lecture 06 tension members

N.W.F.P. University of Engineering and Technology PeshawarTechnology Peshawar

Lecture 06: Tension Members

By: Prof Dr. Akhtar Naeem Khan

1

[email protected]

Topics to be AddressedTypes of Steel Structures

Introductory conceptsIntroductory concepts

Design Strength

Net Area at Connection

Shear Lag Phenomenon

ASD and LRFD Design of TensionASD and LRFD Design of Tension Members

CE-409: Lecture 06 Prof. Dr Akhtar Naeem Khan 2

Design Examples

Types of steel structures

The form of a tension member is governed to a large extent by

Type of structure of which it is a part

Method of joining it to connecting portionsMethod of joining it to connecting portions.


Sections for Tension Members


Design Stresses Design Stresses for

Base Material


Introductory Concepts

Stress: The stress in an axially loaded tension member is given by Equationmember is given by Equation

The stress in a tension member is uniform throughout the cross-section except:

near the point of application of load, and

at the cross-section with holes for bolts or other discontinuities, etc.



Gusset plateGusset plateGusset plate

b b7/8 in. diameter hole

Section b-bb b7/8 in. diameter hole


Section b-bSection b-b

aa

8 x ½ in. barSection a-a

aa

8 x ½ in. bar

aa

8 x ½ in. barSection a-aSection a-a



Gusset plateGusset plateGusset plate


Section b-bb b7/8 in. diameter hole


Section b-bSection b-b

aa

Section a-a

aa aa

Section a-aSection a-a

Area of bar at section a – a = 8 x ½ = 4 in2

A f b t ti b b (8 2 7/8 ) ½ 3 12 i 2

8 x ½ in. barSection a a

8 x ½ in. bar8 x ½ in. barSection a aSection a a

Area of bar at section b – b = (8 – 2 x 7/8 ) x ½ = 3.12 in2

The unreduced area of the member is called its gross area = Ag


The reduced area of the member is called its net area = An

Design strength

• A tension member can fail by reaching one of two limit states:of two limit states:

1. Excessive deformation• Yielding at the gross area

2 Fracture2. Fracture • Fracture at the net area


Design strength1. Excessive deformation can occur due to the

yielding of the gross section at section a-a

b bb bb b7/8 in. d7/8 in. d7/8 in. d

aa aa aa


8 x ½ i8 x ½ i8 x ½ i

Design strength2. Fracture of the net section can occur if the stress

at the net section (section b-b) reaches the ultimate stress Fuultimate stress Fu

b bb bb bb b7/8 in. d

b b7/8 in. d

b b7/8 in. d

aa aa aa


8 x ½ i8 x ½ i8 x ½ i

Design strengthYielding of the gross section will occur when the stress f reaches Fy

yg

FAP

==f

Nominal yield strength = Pn = Ag Fy

g

• Fracture of the net section will occur after the stress on the net section area reaches the ultimate stress Fu

ue

FAP

==f


Nominal fracture strength = Pn = Ae Fu

Design strength• AISC/ASD

Ft = 0 6 Fy on Gross AreaFt = 0.6 Fy on Gross AreaFt = 0.5 Fu on Effective Area

• AISC/LRFDDesign strength for yielding on gross areaDesign strength for yielding on gross area

øtPn =øt Fy Ag = 0.9 Fy AgD i t th f f t f t tiDesign strength for fracture of net section

øtPn = øtFu Ae = 0.75 Fu Ae


Effective Net Area• The connection has a significant influence on the performance of a tension member.

A i l l k h b• A connection almost always weakens the member and a measure of its influence is called joint efficiencyefficiency.


Effective Net Area

•Joint efficiency is a function of:y

(a) Material ductility

(b) F t i(b) Fastener spacing

(c) Stress concentration at holes

(d) Fabrication procedure

( ) Sh l(e) Shear lag.


Effective Net AreaResearch indicates that shear lag can be accounted for by using a reduced or effective net area Ae

CGFor Bolted Connections

1x

LxU −= 1

2x

• 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• For W, M, and S shapes with width-to-depth ratio of at least

2/3 and for Tee shapes cut from them, if the connection is p ,through the flanges with at least three fasteners per line in the direction of applied load ,

U= 0.9

• For all other shapes with at least three fasteners per line , U= 0.85

• For all members with only two fasteners per line U= 0 75


U= 0.75

Net Area ExampleE l A 5 ½ b f A572 G 50 l i d iExample : A 5 x ½ bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with six 7/8 in. diameter bolts as shown in below. Assume that the effective net area Ae equals e qthe actual net area An and compute the tensile design strength of the member.

b

Gusset plate

b

Gusset plate

b

Gusset plate

b b7/8 in. diameter bolt



aa

5 x ½ in. bar

aa

5 x ½ in. bar

aa

5 x ½ in. bar


5 x ½ in. barA572 Gr. 505 x ½ in. bar5 x ½ in. bar

A572 Gr. 50

Net Area Example

Gross section area (Ag):

Ag = 5 x ½ = 2.5 in2

Net section area (A ):Net section area (An):Bolt diameter = db = 7/8 in.

Nominal hole diameter = d = 7/8 + 1/16 in = 15/16 inNominal hole diameter = dh = 7/8 + 1/16 in. = 15/16 in.

Hole diameter for calculating net area = 15/16 + 1/16 in. = 1 in.

N i A (5 2 (1)) ½ 1 5 i 2Net section area = An = (5 – 2 x (1)) x ½ = 1.5 in2


Net Area Example

Gross yielding design strength:Gross yielding design strength:ft Pn = ft Fy Ag

= 0.9 x 50 ksi x 2.5 in2 = 112.5 kipsFracture design strength:

ft Pn = ft Fu Ae= 0.75 x 65 ksi x 1.5 in2 = 73.125 kips 0.75 x 65 ksi x 1.5 in 73.125 kips

Assume Ae = An (only for this problem)

Therefore design strength = 73 125 kips (net section fractureTherefore, design strength = 73.125 kips (net section fracture controls).


Shear Lag in Tension M bMembers

• Shear lag in tension members arises when all the elements of a cross section do not participate in the load transfer at a connection. Th t i h th t i i•There are two primary phenomena that arise in

these cases:(i) N if t i i f th b lti i(i) Non-uniform straining of the web resulting in

biaxial stress states

(ii) Effective area reduction.



Effective area reduction ect ve a ea educt o



D i B tt LiDesign Bottom LineShear lag can have a large influence on the t th f t i b istrength of tension members , in essence

reducing the effective area of the section. The amount of the reduction is related to the length of the connection and the arrangement of cross-section elements that do not participate directly in the connection load transfer.


Block Shear in Tension M bMembers

Bl k h i bi d t il / h t iBlock shear is a combined tensile/shear tearing out of an entire section of a connection.



For such a failure to occur, there are two possible mechanisms: p

(1) Shear rupture + tensile yielding; and

(2) Shear yielding + tensile rupturing.



Design Bottom Line

As a likely limit state for connectionsAs a likely limit state for connections, block shear must be considered in design This can be accomplished bydesign. This can be accomplished by considering the strength limit states of the two failure mechanisms outlinedthe two failure mechanisms outlined above.


Design Example 1-ASD


Design Example 1-LRFD



Design Alternative 2


lecture 06 tension members

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