chapter 4: tension members - steel design 4300:401 chap04 - tension... · chapter 4: tension...

4.1

Chapter 4: Tension Members The following information is taken from “Unified Design of Steel Structures,” Second Edition, Louis F.

Geschwindner, 2012, Chapter 4.

4.1 Introduction

Tension members are one of the simplest type of structural member to design and

a good starting point for studying structural steel design.

• Tensile forces are assumed to be applied concentrically so that other effects,

such as buckling or bending, are not normally present to reduce the load

carrying capability of the member.

4.2 Tension Members in Structures

Tension members are found in bridge and roof trusses, towers, hangers, bracing

systems, and sag rods.

• Tension members used as bracing for structures are normally long and slender.

• These slender members are relatively flexible and must be carefully designed

and installed, particularly if there is a chance of load reversal which would

require them to carry a compression load.

4.3 Cross Sectional Shapes for Tension Members

Tension members may be structural steel shapes, plates, or combinations of shapes

and plates; eyebars and pin-connected plates; rods and bars, or wire rope and steel

cables.

One of the simplest forms of tension members is the circular rod.

• The rod has been used frequently in the past, but finds only occasional use

today in bracing systems, light trusses, and timber construction.

- There is some difficulty in connecting circular rods to many structures.

- The average-size rod has very little bending stiffness and may easily sag

under its own weight, diminishing the appearance of the structure.

Rolled shapes such as angles have replaced rods for most applications.

• Although the use of cables is increasing for suspended-roof structures, tension

members usually consist of single angles, double angles, tees, W-sections,

channels, or sections built up from plates or rolled shapes.

• Rolled shapes look better than a sagging rod, they are stiffer, they may be able

to carry small compression loads resulting from load reversals, and they are

easier to connect.

4.2

- For example, two angles back-to-back permit the insertion of a plate (called

a gusset plate) for connection purposes.

• Single angles and double angles are the most common types of tension members

that are in use.

• Structural tees make very good chord members for welded trusses, because

angles can be conveniently connected to the webs of the tees.

• Square, rectangular, and circular HSS have become more common as tension

members over the past few years.

- HSS sections have become popular because of their attractive appearance

and ease of maintenance.

• Built-up sections are used occasionally when the designer is unable to obtain

sufficient area or stiffness from a single structural shape.

- Built-up sections are not as economical as a single structural shape.

4.4 Behavior and Strength of Tension Members

A tension member may be subject to one of two limit states: tensile yielding in

the gross section or tensile rupture in the net section.

• Based on these two limit states, Section D2 of the AISC Specification (entitled

“Tensile Strength”) states that the nominal strength of a tension member (Pn) is

the smaller of the values from the following two conditions.

Yielding

A ductile steel member without bolt holes and subject to a tensile load can resist

without fracture a load larger than its “gross cross-sectional area times its yield

stress” (i.e. Ag x Fy) because of strain-hardening.

• However, the excessive yielding will likely end the usefulness of the member.

• Also, the excessive yielding may cause failure of the structural system of which

the member is a part.

• The limit state of significance is tensile yielding rather than fracture.

For tensile yielding in the gross section (which is intended to prevent excessive

elongation of the member):

Pn = Fy Ag AISC Equation D2-1

LRFD (φt = 0.90): Design tensile strength = φt Pn = φt Fy Ag

ASD (Ωt = 1.67): Allowable tensile strength = Pn/Ωt = Fy Ag/Ωt

where

Pn = the nominal tensile yield strength of the tension member

4.3

Fy = the specified minimum yield stress of the type of steel being used

(ref. AISC Table 2-4, or Figure 3.10, p. 63 of the textbook)

Ag = the gross area of the member

Rupture

A tension member with bolt holes may fail by fracture at the net section through

the bolt holes at a load smaller than the load required to yield the gross section

away from the holes due to stress concentrations.

• Although strain hardening is quickly reached at the net section, yielding may not

be the limit state of significance because the overall change in length of the

member may be negligible since the yielding is confined to such a short section

of the member.

• The limit state of significance is tensile rupture (fracture) rather than tensile

yielding.

For tensile rupture in the net section (where bolt or rivet holes are present):

Pn = Fu Ae AISC Equation D2-2

LRFD (φt = 0.75): Design tensile rupture strength = φt Pn = φt Fu Ae

ASD (Ωt = 2.00): Allowable tensile rupture strength = Pn/Ωt = Fu Ae/Ωt

where

Pn = the nominal tensile rupture strength of the tension member

Ae = effective net area of the member

Fu = the specified minimum tensile strength of the type of steel being used

(ref. AISC Table 2-4, or Figure 3.10, p. 63 of the textbook)

For tension members consisting of rolled steel shapes, there is actually a third

limit state, block shear, presented in Section 4.7 of the textbook.

4.5 Computation of Areas

The design of tension members uses the following cross-sectional area definitions.

1. Gross area, Ag

2. Net area, An

3. Effective net area, Ae

The criteria governing the computation of the various areas required for the

analysis and design of tension members are given in Sections B4.3 (Gross and Net

Area Determination) and D3 (Effective Net Area) of the Specification.

4.4

Gross Area

The gross area of a member is taken as the full cross-sectional area taken

perpendicular to the longitudinal axis of the member along which the tensile force

is acting.

• No holes or other area reductions can be present where the section is taken.

• For common structural shapes, Part 1 of the AISC Manual provides tabulated

values for gross areas.

• For shapes composed of flat plate components, the gross area may be

determined as follows.

Ag = ∑wi ti

where

wi = the width of the rectangular cross-sectional element

ti = the thickness of the rectangular cross-sectional element

Net Area

The term “net cross-sectional area” (or “net area”) refers to the gross cross-

sectional area of a member minus any holes, notches, or other indentations.

Standard Holes

• A long-used practice was to consider the diameter of the bolt holes 1/8” larger

than the diameter of the bolt.

- The holes were punched to be 1/16” larger than the diameter of the bolt.

- The punching of the hole was assumed to damage 1/16” more of the

surrounding material.

• Today a large number of holes are drilled and it seems reasonable to add only

1/16” to the bolt diameters for such holes.

- However, because the decision to punch or drill a hole is based on the steel

fabricator’s equipment capability, it is standard practice to add 1/8” to the

bolt diameter for all standard bolt holes.

• For steel that is thicker than the bolt diameters, holes may be subpunched

(with diameters 3/16” smaller than the required diameters) and then the holes

are reamed out to full size.

- Very little material adjacent to the hole is damaged by this process, but this

process is expensive.

- The holes are even and smooth, and it is considered unnecessary to subtract

the additional 1/16” for damage to the sides of the holes.

4.5

Example Problems – Net Area

Example

Given: The plate connection shown.

Bolts: ¾” diameter

Find: The net area of the 3/8” x 8”

plate.

Solution

Net area: An = (3/8) [8 – 2 (3/4 + 1/8)] = 2.34 in2

Example

Given: An 8” x 3/4” plate with a single line of standard holes for 7/8” bolts.

Find: Gross and net areas.

Solution

Gross area: Ag = 8 (0.75) = 6.0 in2

Net area: An = (3/4) [8 – 1 (7/8 + 1/8)] = 5.25 in2

Example

Given: A 4” x 4” x 1/2” angle with two lines of holes (one in each leg) of standard

holes for 3/4” bolts.

Find: Gross and net areas.

Solution

4” x 4” x 1/2” angle (A = 3.75 in2)

Gross area: Ag = 3.75 in2

Net area: An = 3.75 – 2 (1/2) (3/4 + 1/8) = 2.875 in2

4.6

Oversize and Slotted Holes

Section J3.2 of the Specification prescribes the maximum size holes used for

standard and oversize bolt holes, and for short-slotted and long-slotted holes.

• Oversize bolt holes, short-slotted holes, and long-slotted holes are used to

facilitate construction or to permit larger rotations or deformations under

loading.

