LRFDLRFD--Steel DesignSteel DesignLRFDLRFD--Steel DesignSteel Design

Chapter 6Chapter 6

6.1 INTRODUCTION6.1 INTRODUCTION

Most beams and columns are subjected to some degree of both Most beams and columns are subjected to some degree of both

bending and axial load especially in statically indeterminate bending and axial load especially in statically indeterminate

structures.structures.

Many columns can be treated as pure compression members with Many columns can be treated as pure compression members with

negligible error.negligible error.

For many structural members, there will be a significant amount of For many structural members, there will be a significant amount of

both bending moment and axial load.both bending moment and axial load.

Such members are called beam-column.Such members are called beam-column.

Consider the rigid frame shown in the Figure:Consider the rigid frame shown in the Figure:

For the given loading condition, For the given loading condition,

The horizontal member AB must not only support the vertical The horizontal member AB must not only support the vertical

uniform load but must also assist the vertical members in uniform load but must also assist the vertical members in

resisting the concentrated load Presisting the concentrated load P11..

Member CD is a more critical case, because it must resist the load Member CD is a more critical case, because it must resist the load

PP11 + P + P22 without any assistance from the vertical members. without any assistance from the vertical members.

The reason is that the bracing members, prevents sidesway in the The reason is that the bracing members, prevents sidesway in the

lower story. lower story. (in the direction of P, ED will be in tension and CF will be slack)(in the direction of P, ED will be in tension and CF will be slack)

Member CD must transmit the load PMember CD must transmit the load P11 + P + P22 from C to D. from C to D.

The vertical members of this frame must also be treated as beam-The vertical members of this frame must also be treated as beam-

column.column.

In addition, at A and B, B.M. are transmitted from the horizontal In addition, at A and B, B.M. are transmitted from the horizontal

member through the rigid joints. member through the rigid joints.

This is also occur at C and D and is true in any rigid frame.This is also occur at C and D and is true in any rigid frame.

Most columns in rigid frame are actually beam-columns, and the Most columns in rigid frame are actually beam-columns, and the

effects of bending should not be ignored.effects of bending should not be ignored.

Another example of beam-column can sometimes be found in roof Another example of beam-column can sometimes be found in roof

trusses if purlins are placed between the joints of the top chord. trusses if purlins are placed between the joints of the top chord.

6.2 INTRODUCTION FORMULAS6.2 INTRODUCTION FORMULAS

The inequality Equation may be written in the following form:The inequality Equation may be written in the following form:

If both bending and axial compression are acting the interaction If both bending and axial compression are acting the interaction

formula would beformula would be

WhereWhere

PPuu is the factored axial compressive load. is the factored axial compressive load.

ФФcc P Pnn is the compressive design strength. is the compressive design strength.

MMuu is the factored bending moment. is the factored bending moment.

ФФcc M Mnn is the flexural design strength. is the flexural design strength.

1.0resistance

effectsload

φR

Qγ

n

ii

1.0Mφ

M

Pφ

p

nb

u

nc

u

For biaxial bending, there will be two bending ratios:For biaxial bending, there will be two bending ratios:

Two formulas are given in the specification:Two formulas are given in the specification:

One for small axial load and one for large axial load.One for small axial load and one for large axial load.

1.0Mφ

M

Mφ

M

Pφ

p

nyb

uy

b

ux

nc

u

nx

1.0Mφ

M

Mφ

M

P2φ

p

Pφ

1.0Mφ

M

Mφ

M

Pφ

p

Pφ

nyb

uy

b

ux

nc

u

nc

nyb

uy

b

ux

nc

u

nc

nx

u

nx

u

PFor

PFor

2.0

9

8

2.0

Example 6.1Example 6.1

Determine whether the member shown in the Figure satisfies the Determine whether the member shown in the Figure satisfies the

appropriate AISC Specification interaction equation if the bending appropriate AISC Specification interaction equation if the bending

is about the strong axis.is about the strong axis.

