05 bending

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Local Buckling in Bending

Read before the lecture

Sections 2.5 & additional material included in this lecture

After this lecture you should

Be able to compute flexural stresses and plastic moment for any arbitrary section.

Understand local buckling in members subjected to bending

Be able to compute flexural strength of various sections based on the flange and web local buckling.

Review Activity

An I-beam is made by welding three plates as shown in the figure. Determine elastic and plastic section moduli withrespect to horizontal axis of the cross-section. Also compute the yield moment and the plastic moment if F y = 50 ksi.

PL20"μ½"

PL12"μ2"

PL9"μ2"

Compression

Tension

Tension flange, compression flange, and web plate dimensions

bft 9; tft 2;

bfc 12; tfc 2;

hw 20; tw 1 2;

h tft hw tfc

24

Area of cross section, in2

A bft tft bfc tfc hw tw N

52.

(a) Computation of elastic section moduli with respect to x-axis

University of Iowa Instructor: M. Asghar Bhatti



PL20"μ½"

PL12"μ2"

PL9"μ2"

y t

y c

Centroidal axis

Centroid location from the tension flange, in

yt bfc tfc tft hw .5 tfc bft tft .5 tft hw tw tft 0.5 hw A

13.2692

Centroid location from the compression flange, in

yc bfc tfc .5 tfc bft tft tft hw .5 tft hw tw tfc 0.5 hw A

10.7308

Moment of inertia about x-axis, in4

Ix 1 12 bfc tfc^3 bfc tfc yc .5 tfc ^ 2

1 12 bft tft^3 bft tft yt .5 tft ^ 2

1 12 tw hw^3 hw tw yt .5 hw tft ^ 2

5345.56

Elastic section modulus with respect to tension and compression faces, in3

Sxc, Sxt Ix yc, Ix yt

498.153, 402.854

(b) Computation of plastic section modulus with respect to x-axis

PL20"μ½"

PL12"μ2"

PL9"μ2"

y p t

y p c

Plastic neutral axis

Area under compression must be equal to the area under tension

Clearypc

ypt d ypc

24 - ypc

53:134 Design of Steel Structures 2




Area under compression

Ac bfc tfc tw ypc tfc

ypc - 2

2+ 24

Area under tension

At bft tft tw ypt tft

22 - ypc

2+ 18

Solve Ac At, ypc

ypcØ 6ypc 6;

ypt

18

Plastic section modulus, in3

Zx bfc tfc ypc 0.5 tfc tw ypc tfc ypc tfc 2

bft tft ypt 0.5 tft tw ypt tft ypt tft 2

494.

Yield moment M Y and plastic moment M p (kip-ft)

Fy 50;

Sxt, Zx

402.854, 494. Myx, Mpx Sxt, Zx Fy 12

1678.56, 2058.33

Additional Material that is part of Section 2.4

Web Local Buckling (WLB)

M M

A

A

Web slenderness parameter

lw = h t





the width/thickness ratio of stiffened elements acting as webs of beams. As shown in the following figure for W and C

shapes h is the clear depth of the web and t its thickness. For rectangular HSS shapes h = H - 3 t where H is the

dimension of the section acting as web (typically the longer dimension) and t its thickness. For round shapes h = D the

outside diameter and t is the wall thickness. For standard sections h t values are given in the AISC tables. They are

more accurate and should be used when available.

AISC specifications do not give explicit WLB criteria for unstiffened web elements such as the stems of Tee shapesacting as webs in flexure. Also the web local buckling does not apply to I and C shapes bent about their minor axis.

Limiting values lpw and lrw. AISC Table B4.1b p. 16.1-17.

