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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

PLATE GIRDER 1. INTRODUCTION Welded plate girders, which are the most common form of plate girders, are built-up structural steel members that consists of flange plates welded to a web plate with fillet welds. They are used to support loads over long spans (60 ft. to 200 ft.) and to support structural loads that are too large to be supported by the rolled steel shapes shown in the AISCM. Plate girders are rarely used in building structures, but are commonly used in bridge structures. They are used as transfer girders in building structures to support columns above large column-free areas. Plate girders may also be used in the retrofitting of existing building structures where column-free areas are needed and existing columns have to be cut off or removed below a certain floor level. Plate girders are also used as crane support girders in heavy industrial structures with long spans. All the standard hot-rolled shapes in the Manual, the webs are compact. Some have noncompact flanges, but none have slender flanges. With shapes built up from plates, however, both flanges and webs can be compact, noncompact, or slender. These built-up shapes usually are used when the bending moments are larger than standard hot-rolled shapes can resist, usually because of a large span. These girders are invariably very deep, resulting in noncompact or slender webs. A plate girder cross section can take several forms. The usual configuration is a single web with two equal flanges, with all parts connected by welding. The box section, which has two webs as well as two flanges, is a torsionally superior shape and can be used when large unbraced lengths are necessary. Hybrid girders, in which the steel in the flanges is of a higher strength than that in the web or webs, are sometimes used.

Lecture 7 ....... Page 1

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

The term “plate girder” no longer exists in the most recent AISC specification. Instead, this term has been replaced by the term “built-up sections,” and the design requirements for flexure and shear for these sections are found in Sections F5 and G of the AISC specification, respectively. Stiffened plate girders may or may not be designed for tension-field action to resist the shear. A stiffened plate girder with tension-field action acts like a Pratt truss, with the web plate resisting diagonal tension; the vertical stiffeners are in compression and also add stability to the web plate. Tension-field action is this truss-like behavior, and designing for it is an economical way to increase the strength of the girder because the stability added by the stiffeners allows the web plate to be thinner and lighter than would otherwise be necessary. When tension-field action is used, however, the stiffeners must then be designed to have a larger moment of inertia.

Tension-field action is not permitted in the design of end panels. When designing or analyzing rolled sections, the overall depth of the member (measured between the outside faces of the flanges) is used to resist the shear force. For plate girders, only the girder web (measured between the inside faces of the flanges) is used to resist shear.

Lecture 7 ....... Page 2

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

2. STIFFENED AND UNSTIFFENED PLATE GIRDER At a location of high shear in a girder web, usually near the support and at or near the neutral axis, the principal planes will be inclined with respect to the longitudinal axis of the member, and the principal stresses will be diagonal tension and diagonal compression. The diagonal tension poses no particular problem, but the diagonal compression can cause the web to buckle. This problem can be addressed in one of three ways: (1) The depth-to-thickness ratio of the web can be made small enough that the problem is eliminated, (2) web stiffeners can be used to form panels with increased shear strength, or (3) web stiffeners can be used to form panels that resist the diagonal compression through tension-field action. Plate girders may be stiffened (with transverse stiffeners) or unstiffened. If an unstiffened web is incapable of resisting the applied shear, appropriately spaced stiffeners are used to develop tension-field action. Cross-sectional requirements for these stiffeners, called intermediate stiffeners, are minimal because their primary purpose is to provide stiffness rather than resist directly applied loads.

additional stiffeners may be required at points of concentrated loads for the purpose of protecting the web from the direct compressive load. These members are called bearing stiffeners, and they must be proportioned to resist the applied loads. They can also simultaneously serve as intermediate stiffeners.

Lecture 7 ....... Page 3

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

3.AISC REQUIREMENTS FOR PROPORTIONS OF PLATE GIRDERS 3.1- Web Proportions Whether a girder web is noncompact or slender depends on h/tw , the width-to-thickness ratio of the web, where h is the depth of the web from inside face of flange to inside face of flange and tw is the web thickness.

