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'

SCHAUM 'S OUTLINE OF

THEORY AND PROBLEMS OF

STRUCTURAL STEEL DESIGN (Load and Resistance Factor Method )

• ABRAH Al\1 J. ROK AC II. :\I SCE Direcwr l ( Building f)e.\ign aiiCI Su{t11 an•

\mericm1 lnsritwe uf Steel Conltmcrinn. Inc.

AUNG MYAT KYAW EFFECTIVE ENGINEERING GROUP.

SCHAUM'S OU TLINE SERIES i\IcGraw -HiU No:'' York San Franl'i,c.:o \\'a.,hinglon. D. C. AudlanJ Bogm i Carat'U'Li bon London :JadriJ \ lc\lco Cil\ :\Illan lon1real a Fellov. of the Ame rica n Societ) oi' C i\'i l Eng i neer Loads Load Combinations

Chapter J

TENSION MEMBERS .

. . . . . . . .

.

.

. .

.

.

. . . . .

. .

14

. .

2 3

Notation Introduction Cross-Sectional Areas De ign Tensile Strength Displacement

Chapter 4

COLUMNS AND OTHER COMPRESSION MEMBERS NoHHion fntroduction Local Buckling Column Buckling Effective Length FactOr : Judgmental Method Effecuve Length Factor: Analytical Method Design Compressive Strength Column Destgn Dtsplacement

v

. . .

..

.

.

CONTENTS

VI

Chapter

CO)'IPACT BEAMS AND OTHER FLEXURAL MEMBERS .

5

\lotation lntroductton Cumpactnc-, Fl.:,ur!! s\

A441

50 46 42 40

A572-Grade 65 - Grade 60 - Grade 50 -Grade 42

65 60

s I ' -I zI I -4 4 8

A242

50

A588

46 42 50

ASl-t

< I! s I! "S4

42

46 42 Quenched and tempered alloy

Plate Thicknes . in

36 32

A529

High-mengt h low-alloy

F,. hi

100

90

«

-· '

I -4 s4 4-5

S-8 -

-·

:H-6

STRUCTURAL SHAPES A structural member can be a rolled shape or ca n be built up from two or m ore rolled shapes or plates, connected by welds or bolts. The more economical rolled shapes arc utilized when ever possible. However. special conditi ons (such as the need for heavier members or particular crosssect ional geomet ries) may dicta t e the usc of built-up members. Available rolled shapes arc catalogued in Part 1 of the A I SC Manual. Those most common ly used in building construction include wide flange (or W), angle (or L), channel (or C), and tee (or WT). They are hown in Table 1-2 with examples of their nomenclature. Examples of common built-up shapes are given in Fig. 1-3.

6

STRUCTURAL

STEEL

[CHAP.

I

Table 1-2 RoUed Structural Steel Shapes and Their Designations Type of Shape

Example of Designation

Cross Section

W (wtde flange)

,-...J

T .

,........-'

C (channel)

\.......,

_.......,I

-

-...

Explanation of Designation

Wl4x9Q•

Nommal depth. 14m; wetght . 90 lb/ft

Cl2X30

Depth, 12 in; weight. 30 lb/ft

L4x3x!

Long leg. 4 in; hort leg. 3 in ; thtckness. !in

WT7x4s·

Nommal depth. 7 in. wctghl. 45 lb/ft

r;

c . 0

..L .r;

c..

0 -r

L (angle)

-!..

·

c ..".J"·:;.

E '5

-

wr

Thickne > j.-

t l f. Leg..

dimension

(structural tee cut from W shape)

,...-' .CT

c..

' Cu ttmg a Wl-lx90 in half longitudinally results m two \vnx-IS .

•

Welded plate girder

Cover plated W shape

W hape with helf angle

Fig. 1-3 Common built-up structural sh from load statistics. The resistance factor cp depend' on .A a-. well as on the uncertainty in the resistance R,. The selection of a reliability tnde\ Jl determtne' the value of cp for each limit state. In general. to reduce the probability of failure . .A would be wcrca-,ed. resulting in a lower value for

cp

LOADS Structural load arc cla sificd as follows. Dead load (D)-The weight of the structure and all other permanently installed features in the building. including built-in partitions. Liue load (L)-The gravity load due to the intended usage and occupancy: includes the weight ot people. furniture. and movable equipment and partitions. In LRFD. the notation L refers to tlnnr live loalh and /.,. to roof live loads. Ram load (/?)-Load due lO the initial rainwater or ice. excluding the contribution of ponding. Snow load (S)

W111d load ( W)

t.artltquak.e

load

(

£). In dc,1gn. the dead load is calculated from the actual we1ghts of the 'ariou-. structural and non'>tructural clement,. All the orher design loads are specified by the governing building code. When beam" ... upport large floor areas or columns support several floors, building codes generally allow a hvc-lnad reduction. The reduced live load is used in LRFD.

LOAD COMBINATIONS The required mength i'> defined in the AISC LRFD Specification as the maximum (absolute value) force obtained !rom the following load combinations.

1.40 1.2D

(A 4-1)

+ 1.6L + 0.5 (L, or S orR)

(A4-2)

1.2D + 1.6 (L, or S orR)+ (0.5L or 0.8W)

(A4-3)

1.2D + 1.3W + 0.5L + 0.5 (L, or S orR)

(A4-4)

1.2D + 1.5£ + (O.SL or 0.2S)

(A4-5)

0.9D- ( 1.3W or 1.5£)

(A-l-6)

[t:xcepuon : 'I he load factor on Lin combinations (A4-J). (A4-4). and (A4-5) shall equal l.O for

garage\. areas occupied as places of public assembly. and all areas where the live load is greater than J(Xl lh/ ft•

I

Load D. L. L,. S. R. W. and £represent either the loads themselves or the load effects (i.e.. the force\ or moments cau.,ed by the loads). In the preceding expressions. only one load assumes its max1mum ltfettme \'alue at a time. while the others are at their "arbitrary point-in-time" values. Each combinatmn model'> the design loading condition when a different load I!> at 1ts maximum.

CHAP. :j

ll

INTRODUCTION TO LRFD

Load Combination

Load at I ts Lifetime Maximum

(A4-J)

D (during const ruction; other loads not present)

(A4-2)

L

(A 4-3)

L, or S or R (a roof load)

(A4--1) (A4-5)

W (acting tn the direction of D)

(A4-6)

E (acting in the direction of D) WorE (opposing D)

Load combinations (A4-l) to (A4-6) are for computing l> trcngth limit states. In determining serviceability l imit states (e.g., deflections) the unfactorcd (service) loads arc used.

Solved Problem s 2.1.

The moments acting on a floor beam are a dead-load moment of 50 kip-ft and a li\e-load moment of 35 kip-ft. Determine the required strength. Becau e dead load and floor hve load are the onI} load acting on the member , L,= S = R = W = £ = 0. By tn\peCtiOn of formulas (A-l-I) to (A-1-6). it is obviou that one of the firM t\\O formula mu t go, ern. a folio""'

UD = I Ax SO kip-ft = 70 k1p-ft

(A -1-1)

1.2D + 1.6L = l.2 X 50 kip-ft + 1 .6 X 35 klp-ft = 116 kip-ft

(A-1-1)

Because It produces the max1mum requ1rcd Mrcngth 1s 116 k1p-ft.

2.2.

required strength, the econd

load combination

governs. The

Floor beams W21 x50, spaced I 0 ft 0 in center-to-center. support a superimposed dead load of 651b/ft 2 and a live load of 40 lb/ft2 . Determine the governi ng l oad combinatio n and the correspon ding factored load. Total dead load D =50 lb/f t + 651b/ft! x 10.0 ft = 700 lb/ft Total live load L = 40 lb/ft 2 X 10.0 rt = 400 lb/ft As in Prob. 2.1. L, = S = R = W = E = 0. The two relevam load combinations are l.4D = 1.4 X 700 lb/ ft = 980 lb/ ft

(A-l-l)

1.2D + 1.6L = 1.2 x 700 lb/ ft + 1.6 X 400 lb/ ft = 1480 lb/ ft

(A -1-2)

The l>econd load combination. 1.48 kips/ft), governs. 2.3.

which gives the maximum factored load. 1480 lb/ft (or

Roof design loads include a dead load of 35lb/ft!, a live (or now) load of 251b/ft . and a wind pressure of 151b/ft! (upward or downward}. Determme the governing loading .

12

1:-ITRODUCflON TO LRFD

(CIIAP. 2

The six load combinations are

Load Combination

Factored Load, lb/ft1

(A-1-/)

I 4 X 35

= 49

(A4-2)

1.2 X 35 + 0 + 0. 5 X 25

= 55

(A4-J)

1.2x35+ J.6x25+0.8x I S

=94

(A4-4)

1.2 X 35 + 1.3 X IS + 0 + 0.5 X 25

= 74

(A4-5)

1.2x35+0+0.2x25

=47

(A-1-6)

0.9

= 12

X

35- 1.3 X 15

The thard load combanauon go\'erns; it has a total factored load of 94 lb/f t .

2.4. The axial forces on a building column from the code-specified loads have been calcu lated as 200 kips of dead load, 150 kips (reduced) noor live load. 25 kips fro m the roof ( L , or S or R), 100 kips from wind, and 40 kips from earthquake. Determine the required strength of the column. Load Combination

Factored Axial Force, kips

(A4-1)

1.4 X 20()

(A4·.?)

1.2 X 2(Xl +

(A4-Ja)

=280 1.6 X ! 50+ U.S X 2S

= 493

1.2 x 200 +

1.o x 2s + o.s x ISO

= 355

(A-1-Jb)

1.2 X 200+

I 6 X 2S +0.8X HXl

= 360

( A-1-·1)

l.2x200+

I 3x IOO+O.Sx 150 +0.Sx25

=458

(A .J-5a)

1.2 X 200 +

I 5 X 40 + 0.5 X ISO

= 375

(A4-5b}

1.2 X 200 +

1.5 X 40 + 0.2 X 2S

= 305

(A4·oo)

0.9 X 200 -

1.3 X I 00

-

(A4-6b)

0.9 X 200-

I.S X 4()

= 120

The requared strength for the column is 493 kips. ba ed on the second load combination.

2.5.

Repeat Prob. 2.4 for a garage column. According to the A ISC LRFD Specification. load combin ations (A 4-J) to (A 4-5) are modified for garage . areas of public assembly. and areas with live load exceeding 100 lb/ftz. as follows. 1.20 + 1.6 (L, or S or R} + ( l.OL or O.RW)

(A-1-J') (A -1-·1')

1.20 + 1.3W + I.OL + 0.5 (L, or S orR)

(A 4-5')

1.20 + 1.5£ + ( I.OL or 0.2S)

The solution to Prob. 2.4 i s still valid for ga rages except for load combi nation s (A4-Ja), (A4-4), and (A4-5A), which become

Load Combination (A4 -Ja ')

Factored Axial Force , kips 1.2 X 2(Xl + I 6 X 25 + 1.0 X ISO

=-BO

( A4-4 ')

l.2 X 200 + 1.3 X 100 +].()X 15() + 0.5 X 2S

(A4-5a ')

1.2 X 200 + I.S x 40 + 1.0 X ISO

= 533

= 4SO

so

CHAP. 2)

INTRODUCfiON TO LRFD

13

Because 533 kips is gremer than 493 kips. the required strength for the garage column is 533 kips. which is obtained from modified load combination (A4-4).

Supplementary Problems 2.6.

A beam-column IS subjected to the following forces by the service loads indicated. Axial compression, P = 60 kips (dead load). Skips (live load). Bending. M = 10 kip-ft (dead load), 3 kip-ft (live load). Determine the governing load combination and the required axial compressive and bending strengths. Ans. Load combination (A4-J) governs for axial compression; the required strengths are P,, = 84 kips , M., = 14 kip-ft. Load combination (A4-2) governs for bending moment; the required strengths are P,, = 80 kips, M., = 17 kip-ft. Both of the preceding P,,-M., pairs should be checked in the design of the beam-column.

2.7.

A member is subjected to the following axial forces: 35 kips (axml compression from dead load) and 30 k ip (axial compression or tension from wind) . Determine the governing load combinations and the required strengths. A tu. Axial compression: P,, = 81 kips ; load combination (A+-4). Axial tension: P,, = 8 kips; load combination (A46).

2.8.

The axial forces on d building column are as follows: SO kips dead load. 40 kips floor live load, 10 kips roof live load. and SS kips wind. Determine the required strength. Ans. Axial compression: load combination (A -1-6).

P,, = 157 kips; load combination (A-1--1). Axwl tension : P.= 27 kips;

Chapter 3 Tension Members NOTATION A, =effective net cross-sectional area of member. in A. = gross cross-sectional area of member. in

A, = net cross-sectional area of member. in £=modulus of elasticity of steel= 29,000 ksi F.,= specified minimum tensile strength, ksi ·=specified min i mum yield stress. ksi g =gage (i.e .. the transverse center-to-center spacing betwee n fastener gage lines), in l =member length, in P = (unfactored) axial force in member, kips P, =nominal axial strength of member, kips s =pitch (i.e .. the longitudinal center-to-center spacing of any two consecutive holes). in U = reduction coefficient =axial elongation of member. in cp,P,, =design strength of tension member, in ¢, = resistance factor for tension = 0.90 or 0.75

INTRODUCTION

This chapter covers members subjected to pure tension. such as hangers and truss members. When a tensile force is applied through the centroidal axis of a member. the result is a uniform tension stress at each cross section. Tensile forces not acting through the centroid cause bending in addition to tension; lateral forces also cause bending. Members with combined bending and tension are discussed in Chap. 7.

CROSS-SECTIONAL EAS

AR

The design tensile strength of a structural steel member depends on the appropriate crosssectional area. The three cross-sectional areas of interest are the gross area A 11 • the net area A,, and the effective net area A ,. The gross area of a member at any point is the total area of the cross section. with no deductions for holes. The net area is the gross area minus the area of the holes. In computing the net area for tension, the width of a hole is taken as it; in greater than its specified dimension. Since tolerances require that a bolt hole be ft, in greater than the diameter of the bolt, the width of a hole is assumed for design purposes to be twice in in. or A in, greater than the diameter of the bolt. The net area of an clement is its net width multiplied by its thickness. For one hole, or two or more holes running perpendicular to the axis of the member, the net width is the gross width minus the sum of the widths of the holes. However, if a chain of holes extends across a part in a diagonal or zigzag fashion. the net width is the gross width minus the sum of the hole dimensions plus the 14

15

TENSION MEMBERS

CHAP . 3j

quantity J '!4g for each gage pace in the chain. where

s = pitch (i.e.. the longitudinal center-to-center spacing of any two consecu t ive hole ). in g =gage (i.e .. the tranwerse center-to-center spacing be tween fa!>tener gage lines). in (See Fig. 3-1.) It may be necessary to examine several chain s to determine which chain has the lea'>t net width.

/

t

P ----;} •

--- P

s

I• •I

Fig. 3-1 Definitions of.f and g

The concept of effecuue net area account s for shear lag in the vicinity of connections. When the member end connection transmits tension directly to all cross-sectional element of the member. A equa ls A ,. But if the endconnection transmits tension through some. but not all. of the cross-sectional clemen t s. a reducedeffective net area is m.ed instead. For bolted and riveted member (83-1)

For welded members

(83-2) Desi gn values for U and A, are given in Sec. 83 of the AlSC LRFD Specification. For W , M , or S shapes and structural tees cut from these shapes: If the tensile force is tran smitted by transverse weld . A, equa ls the area of the direct ly conn ect ed clement s. I f th e force is transmitted by bolts. the va lue of U d epends on the cri teria listed in Table 3- L

Table 3- 1 Valu es of U for Bolted W, M , S. Wf , MT. and ST Shapes Critena

u

(a) Flange v.tdth 2: ;x depth: connection is to the flanges: mtmmum of three faMene rs per hne m

the direction of Mre (h) Mmunum of three fasteners per line m the direction of Mre s otherwtM not meeting criteria (a) (c) Two fa rcncr per line in the direction of tress

0.90 O.R5 0.75

[CHAP. 3

TENSION MEMBERS

16

DESI G"' TENSILE STRENGTH lwo cntcna hmll the design tensile strength a

cp,P""

l-or yiel,P. = minimum of 0.90FA.=0.90X36hJX 14..1m'=467klp

() 75F.A,= 0.75 X 58 ksi X 14.4 In:= 626 k1p upport. A.= 14.4 in:

For welded connection . effective net area A, =area of directly connect ed elemenb = area of the two flanges = 2( I0.0 in x 0.560 in)= 1 1.20 in: Design \trcngth in-diameter bolt joins the ch anneb. as in Fig. :uo. 10 form a built-up ection. All

Ans.

1/J,P,, = 200 kip . 0.314'"

-II-

' b

X

1 C

10.5

Fig. 3-10

3.15.

Only the web of the channels are welded to the upporl. Away from the support. orne ections have a -m -d1ameter bolt . a in Fig. 3-9. 10 form a built-up ection. A11s. rp,P. = 16-' kips.

22

TENSION MEMBERS

(CH AP . 3

3.16.

The connecuon of the channels to their support ts as shown tn Fig. 3-10 with three -tn-diameter bolts in the dtrection of stres!.. Ans. lj>,P" = 200 kips .

3.17.

. Calculate the increa e tn length of the 3-ft-long tension hanger tn Fig. 3-9 (2 C6x 10.5) under an axial ervice load of 100 ktps. Ans . = 0.020 in.

Chapter 4 Columns and Other Compression Members NOTATIO N A 11 =gross cross-sectional area of member, in2 b=width, in

b1 =width of flange, in d =depth, in E =modulus of elasticity of steel= 29,000 ksi F,, =cri tical comprehensive stress, ksi

F,. =compressive residual stress in flange. ksi F.,. = specified minimum yield stress, ksi G = alignment chart parameter defined in Eq. f .J. 2] G' =alignment chart parameter defined in Eq. [-1. 11 It,, h"' =web dimensions defined in Fig. 4-1 in I= moment of inertia. in 4

K =effective length factor KL =effective length, ft Kl =effective length. in L = length of member. ft

I= length of member, in

P = (unfactored) axial force in member, kips P,, =nominal axial strength of member, kips P,, = required axial strength. kips r =radius of gyration of the cross section. in

r =th ickness, in 1,.. =th

ickness of web, in

t1 =axial shorteni ng of member, in

A,.-column slenderness parameter A,.= limiting width-thickness ratio for compact section A,= limiting width-thickness ratio for column design

q>J . =design

strength of compression member, k1ps ¢, =resistance factor for compression= 0.85

INTROD UCTION This chapter covers members subjected to pure compress1on '>UCh as columns and truss members. When a compres!>ive force is applied through the centr01dal axi!> of a member. a uniform compression stress develops at each cross section. Bending is caused by compressive forces not acting through the centroid or by lateral forces. Bending combined "ith compression is discussed in Chap. 8.

