CE 405: Design of Steel Structures – Prof. Dr. A.

Varma Tension Member Design

Chapter 4. TENSION MEMBER DESIGN

4.1 INTRODUCTORY CONCEPTS

• Stress: The stress in an axially loaded tension member is given by Equation (4.1)

P
f = (4.1)
A

where, P is the magnitude of load, and

A is the cross-sectional area normal to the load

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

- near the point of application of load, and

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

• For example, consider an 8 x ½ in. bar connected to a gusset plate and loaded in tension as

shown below in Figure 4.1

Gusset plate
b Section b-b
b
7/8 in. diameter hole

a a
Section a-a
8 x ½ in. bar

Figure 4.1 Example of tension member.

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

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

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Therefore, by definition (Equation 4.1) the reduced area of section b – b will be subjected to

higher stresses

• However, the reduced area and therefore the higher stresses will be localized around section

b – b.

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

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

4.2 STEEL STRESS-STRAIN BEHAVIOR

• The stress-strain behavior of steel is shown below in Figure 4.2

Fu

Fy
Stress, f

εy εu
Strain, ε

Figure 4.2 Stress-strain behavior of steel

• In Figure 4.2, E is the elastic modulus = 29000 ksi.

Fy is the yield stress and Fu is the ultimate stress

εy is the yield strain and εu is the ultimate strain

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Deformations are caused by the strain ε. Figure 4.2 indicates that the structural deflections

will be small as long as the material is elastic (f < Fy)

• Deformations due to the strain ε will be large after the steel reaches its yield stress Fy.

4.3 DESIGN STRENGTH

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

(1) excessive deformation; or (2) fracture

• Excessive deformation can occur due to the yielding of the gross section (for example section

a-a from Figure 4.1) along the length of the member

• Fracture of the net section can occur if the stress at the net section (for example section b-b in

Figure 4.1) reaches the ultimate stress Fu.

• The objective of design is to prevent these failure before reaching the ultimate loads on the

structure (Obvious).

• This is also the load and resistance factor design approach recommended by AISC for

designing steel structures

4.3.1 Load and Resistance Factor Design

The load and resistance factor design approach is recommended by AISC for designing steel

structures. It can be understood as follows:

Step I. Determine the ultimate loads acting on the structure

- The values of D, L, W, etc. given by ASCE 7-98 are nominal loads (not maximum or

ultimate)

- During its design life, a structure can be subjected to some maximum or ultimate loads

caused by combinations of D, L, or W loading.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- The ultimate load on the structure can be calculated using factored load combinations,

which are given by ASCE and AISC (see pages 2-10 and 2-11 of AISC manual). The

most relevant of these load combinations are given below:

1.4 D (4.2 – 1)
1.2 D + 1.6 L + 0.5 (Lr or S) (4.2 – 2)
1.2 D + 1.6 (Lr or S) + (0.5 L or 0.8 W) (4.2 – 3)
1.2 D + 1.6 W + 0.5 L + 0.5 (Lr or S) (4.2 – 4)
0.9 D + 1.6 W (4.2 – 5)

Step II. Conduct linear elastic structural analysis

- Determine the design forces (Pu, Vu, and Mu) for each structural member

Step III. Design the members

- The failure (design) strength of the designed member must be greater than the

corresponding design forces calculated in Step II. See Equation (4.3) below:

φ Rn > ∑ γ i Qi (4.3)

- Where, Rn is the calculated failure strength of the member

- φ is the resistance factor used to account for the reliability of the material behavior and

equations for Rn

- Qi is the nominal load

- γi is the load factor used to account for the variability in loading and to estimate the

ultimate loading condition.

4.3.2 Design Strength of Tension Members

• Yielding of the gross section will occur when the stress f reaches Fy.

P
f = = Fy
Ag

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Therefore, nominal yield strength = Pn = Ag Fy (4.4)

Factored yield strength = φt Pn (4.5)

where, φt = 0.9 for tension yielding limit state

• See the AISC manual, section on specifications, Chapter D (page 16.1 –24)

• Facture of the net section will occur after the stress on the net section area reaches the

ultimate stress Fu

P
f = = Fu
Ae

Therefore, nominal fracture strength = Pn = Ae Fu

Where, Ae is the effective net area, which may be equal to the net area or smaller.

The topic of Ae will be addressed later.

Factored fracture strength = φt Ae Fu (4.6)

Where, φt = 0.75 for tension fracture limit state (See page 16.1-24 of AISC manual)

4.3.3 Important notes

• Note 1. Why is fracture (& not yielding) the relevant limit state at the net section?

Yielding will occur first in the net section. However, the deformations induced by yielding

will be localized around the net section. These localized deformations will not cause

excessive deformations in the complete tension member. Hence, yielding at the net section

will not be a failure limit state.

• Note 2. Why is the resistance factor (φt) smaller for fracture than for yielding?

The smaller resistance factor for fracture (φt = 0.75 as compared to φt = 0.90 for yielding)

reflects the more serious nature and consequences of reaching the fracture limit state.

• Note 3. What is the design strength of the tension member?

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

The design strength of the tension member will be the lesser value of the strength for the two

limit states (gross section yielding and net section fracture).

• Note 4. Where are the Fy and Fu values for different steel materials?

The yield and ultimate stress values for different steel materials are noted in Table 2 in the

AISC manual on pages 16.1–141 and 16.1–142.

• Note 5. What are the most common steels for structural members?

See Table 2-1 in the AISC manual on pages 2–24 and 2-25. According to this Table: the

preferred material for W shapes is A992 (Fy = 50 ksi; Fu = 65 ksi); the preferred material for

C, L , M and S shapes is A36 (Fy = 36 ksi; Fu = 58 ksi). All these shapes are also available in

A572 Gr. 50 (Fy = 50 ksi; Fu = 65 ksi).

• Note 6. What is the amount of area to be deducted from the gross area to account for the

presence of bolt-holes?

- The nominal diameter of the hole (dh) is equal to the bolt diameter (db) + 1/16 in.

- However, the bolt-hole fabrication process damages additional material around the hole

diameter.

- Assume that the material damage extends 1/16 in. around the hole diameter.

- Therefore, for calculating the net section area, assume that the gross area is reduced by a

hole diameter equal to the nominal hole-diameter + 1/16 in.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Example 3.1 A 5 x ½ bar of A572 Gr. 50 steel is used as a tension member. It is connected to a

gusset plate with six 7/8 in. diameter bolts as shown in below. Assume that the effective net area

Ae equals the actual net area An and compute the tensile design strength of the member.

Gusset plate

b b
7/8 in. diameter bolt

a a

5 x ½ in. bar
A572 Gr. 50

Solution

• Gross section area = Ag = 5 x ½ = 2.5 in2

• Net section area (An)

- Bolt diameter = db = 7/8 in.

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

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

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

• Gross yielding design strength = φt Pn = φt Fy Ag

- Gross yielding design strength = 0.9 x 50 ksi x 2.5 in2 = 112.5 kips

• Fracture design strength = φt Pn = φt Fu Ae

- Assume Ae = An (only for this problem)

- Fracture design strength = 0.75 x 65 ksi x 1.5 in2 = 73.125 kips

• Design strength of the member in tension = smaller of 73.125 kips and 112.5 kips

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

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

Example 3.2 A single angle tension member, L 4 x 4 x 3/8 in. made from A36 steel is connected

to a gusset plate with 5/8 in. diameter bolts, as shown in Figure below. The service loads are 35

kips dead load and 15 kips live load. Determine the adequacy of this member using AISC

specification. Assume that the effective net area is 85% of the computed net area. (Calculating

the effective net area will be taught in the next section).

• Gross area of angle = Ag = 2.86 in2 (from Table 1-7 on page 1-36 of AISC)

L 4 x 4 x 3/ 8
d b = 5/8 in.

Section a-a
a
Gusset plate

• Net section area = An

- Bolt diameter = 5/8 in.

- Nominal hole diameter = 5/8 + 1/16 = 11/16 in.

- Hole diameter for calculating net area = 11/16 + 1/16 = 3/4 in.

- Net section area = Ag – (3/4) x 3/8 = 2.86 – 3/4 x 3/8 = 2.579 in2

• Effective net area = Ae = 0.85 x 2.579 in2 = 2.192 in2

• Gross yielding design strength = φt Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips

• Net section fracture = φt Ae Fu = 0.75 x 2.192 in2 x 58 ksi = 95.352 kips

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Design strength = 92.664 kips (gross yielding governs)

• Ultimate (design) load acting for the tension member = Pu

- The ultimate (design) load can be calculated using factored load combinations given on

page 2-11 of the AISC manual, or Equations (4.2-1 to 4.2-5) of notes (see pg. 4)

- According to these equations, two loading combinations are important for this problem.

These are: (1) 1.4 D; and (2) 1.2 D + 1.6 L

- The corresponding ultimate (design) loads are:

1.4 x (PD) = 1.4 (35) = 49 kips

1.2 (PD) + 1.6 (PL) = 66 kips (controls)

- The ultimate design load for the member is 66 kips, where the factored dead + live

loading condition controls.

• Compare the design strength with the ultimate design load

- The design strength of the member (92.664 kips) is greater than the ultimate design load

(66 kips).

- φt Pn (92.664 kips) > Pu (66 kips)

• The L 4 x 4 x 3/8 in. made from A36 steel is adequate for carrying the factored loads.

4.4 EFFECTIVE NET AREA

• The connection has a significant influence on the performance of a tension member. A

connection almost always weakens the member, and a measure of its influence is called joint

efficiency.

• Joint efficiency is a function of: (a) material ductility; (b) fastener spacing; (c) stress

concentration at holes; (d) fabrication procedure; and (e) shear lag.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• All factors contribute to reducing the effectiveness but shear lag is the most important.