Doubly Symmetric I-Shaped section Noncompact

3.76

web

Singly Symmetric I-Shaped section

E h E   5.70 Fy t w Fy

hc hp

E Fy

  M  0.54 p  0.09    My  

Slender web

h tw

 5.70

2



hc E  5.70 tw Fy

hc E  5.70 tw Fy

E Fy

hc= twice the distance from the elastic neutral axis (the centroidal axis) to the inside face of the compression flange. (hc/2 defines the part of the web that is in compression for elastic bending. hc = h for girders with equal flanges).

hp= twice the distance from the plastic neutral axis to the inside face of the compression flange. (hp/2 defines the part of the web in compression for the plastic moment. hp= h for girders with equal flanges). Mp= plastic moment = Fy Zx My= yield moment = Fy Sx

To prevent web buckling (to prevent vertical buckling of the compression flange into the web), web stiffeners can be added for stability. AISC F13.2 imposes an upper limit on the web slenderness. The limiting value of h/tw is a function of the aspect ratio, a/h, of the girder panels, which is the ratio of intermediate stiffener spacing to web depth.

Lecture 7 ....... Page 4

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

Thus I-shaped members must also satisfy the following limits, where a is the clear distance between transverse stiffeners, and h is the height of the web between flanges.

Without transverse Web

With transverse Web Stiffeners

Stiffeners (Unstiffened Girder) h tw

 260 and

a 1.5 h

Aw 10 A fc  h   tw

 E   12 Fy  max

a 1.5 h  h   tw

 0.42 E   Fy  max

Aw = Area of the web = h*tw Afc = Area of the compression flange = bfc * tfc

3.2- Flange Proportions Singly symmetrical I-shaped members must satisfy : 0.1 

I yc Iy

 0.9

Iy = is the moment of inertia about the y-axis, and I Iyc = is the moment of inertia about the y-axis referred to the compression flange.

Lecture 7 ....... Page 5

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

4. FLEXURAL STRENGTH When a plate girder section has a noncompact or slender web, the nominal moment capacity, Mn , will be less than the plastic moment capacity, M p , of the section because of several limit states that are attained before the section can reach its full plastic moment capacity. There are four possible limit states that may occur in built-up sections in bending and Mn ,will be the smallest strength obtained for the following four limit states.: • Tension flange yielding. • Compression flange yielding, • Compression flange local buckling, and • Lateral torsional buckling, For all states   0.9

4.1. Tension Flange Yielding , TFY For S xt  S xc , the design moment capacity is given as  M n   Fy S xt

This limit state does not apply to built-up sections when the section modulus of the built-up section with respect to the tension face, Sxt , is greater than or equal to the section modulus with respect to the compression face, Sxc (i.e., when Sxt  Sxc ).

4.2. Compression Flange Yielding, CFY  M n   R pg Fy S xc   aw  R pg  1    1200  300 a w  

h   c  5.7 E   1.0  tw Fy   

where

aw 

hc t w  10 b fc t fc

S xc 

Ix yc

Lecture 7 ....... Page 6

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

4.3. Lateral Torsional Buckling, LTB Lateral torsional buckling is a function of the lateral unbraced length, L b , of the compression flange of the plate girder.  M n   R pg Fcr S xc

L p 1.1 rt

E Fy

L r   rt

E 0.7 Fy

rt 

b fc  1  12 1  a w   6 

Unbraced Length, Lb

Fcr

Lb  Lp

lateral torsional buckling limit state does not apply

Lp  Lb  Lr

 Fcr  C b Fy  0.3 Fy 

Lb  Lr

Fcr 

Cb 2 E  Lb   rt

  

2

 L b  L p     Fy  L r  L p   

 Fy

Cb = Bending moment coefficient. It is conservative to assume a Cb = 1.0

4.4. Compression Flange Local Buckling, FLB  M n   R pg Fcr S xc

If the section has compact flanges, the limit state of compression flange local buckling doesn’t apply. 

b fc 2 t fc

kc 

 p  0.38

4 h / tw

E Fy

 r  0.95

kc E FL

, (0.35  k c  0.76)

FL  0.7 Fy for symmetric girders with slender or noncompact webs.