24

COLUMi'IS AND OTHER COMPRESSION MEMBERS

[CHAP. 4

The strength of comprcs ion members is limited by instability. The instability can be either local buckling or overall (column) buckling. LOCAL BUCKLING

The cross sections of structural steel members arc classified as either compact. noncompact, or slender-element sectiOn!., depending on the width -thi ckness ratios of their elements. A section is compact if the Ranges arc continuously connected to the web. and the width-thickness ratiOS of all its compression elements are equal to or less than A,.. A section is noncompact if the width-thjcknc ss rau o of at least one clement is greater than A,., provided the width-thickness ratios of all compression elements are equal to or less than A,. ff the width-thicknes s ratio of a compression clement is greater than A,. that clement is a slender compression element; the cross section is called a slender-element section. Steel members with compact sections can develop their full compressive strength without local instability. Noncompact shapes can be stressed to initial yielding before local buckling occurs. I n members with slender element s. clastic local buckling i s the limitation on strength. Columns with compact and noncompact sections are designed by the method described herein (and in Chap. E of the A I SC LRFD Specification). Nearly all building columns arc in this category. For the occasional case of a slender-element column, the specia l design procedur es listed in App . 85.3 of the AISC LRFD Specification are required, to account for local buckling. Because of the penalties imposed by App . 85.3. it is generally more economical to avoid slender element s by increasing thicknesses. To summarize: if. for all elements of the cross section , the width-thkkness ratios (b/t, d/t,... or hef t.,.) are equal to or lcs than A,. column design should be by the method of this chapter. Otherwise, the method given in App. 85.3 of the LRFD Specification must be used. The width -thjckness ratios for columns and the corresponding values of A, are defined in Table 4-1 and Fig. 4-1, which are based on Sec. B5 of the AISC LRFD Specification. Table 4-1 Limiting Width-Thickness Ratios ror Columns

Column Element

WidthThickne Ratio

Limiting Width-Tht ckne'' Ratio. i., General

A36 Steel

Flanges of W and other I shapes and channels; outstanding legs of pairs of angles in continuous contact

/J / t

95/\(F,

15.8

Flanges of quare and rectangular box sections; flange cover plates and diaphragm plate!> between lines of fasteners or weld

b/r

238/VF.,. -F,*

46.7 (rolled) 53.9 (welded)

Legs of ingle angle Mrut and double angle struts wtth epara10rs; unstiffcncd element' (1.e .. upported along one edge)

h/r

76/#.

12.7

Stems of t ees

d It

127/,fi{

2 1.2

All other stiffened elements (i.e..

b/t

253/

42.2

supported along two edges)

h, lt.

•F, = compres\tvc rc,•tlual trC\ in !lange: W ksi for rolled \hapc'>. 16.5 ksi for welded secuon\

25

COLUMNS AND OTHER COM PRESSION M EMB ERS

CHAP. 4)

I•

b = b,i2

h = hl '2

h

h,

1

b = b,

..,

I•

b.

•I

I• h. h

.L I> = I>, 2

b

h

d I

h.

1

\

I

h - b, - 31

It - It - 31

Fig. 4-1 Definmon of width (h d. and h.) and th•cknes' (llange or leg thickness 1 and web thlcl..ne., 1.) for use in Table 4-1

COLUMN BUCKLING

The most significant parameter affecting column !.lability is the !>lendemess ratio Kl I r, where I is the actual unbraced length of the column. in: Kl i the effecti\'C length of the column . in: and r is the radiu of gyration of the column cross section. m . Column strength equations are normally writren for ideal "pin-ended'' column . To make the '>trcngth equations applicable to all columns, an effective length facwr K is used to account for the influence of end cond itions on column stabilit y. Two methods for determining K for a colum n are presented in Sec. C2 of the Commentary on the AISC LRFD Specification : a judgmental method and an approxi mate analytical meth od . A discussion of the two method foliO\\ .

EFFECTI VELENGTH FACTOR: JUDGMENTAL METHOD

Six ca cs arc show n in Table -1-2 for individual colum n!>. with their corresponding K values. both theoretical and recommended. The more conservative recommendations (from the Structural Stabilit) Research Council) reflect the fact that perfect fixity cannot be achieved in real tructu res. The LRFD Specification distmguishes between columns tn braced and unbraccd f rames. In braced frame . sideswa} i inhibited by diagonal bracing or shear walls. In Table 4-2. case d ( the classical pin-ended column. K = 1.0) a well as ca es a and b repre ent colum ns in braced frames: K 1.0. AISC' recomme nd!. that K for compression members in braced frames ··shall be taken as unit y, unlc!>s structural analysis show!> that a sm aller value may be used." It is common practice to assume K = 1.0 for columns in braced frames.

26

COLUMNS AND OTH ER COMPR ESSI ON MEMBE RS

Table 4-2

Effecti"e Length Factors K Cor Columns b)

WI

Buckledby Shape of Column Shown Dashed Lme

[CH A P. 4

L

I

I

I

'

I \

n7

t

-.;;t1'"

t. ,.

cJ

ct •.,

II

'

n7

,.,

·r·

I

I

I

I

I

"'"" t t

E

I

I

I

I I

If)

p

'\

I I

I I I

t

)

d)

I I I

,+. it

I I I

I I I

'

n-r

t

.

I I

.,

I

Theoretical K value

0.5

0.7

1.0

1.0

2.0

2.0

Recommen ded design values when ideal conditiom are approximated

0.65

0.80

1.2

1.0

2.10

2.0

4" End condition code

Rotation fixed and translation fixed

Rotation free and translation fixed fZ{J Rotation fixed and tra nslat ion free

?

Rotation free and transla t ion free

Reproduced with perm''"on rrom the A ISC L RFD Manual

Cases c, e, and fin Table 4-2 cover columns in unbraccd frames (sidesway u ni nhibited); K 1.0. The K values recommended therein may be u ed in column design.

EFFECTIVE LENGTH FACTOR: ANALYTICAL METHOD If beams are rigidly connected to a column, nomographs are available for approximating K for that column. Two such "alignment charts.. have been developed: one for "'sidesway i nhibited" (i.e., braced frames, K s 1.0): the other. for "'sidesway uninhibited" (i.e., u nbraced frames, K 1.0}. Again, for col umns in braced frames, it is customary to conservatively l et K = l.O. For columns in u nbraced frames, the align ment chart in Fig. 4-2 may be used to determin e K. Beca u se the alignme nt charts were developed with the assumption of purely clastic act ion, t h e stiffness red uction factors (SRF) in Table 4-3 arc available to account for inelast ic column be havior. (Figure 4-2 has been reproduced with permission from the Commentary on the A I SC LRFD Specification . Table 43 is a corrected version of Table A in the A ISC LR FD Manual. Part 2.) Th e procedure for obtaining K from Fig. 4-2 is as follows. I.

At each of the two joints (A and B) at the ends of the col umn. det ermine I (th e moment of inertia, in ) and I (the unbraced length, in) of each column ci and each beam gi rigidly connected to that joint and lying in the plane in which buckling of the column is being considered.

2.

At each end of t he column, A an d B

c· = (1/l}c + (1/1),.2 CHAP. 4]

(/ / I}!< I+ (I/ f)t:2 COLUMNS AND OTIIER COMPRESSION MEMBERS

( 4.11

27

c. G.

I U II

...

..'• '

' II.. I:II II

'

illI ., I

"I

rr-J l'1

,!\) J

• I

I

0

)

'" "

"'

"

'

)

cl

"

111 1

:!

"

'

c

E

:>

0

...

'

v

!••

.

I '

·,ec Jional a rea or 1hc ,UOJCCI column

Adjust for inelastic column action G 1 =G xSRF

[4. 2)

GH = G1 x

SRF where SRF is the stiffncs& reduction factor for the column obtained from Table 4-3. 4.

5.

. For a column end attached t o a foundation. G = 10 for a ··pin '' support and G = 1 for a rigid support are recommend ed. Determin e K by drawmg a straight line from G" to Gu on the alignment chart in Fig. 4·2.

0.81

28

COLUMNS AND OTHER COMPRESSION MEMBERS

(CIIAP. 4

DESIGN COMPRESSI VE STR ENGTH Column buckling can be either clastic or inelastic . For design purposes. )., = I.5 b taken as the boundary between elastic and inelastic column buckling. .

Kl

A,.=

(£2-4)

r;r

For columns with cross-sectional elements having width-thickness ratios equal to or less than ),, , the design compressive strength is 1.5, column buckling ill clastic.

F. r = The terms in these equations include

[

0.877] ).2

(£2-3}

F,. '

),, = slenderness parameter F, =specified minimum yield stress, ksi E = modulus of elasticity of steel= 29.000 kst cp,. = resistance factor for compression P,= nominal compressive strength, kips 1

A 11 =gross cross-sectional area. in

F,., =critical compressive stress. ksi

Equation (£2-J) is the Euler equation for column instability multiplied by 0.877 to account for the initial out-of-waightnes of actual columns. Equation (£2-2) and (its equivalent) Eq . (C-£2-J) are empirica l equations for inelastic column buckling, providing a transition from F,, =F. at ;,, = 0 (i.e.. Kl/r = 0) to the modified Euler equation [Eq . (£2-3)1 for elastic buckling at ),r > 1.5. For A36 steel )., = 1.5 corresponds to a slenderness ratio Kl /r of 133.7.

COLUMN DESIGN According to Sec. B7 of the A ISC LR FD Specification, for compression members Kl /r "preferably should not exceed 200... In design. selection of an appropriate column can be facilitated by referral to tables in one of two ways. The design compressive strengths cp, P,, of W and other rolled shapes are tabulated in the A I SC LRFD Manu al. Part 2. Column shapes can be selected directly from those tables. For bui l tup sections and rolled shapes not tabulated. Table 4-4 for A36 steel (and similar tables for other grades of steel in the AISC LRFD Specification) can be used in iterative design. I n both cases. reference to tables replaces the need to solve the column strength equations [Eqs. (£2-1) to (£2-4)).

CHAP. 4]

COLUMNS A:-10 OTHER COMPRESSION

Table -' -'

29

MEMBERS

Design Compre ive Stresses for A36 Steel

De tgn Stress for Compre-.-.ton 'v1embe of 36 kst Specified Yield-Stre\ Steel,

Kl r

. I,

-3 4 5

6 7

cp, f;_,. ksi 30.60 30.59 30.59 30.57 30.56

Kl r 41 42 43

M 45

,.F,,.

r

ksi

14.16 13.98 13.1\0 13.62 13.44

161 162 163 164 165

8.23 8.13 8.03 7.93 7.84

)

126 127 128 129 130

13.27 13.09 12.92 12.74 12.57

166 167 168 169 170

7.74 7.65 7.56 7.47 7.38

19.79 19.60 19.41 I \J .22 19.03

131 132 133 134 135

12.40 12.2:\ 12.06 1 1.71

171 172 173 174 175

7.30 7.21 7.13 7.05 6.97

O- 16.5 for the web 253 - ') if v'36 -42._ ": = 36 ksi

253

y'SO = 35.8 For t h e IX in

if

1';.

·

5(1 hi

x Ill in box section in Fig. 4-3. b = h, = 18 in- 2 x ' in= 17 in

.

I .

lr=I=I= • Ill

(a)

b h 17111 -=-= =3-1 in c '· In A3o -.teel. bit and h /t. < i., in all ca!>es: there arc no -.J.:nder element\.

(b)

H F =50 k 1. there arc abo no slender elemenh. bccau'c />It and h /1••, i., 111 all ca es. 18 In

18 10

--.

f-

1=

..I

t in (typical)

b

h,

- -1

Fig. 4-3

I n Probs. 4.4 to 4.7, determine the effective length factor K. from !'able 4-2. tor the given columns. 4.4.

A building column free to rotate at each end. in a braced frame. As a rc,ult of the hracmg. lateral translation of the end, ot the column '' tnh1b11cd . "Rotation free and tran latton fixed" at hoth end!> i cased 111 Table 4-2: K = 1.0.

4.5.

A buildmg column in a braced frame; deep beam rigidl) connected to the column restrict rotation of it end . Th1s corresponds to case a. "rotation fixed and tran..lation fixed" at each end Although K = 0.65 is tndtcatcd for thl\ ca-.e 111 Table -1-::! . it i-; customary to let K = 1.0 '"a con,erntttve mtnimum .

32

4.6.

COLUMNS AND OTHE R COMPRESSION MEMBERS

A building column in a ngtd frame (not braced): end rotation is inhibited by deep beams.

c: K = 1.2 i recommended.

"Rotation fixed and translation free" i case

4.7.

(CHAP . 4

The sam e footing.

as in Prob. 4 .6, except that th e base of the column is ''pin-connected" to a

"Rotation fixed and translation free" at the top and ·•rotation free and translati on fixed" at the bottom i case f: K = 2.0.

In Probs. 4.8 to 4.10, use the alignment chart to determine K . All steel is A36 .

4.8.

The column shown in Fig. 4-4. W21 X 50

(typ1cal beam) -;:

T

A

& x 8]

.; Column requ1red strength:

-

= 3·!I B

.:::

p = 750 I

,_.. •• ·"' 0c-

--

.

X

!:! 0... OliO/>

·-- - 22 24 0

"0 '

0 > -

G)

i not adequate. Usc a WI Ox 54 column with a design trcngth , P. = 385 kip (KL = 14.0 ft) > 360 kips required. Stnce r,/ r, = 1.71 for the Wl0x54, as originally a, umed, recomputation of (K,L,)..... is not necessary.

4.15 . A Wl0X49 column. lOft long. carries a ervice load of 250 kips. Calculate its

axial shortening. PI Shortemng. 6 = =

EA.

=

250 ktp X ( \().() ft X 12 tn/ft) . , , 29.(X)() ktp/ tn"14. 111 4

o.on m

CHAP. 4]

37

COLUMNS AND OTIIER COMPRESS I ON MEMBERS

4.16. The section shown in Fig. 4-3 is used for a 40-ft column: K, = K, = 1.0. Determine t he design compressive strength i r the steel is A36. The design compressive strength ( £2-/)

The value of E N

- "'

.c: o

"x"g"'

I

I

z I

Column requared strength: P. ft 150 kips

B

Fig. 4-7 4.24.

Calculate the decrease in length of the 24-ft column in Prob. 4.22 under an axial load of 200 kips. AilS.

D. = 0. 1 5 i n .

Chapter 5 Compact Beams and Other Flexural Members NOTATION A = cross-sectional area of member, i n 2

A,..= area of the web, in

2

b =width. in b1 =width of flange, in

Ct> = bending coefficient. defined in Eq. [5.10] C,.. =warping constant , ino

c =distance from the centroid to the extreme fiber, in d =overall depth, in

E =modulus of elasticity of steel= 29,000 ksi F,. =compressive residual stress in flange. ksi Fv =specified minimum yield stress, ksi ji, =max imum normal stress due to bending, ksi G =shear modulus of elasticity of steel = 11.200 ksi h = web dimension defined in Fig. 5-7, in he , h...,= web dimensions defined in Fig. 5-2, in I= moment of inertia ,

in• J =torsional constant , in 4 Lh = unbraced length , ft Lm =limiting unbraced length for full plastic bending capacity (Cb > 1.0). ft L,, =limiting unbraccd length for full plastic bending capacity \Co= l.O), ft L, = unbraccd length which is the boundary between ela,tic and inelastic lateral-torsional buckling. ft I = length of member. in M =bending moment, kip-in Ma =elastic buckling moment . kip-in

M,. =nominal flexural strength of member, kip-in MP =plastic moment, kip-in M, =buck ling moment at L,. = L, and Ch = 1.0. kip-in M 1 =smaller end moment in an unbraced length of beam, kip-in M2 =larger end moment in an unbraced length of beam, kip-in P =concentrated load on member, kips r = radius of gyration. in

S =elastic section modulus. in ·' t = thickness, in 1,.. =thickness of web, in V =shea r force, kips

39

40

COMPACT BEAMS A

D OTHER FLE:XURAL MEMBERS

[CHAP. 5

vn =nominal shear strength. kip w =unit load. kips per linear ft X 1 =parameter defined in Eq. (FI-8) X 2 =parameter defined tn Eq. (Fl-9) x = subscript relating symbol to the major principal centroida l axil. y =subscript relating ymbol to the minor princ1pal centroidal axis Z = plastic section modulus. in 1:1 =deflection of beam. in

).."=limiting width-thickness ratio for compact section lJ>hM, =design flexural strength, kip-in tp11 =resistance factor for flexure= 0.90 cp,, V,, = design shear strength. kips ¢,. = resistance factor for shear= 0.90

INTR ODUCTION This chapter covers compact flexural members not subjected to torsion or axial force. Compactness criteria as they relate to beams are described in the next section ; noncom pact flexural memberlo are covered in Chap. 6. Axial tension combined with bending is the subject of Chap. 7; axial compression combined with bending IS discussed in Chap. 8. Torsion and the combination of torsion with flexure are covered in Chap. 9. The strength of flexural members is limited by local buckling of a cross-sectional element (e.g.. the web or a flange), lareral-torsional bucklmg of the entire member, or the development of a plastic hinge at a particular cross section. The equations given in this chapter (and in Chap. F of the A I SC LRFD Specification) arc va lid for flexural members With th e following kinds of compact cross sections and loadings: doubly

! r::::- -•

--

1

............

•

I· (tl)

Q

\

\

I

Shear center

Fig. S- 1 Example of beams covered in Chap. 5: (a) W shape (doubly symmetric) loaded in a plane of symmetry: (b) channel shape (singly ymmetric) loaded through hear center in plane of symmet ry or parallel to web

CI IAP. 5J

COMPACT BEAMS AND OTHER FLEXURAL MEMBERS

41

symmetric (e.g., W, box. and solid rectangular shapes), loaded in a plane of symmetry [as in Fig. 5l(a)) and singly symmetric (e.g., channel shapes), loaded in the plane of symmetry or through the shear center parallel to the web las in Fig. 5-l (b)]. The shear center is defined and its significance is explained in Chap. 9. Shear center location for channels arc given in the Properties Tables in Part 1 of the AISC LRFD Manual. Load:. not applied as shown in Fig. 5-l(a) and (b) will cause torsion. or t\\iSttng of the member. However, if restraint against torsion is provided at the load points and points of support, the equations of this chapter are still valid.

COMPACfNESS The concept of compactness relates to local buckling. As described in more detail in Chap. 4. cross sections arc classified as compact. noncompact, or !.lender-element sections. A section is compact if the flanges are continuously connected to the web, and the width-thickness ratios of all its compression elements are equal to or less than ;.,,. Structural steel members with compact sections can develop their full strength without local inc;tability. In design. the limit state of local buckling need not be considered for compact members. Compactness criteria for beams (as stated in Sec. 85 of the AISC LRFD Specification) arc given in Table 5-1 and Fig. 5-2. Jf the width-thickness ratios of the web and flange in flexural compression arc equal to or less than ).P, beam design is by the standard method described in this chapter. Otherwise the special provisions of Chap. 6 (taken from the appendixes of the A1SC LRFD Specification) are required. Tabl e 5-1 Limiting Width -Thickness Ratios for Beams Limiting Width-Thickness Ratio. ;.,. Beam l.:.lcment Flanges of W and other I shapes and channels Flanges of 'quare and rectangular box secuon\: flange cover plate and diaphragm plates between lines of fasteners or welds Webs in ncxural compression

Width-Thickness Ratio

b/r b/ r

General

A36 Steel

65/#.

