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68402 Slide # 1

Monther Dwaikat Assistant Professor

Department of Building Engineering An-Najah National University

Tension Member Design

68402: Structural Design of Buildings II

61420: Design of Steel Structures

62323: Architectural Structures II

68402 Slide # 2

� Typical Tension Members

� Introductory Concepts

� Design Strength

� Effective and Net Areas

� Staggered Bolted Connections

� Block Shear

� Design of Tension Members

� Slenderness Requirements

Table of Contents

68402 Slide # 3

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

68402 Slide # 4

Typical Tension Members

� Tension chord in a

truss

DiagonalTension""

ChordTensionBottom ""

68402 Slide # 5

Typical Tension Members

� Cables

� Ties TieTension""

68402 Slide # 6

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

68402 Slide # 7

Introductory Concepts

� Stress: The stress in the column cross-section can be

calculated as

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.

A

P=f

68402 Slide # 8

Design Strength

A

P=σ

AA−

Average Stress distribution

P

BB − AA− BB −

Net Area Gross Area

68402 Slide # 9


� For example, consider an 200 x 10 mm. bar connected to a

gusset plate & loaded in tension as shown below in Fig. 2.1

b b

aa

8 x ½ in. bar

Gusset plate

7/8 in. diameter hole

Section a-a

Section b-bb b

aa

8 x ½ in. bar

Gusset plate


b b

aa

8 x ½ in. bar

Gusset plate


Section a-a

Section b-b

Section a-a

Section b-b

Fig. 2.1 Example of tension member

20 mm hole diameter

200 x 10 mm plate

68402 Slide # 10


� 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

68402 Slide # 11

Steel Stress-strain Behavior

� The stress-strain behavior of steel is shown below in

Fig. 2.2

Strain, εεy εuεy εu

Stress, f

Fy

Fu

E

Strain, εεy εuεy εu

Stress, f

Fy

Fu

εy εuεy εu

Stress, f

Fy

Fu

E

Fig. 2.2 Stress-strain behavior of steel

E = 200 GPa

Fy = 248 MPa

68402 Slide # 12

Steel Stress-strain Behavior

� In Fig. 2.2:

• E - the elastic modulus = 200 GPa.

• Fy the yield stress Fu - the ultimate stress

• εy is the yield strain εu the ultimate strain

• Deformations are caused by the strain ε. 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 ε will be large after the steel

reaches its yield stress Fy.

68402 Slide # 13

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

� 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.

A

Pf =

68402 Slide # 14

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

68402 Slide # 15

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:

68402 Slide # 16


• 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

68402 Slide # 17


• 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 - the calculated failure strength of the member

φφφφ - the resistance factor used to account for the reliability of the material behavior & equations for Rn

Qi - the nominal load

γγγγi - the load factor used to account for the variability in loading & to estimate the ultimate loading

∑> iin QR γφ

68402 Slide # 18

Design Strength of Tension Members

� Yielding of the gross section will occur when the stress f

reaches Fy.

� Therefore, nominal yield strength = Pn = Ag Fy

� Factored yield strength = φt Pn

φt = 0.9 for tension yielding limit state

y

g

FA

P==f

68402 Slide # 19

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

� 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 = φt Ae Fu

where: φt = 0.75 for tension fracture limit state

u

e

FA

P==f

68402 Slide # 20

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 3.2Hole boltd d mm= +

[ ] tdnbA holeholesn )(−=

b

t

dhole

dbolt

68402 Slide # 21

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

gytntAFP φφ =

nutnt AFP φφ =

9.0t =φ

75.0=tφ≤uP

Load Effect YIELD

� The difference in the φ 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

FRACTURE

68402 Slide # 22

� 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 (φt) smaller for fracture than

for yielding?

The smaller resistance factor for fracture (φt = 0.75 as compared to

φt = 0.90 for yielding) reflects the more serious nature &

consequences of reaching the fracture limit state.

Important Notes

68402 Slide # 23

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).

� 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.

pypypy

68402 Slide # 24

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.

b b

aa

5 x ½ in. bar

Gusset plate

7/8 in. diameter bolt

A572 Gr. 50

b b

aa

5 x ½ in. bar

Gusset plate


b b

aa

5 x ½ in. bar

Gusset plate


A572 Gr. 50

20 mm hole diameter

200 x 10 mm plate

A572 (Fy = 344 MPa)

68402 Slide # 25

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 = φt Pn = φt Fy Ag

• Gross yielding design strength = 0.9 x 344 x 1250/1000 = 387 kN

68402 Slide # 26

Ex. 2.1 – Tensile Strength

� Fracture design strength = φt Pn = φt 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).

68402 Slide # 27

� 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.

Effective Net Area

68402 Slide # 28

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.

68402 Slide # 29

� 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

Effective Net Area

68402 Slide # 30

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

Over stressed

Under stressed

ge AUA =ne

AUA =Bolted Connections

Welded Connections

x

68402 Slide # 31

� Where, the reduction factor U is given by:

U = 1- ≤ 0.9 (4.7)

- 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.

