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Chapter 12. Combined Forces on Steel Members

Chapter 12. Combined Forces on Steel Members

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214

Principles of Structural Design: Wood, Steel, and Concrete

WD = 2 k/ft

100 k

PL = 100 k

12 ft

Figure 12.1  A tensile and flexure forces member.

The analysis, where a member size is known and the adequacy of the member to handle a certain

magnitude of force is to be checked, is a direct procedure by Equations 12.1 and 12.2. However,

the design of a member that involves selecting of a suitable size for a known magnitude of load is a

trial-and-error procedure by the interaction equations, Equations 12.1 and 12.2. The AISC Manual

2005 presents a simplified procedure to make an initial selection of a member size. This procedure,

however, necessitates the application of the factors that are available from the special tables in the

manual. Since the manual is not a precondition for this text, that procedure is not used here.

Example 12.1

Design a member to support the load shown in Figure 12.1. It has one line of four holes for 7/8 in.

bolt in the web for the connection. The beam has adequate lateral support. Use Grade 50 steel.

Solution

A. Analysis of structure

1. Assume a beam weight of 50 lb/ft

2. Wu = 1.2(2.05) = 2.46 k/ft

2

WuL2 ( 2.46)(12)

=

= 44.28 ft-k or 531.4 in.-k

3. Mu =

8

8

4. Pu = 1.6(100) = 160 k

B. Design

1. Try W 10 × 26 section*

2. Ag = 7.61 in.2

3. Ix = 144 in.4

4. Zx = 31.3 in.3

5. tw = 0.26 in.

6. bf /2tf = 5.70

7. h/tw = 29.5

C. Axial (tensile) strength

1. U = 0.7 from “Shear Lag” section of Chapter 9, h = 7/8 + 1/8 = 1, Ah = 1(0.26) = 0.26 in.2

2. An = Ag − Ah = 7.61 − 0.26 = 7.35 in.2

3. Ae = 0.7(7.35) = 5.15 in.2

4. Tensile strength

ϕFy Ag = 0.9(50)(7.62) = 342.9 k

ϕFu Ae = 0.75(65)(5.15) = 251.06 k ← Controls

D. Moment strength

E

= 9.15 > 5.7 , it is a compact flange

1. 0.38

Fy

3.76

E

= 90.55 > 29.5, it is a compact web

Fy

P

160

* As a guess, the minimum area for axial load alone should be Au = u =

= 3.55 in.2 The selected section is twice

F

0

.

9(50)

φ

y

this size because a moment Mu is also acting.

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Combined Forces on Steel Members

3. Moment strength

ϕbFy Z = 0.9(50)(313) = 1400.5 in.-k

E. Interaction equation

Pu

160

=

= 0.64 > 0.2, Use Equation 12.1

1.

φPn 251.06

2.

Pu

8  Mux 

8  531.4 

+

= (0.64) + 

 = 0.97 < 1 OK

φPn 9  φbMnx 

9  1400.5 

Combination of Compression and Flexure

Forces: The Beam-Column Members

Instead of axial tension, when an axial compression acts together with a bending moment, which

is a more frequent case, a secondary effect sets in. The member bends due to the moment. This

causes the axial compression force to act off center resulting in an additional moment equal to the

axial force times the lateral displacement. This additional moment causes further deflection, which

in turn produces more moment, and so on until an equilibrium is reached. This additional moment

known as the second-order moment is not as much of a problem with the axial tension, which tends

to reduce the deflection.

There are two kinds of the second-order moments, as discussed below.

Members without Sidesway

Consider an isolated beam-column member AB of a frame in Figure 12.2. Due to load wu on the

member itself a moment Mnt results assuming that the top joint B does not deflect with respect to

the bottom joint A (i.e., there is no sway). This causes the member to bend, as shown in Figure 12.3.

The  total moment consists of the primary (the first order) moment, Mnt and the second-order

moment, Puδ. Thus,

Mu1 = Mnt + Puδ

(12.3)

where Mnt is the first-order moment in a member assuming no lateral movement (no translation).

Pu

B

Pu

C

wu

A

D

Figure 12.2  Second-order effects on a frame.