• Shear lag occurs when the tension force is not transferred simultaneously to all elements of

the cross-section. This will occur when some elements of the cross-section are not connected.

• For example, see Figure 4.3 below, where only one leg of an angle is bolted to the gusset

plate.

Figure 4.3 Single angle with bolted connection to only one leg.

• A consequence of this partial connection is that the connected element becomes overloaded

and the unconnected part is not fully stressed.

• Lengthening the connection region will reduce this effect

• Research indicates that shear lag can be accounted for by using a reduced or effective net

area Ae

• Shear lag affects both bolted and welded connections. Therefore, the effective net area

concept applied to both types of connections.

- For bolted connection, the effective net area is Ae = U An

- For welded connection, the effective net area is Ae = U Ag

• Where, the reduction factor U is given by:

x
U = 1- ≤ 0.9 (4.7)
L

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- Where, x is the distance from the centroid of the connected area to the plane of the

connection, and L is the length of the connection.

- If the member has two symmetrically located planes of connection, x is measured

from the centroid of the nearest one – half of the area.

- Additional approaches for calculating x for different connection types are shown in

the AISC manual on page 16.1-178.

- The distance L is defined as the length of the connection in the direction of load.

- For bolted connections, L is measured from the center of the bolt at one end to the

center of the bolt at the other end.

- For welded connections, it is measured from one end of the connection to other.

- If there are weld segments of different length in the direction of load, L is the length

of the longest segment.

- Example pictures for calculating L are given on page 16.1-179 of AISC.

• The AISC manual also gives values of U that can be used instead of calculating x /L.

- They are based on average values of x /L for various bolted connections.

- For W, M, and S shapes with width-to-depth ratio of at least 2/3 and for Tee shapes cut

from them, if the connection is through the flanges with at least three fasteners per line in

the direction of applied load ………………………………………………...U = 0.90

- For all other shapes with at least three fasteners per line …………………... U = 0.85

- For all members with only two fasteners per line …………………………… U = 0.75

- For better idea, see Figure 3.8 on page 41 of the Segui text-book.

- These values are acceptable but not the best estimate of U

- If used in the exam or homeworks, full points for calculating U will not be given

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Example 3.3 Determine the effective net area and the corresponding design strength for the

single angle tension member of Example 3.2. The tension member is an L 4 x 4 x 3/8 in. made

from A36 steel. It is connected to a gusset plate with 5/8 in. diameter bolts, as shown in Figure

below. The spacing between the bolts is 3 in. center-to-center.

- Compare your results with those obtained for Example 3.2.

x
L 4 x 4 x 3/ 8
d b = 5/8 in.

a
Gusset plate L 4 x 4 x 3/ 8

• Gross area of angle = Ag = 2.86 in2 (from Table 1-7 on page 1-36 of AISC)

• Net section area = An

- Bolt diameter = 5/8 in.

- Hole diameter for calculating net area = 11/16 + 1/16 = 3/4 in.

- Net section area = Ag – (3/4) x 3/8 = 2.86 – 3/4 x 3/8 = 2.579 in2

• x is the distance from the centroid of the area connected to the plane of connection

- For this case x is equal to the distance of centroid of the angle from the edge.

- This value is given in the Table 1-7 on page 1-36 of the AISC manual.

- x = 1.13 in.

• L is the length of the connection, which for this case will be equal to 2 x 3.0 in.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- L = 6.0 in.

x 1.13
• U = 1- = 1- = 0.8116 in.
L 6.0

• Effective net area = Ae = 0.8116 x 2.579 in2 = 2.093 in2

• Gross yielding design strength = φt Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips

• Net section fracture = φt Ae Fu = 0.75 x 2.093 in2 x 58 ksi = 91.045 kips

• Design strength = 91.045 kips (net section fracture governs)

• In Example 3.2

- Factored load = Pu = 66.0 kips

- Design strength = φt Pn = 92.66 kips (gross section yielding governs)

- Net section fracture strength = φt Pn = 95.352 kips (assuming Ae = 0.85)

• Comparing Examples 3.2 and 3.3

- Calculated value of U (0.8166) is less than the assumed value (0.85)

- The assumed value was unconservative.

- It is preferred that the U value be specifically calculated for the section.

- After including the calculated value of U, net section fracture governs the design

strength, but the member is still adequate from a design standpoint.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Example 3.4 Determine the design strength of an ASTM A992 W8 x 24 with four lines if ¾ in.

diameter bolts in standard holes, two per flange, as shown in the Figure below.

Assume the holes are located at the member end and the connection length is 9.0 in. Also

calculate at what length this tension member would cease to satisfy the slenderness limitation in

LRFD specification B7

¾ in. diameter bolts

W 8 x 24

3 in. 3 in. 3 in.

Holes in beam flange

Solution:

• For ASTM A992 material: Fy = 50 ksi; and Fu = 65 ksi

• For the W8 x 24 section:

- Ag = 7.08 in2 d = 7.93 in.

- tw = 0.285 in. bf = 6.5 in.
- tf = 0.4 in. ry = 1.61 in.
• Gross yielding design strength = φt Pn = φt Ag Fy = 0.90 x 7.08 in2 x 50 ksi = 319 kips

• Net section fracture strength = φt Pn = φt Ae Fu = 0.75 x Ae x 65 ksi

- Ae = U An - for bolted connection

- An = Ag – (no. of holes) x (diameter of hole) x (thickness of flange)

An = 7.08 – 4 x (diameter of bolt + 1/8 in.) x 0.4 in.

An = 5.68 in2

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

x
- U =1- ≤ 0.90
L

- What is x for this situation?

x is the distance from the edge of the flange to the centroid of the half (T) section

tf d −2tf d + 2t f
(b f × t f ) × +( × tw ) × ( )
2 2 4 6.5 × 0.4 × 0.2 + 3.565 × 0.285 × 2.1825
x= = = 0.76
d 6.5 × 0.4 + 3.565 × 0.285
bf × t f + × tw
2

- x can be obtained from the dimension tables for Tee section WT 4 x 12. See page 1-50

and 1-51 of the AISC manual:

x = 0.695 in.

- The calculated value is not accurate due to the deviations in the geometry

x 0.695
- U = 1- = 1- = 0.923
L 9 .0

- But, U ≤ 0.90. Therefore, assume U = 0.90

• Net section fracture strength = φt Ae Fu = 0.75 x 0.9 x 5.68 x 65 = 249.2 kips

• The design strength of the member is controlled by net section fracture = 249.2 kips

• According to LRFD specification B7, the maximum unsupported length of the member is

limited to 300 ry = 300 x 1.61 in. = 543 in. = 40.3 ft.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

4.4.1 Special cases for welded connections

• If some elements of the cross-section are not connected, then Ae will be less than An

- For a rectangular bar or plate Ae will be equal to An

- However, if the connection is by longitudinal welds at the ends as shown in the figure

below, then Ae = UAg

Where, U = 1.0 for L ≥ w

U = 0.87 for 1.5 w ≤ L < 2 w
U = 0.75 for w ≤ L < 1.5 w
L = length of the pair of welds ≥ w
w = distance between the welds or width of plate/bar

• AISC Specification B3 gives another special case for welded connections.

For any member connected by transverse welds alone,

Ae = area of the connected element of the cross-section

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Example 3.5 Consider the welded single angle L 6x 6 x ½ tension member made from A36 steel

shown below. Calculate the tension design strength.

Solution

• Ag = 5.00 in2

• An = 5.00 in2 - because it is a welded connection

x
• Ae = U An - where, U = 1 -
L

- x = 1.68 in. for this welded connection

- L = 6.0 in. for this welded connection

1.168
- U = 1- = 0.72
6.0

• Gross yielding design strength = φt Fy Ag = 0.9 x 36 x 5.00 = 162 kips

• Net section fracture strength = φt Fu Ae = 0.75 x 58 x 0.72 x 5.00 = 156.6 kips

• Design strength = 156.6 kips (net section fracture governs)

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

4.5 STAGGERED BOLTS

For a bolted tension member, the connecting bolts can be staggered for several reasons:

(1) To get more capacity by increasing the effective net area

(2) To achieve a smaller connection length

(3) To fit the geometry of the tension connection itself.

• For a tension member with staggered bolt holes (see example figure above), the relationship f

= P/A does not apply and the stresses are a combination of tensile and shearing stresses on

the inclined portion b-c.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Net section fracture can occur along any zig-zag or straight line. For example, fracture can

occur along the inclined path a-b-c-d in the figure above. However, all possibilities must be

examined.

• Empirical methods have been developed to calculate the net section fracture strength

According to AISC Specification B2

s2
- net width = gross width - ∑ d + ∑
4g

- where, d is the diameter of hole to be deducted (dh + 1/16, or db + 1/8)

- s2/4g is added for each gage space in the chain being considered

- s is the longitudinal spacing (pitch) of the bolt holes in the direction of loading

- g is the transverse spacing (gage) of the bolt holes perpendicular to loading dir.

- net area (An) = net width x plate thickness

- effective net area (Ae) = U An where U = 1- x /L

- net fracture design strength = φt Ae Fu (φt = 0.75)

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

EXAMPLE 3.6 Compute the smallest net area for the plate shown below: The holes are for 1 in.

diameter bolts.

i a
3 in.
j b

5 in.

5 in.
d
3 in. f

h e

3 in. 3 in. 3 in. 3 in. 3 in. 3 in.

• The effective hole diameter is 1 + 1/8 = 1.125 in.

• For line a-b-d-e

wn = 16.0 – 2 (1.125) = 13.75 in.

• For line a-b-c-d-e

wn = 16.0 – 3 (1.125) + 2 x 32/ (4 x 5) = 13.52 in.