Effective Net Area

L

x

x

x

68402 Slide # 32

Effective Net Area

� Increasing the connection length reduces the shear lag effect

� Some special cases govern bolted and welded connections

Bolted Connections

LL

Welded Connections

68402 Slide # 33

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.

68402 Slide # 34

� 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

• All other shapes

• Connections with only two bolts per line

U for Bolted Connections

9.0=U

3

2≥

d

bf

9.01 ≤−=L

xU

75.0=U

85.0=U

OR

68402 Slide # 35

Ex. 2.2 – Design Strength

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

� 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.

x

L 4 x 4 x 3/8

x

L 4 x 4 x 3/8

L 4 x 4 x 3/8

db = 15 mm

68402 Slide # 36

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

• 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.

• = 28.7 mm.

x

x

68402 Slide # 37

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.

•

• Effective net area = Ae = 0.809 x 1677.1 in2 = 1357 mm2

• Gross yielding design strength = φφφφ t Ag Fy = 0.9 x 1850 x

248/1000 = 412.9 kN

• Net section fracture = φφφφ t Ae Fu = 0.75 x 1357 x 400/1000 = 407.1

kN

• Design strength = 407.1 kN (net section fracture governs)

• (Lower of the two values)

28.71 1 0.809

150

xU

L= − = − =

68402 Slide # 38

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.

W 8 x 24

¾ in. diameter bolts

3 in. 3 in. 3 in.

Holes in beam flange

W 8 x 24


W 8 x 24


3 in. 3 in. 3 in.


3 in. 3 in. 3 in.3 in. 3 in. 3 in.


db = 20 mm

75 mm 75 mm 75 mm

68402 Slide # 39

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 = φφφφ t Pn = φφφφ t Ag Fy = 0.90 x 4570 x

344/1000 = 1414.9 kN

• Net section fracture strength = φφφφ t Pn = φφφφ 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

68402 Slide # 40


•

• What is for this situation?

• is the distance from the edge of the flange to the centroid of

the half (T) section

•

•

9.01 ≤−=L

xU

x

x

2 2( ) ( ) ( )

2 2 42

2

f f f

f f w

f

f f w

t d t d tb t t

xd t

b t t

− +× × + × ×

=−

× + ×

201 2 10.2 201 2 10.2165 10.2 5.1 6.2

2 417.6 .

201 2 10.2165 10.2 6.2

2

x mm

− × + × × × + × × = =

− × × + ×

68402 Slide # 41

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

68402 Slide # 42


• The calculated value is not accurate due to the deviations

in the geometry

•

• But, U ≤ 0.90. Therefore, assume U = 0.90

17.61 1 0.922

225

xU

L= − = − =

68402 Slide # 43


• Net section fracture strength = φφφφ 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.

68402 Slide # 44


68402 Slide # 45


� For any member connected by transverse welds alone,

Ae = area of the connected element of the cross-section

68402 Slide # 46

� Two major groups of welded connections • General case

• W,M and S shapes and T cut from them connected in flange with

• All other shapes

• Special case for plates welded at their ends

• Any member with transverse welds all around ONLY

U for Welded Connections

9.0=U

3

2≥

d

bf

9.01 ≤−=L

xU

75.05.1

87.025.1

0.12

=≤≤

=≤≤

=≥

UWLW

UWLW

UWL

KKK

KKK

KKKKKK

85.0=U

OR

W

L

0.1=U

68402 Slide # 47

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.

140 mm

42.4 mm

68402 Slide # 48

Ex. 2.4 – Tension Design Strength

• Ag = 3720 mm2

• An = 3720 mm2 - because it is a welded connection

• Ae = U An

• = 42.4 mm for this welded connection

• L = 152 mm for this welded connection

•

• Gross yielding design strength = φφφφ t Fy Ag = 0.9 x 248 x 3720/1000 = 830 kN

• Net section fracture strength = φφφφ t Fu Ae = 0.75 x 400 x 0.72 x 3720/1000 = 803 kN

• Design strength = 803 kN (net section fracture governs)

42.41 1 0.72

152

xU

L= − = − =

x

68402 Slide # 49

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 (φPn) greater than or equal

to the maximum factored design tension load (Pu) acting

on it.

• φ Pn ≥ Pu

• Pu is determined by structural analysis for factored load

combinations

• φ Pn is the design strength based on the gross section yielding,

net section fracture & block shear rupture limit states.

68402 Slide # 50


� For net section fracture limit state, Pn = 0.75 x Ae x Fu

• Therefore, 0.75 x Ae x Fu ≥ Pu

• Therefore, Ae ≥

• 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.

u

u

F75.0

P

×

68402 Slide # 51


� 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.

68402 Slide # 52


� 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.

68402 Slide # 53


� 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.

68402 Slide # 54

Slenderness Requirements

� Although tension elements are not likely to buckle, it is

recommended to limit their slenderness ratio to 300

� 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.