Short Connecting Elements

When a short connecting element (e.g. link, splice plate, flange plate, gusset plate)

is used as statically loaded tensile connecting element, its strength is determined

as follows.

a) For tensile yielding of connecting elements

Rn = Fy Ag AISC Equation J4-1

φt = 0.90 (LRFD) Ωt = 1.67 (ASD)

b) For tensile rupture of connecting elements

Rn = Fu Ae AISC Equation J4-2

φt = 0.75 (LRFD) Ωt = 2.00 (ASD)

When a short connecting element is used, and the net area and gross area are equal

or nearly equal, there may not be sufficient length for the entire cross section to

yield uniformly.

• Tests have shown that bolted tension elements rarely have an efficiency

greater than 85%, even if the holes represent a very small percentage of the

gross area of the elements.

• The area that is first to yield may reach rupture at an early stage and the

rupture limit state reached prematurely.

- This is an undesirable mode of failure since such failure would likely occur

suddenly with little or no warning.

• Section J4.1 of the Specification prescribes the net area used in Eq. J4-2 for

bolted splice plates to be determined as follows.

Ae = An ≤ 0.85 Ag

Influence of Hole Placement

Members connected by bolts that are staggered may fail along a section other than

the straight transverse section (i.e. line AB shown in the figure below).

• The critical section is the one with the critical (i.e. the smallest) net area.

4.7

• Failure along section ABCD is possible unless the holes are a large distance

apart.

- Along the diagonal line from B to C there is a combination of tensile stress

and shear stress complicating the analysis.

Section B4.3b of the AISC Specification offers a simple empirical method for

computing the net width of a tension member along a zigzag section.

• The method is to take the gross width of the member, regardless of the line

along which failure might occur, subtract the number of holes along the zigzag

section being considered, and add for each inclined line the quantity given by

the expression s2/4g.

An = t [ w – n (d + 1/8) + ∑s2/4g ]

where

t = the thickness of the member

w = the gross width of the member

n = the number of bolts through the section considered

d = the diameter of the bolts

s = the longitudinal spacing (or pitch) of any two holes

g = the transverse spacing (or gage) of the same holes

• For each diagonal line, the quantity s2/4g is added back into the net width to

account for the overestimation of the required deduction when a full adjacent

hole has been deducted.

• There may be several paths to consider, any one of which may be critical.

- The path giving the least net area should be used.

- The s2/4g rule is merely an approximation or simplification of the complex

stress variations that occur in members with staggered arrangements of

bolts.

4.8

Example Problems – Effect of Staggered Holes

Example

Given: ½” thick plate shown.

Holes are punched for ¾” bolts.

Find: The critical net area using AISC

Specification (Section B4.3b).

Solution:

Section ABCD: An = (1/2) [11.0 – 2(3/4 + 1/8)] = 0.5 (9.25) = 4.625 in2

Section ABCEF: An = (1/2) [11.0 – 3(3/4 + 1/8) + 32/4(3)] = 0.5 (9.125) = 4.56 in2

Section ABEF: An = (1/2) [11.0 – 2(3/4 + 1/8) + 32/4(6)] = 0.5 (9.625) = 4.81 in2

Section ABCEF controls: An = 4.56 in2

Example

Given: W12 x 16 section shown.

Ag = 4.71 in2, tw = 0.220”

Holes are punched for 1” bolts.

Find: The critical net area using

AISC Specification (Section

B4.3b).

Solution:

Section ABDE: An = 4.71 – 0.220 [2(1 + 1/8)] = 4.21 in2

Section ABCDE: An = 4.71 – 0.220 [3(1 + 1/8)] + 0.220 (2) [22/4(3)] = 4.11 in2

Section ABCDE controls: An = 4.11 in2

4.9

Effective Net Area

When a member (other than a flat plate or bar) is loaded in axial tension until

failure across the net section occurs, the computed tensile failure stress (based on

the net section) is likely less than the tensile strength of the steel, unless all of

the various elements which make up the section are so connected so that stress is

transferred uniformly across the entire section.

• If the forces are not transferred uniformly across the cross section of a

member, there will be a transition region of uneven stress running from the

connection out into the member some distance.

- For example, the stresses are not evenly distributed in a single angle tension

member that is connected by one leg only (ref. figure below).

• The flow of tensile stress between the full member cross section and the

smaller connected cross section is not 100 percent effective.

- The behavior is called shear lag.

- As a result, Section D3.3 of the AISC Specification states that the

effective net area Ae of such a member is determined by multiplying an area

(i.e. the net area, the gross area, or the directly connected area, as

described below) by a reduction factor (i.e. shear lag factor, U).

Ae = An U AISC Equation D3-1

• The value of the reduction coefficient U is affected by the cross section of

the member and by the length of the connection.

- For example, if an angle is connected at its end by only one leg, the effective

area in resisting tension can be increased by using the longer leg as the

connected leg and the shorter leg as the unconnected leg.

• The reduction factor U can be expressed empirically by the following formula.

U = 1 - x /L

4.10

- One measure of the effectiveness of a member (e.g. an angle connected by

one leg) is the distance x measured perpendicular from the plane of the

connection to the centroid of the area of the whole section.

◦ The smaller the value x , the larger is the effective area of the member.

- Another measure of the effectiveness of a member is the length of the

connection L.

◦ The greater the length L, the smoother is the transfer of stress to the

members unconnected parts.

- The smaller the value x and the larger the value L, the larger will be the

value of U, and thus the larger will be the effective area of the member.

Bolted Members

If a tension load is transmitted by bolts, the area used in AISC Equation D3-1 is

the net area An of the member, and U is computed as follows.

U = 1 - x /L

• The length L is equal to the distance between the first and last bolts in the line.

- When there are two or more lines of bolts, L is the length of the line with

the maximum number of bolts.

- If the bolts are staggered, L is the out-to-out dimension between the

extreme bolts in a line.

• If only one bolt is used in each line, the conservative approach is to let Ae = An

of the connected element.

Table D3.1 of the Specification provides a detailed list of the U factors for

various situations.

For some problems, the value U may be calculated with the 1 - x /L expression and

compared with the value given in Table D3.1 of the Specification.

• The use of the larger of the two U values for the effective area Ae is

permitted by Section D3 of the AISC Specification Commentary (p. 16.1-286).

- Table D3.1 (Cases 7 and 8) also contains the following statement: “If U is

calculated per Case 2, the larger value is permitted to be used.”

4.11

Example Problems – Effective Net Area for Bolted Members

Example

Given: Tension member consisting of a W10 x 45 with two lines of ¾” bolts in each

flange: three bolts in each line spaced at 4” c/c.

Steel: A572 Grade 50 (Fy = 50 ksi, Fu = 65 ksi)

Find: LRFD tensile strength and ASD allowable tensile design strength using the

AISC Specification.

Solution

W10 x 45 (Ag = 13.3 in2, d = 10.1”, bf = 8.02”, tf = 0.620”)

Tensile yield strength in the gross section

Compute the nominal tensile strength of the section.

Pn = Fy Ag = 50 (13.3) = 665 kips

Compute the LRFD design tensile strength and ASD allowable tensile design

strength.

LRFD (φt = 0.90): φt Pn = 0.90 (665) = 598.5 kips

ASD (Ωt = 1.67): Pn/Ωt = 665/1.67 = 398.2 kips

Tensile rupture strength in the net section

Compute the net area.

An = 13.3 – 0.620 [4 (3/4 + 1/8)] = 11.13 in2

Compute the effective area.

• To determine the value of U for a W-section connected by its flanges only,

assume that the section is split into two structural tees.

• Using AISC Table D-3.1 (Case 2), compute a value for U.