Limiting values of lw:

For W and C shapes: lpw = 3.76E

F y; lrw = 5.70

E

F y

For an unsymmetric I shape: lpw =

hc

h p

E

F y

0.54 M p

M y-0.092

; lrw = 5.70E

F y

where M y is the yield moment and hc and h p are twice the dimensions shown in the following figure.





h

h c

2

h p

2

Plastic NACentroid

C o m p r e s s i o n

T e n s i o n

x

For rectangular HSS shapes: lpw = 2.42E

F y; lrw = 5.70

E

F y

For round HSS shapes or pipes: lpw = 0.07E

F y; lrw = 0.31

E

F y

Web is

compact if lw § lpw : No WLB

non - compact if lpw § lw § lrw : Inelastic WLB

slender if lw > lrw : Elastic WLB

lpw lrw

lw

M p

M n

WLB Flexural Strength

Compact Non-compact Slender

For sections with compact webs: fMn = 0.9 M p.

All standard W and C shapes for F y § 50 ksi have compact webs.

For I shapes with non-compact or slender webs the web local buckling strength (WLB) in not explicitly computed.

Instead web plastification factors Rpc and Rpt in compression and tension respectively are introduced in calculations for

compression flange local buckling and tension flange yielding.For rectangular HSS and built-up box-shape members with non-compact or slender webs (lw § lrw)

fMn = 0.9μ M p - M p - F y S x -0.738 + 0.305h

t

F y

E

M p = Plastic moment.

S x = elastic section modulus about x-axis.





For round HSS and pipe shapes

For non-compact sections (lpw § lw § lrw)

fMn = 0.9μ 0.021 E

Dt + F y S

For slender sections (lw > lrw)

fMn = 0.9μ 0.33 E

Dt S

S = elastic section modulus

Class Activity

Determine flexural strength about x-axis based on WLB for a rectangular HSS18x6x1/4 section made of A500 Gr. B

steel.

6"

1 8 "

x

y

t, w, Zx, Sx AISCshapeProps"HSS18x6x14", "tdes", "Htt", "Zx", "Sx"

0.233, 74.3, 59.4, 46.5Fy 46; Es 29 000;

Mp Zx Fy

2732.4

For rectangular HSS shapes: lpw = 2.42E

F y; lrw = 5.72

E

F y

pw, rw 2.42, 5.72 SqrtEs Fy

60.7625, 143.62Since lw > lpw the web is non-compact

fMn = 0.9μ M p - M p - F y S x -0.738 + 0.305h

t

F y

E

Mn Mp Mp Fy Sx 0.738 0.305 w SqrtFy Es

2634.76

Beam strength (kip - ft)





MnWLB 0.9 Mn 12

197.607

AISC Table 3-12 p. 3-140 lists the flexural strength for this section = 198 kip-ft.

Compression Flange Local Buckling (FLB)

MM

A

A

Flange slenderness parameter

l f = b t

the width/thickness ratio of unstiffened elements acting as flanges of beams. For W and T shapes b is half the width of

the flange and t its thickness. For C shapes b is the full width of the flange and t its thickness. For angles b is the width

of the leg that is acting as flange and t its thickness.

The only section with stiffened elements explicitly covered in the specifications for FLB is rectangular HSS section.

Here b is based on the dimensions that is acting as the flange (typically the shorter dimension).





Limiting values lpf and lrf . AISC Table B4.1 p. 16.1-16 values for unstiffened elements.

Limiting values of l f :

For W, C, and T shapes lpf = 0.38E

F y; lrf = 1.0

E

F y

For angles lpf = 0.54E

F y; lrf = 0.91

E

F y

For rectangular HSS shapes: lpf = 1.12E

F y; lrf = 1.4

E

F y

Flange is

compact if l f § lpf : No F LB

non - compact if lpf § l f § lrf : Inelastic FLB

slender if l f > lrf : Elastic FLB

lpf lrf

l f

§ M p

M n

Flexural strength based on FLB

Compact Non-compact Slender





FLB Flexural strength for symmetric I shapes bent about major axis

b f

t f

d 2

h

t w

Centroid

C o m p r e

s s i o n

T e n s i o n

x

l f = 0.5 b f t f

Flange slenderness parameters: lpf = 0.38E

F y; lrf = 1.0

E

F y

Web slenderness parameters: lw =h

t w; lpw = 3.76

E

F y; lrw = 5.70

E

F y

For symmetric I shapes with compact webs and non-compact flanges (lw § lpw and lpf § l f § lrf )

fMn = 0.9μ M p - M p - 0.7 F y S x l f -lpf

lrf -lpf

M p = Plastic moment.

