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
Compute the nominal tensile strength of the section.
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”)
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 (4.38) = 157.7 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
Compute the net area.
An = 4.38 – (3/8) [1 (7/8 + 1/8)] = 4.00 in2
Compute the effective area.
• 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”
L = 3 spaces (3”/space) = 9”
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
Effective area: Ae = U An = 0.820 (4.00) = 3.28 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58 (3.28) = 190.2 kips
4.14
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Answers: LRFD = 141.9 kips and ASD = 94.4 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
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 50 ksi (1” x 6”) = 300 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
LRFD (φt = 0.90): φt Pn = 0.90 (300) = 270.0 kips
ASD (Ωt = 1.67): Pn/Ωt = 300/1.67 = 179.6 kips
Tensile rupture strength in the net section
Compute the effective area.
• 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
Effective area: Ae = U An = 0.75 (6.00) = 4.50 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 65 (4.50) = 292.5 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Answers: LRFD = 219.4 kips and ASD = 146.2 kips
Tensile rupture strength in the net section controls.
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.
Steel: Fy = 36 ksi, Fu = 58 ksi
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”)
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 ksi (9.99) = 359.6 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
Compute the effective area.
• 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
Effective area: Ae = U An = 0.740 (9.99) = 7.39 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58 (7.39) = 428.6 kips
4.18
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Answers: LRFD = 321.5 kips and ASD = 214.3 kips
Tensile rupture strength in the net section controls.
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
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 [2 (3/8 x 12)] = 324.0 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
Compute the effective area.
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
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58 (7.65) = 443.7 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Answers: LRFD = 291.6 kips and ASD = 194.0 kips
Tensile yield strength in the gross section controls.
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):
Pn = Fy Ag AISC Equation D2-1
Pu = φt Fy Ag
Tensile rupture strength in the net section (φt = 0.75):
Pn = Fu Ae AISC Equation D2-2
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):
Pn = Fy Ag AISC Equation D2-1
Pa = Pn/Ωt = Fy Ag/Ωt
Tensile rupture strength in the net section (Ωt = 2.00):
Pn = Fu Ae AISC Equation D2-2
Pa = Pn/Ωt = Fu Ae/Ωt
Block shear strength
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)
Steel: A992 (Fy = 50 ksi, Fu = 65 ksi)
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
Tensile yield strength in the gross section (φt = 0.90):
Pn = Fy Ag = 50 (10.3) = 515.0 kips
φt Pn = 0.90 (515.0) = 463.5 kips > Pu = 332 kips OK
Tensile rupture strength in the net section (φt = 0.75):
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”
L = 2 spaces (4”/space) = 8”
U = 1 - x /L = 1 – 1.30/8 = 0.838
Using AISC Table D3.1, Case 7:
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
φt Pn = 0.75 (454.4) = 340.8 kips > Pu = 332 kips OK
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 U = 0.85 (AISC Table D3.1, Case 7).
Assume that the flange thickness is about 0.440” after looking at W12
sections in the Steel Manual that have areas of 8.02 in2 or more.
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
Determine the preferable minimum r.
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
Tensile yield strength in the gross section (Ωt = 1.67):
Pn = Fy Ag = 50 (11.7) = 585.0 kips
Pn/Ωt = 585.0/1.67 = 350.3 kips > Pa = 240 kips OK
Tensile rupture strength in the net section (Ωt = 2.00):
Using AISC Table D3.1, Case 2:
x (listed as y ) for half W12 x 40 (i.e. WT 6 x 20) = 1.09”
L = 2 spaces (4”/space) = 8”
U = 1 - x /L = 1 – 1.09/8 = 0.864
Using AISC Table D3.1, Case 7:
bf = 8.01” > (2/3) d = (2/3) 11.9 = 7.93” and U = 0.90
Use U = 0.90 (the larger value).
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
Pn/Ωt = 564.2/2.00 = 282.1 kips > Pa = 240 kips OK
Check the slenderness ratio.