10.8

190/ Vf.

31.7

640/#.

106.7

h./r •.

FL EX URAL BEHAVIOR The distribution of internal normal strains and stresses on the cross section of a beam is shown in Fig. 5-3. It is based on the idealized stress-strain diagram for structural steel in Fig. 5-4, which is a simplified version of the actual stress-strain curves in Fig. 1-2. A shown in Ftg . 5-3. the normal strain di tribution is alwayc; linear. The magnitude of train is proportional to the distance from the neutral (or ccntroidal) axts. On one side of the neutral axis. the fibers of the flexural member are in tension (or elongation); on the other side. in compression (or shortening). The distribution of normal stresses depends on the magn.itude of the load. Under working loads and until initial yielding, stresses (which are proportional to strains in Fig. 5-4) are also linearly distributed on the cross ser.:tion. Beyond initial yielding. the strain will increase under additional load. The maximum stress. however. is the yield stress f".. Yielding will proceed inward. from the outer fiber to the neutral axis. as the load is increased, until a plasuc lunge is formed.

42

COMPACf BEA MS AND OTH ER FLEXU RAL MEMBERS

. .

,

(CHAP . 5

h,

I

b • b1 - 3t

h,

h. - 31

Fie.5-2 Definitions of widths (b and h,) and thickness (flange thickness in Table 5-1

r and web thickness r.,) for use

l Cross

1\

J

Beam

sce1ion

Stresses

Strains

F•.--.

Compre$sion

+

Tension

Workin g load

Initial yielding

Fig. S-3 Aexural strains and stresses

Plastic

hinge

COMPACT BEAMS AND OTHER FLEXURAL MEMBER S

CHAP. 5)

43

Strain. in/in

Fig. 5-4 Idealized stress-strain diagram for structu ral steel

The plastic hinge condition (under which the entire cross section has yielded) represents the absolute limit of usefulnes s of the cross section. Only beams which are compact (i.e., not susceptible to local buckling) and adequately braced (to prevent lateral-torsional buckling) can attain this upper limit of nexural strength . The relationships between moment and maximum (extreme fiber) bending stresses, tension or compression , at a given cross section have been derived in a number of engineering mechanics textbooks . At the various stages of loading, they are as follows: Until i nit ial yielding

(5.1)

M =Sf,.

At initial yielding [5.2) At full plastification (i.e.• plastic hinge) (5.3) Because of the presence of residual stresses (prior to loading, as a result of uneven cooling after rolling of the steel member). yielding actually begins at an applied stress of F,- - Equation [5.2] should be modified to

[5.4) Equation [5.3] is still valid. however. The plastic moment is not affected by residual stresses. (Because of their existence in a zero-moment condition before the application of loads, the tensile and compressive residual stresses must be in equilibrium .) The terms in Eqs. [5.I] to [5.4] are defined as

M = bending moment due to the applied loads, kip-in M, = bending moment at initial yielding. kip-in

MP = plastic moment. kip-in S = elastic section modulus, in 3 Z = plastic section modulus , in 3

f ,. = maximum normal stress due to bending, ksi

F, = specified minimum yield stress, ksi F, = the maximum compressive residual stress in either flange; 10 ksi for rolled shapes; 16.5 ksi for welded shapes Elastic section modulus S =

c

(5.5)

44

COMPACT BEAMS A:-10 OTHER FLEXURAL MEMBERS

(CHAP. 5

where I is the moment of inertia of the cross section about its centroidal axis. in ; and c is the distance from the centroid to the extreme fiber, in. The Properties Tables in Part I of the A I SC LRFD Manual i nclude the values of/, S, and Z for all the rolled shapes listed.

ELASTIC ANALYSIS

VERS

US

PLA

STIC

Design by either clastic or plastic analysis is permiucd by the A1SC LRFD Specification (Sec. A5.l). The more popular elastic analysis has been adopted throughout this text. When an elastic analysis procedure (such a!. moment distribution or a typical frame analysis computer program) is used. the factored moments are obtained assuming linear elastic behavior. Although thi assumption i incorrect at the strength limit states. the fact that clastic analysis is less complex and is valid under normal service loads has led to its widespread use. Several restrictions have been placed on plastic design. They are stated in the AISC LR FD Specification in Sees. A5.1. 85.2, C2.2, El.2. Fl. l . H 1.2, and lJ.

DESIGN FLEX URAL STR ENGTH: Cb = 1.0, Lb s L, The design strength of flexural members is >Mn, where t> = 0.90. For compact sections. the design bending strength is governed by the limit state of lateral-torsional buckling. As the name implie!., lateral-torsional buckling is an overall instability condition of a beam involving the simultaneous twisting of the member and lateral buckling of the compression flange. To prevent lateral-torsional buckling. a beam must be braced at certain intervals against either twisting of the cross section or lateral displacement of the compression flange. Unlike the bracing of columns (which requires another structural member framing into the column), the bracing of beams to prevent lateral-torsional buckling can be minimal. Even the intermittent welding of a metal (floor or roof) deck to the beam may be sufficient bracing for this purpose. The equations for the nominal flexural strength M, follow from the preceding discussion of flexural behavior. Length L, is defined as the distance between points of bracing. Compact shapes bending about their minor (or y) axes will not buckle before developing a plastic hinge.

[5.6} for bending about the minor axis regardless of Lh. Compact sections bending about their major (or x) axes will also develop their full plastic moment capacity without buckling. if Lh s L,,. M,x = M,,. = Z, F,

[5. 7)

for bending about the major axis if Lb s Lp. Jf L, = L.. lateral-torsional buckling occurs at initial yielding. From Eq. [5.-lj.

M,_. = M.. = S,(F,- F,)

[5.8}

for bending about the major axis if Lh = L,. lf Lp < Lt> < L., M, for bending about the maJor axis is determined by linear interpolation between Eqs. (5. 7} and

[5.81: [5.9}

CHAP.

51

45

COMPACT BEAMS AND OTHER FLEXURAL MEMBERS

In the foregoing z, = plastic section modulus with respect to the minor centroidal (or y) axi.,, in

3 3

Zx = plastic section modulus with respect to the major centroidal (or x) axis, in 1 S, =elastic sect ion modulus with respect to the major centroidal (or x) axis, in Lengths LP and L, are defined in Sec. Fl.2 of the AISC LRFD Specification as follows. For !-shaped sections and channels bending about their major axis

300ry LP =Vf:

(FI-4)

_ 3750r,. ,,..-:; LP- M viA

(Fl-5)

For solid rectangular bars and box beams

p

where '• = the radius of gyration with respect to th e minor ccntroidal (or y) axis, in

A =cross-sectional area, in 1= torsional constant, in• The limiting laterally unbraced length L, and the corresponding buckling moment M, are determined as follows . For 1-:.ha ped section!>, doubly symmetric and singly symmetric with the compres:.ion nange larger than or equal to the tension nange. and channels loaded in the plane of the web L,=

;J >

Vl+Yl+X 2(f; -F,}''

M,= (f;- F_)Sx

{EGjA

- Jt

where

X I-

s.

v ----:z-

(FI-7)

( F1-8)

(FJ-9)

=4S( )z

x 2

where

(FI-6)

(v GJ

E = modulus of elasticity of steel= 29,000 ksi G =shear modulus of elasticity of steel = 11,200 k si

ly = moment of inertia about the minor centroidal (or y) axis, in4 C,.. = warping constant , in 6 For symmetric box sections bending about the major axis and loaded in the plane of symmetry, M,and L, shall be determined from formulas (Fl-7) and (Fl-10), respectively . For solid rectangular bars bending about the major axis 57,000r, Y.fA L = .

'

M

(FI-10)

' M, = F;S. ( Fl -11) Values of 1and C.. for many structural shapes arc listed in Torsion Properti es Tables in Part I of the A ISC LRFD Manual. The practical design of steel beams (Cb = 1.0) can best be done graph1cally by ( I} reference to the beam graphs in the section entitled Design Moments in Beams, in Part 3 of the AISC LRFD Manual , where q>hM, is plotted versus Lb for F; = 36 and SO ksi or (2) constructing a graph similar to Fig . 5-S from data in the Load Factor Design Selection Table, also in Part 3 of th e AISC LRFD Manual.

46

CO'v1PA 1.0, there is a twofold advantage in including C, > I.0 in Eqs. [5. I 1] and ( FI-3 ), and noi conservatively letting C, = 1.0 (as in the graphs in Part 3 of the AlSC LRFD Manual): (1) the unbraced length for which M, = MP is extended from L,. to L,,. and (2) for Ln > L,, the moment capacity M, is multiplied by Cn. The reader can find these facts depicted in Fig. S-6.

DESIGN FLEXURAL STRENGTH: Lb > L, If ihe unbraced length L1, > L, and C, = L.O. elastic lateral-torsional buckling occurs. Th ere is a significant reduction in the flexural design strength ¢nM, as Lb increases beyond L,. Intermediate bracing should be provided. if possible. to avoid such uneconomical designs. However. if L, > L, (FJ-12) for bending of a compact section about its major axis. The critical elastic moment M e, is defined as follows. For doubly symmetric 1-shaped members and channels loaded in the plane of the web

(Fl-13) For solid rectangular bars and symmetric box sect ions

M = 57.000C, VM rr L,/ r,.

(Fl-14)

48

COMPACT BEAMS AND OTHER FLEXURAL MEMBERS

[CHAP. 5

GROSS AND NET CROSS SECTIONS Flexural members are usually designed on the basis of their gross sections. According to Sec. B I of the AISC LRFD Specification. the rules for beams with holes in the flanges are as follows: (I) No deducation is made for holes in a given flange if the area of the hol es is equal to or less than 15 percent of the gross area of the flange. (2) For holes exceeding this limit, only the area of holes tn excess of 15 percent is deducted .

DESIGN SHEAR STRENGTH The shear strength of beams c;hould be checked. Although flexural strength usually controls the selection of rolled beams. shear strength may occasionally govern. particularly for short-span members or those supportin g concentrated loads. In built-up members, the thickness of the web plate is often determined by shear. For rolled shapes and built-up members without web stiffeners. the equations in Sec. F2 of the AISC LRFD Specificat ion can be somewhat stmplified. as follows. The design shear strengt h is 1/J,, vn' where 1/J.. = 0.90. h

418

For-< t.., v fly

V,, = 0.6f;A

[5. 12]

418/..[i\ Vn = 0.6F,A I I

[5. 13]

418 h 523 or -:7"::

132,000 V,, =A... (h/t../

vF,

t,.

(5.14)

V,, = nominal shear strength, kips

where

2

A ... = area of the web. in = dt...

d =overall depth . in t,. =thickness of web . in

It = the following web dimensions . in: clear distance between fillets, for rolled shapes; clear distance between flanges for welded sections (See Fig . 5-7.)

The limit states for shear strength are yielding of the web in Eq. [5.121, inelastic buckling of the web in Eq. (5.13], and elastic buckling of the web in Eq. (5.14].

DISPLACEM VIBRATION

ENT

AND

The two primary serviceability considerations for beams are displacement and vibration. Traditionally , the maximum deflections of floor beams have been limited to W, of the span under the service live load specified in the governi ng building code. Depending on the use of the member and its span , other deflection criteria (stated in inches or in fractions of the span) may be more appropriate . Formulas for maximum beam deflections under various loading conditions are given in many textbook s on engineering mechanics and in th e AlSC LRFD Manual , Part 3. under the heading Beam Diagram s and Formul as. The most common beam loadings are shown here in Table 5-2. together with the resultin g m aximum shears. moments, and deflections.

Table 5-2 Beam Formulas Maximum Value

Loading Condition Simple beam-uniform load

w

jj

A

I

wl M =8 wl V=-

2

I

' 6=

M

swr 384£/

Location

Midspan Ends Midspan

Simple beam-concentrated loa d at center

r I• M

{12

•I•

PI M =-

/12

#/ ..I

4

p

V=-

2

6=

PI'

48£/

Midspan Ends Midspan

50

COMPACf BEAMS AND OTHER FLEXU RAL MEM BERS

[CH AP. 5

Table 5-2.--(.A."= 10.8). the W6x 15 beam i noncompact 11 t:.

n oncompact if

5.2.

F.

in A36 :.teel. Like\\t c.

=50 k:.t.

WJ 2x65. From the Properties Tables for W Shapes. for a Wl2x65 flange

b 1

b,

=-

9.9

21,

lr, web =24.9

r.

(a)

fn A36 steel flange;.,, = I 0.8 web !.,, = I 06.7

(See Prob. 5.1.)

Since flange (b/t = 9.9) < (}.,. = 10.8), and web (hJt. = 2-'.9) < (}..P = 106.7), a W 12x65 beam t\

compact (b)

tn

Howevc:r. if

A36 steel.

F.. = 50 ksi flange;.,,= 9.2 web i..l' = 90.5

(See Prob. 5.1.)

Because flange (b/t=9.9)>(}.,.=9.2), a WI2X65 beam i;, noncompact if F.. =50 ksi.

COMPAt"T BEAMS AND OTHER Fl FX U RAL MEMBERS

5.3.

(CIIAP. 5

The bu ilt -up beam 'ectmn in Fig. 5-8. Referring to Fig. 5-2: b b, Jilin flange-= I 2t1 2 X I in web h, = 40 in

RO.O

0.5

I.

9.0

Ill

(a) The beam 1., comp< ct in A36 steel because flange (b/t = 9.0) < (J.P = 10.8) and web (h f t.= XO. II l he d1sregarded; the grm.> eros; \Ct:llon 1\ u\ed in flexural design. (a) S, = 1,/c. v.hcn.: r 1s tho: major ce/1/rordal ax1s. f'or the symmetric secuon in Fig . 5-7. the centroid ca n be located by inspection. (Otherwi t:. ct lculali on W(lulu be required.) Abo

d --12 in c=.., =.. , lin ")

-

-

-

•

CJIAP. 5]

53

COMPACT BEAMS AND OTHER FLEXURAL MEMBERS

The contributions of the two flange& and the web to the moment of inertia /, are Elements 18inx(l m)1

.

(

[

12

+

I

10 X

I10

(

.)

tn)" X

0 .5 i n

x

32 = b,l

.• 10

( 4 0 i n) '

- - - -- -=+ - - - - In' -

Web

667

'· (b)

2

20

8 2 Flange

_ _ ·]

. )

.•

0. = 2 ,

L

. 17,799 = 84 8 10 S = in' ' 21 in

1

Z, =!AD, where A is the cross-sectional area of each element and D represents its distance from the centroidal axis. In calculating z.. the upper half of the web (in flexural compression) and the lower half (in flexural ten ion) are taken \Cparately .

AD

Elements 2 Flanges 2 half-Webs

((18in xI in) X 20.5 in] x 2= 738 in' 1(20in x0.5in) x lOin X2=200in'

z.

938in'

Z, = 938 in'

5.5.

Repeat Prob. 5.4 for four I ; -in-diameter holes. For each flange, gross area= 18m (as in Prob. 5.4). 15 percent of gross area = 0.15 x 18in 2 = 2.70 m .

hole area = 2 x (Ij!, + 1 )in x I 10 = 3.25 in . In flexural design, only the hole area in excess of 15 percent of the flange area is deducted. Design hole area= 3.25 in - 2.70 in 2 = 0.55 in 2 for each flange. (a)

Adjusting /, in Prob . 5.4: Hole/, = !AD = (0.55 in x (20.5 m) ] x 2 = 462 in' Net section /, = gross section /, hole /, 4

= 17,799in -462in' = 17.337 in' . I, 17.337 in , Net sechon S, = -= = 826 m c 21 in

(b) Adjusting Z, in Prob . 5.4: Hole Z, =!AD = (0.55 in 2 x 20.5 in) x 2 = 23 in ' Net section Z, = gross section Z, - hole z.

= 938 in ' - 23 m ' = 915in·'

5.6.

For a simply supported W24x76 beam, l aterally braced only at the supports. determine the flexural design strength for (a) minor-axis bending and (b) major-axis bending. Use the Load Factor De ign Selection Table for Beams in Part 3 of the A lSC LRFD Manual , an excerpt from which appears herein (with permi ssion) as Table 5-3. Steel is A36.

(CHAP. 5

COMPACf BEAMS AND OTHER FLEXURAL M EMBERS

54

+.M.= 540 kip-ft 1+.M, 343 kip-fi

t

---------- c . W

2

4

X

76

1.0

L-----------+--------------- --------4--------L. L, = S.O ft

L.

= 23.4 ft

Fig. 5-9

The W24x76 is a compact section. This can be verified by noung that in the Properlies Tables in Part 1 of the A ISC LRFD Manua l , both b1/2tr and h,/t •. for a W24X76 beam arc less than the respective fl ange and web values of A1, for F,. = 36 ksi (Table 5- 1 ). (a)

For minor- (or y-) axis bending. M., = M,,, = Z,.F., regardless of unbraced length (Eq. [5.61). The flexural design strength for minor-axis bending of a W24 x 76 i always equal to rp,M., = rp,z, F, = 0. 90 x 28.6 in ' x 36 ksi = 927 kip-in = 77 kip-ft. ·

(b)

The flexural design strength for major-axis bending depends on C, and L,. For a simply supported member. the end moments M, = M,= 0; C,= 1.0. Figure 5-9 can be plotted from the information in Table 5-3: For 0 < L.< ( L,. = 8.0 ft). rp,M.= rp.M,. = 540 kip-ft

At L,

= L, = 23.4 ft . q>,.M•• = q>,.M, = 343 kip-ft. Lmcar mterpolation •

required for L,. < L,
L,. refer to the beam graphs tn Part 3 of the AISC LRFD \ 1anual.

Table S-3 Excerpt from Load Factor Design Selection Table (AISC LRFO Manual, Part 3) For F, = 36 ksi

z..

Shape

tn

rp,M". kip-ft

rp,.M,.

L,,

kip-ft

ft

L, ft

W24X 84 W21 x93 WI4X120 W18X97

605 51.)7 572 570

382 374 371 367

8.t 7.7 15.6 11.0

24.5 26.6 67.9 38.1

200 198 196 192 186 186

W24X 76 WI6X100 W21X83 WI4X 109 Wl8x86 Wl2Xl20

540

343 341 333 337 324 318

8.0 10.5 7.6 15.5 ll.(J 13.0

23.4 42.1 24.9 62.7 35.5 75.5

177

W24 X 68 Wl6x89

300

7.8

224 221 212 211

175

.

535 529 518 502 502

478

473

Nott! : nexural design strength q,,.M .= q,.MP as tabulated L., -t.,.: othcnnse . L,. > LP . Here q, = 0.90.