S x = I x 0.5 d elastic section modulus

For symmetric I shapes with non-compact webs and non-compact flanges (lw > lpw and lpf § l f § lrf )

fMn = 0.9μ R p M y - R p M y - 0.7 F y S x l f -lpf

lrf -lpf R p =

M p

M y- M p

M y- 1 lw-lpw

lrw-lpw

web plastification factor

M y = F y S x yield moment for compression flange.

For all symmetric I shapes with slender flanges (l f > lrf )

fMn = 0.9μ0.9 E k c S x

l f 2

k c =Min0.76, Max 0.35,4

ht w

Class Activity

Determine flexural strength based on FLB for a W21x48 A992 beam.

Fy 50; Es 29 000;

Limiting values of local buckling parameters

w, f, Zx, Sx AISCshapeProps"W21x48", "htw", "bf2tf", "Zx", "Sx"

53.6, 9.47, 107., 93.





pf, rf, pw, rw 0.38, 1.0, 3.76, 5.70 SqrtEs Fy

9.15161, 24.0832, 90.5528, 137.274Since lpf § l f § lrf the flange is non-compact. Since lw § lpw the web is compact.

FLB flexural strength: fMn = 0.9μ M p - M p - 0.7 F y S x l f -lpf

lrf -lpf

Plastic moment

Mp Zx Fy

5350.

Mn Mp Mp 0.7 Fy Sx f pf rf pf

5305.33


MnFLB 0.9 Mn 12

397.9

FLB Flexural strength for I or C shapes bent about minor axis (y-axis)

l f = 0.5 b t

lpf = 0.38E

F y; lrf = 1.0

E

F y

S y = I y0.5 b f elastic section modulus for I shape about y-axis.

S y = I y

b f - x

elastic section modulus for C shape about y-axis.

M p = F y Z y Plastic moment about y axis.

For compact flanges: fMn = 0.9μ M p

For non-compact flanges

fMn = 0.9μ M p - M p - 0.7 F y S y l f -lpf

lrf -lpf

For slender flanges





fMn = 0.9μF cr S yF cr =

0.69 E

l f 2

Class Activity

Determine minor axis flexural strength based on FLB for a W21x48 A992 beam.

Fy 50; Es 29 000;

w, f, Zy, Sy AISCshapeProps"W21x48", "htw", "bf2tf", "Zy", "Sy"

53.6, 9.47, 14.9, 9.52Limiting values of local buckling parameters

pf, rf 0.38, 1.0 SqrtEs Fy

9.15161, 24.0832Since lpf § l f § lrf the flange is non-compact.

FLB flexural strength: fMn = 0.9μ M p - M p - 0.7 F y S x l f -lpf

lrf -lpf

Plastic moment

Mp Zy Fy

745.

Mn Mp Mp 0.7 Fy Sy f pf rf pf

736.219


MnFLB 0.9 Mn 12

55.2164

FLB Flexural strength for rectangular HSS and built-up box-shape members


F y; lrf = 1.4

E

F y

For members with non-compact flanges (lpf § l f § lrf )

fMn = 0.9μ M p - M p - F y S -4 + 3.57b

t

F y

E

M p= Plastic moment.

S = elastic section modulus.

For members with slender flanges (l f > lrf )

fMn = 0.9μF y Seff Seff = elastic section modulus based on the effective width be of the compression flange defined as follows.





be = 1.92 t E

F y1 - 0.38

bt E

F y

B

H

be

x

Class Activity

Determine flexural strength about x-axis based on FLB for a rectangular HSS10x6x3/16 section made of A500 Gr. B

steel.

6"

1 0 "

x

y

Fy 46; Es 29 000;

t, f, Zx, Sx AISCshapeProps"HSS10x6x316", "tdes", "bt", "Zx", "Sx"

0.174, 31.5, 18., 14.9 Mp Zx Fy

828.