- Refer to AISC Table 1-8 for half of a W10 x 45 (i.e. WT5 x 22.5):

x (listed as y in the table) = 0.907”

L = 2 spaces (4”/space) = 8”

U = 1 - x /L = 1 – 0.907/8 = 0.887

4.12

• Using AISC Table D-3.1 (Case 7), look up the value for U.

bf = 8.02” > (2/3) d = (2/3) 10.1 = 6.73” and U = 0.90

Use the larger value: U = 0.90

Effective area: Ae = U An = 0.90 (11.13) = 10.02 in2


Pn = Fu Ae = 65 (10.02) = 651.3 kips

Compute the LRFD tensile strength and ASD allowable tensile design strength.

LRFD (φt = 0.75): φt Pn = 0.75 (651.3) = 488.5 kips

ASD (Ωt = 2.00): Pn/Ωt = 651.3/2.00 = 325.6 kips

Answers: LRFD = 488.5 kips and ASD = 325.6 kips

Tensile rupture strength in the net section controls.

4.13

Example

Given: Tension member consisting of an L6 x 6 x 3/8 with one line of 7/8” bolts in

one leg with four bolts in each line spaced at 3” c/c.

Steel: A36 (Fy = 36 ksi, Fu = 58 ksi)

Find: LRFD tensile strength and ASD allowable tensile design strength using the

AISC Specification.

Solution

L6 x 6 x 3/8 (Ag = 4.38 in2, x = y = 1.62”)



Pn = Fy Ag = 36 (4.38) = 157.7 kips


LRFD (φt = 0.90): φt Pn = 0.90 (157.7) = 141.9 kips

ASD (Ωt = 1.67): Pn/Ωt = 157.7/1.67 = 94.4 kips


Compute the net area.

An = 4.38 – (3/8) [1 (7/8 + 1/8)] = 4.00 in2


• Using AISC Table D-3.1 (Case 2), compute a value for U.

- Refer to AISC Manual Table 1-7 for the angle.

x = 1.62”


U = 1 - x /L = 1 – 1.62/9 = 0.820

• Using AISC Table D-3.1 (Case 8), look up the value for U.

- A single angle with four or more fasteners per line: U = 0.80

Use the larger value: U = 0.820



Pn = Fu Ae = 58 (3.28) = 190.2 kips

4.14


LRFD (φt = 0.75): φt Pn = 0.75 (190.2) = 142.6 kips

ASD (Ωt = 2.00): Pn/Ωt = 190.2/2.00 = 95.1 kips


Tensile yield strength in the gross section controls.

4.15

Welded Members

When tension loads are transferred by welds, the following rules from AISC Table

D3.1 (Table 3.2 of the textbook) are to be used to determine values for A and U.

1. If the tension load is transmitted directly to each of the cross-sectional

elements by welds, then U is equal to 1.0 and the effective area Ae is to equal

the gross area of the member, Ag (AISC Table D3.1, Case 1).

2. If the tension load is transmitted to some but not all of the cross sectional

elements by longitudinal welds, or by longitudinal welds in combination with

transverse welds, then U = 1 - x /L (AISC Table D3.1, Case 2).

3. If a tension load is transmitted only by transverse welds to some but not all of

the cross-sectional elements, then the area A is to equal the area of the

directly connected elements and U is equal to 1.0 (AISC Table D3.1, Case 3).

4. For plates where the tension load is transmitted by longitudinal welds only, the

effective net area is Ae = U A, where A = the area of the plate and the value of

U is determined as follows (AISC Table D3.1, Case 4).

When ℓ ≥ 2w U = 1.0

When 2w > ℓ ≥ 1.5w U = 0.87

When 1.5w > ℓ ≥ w U = 0.75

where

ℓ = weld length (inches)

w = plate width (distance between welds) (inches)

5. Tension members may fail prematurely by shear lag at the corners if welds are

too far apart.

• As a result, when flat plates or bars used as tension members are connected

by longitudinal fillet welds, the length of the welds may not be less than the

width of the plates or bars.

• For combinations of longitudinal and transverse welds, ℓ is equal to the

length of the longitudinal weld only, because the transverse weld has little or

no effect on the shear lag.

- In other words, the transverse weld does little to get the load into the

unattached parts of the member.

4.16

Example Problems – Effective Net Area for Welded Members

Example

Given: A 1 x 6 plate connected to a 1 x 10

plate with 8” longitudinal fillet welds

to transfer a tensile load.

Steel: Fy = 50 ksi, Fu = 65 ksi

Find: LRFD tensile strength and ASD

allowable tensile design strength using

the AISC Specification.

Solution



Pn = Fy Ag = 50 ksi (1” x 6”) = 300 kips


LRFD (φt = 0.90): φt Pn = 0.90 (300) = 270.0 kips

ASD (Ωt = 1.67): Pn/Ωt = 300/1.67 = 179.6 kips



• Using AISC Table D-3.1 (Case 4), determine the value for U.

1.5 w = 1.5 (6) = 9” > ℓ = 8” > w = 6” and U = 0.75



Pn = Fu Ae = 65 (4.50) = 292.5 kips


LRFD (φt = 0.75): φt Pn = 0.75 (292.5) = 219.4 kips

ASD (Ωt = 2.00): Pn/Ωt = 292.5/2.00 = 146.2 kips



4.17

Example

Given: L8 x 6 x ¾ connected to a plate with longitudinal fillet welds on the ends and

sides of the 8” leg to transfer a tensile load.


Find: LRFD tensile strength and

ASD allowable tensile design

strength using the AISC

Specification.

Solution

L8 x 6 x ¾ (A = 9.99 in2, y = 2.55”, x = 1.56”)



Pn = Fy Ag = 36 ksi (9.99) = 359.6 kips


LRFD (φt = 0.90): φt Pn = 0.90 (359.6) = 323.6 kips

ASD (Ωt = 1.67): Pn/Ωt = 359.6/1.67 = 215.3 kips



• Since only one leg of the angle is connected, a reduced effective area must be

computed.

- Using AISC Table D-3.1 (Case 2), determine the value for U.

◦ Refer to AISC Manual Table 1-7 for the angle: x = 1.56” (the dimension

measured perpendicular from the connecting plane, in this case the longer

leg), L = 6”

U = 1 - x /L = 1 – 1.56/6 = 0.740



Pn = Fu Ae = 58 (7.39) = 428.6 kips

4.18


LRFD (φt = 0.75): φt Pn = 0.75 (428.6) = 321.5 kips

ASD (Ωt = 2.00): Pn/Ωt = 428.6/2.00 = 214.3 kips



4.19

Example Problem – Short Connecting Elements for Tension Members

Example

Given: Tension connection shown.

Steel (plates): Fy = 36 ksi, Fu = 58 ksi

Bolts: Two lines of ¾” bolts are

used in each plate.

Find: LRFD tensile strength and

ASD allowable tensile design

strength using the AISC

Specification.

Solution



Pn = Fy Ag = 36 [2 (3/8 x 12)] = 324.0 kips


LRFD (φt = 0.90): φt Pn = 0.90 (324.0) = 291.6 kips

ASD (Ωt = 1.67): Pn/Ωt = 324.0/1.67 = 194.0 kips



Net area of two plates: An = 2 (3/8 x 12) – 4 (3/4 + 1/8) (3/8) = 7.69 in2

≤ 0.85 Ag = 0.85 [2 (3/8 x 12)] = 7.65 in2 (controls)

Effective area of two plates: Using Table D-3.1 (Case 1), U = 1.0.

Ae = U An = 1.0 (7.65) = 7.65 in2


Pn = Fu Ae = 58 (7.65) = 443.7 kips


LRFD (φt = 0.75): φt Pn = 0.75 (443.7) = 332.8 kips

ASD (Ωt = 2.00): Pn/Ωt = 443.7/2.00 = 221.9 kips



4.20

4.6 Design of Tension Members

Although the designer has considerable freedom in the selection of tension

members, the resulting members should have the following properties.

• Compactness.

• Dimensions that compatible with other members of the structure.

• Connections to as many parts of the sections as possible to minimize shear lag.

The type of tension members selected is often influenced by the type of

connections used within the structure.

• Tension members consisting of angles, channels, and S or W sections are

generally used when the connections are made with bolts.