302 I'>

•ahd for t.s L..,

10.4

If ( • = 1.0,

22.4 38.6

CHAP. 51

5.7.

CO PA0; M ,/ M 2 = 0. In Eq [5-/0[, = ( 1.75 + LOS X 0 + 0.3 X 0) =

c

1.75. Figure 5-10 can be denved from Fig. 5-9 aJ. folloo.>.s . For all L,. the design flexural \lrcngth for C, = 1.75, rp.M••(C,= I.75) = I.75 x rp,M,,(C,= 1.0) :s rp.M,... The previous (C, = 1.0) dcstgn flexural strengths are multiplied by ( C,. = I.75): however. the pht!.tic moment strength ( rp,.M,.. = 540 kip-ft) cannot be exceeded.

c.;,M, h

I 75

X

W24 X 76 C, = 1.75

540 ktp-ft = 945 ktp-

- --

-- -- --- -- --- --

--

C ;,.M. = = 600 I 75kipX 343 kip-ft

-.._,

'..... .....

L---------------------------------1-r-------l., = 8.0 ft

L, = 23.4 ft

L,

fig. 5-10

5.8.

Select the most economical rolled shape for a 27-ft simply supported floor beam. The upper (compression) Range of the beam is adequately welded to the floor deck at I ft 0-in intervals. Dead load supported by the beam (including its own weight) is 1.3 kips per linear foot; live load is 2.6 kips per linear foot. Steel is A36 . Assume: (t1) There il> no member depth limitation . (b)

The deepe t (architecturally allowable) member is a W21.

(c)

The deepel>t desired member is a Wl8.

For the case of dead load and floor live load only. th e critical load combination tn Chap. 1

tS

formula

(A-1-2):

1.20 + I .6L + 0.5( L, or S orR)= 1.2 x U kip/ft + 1.6 x 2.6 kips/ft + 0 = 5.7 kips/ ft For uniformly di,tributcd loads. maximum M = wl !/ 'ro and V = wl /2. (See Table 5-2.) 5.7 kips/ft x ( 7ft)! Required M., = = 521 kip-ft

8

kip> 27ft . Requtred V,, = 5.7 f x --;- = 77 ktps t

-

Here . Lb = 1.0 < Lr (all rolleJ shapes).

56

COMPACT BEAMS AND OTH E R FLEXURAL MEMBER S

(a)

[CHAP . 5

In Table 5-3, as in the beam Selection Table m the LRFD Manual. the most economical bea ms appear in boldface print . Of those beam . the one of least weight for which tJ> M = tJ> Mv 521 kip-ft is a W24x76.

Check ing shear strength with Eq . [ 5.12] , for hIt. < (418/VF. = 418/\136 = )69.7 V. = 0.6F,A. = 0.6 X 36 ksi X dt. tj>,. V. = 0.90 X 0.6 X 36 ksi dt. = 19.4 kSI X dt•. For a W24X76, h/ t..,. = 49.0 < 69. 7. (See Properties Tables for W Shapes in the AISC LRFD M anual. Part 1.) Then 77 kips required . Use a W24x76. (b)

By im.pection of Table 5-3. the least-weight W21 for which tJ>.M.. = tJ> MP,?: 521 kip/ft is a W2 1X83. Checking \hear: q>,.V. = 19.4 k 1 x dt•. For a W21X83, tj>,.V. = 19.4 ksi x 21.43 in x 0.515 m = 214 kips > 77 k1ps required . U\e a W21 x83.

(c) By inspection of Table 5-3. the lca t-weight Wl 8 for whi ch

•MP, 2:: 521 kip-ft is a W18X97. Checkin g sh ear: tj>,.V,,=l 9.4ksiXdt•. For a W18x97. tj>,.V,, c= 19.4ksi x 18.59inx 0.535in = 193kips >77 kips required. Use a WI 8x97. (Note. In lieu of calculations, the design hear strengths tj>, v. for W shapes can be found tabulat ed in the \ection Uniform Load Constams in Part 3 of the A I SC LRFD M anual.)

5.9.

Repeat Prob. 5.8 assuming th at the floor deck is not pr esent and the beam i s laterally braced only at midspan and the supports. 27 ft L,. = = 13.5 ft

., -

For this case, C, = I.75, '" m Prob. 5.7. From Fig. 5- 10, 11 b evident that L. = 13.5 ft < Lm for a W24 x76. Similar plots wi ll show the same values for the other bea m sections in Table 5-3. For these shapes. the design tlexural \trength tj>,.M,,. = 1/>,,M,,.•• as for the fully braced case. Accordingly. the results of Prob. 5.8 are still valid.

5.10. Repeat Prob. 5.8 for a beam braced only at its end supports. Here, C,. = 1.0. For som e of the W shapes in quest i on . L,. > L,. The beam graph s in Part 3 of the A ISC LRFD M anua l (for 1.0. F. = 36 ksi) arc helpful in this ca e. In the graphs, the solid lines

c.=

denote the most economical W shape. the dashed line\ mdicate alternate . One page of the AISC beam graphs is repr::>duced (with permi, IOn) as Fig. 5-11. where it can be een that at L, =27 ft .among the members with cp.M. 2:52 1 kip-ft. a W21 x 101 (solid line) is most economical: a W 18x 119 (dashed line) can be used if beam depth i s limited to 18 in .

5.11. Determine C1, for the span of the continuous beam shown in Fig. 5-12. (a)

Lateral braces arc pro\ ided only at the supports.

(b) Lateral braces are provided at midspan and the supports.

(a) In Eq . 15.10]. M ,! M = -(500 kip-ft/500 kip-ft) = - 1.0 where M ,/ M 1 i s negative because the end moment .. M, and M! cau..e rotations in opposite direction C,= I 75+ I

U5-'+0.3M(=

1.75 +

1

M

t\1. .. 1.05( -1.0)

M ' )

M, = 1/>bS,( F, - F,) = 0.90 X 848 in 'X (36- 16.5)ksi = 14.882 k•p-in = 1240 kip-ft

According

10 Eq.

(FI ·•1), for 1-shapcd member

L

,,

= 3 0 0r ,

VF .

The radius of gyration r:• -

The contributions of the two flanges and the web to the moment of inertia /, are Elements

I in X (18in)\ [ 12

2 Flanges Web

o]

X

2 = 972 10•

40inx(0.5in) ' . • 0 .410 12 +0=

'·

Cro s-sectional area A= ( IR in x l10) x 2 + 40 in x 0.5 in= 56 in" and

r, =

=4.1710

300x4.17in

VJ6

LP=

=208in = 17.4ft

According to Eq . (FJ-6), for !-shaped member L, wher e

=;·: VI+ VI+ X !( F, - F,)!

.

'

/ £GJA X, - S, \ 2 _ .'r

x.= 4 s -

I•

( )! GJ

Here, r,= 4.17in. F;-F.=(36 - 16.5)ksi = 19.5 ks•. S,=848in '. A=56in", £=29.000ks•. G= 11,200 ksi, and 'kbt ;.... 1 =i{ll8inx(lin)'J2+[40inx(O.Sin)']} 3

= 13.67 in•

60

COMPACT BEAMS AND OTHER FLEXURAL

For l- hapcd members.

MEMBERS

(CHAP. S

c. = (JJ )(d- r,) . Then

X ,= 4((/J4)(d·

r, )' l (

I,

):= [S,(d- r1 )]

GJ

GJ 2

8 8in'x (42-l)m ] =[ =0.05 11.200 k i X 13.67 In

29.000 k i

:rr

X

X, =R48in' \ L,=

11.200 k i 2

X

13.67 in

4

X

56 in2

= 1306

4.17 in x 1306" . , _ . yl +V 1+0.05( 19.5 ksi)" I9 .) hi

= 658 in= 55ft

5.14. Simply supported 30-ft-long floor beams. Wl8X35. are spaced 10ft 0 in cen ter-to-center. What is their maximum deflection under a live load of 50 lb/ft 2? - lb ft ·

w = :>0-, X IO.Oft = 500- = 0.5-

lb ft

kip\ ft

For a WIRx35 beam. I,= 510 in. From Table 5-2. for a uniformly loaded simply supported beam. the maximum deflection

kips

5x0.5-x(30ft)

---- 19 31\4 X-

fl

.----x kipS .

.000

X

m·

Since live load deflecuon

510

( 12 in / fl)'

4 In

= 0.62 in L 30 ft X 12 in/ft 6 = 0.62in imply ..upported beam 111 Fig. 5-14 b subjected 10 a concentnued factored force of 50 k1p . Steel1s A36. Assume continuous latl!ral bracing. (a)

Determine the re4uired !lexural strength.

(b)

Select the most economical W shape.

(c)

Select the most econom1cal \\'21.

(d)

Select the most econom1cal \\16.

An.s . (a) M., = 400kip-ft. (b) W:!4X62 . (c) W21Xn!t (d) Wl6X77. P, =

50 k1p

I 16ft

16 ft

,/9,.

Fig. 5-14 5.20.

The beam in Fig . 5-14 " braced at only. Determine C, for each unbraced length. Ans .

the suppom and quarter point.,

(See Fig. 5-15.)

C.-

I 75

1.30

l

1.30

I 75

# --4-----rFig. 5-15

5.21.

Sel ect th e most economical W ;ection for the beam in Fig. 5- 14. hraced at the (a) supports and 4uarter points only: (b) support> and m1d pan on ly: (c) '>upports only. An .

5.22.

(a) W:!4x62. (b) W24X62. (c) WlRX97

The unfactored concentrated Ji, e load for the \\'24 X62 beam in F1g. 5-14 is 20 kips. Determmc the mmomum hvc-load deflect1on Am.

t:.. = 0.61 in.

Chapter 6 Noncompact Beams and Plate Girders NOTATION A= cross-sectional area of member. in 1 A,,= cross-sectional area of stiffener or pair of stiffeners. i n 1

A .. = web area. in =

2

cit.,

a= clear distance between transverse stiffener =bending coefficient. defined in Eq. [5.10J C.,= shear parameter defined in Eqs. (A -GJ-5) and (A -GJ-6) D=coefficient for usc in Eq. (A-G4-2) d =overall depth. in

F,, =critical plate girder compression flange stress. ksi F, =compressive minimum yie ld stress. ksi

F..

11

=specified minimum yield stress of the stiffener material. ksi

lr. lr , = ''eb dimensio.1s dd1ned in Fig. 6·2. in I = moment of inerti.a. ir." 1,= mom.:nt of inertia nf stiffener or pai r of stiffener . in• J = torsional c...>nstant. in•

j =coefficient defined m Eq. (A-G4-I) k = coefficient defined in Eq. (A -G3-·+)

L,= unbraced length. It M, =nominal flexural strength of member . kip-in

M,, =plastic moment. "1p-in M, =limiting buck l ing moment when ). =A, kip-in

M., =required flexural strength, kip-in R,.c;= plate girder flexural coefficient. defined in Eq . (A-G2-3) r = radius of gyration. 111 rT

=radius of gyration of the compression flange plus one-t h ird of the compression portion of the web taken about an axis in the plane of the web. in

S =elastic

62

NONCOMPACT BEAMS AND PLATE GIRDERS

CI lAP. 6)

63

V,, =required shear trcngth. kips x =subscript relating symbol to the major principal ccntroi da l axis

y = sub cript relating ymbol to the minor principal ccntroidal axis Z =plastic section modulu . in' A = tenderness parameter (e.g., w1dth-th1ckness ratio) ;,,. = large t value of). for which M., - M, A,= largest val ue of ,1. for which buckling i inelastic

A,.. The subject of the next cction is noncom pact beams with a width -t hickness ratio (.A): A,.< A .A,. Plate girders w11h slender webs ().> ).,), u ually stiffeneu. are covered .n the following section.

1\0NCOMPACT BEAMS The flexural design strength b ¢hM,, where cp,. = 0.90. For noncom pact beams, the nominal flexural strength Mn i the lowe t \'alue determmed from the limit '>tatcs of lateral-torsional buckling ( LTB) flange loca l buck ling ( FLB) web loca l buckling ( WLB). For .A,.< A =laterally unbraeed length. in r, = radiu of g) ration about the minor ax1s. in b, t, h,. t = dimen-;ions of cross section. defined 111 1-'ig. 5-2. in A= eross-sect10nal area. m'· J =torsional con tant. in•

F, =compressive residual stress in the flange = 10 ksi for rolled shap 970/ .

or both

Stiffeners arc discussed later in this chapter. Web stiffeners arc not required if web h)t ... < 260 and adequate shear stn.:ngth is provided by the web in accordance with Eqs. (5.12] to (5. 14]. (Please note: Two different parameters in the A I SC LRFD Specification refer to the clear height of the web: h and h,. In Sec. B5 of the LRFD Specification they are thus defined: For webs of rolled or formed secuons. h is the clear distance between flange les!l the fillet or

66

NONCOMPACT BEAMS AND PLATE GIR DE RS

!CHAP. 6

corner radius at each flange; h, is twice the distance from the neutral axis to the inside face of the compression flange less the fillet or corner radiu . For webs of bui lt-up sections. h is the distance between adJacent lanes of fasteners or the clea r distance between flanges when welds are used and h, i t\\ icc the distance from the neutral axis to the nearest line of fasteners at the compression flange or the inside face of the compression flange when welds are used. The distinction between h and h, is shown in Fig. 6-2. where it can be seen that for doubly symmetric cross sections. h = h,.)

c=;::::==J --h

--

•I I

Compression flange'>--

lr.

Neutral ax"

c=::::: :::J-----

--

"

r-'

Tension

......

flanges ---

I

l_

(a)

.f_

h

, ..L

Fig.6-2

Definitions of J, and 11,: (a) singly symmetric built-up sections; (b) doubl) ymmetric bui lt-up sections

For plat e girders, the maximum permissible web sl enderness 11 / t,.. depends on the spacing of the stiffen ers. If

a


,

where

h 14.000 t:tre!!Sth is 970/ VF,. (i.e.. the web 1s slender), M, is governed by the limit states of tension flange yielding and compression flange buckling, as follows. For yielding of the tension flange (A-G2-1)

For buckling of the compression flange (A-G2-2)

The nominal flexural strength M" is the lower value obtained from these equations, where

/r,

970 )

R 1.c. = 1 -0.000Sa, (---:7":!"' < 1., V r 1.0 where

(A-G2-3)

a,= ratio of web area to compression flange area

F;, =critical compression flange stress, ksi

F;. =minimum

specified yie ld stress, ksi S,c =elastic section modulus referred to compression flange. in 3

s.., =elastic section modulus referred to tension flange, in

1

The critical stress F;, in Eq . (A-G2-2) depends on the slenderness parameters)., ).1,

)."

and Crc·

For), ).,

(A-G2-6)

The slendern ess parameter are determined for both the limit state of lateral-torsional buckling and the limit state of flange local buckling: the lower value of F;, governs . For th e limit state of lateral-torsional buckling ), = L,

r, ).

3 0 0

(A-G2-7)

,, - \ 1 'F ;

(A-G2-8)

).

(A-G2-9)

75 6

, -V F . Cpr, = 286.000C,

(A-G2-JO)

68

NONCOMPACT BEAMS AND PLATE GIRDERS

[CHAP. 6

where C, is determined from Eq. [5.10] and rr is the radius of gyration of compression flange plus one-third of the compression portion of the web taken about an axis in the plane of the web, in. For the limit state of flange local buckling J... =

bf

(A-G2-11)

2t1

= 65 PVF_,

J...

J...

(A-G2-12)

= 150

(A-G2-13)

r

c

Pf"i

= I I '200

(A-G2-14)

C,= I The limit state of web local buckling is not applicable .

DESIGN SHEAR STRENGTH OF PLATE GIRDERS The design shear strength is ¢v V,, where ¢u = 0.90. For h lt...

$

187Yk!i\.

(A-G3-1) For hit_. > 187

vn = 0.6A...Fy (C,, +

1 - C,, ) 1.15Vl + (a/h)2

(A-GJ-2)

except for end panels and where

3.0 a

or

h

260

->

[6. 1]

(hit S In such cases tension field action does not occur and

Vn = 0.6AwFvC,,

(A-G3-3)

I n the preceding equations 5

(A-G3-4)

k=5+(a/h)2

except that k = 5.0 if Expression [6.1) is true or if no stiffeners are present; A. is the area of the web. 2 in = dt..,; and dis the overall depth, in. If 187

C,. =

h/

If -h>234 -. t.., Fv

44.000 k C,, = (h/t..,)2Fv

1,..

(A-GJ-5)

(A-GJ-6)

ONCOMPACT BEAMS AND

CIIAP. 6]

PLATE GIRDL:.RS

69

W EB STIFFEN ERS

Transverse stiffeners arc required if web h/r .. > 260 or \\eb shear trcngth, as determined from Chap. 5 (for unstiffcncd beams). is inadequate. The stiffeners hould be spaced to provide sufficient shear '>trength tn accordance with the preceding provision!> for plate girderl.. Additional requirement for stiffeners are I,>at.,] .

whenever stiffeners are required A., 2:

and

!- lo. 15Dht,..( I- C,,)¢,,

f,_.,

[6. 2]

V,,

-

(A -G.J-2)

IHt ] 2:0

for tcn i on field action j=

where and

2.5 ,-2> 0.5

(A-G4-1)

(a/h)·

/,,=moment of inertia of a transverse web stiffener about an ax1s 1n the web center for stiffener pairs or abou t the face in cont, in 4 A,, = cross-sectional area of a transverse web stiffeners. in F, =specified minimum yield stress of the girder steel. ksi F. ..,= specified mmimum yield stress of the stiffener material. k-.i D = 1.0 for tiffenerl> in pairs 1.8 for l.ingle angle stiffeners 2.4 for single plate stiffeners V,, = required shear strength at the location of the tiffener. k i

and C,. and V,, are a defined above. Plate girders with web!> that depend on tension field action [i.e.. their !>hear strength b governed by Eq. (A -GJ-2)J, must satisfy an additional criterion, flexure-shear interaction.

If

t h en (A-G5-1)

must be true. Here, V,, and Mu arc the required shear and moment trcngths at a cross section calculated from t he factored loads; V,, and Mn are the nominal shear and moment strengths ( V,, :5 V,, and M,:5 M11 ; = 0.9}.

STU'FENE DETAILS

R

Spcc1al requirements appl> to stiffeners at concentrated loads or reaction-.; '>CC Chap. 12. The web stiffeners provided in accordance wi t h the prO\ i'>ions cued in thl\ chapter may be onc-s1ded or two-s1ded. If a pair of stiffeners is used. they can be welded to the web only. Smgle

stiffeners are also welded to the compre ion flange. a are '

A indicated in the text of thi-, chapter. the ca c of no tiffeners corre ponds to k = 5. (This can be verified by comparing the ju t-derived expressiOn for V,, with Eq . [5. /4].) To double the shear strength. let k = 2 x 5 = 10. In Eq. (A-GJ-4)

5 k =5+

· '= 10 (a/h) Thi implies a/h = 1.0 or a = h: thu . the clear di tance between tran vcr e web stiffeners a = h =56 in . Checking the origmal a-. umption. we obtain (

h= 56 In. = 121!.0). t.