Solution:Solution:

From the column load tables:From the column load tables:

ФФcc P Pn n = 382 kips.= 382 kips.

Since bending is about the strong axis,Since bending is about the strong axis,

ФФbb M Mn n for Cfor Cb b can be obtained from the beam can be obtained from the beam

Design chart in Part 5 of the Manual.Design chart in Part 5 of the Manual.

For LFor Lbb = 17 ft, = 17 ft, ФФbb M Mn n = 200 ft.kips. For the end condition and loading = 200 ft.kips. For the end condition and loading

of this problem, Cof this problem, Cb b = 1.32.= 1.32.

ФФbb M Mn n = C= Cb b * 200 = 1.32 * 200 = 264.0 ft-kips.* 200 = 1.32 * 200 = 264.0 ft-kips.

This moment is larger than This moment is larger than ФФbb M Mp p = 227 ft-kips (also from Manual)= 227 ft-kips (also from Manual)

So the design moment must be limited to So the design moment must be limited to ФФbb M Mp.p. Therefore, Therefore,

ФФbb M Mn n = 227 ft-kips.= 227 ft-kips.

Max. B.M. occurs at midheight, so Max. B.M. occurs at midheight, so MMu u = 25*17/4 = 106.3 ft-kips.= 25*17/4 = 106.3 ft-kips.

Determine which interaction equation controls:Determine which interaction equation controls:

This member satisfies the AISC Specification.This member satisfies the AISC Specification.

1.00106.3

382

200

Mφ

M

Mφ

M

Pφ

p

Pφ

nyb

uy

b

ux

nc

u

nc

94.02279

8

9

8

2.005236382

200

nx

uP

6.3 MOMENT AMPLIFICATION6.3 MOMENT AMPLIFICATION

The presence of the axial load produces secondary moment.The presence of the axial load produces secondary moment.

The total moment =The total moment =P

8

2wL

The second term may be neglected if P is small.The second term may be neglected if P is small.

Because the total deflection cannot be found Because the total deflection cannot be found

directly, this problem is nonlinear, and without directly, this problem is nonlinear, and without

knowing the deflection, we cannot compute the knowing the deflection, we cannot compute the

moment.moment.

Ordinary structural analysis methods that do not take the displaced into Ordinary structural analysis methods that do not take the displaced into

account are referred to as first-order methods.account are referred to as first-order methods.

Iterative numerical techniques, called second-order methods, can be Iterative numerical techniques, called second-order methods, can be

used to find the deflection and secondary moments. used to find the deflection and secondary moments.

These method are usually implemented with a computer program. These method are usually implemented with a computer program.

Most current design codes permit the use of either a second-order Most current design codes permit the use of either a second-order

analysis or the moment analysis or the moment amplification method.amplification method.

This method entails computing the maximum B.M. resulting from This method entails computing the maximum B.M. resulting from

flexural loading by a first-order analysis, then multiplying it by a flexural loading by a first-order analysis, then multiplying it by a

moment amplification factor to account for the secondary moment amplification factor to account for the secondary

moment.moment.

Where, MWhere, M00 is the unamplified maximum moment. is the unamplified maximum moment.

PPee is the Euler buckling load = and P is the Euler buckling load = and Puu is factored load. is factored load.

)eu

0max /P(P1

1MM

2

2

(kL)

EIπ

As we describe later, the exact form of the AISC moment As we describe later, the exact form of the AISC moment amplification factor can be slightly different.amplification factor can be slightly different.

Example 6.2. Example 6.2. Compute the amplification factor for the beam-Compute the amplification factor for the beam-column of example 6.1. column of example 6.1.

This represents a 12 % increase in B.M.This represents a 12 % increase in B.M.

MMmaxmax = 1.12 * 106.3 = 119 ft-kips = 1.12 * 106.3 = 119 ft-kips

kips187412)/4.35)*(17*(1.0

14.4*29000*πP

(kL/r)

EAπ

(kL)

EIπP

2

2

e

2

2

2

2

e

1.12(200/1874)1

1factorAmp.