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Principles of Structural Design: Wood, Steel, and Concrete

Members with Sidesway

Pu

Now consider that the frame is subject to a sidesway where the ends

of the column can move with respect to each other, as shown in

Figure 12.4. Mlt is the primary (first order) moment caused by the

lateral translation of the frame only. Since the end B is moved by Δ

with respect to A, the second-order moment is Pu Δ.

Therefore, the total moment is

Mu 2 = Mlt + Pu ∆

A

δ

(12.4)

where Mlt is the first-order moment caused by the lateral translation.

It should be understood that the moment Mu1 (Equation 12.3) is the

property of the member and Mu2 (Equation 12.4) is a characteristic

of a frame. When a frame is braced against sidesway, Mu2 does not

exist. For an unbraced frame, the total moment is the sum of Mu1 and

Mu2. Thus,

Mnt

Mu = ( M nt + Puδ ) + ( Mlt + P ∆ )

(12.5)

Mnt

B

Pu

Figure 12.3  Second-order

moment within a member.

Pu

B ∆

Mlt

The second-order moments are evaluated directly or through the factors that magnify the primary moments. In the second case,

Mu = B1M nt + B2 Mlt

(12.6)

where B1 and B2 are the magnification factors when the first-order

moment analysis is used.

For braced frames, only the factor B1 is applied. For unbraced

frames, the factors B1 and B2 both are applied.

A

Pu

Figure 12.4  Second-order

moment with sideway.

Magnification Factor, B1

This factor is determined assuming the braced (no sway) condition. It can be demonstrated that for

a sine curve, the magnified moment directly depends on the ratio of the applied axial load to the

elastic (Euler) load of the column. The factor is expressed as follows:

B1 =

Cm

≥1

1 − ( Pu Pe1 )

(12.7)

where Cm is an expression that accounts for the nonuniform distribution of the bending moment

within a member as discussed in the “Moment Modification Factor, Cm” section below.

Pe1 = Euler buckling strength

=

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π 2 EA

( KL r )2

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Combined Forces on Steel Members

The slenderness ratio ((KL)/r) is along the axis in which the bending occurs and Pu is the applied factored axial compression load.

Equation 12.7 suggests that B1 should be greater or equal to 1; it

is a magnification factor.

M1

M1

δ

δ

Moment Modification Factor, Cm

In Equation 12.7 for B1, a factor Cm appears. Without this factor,

B1 may be over-magnified. When a column is bent in a single curvature with equal end moments, the deflection occurs, as shown in

Figure  12.5a. In this case, Cm = 1. When the end moments bend a

member in a reverse curvature, as shown in Figure 12.5b, the max

deflection that occurs at some distance out from the center is smaller

than the first case; using Cm = 1 will overdo the magnification. The

purpose of the modifier Cm is to reduce the magnified moment when

the variation of the moment within a member requires that B1 should

be reduced. The modification factor depends on the rotational restraint

placed at the member ends. There are two types of loading for Cm.

M2

(a)

M2

(b)

Figure 12.5  Deflection of

a column under different end

moment conditions.

1. When there is no transverse loading between the two ends of a member:

The modification factor is given by

M 

Cm = 0.6 − 0.4  1  ≤ 1

 M2 

(12.8)

where

M1 is the smaller end moment

M2 is the larger of the end moments

The ratio (M1/M2) is negative when the end moments have the opposite directions causing the member to bend in a single curvature. (This is opposite of the sign convention for

concrete column in the “Short Columns with Combined Loads” section in Chapter 16.) The

ratio is taken to be positive when the end moments have the same directions causing the

member to bend in a reverse curvature.

2.When there is a transverse loading between the two ends of a member:

The Cm value is as follows:

a. Cm = 0.85 for a member with the restrained (fixed) ends

b. Cm = 1.0 for a member with the unrestrained ends

Example 12.2

The service loads on a W12 × 72 braced frame member of A572 steel are shown in Figure 12.6.

The bending is about the strong axis. Determine the magnification factor B1. Assume the pinned

end condition.