• The line a-b-c-d-e governs:

• An = t wn = 0.75 (13.52) = 10.14 in2

Note

• Each fastener resists an equal share of the load

• Therefore different potential failure lines may be subjected to different loads.

• For example, line a-b-c-d-e must resist the full load, whereas i-j-f-h will be subjected to 8/11

of the applied load. The reason is that 3/11 of the load is transferred from the member before

i-j-f-h received any load.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Staggered bolts in angles. If staggered lines of bolts are present in both legs of an angle,

then the net area is found by first unfolding the angle to obtain an equivalent plate. This plate

is then analyzed like shown above.

- The unfolding is done at the middle surface to obtain a plate with gross width equal to the

sum of the leg lengths minus the angle thickness.

- AISC Specification B2 says that any gage line crossing the heel of the angle should be

reduced by an amount equal to the angle thickness.

- See Figure below. For this situation, the distance g will be = 3 + 2 – ½ in.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

4.6 BLOCK SHEAR

• For some connection configurations, the tension member can fail due to ‘tear-out’ of material

at the connected end. This is called block shear.

• For example, the single angle tension member connected as shown in the Figure below is

susceptible to the phenomenon of block shear.

(a)

(b)

Shear failure
(c)

Tension failure

Figure 4.4 Block shear failure of single angle tension member

• For the case shown above, shear failure will occur along the longitudinal section a-b and

tension failure will occur along the transverse section b-c

• AISC Specification (SPEC) Chapter D on tension members does not cover block shear

failure explicitly. But, it directs the engineer to the Specification Section J4.3

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Block shear strength is determined as the sum of the shear strength on a failure path and the

tensile strength on a perpendicular segment.

- Block shear strength = net section fracture strength on shear path + gross yielding

strength on the tension path

- OR

- Block shear strength = gross yielding strength of the shear path + net section fracture

strength of the tension path

• Which of the two calculations above governs?

- See page 16.1 – 67 (Section J4.3) of the AISC manual

- When Fu Ant ≥ 0.6Fu Anv; φt Rn = φ (0.6 Fy Agv + Fu Ant) ≤ φ (0.6 FuAnv + Fu Ant)

- When Fu Ant < 0.6Fu Anv; φt Rn = φ (0.6 Fu Anv + Fy Agt) ≤ φ (0.6 FuAnv + Fu Ant)

- Where, φ = 0.75

Agv = gross area subject to shear

Agt = gross area subject to tension

Anv = net area subject to shear

Ant = net area subject to tension

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

EXAMPLE 3.8 Calculate the block shear strength of the single angle tension member

considered in Examples 3.2 and 3.3. The single angle L 4 x 4 x 3/8 made from A36 steel is

connected to the gusset plate with 5/8 in. diameter bolts as shown below. The bolt spacing is 3

in. center-to-center and the edge distances are 1.5 in and 2.0 in as shown in the Figure below.

Compare your results with those obtained in Example 3.2 and 3.3

x
L 4 x 4 x 3/ 8
2 .0 d b = 5/8 in.
.0 3 .0
1 .5 3
a
Gusset plate L 4 x 4 x 3/ 8

• Step I. Assume a block shear path and calculate the required areas

2 .0 d b = 5/8 in.
.0 3 .0
1 .5 3
a
Gusset plate

- Agt = gross tension area = 2.0 x 3/8 = 0.75 in2

- Ant = net tension area = 0.75 – 0.5 x (5/8 + 1/8) x 3/8 = 0.609 in2

- Agv = gross shear area = (3.0 + 3.0 +1.5) x 3/8 = 2.813 in2

- Anv = net tension area = 2.813 – 2.5 x (5/8 + 1/8) x 3/8 = 2.109 in2

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Step II. Calculate which equation governs

- 0.6 Fu Anv = 0.6 x 58 x 2.109 = 73.393 kips

- Fu Ant = 58 x 0.609 = 35.322 kips

- 0.6 Fu Anv > Fu Ant

- Therefore, equation with fracture of shear path governs

• Step III. Calculate block shear strength

- φt Rn = 0.75 (0.6 Fu Anv + Fy Agt)

- φt Rn = 0.75 (73.393 + 36 x 0.75) = 75.294 kips

• Compare with results from previous examples

Example 3.2:
Ultimate factored load = Pu = 66 kips
Gross yielding design strength = φt Pn = 92.664 kips
Assume Ae = 0.85 An
Net section fracture strength = 95.352 kips
Design strength = 92.664 kips (gross yielding governs)

Example 3.3
Calculate Ae = 0.8166 An
Net section fracture strength = 91.045 kips
Design strength = 91.045 kips (net section fracture governs)
Member is still adequate to carry the factored load (Pu) = 66 kips

Example 3.8
Block shear fracture strength = 75.294 kips
Design strength = 75.294 kips (block shear fracture governs)
Member is still adequate to carry the factored load (Pu) = 66 kips

26
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Bottom line:

- Any of the three limit states (gross yielding, net section fracture, or block shear failure)
can govern.
- The design strength for all three limit states has to be calculated.
- The member design strength will be the smallest of the three calculated values
- The member design strength must be greater than the ultimate factored design load in
tension.

Practice Example Determine the design tension strength for a single channel C15 x 50

connected to a 0.5 in. thick gusset plate as shown in Figure. Assume that the holes are for 3/4 in.

diameter bolts and that the plate is made from structural steel with yield stress (Fy) equal to 50

ksi and ultimate stress (Fu) equal to 65 ksi.

gusset plate

3 @ 3” = 9” T
T center-to-center

C15 x 50

1.5” 3” 3”

▪ Limit state of yielding due to tension:

φTn = 0.9 * 50 *14.7 = 662kips
▪ Limit state of fracture due to tension:
⎛7⎞
An = Ag − nd e t = 14.7 − 4⎜ ⎟(0.716 ) = 12.19in 2
⎝8⎠
⎛ x⎞ ⎛ 0.798 ⎞
Ae = UAn = ⎜1 − ⎟ An = ⎜1 − ⎟ * 12.19 = 10.57in
2

⎝ L⎠ ⎝ 6 ⎠
Check: U = 0.867 ≤ 0.9 OK.

27
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Note: The connection eccentricity, x, for a C15X50 can be found on page 1-51 (LRFD).

φTn = 0.75 * 65 *10.57 = 515kips

▪ Limit state of block shear rupture:

⎡ ⎛ 7 ⎞⎤
0.6 Fu Anv = 0.6 * 65 * ⎢2 * ⎜ 7.5 − 2.5 * ⎟⎥ * 0.716 = 296.6925
⎣ ⎝ 8 ⎠⎦

⎡ ⎛ 7 ⎞⎤
Fu Ant = 65 * ⎢9 − 3⎜ ⎟⎥ * 0.716 = 296.6925
⎣ ⎝ 8 ⎠⎦

Fu Ant ≥ 0.6 Fu Anv

[ ]
⎡
∴ φRn = φ 0.6 Fy Agv + Fu Ant = 0.75⎢0.6 * 50 * 15 * 0.716 + 65 *
296.6925 ⎤
65 ⎥⎦
= 464kips
⎣

Block shear rupture is the critical limit state and the design tension strength is 464kips.

28
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

4.7 Design of tension members

• The design of a tension member involves finding the lightest steel section (angle, wide-

flange, or channel section) with design strength (φPn) greater than or equal to the maximum

factored design tension load (Pu) acting on it.

- φ Pn ≥ Pu

- Pu is determined by structural analysis for factored load combinations

- φ Pn is the design strength based on the gross section yielding, net section fracture, and

block shear rupture limit states.

• For gross yielding limit state, φPn = 0.9 x Ag x Fy

- Therefore, 0.9 x Ag x Fy ≥ Pu
Pu
- Therefore, Ag ≥
0.9 × Fy

• For net section fracture limit state, φPn = 0.75 x Ae x Fu

- Therefore, 0.75 x Ae x Fu ≥ Pu
Pu
- Therefore, Ae ≥
0.75 × Fu
- But, Ae = U An

- Where, U and An depend on the end connection.

• Thus, designing the tension member goes hand-in-hand with designing the end connection,

which we have not covered so far.

• Therefore, for this chapter of the course, the end connection details will be given in the

examples and problems.

• The AISC manual tabulates the tension design strength of standard steel sections

- Include: wide flange shapes, angles, tee sections, and double angle sections.

29
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- The gross yielding design strength and the net section fracture strength of each section is

tabulated.

- This provides a great starting point for selecting a section.

• There is one serious limitation

- The net section fracture strength is tabulated for an assumed value of U = 0.75, obviously

because the precise connection details are not known

- For all W, Tee, angle and double-angle sections, Ae is assumed to be = 0.75 Ag

- The engineer can first select the tension member based on the tabulated gross yielding

and net section fracture strengths, and then check the net section fracture strength and the

block shear strength using the actual connection details.

• Additionally for each shape the manual tells the value of Ae below which net section fracture

will control:

- Thus, for W shapes net section fracture will control if Ae < 0.923 Ag

- For single angles, net section fracture will control if Ae < 0.745 Ag

- For Tee shapes, net section fracture will control if Ae < 0.923

- For double angle shapes, net section fracture will control if Ae < 0.745 Ag

• Slenderness limits

- Tension member slenderness l/r must preferably be limited to 300 as per LRFD

specification B7

30
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

Example 3.10 Design a member to carry a factored maximum tension load of 100 kips.

(a) Assume that the member is a wide flange connected through the flanges using eight ¾ in.

diameter bolts in two rows of four each as shown in the figure below. The center-to-center

distance of the bolts in the direction of loading is 4 in. The edge distances are 1.5 in. and 2.0

in. as shown in the figure below. Steel material is A992

¾ in. d iameter bolts

2 in. 4 in.

1.5 in.

W
1.5 in.