0.44 Ill

> (234 -=234 123.3)

F,

-=

O.k.

36

Stiffener design can be determined as follows. Because tension field action b not utilized , Eq . (A -G42) and (A·G5-1) can be ignored. However, Formula (6.21 must be satisfied: 1., ?.at:.j 2.5 j = (a Ih )' - 2 2: 0. 5

where

.., -

-·'

j = )T- 2 = 0.5 1,, 2:56 in x (0.44 in) ' x 0.5 = 2.34 in"

Try a pair of stiffener plates. 2 in x !an as in Fig. 6-6

--

'"'"

5 44 in

I/ Fig. 6-6

The moment of inertia of the stiffener pai r abou t the web centerline _ 0.25 in x (5.44 in)'_ ." ? . " /.,-:us m >_.34m o.k. 12 Try a smgle Miffener plate. 3 m

x! in. as in

Fig. 6-7.

/Web 3.5 an

.----- Stiffener

- '-

--J f-- 0.25 in Fig . 6-7

74

NONCOMPACT BEAMS AND PLATE GIRDERS

[CHAP. 6

The moment of inertia of the stiffener about the face of the web

I,, •m

6.3.

0.25 in X (3.5 in)' ' 7 • -• ·'··' an

"c> '

1

_) .3... ,

o.k.

3

See Prob. 6.1 and Fig. 6·3. Change the web thickness to r.. = moment and shear strengths.

1in. Determine

the design

Checking web lendcrne '· we obtain

(

h. = 56 in = 2.,4 ) > (. _ 970 _ 970 ) .,- · A, ·I.-r= = 161.7 0 '· _) m v /-, \'36

Because the web is \l ender. the member is cla \incd a a plate girder. and the flexural de ign provt ion of this chapter govern [Eq. (A-G2-1) to (A-G2-/,M., = 0. 90 x 3337 kip-ft = 3003 kap-ft. Shear strength for an unstiffcncd web is governed by Eq. [5. J2f, [5./J], or [5. J.lj. on h/ r•. Here , h/ r. = 56 in/0.25 in = 224.0

depending

523 523 ) ! = = l\7? 224 > \ F, \136 ·( Equation [5. J.Jf go,ern\ 132.000 (58 in x 0.25 in) 132.000 V =A =-------• • (h/t.)' (224.0) = 38.1 kips

1he de\agn hear \trength ¢,. V., = 0.90 x 38. I kips= 34.3 l.ap .

6.4.

Design web stiffeners for the end panels of the plate girder in Prob. 6.3, to increase shear strength. As umc a= 24 in. Ten aon field action i not permitted for end panels of girder\. The nomanal shear \trength as obtained from Eq. (A ·GJ-J) in thi chapter: V., = 0.6A•.F,C,,. To determine C,.:

5

k =5 + Since h

5 6 in ( ( 234 r. . an

=0 2 5 .

c:.

5

I •).=5+ (24 .an/56 . )',= 32.2 (a h an

= 224

)

> -= 234 F,

+1,000 k ... = (hIr., )"F,

ffj.,

?

:...:.:.: ;; 221 .4 36

by Eq. (A -GJ-6)

SuhMIIuttng for C, an Eq. (A-GJ-3), we have +1, !XXJ J.

vn = 0. 6A.F, X I '

(h r.) F

26.4!Xl k

= r1..,

-

... (11/r.)

(58 an x 0.2510)26.400 x 32.2 (56 in /(1.15 an)

= 246 kip

76

;-JONCOMPACT BEAMS A 'JD PLATE GIRDERS

(CHAP. 6

1he de.,1gn 'hear 'trength becomes 1/>. v.= 0.90 x 246 kip = 221 l..1p\. a large mcrea'e over the:U.3kip 'trcngth ot an un.,Hff.:n.:d ''eb (m Prob. 6.3). S11ftcncr dc\lgn (\ ith no tension field action) conml\ of compl}'mg w11h FomlUia (1\..?l . /, > m j

..

....,.._'i

_.,Jn-.

J=

(a /11):-

·•

....-

-----:- 2 = II ll (24 in / 56 in): · 1,? 24 in x (0.25m) x 11.6 = 4.35 m

1

Among the po''1ble \tiffener configurations are (u)

A 'inglc ...urfener plato: 4 in x! in !The moment of inertia of the single stiffener about the face of the web U.25inx(4in) ' _

. .J). 1 = .JJ ll ..,_

f,, =

3 (h) 1\ pair of tiffener plates 3 i n

O.k .l

·

4 ':l

Ill

I

X! in

!The moment of inertia of t he stiffener pair about the web .:cntcr l inc 0.25x(6.25in) ' _ , = ':1.09m /,, = 12

6.5.

, 4.3510

o.k.l

Repeat Proh . 6.4 for an intermediate web panel. tncluding tension field action. f\, 111 Prob . 6.4, 1.. -= 32.2. hit.= 224, and C, i determined from Ell (A-GJ-(1) .

M.OOO k 44.000 x 3.2

C,. =

, =

(hft.) f;

(224)'

-=

X

6

o 7S

The 1Hlm1nal ...hear •arcngth (mcludmg tension field acttonl "go,erncd by Eq . (A-GJ-.?). V,= 0.6A F. [ . ( ,.

+

I- (,. ] I 15\11 + (u / h)'

k ip s [ ] V.- 0.6 X (58 111 X 0.25 in) X 36 .- , 0.71i +,r; I - 0 = 300 k1p 7!! · _ , m· l.ISv I+ (24m/':16m) · fhc dC\Ign '>hear trength IS 1/J, V. = 0.90 X 300 kip = 270 kip St1tfcner de ign taking advantage of tension field action muM comply wirh formulas (6..?], (A-G.J-2), and (A-G5-1). The de igm in the solution to Prob. foA comply with formu la 111.21. Check ing lormuh1 (A -G'-1-2), we obtai n A, > -FF.

[o

s D1lt.,( l v.

.I

- (.•.) A \UOlC

v.. = tp,. v.. = 270

·1

- l!!t

.

d)

cj>,.

\ ' t

kipS. (a)

A '>Ingle tiffcncr plate 4 1n '< ', in A , ·36 ""

r 0.15x2.4X56inx0.25inx(1-

36 bi

1\n) '>lll!!le ,llffener plate (A,> 0)

.

·J

0.7!!)X .

270 ktp... - IX(025m) = - 0.02in 270 kip\

ts

satisfactor)

(h) A pa1r of \llffencr plate .3 111 x in 16 k" [ A (} 15 X 1.0 X 56 in X 0.25 in X (10.78) X

70

18(0 25m)· -

]

• = -11.66111 -

36 bi An p;ur of \llffcner plates (A ,> 0) is okay.

270l..1p'>

CIIAP. 6]

NONCOMPACT BEAMS AND PLATE GIRDLRS

77

Regardmg cntcnon (A-G5-I) (flexure-shear tntcracuon) . 1', = :!70 l-.1p,, V.. = 3CXJ kip . .H.= 3337 k1p-ft (from Prob . 6.3). Let the requtred flexural 'trcngth \1.. 15110 k1p-lt at the .,arne cro ection Becau'e

V "70 k1ps ) ( v 100 k1p' ) ( "= - . = 0.18/ft > 1.33 : = 1.33 X ) J7 k ..f = 0.1:!/ ft M.. 1500 k1p-ft M. .... .. 1p t

ernenon (II -G5- I) ( ftcxurc-,hear interaction) need not be '>ati,hed

6.. 6.. Determine the minimum web thickness for the plate gin.kr in Fig. 6-3, both with and without web sti ffcner': as u me A36 steel. According to the AISC LRFD Specification (App. G), in un tiffcnctl g1rdcr' II

-< 260 implies that 1., .,.,,

1.,

II

56in

> - = 260 _,nl

11/t., muM be le> than 260.

.,., • (I --In

In stiffened girder (a / 11 s

1.5):

:woo

II

r;:

-
with Tabl e 6- 1 (for a doubly >ymmetri c l·\hape bend ing ubou 1 it\ maJor axis):

b b, 30 in = I SO .1. =-=-= ., I . ..

Flange

I

-11

-X

Ill

65 65 ;," -.. =- IO.X = = \IF;. V'!.6

Flange

106 vF;- 16.5

}., =

Flange

... 106 = 2-Ul \36 - 16.5

For the flange. (I.,. = 10.8) < (). = 1.".0) < 0.. = 24.0). f-rom Prob. 6.1. for the web (\ hich ha' not changed)

(.I.,= 106.7) < (I. = 12RU)
e . Mn, 1\ dctermtncd ll\tng the hnear interpolation Eq . (A -F/3)

Regardtng lateral bracing and the lim1t tate of Ll B· "- 1. / r,. ''here r, =\I)A The cro ·\cCtiOnal area A = (30 111 x l1n)::! +(56 in x 0.-W Ill)= X-1 .5tn The contribution of the

tv.o flanges

78

NONCOMPACT BEAMS AND PLATE GIRDERS

and the web to the moment of inertia I are

[CH A P. 6

BT ' -+AD ' 12

Elements

J

Iin x (30in)'

2 Flanges

2 = 450Chn'

---'---..:....+II

[

12 56 in X (0.44 in)'

Web

, ._ ,

+ () = 0.4 m• 4500 in•

I,

4500in'

r, = For LTB.

S . = 7.30 in .,, o-..• 10

?.. = L1, = 40ft x 12 in/ft = 65.1! r, 7.3m

For LTB

300

300

?.. =-=-=SOO Vi\ v'36 .. p

For LTB. ?.., can be determined (as indicated in Table 6-1) !rum Eq . (F/-6). (FI-H). and (FI-9) in Chap. 5. as follows.

)., = L. = X I vI r, F.-F.

+ vI + >. J,. - f; )'

where _ .T ,

X ,- S, V1

EGJA

2

. A= H4.5111 '

F. -F.= (36- 16.5) ks1 = 19.5 k\1. E = 29.000 ksi.

G = 11.200 ksi.

I::; {[ =3011.1 X (1 1.11)']"- + [."161.1l X (O,.,,.,,.m)']) = 2 l.,lll. 1• 3 _,

J='fb1' For (-shaped members,

C = l./4(d -td. Then

x, = 4[1./4(d -t, ) ] ( )2 -

S '

I,

GJ

=!.c= = C

d/2

I,

_

=I S,(d - t, )]

2

(iJ

I,

58 111/2 2lJ I ll

The contributions of the rwo flanges and the web to the moment of inertia I, arc

BT"' -+AD '

Elements

2 Flanges Web

3()

in

[

x {lin)' 12

-)-,)-

I,

S = '

12 ·]

+

(30

111

x l1n )(2!!.5 an)' 2 4!!,740 111

0.44 m x,.(.:5..6-_i.n.:.)_ + () 12

·. . ..

) 29 111

111

=

(90( Ill

1

•

"' 6.403

55, 143 tn'

Ill

1

79

NONCOMPACT BEAMS AND PLATE GIRDERS

CHAP. 6j

X ,=

29,000 kips/in x 11.200 kip/in 2 x 21.6 in' x 84.5 in 2

lT

=

1901m'

. .,

x, = [ • in•

For LTB.

900

)., =

19.5

-

1901in'x(58-l)in]

900

2

=

11.200 ksi x 21.6

0.20

1

V l +VI+ 0.20(19.5) = 144.3

For the hmit state of LTB = 50.0) < (A= 65.8) < ()., = 14-U)

(}.p

In summary, for all three limit tates (LTB. FLB. and WLB). A,, < A < A,; that is, the member is a noncompact beam. and the "noncompact beam " provisions of this chapter apply. The equations for , and M, are given in Table 6-1 (for a doubly symmetric Ishape bending about M1 it:. major axis)

M, = { F.S.

(r,

F, )S,

for LTB and FLB for WLB

To determine z.. we obtain ! AD. In calcu lating Z,, the upper and lower halves of the web are taken separa t ely. Elements

AD

2 Flanges

((30 in x lm) x 28.5 in]2 = 1710

2 half-Webs

in' 1(2!! in x 0.44 in) x l.t in]2 = 3.t3 m'

Z,

2053in' Z, = 2053 in'

Determming flexural Mrength:.. we obtain

= F. Z= 36 kips /In: x 2053 in'= • k ' .f 1 ' ' 12 in / ft 'P 6 ()9

1,A .,,..

For LTB and FLB 1

_ 19.5 kips /in x 1901 in'_ . ,)5x in/ft -30H9 k1p-ft 12

M _(F. _F. , -

,

For WLB

= F.S = 36 kips/in: x 1901 in M,.

=

,. '

3

.f 570 ktp t

12 in/ft

The various results for). and Mare plotted in Fig. 6-8. From the figure . or by solving Eqs. (A-FI-2) and F -3), it is evident that FLB governs for minimum M,.,; M.,= 5182 kip-ft. The de ign ftexural strength ,.M•• = 0.90 x 5182 kip-ft = 4664 kip-ft . The design shear strength i 184 kips as in Prob . 6.1.

6.8.

Repeat Prob . 6.7 with an additional change. The thickness of the web is lw =

1in.

The design shear strength i a' in Probs. 6.3 (for an unstiffened web) .6.4. and 6.5 (for a stiffened web).

80

NONCOMPACT BEAMS AND PI.ATE GIRDERS

[CHAP. 6

---...:.:.:.:.:.:_::::-':f: -i--- ..._-------t--====r=' ""-"'"--• M _, = 5703 kip-ft (V

M., - 5 1 8 !up-It

1

I

I

I I

I

I

A,= 10.8

.11 = '089 1p-ft

1 (LTB. C, • I

I I

!I·LBI

I

1=658

1.75) I

I ----I

;

I

t

........._ --..._ --..._

I

II

1 M., - 5406 kip-ft

I 1

I I

L---+

-......_

-. M., 089 kip fl t LTB. C. = I

LTB I I

A = 50.0 I A .- 1-1-' 3 ---- -- A --+l ---------r----*-------------------rI 1 ------

A , - 140

t

A,- 106.7

A = 15

A ,= 161.7

A = 128

WLB

FLB

Fig. 6-8

Regardang flexural trength. the plate girder proviston mu t be applied because

( = 5 i =224) > (1)70 = 97 =161.7)

VF.

r. 0._5 10

y36

The appropriate equation are (A-G2-J) to (A-G2·1 ). For the limit tate of LTB,).. = L./r" ). = 300 = 300 = p

.

-

)...,

so ()

VF.

\(36 ..

756

756

--

vF

y3n '·

j"!L).()

Determ10ing r 1 , we obtain r1 =\)I,/ A of a \cgment consisting of the compres.,aon flange plu one-sixth of the web . (See Fig. 6-9.)

BT·' -+AD1

Element

12

linx(30in) '

---..:..-,....:... + 0 = 2250 in4

Flange

1-

9.3 in x (0.25 in) '

;, Web

12

+0 =0 2250 in

I

A = (30 in x I in) + (9.3tn x 0.25 111) = 32.3 in1 2250 in• _ 32.3

.

0 3 10· . ' - 0.. 10

). = L. =40ft X 1_2 in / ft =

r1

8.3 111

_ 57 5

For LTB, (.l.,, = 50.0) < (). = 57.5) < ().,= 126.0). The value of

c.is

determined [5.10]; however, Co = 1.75 is stated in Prob. 6.7 .

E4.

from

normally

NONCOMPAC'T BEAMS AND PLATE GIRDERS

CHAP. 6]

81

I in

6

h

56 in

=

= 9.3 in

6

-1

0.25tn

Fig. 6-9

Regarding the lim i t state of FLB

).=!?.£.= 30in =IS.O 2t1 2 x l i n 65

65

;.,,, = \(F, = v'J6 = I 0.8

)., =

15 0 =

V F.

1 5 0=? V-5.3 6

0

For FLB, Up= 10.8) T"'P'

g M.

+9

.P.M.

=I

-----------------

M.

Fig. 7-2 Interaction formula (H 1-Ja) and (HJ-Jb) modified for axial load combined with bending about one axis only

Solved Problems 7.1. Find the lightest W8 in A36 steel to suppon a factored load of 100 kips in tension with an eccentricity of 6 in. The member is 6ft long and is laterally braced onl y at the suppons ; C, = 1.0. Try orien tations (a) to (c) in Fig. 7-3.

86

MEMBERS IN FLEXURE AND TENSION

[CHAP. 7

• 100 kips

T t• • bm l

e

----

t

T 6in

=

r-6m

Fig . 7-3

P,, = 100 kips;

M,. = P,,e

100 kip x 6 in _ S(J . 12 in/ft - klp-fl

= For orientation (a) in Fig. 7.3

P,, = 100 kips. M," =50 kip-ft.

M,.. = 0

Try a W8x28: the design tensile strength (for a cross section with no holes)

,Pft = ,F.A,

[Chap. 3. Eq. (01-I)l

= 0.90 X 36 k i X 8.25m'= 267 kip

For (L = 6.0 ft) < (LP = 6.8 ft). the design flexural strength for x-axi bendmg

(Chap. 5, Eq . [5. 7))

cp .w.... = cp,MP = ,Z,F which IS also the tabulated value for AISC LRFD Manual.

0.90 x 27.2 in ' x 36 ksi f 73 .4 kipt = 12 in/ ft

l/>.M" for a W8x 28 in the Beam Selection Table in Part 3 of the

_!l_ = 100 k p = 0.37 > 0.2

Since

lJ>,P,, 267 kipS

the fh t of the two interaction formu l as applies.