• Plates, channels, and structural tees are generally used for welded structures.

Steel specifications give preferable maximum values of slenderness ratios for both

tension and compression members.

• The slenderness ratio of a member is the ratio of its unsupported length to its

least radius of gyration (i.e. L/r).

• For steel compression members the maximum slenderness ratio is 200.

• For tension members the AISC Specification does not specify a maximum

slenderness ratio, but Section D1 of the specification suggests that a maximum

value of 300 be used (except for rods and hangers in tension).

- Maximum slenderness ratios for rods are left to the designer’s judgment.

- If a maximum value of 300 were specified, rods would seldom be used

because of their extremely small radii of gyration, and thus very high

slenderness ratios.

• The AASHTO (American Association of State Highway and Transportation

Officials) Specifications provide mandatory maximum slenderness ratios of 200

for main tension members, 240 for secondary members, and 140 for members

subjected to stress reversal.

- A main member is defined by AASHTO as one in which the stresses result

from dead and/or live loads.

- A secondary member is one used to brace structures or to reduce the

unbraced length of other members whether that member is a main or

secondary member.

- The AISC Specification makes no distinction between main and secondary

members.

4.21

• Specifications usually recommend that slenderness ratios not exceed certain

maximum values so that some minimum compressive strength is available in the

members.

- Stress reversals may occur during shipping and assembly and perhaps due to

wind and earthquake.

- The purpose of the maximum slenderness ratio is also to ensure the use of

sections with stiffness sufficient to prevent undesirable lateral deflections

or vibrations.

The goal of the design process is to size members so that they are safe by

satisfying the limit states listed in the AISC Specification.

• The design is generally a trial-and-error process, although tables in the Steel

Manual often enable the designer to select a desirable section directly.

• For a tension member, we can estimate the area required, select a section from

the Steel Manual providing that area, and then check the section’s strength.

LRFD

Based on the LRFD equations, the design strength of a tension member is the least

of the following.

Tensile yield strength in the gross section (φt = 0.90):


Pu = φt Fy Ag

Tensile rupture strength in the net section (φt = 0.75):


Pu = φt Pn = φt Fu Ae

Block shear strength

a. To satisfy the first of these equations (i.e. Pu = φt Fy Ag), the minimum gross

area Ag must be at least equal to

min Ag = Pu/φt Fy

b. To satisfy the second of these equations (i.e. Pu = φt Fu Ae), the minimum

effective area Ae must be at least equal to

min Ae = Pu/φt Fu

Since Ae = U An for a bolted member, the minimum net area An must be at least

equal to

min An = min Ae/U = Pu/φt Fu U

4.22

Then the minimum gross area must be at least equal to

min Ag = min An + estimated area of holes

= Pu/φt Fu U + estimated area of holes

Use the larger of the two values for “min Ag” determined in parts “a” and “b”

above as an initial estimate for the size of the tension member.

c. The third criteria, evaluating the block shear strength, can be performed once

a trial shape has been selected and the other parameters related to block shear

strength are known.

ASD

Based on the ASD equations, the allowable strength of a tension member is the

least of the following.

Tensile yield strength in the gross section (Ωt = 1.67):


Pa = Pn/Ωt = Fy Ag/Ωt

Tensile rupture strength in the net section (Ωt = 2.00):


Pa = Pn/Ωt = Fu Ae/Ωt


a. To satisfy the first of these equations (i.e. Pa = Fy Ag/Ωt), the minimum gross

area Ag must be at least equal to

min Ag = Ωt Pa/Fy

b. To satisfy the second of these equations (i.e. Pa = Fu Ae/Ωt), the minimum

effective area must be at least equal to

min Ae = Ωt Pa/Fu

Since Ae = U An for a bolted member, the minimum net area An must be at least

equal to

min An = min Ae/U = Ωt Pa/Fu U

Then the minimum gross area must be at least equal to

min Ag = min An + estimated area of holes

= Ωt Pa/Fu U + estimated area of holes

Use the larger of the two values for “min Ag” determined in parts “a” and “b”

above as an initial estimate for the size of the tension member.

c. The third criteria, evaluating the block shear strength, can be performed once

a trial shape has been selected and the other parameters related to block shear

strength are known.

4.23

The estimated areas required by these two methods (LRFD and ASD methods) will

normally vary a little from each other.

• In LRFD Pu represents the result of the factored LRFD load combinations.

• In ASD Pa represents the result of the ASD load combinations.

4.24

Example Problems – Design of Tension Members

Example

Given: 30’ long tension member with two

lines of 7/8” bolts in each flange (at

least 3 in a line at 4” c/c)


Service loads: PD = 130 kips, PL = 110 kips

Find: Select a W12 section to support the

given loads. Neglect block shear.

Solution

LRFD

Determine the critical factored load.

Pu = 1.4 D = 1.4 (130) = 182 kips

Pu = 1.2 D + 1.6 L = 1.2 (130) + 1.6 (110) = 332 kips (controls)

Compute the minimum Ag required for tensile yield in the gross section (φt = 0.90).

min Ag = Pu/φt Fy = 332/[0.9 (50)] = 7.38 in2

Compute the minimum Ag required for tensile rupture in the net section (φt = 0.75).

min Ag = Pu/φt Fu U + estimated area of holes

Assume that U = 0.85 (AISC Table D3.1, Case 7).

Assume that the flange thickness is about 0.380” after looking at W12

sections in the Steel Manual that have areas of 7.38 in2 or more.

min Ag = 332/[0.75 (65) (0.85)] + 4 (7/8 + 1/8) (0.380)

= 8.01 + 1.52 = 9.53 in2 (controls)

Determine the preferable minimum r.

min r = L/300 = 30(12)/300 = 1.20”

Try W12 x 35 (Ag = 10.3 in2, d = 12.5”, bf = 6.56”, tf = 0.520”, ry = 1.54”)

4.25

Checking


Pn = Fy Ag = 50 (10.3) = 515.0 kips

φt Pn = 0.90 (515.0) = 463.5 kips > Pu = 332 kips OK


Using AISC Table D3.1, Case 2:

x (listed as y ) for half W12 x 35 (i.e. WT 6 x 17.5) = 1.30”


U = 1 - x /L = 1 – 1.30/8 = 0.838


bf = 6.56” < (2/3) d = (2/3)(12.5) = 8.33” and U = 0.85

Use U = 0.85 (the larger value).

An = 10.3 – 4 (7/8 + 1/8) (0.520) = 8.22 in2

Ae = U An = 0.85 (8.22) = 6.99 in2

Pn = Fu Ae = 65 (6.99) = 454.4 kips


Check the slenderness ratio.

Ly/ry = 30 (12)/1.54 = 233.8 < 300 OK

Select W 12 x 35

ASD

Determine the critical load combination.

Pa = D + L = 130 + 110 = 240 kips

Compute the minimum Ag required for tensile yield in the gross section (Ωt = 1.67).

min Ag = Ωt Pa/Fy = 1.67(240)/50 = 8.02 in2

Compute the minimum Ag required for tensile rupture in the net section (Ωt = 2.00).

min Ag = Ωt Pa/Fu U + estimated area of holes


Assume that the flange thickness is about 0.440” after looking at W12


min Ag = 2.00(240)/65(0.85) + 4 (7/8 + 1/8) (0.440)

= 8.69 + 1.76 = 10.45 in2 (controls)

4.26


min r = L/300 = 30(12)/300 = 1.20”

Try W12 x 40 (Ag = 11.7 in2, d = 11.9”, bf = 8.01”, tf = 0.515”, ry = 1.94”)

Checking


Pn = Fy Ag = 50 (11.7) = 585.0 kips

Pn/Ωt = 585.0/1.67 = 350.3 kips > Pa = 240 kips OK



x (listed as y ) for half W12 x 40 (i.e. WT 6 x 20) = 1.09”


U = 1 - x /L = 1 – 1.09/8 = 0.864


bf = 8.01” > (2/3) d = (2/3) 11.9 = 7.93” and U = 0.90


An = 11.7 – 4 (7/8 + 1/8) (0.515) = 9.64 in2

Ae = U An = 0.90 (9.64) = 8.68 in2

Pn = Fu Ae = 65 (8.68) = 564.2 kips



Ly/ry = 30 (12)/1.94 = 185.6 < 300 OK

Select W 12 x 40

Answers: By LRFD, use W12 x 35

By ASD, use W12 x 40

4.27

Example Problem

Given: 20’ long, single angle member connected with one line of four 7/8” bolts at

3½” c/c (on the longer leg if an unequal leg angle is used).



