MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE VIII

Dr. Jason E. Charalambides

Introduction to Steel Design,Tensile Steel Members

Modes of Failure & Effective Areas

2

Failure in Flexure

! Consider an experimentwhere a point load is appliedon a W-section at mid-pointalong the strong axis:"

3

Failure in Flexure

! The load was increased until P_max is reached, i.e.flexural failure is experienced.

! Mn is the nominal flexural strength that the beam cansustain without failure, and it is considered based on thefollowing failure modes

" Local Buckling• Flange• Web

" Lateral torsional buckling" Development of fully plastic cross

section failure by excessivedeformation

4

Local Buckling

! Buckling occurs on the upper flange and the upperportion of the web that are subjected to compression" Buckling would not occur in any portion below the NA as it is

subject to Tensile stresses

5

Local Buckling

! Flange" In this case the flange starts

twirling, with one side bendingdownwards and the otherupwards to compensate thecompressive stress that it issubjected to, which exceeds itsmaximum capacity.

6

Local Buckling

! Web" In this case the web behaves in

a manner similar to the way wesaw the flange behaving in theprevious slide to compensatethe compressive stress that it issubjected to, which exceeds itsmaximum capacity.

7

Lateral Torsional Buckling

! Lateral Torsional Buckling" The center line of top flange shifts off

axis, and the beam rotates, minimizingthe exposure of the load on its strong xxaxis, and compensating the stress takenby rotating toward the weaker yy axis.The compression flange buckles laterally,like a column.

" The tension flange resists the bucklingand the lateral movement of thecompression flange, resulting in acombined effect of lateral movement androtation.

" To prevent this effect, lateral braces mustbe provided along the length of the beam

8

Compact Section

! What is defined as a compact section?" It is thick and not too long, practically the opposite of a “slender”

element, that can develop a plastic stress distribution beforebuckling

! What parameters define a section as compact? " It has to have width-thickness rations equal to or less than the

limiting values of λp given in Table B4.1 (pp. 16.1-16 &17), and " The compressive portions have to be connected to the web or

webs

9

Bracing

! Limit to moment that a beam can sustain:" A beam's strength against LTB depends on the unbraced

length Lb:" The shorter the Lb, the higher the strength of the beam against

LTB:• Think of it in a way similar to a column's slenderness. The

longer the unbraced length, the higher the slenderness factor

10

Beam Lateral Braces

! In a building it is typical to have bracing by cross beams or by aconcrete floor slab. A lateral brace needs to satisfy the following:" Prevent lateral displacement of the compression flange, OR" Prevent twist of the element

11

Beam Lateral Braces

! In this example with the shear studs applied on the top flange. In thescenario where the bottom flange would be in compression, the concretefloor slab would not serve as a lateral brace as it would not control thelateral movement of the compression flange.

12

Beam Lateral Braces

! An example where the lateral displacement of compression flange isprevented by diagonal bracing. " A typical scenario is the use of angles and gussets.

13

Ranges Of Beam Behavior inFlexure

! In all three previously mentioned scenarios of instability (FLB, WLB, LTB)occurs, it is safe to state that the failure occurred in one of the followingthree possible ranges of beam behavior. " Failure in Plastic range" Failure in Inelastic range" Failure in Elastic range

14

Ranges Of Beam Behavior inFlexure

! Summarizing in a tabular format the characteristics of failure and ranges:

"

"

"

"

"

"

! The AISC provides different equations that govern the design of flexuralmembers for each of these ranges of behavior

Failure mode Mn Ductility

● Elastic Mn ≤ 0.7 My Little or none

● Inelastic 0.7<Mn≤Mp Little or none

● Plastic Mn = Mp Large

15

Designing for Flexure

! At any point along the length of a flexural member, the following equationhas to be satisfied:

! Mu is the moment generated due to design loads! Mn is the nominal flexural strength of the beam

" Mn = Lesser of:

