does not cover
block shear failure explicitly. But, it directs the engineer to
the Specification on connections
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Slide # 60
Block Shear
(a)
T
(b)
T
Shear failure
(c)
Tension failure
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Fig. 2.3 Block
shear failure of
single angle
tension member
Slide # 61
Block Shear

Block shear strength is determined as the sum of the
shear strength on a failure path & the tensile strength on a
perpendicular segment.
•
•
•

Block shear strength = net section fracture strength on shear path
+ net section fracture strength of the tension path
OR
Block shear strength = gross yielding strength of the shear path +
net section fracture strength of the tension path
Which of the two calculations above governs?
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Slide # 62
Block Shear
 Rn =  (0.6 Fu Anv + UbsFu Ant) ≤  (0.6 FyAgv + UbsFu Ant)
 = 0.75
Ubs = 1.0 for uniform tensile stress; = 0.5 for nonuniform tensile stress
Agv - gross area subject to shear
Agt - gross area subject to tension
Anv - net area subject to shear
Ant - net area subject to tension
Fu - ultiamte strength of steel
Fy - yield strength of steel
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Slide # 63
Block Shear

Failure happens by a
combination of shear
and tension.
Area failing by shear
Shear
Fracture
Shear Yield
Tension
Fracture
Tension
Fracture
Failure Mode 2
Failure Mode 1

Area failing by tension
The two possible failure modes shall be investigated
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Slide # 64
Ex. 2.5 – Block Shear

Calculate the block shear strength of the single angle tension
member shown bellow. The single angle L 4 x 4 x 3/8 made
from A36 steel is connected to the gusset plate with 15 mm
diameter bolts as shown below. The bolt spacing is 75 mm
center-to-center & the edge distances are 40 mm & 50 mm as
shown in the Figure below.
a
x
L 4 x 4 x 3/ 8
dd
b ==5/8
15in.
mm
b
Section a-a
L 4 x 4 x 3/ 8
a
Gusset plate
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Slide # 65
Ex. 2.5 – Block Shear
•
Assume a block shear
apath & calculate the required areas
L4
.0
250
d b d=
in.
15 mm
b =5/8
3 .0
.5 75
140
.0
375
a
Gusset plate
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Slide # 66
Ex. 2.5 – Block Shear
•
•
•
•
•
•
Agt = gross tension area = 50 x 9.5 = 475 mm2
Ant = net tension area = 475 - 0.5 x (15 + 3.2) x 9.5 = 388.5
mm2
Agv = gross shear area = (75 + 75 +40) x 9.5 = 1805 mm2
Anv = net shear area = 1805 - 2.5 x (15 + 3.2) x 9.5 = 1372.8
mm2
Ubs = 1.0
Calculate block shear strength
• t Rn = 0.75 (0.6 Fu Anv + UbsFu Ant)
• t Rn = 0.75 (0.6 x 400 x 1372.8 + 1.0 x 400 x 388.5)/1000 =
363.7 kN
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Slide # 67
Ex. 2.5 – Block Shear
•
Check upper limit
• t Rn ≤  (0.6 FyAgv + UbsFu Ant)
• t Rn ≤ 0.75 (0.6 x 248 x 1805 + 1.0 x 400 x 388.5)/1000
• t Rn ≤ 318 kN
• Block shear strength = 318 kN
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Slide # 68
Ex. 2.6 – Design Tensile Strength

Determine the design tension strength for a single
channel C15 x 50 connected to a 15 mm thick gusset
plate as shown in Figure. Assume that the holes are for
20 mm diameter bolts. Also, assume structural steel with
yield stress (Fy) equal to 344 MPa & ultimate stress (Fu)
equal to 448 MPa.
gusset plate
3 @ 75 mm = 225
mm center-tocenter
T
T
C15 x 50
40
75
75
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Slide # 69
Ex. 2.6 – Design Tensile Strength
•
Limit state of yielding due to tension:
Tn  0.9*344*9480 /1000  2935 kN
•
Limit state of fracture due to tension:
An  Ag  ndet  9480  4 18.2 23.2  7791 mm2
•
•
 x
 20.3 
2
Ae  UAn  1   An  1 
*7791

6736.6
mm

L
150




Check: U  0.867  0.9 OK.
Note: The connection eccentricity, x, for a C15X50 can be found
in section property tables.
Tn  0.75*448*6736.6 /1000  2263.5 kN
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Slide # 70
Ex. 2.6 – Design Tensile Strength
• Limit state of block shear rupture:
• Agt = gross tension area = 225 x 18.2 = 4095 mm2
• Ant = net tension area = (225 - 3*(23.2))*18.2 = 2828 mm2
• Agv = gross shear area = 2*(190*18.2) = 6916 mm2
• Anv = net shear area = 2*((190 - 2.5*23.2) *18.2) = 4805 mm2
• Ubs = 1.0
• Calculate block shear strength
•
•
t Rn = 0.75 (0.6 Fu Anv + UbsFu Ant)
t Rn = 0.75 (0.6 x 448 x 4805 + 1.0 x 448 x 2828)/1000 = 1919
kN
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Slide # 71
Ex. 2.6 – Design Tensile Strength
• Check upper limit
•
•
•
•
•
t Rn ≤  (0.6 FyAgv + UbsFu Ant)
t Rn ≤ 0.75 (0.6 x 344 x 6916 + 1.0 x 448 x 2828)
t Rn ≤ 2021 kN
Block shear strength = 1919 kN
Block shear rupture is the critical limit state & the design
tension strength is 1919 kN.
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Slide # 72
Staggered Bolts
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Slide # 73
Staggered Bolts

For a bolted tension member, the connecting bolts can be
staggered for several reasons:
•
•
•
To get more capacity by increasing the effective net area
To achieve a smaller connection length
To fit the geometry of the tension connection itself.

For a tension member with staggered bolt holes (see example
figure above), the relationship f = P/A does not apply & the
stresses are a combination of tensile & shearing stresses on
the inclined portion b-c.

Net section fracture can occur along any zig-zag or straight
line. For ex., fracture can occur along the inclined path a-b-c-d
in the figure above. However, all possibilities must be
examined.
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Slide # 74
Staggered Bolts

Empirical methods have been developed to calculate the net
section fracture strength.
•
s2
net width = gross width -  d  
4g
d - the diameter of hole to be deducted (db + 3.2 mm)
s2/4g - added for each gage space in the chain being considered
s - the longitudinal spacing (pitch) of the bolt holes in the direction of loading
g - the transverse spacing (gage) of the bolt holes perpendicular to loading
direction.
net area (An) = net width x plate thickness
x
effective net area (Ae) = U An
where U  1 
net fracture design strength = t Ae Fu
(t = 0.75)
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L
Slide # 75
Staggered Bolted Connections
BB
A A
g
P
S

Stresses on inclined planes are a mix of tension and shear
and thus a correction is needed.
S2
Wn  Wg   d  
4g

An  Wn t
All possible failure paths passes shall be examined. The path
that yields the smallest area governs.
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Slide # 76