2 in. 4 in.

Holes in beam flange

SOLUTION

• Step I. Select a section from the Tables

- Go to the TEN section of the AISC manual. See Table 3-1 on pages 3-17 to 3-19.

- From this table, select W8x10 with Ag = 2.96 in2, Ae = 2.22 in2.

- Gross yielding strength = 133 kips, and net section fracture strength=108 kips

- This is the lightest section in the table.

- Assumed U = 0.75. And, net section fracture will govern if Ae < 0.923 Ag

• Step II. Calculate the net section fracture strength for the actual connection

- According to the Figure above, An = Ag - 4 (db + 1/8) x tf

- An = 2.96 - 4 (3/4 + 1/8) x 0.205 = 2.24 in2

- The connection is only through the flanges. Therefore, the shear lag factor U will be the

distance from the top of the flange to the centroid of a WT 4 x 5.

31
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- See DIM section of the AISC manual. See Table 1-8, on pages 1-50, 1-51

- x = 0.953

- U = 1- x /L = 1 - 0.953 / 4 = 0.76

- Ae = 0.76 An = 0.76 x 2.24 = 1.70 in2

- φtPn = 0.75 x Fu x Ae = 0.75 x 65 x 1.70 = 82.9 kips

- Unacceptable because Pu = 100 kips; REDESIGN required

• Step III. Redesign

Many ways to redesign. One way is shown here:

- Assume φt Pn > 100 kips

- Therefore, 0.75 x 65 x Ae > 100 kips

- Therefore, Ae > 2.051 in2

- Assume, Ae = 0.76 An (based on previous calculations, step II)

- Therefore An > 2.7 in2

- But, Ag = An + 4 (db + 1/8) x tf (based on previous calculations, step II)

- Therefore Ag > 2.7 + 3.5 x tf

- Go to the section dimension table 1-1 on page 1-22 of the AISC manual. Select next

highest section.

ƒ For W 8 x 13, tf = 0.255 in.

ƒ Therefore, Ag > 2.7 + 3.5 x 0.255 = 3.59 in2

ƒ From Table 1-1, W8 x 13 has Ag = 3.84 in2 > 3.59 in2

ƒ Therefore, W8 x 13 is acceptable and is chosen.

32
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Step IV. Check selected section for net section fracture

- Ag = 3.84 in2

- An = 3.84 - 3.5 x 0.255 = 2.95 in2

- From dimensions of WT4 x 6.5, x = 1.03 in.

- Therefore, U = 1- x /L = 1-1.03/4 = 0.74

- Therefore, Ae = U An = 0.74 x 2.95 = 2.19 in2

- Therefore, net section fracture strength = 0.75 x 65 x 2.19 = 106.7 kips

- Which is greater than 100 kips (design load). Therefore, W 8 x 13 is acceptable.

• Step V. Check the block shear rupture strength

o Identify the block shear path

2 in. 4 in.

1.5 in.

1.5 in.

2 in. 4 in.

- The block shear path is show above. Four blocks will separate from the tension

member (two from each flange) as shown in the figure above.

- Agv = [(4+2) x tf ] x 4 = 6 x 0.255 x 4 = 6.12 in2 - for four tabs

- Anv = {4+2 - 1.5 x (db+1/8)} x tf x 4 = 4.78 in2

- Agt = 1.5 x tf x 4 = 1.53 in2

- Ant = {1.5 - 0.5 x (db+1/8)}x tf x 4 = 1.084 in2

o Identify the governing equation:

33
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- FuAnt = 65 x 1.084 = 70.4 kips

- 0.6FuAnv = 0.6 x 65 x 4.78 = 186.42 kips , which is > FuAnt

o Calculate block shear strength

- φtRn = 0.75 (0.6FuAnv + FyAgt) = 0.75 (186.42 + 50 x 1.53) = 197.2 kips

- Which is greater than Pu = 100 kips. Therefore W8 x 13 is still acceptable

• Summary of solution

Mem. Design Ag An U Ae Yield Fracture Block-shear

load strength strength strength
W8x13 100 kips 3.84 2.95 0.74 2.19 173 kips 106.7 kips 197.2 kips

Design strength = 106.7 kips (net section fracture governs)

W8 x 13 is adequate for Pu = 100 kips and the given connection

34
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

EXAMPLE 3.11 Design a member to carry a factored maximum tension load of 100 kips.

(b) The member is a single angle section connected through one leg using four 1 in. diameter

bolts. The center-to-center distance of the bolts is 3 in. The edge distances are 2 in. Steel

material is A36

x
3.0 in.

4.0 in.
2.0 in.

2.0 in. 3.0 in. 3.0 in. 3.0 in.

• Step I. Select a section from the Tables

- Go to the TEN section of the AISC manual. See Table 3-2 on pages 3-20 to 3-21.

- From this table, select L4x3x1/2 with Ag = 3.25 in2, Ae = 2.44 in2.

- Gross yielding strength = 105 kips, and net section fracture strength=106 kips

- This is the lightest section in the table.

- Assumed U = 0.75. And, net section fracture will govern if Ae < 0.745 Ag

• Step II. Calculate the net section fracture strength for the actual connection

- According to the Figure above, An = Ag - 1 (db + 1/8) x t

- An = 3.25 - 1(1 + 1/8) x 0.5 = 2.6875 in2

- The connection is only through the long leg. Therefore, the shear lag factor U will be the

distance from the back of the long leg to the centroid of the angle.

- See DIM section of the AISC manual. See Table 1-7, on pages 1-36, 1-37

- x = 0.822 in.

- U = 1- x /L = 1 - 0.822 /9 = 0.908

- But U must be ≤ 0.90. Therefore, let U = 0.90

35
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

- Ae = 0.90 An = 0.90 x 2.6875 = 2.41 in2

- φtPn = 0.75 x Fu x Ae = 0.75 x 58 x 2.41 = 104.8 kips

- Acceptable because Pu = 100 kips.

• Step V. Check the block shear rupture strength

o Identify the block shear path

2.0 in.

2.0 in. 3.0 in. 3.0 in. 3.0 in.

- Agv = (9+2) x 0.5 = 5.5 in2

- Anv = [11 - 3.5 x (1+1/8)] x 0.5 = 3.53 in2

- Agt = 2.0 x 0.5 = 1.0 in2

- Ant = [2.0 - 0.5 x (1 + 1/8)] x 0.5 = 0.72 in2

o Identify the governing equation:

- FuAnt = 58 x 0.72 = 41.76 kips

- 0.6FuAnv = 0.6 x 58 x 3.53 = 122.844 in2, which is > FuAnt

o Calculate block shear strength

- φtRn = 0.75 (0.6FuAnv + FyAgt) = 0.75 (122.84 + 36 x 1.0) = 119.133 kips

- Which is greater than Pu = 100 kips. Therefore L4x3x1/2 is still acceptable

36
CE 405: Design of Steel Structures – Prof. Dr. A. Varma Tension Member Design

• Summary of solution

Mem. Design Ag An U Ae
Yield Fracture Block-shear
load strength strength strength
L4x3x1/2 100 kips 3.25 2.69 0.9 2.41 105 kips 104.8 kips 119.13 kips
Design strength = 104.8 kips (net section fracture governs)
L4x3x1/2 is adequate for Pu = 100 kips and the given connection

• Note: For this problem Ae/Ag = 2.41/3.25 = 0.741, which is < 0.745. As predicted by the

AISC manual, when Ae /Ag < 0.745, net section fracture governs.

37
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

CHAPTER 3. COMPRESSION MEMBER DESIGN

3.1 INTRODUCTORY CONCEPTS

• Compression Members: Structural elements that are subjected to axial compressive forces

only are called columns. Columns are subjected to axial loads thru the centroid.

• Stress: The stress in the column cross-section can be calculated as

P
f = (2.1)
A
where, f is assumed to be uniform over the entire cross-section.

• This ideal state is never reached. The stress-state will be non-uniform due to:

- Accidental eccentricity of loading with respect to the centroid

- Member out-of –straightness (crookedness), or

- Residual stresses in the member cross-section due to fabrication processes.

• Accidental eccentricity and member out-of-straightness can cause bending moments in the

member. However, these are secondary and are usually ignored.

• Bending moments cannot be neglected if they are acting on the member. Members with axial

compression and bending moment are called beam-columns.

3.2 COLUMN BUCKLING

• Consider a long slender compression member. If an axial load P is applied and increased

slowly, it will ultimately reach a value Pcr that will cause buckling of the column. Pcr is called

the critical buckling load of the column.

1
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

P
(a) Pcr (b)

What is buckling?

Buckling occurs when a straight column

subjected to axial compression suddenly

undergoes bending as shown in the Figure 1(b).

Buckling is identified as a failure limit-state for

columns.

P
Pcr

Figure 1. Buckling of axially loaded compression members

• The critical buckling load Pcr for columns is theoretically given by Equation (3.1)

π2 E I
Pcr = (3.1)
( K L )2
where, I = moment of inertia about axis of buckling

K = effective length factor based on end boundary conditions

• Effective length factors are given on page 16.1-189 of the AISC manual.

2
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• In examples, homeworks, and exams please state clearly whether you are using the

theoretical value of K or the recommended design values.

3
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

EXAMPLE 3.1 Determine the buckling strength of a W 12 x 50 column. Its length is 20 ft. For

major axis buckling, it is pinned at both ends. For minor buckling, is it pinned at one end and

fixed at the other end.

Solution

Step I. Visualize the problem

Figure 2. (a) Cross-section; (b) major-axis buckling; (c) minor-axis buckling

• For the W12 x 50 (or any wide flange section), x is the major axis and y is the minor axis.