Lecture 7 ....... Page 7

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

Flange Compactness Compact Flange

  p

Fy

 Fcr  Fy  0.3 Fy 

  r

Slender Flange

Example 1: Determine the design flexural strength  b M n of the following welded I-shaped plate girder. The flanges are 15*1.25 in, the web is 50*0.25 in, and the member is uniformly loaded and simply supported. Use A36 steel and assume the girder has continuous bracing for its compression flange.

Fcr 

b fc 15   6.0 2 t fc 2 * 1.25

   p  Compact Flange

0.9 E k c  b fc   2t fc

  

2

h c t w 50 * 0.25   0.667  10  OK bfc t fc 15 *1.25

  0.667  R pg  1    1200  300 * 0.667   0.982  1.0  OK

E 29000  0.38  10.79 Fy 36

   

4- CFY  M n   R pg Fy S xc

  aw  R pg  1    1200  300 a w  

Determine the flange compactness

   p   r  p 

3-LTB Since Lb=0, LTB does not apply

aw 

Solution

 p  0.38

Fcr

p    r

Noncompact Flange



Limit

h   c  5.7 E   1.0  tw Fy   

   200  5.7 29000   36  

15 *1.25 3  0.25 * 50 3  2  1.25 *15 * 25.625 2  12  12   27,233 Ix 

Determine the flange compactness h 50 E 29000   200  5.70  5.70  161.78 t w 0.25 Fy 36

Thus slender web

Sxc 

I x 27233   1037.4 yc 26.25

 Mn  0.9 * 0.982 * 36 *1037.4 / 12  2750 kips.ft

 b M n is the lowest value of TFY,CFY, LTB, FLB

1- TFY Since member is symmetric about x-axis S xt  S xc  TFY does not apply

2- FLB Since flange is compact, FLB does not apply

Lecture 7 ....... Page 8

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

Example 2: Determine the design flexural strength  b M n of the following welded I-shaped plate girder. The flanges are 24*1.0 in, the web is 45*5/16 in, and the member is uniformly loaded and simply supported 100 ft span. Use A36 steel and the unbraced length of compression flange is 20 ft.

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 25 *13  0.3125 * 453  2  1* 24 * (22.5  0.5)2  12  12   27,777 Ix 

S xc 

I x 277777   1182 yc 23.5

 M n  0.9 *1.0 * 36 *1182 / 12  3191.4 kips.ft

Solution

3-LTB

Determine the flange compactness

Lb  20 ft

b 24 E 29000   fc   12  p  0.38  0.38  10.79 2 t fc 2 *1 Fy 36

rt 

bfc  1  12 1  a w   6 

24



 1  12 1  * 0.586   6 

 6.61

  p

kc 

4 h / tw

 r  0.95



4 45 / 0.3125

 0.333

kc E 0.333 * 29000  0.95  18.59 FL 0.7 * 36

Lp  1.1 rt

E 29000  1.1* 6.61  206.36 in  17.2 ft Fy 36

Lr   rt

E 29000   * 6.61  704.45in  58.7 ft 0.7 Fy 0.7 * 36

p    r  Non  compact flange

Lp  Lb  Lr

Determine the flange compactness h tw

45 E 29000   144 }  5.70  5.70  161.78 0.3125 Fy 36 

E 29000  3.76  161.78 Fy 36

Thus Non-compact web  b M n is the lowest value of TFY,CFY, LTB, FLB

1- TFY Since member is symmetric about x-axis

  L b  L p    F Fcr  C b Fy  0.3 Fy   L r  L p  y      20  17.2  Fcr  1.0 36  0.3 * 36    35.27  36  OK  58.7  17.2  