Find: Select a single angle section to support the given loads. Neglect block shear.

Solution

LRFD


Pu = 1.4 D = 1.4 (50) = 70 kips

Pu = 1.2 D + 1.6 L = 1.2 (50) + 1.6 (100) = 220 kips (controls)


min Ag = Pu/φt Fy = 220/[0.9 (36)] = 6.79 in2

Compute the minimum Ag required for tensile rupture in the net section (φt = 0.75).

min Ag = Pu/φt Fu U + estimated area of holes


Assume that the angle thickness is about 0.750” after looking at single angle


min Ag = 220/[0.75 (58) (0.80)] + 1 (7/8 + 1/8) (0.750)

= 6.32 + 0.75 = 7.07 in2 (controls)


min r = L/300 = 20(12)/300 = 0.80”

Possible selections

L5 x 5 x 7/8 A = 8.00 in2 Wt = 27.2 lb/ft

L6 x 4 x 7/8 A = 8.00 in2 Wt = 27.2 lb/ft

L6 x 6 x 5/8 A = 7.13 in2 Wt = 24.2 lb/ft

L7 x 4 x 3/4 A = 7.74 in2 Wt = 26.2 lb/ft

L8 x 4 x 5/8 A = 7.16 in2 Wt = 24.2 lb/ft

L8 x 6 x 9/16 A = 7.61 in2 Wt = 25.7 lb/ft

L8 x 8 x 1/2 A = 7.84 in2 Wt = 26.4 lb/ft

Try L8 x 4 x 5/8 (one of the two lightest) (Ag = 7.16 in2, rz = 0.856”, x = 0.902”)

4.28

Checking


Pn = Fy Ag = 36 (7.16) = 257.8 kips




x for longest leg connected = 0.902”

L = 3 spaces (3.5”/space) = 10.5”

U = 1 - x /L = 1 – 0.902/10.5 = 0.914

Using AISC Table D3.1, Case 8: U = 0.80


An = 7.16 – 1 (7/8 + 1/8) (5/8) = 6.54 in2

Ae = U An = 0.914 (6.54) = 5.98 in2

Pn = Fu Ae = 58 (5.98) = 346.8 kips



Lz/rz = 20 (12)/0.856 = 280.4 < 300 OK

ASD


Pa = D + L = 50 + 100 = 150 kips

Compute the minimum Ag required for tensile yield in the gross section (Ωt = 1.67).

min Ag = Ωt Pa/Fy = 1.67(150)/36 = 6.96 in2

Compute the minimum Ag required for tensile rupture in the net section (Ωt = 2.00).

min Ag = Ωt Pa/Fu U + estimated area of holes


Assume that the angle thickness is about 0.875” after looking at single angle


min Ag = 2.00(150)/[58(0.80)] + 1 (7/8 + 1/8) (0.875)

= 6.47 + 0.875 = 7.35 in2

4.29


min r = L/300 = 20(12)/300 = 0.80”

Possible selections

L5 x 5 x 7/8 A = 8.00 in2 Wt = 27.2 lb/ft

L6 x 4 x 7/8 A = 8.00 in2 Wt = 27.2 lb/ft

L6 x 6 x 3/4 A = 8.46 in2 Wt = 28.7 lb/ft

L7 x 4 x 3/4 A = 7.74 in2 Wt = 26.2 lb/ft

L8 x 4 x 3/4 A = 8.49 in2 Wt = 28.7 lb/ft

L8 x 6 x 9/16 A = 7.61 in2 Wt = 25.7 lb/ft

L8 x 8 x 1/2 A = 7.84 in2 Wt = 26.4 lb/ft

Try L8 x 6 x 9/16 (lightest) (Ag = 7.61 in2, rz = 1.30”, x = 1.49”)

Checking


Pn = Fy Ag = 36 (7.61) = 274.0 kips




x for longest leg connected = 1.49”

L = 3 spaces (3.5”/space) = 10.5”

U = 1 - x /L = 1 – 1.49/10.5 = 0.858

Using AISC Table D3.1, Case 8: U = 0.80


An = 7.61 – 1 (7/8 + 1/8) (9/16) = 7.05 in2

Ae = U An = 0.858 (7.05) = 6.05 in2

Pn = Fu Ae = 58 (6.05) = 350.9 kips



Lz/rz = 20 (12)/1.30 = 184.6 < 300 OK

Answers: By LRFD, use L8 x 4 x 5/8

By ASD, use L8 x 6 x 9/16

Note: The L8 x 4 x 5/8 (the result of the LRFD method) is also satisfactory by the ASD method; however, you

need to be a little less conservative on the assumption regarding the angle thickness to achieve such a result.

4.30

4.7 Block Shear

The LRFD design strength and the ASD allowable strength of tension members are

not always controlled by φt Pn or Pn/Ωt or by the strength of the bolts or welds by

which they are connected.

• The LRFD design strength and the ASD allowable strength may be controlled by

block shear.

The failure of a member may occur along a path involving tension in one plane and

shear on a perpendicular plane (as shown in the figures below).

• For these situations, it is possible for a “block” of steel to tear out.

Theoretically, the total strength of the connection equals the fracture strength of

the stronger plane plus the yield strength of the weaker plane.

• When a tensile load applied to a particular connection is increased, the fracture

strength of the weaker plane will be approached.

- However, the weaker plane will not fail because it is restrained by the

stronger plane.

- But during this time the weaker plane is yielding.

• The load can be increased until the fracture strength of the stronger plane is

reached.

4.31

According to the AISC Specification, tensile rupture will always be the failure

mode on the tension plane of the failure block due to the relatively short length of

material that will be available to yield.

• The controlling limit states on the shear plane will be either yielding or rupture,

whichever has the lower strength.

The block shear design strength of a particular member is determined by the

smaller of the two following values.

1. The shear fracture strength on the net area subject to shear plus the tensile

rupture strength on the net area subject to tension on the perpendicular plane.

2. The shear yield strength on the gross area subject to shear plus the tensile

rupture strength on the net area subject to tension on the perpendicular

segment.

According to Specification Section J4.3, the available strength Rn for the block

shear rupture design strength is determined by the following equation.

Rn = 0.6 Fu Anv + Ubs Fu Ant ≤ 0.6 Fy Agv + Ubs Fu Ant AISC Equation J4-5

where

Agv = gross area subject to shear

Anv = net area subject to shear

Ant = net area subject to tension

Ubs = reduction factor

= 1.0 if the tensile stress is uniform. The tensile stress is generally

considered uniform for angles, gusset (or connection) plates, and for

coped beams with one line of bolts.

= 0.5 if the tensile stress is not uniform. The tensile stress is not

considered uniform in coped beams with two lines of bolts.

According to the AISC Commentary, “block shear is a rupture or tearing

phenomenon, not a yielding limit state. However, gross yielding on the shear plane

can occur when tearing on the tensile plane commences if 0.6FuAnv exceeds

0.6FyAgv. Hence, Equation J4-5 limits the term 0.6FuAnv to not greater than

0.6FyAgv.”

If the block shear strength of a connection is not sufficient, it may be increased

by increasing the edge distance and/or increasing the bolt spacing.

4.32

Example Problems – Block Shear

Example

Given: The L6 x 4 x ½ tension member shown.