Major axis means axis about which it has greater moment of inertia (Ix > Iy)

Figure 3. (a) Major axis buckling; (b) minor axis buckling

4
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Step II. Determine the effective lengths

• According to Table C-C2.1 of the AISC Manual (see page 16.1 - 189):

- For pin-pin end conditions about the minor axis

Ky = 1.0 (theoretical value); and Ky = 1.0 (recommended design value)

- For pin-fix end conditions about the major axis

Kx = 0.7 (theoretical value); and Kx = 0.8 (recommended design value)

• According to the problem statement, the unsupported length for buckling about the major (x)

axis = Lx = 20 ft.

• The unsupported length for buckling about the minor (y) axis = Ly = 20 ft.

• Effective length for major (x) axis buckling = Kx Lx = 0.8 x 20 = 16 ft. = 192 in.

• Effective length for minor (y) axis buckling = Ky Ly = 1.0 x 20 = 20 ft. = 240 in.

Step III. Determine the relevant section properties

• For W12 x 50: elastic modulus = E = 29000 ksi (constant for all steels)

• For W12 x 50: Ix = 391 in4. Iy = 56.3 in4 (see page 1-21 of the AISC manual)

Step IV. Calculate the buckling strength

π2 E I x π 2 × 29000× 391
• Critical load for buckling about x - axis = Pcr-x = =
(K x L x )2 (192)2
Pcr-x = 3035.8 kips

π2 E I y π 2 × 29000× 56.3
• Critical load for buckling about y-axis = Pcr-y = =
(K y L y )2 (240)2

Pcr-y = 279.8 kips

• Buckling strength of the column = smaller (Pcr-x, Pcr-y) = Pcr = 279.8 kips

Minor (y) axis buckling governs.

5
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Notes:

- Minor axis buckling usually governs for all doubly symmetric cross-sections. However, for

some cases, major (x) axis buckling can govern.

- Note that the steel yield stress was irrelevant for calculating this buckling strength.

3.3 INELASTIC COLUMN BUCKLING

• Let us consider the previous example. According to our calculations Pcr = 279.8 kips. This Pcr

will cause a uniform stress f = Pcr/A in the cross-section

• For W12 x 50, A = 14.6 in2. Therefore, for Pcr = 279.8 kips; f = 19.16 ksi

The calculated value of f is within the elastic range for a 50 ksi yield stress material.

π2 E I y
• However, if the unsupported length was only 10 ft., Pcr = would be calculated as
(K y L y )2
1119 kips, and f = 76.6 kips.

• This value of f is ridiculous because the material will yield at 50 ksi and never develop f =

76.6 kips. The member would yield before buckling.

• Equation (3.1) is valid only when the material everywhere in the cross-section is in the

elastic region. If the material goes inelastic then Equation (3.1) becomes useless and

cannot be used.

• What happens in the inelastic range?

Several other problems appear in the inelastic range.

- The member out-of-straightness has a significant influence on the buckling strength in

the inelastic region. It must be accounted for.

6
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

- The residual stresses in the member due to the fabrication process causes yielding in the

cross-section much before the uniform stress f reaches the yield stress Fy.

- The shape of the cross-section (W, C, etc.) also influences the buckling strength.

- In the inelastic range, the steel material can undergo strain hardening.

All of these are very advanced concepts and beyond the scope of CE405. You are welcome

to CE805 to develop a better understanding of these issues.

• So, what should we do? We will directly look at the AISC Specifications for the strength of

compression members, i.e., Chapter E (page 16.1-27 of the AISC manual).

3.4 AISC SPECIFICATIONS FOR COLUMN STRENGTH

• The AISC specifications for column design are based on several years of research.

• These specifications account for the elastic and inelastic buckling of columns including all

issues (member crookedness, residual stresses, accidental eccentricity etc.) mentioned above.

• The specification presented here (AISC Spec E2) will work for all doubly symmetric cross-

sections and channel sections.

• The design strength of columns for the flexural buckling limit state is equal to φcPn
Where, φc = 0.85 (Resistance factor for compression members)
Pn = Ag Fcr (3.2)

- For λc ≤ 1.5 (
Fcr = 0.658 λ c Fy
2
) (3.3)

 0.877 
- For λc > 1.5 Fcr =  2  Fy (3.4)
 λ c 

K L Fy
Where, λc = (3.5)
rπ E

7
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Ag = gross member area; K = effective length factor

L = unbraced length of the member; r = governing radius of gyration

1.0
(
Fcr = 0.658 λ c
2
)Fy

Fcr/Fy
 0.877 
0.39 Fcr =  2  Fy
 λ c 

K L Fy 1.5
λc =
rπ E

π2E I
• Note that the original Euler buckling equation is Pcr =
(K L )2
Pcr π2E I π2E 2 π2E
∴ Fcr = = × = × r =
A g (K L )2 A g (K L )2 K L
2
 
 r 
F π2E 1 1
∴ cr = 2
= 2
= 2
Fy  K L   F  λc
  × Fy  K L × y 
 r   rπ E 

1
∴ Fcr = Fy × 2
λc
0.877
• Note that the AISC equation for λc < 1.5 is Fcr = Fy ×
λ2c
- The 0.877 factor tries to account for initial crookedness.

• For a given column section:

- Calculate I, Ag, r

- Determine effective length K L based on end boundary conditions.

- Calculate λc

- If λc is greater than 1.5, elastic buckling occurs and use Equation (3.4)

8
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

- If λc is less than or equal to 1.5, inelastic buckling occurs and use Equation (3.3)

• Note that the column can develop its yield strength Fy as λc approaches zero.

3.5 COLUMN STRENGTH

• In order to simplify calculations, the AISC specification includes Tables.

- Table 3-36 on page 16.1-143 shows KL/r vs. φcFcr for steels with Fy = 36 ksi.

- You can calculate KL/r for the column, then read the value of φcFcr from this table

- The column strength will be equal to φcFcr x Ag

- Table 3-50 on page 16.1-145 shows KL/r vs. φcFcr for steels with Fy = 50 ksi.

• In order to simplify calculations, the AISC specification includes more Tables.

- Table 4 on page 16.1-147 shows λc vs. φcFcr/Fy for all steels with any Fy.

- You can calculate λc for the column, the read the value of φcFcr/Fy

- The column strength will be equal to φcFcr/Fy x (Ag x Fy)

EXAMPLE 3.2 Calculate the design strength of W14 x 74 with length of 20 ft. and pinned ends.

A36 steel is used.

Solution

• Step I. Calculate the effective length and slenderness ratio for the problem

Kx = Ky = 1.0

Lx = Ly = 240 in.

Major axis slenderness ratio = KxLx/rx = 240/6.04 = 39.735

Minor axis slenderness ratio = KyLy/ry = 240/2.48 = 96.77

• Step II. Calculate the buckling strength for governing slenderness ratio

9
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

The governing slenderness ratio is the larger of (KxLx/rx, KyLy/ry)

K y Ly Fy
KyLy/ry is larger and the governing slenderness ratio; λc = = 1.085
ry π E

λc < 1.5; (
Therefore, Fcr = 0.658 λ c Fy
2
)
Therefore, Fcr = 21.99 ksi

Design column strength = φcPn = 0.85 (Ag Fcr) = 0.85 (21.8 in2 x 21.99 ksi) = 408 kips

Design strength of column = 408 kips

• Check calculated values with Table 3-36. For KL/r = 97, φcFcr = 18.7 ksi

• Check calculated values with Table 4. For λc = 1.08, φcFcr = 0.521

10
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

3.6 LOCAL BUCKLING LIMIT STATE

• The AISC specifications for column strength assume that column buckling is the governing

limit state. However, if the column section is made of thin (slender) plate elements, then

failure can occur due to local buckling of the flanges or the webs.

Figure 4. Local buckling of columns

• If local buckling of the individual plate elements occurs, then the column may not be able to

develop its buckling strength.

• Therefore, the local buckling limit state must be prevented from controlling the column

strength.

• Local buckling depends on the slenderness (width-to-thickness b/t ratio) of the plate element

and the yield stress (Fy) of the material.

• Each plate element must be stocky enough, i.e., have a b/t ratio that prevents local buckling

from governing the column strength.

11
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• The AISC specification B5 provides the slenderness (b/t) limits that the individual plate

elements must satisfy so that local buckling does not control.

• The AISC specification provides two slenderness limits (λp and λr) for the local buckling of

plate elements.

Compact
Fy

Non-Compact
Axial Force, F

Slender

F b

Axial shortening, ∆

Figure 5. Local buckling behavior and classification of plate elements

- If the slenderness ratio (b/t) of the plate element is greater than λr then it is slender. It will

locally buckle in the elastic range before reaching Fy

- If the slenderness ratio (b/t) of the plate element is less than λr but greater than λp, then it

is non-compact. It will locally buckle immediately after reaching Fy

- If the slenderness ratio (b/t) of the plate element is less than λp, then the element is

compact. It will locally buckle much after reaching Fy

• If all the plate elements of a cross-section are compact, then the section is compact.

- If any one plate element is non-compact, then the cross-section is non-compact

- If any one plate element is slender, then the cross-section is slender.

• The slenderness limits λp and λr for various plate elements with different boundary

conditions are given in Table B5.1 on pages 16.1-14 and 16.1-15 of the AISC Spec.

12
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Note that the slenderness limits (λp and λr) and the definition of plate slenderness (b/t) ratio

depend upon the boundary conditions for the plate.

- If the plate is supported along two edges parallel to the direction of compression force,

then it is a stiffened element. For example, the webs of W shapes

- If the plate is supported along only one edge parallel to the direction of the compression

force, then it is an unstiffened element. Ex., the flanges of W shapes.

• The local buckling limit state can be prevented from controlling the column strength by using

sections that are non-compact

- If all the elements of the cross-section have calculated slenderness (b/t) ratio less than λr,

then the local buckling limit state will not control.

