Shear Lag in Tension Members

1

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

Lecture 06: Tension Members

1

By: Prof Dr. Akhtar Naeem [email protected]

Types of Steel Structures

Introductory concepts

Topics to be Addressed

Design Strength

Net Area at Connection

Shear Lag Phenomenon

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

ASD and LRFD Design of Tension Members

Design Examples

2

The form of a tension member is

Types of steel structures

governed to a large extent by Type of structure of which it is a part

Method of joining it to connecting portions.




3





4





5



Sections for Tension Members


6

Sections for Tension Members


Design Stresses for

Base Material


7

Introductory Concepts

Stress: The stress in an axially loaded tension member 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


discontinuities, etc.


b b

Gusset plate

7/8 in. diameter hole

Section b-bb b

Gusset plate

7/8 in. diameter holeb b

Gusset plate


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

8


b b

Gusset plateSection b-bb b

Gusset plate

b b

Gusset plateSection b-bSection b-b

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

aa

8 x ½ in. bar


Section a-a

aa

8 x ½ in. bar


aa

8 x ½ in. bar


Section a-aSection a-a


Area of bar at section a – a = 8 x ½ = 4 inArea 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 = AgThe 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:

1. Excessive deformation• Yielding at the gross area

2. Fracture • Fracture at the net area


9

Design strength1. Excessive deformation can occur due to the

yielding of the gross section at section a-a

b b7/8 in. d

b b7/8 in. d

b b7/8 in. d


aa

8 x ½ i

aa

8 x ½ i

aa

8 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 Fu

b b7/8 in. d

b b7/8 in. d

b b7/8 in. d


aa

8 x ½ i

aa

8 x ½ i

aa

8 x ½ i

10

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

P

Nominal yield strength = Pn = Ag Fy

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

yg

FAP

==f


on the net section area reaches the ultimate stress Fu

Nominal fracture strength = Pn = Ae Fu

ue

FAP

==f

Design strength• AISC/ASD

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

• AISC/LRFDDesign strength for yielding on gross area

øtPn =øt Fy Ag = 0.9 Fy Ag


øt n øt y g y gDesign strength for fracture of net section

øtPn = øtFu Ae = 0.75 Fu Ae

11

Effective Net Area• The connection has a significant influence on the performance of a tension member. • A connection almost always weakens the member and a measure of its influence is called joint efficiency.


Effective Net Area

•Joint efficiency is a function of:

( ) M i l d ili(a) Material ductility

(b) Fastener spacing

(c) Stress concentration at holes

(d) Fabrication procedure


(e) Shear lag.

12

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

CG

2x

1x

F b lt d ti h ff i i A U A

LxU −= 1

For Bolted 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• 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 through the flanges with at least three fasteners per line inthrough 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

13

Net Area ExampleExample : 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 the actual net area An and compute the tensile design strength of the n p g gmember.

b b

Gusset plate

7/8 in. diameter boltb b

Gusset plate

7/8 in. diameter boltb b

Gusset plate

7/8 in. diameter bolt


aa

5 x ½ in. barA572 Gr. 50

aa

5 x ½ in. bar

aa

5 x ½ in. barA572 Gr. 50

Net Area Example

Gross section area (Ag):

Ag = 5 x ½ = 2.5 in2g 5 ½ .5

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

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

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


g

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

14

Net Area Example

Gross yielding design strength:f P f F Aft 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

Assume Ae = An (only for this problem)


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

Shear Lag in Tension Members

• Shear lag in tension members arises when all the elements of a cross section do not participate in the p pload transfer at a connection. •There are two primary phenomena that arise in these cases:(i) Non-uniform straining of the web resulting in

bi i l t t t


biaxial stress states

(ii) Effective area reduction.

15




Effective area reduction


16


Design Bottom LineShear lag can have a large influence on the strength 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


the connection and the arrangement of cross-section elements that do not participate directly in the connection load transfer.

Block Shear in Tension Members

Block shear is a combined tensile/shear tearing out of an entire section of a connection.


17


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

(1) Shear rupture + tensile yielding; and

(2) Shear yielding + tensile rupturing.



Design Bottom LineDesign Bottom Line

As a likely limit state for connections, block shear must be considered in design. This can be accomplished by considering the strength limit states of


g gthe two failure mechanisms outlined above.

18

Design Example 1-ASD




19





20





21





22

Design Example 1-LRFD




23





24





25





26





27





28





29





30



Design Alternative 2



31





32





33





34



Shear Lag in Tension Members

Documents

Transcript of Shear Lag in Tension Members