CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 1

MODULE B: TENSION MEMBERS

This module of CIE 428 covers the following subjects

Introduction

Design strength

Net area

Staggered fasteners

Block shear

Design of tension members

Rods, pin-connected members

READING: Chapter 3 of the textbook by Segui

INTRODUCTION

Tension members are structural elements that are subjected to axial tensile forces. Examples include:

Members in trusses

Cables in cable-stayed and suspension bridges

Bracing in frames to resist lateral forces from blast, wind, and earthquake

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 2

Stresses (f) in axially loaded members are calculated using the following equation

PfA

=

where P is the load and A is the cross-sectional area normal to the load.

Consider the tension member below (F3.2 from the text).

Design of this component involves calculations for

Bar (gross area)

Bar at connection (net area)

Gusset plate at connection (net area)

Gusset plate at support

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 3

What are the gross and net areas for the bar?

Gross area:

Net area:

The design of tension members and their connections is covered by Section B (Design Requirements), D (Tension Members), and J (Connections) of the AISC LRFD Specification, 3rd Edition.

DESIGN STRENGTH

A tension member can fail by

Excessive deformation (yielding under gravity loads)

Fracture

Excessive deformation is prevented (for gravity loading) by limiting stresses on the gross section to less than the yield stress. For yielding on the gross section, the nominal strength is:

n y gP F A=

Fracture is avoided by limiting stresses on the net section to less than the ultimate tensile strength. For fracture on the net section, the nominal strength is:

n u nP F A=

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 4

Resistance (or phi) factors are applied to these nominal strengths

0.9φ = for yielding

0.75φ = for fracture

to calculate design strengths. The smaller of the two calculated strengths:

0.9 ; 0.75y g u nF A F A

is taken as the design strength.

Consider the example from the text that is reproduced below. The gusset plate connection is subjected to service dead and live loads of 35 and 15 kips, respectively. Steel is A36. Bolts are 0.875-inch in diameter. Assume that the effective net area is 85% of the computed net area.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 5

NET AREA

The performance of a tension member is often governed by the response of its connections. The LRFD Specification introduces a measure of connection performance known as joint efficiency, which is a function of

Material properties (ductility)

Fastener spacing

Stress concentrations

Shear lag

Most important of the four

See F3.4 on page 4

Addressed specifically by the LRFD Specification

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 6

The LRFD Specification introduces the concept of effective net area to account for shear lag effects.

For welded connections:

e gA UA=

For bolted connections:

e nA UA=

where

1 xUL

= −

and x is the distance from the plane of the connection to the centroid of the connected member and L is the length of the connection in the direction of the load. Figures from the text are reproduced below to illustrate these definitions.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 7

For welded connections, the commentary to the LRFD Specification (Section B.3) gives values for U that can be used in lieu of detailed calculation. Different values are given depending on whether there are more or less than two fasteners per line in the direction of loading as follows.

1. For W, M, and S shapes that have a width-to-depth ratio of 0.67 or more and are connected through the flanges with at least three fasteners per line in the direction of the applied load, U = 0.9.

2. For other shapes with at least three fasteners per line, U = 0.85.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 8

3. For all members with only two fasteners per line, U = 0.75.

The average values for welded connections can be taken as:

1. For W, M, and S shapes that have a width-to-depth ratio of 0.67 or more and are connected through the flanges, U = 0.9.

2. For all other shapes, U = 0.85.

The figure below from the textbook illustrates the calculation for bolted connections.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 9

The effective net area will only be less than the net area if some elements of the cross-section are not connected. For tension members such as single plates and bars connected by longitudinal fillet welds,

the LRFD Specification writes that e gA UA= , where U = 1.0, 0.87,

and 0.75 for 2 ,1.5 2 , 1.5 ,l w w l w w l w≥ ≤ ≤ ≤ ≤ respectively.

For transverse welds, the LRFD Specification writes that the effective net area is equal to the area of the connected element of the cross section.

Calculations for effective net area are presented in the two examples from the textbook that are reproduced on the following two pages.

Bolted connection

Welded connection

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 10

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 11

STAGGERED FASTENERS

The presentation to date has focused on regular, non-staggered fastener geometries.

Net area maximized with fasteners placed in a straight line.

Often, for reasons of connection geometry, fasteners must be placed in more than one line

Staggered fasteners

See F3.13 from the textbook that is reproduced below for examples of staggered connections

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 12

Consider part (c) of the figure above. One cross-section to be considered for calculation of the effective net area is shown in red; another is given by line abcd. For fracture plane abcd, the equation f=P/A does not apply because stresses on bc are a combination of normal and shearing stresses.