)/P(P1

1factorionAmplificat

eu

6.4 WEB LOCAL BUCKLING IN BEAM-COLUMNS6.4 WEB LOCAL BUCKLING IN BEAM-COLUMNS

The determination of the design moment requires that the cross The determination of the design moment requires that the cross

section be checked for compactness.section be checked for compactness.

The web is compact for all tabulated shapes if there is no axial load.The web is compact for all tabulated shapes if there is no axial load.

If λ ≤ λIf λ ≤ λpp, the shape is compact, the shape is compact

If λIf λpp ≤ λ ≤ λ ≤ λ ≤ λrr, the shape is noncompact; and, the shape is noncompact; and

If λIf λrr ≤ λ, the shape is slender ≤ λ, the shape is slender

AISC B5, Table 5.1, prescribes the following limits:AISC B5, Table 5.1, prescribes the following limits:

. .

yb

u

yp

yb

u

Pφ

2.75P1

F

E3.76λ0.125,

Pφ

PFor

yyb

u

yp

yb

u

FPφ

P2

F

E.λ0.125,

Pφ

PFor

E49.133.121

For any value ofFor any value of

Where PWhere Pyy = A = Agg F Fyy

Because PBecause Pyy is variable, compactness of the web cannot be is variable, compactness of the web cannot be

checked and tabulated. checked and tabulated.

Some rolled shapes satisfy the worst case limit of Some rolled shapes satisfy the worst case limit of

Shapes listed in the column load tables in Part 4 of the Manual Shapes listed in the column load tables in Part 4 of the Manual

that do not satisfy this criterion are marked, and these shapes that do not satisfy this criterion are marked, and these shapes

need to be checked for compactness of the web.need to be checked for compactness of the web.

Shapes whose flanges are not compact are also marked Shapes whose flanges are not compact are also marked

yb

u

yr

yb

u

Pφ

P

F

Eλ,

Pφ

P74.00.170.5

yFE /49.1

Example 6.3Example 6.3A W12 x 58 of A992 steel is subjected to a bending moment A W12 x 58 of A992 steel is subjected to a bending moment

and a factored axial load of 300 kips. Check the web for and a factored axial load of 300 kips. Check the web for compactness. compactness.

The upper limit is The upper limit is

From the dimension and properties tables From the dimension and properties tables λλ = h/t = h/tw w = 27.0 < λ= 27.0 < λpp

The web is therefore compactThe web is therefore compact

0.125Aφ

P

Pφ

P

b

u

yb

u 3922.050*0.17*90.0

300

ygF

27.523922.33.12133.121

02

50

29000.

Pφ

P2

F

E.λ

yb

u

yp

27.5227.5288.3529000

49.149.1 p

E 50Fy

6.5 BRACED VERSUS UNBRACED FRAMES6.5 BRACED VERSUS UNBRACED FRAMES

Two amplification factors are used in LRFDTwo amplification factors are used in LRFD

One to account for amplification resulting from the member deflection and the other to One to account for amplification resulting from the member deflection and the other to

account for the effect of sway when the member is part of unbraced frame. The following account for the effect of sway when the member is part of unbraced frame. The following

Figure illustrates these two components.Figure illustrates these two components.

. .

In Figure a, the member is restrained against In Figure a, the member is restrained against

sidesway, and the max. secondary moment is Psidesway, and the max. secondary moment is Pδδ..

If the frame is unbraced, there is an additional If the frame is unbraced, there is an additional

component of the secondary moment, shown in component of the secondary moment, shown in

Figure b, and the max value of it is PFigure b, and the max value of it is PΔΔ, which , which

represents an amplification of the end moment.represents an amplification of the end moment.

The amplified moment to be used in design is:The amplified moment to be used in design is:

MMuu = B = B11 M Mntnt + B + B22 M Mltlt

WhereWhere

MMnt nt = maximum moment assuming that no sidesway occurs.= maximum moment assuming that no sidesway occurs.