F y; lrf = 1.4

E

F y


28.1215, 35.1518Since l f > lpf but less than lrf the flange is non-compact

fMn = 0.9μ M p - M p - F y S -4 + 3.57b

t

F y

E

Mn Mp Mp Fy Sx 4 3.57 f SqrtFy Es

759.728






MnFLB 0.9 Mn 12

56.9796

FLB Flexural strength for Tees and Double angles loaded in the plane of symmetry

b

x

b

t

x

l f = b t

lpf = 0.38E

F y; lrf = 1.0

E

F y

Plastic and yield moments: M p = F y Z x M y = F y S x

If stem is in tension: fMn = 0.9μMin M p, 1.6 M y, F cr SxcIf tip of the stem is in compression: fMn = 0.9μMin M p, M y, F cr SxcSxc = elastic section modulus with respect to compression flange. Note S x given in the AISC manual is based on the

maximum distance to an extreme fiber. Thus in case of a Tee section it is with respect to tip of the stem and is not the

value needed in this calculation.

For members with non-compact flanges (lpf § l f § lrf )

F cr = F y 1.19 - 0.50 l f F y

E

For members with slender flanges (l f > lrf )

F cr =0.69 E

l f 2

Example

Determine flexural strength based on flange local buckling of a WT5x6 bent about its x -axis as shown in the figure.A992 Gr 50 steel.





Tension

Compression

x

y

Note as loaded the flange of the Tee is in compression. Thus we need to check FLB.

Fy 50; Es 29 000;

d, f, Zx, Sx, Ix, yb AISCshapeProps"WT5x6", "d", "bf2tf", "Zx", "Sx", "Ix", "y"

4.94, 9.43, 2.2, 1.22, 4.35, 1.36 Mp Fy Zx

110.

My Fy Sx

61.


9.15161, 24.0832Sxc Ix yb

3.19853

The flange is non-compact

fMn = 0.9μMin M p, 1.6 My, F cr Sxc

F cr = F y 1.19 - 0.50 l f F y

E

FLB strength (kip-ft)

MnFLB 0.9 Min Mp, 1.6 My, Fy Sxc 1.19 0.50 f SqrtFy Es 12

7.32

Leg local buckling (LLB) Flexural strength for single angles

The local leg buckling applies when the toe of the angle is in compression.





b

t

Tension

Compression

x

Leg local buckling

b

t

Tension

Compression

x

No leg local buckling

Compression

Tension

x

l f = b t

lpf = 0.54E

F y; lrf = 0.91

E

F y

For members with non-compact legs (lpf § l f § lrf )

fMn = 0.9μF y Sc 2.43 - 1.72 l f F y

E

For members with slender legs (l f > lrf )

fMn = 0.9μF cr Sc F cr = 0.71 E

l f 2

Sc = elastic section modulus to the toe in compression relative to the axis of bending. For bending about one of

the geometric axes of an equal-leg angle with no lateral torsional restraint Sc is equal to 0.80 of the geometric

axis section modulus.

Example

Determine flexural strength based on leg local buckling of an L4x4x1/4 bent about its geometric x-axis as shown in

the figure. A36 steel.

4"

1 4 "

Tension

Compression

x





Note as loaded the toe of the angle is in compression.

Fy 36; Es 29 000;

b 4; t 1 4;

Sx, Ag AISCshapeProps"L4x4x14", "Sx", "Ag"

1.03, 1.94Leg local buckling (LLB) Flexural strength for single angles

f b t

16


15.3264, 25.8279The angle leg is non-compact

Sc 0.8 Sx;

MnLLB 0.9 Fy Sc 2.43 1.72 f SqrtFy Es 12

3.24906

FLB Flexural strength for unsymmetric I shapes bent about major axis

bfc

bft

t f c

t f t

y c

y t

t w

h

h c

2 h

p 2

Plastic NA

Centroid


T e n s i o n

x

l f = 0.5 bfc t fc

lpw =Minlrw,

hc

h p

E

F y

0.54 M p

M y-0.092

; lrw = 5.70E

F y





lpf = 0.38E

F y; lrf = 1.0

E

F y

Sxc = I x yc elastic section modulus with respect to compression flange.