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).
Steel: A36 (Fy = 36 ksi, Fu = 58 ksi)
Service loads: PD = 50 kips, PL = 100 kips
Find: Select a single angle section to support the given loads. Neglect block shear.
Solution
LRFD
Determine the critical factored load.
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)
Compute the minimum Ag required for tensile yield in the gross section (φt = 0.90).
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 U = 0.80 (AISC Table D3.1, Case 8).
Assume that the angle thickness is about 0.750” after looking at single angle
sections in the Steel Manual that have areas of 6.79 in2 or more.
min Ag = 220/[0.75 (58) (0.80)] + 1 (7/8 + 1/8) (0.750)
= 6.32 + 0.75 = 7.07 in2 (controls)
Determine the preferable minimum r.
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
Tensile yield strength in the gross section (φt = 0.90):
Pn = Fy Ag = 36 (7.16) = 257.8 kips
φt Pn = 0.90 (257.8) = 232.0 kips > Pu = 220 kips OK
Tensile rupture strength in the net section (φt = 0.75):
Using AISC Table D3.1, Case 2:
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
Use U = 0.914 (the larger value).
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
φt Pn = 0.75 (346.8) = 260.1 kips > Pu = 220 kips OK
Check the slenderness ratio.
Lz/rz = 20 (12)/0.856 = 280.4 < 300 OK
ASD
Determine the critical load combination.
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 U = 0.80 (AISC Table D3.1, Case 8).
Assume that the angle thickness is about 0.875” after looking at single angle
sections in the Steel Manual that have areas of 6.96 in2 or more.
min Ag = 2.00(150)/[58(0.80)] + 1 (7/8 + 1/8) (0.875)
= 6.47 + 0.875 = 7.35 in2
4.29
Determine the preferable minimum r.
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
Tensile yield strength in the gross section (Ωt = 1.67):
Pn = Fy Ag = 36 (7.61) = 274.0 kips
Pn/Ωt = 274.0/1.67 = 164.1 kips > Pa = 150 kips OK
Tensile rupture strength in the net section (Ωt = 2.00):
Using AISC Table D3.1, Case 2:
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
Use U = 0.858 (the larger value).
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
Pn/Ωt = 350.9/2.00 = 175.5 kips > Pa = 150 kips OK
Check the slenderness ratio.
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”)
Block shear strength
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
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 50 (4.75) = 237.5 kips
4.33
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
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”
L = 2 spaces (4”/space) = 8”
U = 1 - x /L = 1 – 0.981/8 = 0.877
Effective area: Ae = U An = 0.877 (4.31) = 3.78 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 65 (3.78) = 245.7 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Answers: LRFD = 162.7 kips and ASD = 108.5 kips
Block shear controls.
4.34
Example
Given: The welded plate tension
members shown.
Steel: A36 (Fy = 36 ksi, Fu = 58 ksi)
Find: LRFD design strength and the
ASD allowable design strength of
the plates.
Solution
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 [0.5 (10.0)] = 180.0 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
Compute the effective area.
• Using AISC Table D3.1 (Case 1), determine the value for U.
U = 1.0
Effective area: Ae = U An = 1.0 (0.5 x 10) = 5.00 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58 (5.00) = 290.0 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Block shear strength
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
Compute the block shear rupture design strength.
Rn = 0.6 Fu Anv + Ubs Fu Ant = 0.6 (58) (4.00) + 1.0 (58) (5.00) = 429.2 kips
≤ 0.6 Fy Agv + Ubs Fu Ant = 0.6 (36) (4.00) + 1.0 (58) (5.00) = 376.4 kips
Rn = 376.4 kips
Compute the LRFD block shear strength and ASD allowable block shear design
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
Answers: LRFD = 162.0 kips and ASD = 107.8 kips
Tensile yield strength in the gross section controls.
4.36
Example
Given: W12 x 30 tension member shown.