- For the definitions of b/t, λp, λr for various situations see Table B5.1 and Spec B5.

EXAMPLE 3.3 Determine the local buckling slenderness limits and evaluate the W14 x 74

section used in Example 3.2. Does local buckling limit the column strength?

Solution

• Step I. Calculate the slenderness limits

See Table B5.1 on page 16.1 – 14.

- For the flanges of I-shape sections in pure compression

E 29000
λr = 0.56 x = 0.56 x = 15.9
Fy 36

- For the webs of I-shapes section in pure compression

E 29000
λr = 0.56 x = 0.56 x = 15.9
Fy 36

13
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

E 29000
λr = 1.49 x = 1.49 x = 42.3
Fy 36

• Step II. Calculate the slenderness ratios for the flanges and webs of W14 x 74

- For the flanges of I-shape member, b = bf/2 = flange width / 2

Therefore, b/t = bf/2tf. (See pg. 16.1-12 of AISC)

For W 14 x 74, bf/2tf = 6.41 (See Page 1-19 in AISC)

- For the webs of I shaped member, b = h

h is the clear distance between flanges less the fillet / corner radius of each flange

For W14 x 74, h/tw = 25.4 (See Page 1-19 in AISC)

• Step III. Make the comparisons and comment

For the flanges, b/t < λr. Therefore, the flange is non-compact

For the webs, h/tw < λr. Therefore the web is non-compact

Therefore, the section is compact

Therefore, local buckling will not limit the column strength.

3.7 COLUMN DESIGN

• The AISC manual has tables for column strength. See page 4-21 onwards.

• For wide flange sections, the column buckling strength (φcPn) is tabulated with respect to the

effective length about the minor axis KyLy in Table 4-2.

- The table takes the KyLy value for a section, and internally calculates the KyLy/ry, then λc

K y Ly Fy
= ; and then the tabulated column strength using either Equation E2-2 or
ry π E

E2-3 of the specification.

14
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• If you want to use the Table 4-2 for calculating the column strength for buckling about the

major axis, then do the following:

K xLx
- Take the major axis KxLx value. Calculate an equivalent (KL)eq =
rx / ry

- Use the calculated (KL)eq value to find (φcPn) the column strength for buckling about the

major axis from Table (4-2)

• For example, consider a W14 x 74 column with KyLy = 20 ft. and KxLx = 25 ft.

- Material has yield stress = 50 ksi (always in Table 4-2).

- See Table 4-2, for KyLy = 20 ft., φcPn = 467 kips (minor axis buckling strength)

- rx/ry for W14x74 = 2.44 from Table 4-2 (see page 4-23 of AISC).

- For KxLx = 25 ft., (KL)eq = 25/2.44 = 10.25 ft.

- For (KL)eq = 10.25 ft., φcPn = 774 kips (major axis buckling strength)

- If calculated value of (KL)eq < KyLy then minor axis buckling will govern.

EXAMPLE 3.4 Determine the design strength of an ASTM A992 W14 x 132 that is part of a

braced frame. Assume that the physical length L = 30 ft., the ends are pinned and the column is

braced at the ends only for the X-X axis and braced at the ends and mid-height for the Y-Y axis.

Solution

• Step I. Calculate the effective lengths.

For W14 x 132: rx = 6.28 in; ry = 3.76 in; Ag =38.8 in2

Kx = 1.0 and Ky = 1.0

Lx = 30 ft. and Ly = 15 ft.

KxLx = 30 ft. and KyLy = 15 ft.

• Step II. Determine the governing slenderness ratio

15
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

KxLx/rx = 30 x 12 in./6.28 in.= 57.32

KyLy/ry = 15 x 12 in./3.76 in. = 47.87

The larger slenderness ratio, therefore, buckling about the major axis will govern the column

strength.

• Step III. Calculate the column strength

K xLx 30
KxLx = 30 ft. Therefore, (KL)eq = = = 17.96 ft.
rx / ry 6.28 / 3.76

From Table 4-2, for (KL)eq = 18.0 ft. φcPn = 1300 kips (design column strength)

• Step IV. Check the local buckling limits

E
For the flanges, bf/2tf = 7.15 < λr = 0.56 x = 13.5
Fy
E
For the web, h/tw = 17.7 < λr = 1.49 x = 35.9
Fy
Therefore, the section is non-compact. OK.

EXAMPLE 3.5 A compression member is subjected to service loads of 165 kips dead load and

535 kips of live load. The member is 26 ft. long and pinned at each end. Use A992 (50 ksi) steel

and select a W shape

Solution

• Calculate the factored design load Pu

Pu = 1.2 PD + 1.6 PL = 1.2 x 165 + 1.6 x 535 = 1054 kips

• Select a W shape from the AISC manual Tables

For KyLy = 26 ft. and required strength = 1054 kips

- Select W14 x 145 from page 4-22. It has φcPn = 1160 kips

16
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

- Select W12 x 170 from page 4-24. It has φcPn = 1070 kips

- No no W10 will work. See Page 4-26

- W14 x 145 is the lightest.

• Note that column sections are usually W12 or W14. Usually sections bigger than W14 are

usually not used as columns.

3.8 EFFECTIVE LENGTH OF COLUMNS IN FRAMES

• So far, we have looked at the buckling strength of individual columns. These columns had

various boundary conditions at the ends, but they were not connected to other members with

moment (fix) connections.

• The effective length factor K for the buckling of an individual column can be obtained for the

appropriate end conditions from Table C-C2.1 of the AISC Manual .

• However, when these individual columns are part of a frame, their ends are connected to

other members (beams etc.).

- Their effective length factor K will depend on the restraint offered by the other members

connected at the ends.

- Therefore, the effective length factor K will depend on the relative rigidity (stiffness) of

the members connected at the ends.

The effective length factor for columns in frames must be calculated as follows:

• First, you have to determine whether the column is part of a braced frame or an unbraced

(moment resisting) frame.

- If the column is part of a braced frame then its effective length factor 0 < K ≤ 1

- If the column is part of an unbraced frame then 1 < K ≤ ∞

17
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Then, you have to determine the relative rigidity factor G for both ends of the column

- G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming

together at an end to the summation of the rigidity (EI/L) of all beams coming together at

the same end.

E Ic
Lc
∑
- G= - It must be calculated for both ends of the column.
EI
∑ Lb
b

• Then, you can determine the effective length factor K for the column using the calculated

value of G at both ends, i.e., GA and GB and the appropriate alignment chart

• There are two alignment charts provided by the AISC manual,

- One is for columns in braced (sidesway inhibited) frames. See Figure C-C2.2a on page

16.1-191 of the AISC manual. 0 < K ≤ 1

- The second is for columns in unbraced (sidesway uninhibited) frames. See Figure C-

C2.2b on page 16.1-192 of the AISC manual. 1 < K ≤ ∞

- The procedure for calculating G is the same for both cases.

18
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

EXAMPLE 3.6 Calculate the effective length factor for the W12 x 53 column AB of the frame

shown below. Assume that the column is oriented in such a way that major axis bending occurs

in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane

buckling. Assume that the same column section is used for the stories above and below.

10 ft.
W14 x 68

10 ft.
W14 x 68 A

12 ft.
W14 x 68
B
W12 x 79

W12 x 79

W12 x 79
15 ft.

18 ft. 18 ft. 20 ft.

Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible.

• It is an unbraced (sidesway uninhibited) frame.

• Lx = Ly = 12 ft.

• Ky = 1.0

• Kx depends on boundary conditions, which involve restraints due to beams and columns

connected to the ends of column AB.

• Need to calculate Kx using alignment charts.

Step II - Calculate Kx

• Ixx of W 12 x 53 = 425 in4 Ixx of W14x68 = 753

19
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Ic 425 425
∑ +
L c 10 × 12 12 × 12 6.493
• GA = = = = 1.021
Ib 723 723 6.360
∑ +
L b 18 × 12 20 × 12

Ic 425 425
∑ +
L c 12 × 12 15 × 12 5.3125
• GB = = = = 0.835
Ib 723 723 6.360
∑ +
L b 18 × 12 20 × 12

• Using GA and GB: Kx = 1.3 - from Alignment Chart on Page 3-6

Step III – Design strength of the column

• KyLy = 1.0 x 12 = 12 ft.

• Kx Lx = 1.3 x 12 = 15.6 ft.

- rx / ry for W12x53 = 2.11

- (KL)eq = 15.6 / 2.11 = 7.4 ft.

• KyLy > (KL)eq

• Therefore, y-axis buckling governs. Therefore φcPn = 518 kips

3.8.1 Inelastic Stiffness Reduction Factor – Modification

• This concept for calculating the effective length of columns in frames was widely accepted

for many years.

• Over the past few years, a lot of modifications have been proposed to this method due to its

several assumptions and limitation. Most of these modifications have not yet been accepted

in to the AISC provisions.

• One of the accepted modifications is the inelastic stiffness reduction factor. As presented

earlier, G is a measure of the relative flexural rigidity of the columns (EIc/Lc) with respect to

the beams (EIb/Lb)

20
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

- However, if column buckling were to occur in the inelastic range (λc < 1.5), then the

flexural rigidity of the column will be reduced because Ic will be the moment of inertia of

only the elastic core of the entire cross-section. See figure below

σrc = 10 ksi
σrt = 5 ksi
Yielded zone

σrt = 5 ksi
Elastic core, Ic

σrc = 10 ksi
σrt = 5 ksi

(a) Initial state – residual stress (b) Partially y ielded state at buckling

- The beams will have greater flexural rigidity when compared with the reduced rigidity

(EIc) of the inelastic columns. As a result, the beams will be able to restrain the columns

better, which is good for column design.