8 ( 50 kip-ft ) 0.37+ +0 =0.37+0.o l _ kip-ft 9 73 4 Fur oricntauon (b) in Fig . 7-3

O.lJ!\ .URE AND TE:-ISION

CI IAP. 7]

The de 1gn flexural \trcngth' arc determmt!d a\ m Chap. 5. For a •mph.! beam. C, = 1.0. h1r \-axis bendmg. cp,..Mn. (L, = 12.0 It C.= 1.0) can be deternuned either dm:cth from the beam graph' m Part 3 of the AISC LRFD Manual or b} interpolauon of the data m the B.:am Sclccuon Table prC\Cntcd ther.:m The latter procedure i-. 'hown m F1g. 7-5

+,M -

83.-1 klp-ft •..If.,

= 63 2 k•p-fl \\'10 " 30 C.= I 0

L,.- 5.7 f1

L = 20 .1 II L = Itl II

f'ig. 7-5

U•mg either method one ohtnins cp,M,. = X3.4 k1p-lt tor the W IOxJO. For y-axis bending ( r.:gard lcs' of L,,) 0.1)0 X X .H4 in ' X 36 k i

ip1.M,. = q>,Z, F. =

,

l_m

/f

t

= 23.1J k1p-tt

Suh,tlluttng m the mteractllln formula for hlil,t!>ion = O.XS cp1, =resistance factor for flexure = 0.90

The subscript x rclcr'> to bendmg about the maJOr principal ccntroidal (or x) ax1s; y refers to the m10or principal ccntroidal (or I' ) ax1\ . SIMPLIFIED SECOND-ORDER ANALYSIS

Second-order moments m heam-columns arc the add1t10nal moment'> caused by the ax1al comprcss1ve forces acung on a displaced structure.:. Normally. structural analysi'> i'> fibt-order; that 1s. the everyday methods used in practice (whether done manually or by one of the popular computer programs) assume the forces as acting on 1he original undeflected Mructure. S '::ond-order effects are neglectt•d. T ,ati fy the AISC LRFD Specificatio , second-order mlntion . In a ') - braced frame. M, =0. In an unhraccd frame. M1, inclu des tht.: moment5 from the lateral loads. If both the frame and its vertical load arc yrnmetric . M1, from the vertical load is · i'cro. However. it either the vertical load\ (i.e . dead and ltve loads) or the frame geometry is asymmetnc and the frame 1\ nm t-raced. lateral tran latmn occurs and .'111, 0. To detcrmtne .\41, (a) appl) ficutiou hortLontal reaction· at each floor lcvcl to prevent lateral tr an\la t ion and (b) u c the rcver e ot the'e rcact101:' a "sway force!\.. to obta111 A11,. This prot·cdure is illu strated in Fig. X-2. A b indicated there. M,, for an dnbraced frame i the sum of the moment due to the lateral loads and the ..,way force\ ."

*

p

p

l

R

I I

I

p

I.

I I

r

--

R I

R

i

I'. f R .

I

p

R ,

I

-...

I •

-

I +

R.

- ... - ... nt'"'".t l rJm\! fnr

It"

Fig . 8-l f ram;.: models for M.,, and A·/ 1,

I "'a l ramc lor .If

--

Once M111 and /1./ 1, have been obtained. the) are multiplied b) their rc pectivc magn1ticauon factor\. 8 1 and 8 '· and ot the AISC LRFD Manual (p. :!-1\1) If 1.; = 36 k i and KL = 15 fl. 1/J, P.. = 1430 krps ( >P..'" = 14011 krp') for a W 14x 176. Try a WI4X 176 To determine M,., (the econd-order momenl). u-.e Eq. (H 1-2), which. for a braced frame (M,, = 0). hccomcs M,. = 8 1M,,. or AI... = 81 x 200 kip-ft. "' 111 Proh. !U. C,., = 1.0 for equa l end moment' in ingl c cunaturc: bendtng (i.l!.. t!l1d rotatinn' rn oppo itc direction ) . 4 Determining P, for y-axi bend ing of a WI4 X 176. (1, = 838 in )

;rz £1 ,,.z X :!9.00() k i/in" p = ''

In Gq. (II 13)

(Kl)

X

X3X 111 1

'

(1.0 X 15ft X 1:! tn/fl) ·

- 7403 kip>

c..

1.0 B1 =1.1:! - l - P)P I - XtJO !..rp/7403 ktp'

The \t:cond-order requrred flexural strength J/,., = I 12 x 200 l..rp-11 = 22-1 l..tp-tt. Sub,tlluttng M., = :!24 k1p-ft 111 preliminar) de,tgn Eq ( .11 P.cr = XOU + II + :!24 X 2 .II X I 5 = 1472 kip'>

> 1430 k1p' = l/1. Pn

for

WI4X 1

< 157Ukip, = l/JP.,

for

Wl4xl9

(Sec p. 2-19 of the AISC LRFD \lanu al.)

9R

BEAM-COLUMNS: COMBINED FLEXURE AND COMPRESSION

[CHAP. 8

Tn a Wl4x 193: /, = 931 in' P lnEq .(lll3)

=,T'£[

•-

(KJ)·

8 -

.T'X29,{)()()kip /in'x931 in'_.,.,., k . • -n -4 tps ( 1.0 X 15ft X !2tn/ft)·

C,..

-

= I II

I.O

( J -P..l P,.) l- 800 kip/8224 ktJh .

'Iht• \ccund-nrdcr required fle-..;ural strength M ,., = I II x 200 kip-ft = 222 kip-ft. Sclccttng the appropnatc beam-column interaction formula, (f/1-lu) or ( II l-Ib)

P,,

_8 tl ?

U\t: furmula (II 1-la), which. forM..,= 0. reduce to P,, 8 M,.,

---''-+< 1.0 cp, P,, 9 ljJpM.,,

To dctcrmtnc rp,,M... (the dc;.ign flexural strength), refer to Chap. 5 of M... = M,. = /, (f(lr nHnor-aXI\ bending) regardless of the unbraccd length For a Wl4X JC)3, L, = 180 in ' 180in' x 36 kip /in ,..f... = 12 tn/ft 540 ktp-ft

r,

thi text. From Eq. [5.6), L,,.

and ljJ •.•\1., =II YO x 540 ktp-ft = 486 kip-ft. Sub,tttuttn tn the tnteraction formula. "e obtain ')')? k' f 8 -0. :I+·tp - I ' 9 486 ktp-ft

-

=

8.3.

().5)

T

0,4( = () 92 < (.0

0, J..,

Select a Wl4 cellon (A36 :.teel) for a beam-column 111 a braced frame with the factored load-.: P,, = 200 kips: fim-order moments M, = 200 kip-ft. M, -200 kip-ft. The 15-ft-long beam-column is ubjccted to transverse loads: its ends arc ..pinned ... f-or a braced frame. K = 1.0.

K, L, = K, L, = l.0

X

15 ft - 15 ft

Select a trwl Wl4 hape using Eq. (8.2]:

P...dt = P,, + M,.,m + M,.,m U

Fnr a Wl4 with f.../.= 15ft: 111 = 2.0 and U = 1.5. in Table 8-1. Sub\tituting in Eq. [8.2). we obtain P.. do = 200+ 200

X

1.0+ 200 X 2.0

X

1.5 = 1200

ktps In the Column Load Tables of rhe AISC LRFD M anual (p. 2-19). if 1-; = 36 k i and KL =15f t. cp, P,, = 1280 ktp ( -'"P., "' = 1200 kips) for a Wl4X 15Y. Trv a Wl4 x 15C) To delermme the required .:cond-order moments, M.,, and M,.,, use Eq . (H1-2). whtch for .1 braced frame (.'v!,, = 0). reduce to

Accordmg to Eq ( H 1-3) C,".

8,.

=

/P

1- P , ,.

8 .. -

c.

I-P./P.,

2:

1.0

">

1.0

CHAP. 8]

BEAM-COLUMNS: COMBINED FLEXURE AND COMPRESSION

99

For a beam-column subjected to transver e l oad unrestrained against rotation, C,., = 1.0. Therefore. C,.,. = C,.., = I.0. For a Wl4x 159, I, = 1900 in" Jr:x

29.000 kips/ in:x I/in X 748 in"=

,., (KJY

(

660

1.0 x 15ft x 12 in /ftf

!\ ki s p

In Eq. (H 1-3) 1.0

B,. =

. = l .03

I- 200 kips/660!\ kips

M,., = 1.03 X 200 kip-ft = 206 kip-ft The second-order required flexural strengths arc M,., = 202 kip-ft and M,.,. = 206 kip-ft. (Substituting these values in preliminary de ign Eq. [8.2] reconfirms a WJ4Xl59 as the trial section.) Selecting the appropriate beam-column interaction formula. ( H 1-Ja) or (H l-Ib), we obtain

P,,

---' =

200 kips .

P.... =683 kips} for a Wl2 x 6. Selectmg the appropriate beam-column interaction formula. (Hl-In} or (H 1-/h). we obtain

P,, = 400 ktpS = 0.55 > 0.2

rp P. 732 ktps

Uo;c formula ( 111-/a). which. for J-1., = U, reduces to _,P.:.... 8 +-

cp, P,,

.w.,

Cl s I

9 rp,M.,

The design flexural strength rp,.M , for a WI2X96 can be determined from the Beam Selection Table on page 3-15 of the AISC LRFD Manual : because (L, =12ft) < (1.1,- 12.9 ft). rp,M,.. = rp"M,. = 397 kip-fl. as tabulated . Substituting in the interaction formula :

s

118 kip-ft

0.55 +-X < 1.0 9 31.}7 kip-ft 0.55 + 0.26 = 0.81 < 1 .0

O.k .

By a simihtr &o luti on of interaction formula (H l-Ja). it can be sh own that a WI 2X87 a nd a Wl 2x79 are abo adequate.

8.5. 0.

Assume th e beam-co lumn in Prob. 8.4 is turned 90°: t.C .• M, = 0. M1,.. = 100 kip-ft. M,,.. = Select the appropriate W 12 section. Gtven : P,, = 400 ktp . M.., = 0. M,,, = 0. M,, = 100 kip-ft. E P,, = 4li00 ktP'· 6,.,,/ L = 0.0025. H= 80 kip,, 1\L = 1.2 X 12ft = 14.Ht. rrom Eq ( H 1-2). .\1"' = B M 0.2

U;c formula (/:/1-/a). which. lor M,,, = 0, reduce.. to

P,, + M ,., 1/1 P,, I) 1/1,, \f

ror

1.0

To detcrmmc rp,..w... (thc t.Jc.,ignflexural -.trcnj!th). rder to Chap. 5 nt tht text. From f:q (5.6[. •W.,, \lr = Z F (for mmur- P,.,.

I'....

102

BEAM-COLUMNS: COMBINED FLEXURE AND COMPRESSION

From the statement of the problem, SubMituting. we obtain

E 1. = 3000 kips. E P,, I

8=

I -J(KlO klp'>/60.000 kips

60.000 kips. and

[CHAP. 8

l: P, =- 30.0!Xl kip .

= 1 .05

I B., = = \.II l -3000 kip/30.000 kips The second-order required flexural strengths arc ,\fu, = 1.05 X I()() kip-ft = 105 kip-ft

J1.., = I.II X 100 kip-ft = lJ Ikipft Sdecung a trial \.\I:! '>hape \\llh Eq. (8.2J. we obtam

P....rr = P,, + .\1.,,111 + ttl/.,,mU. where for

a Wl2 (KL = 1.::! x 12ft= 14.4 ft), P..

111 = 2.4 and U = 1.5.

,"= 150+ 105X2.4+

Ill X2.-1-X U=RO::!kips

B:v

mterpolation in the Column Load Table'> (on p. ::!-2-1- in the AISC LRFD Manual). if and KL=I4.4ft. it follow' that cp,P"=hll kip ( ..> P.,." = 802k•ps) fora Wl2X 106. Tr} a W 12 x 106. Select the appropriate beam-column interaction formula.

F.. = 36 ks1

P,, 150 kips ---''- = -;- = 0.18 < 0.2

tp, P., 811

kip:,

U\c mtcraction formula (H l -Ib):

The design flexural \trcngtho., for a Wl2 x 106 can be determined a follow . Becau c ( L o 12ft) < (L,, = 13.0 ft), q>,M., = rp,,M,, = 443 kip-ft. a tabulated in the Load Factor Design Selection Table for Beam' (on p. 3-15 of the A I SC LRFD Manu P. = II +I kap' ( >P,< , = 1115 kt p') for il V. I-IX 1-15 Tr) a WI-IX 145 firM. determtnc B,, and 8,, For re\er\c curvature bendtng \\llh equ; l end moment . ,\.1,/ M, +10. From Eq. (H/--4). C.. 0.6-ll.-I(M,/M )=0.6-0.-1(+1.0) 0.2. C.... = = 0.2. In the equallon tor 8, (HJ -3). r.. i ha...ed 011 Kl in the plane of benc.hng With K 1.0.

em, (By

contra !. P. in Eq. (///-(!} for B . i based on the actua l Kl ol each column in it\ plane of hending). Referring toP, in Eq. (111-3). for a WI4X 1-15 I,= 1 7111 in'

.T' x 29,000 ka p,fin' x

f' = .T'f l,

"

. I •

12t n ft) '

I = 6T' tn'

_ :r 0 x :!'),CKKl ktp'o/111 X o77 in'

P = ' (K,I.)' 8,. =

--

.-= 6865 kip

(J.OxJHtxl::!tn/ft)·

0.2 I-300 kip /17.3-11 kap'

1.0

= 1.0 0.2

B .. = ?: 1 .0 I - JO() ktp/6H65 ktp'

= 1.0

. = 17.3-11 k 'P'

(I.Ox 1-Ift x

). torsion will occur when th e line of action of a lateral load does not pass through the shear cemer. The emph asis in thi!> chapter is on torsion of located at the centroid.

106

CHAP. 91

TORSION

107

Smgly 'l}mme tnc cross secuons, such as C shape:-. have thctr -.hear center-. on the axi of S)mmetr). but not at the centroid. (The shear center locattom. fur C -.ccuon'> arc gwen in the Properue-. Tables in Pan I of the A ISC LRFD Manual.)

•

• Fig. 9-1 Shear cent er locaunn'

As shown in Fig. Y-2, the torsion al moment , Tor 1. eq ual s th e m agnitude of the force multiplied by its distance from the sh ear cent er e.

A VOIDI NG OR MINIMIZI NG TORSION * As i s demonstrated lat er in this chapter, open sections. uch a!. W and C shapes, arc very in efficient in rcsi ting torsion; thus, torsional rotati on can be large and torsion a l stre!. es relatively high . It i!> best to avoid torsion by detailin g the l oads and reactions t o act through the shear cen ter of th e 1 tember. In the ea e ol spa ndrel members supporting bu ilding facade cll.!ments. this may not be possibl e. J I eavy exterior masonrv walls and stone panels can unposc !'Ievere t or,iona l !oads on spa ndrel bl.!am!.. The following are suggestion!> for diminatmg or reducing tht-. 1-.ind of torston. I.

Wall element" may '>pan between floors. The moment due to the eccentricity of the wall with respect to the edge beams can be resisted b> lateral Ioree-. acting through the floor diaphragms. No torsion would be imposed on the spandrel beams

2.

If facade panels ext end only a partial stor hetght bclo" the floor ltnc. the usc of diagonal teel "kickers" rna} be possible. These light member'> ''ould prondc lateral -.upport to the wall panel'>. TorsiOn from the panels would be re t ted by force' ortgtnatmg from 'itrurtural clement'> other than the pandrel beams .

• Thl) 'C(IJOn " rcpnmctl \\-Jth p.:rml\>tlln rrum Ihe aulhu(, carhcr wnrl. (iwdt• to / .oud '""' Rc\1\lti//CI' uj Stmourul Stl'rl Bwldm > . Amencan lnsmute ol Steel Conlrucuon (AI ( 1. ( htCa!!o. 1%11.

ft1CItances of the beams can be mobilized to provide the required torsional reactions along the girder.

4.

Closed sections provide considerably better resistance to torsion than do open sections; tor'iional rotations and stresses arc much lower for box beams than for wide-flange members. Fur mcmhcr

160 =

2.T

rad•an,, or 180' =

x 0.00065 radwn = O.o:IH

9.2. For the cross sect ion in Fig. 1:1-4(c). The tor 1onal bchav1m of a hollow circular ;haft ",imilar to tha t of a round bar: St. Yenant torsion w11h no warpmg: no normal se\.

112

TORSION

[CHAP. 9

From Table 9-1. the tor-tonal shear •me,-.e, where

•

[(5 m)'- (4 in)') = 580 in'

1=

-

The maximum hear stre 'c' arc at the out.:r cdg.:, where r = R.. = 5 in. . Tf(, ( 10 kip-ft X 12 in/ft) X 5 in 1··\1 = J-= s_ ono I·ll '

= I IJ..l ksi The angle ot rotation. in rathan

T/

( 10 kip-ft X 12 in / ft)(5 ft X 12 in/ft) 11.:!00 kip/in ' x SHO in'

II= GJ =

= 0.0011 radian !n degree 180' fl = . x 0.(l011 radian\= 0.064° ,T rad1an'

9.3.

For the cross section in rig. 9-4(d). When a \quare tube i twi\tcd. warping i' m111or: the normal and hear stre sce> St. Venant

f.

f,r

Shear \tresses

'liormal \lreS\C\ "arpmgIOf\ton (flanges onl>)

warping torstnn !Oange\ onI> l

tor"on (flanges and web)

Fig. 9-6

As is hown in Fig. 9-6, the maximum shear stre " m the llangev j,.. -f,.,r =

14.9 ksi. The maxtmum normal stress is also in the flang : :j,.,. The maximum angl e of rotation. from Eq. W 7] . is

=f.,,

+ f,... 7 = ( 12.9 + 2.0) ksi

= 4!Ul ht.

n

8= G.l where J is as determined above: J = 9.33 in'. In radians

8

( lO kip-ft x 12 in/ft)(''ft x 12 in/h) = 11.200 kip/ in! x 9 33 111• = 0.069 radtan

In degrees 180"

x 0.1169 radian'= 3 9• radtan' A comparhon of the 'elutions to Probs. 9.1 to 9.4 indtcate' that apr•n H'CIIons are poor 111 torsional resistance . The stresses and rotations of the I haJJ< w (beam assumed)

=

100 lb/ft

w (total)

=

!000 lb/ft

w = 1000 lb/ft = 1.0 kips/ft

As a result of an eccentricity of 9 in for the wall panel, the torsional moment lb

= I we =

900- X ft

9 in

lb-ft

= 675 12 in/ft fl

= 0.675 kip-ft/ft

For the case of dead load only [i.e.. Formula (A4-/) in Chap. 2). the factored loads are w,. = L4w = 1.4 x 1.0 kips/ft = 1.4 kips/ ft 1., = 1.41 = 1.4 X 0.675 kip-ft /ft = 0 945 kip-ft / ft

A W shape with side plates is a box section with negligible warping torsion . The problem of designing this beam for flexure combined with tor ion can be divided into the following components. Flexure is resisted by the W shape: the flexural normal stresses by the flanges: and the flexural shear . by the web . The torsion (which in this case involves mainly St. Yenant shear stresses) is resisted by a ..box " consisting of the flanges and side plates. The shear. moment, and torsion diagrams for this case !corresponding to Fig. 9-7(d)] appear in Fig. 9- ll. The location of the maximum moment (and, hence. the maximum flexural normal tresses) is at midspan. Flexural and torsional shear are maximum at the end supports. Design for flexure. M

wf

l.4 kips/ ft X (20 ft)z

"8

8

= -=

=

. f 70 ktp- t

Because a box shape will not experience lateral torsional buckling. rp,M,. = The ma,imum torsional moment (as ...hown 10 Fie. . 9-11) ... i-1p-tt 7;, = - -=-:;- = 0.945 It I.:> 0 mx-.Xlmxo._ _ m

For the tlange' (where 1 = 0.30) [., ,r

_ -

., 9.45 k1p-ft x 12 in/ft _.,

-X 111

5.50 X Y.!ll

X

IJ.:l()

.· 9

-.

ks1

In both ca es, the tor\Ulll"l 'hear stres;e, arc within the limits 'et hy Fonnula f9.J]: /.•.
...cs. (Both tyJX of normal \trc.,.,c, are maximum in the tlangc., at mu.hpan of the member )

(2) Shear ...tre-....e... due to ' arpmg torsion are \upcnmpo,ed on the shear strcsc' from St. Venant torsion (Both are ma\tmum at the upporl\: the fnrmer occur only in the f1angcs. ''h1le the latter occur m the web a' "ell.) (3) Also. the web 'hear \tre\St:\ from St. Venant tnr\lon :u.ld to the flexural shear strCS\t:\ (Both are

max1mum at the \Upport>: the later are primarily m the web.) Warping Torsion Figure 9-U, which corrc ponds to Fig. 9-3, approximate\ th : effects of warping torsion on the W l4x99 beam. The umfmm ly d1 tributed factored torqut: 1., - 0.945 k1p-ft/ft (obtained in the solution t o Prob. 9.9) is resolved 11110 equal-and-opposite uniform load

f

.,

=::::;-;::::::J•...,.._

-

.