Steel: Fy = 50 ksi, Fu = 65 ksi
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”)
Tensile yield strength in the gross section
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 50 (8.79) = 439.5 kips
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Tensile rupture strength in the net section
Compute the effective area.
Net area of the wide flange: An = 8.79 – 0.440 [4 (7/8 + 1/8) ] = 7.03 in2
• Using AISC Table D3.1 (Case 2), determine the value for U.
x (listed as y ) = 1.27” (for WT6 x 15)
L = 2 spaces (4”/space) = 8”
U = 1 - x /L = 1 – 1.27/8 = 0.841
• Using AISC Table D3.1 (Case 7), determine the value for U.
(2/3) d = (2/3) (12.3) = 8.2” > bf = 6.52” and U = 0.85
Thus, U = 0.85
Effective area: Ae = U An = 0.85 (7.03) = 5.98 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 65 (5.98) = 388.7 kips
4.37
Compute the LRFD tensile strength and ASD allowable tensile design strength.
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
Compute the block shear rupture design strength.
Rn = 0.6 Fu Anv + Ubs Fu Ant = 0.6 (65) (13.20) + 1.0 (65) (1.78) = 630.5 kips
≤ 0.6 Fy Agv + Ubs Fu Ant = 0.6 (50) (17.60) + 1.0 (65) (1.78) = 643.7 kips
Rn = 630.5 kips
Compute the LRFD block shear strength and ASD allowable block shear design
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
Answers: LRFD = 291.5 kips and ASD = 194.3 kips
Tensile rupture strength in the net section controls.
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.
Pn = Fy Ag AISC Equation D2-1
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.
Steel: A36 (Fy = 36 ksi, Fu = 58 ksi)
Service loads: PD = 10 kips, PL = 20 kips
Find: Select a standard threaded rod to support the specified load using the AISC
Specification.
Solution
LRFD
Determine the critical factored load.
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
Determine the critical load combination.
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
Determine the critical factored load.
Pu = 1.2 D + 1.6 L = 1.2 (120) + 1.6 (240) = 528 kips
Checking
Tensile yield strength in the gross section (φt = 0.90):
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 (2) (8.81) = 634.3 kips
Compute the LRFD design tensile strength of the section.
φt Pn = 0.90 (634.3) = 570.9 kips > Pu = 528 kips OK
4.46
Tensile rupture strength in the net section (φt = 0.75):
Compute the effective area.
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)
L = 2 spaces (3”/space) = 6”
U = 1 - x /L = 1 – 0.674/6 = 0.888
Effective area: Ae = U An = 0.888 (15.58) = 13.84 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58(13.84) = 802.7 kips
Compute the LRFD design tensile strength of the section.
φt Pn = 0.75 (802.7) = 602.0 kips > Pu = 528 kips OK
Check the slenderness ratio.
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
Determine the critical load combination.
ASD: Pa = D + L = 120 + 240 = 360 kips
Checking
Tensile yield strength in the gross section (Ωt = 1.67):
Compute the nominal tensile strength of the section.
Pn = Fy Ag = 36 (2) (8.81) = 634.3 kips
Compute the ASD allowable tensile strength of the section.
Pn/Ωt = 634.3/1.67 = 379.8 kips > Pa = 360 kips OK
Tensile rupture strength in the net section (Ωt = 2.00):
Compute the effective area.
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)
L = 2 spaces (3”/space) = 6”
U = 1 - x /L = 1 – 0.674/6 = 0.888
Effective area: Ae = U An = 0.888 (15.58) = 13.84 in2
Compute the nominal tensile strength of the section.
Pn = Fu Ae = 58(13.84) = 802.7 kips
Compute the ASD allowable tensile strength of the section.
Pn/Ωt = 802.7/2.00 = 401.4 kips > Pa = 360 kips OK
Check the slenderness ratio.
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
Steel: A36 (Fy = 36 ksi, Fu = 58 ksi)
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
Compute the minimum Ag required for tensile yield in the gross section (φt = 0.90).
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.