Steel: A572 Grade 50 (Fu = 65 ksi)

Bolts: Three – ¾”

Find:

a) LRFD block shear strength and the

ASD allowable block-shear rupture

strength of the member.

b) LRFD design tensile strength and the ASD allowable tensile design strength

of the member.

Solution

L6 x 4 x ½ (A = 4.75 in2, x = 0.981”)


Compute the areas.

Agv = 0.5 (10.0) = 5.00 in2

Anv = 0.5 [10 – 2.5 (3/4 + 1/8)] = 3.91 in2

Ant = 0.5 [2.5 – 0.5 (3/4 + 1/8)] = 1.03 in2

Compute the block shear rupture design strength.

Rn = 0.6 Fu Anv + Ubs Fu Ant = 0.6 (65) (3.91) + 1.0 (65) (1.03) = 219.44 kips

≤ 0.6 Fy Agv + Ubs Fu Ant = 0.6 (50) (5.00) + 1.0 (65) (1.03) = 216.95 kips

Rn = 216.95 kips

Compute the LRFD block shear strength and ASD allowable block shear design

strength.

LRFD (φ = 0.75): φ Rn = 0.75 (216.95) = 162.7 kips

ASD (Ω = 2.00): Rn/Ω = 216.95/2.00 = 108.5 kips



Pn = Fy Ag = 50 (4.75) = 237.5 kips

4.33


LRFD (φt = 0.90): φt Pn = 0.90 (237.5) = 213.7 kips

ASD (Ωt = 1.67): Pn/Ωt = 237.5/1.67 = 142.2 kips


Compute the effective area of angle.

Net area: An = 4.75 – 0.5 [1 (3/4 + 1/8)] = 4.31 in2

• Using AISC Table D3.1 (Case 2), determine the value for U.

x = 0.981”


U = 1 - x /L = 1 – 0.981/8 = 0.877



Pn = Fu Ae = 65 (3.78) = 245.7 kips


LRFD (φt = 0.75): φt Pn = 0.75 (245.7) = 184.3 kips

ASD (Ωt = 2.00): Pn/Ωt = 245.7/2.00 = 122.9 kips


Block shear controls.

4.34

Example

Given: The welded plate tension

members shown.


Find: LRFD design strength and the

ASD allowable design strength of

the plates.

Solution



Pn = Fy Ag = 36 [0.5 (10.0)] = 180.0 kips


LRFD (φt = 0.90): φt Pn = 0.90 (180.0) = 162.0 kips

ASD (Ωt = 1.67): Pn/Ωt = 180.0/1.67 = 107.8 kips




U = 1.0

Effective area: Ae = U An = 1.0 (0.5 x 10) = 5.00 in2


Pn = Fu Ae = 58 (5.00) = 290.0 kips


LRFD (φt = 0.75): φt Pn = 0.75 (290.0) = 217.5 kips

ASD (Ωt = 2.00): Pn/Ωt = 290.0/2.00 = 145.0 kips


Compute the areas.

Agv = 2 ( ½ x 4 ) = 4.00 in2

Anv = Agv = 4.00 in2

Ant = 0.5 (10.0) = 5.00 in2

4.35




Rn = 376.4 kips


strength.

LRFD (φ = 0.75): φ Rn = 0.75 (376.4) = 282.3 kips

ASD (Ω = 2.00): Rn/Ω = 376.4/2.00 = 188.2 kips



4.36

Example

Given: W12 x 30 tension member shown.


Find: LRFD tensile design strength

and the ASD tensile strength.

Include block shear calculations

for the flanges.

Solution

W12 x 30 (A = 8.79 in2, d = 12.3”,

bf = 6.52”, tf = 0.440”, g = 3.50”)



Pn = Fy Ag = 50 (8.79) = 439.5 kips


LRFD (φt = 0.90): φt Pn = 0.90 (439.5) = 395.5 kips

ASD (Ωt = 1.67): Pn/Ωt = 439.5/1.67 = 263.2 kips



Net area of the wide flange: An = 8.79 – 0.440 [4 (7/8 + 1/8) ] = 7.03 in2


x (listed as y ) = 1.27” (for WT6 x 15)


U = 1 - x /L = 1 – 1.27/8 = 0.841


(2/3) d = (2/3) (12.3) = 8.2” > bf = 6.52” and U = 0.85

Thus, U = 0.85



Pn = Fu Ae = 65 (5.98) = 388.7 kips

4.37


LRFD (φt = 0.75): φt Pn = 0.75 (388.7) = 291.5 kips

ASD (Ωt = 2.00): Pn/Ωt = 388.7/2.00 = 194.3 kips

Block shear strength (both flanges)

Compute the areas.

Agv = 4 (0.440 x 10.0) = 17.60 in2

Anv = 4 (0.440) [10 – 2.5 (7/8 + 1/8) ] = 13.20 in2

Ant = 4 (0.440) [½ (6.52 – 3.50) – 0.5 (7/8 + 1/8)] = 1.78 in2




Rn = 630.5 kips


strength.

LRFD (φ = 0.75): φ Rn = 0.75 (630.5) = 472.9 kips

ASD (Ω = 2.00): Rn/Ω = 630.5/2.00 = 315.2 kips



4.38

4.8 Pin-Connected Member

Sections D2, D5, and J7 of the AISC Specification provide the requirements for

pin-connected members regarding strength and proportions of the pins and plates.

• The design tensile strength φtPn and the allowable tensile strength Pn/Ωt of pin-

connected members shall be the lowest value obtained according to the limit

states of tensile rupture, shear rupture, bearing, and tensile yielding obtained

from the following equations.

1. Tension rupture on the effective net area.

Pn = Fu (2 t be) AISC Equation D5-1

φt = 0.75 (LRFD) Ωt = 2.00 (ASD)

where

t = thickness of the plate (inches)

be = effective edge distance (inches)

= 2t + 0.63, but not more than the actual distance from the edge of

the hole to the edge of the part measured in the direction normal

to the applied force.

2. For shear rupture on the effective area.

Pn = 0.6 Fu Asf AISC Equation D5-2

φsf = 0.75 (LRFD) Ωsf = 2.00 (ASD)

where

Asf = shear area on the failure path = 2 t (a + d/2)

t = thickness of the plate

a = the shortest distance from edge of pin hole

to edge of member measured parallel to the

direction of force

d = pin diameter

3. Strength of surfaces in bearing.

Rn = 1.8 Fy Apb AISC Equation J7-1

φ = 0.75 (LRFD) Ω = 2.00 (ASD)

where

Apb = projected bearing area of the pin = t d

4. Tensile yielding on the gross section.


φt = 0.90 (LRFD) Ωt = 1.67 (ASD)

where

Ag = gross area of the member

4.39

Section D5 of the AISC Specification prescribes certain dimensional requirements

for pin-connected members.

• These values are based on long experience in the steel industry and on

experimental work.

4.9 Eyebars and Rods

Pin-connected eyebars are used occasionally as tension members for long-span

bridges and as hangers for some types of bridges or other structures where they

are normally subjected to very large dead loads.

• Today pin-connected bridges are not frequently used because of the advantage

of bolted and welded connections.

- One problem with the old pin-connected trusses was the wearing of the pins

in the holes which caused looseness of the joints.

• An eyebar, a special type of pin-connected member, has an enlarged end where

the pin holes are located.

- Though just about obsolete today, eyebars at one time were very commonly

used for the tension members of bridge trusses.

Eyebars

Eyebar tension members are not commonly used in new construction.

• Eyebar tension members may be found in special applications where the goal is a

design that has some historical context.

The provisions for the design of eyebars are found in Section D6 of the

Specification.

• Eyebars are designed only for the limit state of yielding on the gross section

because the dimensional requirements preclude the possibility of failure at any

load below that level.

Rods

Rods are commonly used for tension members in situations where the required

tensile strength is small.

• These tension members are generally considered secondary members (e.g. sag

rods, hangers, tie rods).

• Rods may also be used as part of the lateral bracing system in walls and roofs.