A

Ir

r

L

minmin

min

max 300

=

≤=λ

68402 Slide # 55

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

for welded45.0

boltedfor 45.0

9.0

u

u

u

ug

y

ug

F

P

F

PA

F

PA

≈

≥

68402 Slide # 56

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

• Check

• Check slenderness ratio

≤Ue

yg

uFA

FAP

75.0

9.0

300min

max≤=

r

Lλ

68402 Slide # 57

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

W

¾ in. d iameter bolts

W


W


Holes in beam flangeHoles in beam flangeHoles in beam flange

4 in. 2 in .

1.5 in.

1.5 in.

2 in . 4 in.

W


W


W


W


W


W



4 in. 2 in .

1.5 in.

1.5 in.

2 in . 4 in.


4 in. 2 in .

1.5 in.

1.5 in.

2 in . 4 in.

Fy = 344 MPa

Fu = 448 MPa

100 mm 50 mm

100 mm 50 mm

40 mm

40 mm

20 mm

68402 Slide # 58

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

68402 Slide # 59

Extra Slides

68402 Slide # 60

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

68402 Slide # 61

Block Shear

T

T

Shear failure

Tension failure

(a)

(b)

(c)

T

T

Shear failure

Tension failure

Shear failure

Tension failure

(a)

(b)

(c)

Fig. 2.3 Block

shear failure of

single angle

tension member

68402 Slide # 62

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?

68402 Slide # 63

Block Shear

φ Rn = φ (0.6 Fu Anv + UbsFu Ant) ≤ φ (0.6 FyAgv + UbsFu Ant)

φφφφ = 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

68402 Slide # 64

Block Shear

� The two possible failure modes shall be investigated

Area failing by shear Area failing by tension

Shear

Fracture

Shear Yield Tension

Fracture

Tension

Fracture

Failure Mode 1 Failure Mode 2

� Failure happens by a

combination of shear

and tension.

68402 Slide # 65

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.

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

L 4 x 4 x 3/ 8

db= 5/8 in.

Gusset plate

a

a

x

L 4 x 4 x 3/8

x

L 4 x 4 x 3/8

db = 15 mm

68402 Slide # 66


• Assume a block shear path & calculate the required areas

db = 5/8 in.

Gusset plate

a1.5

3.0 3.0

2.0 db = 5/8 in.

Gusset plate

a1.5

3.0 3.0

2.0 db = 5/8 in.

Gusset plate

a

db = 5/8 in.

Gusset plate

db = 5/8 in.

Gusset plate

a1.5

3.0 3.0

1.5 3.0

3.0

2.0db = 15 mm

50

40 75 75

68402 Slide # 67


• 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

• φt Rn = 0.75 (0.6 Fu Anv + UbsFu Ant)

• φt Rn = 0.75 (0.6 x 400 x 1372.8 + 1.0 x 400 x 388.5)/1000 =

363.7 kN

68402 Slide # 68

Ex. 2.5 – Block Shear • Check upper limit

• φt Rn ≤ φ (0.6 FyAgv + UbsFu Ant)

• φt Rn ≤ 0.75 (0.6 x 248 x 1805 + 1.0 x 400 x 388.5)/1000

• φt Rn ≤ 318 kN

• Block shear strength = 318 kN

68402 Slide # 69

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.

40 75 75

3 @ 75 mm = 225

mm center-to-

center

T T

gusset plate

C15 x 50

68402 Slide # 70


• Limit state of yielding due to tension:

• Limit state of fracture due to tension:

• Check: OK.

• Note: The connection eccentricity, x, for a C15X50 can be found

in section property tables.

0.9*344*9480/1000 2935n

T kNφ = =

( ) ( ) 29480 4 18.2 23.2 7791n g e

A A nd t mm= − = − =

220.31 1 *7791 6736.6

150e n n

xA UA A mm

L

= = − = − =

9.0867.0 ≤=U

0.75*448*6736.6 /1000 2263.5nT kNφ = =

68402 Slide # 71


• 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

• φt Rn = 0.75 (0.6 Fu Anv + UbsFu Ant)

• φt Rn = 0.75 (0.6 x 448 x 4805 + 1.0 x 448 x 2828)/1000 = 1919

kN

68402 Slide # 72


• Check upper limit

• φt Rn ≤ φ (0.6 FyAgv + UbsFu Ant)

• φt Rn ≤ 0.75 (0.6 x 344 x 6916 + 1.0 x 448 x 2828)

• φt Rn ≤ 2021 kN

• Block shear strength = 1919 kN

• Block shear rupture is the critical limit state & the design

tension strength is 1919 kN.

68402 Slide # 73

Staggered Bolts

68402 Slide # 74

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.

68402 Slide # 75

Staggered Bolts � Empirical methods have been developed to calculate the net

section fracture strength.

• net width = gross width -

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

effective net area (Ae) = U An where

net fracture design strength = φφφφ t Ae Fu (φφφφ t = 0.75)

∑∑ +g4

sd

2

L

xU −=1

68402 Slide # 76

Staggered Bolted Connections

P

S

g

� Stresses on inclined planes are a mix of tension and shear and thus a correction is needed.

∑∑ +−=g

SdWW gn

4

2

AA−BB −

� All possible failure paths passes shall be examined. The path that yields the smallest area governs.

tWA nn =

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