- This effect is incorporated in to the AISC column design method through the use of Table

4-1 given on page 4-20 of the AISC manual.

- Table 4-1 gives the stiffness reduction factor (τ) as a function of the yield stress Fy and

the stress Pu/Ag in the column, where Pu is factored design load (analysis)

21
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

EXAMPLE 3.7 Calculate the effective length factor for a W10 x 60 column AB made from 50

ksi steel in the unbraced frame shown below. Column AB has a design factor load Pu = 450 kips.

The columns are oriented such that major axis bending occurs in the plane of the frame. The

columns are braced continuously along the length for out-of-plane buckling. Assume that the

same column section is used for the story above

W14 x 74

12 ft.
W14 x 74
A
W12 x 79

W12 x 79

W12 x 79
15 ft.

18 ft. 18 ft. 20 ft.

Solution

Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible.

• It is an unbraced (sidesway uninhibited) frame.

• Ly = 0 ft.

• Ky has no meaning because out-of-plane buckling is not possible.

• Kx depends on boundary conditions, which involve restraints due to beams and columns

connected to the ends of column AB.

• Need to calculate Kx using alignment charts.

Step II (a) - Calculate Kx

• Ixx of W 14 x 74 = 796 in4 Ixx of W 10 x 60 = 341 in4

22
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Ic 341 341
∑ +
L c 12 × 12 15 × 12 4.2625
• GA = = = = 0.609
Ib 796 796 7.002
∑ +
L b 18 × 12 20 × 12
• G B = 10 - for pin support, see note on Page 16.1-191

• Using GA and GB: Kx = 1.8 - from Alignment Chart on Page 16.1-192

• Note, Kx is greater than 1.0 because it is an unbraced frame.

Step II (b) - Calculate Kx– inelastic using stiffness reduction factor method

• Reduction in the flexural rigidity of the column due to residual stress effects

- First calculate, Pu / Ag = 450 / 17.6 = 25.57 ksi

- Then go to Table 4-1 on page 4-20 of the manual, and read the value of stiffness

reduction factor for Fy = 50 ksi and Pu/Ag = 25.57 ksi.

- Stiffness reduction factor = τ = 0.833

• GA-inelastic = τ x GA = 0.833 x 0.609 = 0.507

• GB = 10 - for pin support, see note on Page 16.1-191

• Using GA-inelastic and GB, Kx-inelastic = 1.75 - alignment chart on Page 16.1-192

• Note: You can combine Steps II (a) and (b) to calculate the Kx-inelastic directly. You don’t need

to calculate elastic Kx first. It was done here for demonstration purposes.

• Note that Kx-inelastic< Kx. This is in agreement with the fact that the beams offer better

resistance to the inelastic column AB because it has reduced flexural rigidity.

Step III – Design strength of the column

• KxLx = 1.75 x 15 = 26.25 ft.

- rx / ry for W10x60 = 1.71 - from Table 4-2, see page 4-26

- (KL)eq = 26.25/1.71 = 15.35 ft.

23
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• φcPn for X-axis buckling = 513.9 kips - from Table 4-2, see page 4-26

• Section slightly over-designed for Pu = 450 kips.

Column design strength = φcPn = 513.9 kips

EXAMPLE 3.8:

• Design Column AB of the frame shown below for a design load of 500 kips.
• Assume that the column is oriented in such a way that major axis bending occurs in the plane
of the frame.
• Assume that the columns are braced at each story level for out-of-plane buckling.
• Assume that the same column section is used for the stories above and below.

10 ft.
W14 x 68

10 ft.
W14 x 68 A

12 ft.
W14 x 68
B
W12 x 79

W12 x 79

W12 x 79

15 ft.

18 ft. 18 ft. 20 ft.

Step I - Determine the design load and assume the steel material.
• Design Load = Pu = 500 kips
• Steel yield stress = 50 ksi (A992 material)

Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible.
• It is an unbraced (sidesway uninhibited) frame.

24
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Lx = Ly = 12 ft.
• Ky = 1.0
• Kx depends on boundary conditions, which involve restraints due to beams and columns
connected to the ends of column AB.
• Need to calculate Kx using alignment charts.
• Need to select a section to calculate Kx

Step III - Select a column section

• Assume minor axis buckling governs.
• Ky Ly = 12 ft.
• See Column Tables in AISC-LRFD manual
Select section W12x53
• φcPn for y-axis buckling = 518 kips

Step IV - Calculate Kx-inelastic

• Ixx of W 12 x 53 =425 in4 Ixx of W14x68 = 753 in4
• Account for the reduced flexural rigidity of the column due to residual stress effects
- Pu/Ag = 500 / 15.6 = 32.05 ksi
- Stiffness reduction factor = τ = 0.58
Ic  425 425 
τ×∑ 0.58 ×  + 
• GA =
Lc
=  10 × 12 12 × 12  = 3.766 = 0.592
I 723 723 6.360
∑ Lb +
18 × 12 20 × 12
b

Ic  425 425 
τ× ∑ 0.58 ×  + 
Lc  12 × 12 15 × 12  3.0812
• GB = = = = 0.484
I 723 723 6.360
∑ Lb +
18 × 12 20 × 12
b

• Using GA and GB: Kx-inelastic = 1.2 - from Alignment Chart

Step V - Check the selected section for X-axis buckling

• Kx Lx = 1.2 x 12 = 14.4 ft.
• rx / ry for W12x53 = 2.11

25
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Calculate (KL)eq to determine strength (φcPn) for X-axis buckling

(KL)eq = 14.4 / 2.11 = 6.825 ft.
• From the column design tables, φcPn for X-axis buckling = 612.3 kips

Step VI. Check the local buckling limits

E
For the flanges, bf/2tf = 8.69 < λr = 0.56 x = 13.5
Fy
E
For the web, h/tw = 28.1 < λr = 1.49 x = 35.9
Fy
Therefore, the section is non-compact. OK, local buckling is not a problem

Step VII - Summarize the solution

Lx = Ly = 12 ft. Ky = 1.0
Kx = 1.2 (inelastic buckling - sway frame-alignment chart method)
φcPn for Y-axis buckling = 518 kips
φcPn for X-axis buckling = 612.3 kips
Y-axis buckling governs the design.
Selected Section is W12 x 53 made from 50 ksi steel.

26
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

EXAMPLE 3.9
• Design Column AB of the frame shown below for a design load of 450 kips.
• Assume that the column is oriented in such a way that major axis bending occurs in the plane
of the frame.
• Assume that the columns are braced continuously along the length for out-of-plane buckling.
• Assume that the same column section is used for the story above.
W14 x 74

12 ft.
W14 x 74
A
W12 x 79

W12 x 79

W12 x 79
15 ft.

18 ft. 18 ft. 20 ft.

Step I - Determine the design load and assume the steel material.
• Design Load = Pu = 450 kips
• Steel yield stress = 50 ksi

Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible.
• It is an unbraced (sidesway uninhibited) frame.
• Ly = 0 ft.
• Ky has no meaning because out-of-plane buckling is not possible.
• Kx depends on boundary conditions, which involve restraints due to beams and columns
connected to the ends of column AB.
• Need to calculate Kx using alignment charts.
• Need to select a section to calculate Kx

27
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Step III. Select a section

• There is no help from the minor axis to select a section
• Need to assume Kx to select a section.
See Figure below:
W14 x 74

12 ft.

W14 x 74
A
W12 x 79

W12 x 79

W12 x 79
15 ft.

Kx = 2.0
18 ft. 18 ft. 20 ft.
Best Case Scenario

• The best case scenario for Kx is when the beams connected at joint A have infinite flexural
stiffness (rigid). In that case Kx = 2.0 from Table C-C2.1
• Actually, the beams don't have infinite flexural stiffness. Therefore, calculated Kx should be
greater than 2.0.
• To select a section, assume Kx = 2.0
- KxLx = 2.0 x 15.0 ft. = 30.0 ft.
• Need to be able to calculate (KL)eq to be able to use the column design tables to select a
section. Therefore, need to assume a value of rx/ry to select a section.
- See the W10 column tables on page 4-26.
- Assume rx/ry = 1.71, which is valid for W10 x 49 to W10 x 68.
• (KL)eq = 30.0/1.71 = 17.54 ft.
- Obviously from the Tables, for (KL)eq = 17.5 ft., W10 x 60 is the first section that will
have φcPn > 450 kips
• Select W10x60 with φcPn = 457.7 kips for (KL)eq = 17.5 ft.

28
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Step IV - Calculate Kx-inelastic using selected section

• Ixx of W 14 x 74 = 796 in4 Ixx of W 10 x 60 = 341 in4
• Account for the reduced flexural rigidity of the column due to residual stress effects
- Pu/Ag = 450 / 17.6 = 25.57 ksi
- Stiffness reduction factor = τ = 0.833
Ic  341 341 
τ×∑ 0.833 ×  + 
Lc  12 × 12 15 × 12  3.550
• GA = = = = 0.507
I 796 796 7.002
∑ Lb +
18 × 12 20 × 12
b

• G B = 10 - for pin support

• Using GA and GB: Kx-inelastic = 1.75 - from Alignment Chart on Page 3-6
• Calculate value of Kx-inelastic is less than 2.0 (the assumed value) because GB was assumed to
be equal to 10 instead of ∞

Step V - Check the selected section for X-axis buckling

• Kx Lx = 1.75 x 15 = 26.25 ft.
- rx / ry for W10x60 = 1.71
- (KL)eq = 26.25/1.71 = 15.35 ft.
- (φcPn) for X-axis buckling = 513.9 kips
• Section slightly over-designed for Pu = 450 kips.
• W10 x 54 will probably be adequate, Student should check by calculating Kx inelastic and
φcPn for that section.