When rods and bars are used as tension members, they may be welded at their

ends, or, more commonly, they may be threaded and held in place with nuts.

4.40

• Rods can be threaded in two ways (i.e. standard rods and upset rods) and the

strength of the rod depends on the manner in which the threads are applied.

- Standard rods have threads that reduce the cross-sectional area through

the removal of material.

- Upset rods have enlarged ends, with the threads reducing that area to

something larger than the gross area of the rod.

Section D6.2 of the AISC Specification states that a thickness less than ½” for

eyebars and pin-connected plates is permissible only if external nuts are provided

to tighten the pin plates and filler plates into snug contact.

For a standard threaded rod, the nominal tensile strength is given in Section J3.6

of the Specification as follows.

Rn = Fn Ab Equation J3-1

Rn = 0.75 Fu Ab

where

Fn = the nominal stress, Fnt = 0.75 Fu (AISC Table J3.2)

Ab = nominal unthreaded body area of the rod

LRFD (φ = 0.75): φ Rn = the design tensile strength

ASD (Ω = 2.00): Rn/Ω = the allowable tensile strength

The area required for a particular tensile load can then be calculated as follows.

LRFD (φ = 0.75): Ab ≥ Pu/(φ 0.75 Fu)

ASD (Ω = 2.00): Ab ≥ (Ω Pa)/0.75 Fu

AISC Table 7-17 of the AISC Manual, entitled “Threading Dimensions for High-

Strength and Non-High-Strength Bolts,” presents dimensional properties of

standard threaded rods.

Designers use their own judgment in limiting the slenderness ratio for rods.

• A common practice is to select a rod diameter no less than 1/500th its length to

obtain some rigidity even though the design calculations may permit a smaller

diameter.

• It is desirable to limit the minimum size of sag rods (used to provide lateral

support for purlins) to 5/8” because smaller rods are often damaged during

construction.

4.41

- The threads on smaller rods are quite easily damaged by over-tightening

which seems to be a frequent habit of construction workers.

4.42

Example Problem – Threaded Rod

Example

Given: A standard threaded steel rod.



Find: Select a standard threaded rod to support the specified load using the AISC

Specification.

Solution

LRFD


Pu = 1.2 D + 1.6 L = 1.2 (10) + 1.6 (20) = 44 kips

Determine the required gross area Ab (φ = 0.75).

Ab ≥ Pu/(φ 0.75 Fu) = 44/[0.75 (0.75) 58] = 1.35 in2

Using AISC Table 7-17, select 1-3/8” diameter rod with a gross area of 1.49 in2.

Check the LRFD design strength (φ = 0.75).

Rn = 0.75 Fu Ab = 0.75 (58) 1.49 = 64.8 kips

φ Rn = 0.75 (64.8) = 48.6 kips > Pu = 44 kips OK

Select 1-3/8” diameter rod with 6 threads per inch.

ASD


Pa = D + L = 10 + 20 = 30 kips

Determine the required gross area Ab (Ω = 2.00).

Ab ≥ Ω Pa/0.75 Fu = 2.00(30)/(0.75) 58 = 1.38 in2

Using AISC Table 7-17, select 1-3/8” diameter rod with a gross area of 1.49 in2.

Check the ASD allowable design strength (Ω = 2.00).

Rn = 0.75 Fu Ab = 0.75 (58) 1.49 = 64.8 kips

Rn/Ω = 64.8/2.00 = 32.4 kips > Pa = 30 kips OK

Select 1-3/8” diameter rod with 6 threads per inch.

4.43

4.10 Built-Up Tension Members

The AISC Specification allows tension members that are fabricated from a

combination of shapes and plates.

• The strength of built-up members is determined in the same way as the

strength for single-shape members.

• Section D4 of the Specification provides a set of rules describing how the

different parts of built-up tension members are to be connected.

• Section J3.5 of the Specification prescribes the requirements for the

placement of bolts.

Built-up Members (AISC Specification Section D4)

Limitations on the longitudinal spacing of connectors between elements in

continuous contact consisting of a plate and a shape or two plates are outlined in

Specification Section J3.5.

• By User Note, the Specification recommends that the longitudinal spacing of

connectors between components should preferably limit the slenderness ratio

(L/r) in any component between the connectors to 300.

Tie Plates

Tie plates are used to connect the parts of built-up members on their open sides.

• The use of tie plates results in a more uniform stress distribution among the

various parts.

Section D4 of the AISC Specification provides the empirical rules for the design

of tie plates.

• The minimum length of tie plates shall not be less than 2/3 of the distance

between the lines of connectors.

• The minimum thickness of tie plates shall not be less than 1/50 of the distance

between the lines of connectors.

• The minimum width of tie plates shall be the width between lines of connectors

plus the necessary edge distance on each side to keep the bolts from splitting

the plate.

• The maximum preferable spacing between tie plates is based on the least radius

of gyration of an individual component of the built-up member.

- The maximum spacing may be computed based on a preferred maximum

slenderness ratio of L/r = 300.

4.44

Maximum Spacing and Edge Distance (AISC Specification Section J3.5)

The maximum distance from the center of any bolt or rivet to the nearest edge of

parts in contact shall be 12 times the thickness of the connected part under

consideration, but shall not exceed 6”.

The longitudinal spacing of fasteners between elements in continuous contact

consisting of a plate and a shape or two plates shall be as follows.

a) For painted members not subject to corrosion, the spacing shall not exceed 24

times the thickness of the thinner plate or 12”.

b) For unpainted members of weathering steel subject to atmospheric corrosion,

the spacing shall not exceed 14 times the thickness of the thinner plate or 7”.

4.45

Example Problem – Built-Up Tension Members

Example

Given: A 30’ long built-up tension member consisting of two C12 x 30s.

The channels are connected at their ends using two lines of three 7/8” bolts in

each channel web at 3” c/c.

Steel (channels and plates): A36 (Fy = 36 ksi, Fu = 58 ksi) AISC Table 2-4

Service loads: PD = 120 kips

PL = 240 kips

Find: Using the AISC Specification, determine whether the member is satisfactory

and design the tie plates.

Solution

C12 x 30 (Ag = 8.81 in2, d = 12.0”, tw = 0.510”, gage = 1.75”, x = 0.674”,

Ix = 162.0 in4, Iy = 5.12 in4, rx = 4.29”, ry = 0.762”)

LRFD


Pu = 1.2 D + 1.6 L = 1.2 (120) + 1.6 (240) = 528 kips

Checking



Pn = Fy Ag = 36 (2) (8.81) = 634.3 kips

Compute the LRFD design tensile strength of the section.


4.46



Net area of the channels: An = 2(8.81) – 4(7/8 + 1/8) (0.510) = 15.58 in2

• Using Table D3.1 (Case 2), determine the value for U.

x = 0.674” (for the channel)


U = 1 - x /L = 1 – 0.674/6 = 0.888



Pn = Fu Ae = 58(13.84) = 802.7 kips

Compute the LRFD design tensile strength of the section.



Ix = 2 (162.0) = 324.0 in4

Iy = 2[5.12 + (8.81)(6.0 – 0.674)2] = 2(5.12 + 249.9) = 510.0 in4

A = 2(8.81) = 17.62 in2

rx = (Ix/A)1/2 = (324.0/17.62)1/2 = 4.29” (controls)

ry = (Iy/A)1/2 = (510.0/17.62)1/2 = 5.38”

Lx/rx = 30 (12)/4.29 = 83.9 < 300 OK

Design the tie plates (AISC Specification D4).