 M n   R pg Fcr S xc  Mn  0.9 *1.0 * 35.27 *1182  3126.69 k.ft

2- FLB  M n   R pg Fcr S xc

S xt  S xc  TFY does not apply

p    r  Non  compact flange

2- CFY  M n   R pg Fy S xc

 Fcr  Fy  0.3 Fy 

aw 

h c t w 45 * 0.3125   0.586  10  OK bfc t fc 24 *1.0

  aw  R pg  1     1200  300 a w 

h   c  5.7 E   1.0  tw Fy   

  0.586 29000   144  5.7 R pg  1    36   1200  300 * 0.586    1.0075  1.0  R pg  1.0

   p   r  p 

   

  12  10.79  Fcr  36  0.3 * 36    34.32  18.59  10.79    M n  0.9 *1.0 * 34.32 *1182  3042.47 k.ft

Thus FLB gives smaller value (control) Thus  M n  3042.47 k.ft (FLB)

Lecture 7 ....... Page 9

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

5.SHEAR STRENGTH Due to the relatively thin webs used in plate girders, the design for shear is more complicated than for rolled sections. In fact, there are two approaches available for designing for shear in plate girders. One method accounts only for the buckling strength of the web, while the second method accounts for the post-buckling strength of the web panels between stiffeners as a result of diagonal tension field action. Therefore, unless diagonal tension field action is to be relied on, it is recommended that sufficient web thickness be used to avoid the need for stiffeners. The shear strength of a plate girder is a function of the depth-to-thickness ratio of the web and the spacing of any intermediate stiffeners that may be present. The shear capacity has two components: the strength before buckling and the post-buckling strength. The post-buckling strength relies on tension-field action, which is made possible by the presence of intermediate stiffeners. If stiffeners are not present or are spaced too far apart, tension-field action will not be possible, and the shear capacity will consist only of the strength before buckling.

Whether a plate girder will be unstiffened or stiffened must be decided in the early design stages. Using stiffeners reduces the total steel weight but increases fabrication costs. If the web height-to-thickness ratio h/tw > 260, transverse stiffeners are required.

AISC G3.1 lists all of the conditions under which a tension field cannot be used: a- In end panels  b- When a  3 OR a   260 h h  h / tw

c- When d-

  

2

2Aw  2.5 A fc  A ft 

h h or 6 b fc b ft

Aft= area of the tension flange bft= width of the tension flange

Lecture 7 ....... Page 10

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

Whether the shear strength is based on web shear yielding or web shear buckling depends on the web width-to-thickness ratio

1  If

kv E h  1.10 tw Fy

h tw

, Then the strength is based on shear yielding, and

 Vn   0.6 Fy A w C v

2  If

kv E h  1.10 tw Fy

, Then the strength will be based on shear buckling or shear

buckling plus tension-field action. If tension-field behavior exists,  1 C v  Vn   0.6 Fy A w  C v   1.15 1  a / h 2 

   

OR  Vn   0.6 Fy A w C v   0.6 Fy A w

1 C v 1.15 1  a / h 2

  0.9

Aw = area of the web = d * t d = overall depth of the beam Cv= ratio of critical web stress to shear yield stress

h / t w ratio

Unstiffened web with

kv 5

h  260 tw 2

 Stiffened web with a  3 OR a   260 h h  h / tw

  

 Stiffened web with a  3 OR a   260 h h  h / tw

  

2

5

5

a / h 2 5

Lecture 7 ....... Page 11

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

h / t w ratio

Cv C v 1.0

kv E h  1.10 tw Fy

kv E

1.10

Fy



kv E h  1.37 tw Fy

1.10

kv E Fy

h / tw h  1.37 tw

1.51 E k v

kv E

h / t w 2 Fy

Fy

6.DESIGN OF INTERMEDIATE STIFFENERS WITHOUT TENSION-FIELD ACTION Transverse stiffeners are not required if one of the following conditions is satisfied: kv E h  2.46 tw Fy Vu   Vn For k v  5