·'

d''

=

I

. . .... 1 1

= ::::::: -.. . l.. . . : c.

WI-I

X

99

20ft

kip

....-

'

- 0.945 kip = 0 84 1-.Jp

w•'L/2 "'

fi

0 . 84

. '

1.12ft "" = ( i ·o"." =

0

fl

X

2

= 8.4 kip

.

lt ip'•t

lp fi

...;u

'·.. 0.945 h

8

!tips 20 fi - 0.84 T X -8= 42 kip-ft

M.. 'L Fig . 9-L3

From Fig. Y-13, the eqmvalt::nt uniform lateral flange load 0.84 ktp/ft and the resu lting maximum hear and moment are V = 8.4 kips and M = 42 kip-ft. re pectively . The max1mum warping normal stress is

. M /.r = S'

s·= ,,

where

(section modulu of flange)

' b1

6

= 0.78in

x (14.565m)·

6

= ._7 .,

.6m

,

_ 42 kip-ft x 12 m/ ft_ ., k

f,,,-

tn.

27.6

,

- IlL•

The max1mum warping shear stress 1s

1.s v (for shear on a rectangular ,,, 1.5X8.4 kip .

[.,w 1 = b -

St. Vcnant Torsion

.

14.5

65

X

hapc)

0 8 . = I.Ihi .7 tn

A given in Table 9-1 for open ections, the shear stresses due to St. Venant torsion aref,,,r= Tt./1. For a Wl4X99, J = 5.37 inJ. as tabulated under Torsiona l Properue!> 111 Part I of the ALSC LRFD Manual. For a Wl4x9Y. t1 =0.78tn and t,. =OA8S1n. Max1mum 1;, -Y.45k1p-ft. a> determined in Prob. \1.9. In the flange' 1).45 k1p-ft x 0.78 in x 12 in/ ft [,o•.\T = 5.37 in' = 16 5 k l In the ''eb

Flexu re

I

n=

9.45 kip-ft x m/ ft 0.485 in x 12

,

., k =10.- \1

5.37 in From Prob . 9.9, a a re ult of flexure: maximum M., = 70 k1p·ft and ma•omum V,, = 14 kips. The corrc!>pondmg flexural normal and shear !>tresses are M..J S, and V..l A •.

120

TORSION

[CHAP. 9

From Part 1 of the AISC LRFD \ilanual. the requ1red properties of a W14X99 are

S, = 157 in' A.= dt. = 14 16 Jn x 0.4S5 Jn- 6.1'7 in 1 Then .\11,., 70 k1p-ft X 12 Jn/ft . -= ' = :>.4 k · S, 157in V,, 14 kips -. '= 2.0 k'l

A.

ti.87 m·

Combining Slresses ( I) Normal tresses (maximum in the flange at mid pan) . From Eq. 1\).2] M,.

t\1m

±-s if.,,

f,"'=-±-s A

=o

+

' ' 5.4 ksi + 0 + llU k i 23.7 k i

F, = 19.4 ks1

above)

(a!>

(/,,. = 12.2 ksi) < (0.61/>F. = 19.4 k\i)

o.k.

Regarding the hmit tate of buck ling. because a W 14xlJ\) i compact in A36 \tee I. local buckling need not be considered. However . if the 20-ft member i' not laterally braced. lateral-torsional buck ling should be checked using Formula l9.5]: J,,., < 1/>, F... where 1/>. = 0.85. To tlctcrtmnc F... usc th e ratios r

cp,.M,

F,

rj>,,M,.

-=

For a !>i mply supported member (C, = 1.0). the beam graph' (in Part 3 of the A ISC LRFD Manual) indict tc for a W14 X99 ,M,. = 467 kip-ft

(L,, = 20 h) (L1•

From thi' ratio.

11

•

15. ft)

1.,1

follow that

F.. 36 k!.i

450 k•p-ft 467 kip-ft

F , = 34 7 b• Formula [9.5] becomes f.,. hM, =design flexural strength. kip-in or kip-ft

¢11 = resistance factor for bending ¢0 P,, =design strength of compression member, kips

¢..P,.. =the portion of the design compressive strength of a composite column resisted by the concrete, kips cf>c =resistance factor for axial compression= 0.85

INTRODUCTION Composite members consist of rolled or bui lt-up structural steel shapes and concrete. Examples of composite members shown in Fig. 10-1 (p. 125) include (a) concrete-encased steel columns , (b) concrete-filled steel columns. (c) concrete-encased steel beams. and (d) steel beams interactive with and supporting concrete slabs. ln con trast with classical structural steel design. which considers only the strength of the steel, composite design assumes that the steel and concrete work together in resisting loads. The inclusion of the contribution of the concrete results in more economical designs. as the required quantity of steel can be reduced. The provisions for the design of composite columns. beams. and beam-columns discussed in this chapter are from Chap. I of the AlSC LRFD Specification. Design aids are provided in Part 4 of the AISC LRFD Manual.

COLUMNS EMBERS

ANO

OTHER

COMPR ESSION

M

The design of composite compression members is similar to that of noncomposite columns . The equations for composite design (Eqs. [IO.2] to [10.6], below) are the same as Eq. (£2-L) to (£2-4) in Chap. 4 , with the following exceptions: in the design of the structural steel ection in a composite member, a modified yield stress F,"Y and a modified modulus of elasticity Em are used to account for the contributions of the concrete and the longitudinal reinforcing bars. (12-1)

(12-2)

where

[10. Ij

124

COMPOSITE MF"v1BERS

and

ICHAP. 10

F;." = moth fled ) 1eltl tre

for the des1gn of compo'>lte columns. ksi F. = specilkd mm1mum ) ield stress of thc tructural teel shape, ksi t;,=specified minimum yi :ld stress of the l ongitutlinal reinforcing bars. ksi

r; = speci flcd comprc-.sivc

trcngth llf the concrete. k i £, = modifktl modulus of elasticity for thc de ign of composite columns, k i 1:. = modulu\ of eht,licity of steel= :!Y.OOO 1-.\1 £. 11

motlulu\ of cla\ticitv. of concrcte. k 1 =unit WCII!ht of concrete. lblftJ =

A, = cross--;ect10nal area of concrete. in

1

A,= cro"·sectional arca of longitudmal remforcmg bar.... in

1

A,= cross-sectional area of structura l steel , in 1 For concrete-filled pipe and tubing: c 1 = 1.0, c 1 = o.gs, and c3 = 0.4. For concrete-encased shapes c 1 = 0.7, c1 = 0.6, and c,- 0.2. Utilizi ng F,,, and £,a'> defined above . the design strength of axially loaded composite columns is ¢, P,,. where ¢, = O.R5 and P,, = •\ f.,

[/0.2)

F,, = (0.65W)f;,"

(10.3]

lf i.. s 1.5 (mdast1c column

buckling)

r;., = (exp( -0.4 1

or

(10.-1]

9.A.;)]f,,,

where exp(x) = e'. 1f },, > 1.5 (clastic colum n buckling)

F;.,. = (

0.'8.77)

[ 10.5]

F,mA •

.

where

I.,

rm:r

and

Kt

,..--

r;,,

\

,

[10. 6]

£

-

A,= cros'>·sectional area of structural steel. in K =effective length factor, discus\t:d in Chap. 4

I= unbracecl l ength of the member, in r,., =rad ius of gyration of the steel shape. but not lcs" than 0.3 times th e ovent ll thi ckn t: s of the compo\ ite crpHclng of lateral ties - CCII(1na l arc;1 of all reinforcement (latera l and longitudinnl) shal l he 0.007 in 2 per inch of bar racing. A clear concrete cover of at l east 1.5 in must be provided outside all reinfon:cment at the perimeter.

(c) (d)

Minimum design;: is 3 ksi for normnl-weight concrete. and 4 ksi for lightweight concrete . Maximum design f' is H ksi. For hoth \tructural Jnd reinforcing 'tee!. des1gn f, .-55 ksi.

(e)

125

COMPOSITE MEMBERS

CHAP. 10]

The wall thicknesses of structural steel members fill ed with concrete

rF b \ : /3£

t

t

for each face of width b in rectangular tubes,

f3 D \ _

for pipes of outside diameter D

/8£

(f)

If a compo ite cro s ection includes two or more steel shapes, the shapes shall be connected with batten plates. tie plate . or lacmg to prevent buckling of each shape before hardening of the concrete.

(g)

The part of the de!)ign compressive strength developed by direct bearing at connections.

re isted by the concrete

, P,"

must be

[10. 7J where ections tabulated therein, or the above equations for all cross sections.

BEAMS AND OTHER HEX URAL ME IBERS The most common case of a composite flexural member is a steel beam interacting with a concrete slab, as shown in Fig. 10-l(d). The slab can be either a solid reinforced concrete slab or a concrete slab on a corrugated metal deck. In either case, stud or channel shear connec tors are essential to ensure composit e action. (When designed in accordance with this secti on. a beam is composi t e regardless of the type of deck. A steel deck b de ignated as a composite deck when it contains embossments on its upper surfaces to bond it to the concrete slab; the beams supporting it may or may not be compo.,ite in this case.)

-

·. ' . ....... ....

)"

,. .

...,

I

,...

--

: " ,

•

(b)

(II)

.

. '

•

. ....

.

•

..

. .

r It in minimum . , • ,-.z .

'

•r' .

'..... 2 m nummum

-

..

-

I

.

• '. .' ..L

-

-·

2

-

lD mammum

co\er

(lyp•ca()

(()

Fig. 10-1

(d)

Examples of compo itc member

Three criteria determine the effective width of a concrete slab acting compositely with a steel beam. On either side of the beam centerline, the effective width of concrete slab cannot exceed (a) one-eighth of the beam pin, (b) one-half of the distance to the centerline of the adjacent beam, or (c) the distance to th e edge of the slab.

126

COMPOSITE MEMBERS

(CHAP. 10

The horizontal shear forces between the steel beam and concrete \lab. to be transferred by the shear connectors. are as follows. In regi on!. of positiue momem. between the p01nh of zero and ma:omum po itive moments (e.g.. between a support point and midspan on a untforml) loaded. '>impl) supported beam). the smallest of (l) 0.85[;A, (the maximum possible compres t\e force tn the concrete). (2) Aj F.. (the maximum possible tensile force in the steel). and (3) Q., (the capacity of the shear connectors). In region s of negative moment, between the points of zero and maximum negative moments (e.g., between th e free end and the support on a cantileve r beam). the smaller of (4) A,F.., (the maximum possible tensile force in the reinforcement) and (5) L: Q, (th e capacity of the shear connectors). When sufficient shear connectors are provided (in accordance with the section on shear connectors below) to allow condition 1. 2. or 4 above to govern. th ere i s full composite action. However, if the number of shear connectors i s red u ced a nd conditi on J or 5 governs, the result is partial composite action.

DESIGN FLEX UR A L STR ENG TH For positive moment. th e design flexural strength

rp,.M,

i' determined as follow .

If h,/t"..::;, 640/'vF, (i.e .. th e web of the steel beam t-. compact, which •.., true for all rolled w shape in A36 steel), cp,. = 0.85, and M, is calculated from the plaltic stre.H distribwion on the compo ite section. The as!.umptions are (a) a uniform compn:..."vc stress of 0.85[; and zero tensile strength in the concrete. (b) a uniform steel '>tres' of f ; tn the tension area and compres ion area (if an)) of the steel section. and (c) that the net tensile force in the steel section equals the comprel>sive force in the concrete '>lab If hJ t., > 640/ Vf. (i.e.. the web of the steel beam i not compact). cp,. = 0.90, and M,. is calculat ed from the elastic stress distribution, con,tdering the effects of shoring. The assumptions are (a) the strain'> in the steel and concrete are proportional to the distance from the neutral axis; (b) steel stress. tension or compression. equals \train lime\ £, but cannot exceed

F...;

(c) concrete compressive stress equals strain times£,, but cannot exceed 0.85[:: and (d) tensile strength is zero in the concrete. For negative moments the design flexural treng.th rp,M,. b determined by m ost engineers according to the provisions in Chap. 5. neglecting composite acti on. lloweve r . if the steel beam is compact and adequately braced (i.e .. L,, L" for C" = 1.0, or L 1, L,., for C,, > l.O) and the slab reinforcement i s properly developed, the negative nexur = O.H5, and M,. is calcu l ated from th e plasric srress disrrihwion on the composite section. The assumption are (a) a ten sile stress of F.., in all adequately developed longitudinal reinforcing bars within the effective width of the concrete slab (b) n o tensile concrete

strength

(c) a uniform stress of

in the

F. in the tension and compression area:- of the steel section and

(d) that the net compressive force in the steel section equals the total ten sile force in the reinforcement The i!.sue of shonng is tmportant in composite destgn If temporary shores are used during construction to help the steel beams support the newt) poured "wet " concrete. design is as outlined above, with the composite section resisting the total factored load. dead plu\ live. I f shoring is not anticipated. the bare steel beam must also be checked for adequaC) to support the wet concrete and other construction loads (properly factored) in accordanc..: with the requiremen ts of Chap. 5.

CHAP. 10)

COMPOSITE

MEMBERS

127

Because of beam strel.s redistribution under full plastification. the total factored load for unshored construction can still be assumed to act on the composite section, whenever design with a plastic stress distribution is allowed by the AISC LRFD Specification. However, if an elastic stress distribution is required, (I) the unshored loads applied prior to curing of the concrete (defined as attaining 75 percent of[;) must be taken by the steel beam alone. and (2) only the subsequent loads can be resisted by composite action. In the latter case. the total flexural stress at any point in the steel beam is a superposition of the two effects.

SHEAR CONNECTORS Acceptable as shear connectors are headed steel studs of minimum four stud diameters in length and rolled steel channels. The nominal strength of a single stud shear connector in a solid concrete slab is (15-l)

where A " is the cross-sectional area of the stud, in2 , and F,, is the minimum specified tensile strength of the stud, ksi. The nominal strength of a smgle channel shear connector in a solid concrete slab is

Q, = 0. 3(t1 + 0. St., )L, ...;r:£: where

(15-2)

t1 = flange thickness of the channel. in

tw =web thickness of the channel, in

Lc =length of the channel, in The number of shear connectors required between a section of maximum moment and the nearest section of zero moment is l-j, n=

-

Q,

[ /0. 8]

where Q" =the shear capacity of one connector [as determined from Eq. (15-1) or (/5-2)], kips and vh =the total horizontal shear force to be transferred. kips. As discussed above, in regions of positive moment, l-j, = the minimum of (0.85/;.Ac, A,F,, and E Qn), while in regions of negative moment. l-j, =the minimum of (A,f;.,and E Q,). Shear connectors may be uniformly distributed between the points of maximum and zero moment. However. when a concentrated load is present, enough connectOrs must be placed between the point of concentrated load and the point of zero moment to adequately develop the moment capacity required at the concentrated load. The following restrictions on the placement and spacing of shear connectors are imposed by the A ISC LRFD Specification: (a) Minimum l-in lateral concrete cover. except when in talled in a steel deck

(b) Diameter of Muds s2.5 limes the thickness of the flange to which they are welded, unless they are located over the web

(c) Minimum center-to-center spacing of studs, longitudinally along the supporting beam, six diameters in solid slabs and four diameters in decks; laterally. four diameters in all cases (d) Maximum center-to-center spacing of shear connectors of eight times the total slab thickness

SPECIAL PROVISION S FOR STEEL DECKS When a metal deck il> used. the diameter of the shear tuds must not exceed in. The studs may be welded to the steel beam either through the deck (which is the usual practice) or through holes

128

COMPOSITE MEMBERS

[CHAP. 10

punched in the deck . Additional restrictions affecting the studs and deck (from Sec. 13.5 of the AISC LRFD Specification) are shown in Fig. 10-2, which is reproduced (with permission) from the Commentary on the AlSC LRFD Specification.

.. .. . ..·. '.

-;==;;;=?:

..

H

b.

.

4

H 10 rrummum

I'll

. ..

'

..

,

.

·-- . .p. .

J

.. .

4

. 0 .. b .

.

..!.

. . • . .. t . .. .. . :: 1\

.

2 m' rrummum ¢

'A

.

..

.

..

to

I>

H

.. . . :

..

_,_

4

. '. .

..

•• .

A

w

h,

< 3 in

.

.

.

...

.

ll

. A

.

.

.

.,

.. .

'-

2 m m1mmum h. chapter. under the heading Columns and Other Compression Members. If '>hear connectors arc required for a beam (i.e.. when r.. = 0), they must be provided for th at member whenever (P,./ ¢, P,,) < 0.3.

DESIGN SHEAR STRENGTH The design o;hcar strength for composite beam ts taken a' the \hear !.trength of the steel web. a for noncomposite beam . The equations for shear itc 6-in cxtra trong ptpc was requtrcd for the same condition . Tn a fl·tn han·

;.;.;

-:: = -

.6 in + _x. '

o.t,.

28 9

in

•

131

COM POSITE MEMBERS

CHAP. 10]

In the absence of re10forc10g bar,. Eq . (/!-/)and(/:!-:!) become

Fm, = F

,A ,

+cd,A

Em= E + c,E

'VJ'..

A A

where E. = w' c.= 0.85, c,= OA. The modulu ot elastiCity of the concrete

E. = 145' '\13.5 = 3:!67 kM The modified yaeld

re -. for composi t e design b .

.

?_ 8.9

°1 n·

F.... = 36 ksa + 0.85 X 3.5 k5a _ .6.an · 5 x =51.4 ksi Th e modified modulus of elasticity for compo ite design '' E.,. - 19 •000 kS'l + ().4 , .. , · X. > - 67 k

·

l X

,., (1

0

•

-_o.7• IO-,' ).0111 '

= 35.744 hi

rhe radau of gyration of a hollov. circul 200 k1ps required

(rp, P. 205 k1p for tha case IS abo tabulated on p 4-100 of the AISC LRFD M an ual ) The 6-1n standard-\\Cight concrete-filled pipe-column ,., o,ati'>factor}

10.2. Determine the design compressive strength of a W8x 40 (A36 steel) encased in a 16 in x 16 in = 3.5 ksi) normal-weight concrete column in Fig . 10-4.