Solution

73397_C012.indd 217

1. Pu = 1.2(100) + 1.6(200) = 440 k

2. (Mu)B = 1.2(15) + 1.6(40) = 82 ft-k

3. (Mu)A = 1.2(20) + 1.6(50) = 104 ft-k

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Principles of Structural Design: Wood, Steel, and Concrete

PD = 100 k

PL = 200 k

MD = 15 ft-k

ML = 40 ft-k

14 ft

A

MD = 20 ft-k

ML = 50 ft-k

Figure 12.6  A braced frame for Example 12.2.

B. Modification factor

M1 −82

=

= −0.788

1.

M2 104

2. Cm = 0.6 − 0.4(−0.788) = 0.915

C. Euler buckling strength

1. K = 1

2. For W12 × 72, A = 21.1 in.2

rx = 5.31 in., bending in X-direction

KL (1) (14 × 12)

=

= 31.64

3.

rx

5.31

4. Pe1 =

=

5. B1 =

π 2EA

(KL r )2

π 2 ( 29, 000)( 21.2)

(31.64)2

= 6, 055 k

Cm

1− (Pu Pe1)

=

0.915

= 0.99 < 1

 440 

1− 

 6, 055 

Use B1 = 1

Braced Frame Design

To braced frames, only the magnification factor B1 is applied. As stated earlier, the use of the

interaction equation, Equations 12.1 or 12.2, is direct in analysis when the member size is known.

However, it is a trial-and-error procedure for designing of a member.

Instead of making a blind guess, design aids are available to make a feasible selection prior

to application of the interaction equation. The procedure presented in the AISC Manual 2005 for

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Combined Forces on Steel Members

initial selection needs an intensive input of the data from the special tables included in the manual.

In a previous version of the AISC manual, a different approach was suggested, which was less data

intensive. This approach is described below.

The interaction equations can be expressed in terms of an equivalent axial load. With respect to

Equation 12.1, this modification has been demonstrated below:

Muy 

Pu 8  Mux

+ 

+

=1

φPn 9  φb Mnx φb Mny 

Multiply the both sides by ϕPn,

Pu +

8 φPn  Mux Muy 

+

= φPn

9 φb  M nx M ny 

(12.9)

Treating ϕPn as Peff, this can be expressed as

Peff = Pu + m Mux + mU Muy

(12.10)

where

Pu is the factored axial load

Mux is the magnified factored moment about X-axis

Muy is the magnified factored moment about Y-axis

The values of the coefficient m, reproduced from the AISC manual, are given in Table 12.1.

The manual makes an iterative application of Equation 12.10 to determine the equivalent axial

compressive load, Peff for which a member could be picked up as an axially loaded column only.

However, this also requires the use of an additional table to select the value of U.

This chapter suggests an application of Equation 12.10 just only to make an educated guess for a

preliminary section. The initially selected section will then be checked by the interaction equations.

The procedure is as follows:

1. With the known value of the effective length, KL select a value of m from Table 12.1 for the

shape category proposed to be used, i.e., W12. Assume U = 3.

2.From Equation 12.10, solve for Peff

3.Pick up a section having the cross-sectional area larger than the following:

Ag =

Peff

φFy

4.Confirm the selection by the appropriate interaction equation, Equation 12.1 or

Equation 12.2

Example 12.3

For a braced frame, the axial load and the end moments obtained from structural analysis are

shown in Figure 12.7. Design a W14 member of A992 steel. Use K = 1 for the braced frame.

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14

16

3.6

3.9

3.2

3.0

2.6

2.1

1.8

W, S 4

W, S 5

W, S 6

W8

W 10

W 12

W 14

2.6

3.2

2.7

2.9

2.5

2.1

1.7

2.3

1.9

2.4

2.3

2.8

2.5

2.0

1.7

2.2

1.6

1.9

2.0

2.6

2.4

2.0

1.7

2.2

1.5

1.9

2.3

2.3

2.0

1.7

2.1

2.0

1.9

2.4

2.5

2.8

2.5

2.0

1.7

12

Subsequent approximations

2.7

1.4

3.3

1.6

1.5

3.0

2.0

2.0

3.0

2.1

2.0

2.5

2.0

2.0

2.0

1.7

1.7

1.8

10

2.3

22 and over

First approximation

1.9

2.4

20

Note: Values of m are for Cm = 0.85. When Cm is other than 0.85, multiply the tabular value of m by Cm /0.85.