For all other conditions, transverse stiffeners are required and the spacing, a, and thickness, tst , of the stiffener must be selected to satisfy the following conditions:

I st  b t 2w j

b  min a , h j

2.5

a / h 2

I st 

 2  0.5

t st 2b st  t w 3 12

tst = Thickness of transverse stiffener, and bst= Width of transverse stiffener perpendicular to the longitudinal axis of the girder. Ist = Moment of inertia for a pair of stiffeners (i.e., stiffeners on both sides of the web) about a horizontal axis at the centerline of the web.

Lecture 7 ....... Page 12

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

6.DESIGN OF INTERMEDIATE STIFFENERS WITH TENSION-FIELD ACTION Transverse stiffeners with tension-field action must meet the preceding requirements for stiffeners without tension-field action, and they must meet the following requirements:

E b    0.56 Fy,st  t  st  V  Vc1   I st  I st1  (I st2  I st1 )  u  Vc 2  Vc1 

where (b/t)st = width-to-thickness ratio of the stiffener, Fy,st = yield stress of the stiffener Vu = is the larger of the required shear strengths in the adjacent web panels, Vc1 = the smaller of the available shear strengths (  Vn ) in the adjacent panels, calculated with no tension field action . Vc2

= the smaller of the available shear strengths (  Vn ) in the adjacent panels,

calculated with tension field action.

Ist1 = required moment of inertia as calculated for the no-tension-field case. Ist2 = moment of inertia required to develop the buckling plus post-buckling shear strength. h 4 st1.3 I st2  40

1.5

 Fyw     E   

  Fyw    st  max  ,1   Fyst   Fyw = yield stress of the girder web

Lecture 7 ....... Page 13

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

7.DESIGN OF BEARING STIFFENERS Bearing stiffeners are required when the web has insufficient strength for any of the limit states of web yielding, web crippling, or sidesway web buckling. Although the web can be proportioned to directly resist any applied concentrated loads, bearing stiffeners are usually provided. If stiffeners are used to resist the full concentrated load, the limit states of web yielding, web crippling, and sidesway web buckling do not need to be checked. The nominal bearing strength of a stiffener is given in AISC J7 as R n   1.8 Fy A pb ,   0.75

AISC J10.8 requires that full-depth stiffeners be used in pairs and analyzed as axially loaded columns subject to the following guidelines:

1- The cross section of the axially loaded member consists of the stiffener plates and a length of the web. This length can be no greater than 12 times the web thickness for an end stiffener or 25 times the web thickness for an interior stiffener. 2- The effective length should be taken as 0.75 times the actual length—that is, KL = 0.75h.

Lecture 7 ....... Page 14

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Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

3- The nominal axial strength is - If

KL  25  Pn   Fy A g ,   0.9 r

- If

KL  25  the usual requirements for compression members in AISC E apply. r

E b 4-    0.56 Fy,st  t  st

The spacing of stiffeners could be obtained from Table 3-16 and 3-17

Lecture 7 ....... Page 15

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Misan University Engineering College Civil Department

Example 3: A plate girder of A36 steel shown in figure, has been selected for a 65 ft simple span to support the loads wD=1.1 k/ft (not including the beam weight) and wL=2 k/ft. Bearing stiffeners are provided at the ends of girder. a- Find spacing of intermediate stiffeners if needed b- Design intermediate stiffeners c-Design bearing stiffeners

7

For the end panels, tension-field action is not permitted, and the shear strength must be computed from the Equation:

 Vn   0.6 Fy A w  0.9 * 0.6 * 36 * 31.59 * 0.1268  77.87  Vu  156.98

Thus intermediate stiffeners are required -Select stiffeners for end panel (Tension filed action may not be used) V  Vn 156.98  u   4.97 Aw Aw 31.59

Enter AISC Table 3-16a for (tension filed action not included) with h/tw=219 and  Vn / A w  4.97 and read a / h  1.00

 a  1.00 * 82  82 in

-Select spacing of stiffeners for second panel in which tension field action is permitted.