MMltlt = maximum moment caused by sidesway, = 0.0 in the actuall = maximum moment caused by sidesway, = 0.0 in the actuall

braced frame.braced frame.

BB1 1 = amplification factor for the moment occurring in the member = amplification factor for the moment occurring in the member

when it is braced against sidesway.when it is braced against sidesway.

BB2 2 = amplification factor for the moment resulting from sidesway.= amplification factor for the moment resulting from sidesway.

The following sections explain the evaluation of BThe following sections explain the evaluation of B11 and B and B22 . .

6.6 MEMBERS IN BRACED FRAMES6.6 MEMBERS IN BRACED FRAMES

The amplification factor given in section 6.3 was derived for a The amplification factor given in section 6.3 was derived for a

member braced against sidesway.member braced against sidesway.

The following Figure shows a member of this type The following Figure shows a member of this type

Maximum moment amplification occurs at the Maximum moment amplification occurs at the

center, where the deflection is largest.center, where the deflection is largest.

For equal end moment, the moment is constant For equal end moment, the moment is constant

throughout the length of the member, throughout the length of the member,

So the maximum primary moment also occurs at the center.So the maximum primary moment also occurs at the center.

If the end moments are not equal, the maximum primary and If the end moments are not equal, the maximum primary and

secondary moments will occur near each other.secondary moments will occur near each other.

If the end moments produce reverse-curvature bending as shown. If the end moments produce reverse-curvature bending as shown.

Here the max. primary moment is at one of the ends, and the max. Here the max. primary moment is at one of the ends, and the max.

amplification occurs between the ends. amplification occurs between the ends.

Therefore, the max. moment in a beam-column depends on the Therefore, the max. moment in a beam-column depends on the

distribution of bending moment within the member. distribution of bending moment within the member.

This distribution is accounted for by a factor, CThis distribution is accounted for by a factor, Cmm, applied to the , applied to the

amplification factor given in section 6.3.amplification factor given in section 6.3.

The amplification factor given in section 6.3 was derived for the The amplification factor given in section 6.3 was derived for the

worst case, so Cworst case, so Cmm will never be greater than 1.0. will never be greater than 1.0.

The final form of the The final form of the amplification factor isamplification factor is

2

2

1 )/(1

) rKL

EAwhere g

e1e1u

m P/P(P1

CB

Note:Note:

When computing pWhen computing pe1e1, use KL/r for the axis of bending and an effective length , use KL/r for the axis of bending and an effective length

factor K less than or equal to 1.0 (braced condition)factor K less than or equal to 1.0 (braced condition)

Evaluation of CEvaluation of Cmm

The factor Cm applies only to the braced condition. The factor Cm applies only to the braced condition.

There are two categories of members:There are two categories of members:

1.1.Members with transverse loads applied between it’s ends.Members with transverse loads applied between it’s ends.

2.2.Members with no transverse loads.Members with no transverse loads.

If there are no transverse loads acting on the member,If there are no transverse loads acting on the member,

MM11/M/M2 2 is the ratio of the bending moments at the ends of the is the ratio of the bending moments at the ends of the

member. (member. (MM1 1 is the smallest value and Mis the smallest value and M2 2 is the largest one).is the largest one).

The ratio is positive for member bent in reverse curvature and The ratio is positive for member bent in reverse curvature and

negative for single curvature bending as shown in the Figure.negative for single curvature bending as shown in the Figure.

2

14.06.0M

MCm

For transversely loaded members, CFor transversely loaded members, Cmm can be taken as 0.85 if the can be taken as 0.85 if the

ends are restrained against rotation (fixed) and 1.0 if the ends are ends are restrained against rotation (fixed) and 1.0 if the ends are

unrestrained against rotation (pinned). unrestrained against rotation (pinned).

End restrained will usually result from the stiffness of members End restrained will usually result from the stiffness of members

connected to the beam-column.connected to the beam-column.

Although the actual end condition may lie between full fixity and a Although the actual end condition may lie between full fixity and a

frictionless pin, use of one of the two values given here will give frictionless pin, use of one of the two values given here will give

satisfactory results.satisfactory results.