2.4

All shapes

18

12

KL (ft)

10

36 ksi

Fy

Table 12.1

Values of Factor m

1.6

1.8

2.2

2.5

2.4

2.0

1.7

2.2

14

1.6

1.6

1.9

2.2

2.3

1.9

1.7

2.0

16

1.4

1.8

1.9

2.1

1.9

1.7

1.9

18

50 ksi

1.4

1.5

1.6

1.9

1.8

1.7

1.8

20

1.5

1.6

1.7

1.7

1.7

1.7

22 and over

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Principles of Structural Design: Wood, Steel, and Concrete

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Combined Forces on Steel Members

MDX = 15 ft-k

MLX = 45 ft-k

PD = 80 k

PL = 200 k

A

My = 120 ft-k

14 ft

B

MDX = 20 ft-k

MLX = 50 ft-k

A

14 ft

B

My = 120 ft-k

Figure 12.7  A column member of a braced frame.

Solution

(a) 1.2D + 1.6L

1. Assume a member weight of 100 lb/ft, total weight = 100(14) = 1400 lb or 1.4 k

2. Pu = 1.2(81.4) + 1.6(200) = 417.7 k

3. (Mnt)x at A = 1.2(15) + 1.6(45) = 90 ft-k

4. (Mnt)x at B = 1.2(20) + 1.6(50) = 104 ft-k

(b) 1.2D + L + 1.6W

1. Pu = 1.2(84) + 200 = 297.7 k

2. (Mnt)x at A = 1.2(15) + 45 = 63 ft-k

3. (Mnt)x at B = 1.2(20) + 50 = 74 ft-k

4. (Mnt)y = 1.6(120) = 192 ft-k

B. Trial selection

1. For the load combination (a)

From Table 12.1 for KL = 14 ft, m = 1.7

Peff = 417.7 + 1.7(104) = 594.5 k

2. For load combination (b), Let U = 3

Peff = 297.7 + 1.7(74) + 1.7(3)(192) = 1402.7 k ← Controls

P

1402.7

= 31.17 in.2

3. Ag = eff =

φFy (0.9)(50)

4. Select W14 × 109 A = 32.0 in.2

Zx = 192 in.3

Zy = 92.7 in.3

rx = 6.22 in.

ry = 3.73 in.

bf /2tf = 8.49

h/tw = 21.7

Checking of the trial section for the load combination (b)

C. Along the strong axis

1. Moment strength

ϕMnx = ϕFy Zx = 0.9(50)(192) = 8640 in.-k or 720 ft-k

2. Modification factor: reverse curvature

(Mnt )x @A = 63 = 0.85

(Mnt )x @B 74

Cmx = 0.6 − 0.4(0.85) = 0.26

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Principles of Structural Design: Wood, Steel, and Concrete

3. Magnification factor

K = 1

KL (1) (14 × 12)

=

= 27.0

rx

6.22

(Pe1)x =

4.

π 2EA

(KL rx )2

(B1)x = 1−

=

Cm

(Pu

=

π 2 ( 29,000)(32)

( 27.0)2

= 12,551

Pe1)

0.26

= 0.256 < 1; use 1

1− ( 297.7 12,551)

5. (Mu)x = B1(Mnt)x

= 1(74) = 74 ft-k

D. Along the minor axis

1. Moment strength

ϕMny = ϕFy Zy = 0.9(50)(92.7) = 4171.5 in.-k or 347.63 ft-k

2. Modification factor: reverse curvature

(Mnt )y @A

=

(Mnt )y @B

208

=1

208

Cmx = 0.6 − 0.4(1) = 0.2

3. Magnification factor

K = 1

KL (1) (14 × 12)

=

= 45.0

ry

3.73

(Pe1)y =

π 2EA

(KL r )

2

=

y

4.