Solution

shear at second panel is at 82 in form support.

Computing girder weight

A  2 * 1.125 * 20  3 / 8 * 82  75.75 in2

Self weight / ft  A *  steel 

82  123.97 k 12  Vn  77.87  Vu  123.97

Vu  156.98  4.83 *

75.75 * 490 lb / ft 3 144

 258 lb / ft  0.258 k / ft

Thus

w u  1.2w D  1.6w L  1.2 * 0.258  1.1)  1.6 * 2  4.83 k / ft 

required

Ru 

w u L 4.83 * 65   156.98 2 2

intermediate

-Find kv A w  d t w  84.25 * 3 / 8  31.59

filed action is included) V  Vn 123.97  u   3.92 ksi Aw Aw 31.59

action is included) with h/tw=219 and  Vn / A w  3.92 and read a / h  1.4

 a  1.4 * 82  114.8 in

-Find Cv

1.51 E k v

h / t w 

2

Fy

kv E Fy



 1.37

are

Enter AISC Table 3-16b for (tension filed

h 82   219  260  k v  5 t w 0.375

h  219  1.37 tw

stiffeners

-Select stiffeners for second panel (Tension

a- Find spacing of intermediate stiffeners if needed

C v 

more

5 * 29000  86.95 36

1.511 * 29000 * 5 219 2 * 36

 0.1268

Lecture 7 ....... Page 16

Lec

7

Steel Design Fourth Year Dr.Abbas Oda Dawood

Misan University Engineering College Civil Department

b- Design intermediate stiffeners The first intermediate stiffeners is placed 82 in from the end

Lec

7

The cross section of the axially loaded member consists of the stiffener plates and length of the web of 12 *tw =12*5/8=4.5"

a 82   1.0 h 82 2.5 2.5 j  2  2  2  0.5  0.5  OK 2 a / h  1

b  min a , h 82

I st min

3

3  b t 2w j  82 *   * 0.5  2.16 in 4 8

assume thickness of pair of intermediate stiffeners Design the bearing stiffener as column has the crosshatched section.

tst=0.25 in I st

t st 2b st  t w 3  12

2.16 

5 3 A  2 * ( * 9 )  4.5 *  12.94 in 2 8 8

0.25 * 2b st  0.3753  b st  2.16 12

Use 2PLS 3*0.25 in plate intermediate stiffeners

I

18.375 3 * 5 / 8  323 in 4 12

contribution) I 323   5.0 in A 12.94

r

c-Design bearing stiffeners

b st max 

(neglecting the web

KL  0.75 h  0.75 * 82  61.5 in

20  3 / 8  9.81 2

KL 61.5   12.3 r 5.0

Use bst = 9 in E 29000 b  0.56  15.89    0.56 Fy,st 36  t  st

Enter Table 4-22 of AISCM with KL/r = 12.3 and Fy=36 ksi and read  c Fcr  31.17 ksi  c Pn   c Fcr * A  31.17 *12.94  416.3 kips   c Pn  416.3 kips  R u  156.98 kips  OK

b st 9  15.89   15.89  min t st  0.566 t st t st

Check bearing strength R n   1.8 Fy A pb

Use tst =5/8 = 0.625 in Thus try two plate bearing stiffeners 5/8*9 in

A pb  2 * (9  0.5) * 5 / 8  10.625 R n  0.75 *1.8 * 36 *11.25  516.375 kips

R n  516.375 kips  R u  156.98  OK

Use 2PLS 9*5/8 with depth of 81.75 (a depth of 81.75 used instead of 82 in for fitting purposes)

Lecture 7 ....... Page 17