Example 6.5Example 6.5

The horizontal beam-column shows in the Figure is subjected to the The horizontal beam-column shows in the Figure is subjected to the

service live loads shown. This member is laterally braced at its ends, and service live loads shown. This member is laterally braced at its ends, and

bending is about the x-axis. Check for compliance with the AISC bending is about the x-axis. Check for compliance with the AISC

Specification. Specification.

SolutionSolution

The factored load = PThe factored load = Pu u =1.6*28=44.8 kips, w=1.6*28=44.8 kips, wuu=12*0035=0.042 kips=12*0035=0.042 kips

The maximum moment isThe maximum moment is

The member is braced against end translation, so MThe member is braced against end translation, so M ltlt =0.0 =0.0

kipsft112.58

(10)*0042

4

10*44.8M

2

nt

Compute the moment amplification factorCompute the moment amplification factor

The member braced against sidesway, transversely loaded, and with The member braced against sidesway, transversely loaded, and with

unrestrained ends, Cunrestrained ends, Cmm can be taken as 1.0. can be taken as 1.0.

The amplification factor is:The amplification factor is:

For the axis of bendingFor the axis of bending

kips2522(34.19)

10.3*29000*π

(Kl/r)

EAπP

2

2

2

g2

e1

11.018)(44.8/2522-1

1

)/P(P-1

CB

e1u

m1

kips-ft114.50112.5*1.018MBMBM Lt2nt1u

compactisshapethe,pλλsince

9.15250

290000.38

yF

E0.38pλ8.10,

f2t

fbλ

flange the of scompactnes for check first strength, flexural the For

kips339.138.73*10.3*0.85crFgAcφnPcφ

ksi38.73(50)2(0.7813)

(0.658)y)F2cλ

(0.658crF

inelastic1.5cλ

0.781329000

50

π

29.11

E

yF

rπ

KLcλ

59.112.03

12)*1.0(10

yr

LyK

r

KLMaximum

rL

bL

pLSince

ftinrL

yFX

yF

Xyr

rL

ftin

yF

Eyr

pL

ft

09.241.2892)1050)(610*763(11)1050(

3610*03.2

2)10(2

11)10(

1

17.704.8650

2900003.2*76.176.1

10bL

buckling. torsional Lateral

kipsft122.81473*0.9Mφ

kipsin14731024.09

7.17101248)(17351735M

kipsin124831.2*10)(5010)S(FM

kipsin173534.7*50ZFM

)M(MMM

nb

n

yr

yp

rppn

pL

rL

pL

bL

Because the beam weight is very small in relation to the Because the beam weight is very small in relation to the

concentrated live load, Cconcentrated live load, Cbb may be taken from Figure 5.15c as may be taken from Figure 5.15c as

1.32. This value results in a design moment of 1.32. This value results in a design moment of kips162.1ft122.8*1.32Mφ nb

This moment is greater than the plastic moment = 0.90*1735/12 This moment is greater than the plastic moment = 0.90*1735/12

=130.1 ft-kips, so the design strength must be limited to this value.=130.1 ft-kips, so the design strength must be limited to this value.

Check the interaction formula:Check the interaction formula:

adequateis35xW8aSo

(OK)1.00.9430130.1

114.1

2(339.1)

44.8

Mφ

M

Mφ

M

P2φ

P

0.200.1321339.1

44.8

Pφ

P

nyc

uy

nxc

ux

nc

u

nc

u

Please read the remaining examples, 6.4 and 6.6 from the text Please read the remaining examples, 6.4 and 6.6 from the text

book.book.

6.7 MEMBERS IN UNBRACED FRAMES6.7 MEMBERS IN UNBRACED FRAMES

The amplification factor given in section 6.3 was derived for a The amplification factor given in section 6.3 was derived for a

member braced against sidesway.member braced against sidesway.

The following Figure shows a member of this type The following Figure shows a member of this type