• Distance between line of bolts = 12.0 – 2 (1.75) = 8.50”

• Minimum length of tie plates shall not be less than 2/3 of the distance between

the lines of connectors: (2/3) 8.50 = 5.67” (Use 6”)

• Minimum thickness of tie plates shall not be less than 1/50 of the distance

between the lines of connectors: (1/50) 8.50 = 0.17” (Use 3/16”)

• Minimum width of tie plates shall be the width between lines of connectors plus

the necessary edge distance on each side to keep the bolts from splitting the

plate (ref. AISC Specification, Table J3.4): 8.50 + 2 (1.50) = 11.5” (Use 12”)

• Maximum preferable spacing of the plates (based on the least radius of

gyration of one channel section): rmin = ry = 0.762”

Maximum preferable L/r = 300

L = 300 r = 300 (0.762) = 228.6” (19.05’)

Use 15’, placing tie plates at the ends and mid-point.

4.47

Answer: 6” x 12” x 3/16” tie plates located at the ends and at the mid-point of

the 30’-long member. The actual center-to-center spacing is 14’-9”.

ASD


ASD: Pa = D + L = 120 + 240 = 360 kips

Checking



Pn = Fy Ag = 36 (2) (8.81) = 634.3 kips

Compute the ASD allowable tensile strength of the section.




Net area of the channels: An = 2(8.81) – 4(7/8 + 1/8) (0.510) = 15.58 in2

• Using Table D3.1 (Case 2), determine the value for U.

x = 0.674” (for the channel)


U = 1 - x /L = 1 – 0.674/6 = 0.888



Pn = Fu Ae = 58(13.84) = 802.7 kips

Compute the ASD allowable tensile strength of the section.



Ix = 2 (162.0) = 324.0 in4

Iy = 2[5.12 + (8.81)(6.0 – 0.674)2] = 2(5.12 + 249.9) = 510.0 in4

A = 2(8.81) = 17.62 in2

rx = (Ix/A)1/2 = (324.0/17.62)1/2 = 4.29” (controls)

ry = (Iy/A)1/2 = (510.0/17.62)1/2 = 5.38”

Lx/rx = 30 (12)/4.29 = 83.9 < 300 OK

Design the tie plates (AISC Specification D4).

• Distance between line of bolts = 12.0 – 2 (1.75) = 8.50”

4.48

• Minimum length of tie plates shall not be less than 2/3 of the distance between

the lines of connectors: (2/3) 8.50 = 5.67” (Use 6”)

• Minimum thickness of tie plates shall not be less than 1/50 of the distance

between the lines of connectors: (1/50) 8.50 = 0.17” (Use 3/16”)

• Minimum width of tie plates shall be the width between lines of connectors plus

the necessary edge distance on each side to keep the bolts from splitting the

plate (ref. AISC Specification, Table J3.4): 8.50 + 2 (1.50) = 11.5” (Use 12”)

• Maximum preferable spacing of the plates (based on the least radius of

gyration of one channel section): rmin = ry = 0.762”

Maximum preferable L/r = 300

L = 300 r = 300 (0.762) = 228.6” (19.05’)

Use 15’, placing tie plates at the ends and mid-point.

Answer: 6” x 12” x 3/16” tie plates located at the ends and at the mid-point of

the 30’-long member. The actual center-to-center spacing is 14’-9”.

4.49

4.11 Truss Members

The most common tension members found in building structures are the tension

web and chord members of trusses.

• Depending on the particular load patterns that a truss might experience, a

truss member may need to resist tension in some cases and compression in

other cases.

- Because the compression strength of a member is normally significantly less

than the tension strength of the same member, compression may control the

design.

The typical truss member may consist of single shapes or a combination of shapes.

• When composed of a single shape, truss members are designed in the same way

as other tension members.

• When composed of a combination of shapes, truss members are designed in

accordance with the requirements for built-up tension members.

4.50

Example – Tension Member

Example 4.15 (p. 106 of the textbook)

Given: Tension member used as a brace in an

X-braced frame.

Service Load: W = 120 kips


Find: Design the tension members using a

pair of equal leg angles or a threaded rod

of A36 steel.

Solution

LRFD

Determine the critical factored load in the member AC using a joint analysis at

Joint C.

• Assume that member BD is ineffective since it is in compression.

Critical load combination: Pu = 0.9D + 1.0W = 0 + 1.0 (120) = 120.0 kips

Determine the force in member AC using a joint analysis at Joint C.

∑Fx = 0 = 120.0 – 3/ 13 ACu

ACu = ( 13 /3) 120 = 144.2 kips


min Ag = ACu/φt Fy = 144.2/[0.9 (36)] = 4.45 in2

From AISC Table 1-15 (“Double Angles”): Select 2 L2½ x 2½ x ½ (A = 4.50 in2)

Compute the minimum effective area Ae required for tensile rupture in the net

section (φt = 0.75).

min Ae = Pu/φt Fu = 144.2/[0.75(58)] = 3.31 in2

• The combination of holes and shear lag may not result in an effective area less

than 3.31 in2 (or 73.6% of the gross area; i.e. Ae/Ag = 3.31/4.50 = 0.736).

Determine the slenderness ratio for the selected double angles and compare with

the recommended maximum of 300.

AISC Table 1-15: rmin = rx = 0.735”

L/rx = 72.1 (12)/0.735 = 1177 > 300 NG

4.51

Determine rmin required to satisfy the slenderness limit of 300.

rmin = L/300 = 72.1 (12)/300 = 2.88”

A check of AISC Table 1-15 shows that no pair of double angles will satisfy the

slenderness limit of 300.

Select a standard threaded rod to meet the strength requirements.

• The maximum slenderness limit does not apply to rods.

• Determine the required gross area Ab (φ = 0.75).

Ab ≥ Pu/(φ 0.75 Fu) = 144.2/[0.75 (0.75) 58] = 4.42 in2

Using AISC Table 7-17: Select 2½” diameter threaded rod (Ag = 4.91 in2).

Note: The author considers the limit state of yielding as the basis of determining the required area

rather than using the provisions of Specification Section J3.6 directly. The provisions of Section

J3.6 consider only the limit state of tensile rupture as shown above.

ASD

Determine the critical load in the member AC using a joint analysis at Joint C.

• Assume that member BD is ineffective since it is in compression.

Critical load combination: Pa = D + 0.6W = 0 + 0.6 (120) = 72.0 kips

Determine the force in member AC using a joint analysis at Joint C.

∑Fx = 0 = 72.0 – 3/ 13 ACa

ACa = ( 13 /3) 72.0 = 86.5 kips

Compute the minimum Ag required for tensile yield in the gross section (Ω t = 1.67).

min Ag = Ωt ACa/Fy = 1.67(86.5)/36 = 4.01 in2

AISC Table 1-15 (“Double Angles”): Select 2 L3½ x 3½ x 5/16 (A = 4.18 in2)

Compute the minimum effective area Ae required for tensile rupture in the net

section (Ωt = 2.00).

min Ae = Ωt Pa/Fu = 2.00 (86.5)/58 = 2.98 in2

• The combination of holes and shear lag may not result in an effective area less

than 2.98 in2 (or 71.3% of the gross area; i.e. Ae/Ag = 2.98/4.18 = 0.713).

Determine the slenderness ratio for the selected double angles and compare with

the recommended maximum of 300.

AISC Table 1-15: rmin = rx = 1.08”

L/rx = 72.1 (12)/1.08 = 801 > 300 NG

4.52

Determine rmin required to satisfy the slenderness limit of 300.

rmin = L/300 = 72.1 (12)/300 = 2.88”

A check of AISC Table 1-15 shows that no pair of double angles will satisfy the

slenderness limit of 300.

Select a standard threaded rod to meet the strength requirements.

• The maximum slenderness limit does not apply to rods.

• Determine the required gross area Ab (φ = 0.75).

Ab ≥ Ω Pa/0.75 Fu = 2.00(86.5)/(0.75) 58 = 3.98 in2

Using AISC Table 7-17: Select 2¼” diameter threaded rod (Ag = 3.98 in2).

Note: The author considers the limit state of yielding as the basis of determining the required area

rather than using the provisions of Specification Section J3.6 directly. The provisions of Section

J3.6 consider only the limit state of tensile rupture as shown above.

chapter 4: tension members - steel design 4300:401 chap04 - tension... · chapter 4: tension...

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