Step VI. Check the local buckling limits

E
For the flanges, bf/2tf = 7.41 < λr = 0.56 x = 13.5
Fy
E
For the web, h/tw = 18.7 < λr = 1.49 x = 35.9
Fy
Therefore, the section is non-compact. OK, local buckling is not a problem

29
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• Step VII - Summarize the solution

Ly = 0 ft. Ky = no buckling
Kx = 1.75 (inelastic buckling - sway frame - alignment chart method)
φcPn for X-axis buckling = 513.9 kips
X-axis buckling governs the design.
Selected section is W10 x 60
(W10 x 54 will probably be adequate).

30
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

3.9 DESIGN OF SINGLY SYMMETRIC CROSS-SECTIONS

• So far, we have been talking about doubly symmetric wide-flange (I-shaped) sections and

channel sections. These rolled shapes always fail by flexural buckling.

• Singly symmetric (Tees and double angle) sections fail either by flexural buckling about the

axis of non-symmetry or by flexural-torsional buckling about the axis of symmetry and the

longitudinal axis.

Figure 6(a). Flexural buckling Figure 6(b). Flexural-torsional buckling

Flexural buckling will occur about the x-axis

y

Flexural-torsional buckling will occur about the y and z-axis

x
z Smaller of the two will govern the design strength

Figure 6(c). Singly symmetric cross-section

• The AISC specification for flexural-torsional buckling is given by Spec. E3.

Design strength = φcPn = 0.85 Ag Fcrft (1)

 Fcry + Fcrz  4 Fcry Fcrz H 

Where, Fcrft =   1 − 1 −
  (2)
 2H  (Fcry + Fcrz ) 2 

Fcry = critical stress for buckling about the y-axis, see Spec. E2. (3)

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma

GJ
Fcrz = (4)
A ro 2

Ix + Iy
ro2 = polar radius of gyration about shear center (in.) = y o2 + (5)
A
y o2
H=1- (6)
ro2
yo = distance between shear center and centroid (in.) (7)

• The section properties for calculating the flexural-torsional buckling strength Fcrft are given

as follows:

E
- G=
2 (1 + υ)

- J, ro2 , H are given for WT shapes in Table 1-32 on page 1-101 to page 1-105

- ro2 , H are given for double-angle shapes in Table 1-35 on page 1-108 to 1-110

- J for single-angle shape in Table 1-31 on page 1-98 to 1-100. (J2L = 2 x JL)

• The design tables for WT shapes given in Table 4-5 on page 4-35 to 4-47. These design

tables include the axial compressive strength for flexural buckling about the x axis and

flexural-torsional buckling about the y and z axis.

32
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

EXAMPLE 3.10 Calculate the design compressive strength of a WT10.5 x 66. The effective

length with respect to x-axis is 25ft. 6in. The effective length with respect to the y-axis is 20 ft.

and the effective length with respect to z-axis is 20ft. A992 steel is used.

Solution

• Step I. Buckling strength about x-axis

K x Lx Fy 306 50
λc-x = = = 1.321
rx π E 3.06 × 3.1416 29000

φcPn = 0.85 x (0.658)1.321 x 50 x 19.4 = 397.2 kips

2

Values for Ag and rx from page 4-41 of the manual. Compare with tabulated design strength

for buckling about x-axis in Table 4-5

• Step II. Flexural-torsional buckling about the y and z axes

- Calculate Fcry and Fcrz then calculate Fcrft and φcPn

K y Ly Fy 240 50
- λc-y = = = 1.083
ry π E 2.93 × 3.1416 29000

Fcry = (0.658)1.083 x 50 = 30.6 ksi

2
-

- Fcrz = GJ/A ro2 = 11,153 x 5.62/(4.602 x 19.4) = 152.69

   4 × 30.6 ×152.7 × 0.844 

Fcrft =  Fcry + Fcrz  1 − 1 − 4 Fcry Fcrz H  =  30.6 + 152.7 1 − 1 −

- 
 2 H  (Fcry + Fcrz ) 2   2 × 0.844  (30.6 + 152.7) 2 
   

Fcrft =108.58 x 0.272 = 29.534 ksi

- φcPn = 0.85 x Fcrft x Ag = 0.85 x 29.534 x 19.4 = 487 kips

33
CE 405: Design of Steel Structures – Prof. Dr. A. Varma

Values for J, ro2 , and H were obtained from flexural-torsional properties given in Table 1-32

on page 1-102. Compare the φcPn value with the value reported in Table 4-5 (page 4-41) of

the AISC manual.

• Step III. Design strength and check local buckling

E
Flanges: bf/2tf = 12.4/(2 x 1.03) = 6.02 , which is < λr = 0.56 x = 13.5
Fy

E
Stem of Tee: d/tw = 10.9/0.65 = 16.77, which is < λr = 0.75 x = 18.08
Fy

Local buckling is not a problem. Design strength = 397.2 kips. X-axis flexural buckling

governs.

3.10 DESIGN OF DOUBLE ANGLE SECTIONS

• Double-angle sections are very popular as compression members in trusses and bracing

members in frames.

- These sections consist of two angles placed back-to-back and connected together using

bolts or welds.

- You have to make sure that the two single angle sections are connected such that they do

not buckle (individually) between the connections along the length.

- The AISC specification E4.2 requires that Ka/rz of the individual single angles < ¾ of the

governing KL/r of the double angle.

- where, a is the distance between connections and rz is the smallest radius of gyration

of the single angle (see dimensions in Table 1-7)

• Double-angle sections can fail by flexural buckling about the x-axis or flexural torsional

buckling about the y and z axes.

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma

- For flexural buckling about the x-axis, the moment of inertia Ix-2L of the double angle will

be equal to two times the moment of inertia Ix-L of each single angle.

- For flexural torsional buckling, there is a slight problem. The double angle section will

have some additional flexibility due to the intermittent connectors. This added flexibility

will depend on the connection parameters.

• According to AISC Specification E4.1, a modified (KL/r)m must be calculated for the double

angle section for buckling about the y-axis to account for this added flexibility
2 2
 KL   KL   a 
- 
Intermediate connectors that are snug-tight bolted  r  =   +  
m  r  o  rz 

- Intermediate connectors that are welded or fully tensioned bolted:

2
 KL   KL 
2
α2 a 
  =   + 0.82  
 r m  r o 1+ α2  ry 
 

where, α = separation ratio = h/2ry

h = distance between component centroids in the y direction

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma

3/8
EXAMPLE 3.11 Calculate the design strength of the compression

member shown in the figure. Two angles, 5 x 3 x ½ are oriented with the
5 x 3 x½
long legs back-to-back and separated by 3/8 in. The effective length KL is
0.746 0.746
16 ft. A36 steel is used. Assume three welded intermediate connectors

Solution

Step I. Determine the relevant properties from the AISC manual

Property Single angle Double angle

Ag 3.75 in2 7.5 in2

rx 1.58 in. 1.58 in.

ry 0.824 in. 1.24 in.

rz 0.642 in. -----

J 0.322 in4 0.644 in4

ro2 2.51 in.

H 0.646

AISC Page no. 1-36, 1-37, 1-99 1-75, 1-109

Step II. Calculate the x-axis buckling strength

• KL/rx = 16 x 12 /1.58 = 120.8

K x Lx Fy 120.8 36
• λc-x = = = 1.355
rx π E 3.1416 29000

φcPn = 0.85 x (0.658)1.355 x 36 x (2 x 3.75) = 106 kips

2
•

Step III. Calculate (KL/r)m for y-axis buckling

• (KL/r) = 16 x 12/1.24 = 154.8

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma

• a/rz = 48/0.648 = 74.07

a/rz = 74.07 < 0.75 x KL/r = 0.75 x 154.8 = 115.2 (OK!)

• α = h/2ry = (2 x 0.75 + 0.375)/(2 x 0.829) = l.131

2
 KL   KL  α2
2
a 
  =   + 0.82  
•  r m  r o 1+ α2  ry 
 

2
1.1312  48 
= (154.8)o2 + 0.82  
1 + 1.1312  0.829  =158.5

Step IV. Calculate flexural torsional buckling strength.

 KL  1 Fy
• λc-y =   × × =1.778
 r m π E

0.877 0.877
• Fcry = 2
× Fy = × 36 = 9.987 ksi
λ c− y 1.778 2

GJ 11,200 × 0.644
• Fcrz= = = 151.4 ksi
Aro2 7.5 × 2.512

    
• Fcrft =  Fcry + Fcrz  1 − 1 − 4 Fcry Fcrz H  =  9.987 + 151.4 1 − 1 − 4 × 9.987× 151.4 × 02.646 
 2H  2
(Fcry + Fcrz )   2 × 0.646  (9.987 + 151.4) 
  

Fcrft = 9.748 ksi

• φcPn = 0.85 x Fcrft x Ag = 0.85 x 9.748 x 7.50 = 62.1 kips

Flexural torsional buckling strength controls. The design strength of the double angle member is

62.1 kips.

Step V. Compare with design strengths in Table 4-10 (page 4-84) of the AISC manual

• φcPn for x-axis buckling with unsupported length = 16 ft. = 106 kips

• φcPn for y-z axis buckling with unsupported length = 16 ft. = 61.3 kips

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CE 405: Design of Steel Structures – Prof. Dr. A. Varma

These results make indicate excellent correlation between the calculations in steps II to IV and

the tabulated values.

Design tables for double angle compression members are given in the AISC manual. See

Tables 4-9, 4-10, and 4-11 on pages 4-78 to 4-93

- In these Tables Fy = 36 ksi

- Back to back distance = 3/8 in.

- Design strength for buckling about x axis

- Design strength for flexural torsional buckling accounting for the modified slenderness ratio

depending on the number of intermediate connectors.

- These design Tables can be used to design compression members as double angle sections.

38