For calculation of the effective net area, the Section B2 of the LRFD Specification makes use of the product of the plate thickness and the net width. The net width is calculated as

2

4n gsw w dg

= − +∑ ∑

where all terms are defined above.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 13

All possible failure patterns should be considered. For example, consider the plate shown below for use with 1-inch diameter bolts.

What is the smallest net area?

Are the loads on abcde the same as ijfh?

What calculations are needed for design?

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 14

How is the net area calculated if connections are made in both legs of an angle?

The angle is first unfolded along the middle surface.

The gross width is the sum of the leg lengths minus the thickness.

Gage line crossing the heel of the angle must be reduced by the thickness of the angle.

Consider the angle connection below. The holes are for 0.875-inch diameter bolts. Steel is A36.

What is the net width?

What is the smallest net area? abdf? abceg? Others?

Are the loads on the above two sections the same?

Because both legs of the angle are connected, what is the effective net area?

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 15

What are the design strengths for:

Yielding?

Fracture?

BLOCK SHEAR

Block shear is an important consideration in the design of steel connections. Consider the figure below that shows the connection of a single-angle tension member. The block is shown shaded.

In this example, the block will fail in shear along ab and tension on bc. The LRFD procedure is based on one of the two failure surfaces yielding and the other fracturing.

Fracture on the shear surface is accompanied by yielding on the tension surface

Fracture on the tension surface is accompanied by yielding on the shear surface

Both surfaces contribute to the total resistance.

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 16

The nominal strength in tension is u ntF A for fracture and y gtF A for yielding where the second subscript t denotes area on the tension surface (bc in the figure above). The yield and ultimate stresses in shear are taken as 60% of the values in tension. The LRFD Specification considers two failure modes:

For shear yield and tension fracture:

0.75[0.6 ]n y gv u ntR F A F Aφ = +

For tension yield and shear fracture:

0.75[0.6 0.6 ]n u nv y gtR F A F Aφ = +

Because the limit state is fracture, the equation with the larger of the two fracture values controls. Consider the single angle in tension shown below.

What are the shear areas?

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 17

0.375 (7.5); 0.375[7.5 2.5 1]gv nvA A= × = − ×

What are the tension areas?

0.375 (1.5); 0.375[1.5 0.5 1]gt ntA A= × = − ×

DESIGN OF TENSION MEMBERS

The design of a tension member involves selecting a member from the LRFD Specification with adequate

Gross area

Net area

Slenderness

300Lr

≤ to prevent vibration, etc; does not apply to

cables.

If the member has a bolted connection, the choice of cross section must account for the area lost to the boltholes.

Because the section size is not known in advance, the default values of U are generally used for preliminary design.

Detailing of connections is a critical part of structural steel design. Connections to angles are generally problematic if there are two lines of bolts. Consider the figure below that provides some guidance on sizing angles and bolts.

Gage distance 1g applies when there is one line of bolts

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 18

Gage distances 2g and 3g apply when there are two lines

As an example, design an equal angle tension member, 12 feet long to resist the loads shown. Use A36 steel.

What is the factored load?

What is the minimum value of gA ? nA ?

What is minimum value of radius of gyration, r?

Pick a section from the LRFD manual based on gA , r, ig :

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 19

Assume that the angle is connected through the long leg and estimate a value of U (0.85), calculate values of nA and eA .

THREADED RODS

The figure below presents applications of threaded rods and cables.

The nominal tensile strength of a threaded rod can be written as

0.75n s u b uP A F A F=

where

sA is the stress area (threaded portion),

bA is the nominal (unthreaded area),

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 20

and 0.75 is a lower bound (conservative) factor relating sA and

bA . See Section J3.6 of the LRFD Specification for details.

The design strength of a threaded rod is calculated as 0.75n nP Pφ =

PIN-CONNECTED MEMBERS

Pinned connections transmit no moment (ideally) and often utilize components machined to tight tolerances (plus, minus 0.001”). The figure below from the textbook shows failure modes for pin-connected members and each failure mode must be checked for design.

Specifically, the following limit states must be checked.

Tension on the effective net area

0.75(2 )t n eff uP tb Fφ = , where 2 0.63effb t b= + ≤

CIE 428 Module B Instructor: Andrew Whittaker

8/21/2002 9:13 AM 21

Shear on the effective area

0.75(0.6 ) 0.75(0.6[2 { }] )2sf n sf u udP A F t a Fφ = = +

Bearing on projected area

0.75(1.8 ) 0.75(1.8[ ] )n pb y yP A F dt Fφ = =

1.8 pb yA F is based on a deformation limit state under

service loads producing stresses of 90% of yield.

Tension on the gross section

0.90( )n g yP A Fφ =

- END OF MODULE -