(B1)y = 1−

=

Cm

(Pu

π 2 ( 29,000)(32)

( 45.0)2

= 4518.4

Pe1)

0 .2

= 0.21 < 1; use 1

1− ( 297.7 4518.4)

5. (Mu)y = (B1)y(Mnt)y

= 1(192) = 192 ft-k

E. Compression strength

KL (1) (14 × 12)

=

= 27.0

1.

rx

6.22

73397_C012.indd 222

2.

KL (1) (14 × 12)

=

= 45.0 ← Controls

ry

3.73

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Combined Forces on Steel Members

3. 4.71

4. Fe =

E

29,000

= 4.71

= 113.43 > 45 Inelastic buckling

Fy

50

π 2E

(KL r )

2

=

y

π 2 ( 29,000)

( 45.0)2

= 141.2

5. Fcr = (0.65850/141.2)50 = 43.11

6. ϕPn = 0.9Fcr Ag

= 0.9(43.11)(32) = 1241.6 k

F. Interaction equation

Pu

297.7

=

= 0.24 > 0.2

φPn 1241.6

Muy 

8  Mux

Pu

+

+

φPn 9  φbMnx φbMny 

0.24 +

8  74

192 

+

9  720 347.63 

= 0.82 < 1 OK

Magnification Factor for Sway, B2

The term B2 is used to magnify the column moments under sidesway condition. For sidesway to

occur in a column on a floor, it is necessary that all of the columns on that floor should sway simultaneously. Hence the total load acting on all columns on a floor appears in the expression for B2. The

AISC Manual 2005 presents the following two relations for B2:

B2 =

1

ΣPu  ∆H 

1−

ΣH  L 

(12.11)

or

B2 =

1

ΣPu

1−

ΣPe 2

(12.12)

where

ΔH is the lateral deflection of the floor (story) in question

L is the story height

ΣH is the sum of horizontal forces on the floor in question

ΣPu is the total design axial force on all the columns on the floor in question

ΣPe2 is the summation of the elastic (Euler) capacity of all columns on the floor in question,

given by

ΣPe2 = Σ

73397_C012.indd 223

π 2 EA

( KL r )2

(12.13)

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Principles of Structural Design: Wood, Steel, and Concrete

PD = 240 k

W12 × 96

360 k

15 ft

W12 × 120

30 ft

360 k

W12 × 120

30 ft

240 k

W12 × 96

30 ft

Figure 12.8  An unbraced frame.

The term Pe2 is similar to the term Pe1, except that the factor K is determined in the plane of bending

for an unbraced condition (K in Pe1, was for the braced condition).

The designer can use either of the Equation 12.11 or Equation 12.12; the choice is a matter

of convenience. In Equation 12.11, initial size of the members is not necessary since A and r as

a part of Pe2 are not required unlike in Equation 12.12. Further, a limit on ΔH/L, known as the

drift index, can be set by the designer to control the sway. This is limited to 0.004 with factored

Example 12.4

An unbraced frame of A992 steel at the base floor level is shown in Figure 12.8. The loads are

factored dead loads. Determine the magnification factor for sway for the column member (1)

bending in Y-axis.

Solution

A. Exterior columns

1. Factored weight of column = 1.2(0.096 × 15) = 1.7 k

2. Pu = 240 + 1.7 = 241.7 k

3. K = 2

4. For W12 × 96, A = 28.2 in.2

ry = 3.09 in.

KL 2 (15 × 12)

=

= 116.50

5.

ry

3.09

π 2 ( 29,000)( 28.2)

π 2EA

=

= 594.1k

6. Pe 2 =

2

(116.5)2

(KL ry )

B. Interior columns

1. Factored weight of column = 1.2(0.12 × 15) = 2.2 k

2. Pu = 360 + 2.2 = 362.2 k

3. K = 2

4. For W12 × 120, A = 35.3 in.2

ry = 3.13 in.

KL 2 (15 × 12)

=

= 115.0

5.

ry

3.13

6. Pe 2 =

π 2EA

(KL r )

y

73397_C012.indd 224

2

=

π 2 ( 29,000)(35.3)

= 763.2k

(115.0)

C. For the entire story,

1. ΣPu = 2(241.7) + 2(362.2) = 1208 k

2. ΣPe2 = 2(594.1) + 2(763.2) = 2714.6

6/17/2010 2:48:40 PM

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