Mu≤ΦM n Φ=0.9

• Mn based on Flange Local Buckling

• Mn based on Web Local Buckling

• Mn based on Lateral Torsional Buckling

• Mp

16

Nominal Moment on LocalBuckling! There are two major factors that govern the behavior of a steel

flexural member on local buckling:" Slenderness parameter (width to thickness ratio) of flange and/or web" Specified minimum Yield Stress (Fy)

! The symbol λ is used as a generic symbol for that purpose (pg. 16.1-xl)

" For FLB:

" For WLB:

! The values indicated above are tabulated for all rolled W shapes inthe AISC user's manual in Part 1

λ=b f2t f

λ= htw

17

Nominal Moment on LocalBuckling! To determine the Mn based on FLB and WLB, the value of the

slenderness ratio is to be compared to the slenderness ratio for acompact and a non-compact element (λr and λp respectively.)

! Table Β4.1b in section B4 provides all the necessary formulae.See examples for hot rolled W shape:

" For FLB:

" For WLB:

Where λp is the limiting slenderness parameter for compactelement and λr for non-compact

λ=b f2t f

λ= htw

λ p=0.38√ EF yλ p=3.76√ EF y

λr=1.0√ EF y

λr=5.70√ EF y

18

Nominal Moment w.r.t.Unbraced Length

19

Determining Compactness

! For an element to be considered compact the followingconditions must be satisfied:

"

" For FLB:

" For WLB:

! If the above conditions are met, the section is compact, and failureby local buckling will occur in the plastic range.

b f2t f

≤0.38√ EF yhtw

≤3.76√ EF y• Defined as“Compact Section”

λ≤λ p

20

Determining Compactness

! An element that lies in the area of the graph where InelasticLTB is anticipated is Non-Compact:

"

" For FLB:

" For WLB:

! If the above conditions are met, failure by local buckling occurs inthe inelastic range, at a moment between 0.7 My and Mp.

0.38√ EF y< b f2t f

≤1.0√ EF y

3.76√ EF y< htw

≤5.70√ EF y

• Defined as“Non-Compact Section”

λ p<λ≤λ r

21

Determining Compactness

! An element that lies in the area of the graph that Elastic LTBis anticipated is considered Slender:

"

" For FLB:

" For WLB:

! If the above conditions are met, failure by local buckling occurs inthe elastic range, at a moment below 0.7 My.

b f2t f

>1.0√ EF yhtw

>5.70√ EF y• Defined as“Slender Section”

λ>λr

22

Effect of Web LocalBuckling on Mn

! For sections with a compact web, WLB has no effect on Mn, whichis controlled by FLB, LTB, or Mp.

! For sections with a non-compact or slender web:" WLB is considered to reduce the amount of web area

available to resist the applied moment

• Buckled portionconsidered ineffective inresisting moment

23

Effect of Web LocalBuckling on Mn

! For sections with a non-compact or slender web: (cont)" Buckled portion is ineffective," Remaining portions of the cross section can continue to resist

moment until FLB, LTB, or yielding occurs," Equations to account for the effects of WLB on Mn are given in

chapter F, Sections F4 and F5

24

Practical Considerations

! Flange Local Buckling: Most rolled shapes arecompact for FLB at Fy=50ksi.

! Web Local Buckling: All rolled W-shapes arecompact for WLB for Fy=50ksi.

! Local Buckling may control Mn for the followingcases:" Rolled W-shapes with high strength steels (Fy≥50ksi)

" Welded W-shapes made of thin plates" Angles, WT's, and shapes other than standard rolled W-shapes

25

FundamentalNomenclature! Cb - Lateral Torsional Buckling modification factor (F1-1)! Beam Section Properties that LTB depends upon:

" J – St. Venant's torsional constant" Cw – Warping torsion constant" Iy – Moment of inertia about y axis" Sx – Elastic section modulus" h0 – Distance between flange centroids" r_ts -Effective radius of gyration for LTB

! It can be approximated to the radius of gyration of thecompression flange + 1/6 of the web

! The above values are available except of the Cb that can becalculated

rts=√ √ I y CwS x

26

The Lateral Torsional BucklingModification Factor

! The lateral buckling can be substantially restrained by theend conditions and affected by loading patterns.