Sxt = I x yt elastic section modulus with respect to tension flange.

lwc = hc

t wweb slenderness with respect to compression flange

hc = 2μ distance from the centroid to the inside face of the compression flange.

h p = 2μ distance from the plastic neutral axis to the inside face of the compression flange.

M yc = F y Sxc yield moment for compression flange.

M yt = F y Sxt yield moment for tension flange.

Web plastification factor with respect to compression flange

Rpcï

For lwc § lpw compact web Rpc =M p

M yc

For lwc > lpw

noncompactweb

Rpc =

M p

M yc

-

M p

M yc

- 1

lwc-lpw

lrw-l

pw Web plastification factor with respect to tension flange

Rptï

For lwc § lpw compact web Rpt =M p

M yt

For lwc > lpw noncompactweb Rpt =M p

M yt

- M p

M yt

- 1 lwc-lpw

lrw-lpw

For unsymmetric I shapes with compact flanges and compact or non-compact webs

fMnïFor Sxt ¥ Sxc fMn = 0.9μ Rpc M ycFor Sxt < Sxc fMn = 0.9μMin Rpc M yc, Rpt M yt

For unsymmetric I shapes with non-compact flanges and compact or non-compact webs

fMn = 0.9μ Rpc M yc - Rpc M yc - F L Sxc l f -lpf

lrf -lpf

F Lï

For Sxt ¥ 0.7 Sxc F L = 0.7 F y

For Sxt < 0.7 Sxc F L =Max0.5 F y,Sxt

Sxc

F yFor unsymmetric I shapes with non-compact flanges and slender webs

fMn = 0.9μ Rpg F cr Sxc F cr = F y - 0.3 F y l f -lpf

lrf -lpf

Bending strength reduction factor: Rpg = 1 -aw

1200+300 aw

hc

t w- 5.7

E

F y

aw =Min10,hc t w

bfc t fc

For sections with slender flanges

fMn = 0.9μ Rpg F cr SxcF cr =

0.9 E k c

l f 2





k c =Min0.76, Max 0.35,4

ht w

Class Activity

Determine flexural strength based on FLB for the built-up I beam shown in the figure.

20.5"

12"

7 1 6 "

3 4 "

y c

y t

14"

5 0

"

h c

2

Centroid


T e n s i o n

x

Fy 50; Es 29 000;

Compression flange, tension flange, and web dimensions

bfc 20.5; tfc 7 16.;

bft 12; tft 3 4.;

h 50; tw 1 4.;

Overall section depth

d h tft tfc

51.1875

Area of cross section in2 A bfc tfc bft tft h tw

30.4688

Centroid location





yc bft tft d .5 tft bfc tfc .5 tfc h tw .5 h tfc A

25.5095

hc 2 yc tfc

50.144

yt d yc

25.678

Ix 1 12 bft tft^3 bft tft yt 0.5 tft ^ 2

1 12 bfc tfc^3 bfc tfc yc 0.5 tfc ^ 2

1 12 tw h ^ 3 h tw yt 0.5 h tft ^ 2

14 103.6

Sxc Ix yc

552.875

Sxt Ix yt

549.248

Limiting values of local buckling parameters

pf, rf, rw 0.38, 1.0, 5.70 SqrtEs Fy

9.54121, 25.1085, 143.118Web and flange slenderness parameters

f, w 0.5 bfc tfc, h tw

23.4286, 200.Since lpf § l f § lrf the flange is non-compact. Since lw > lrw the web is slender.

fMn = 0.9μ Rpg F cr Sxc F cr = F y - 0.3 F y l f -lpf

lrf -lpf

Bending strength reduction factor: Rpg = 1 -aw

1200+300 aw

hc

t w- 5.7

E

F y

aw =Min10,hc t w

bfc t fc

aw Min10, hc tw bfc tfc

1.39774

Rpg 1 aw 1200 300 aw hc tw 5.7 SqrtEs Fy

0.94536

Fcr Fy 0.3 Fy f pf rf pf

35.6576


fMn = 0.9μ Rpg F cr Sxc





MnFLB 0.9 Rpg Fcr Sxc 12

1397.78

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05 bending

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