! Therefore a factor that helps better define the anticipatedbehavior of an element, based on the factors affect thisbehavior has been developed and refined to today's Cb:

" F1-1

" Where: Mmax is the absolute value of max. moment, MA, MB, andMC is the value at quarter, mid, and three quarter points of theunbraced segment.

Cb=12.5Mmax

2.5Mmax+3MA+4MB+3MC

27

Some Standard Values of Cb

Diagrams copied for educational purposes only

28

Some Standard Values of Cb

Diagrams copied for educational purposes only

Also see Table 3-1 on pg. 3-18 of AISC Manual

29

Determining The Limit State ofthe Unbraced Length

Lp=.76 r y√ EF y" F2-5

" F2-6Lr=1.95 rtsE

0.7 F y √ J cS x h0+√(J cS x h0

)2

+6.79(0.7F y

E)2

The coefficient c (in Jc) is equal to 1 for symmetric shapes.For channels refer to AISC manual F2-8b, pg 16.1-48

30

Determining The Limit State ofthe Unbraced Length

For Lb≤Lp→ΦM n=ΦM p=F y×Z x" F2-1

" F2-2

" F2-3

For Lp<Lb≤Lr→ΦM n=ΦCb [M p−(Mp−0.75F y S x)( Lb−L pLr−Lp )]≤ΦM p

For Lb>Lr→ΦM n=Φ Fr×S x≤ΦM p

31

The critical Stress

! The Critical Stress Fcr, is used in the formulae abovementioned and under these conditions it is given by:

F2-4F cr=

Cbπ2 E

( Lbrts )2 √1+0.078 J c

S x h0 (Lbrts )

2

32

In Class Exercise

Determine the largest value of uniformly distributed load thatcan be applied on a W-shape (W18x50) spanning 20 ft.Assume shear is OK. Use A992 steel:

By M. EngelhardtBeam subjected to LTBProblem Statement:Determine the largest value of uniformly distributed loadthat can be applied on the W-shape (W18x50). Assumeshear is OK.Use A992 steel

Unbraced Length: Lb 20ft:=

Section modulus: Sx 88.9in3:=

Plastic modulus: Zx 101in3:=

Moment of Inertia on yy Iyy 40.1in4:=

radius of gyration y ry 1.65in:=

effective radius of gyration rts 1.98in:=

Dist flange centroids h0 17.4in:=

Warping constant Cw 3040in6:=

Torsional constant J 1.24in4:=

Young's Modulus of Elasticity E 29000ksi:=

Yield Stress: Fy 50ksi:=

Ultimate Strength: Fu 65ksi:=

LTB factor for non-uniformmoment diagrams

Cb 1.14:= Factor of Safety phi ϕ 0.9:=

Solution:1) Determining the plastic moment and the elastic moment

ΦMp ϕZx Fy⋅:= ΦMp 378.75 k'⋅= ΦMr ϕ 0.7⋅ Sx Fy⋅:= ΦMr 233.36 k'⋅=

Check Local buckling, OK per table 3-2, (pg 325)

2) Calculating the Limit State lengths

Lp 1.76 ry⋅EFy

:= Lp 5.83 ft⋅= Lr 1.95 rts⋅E.7 Fy⋅⋅

JSx h0⋅

⋅ 1 1 6.76.7 Fy⋅

E

Sx h0⋅

J⋅

2

++⋅:= Lr 16.95 ft⋅=

OR the above values of Lp and Lr can be obtained from table 3-2 (pg 325)

3) Calculating the Nominal Moment

Since Lb>Lr we need to apply the formulae F2-3 and F2-4

FcrCb π

2⋅ E⋅

Lbrts

21 0.078

JSx h0⋅⋅

Lbrts

2

⋅+⋅:= Fcr 30.76 ksi⋅= ΦMn ϕFcr Sx⋅:= ΦMn 205.1 k'⋅=

4) Calculating the maximum Uniformly Distributed load possible based on Moment capacity:

wmax8 ΦMn⋅

Lb2

:= wmax 4.1kipft

⋅=

OR we could go to charts, p 3-129 at Lb=20ft, read up to W18x50, find ΦMn at Cb=1 that gives us 180k', multiply 180 by 1.14 whichgives us 205k', and then use the formula in step 4 to determine that the maximum happens to be 4.1 kips per foot! Note thatW18x50 is given in a dotted line that indicates that the section is not compact.

By M. EngelhardtBeam subjected to LTBProblem Statement:Determine the largest value of uniformly distributed load that can beapplied on the W-shape (W18x50). Assume shear is OK.Use A992steel. Length is 20 ft with lateral supports at ends and midpoint

Unbraced Length: Lb 10ft:=

Section modulus: Sx 88.9in3:=

Plastic modulus: Zx 101in3:=

Moment of Inertia on yy Iyy 40.1in4:=

radius of gyration y ry 1.65in:=

effective radius of gyration rts 1.98in:=

Dist flange centroids h0 17.4in:=

Warping constant Cw 3040in6:=

Torsional constant J 1.24in4:=

Young's Modulus of Elasticity E 29000ksi:=

Yield Stress: Fy 50ksi:=

Ultimate Strength: Fu 65ksi:=

LTB factor for non-uniformmoment diagrams

Cb 1.3:= Factor of Safety Φ ϕ 0.9:=

Solution:1) Determining the plastic and elastic moments

Mp Zx Fy⋅:= Mp 420.83 k'⋅= My Sx Fy⋅:= My 370.42 k'⋅= Elast_fail_Mom My:= Elast_fail_Mom 370.42 k'⋅=

2) Calculating the Limit State lengths

Lp 1.76 ry⋅EFy

:= Lp 5.83 ft⋅= Lr 1.95 rts⋅E.7 Fy⋅

⋅J

Sx h0⋅⋅ 1 1 6.76

.7 Fy⋅

E

Sx h0⋅

J⋅

2

++⋅:= Lr 16.95 ft⋅=

OR the above values of Lp and Lr can be obtained from table 3-2 (p 325)

3) Calculating the Nominal Moment

Since Lb<Lr we need to apply the formula F2-2

ΦMn ϕCb Mp Mp 0.7My−( )Lb Lp−

Lr Lp−

−

⋅= ... ϕCb Mp Mp 0.7My−( )Lb Lp−

Lr Lp−

−

⋅ 421.451 k'⋅= BUT

ΦMn if ϕCb Mp Mp 0.7My−( )Lb Lp−

Lr Lp−

−

⋅ .9 Mp⋅≤ ϕCb Mp Mp 0.7My−( )Lb Lp−

Lr Lp−

−

⋅, ϕ Mp⋅,

:=ΦMn 378.75 k'⋅=

Check Local buckling, OK per table 3-2

4) Calculating the maximum Uniformly Distributed load possible based on Moment capacity:

wmax8 ΦMn⋅

2 Lb⋅( )2:= wmax 7.58

kipft

⋅= Note that we used 2Lb because the total length of the beam isactually twice the length of the unbraced sections!

ΦMp .9 Mp⋅:= ΦMp 378.75 k'⋅=

OR we could go to charts, p 3-124 at Lb=10ft, read up to W18x50, find PhiMn at Cb=1 that gives us 324k', multiply 324 by 1.3 whichgives us 421k'. That is already larger thatn 0.9 Mp, so 0.9Mp governs at 379 k'... and then use the formula in step 4 to determine thatthe maximum happens to be 7.58 kips per foot!