U:

Reinforcement is fou r No. 7 (Grade 60) bars longitudinally. and No. 3 ties at 10 in horizontally; K,L, = K ,L, = 15.0 ft.

132

COMPOSITE MEMBERS

(CHAP. 10

Io

.

T

an

.

..

·..c

......•

·.

J

. II> an

ll.!m

.

• •

.'

'

--

- -

...

Fig. 10-4 Checking minimum requirements (a)

For a W8 x40. A ,= II.7 in• wtal area= 16 in x 1 6 i., = 256 in 1 1

ll.7in " in = 4.6C:'r > 4"( 256 2 m1mmum (b)

O.k.

Lateral tie 'pacing= 10111 0.l)()7 in· x 10111 \pacmg = 0.07 in·

o.k .

Vertical No. 7 bars· A , = 0.60 in per bar >0.007 in: X II A in -.pacmg = ll.OH 111·

3.0 h1 < (f;- 3.5 hi) < 8.0 ksi for normal weight concrete (d U e F,, =55 ksi for reinforcement in calculation . even though actual

(c)

)

o.k .

O.k. F..= 60 ksi for Grade 60 bars .

Determme F,,., and £,..: F,m

A,

, A,

=F..+ c,F.,,- + c f,-

;\,

;\,

where A, = the cross-sectional a rea of four No. 7 longit udinal bar -4

x 0.6 in'= 2.4 in

A ,= cross-\ecti ona l area of WRx4o = 11.7 in: A, = 16 in

x 16 in -

(II.7 in=

+ 2.4 in:) = 242 in 1

For concrete -encased shapes. c, = 0. 7 and cc = O.o. 2.4 in= f,", = 36 ksi + 0.7 X 55 k'>i X---+ " 11.7 m =

"

(1.u X

•k . >.:>

242 in= \1

X

11.7 in·

,

87.3 ksi A,

Em=E+c,£,

A,

where

c, = 0.::! for concrete-encased shapes £, =

h '

'Vf = 145' '\':ts = 3167 ksi for J 5-k'>i nnrmal-wc1ght

!:.,.. = 29,000 k i + '1.2 x 3267 ksi x 242 in /J1.7m: = 42.513 I..\I

( 1451b/ ft ') concrete

COMPOSITE M EMBERS

CHAP. 10)

133

The mod1fied radiu s of gyration

r,.. = r,(W8x40) > 0.3 X 16 in (overall dimen\lon) =

2.04 in > 4.80 10

= .80 in

The 'lendcrne>> paramet.:r

15.0 ft x 12 in/ ft = 4.80 in X n

The critical stress

87.3 k i 1----.=0.54 2.513 k s1

F" = (0.658"i)F,.,, 1

= 0.6581" in: = 29.000 ksi + 0.2 x 3904 ksi x - . , 11.710" =

45.150 ksi

r.., = 4.80 in as in Prob. 10.2. )., =

E_

j F...,

r,.tc '£'"

= 15.0 ft

X

.

12 in / ft

=

4.80 in x ;r

105.9 ks1 45.150 ksi

0 58 ..

134

(CHAP. 10

COMPOSITE MEMBERS

F;, = {0.0658'·:)Fm,

= 0.658111 "" X 105.9 k i = 9::!.1 K\i rp, P.= rp,A F; . = 0.85 X 11.7 in X 92.1

b1

=916kips (a also tabulated on p. -85 of the AfSC LRFD Manual) . 10.4.

Assume all the column load in Prob. l0.3 enters the composite column at one leveL Determine A 8 , the required bearing area of concrete . In other word . the part of the design compres ive trength re isted by the concrete equals the total design compresive strength of the composite column minus tho: port i on rc i!.tcd by the steel. In this case. 4>, P,, = 916 kips and 1/>, P,. = 238 kips . , P... = 916 kips

- 238 kips = 678 kips

According to formu l a [ JIJ.

71

cp, P•. < 1.7rf>8[ ;A, A

or

> B-

cp,P," =

1.74 f ' >o ,

678 kip

1.7 X 0.00

= Ir

..>In

.

0

X 5 K'>l

The required concrete-bearing area of 133 inz can be satisfied by applymg the load to a 12 in x 12 in bearing plate placed on the column . For Prob!>. I 0.5 to 10.9, determine (a) (b) (c)

The effective width of concrete slab for composite action Vh (the total horizontal shear force to be transferred) for full composite act ion The number of -in-diameter shear studs required tf F,, = 60 ksi

10.5. A Wl8x40 interior beam is shown in Fig. 10-5. Steel is A36. beam span is 30ft 0 in. and beam spaci ng is 10ft 0 in. The beams are to act composi tely with a 5-in normal-weight concrete slab; = 5. 0 ksi.

r:

= ==:;:::========:::::;;;;;;;::::::========:::;;;;;;;;:::::===* 1rs

1 -

.n

}mx40 s=IOftOin

s =IOfiOm

I ...

Fig. 10-5

(a)

For an intenor beam. the effective slab\\ idth on c1thcr •de of the be am centerlme i the minimum of L 30.0 ft - = ft = 4S- In 8 8 = 3.7S

s

10.0 ft

=

=S.OOft

2

2

The effective slab width is 2 x 5 in= 90 in .

COMPOSITE

CHAP. 10]

(b)

135

MEMBERS

In positive moment regiOns, Vh for full compo ile action i the smaller of 0.85[:,.4, = 0.85 X 5 k i

X

(90 in

X

5 in)

= 1913 kip' A,

F, = ll.llm 'x 36 k 1 = -1::!5 kips

vh = 4::!5 (c)

iP'

The nominal 'trength of a ingle hear stud [from Fq (15-I)J i>

Q.= 0.5.A \ j' £, .A F,, For a -m-diameter tud . A , = ;r ( 0.75 in)'= 0.-1-l in .

£, =

w

1

''1/j;. = 1451 'V.'\ .0 = 3YIJI ksi

r,. = 60 hi Q.= U.S X U.-1-1 m 'Y5 () k'>i

X

J\)0-1 ksi S 0.44 in , X 60 '>i

= 30.t) k1ps < 26-1 k1ps = 26.-1 kip per 'tud

The number of hear connectors bet,,een the: pomt of Lcro and maximum momc:nh 1' V,,

-125 k1p'

II=-=

Q. 26.-1 p/stud

= 16.1 or 17 stud For the beam shown in hg. 10-6. the requ1rcd numh.:r of shear tuds is 2n = 2 x 17 = 34. (

I

-

11 sruds

J o4

n )

df 2

in minimum

Shear stud d1ameter = 0.75 in maxi mum !Ieigh t of hear tud H, 2: (h, + l.5 in)= (3.0 + 1.5) in= 4.5 in Use 4\-in-long -in-diameter shear studs.

(a) The effective slab width b 90in. as in Prob. 10.5. (b) Because the deck ribs arc perpendicular to the steel beam. the concrete below the top of the steel deck i ignored in calculating A< and other section propertie . In regiOn of posuive moment. Vh for full compo 11e action is the smaller of

0.85/;A = 0.85 x 5 ksi x (90 in x 2 in) = 765 k1ps A

F.. = 11.8 in x 36 ks1 = 425 k1p

v.= 425 kip (c)

For a sohd slab, the norrunal strength of a single shear stud (as determmed m Prob. 10.5) is

Q. = 0.5A .. Vf:E.< A ,. F,. = 30.7 kips s 26.4 k1p

(15-1)

When the deck ribs are perpendicular to the steel beam. the m1ddle term of Expre s1on (/5-J) is

138

COMPOSI TE \.1EMBCRS

!CHAP. 10

multiplied h} the reduction factor

O.H5(1'')("'- 1 .0) :-" VN, II,

1.0

(13-1)

h,

From the '-Oiution lO th1s problem. w, = 6 in. II, =."\in. H, = 4.5 in. Assume the number of stud connector' in one nh at a beam inteN:ction N, 2. Tht: reduction factor in expre sion (IJ-/) is in ( 4 .5 m

0.X5

6

\ -

.> Ill

X:;-

)

- 1.0 = 0.60

,.

·' Ill

Then

Q.

= 307

ktp

X

0.60 26.-1 kips

= 18.5 kip\ -c at hand (a 1Xx40 beam and a deck with a 4 75·111 nb '"dth at the bottom).

10.9. Repeat Proh . I 0.6 with the following modification: The 4-in lightweight con cret e slab (shown in Fig. I().7) con sist\ of 2 in of concrete on a 2-i n steel deck, with ribs spanning perpendicular to the W24X6H tccl beam. See Fig. 10-9.

,

..

7m

...J

I h 0 tn

-

pocah

'

' '

\

..!..

I

-4 -.t•

5tn '

'

:!m

:!m T

7

z w x 1>8

Fig. 10-9

Verifytng compliance wuh the pecial prO\i'>IOn\ for \tcel deck\ (Fig. 10-2)· Nominal declo. rio he1ght II,= 2 in < 3 in ma\imum Slab thtckncs\ aoovc steel deck= 2 111 man unum Average width of concrete rib II',= (5 + 7) in/2 6 i n ,2 in minimum Sh ear > tud diameto:r = 0.75 in maximum H eight of shear ·1d II

(!J, + 1.5 in) = (2.0 + 1.5) 111 = 1.5 in

Use 3!-in long :-itH.hameter 'hear studs

COMPOSITE

CHAP. 101

139

MEMBERS

(a)

The effective slab ''llllh "-12m. as in Prob 10.6.

(b)

Becau e the deck nb arc perpendicular 10 the \tecl beam . the concrete below the top of the tccl deck i ignored m calculaung \ . In regions of po lll\'l! moment. Vh for full composite action • the mallcr of

o.xs[;A, = 0.85 x 3.S ksi x (-12 in x 2 in)= 250 kip A,F. = 20. I IOZ X 36 k i = 72-1 kip

\ .= 250 kip\ (c)

Fur a \Oiid slab. the nommal 'trength of a -.ingle 'hear Mud (as dctermmcd

10

Prob. 10.6) ''

Q,- 0.'iA., ...rr;£; problem. w, = 6 in, h, The n.:duction factor in cxpres ion (13-1) i'>

1.0

2 in. H, = 3.:'i in .

-U.!l5' {6in ){ l'iJn , \

\, -'"

(/l-1)

)

1.0 s 1.0

- 10

Regardlc 1s n = V.IQ.= 250 k1ps/ IIJ .I k1p' per 'tud = 13.1 or 1-1 "ud-.. As md1cated b) Fig. 10-6. the required mimmum number ot \hear \tuds 15 2n = 2 x I-I = 2h. Becau e the deck ribs are paced at I ft 0 m center-to-center, as ho,,n 10 F1g . 10-\1. there are 32 ribs for the 32-ft beam pan. It IS ad' isable to place one \hear stud per rib, for a total of 32 studs.

10.10. Determine the design Hcxural trength of th .; Wl8x40 beam in Prob. 10.5 with full composite action . Assume the beam is hored during con truction. Becau'c the beam i5 hored , the entire load act\ on a compo ite member From the Properue Tablc:s for W Shapes in Part 1 of the t\ISC LRFD tanu,tl. for a WJ8X-IO

(

) th:,: = 51.0 ) < (16/4F0M( . = '-./36 = 106.7

The dc ign flexural strength 1\ cf! o M . where rfi o = O.ll5 and Mn IS calculated from the plaMic 'tre di-.tribuuon on the compos1tc 'ccuon From the solution to Prob . 10.5: the ma\lmum po,sible comprc"l\1! force in the concrete slab (' = cu sr:A = 1913 kip\ , the maximum passable t en,ilc force in the steel beam T = A.F. = 425 kip . To sau fy cqu1hbrium. It 1 ncce,sar that C = 7 = 425 k•P'· The pla tlc str.: ' distribution b a' \hown in Fig. 10-10. with the pla tlc nl!utral axi (PNA) an the lab. In Fig . J0-10. C = 0.1\S[;ba. where a b the depth of the comprcs ion block (in) and b is the cfl'cctivt: slab width (in).

140

COMPOSITE MEMBERS

[CHAP . 10

a

z

I

+ r---

1

----

---l

'

I I

-

0.85f

1

1 J

- -'

l

T-AF

'

-, '-

.

c = 0.85f ba .., E Q.

I ('

J

.J_

... Q.

F

Fig. 10 10 From Prob . 10.5.1: = 5.0 ksi. b =YO in. Then,

c

425 kipkI' p _ {I=().a;:I"-, = . = O.lb X 5.0- X YO f'b 10. 10

. 1.1110

The nominal flexural trc:ngth

lt4 n

d

= Te = T

/I)

-+

1-

\

-

-

1

= 415 kips x (

17.Y0in

2

1 + 5 1n -

= .QS k1ps X 13.39 1n

= 56Y3 k1p-in = 474 k1p-ft Th.: de\1gn flexural trcngth for full composite action is tJ>hM .. = O.tl5 x 474 kip-ft = 403 kipft. Thi b nearly double the ( rp.M" = ) 212-kip-ft dc ign compresSive strength of a noncomposite Wli!X40 beam ot the same A36 steel (assuming adequate lateral bracing; i.e.. L. s L,). (The composite bc. 0 k'tps /'Ill•' M,=

T. = C,= C

X

90.10 = 0.97 in

d (/) ( 2 + 1- :!

= 370 ktp

17.90in

X

(

:!

+ 5 tn-

0.97in)

2

= 370 kips x 13.47 in

= 983 kip-in= 15 kip-ft

The design flexural strength for partial composite action is IJ> oM, = 0.85 x 415 kip-ft = 353 kipft >350 kip-ft required. This is okay. Try a uniform distribution of shear stud and check the design flexural strength at the concentrated load . where the required Mrength i 300 ktp·ft. (Sec Fig. 10·12.) Propo cd "uniform " shear 1rud dmribuuon' Full cllmposlte acuon

I

Pantal composite acuon

!

10 10

J)

!

8

3@ lOft Om

Total I!

34

10

30ft Din

& ·

Fig. 10-12 At the points of concentrated load (partial composite action).

11

=

10 and

V, =

•Q = 10 Stud X 26.4 Q - ,=II ,

kip s

d = 264

ktpS

stu

'

.

Referring to Fig. 10-10

C = 0.115/'ba s!: Q, = :!64 kips :!6-1 kips

C

.

0 69 a= 0."oS. f,'b = 0.85 x•5.0 kt·ps;·m'·x9(J m = ·

M = Tr = Ce = "

c(d + 2

k. = 264 tp =

0 1-

2 X

10

)

(17.90in 2

+

.

tn-

0.69in) 2

264 ktp' x 13.60 in = 3592 kip-in = 299 ktp·ft •

The design flexural strength for partial composite action is lj>,M, = 0.85 x 299 kip-ft = 254 kipfl < 300 kip-ft requtred. Not adequate. Try n = 12 shear tuds from the end supports to the pomts of concentrated load

ktps

.

V,, = 11Q, = 12 studs x 26.4 -= 317 ktp

stud

CHAP. 10]

COMPOSITE MEMBERS

143

Refernng to Fig 10-10

c

317 kip a= 0.o o:·>.r1•, b = 0.g- X)-.() k.lpS/"1n'· X

t)() 10

= 0.!!3 10

M = c(d + 1-a 2

:!

n

= 317k

. lp

-

X

(17.9010

2

+ 5 1n- _0 ..:...83=-•..:...n) -

2

=317kipsx 13.4in = 4291 kip-in= 358 kip-ft

cp,,M., = 0.85 x 358 kip-ft = 304 kip-ft > 300 kip-ft required. Thi is okay. (See Fig.

10-13.)

The t·orrt•ct hear srud distributiOn'> are

I

Full 1.20 .._ 1.6L.

lb lb w,. = 1.2 X 243 -+ 1.6 X 70 ft ft

= 404 lb/ft

= 0.40 kipi>/ft

M,. =

0.40 l..ip>/ft X (32.0 ft)1

R

52 kip-It

cp,M,

lor th.: W2 x6S t::\'en 1f L,, =the full 32-ft >pan. (Sec A ISC LRFD Manual, p. 3-7L) The umhored n()ncomposire W24 x OR beam b adcquate during con\!ruction even if it i• not laterally braced .
1g.n nc\ural o.,treng.th io., q>.A1•• where cp. = O.X5 and ,\f.. is calculated from the ph.-!lc >In!>' dl'tnhu!lon on the compO>He ...cellon.

(b)

The un...hored noncomposite W24X6S beam can adequate!} >upport the -1n sohd concrete slab in Fig 10-7. (II can sure! carry the lighter 2-in oolid >lab on :!-1n deck.)

Regarding AI•. from the sulu11on to Prob. 10.9 the maximum po»ible compressive force 1n the concrete >lab C, = !l.R5f.'.A, =:!50 kip>: the maximum po>sible ten>Jie force in the >!eel beam T =A,F. =72 kip>. To >atisfy equ1libnum. T = C =( 50+ 7:!4) kips/2 = 487 kips. The pla!.!IC >tre'> diwibution is as shown in Fig. 10-15. Bccau;o;: the net compressive force in the tee! b greater than the beam flange yield force (i.e., [C = (4S7- 250) kip!.= 237 kip )> [brtrF. = 8.965 in x 0.5851n x J6 kip;/in 1= 189 kips]), th e plas t ic neutra l axi (PNA) is located in the web. Ignoring the web lilleb of the beam, the d istance f rom the

1

I

085fbt'

C "'t.uF

I

I

T

c

0.85}'

=

250kip>

= 48 kip>

T

.l-

P:->A

I

F

r.ill t .rtl· l(• ht,F

d (d - 1tl -

U)

r.

1t, - u)F = 298 ki 189 ki

,

....

b

"'

F

fig. 10-15

146

CO:\fPOSITE MEMBERS

bottom of the upper heam flange to the PNA

(CHAP. 10

approx1mately

C. = 4X l-Ip\ =3 . II t,..F; 0.41. m x.36 k.1p.,I .m • . . In =

The contribution to M. from each element of beam or slab = element (tensile or compressive) force x the dit ancc of clement force from the P:-JA

ContributiOns w .\1 frum Compre ion

10

the slah -; C,

(a + t, +

= 250 k ips X ( 3.2 + () 5X5 + 4.0Compression in upper beam flange= - 189 • ki pS X

(

3.-

20

-

1 -)

) 1n = 1696 klfHn

c,(a + )

+ (J. 5!15) .

Ill

2

a

Compress1on m heam weh = C. ::; = 48 k1ps X

32m

= 17

-

1

-

. . b d- 2t1 -a T 1 cns10n 10 cam web- T. ., k'

(23.73- 2

X

= 29o 1ps X

0.5K'i- 3.2) in '

-

= 2XX5 kip-in

J

Tension in lower hcam flange= 'ft(d= 189 kip

11 - a)

x (23 73- !x 0 585- 3.2) in

= 3714

kip-in

9032

klp·ln

M. = 9032 k1p-in = 753 k1pft The design llexural Wt!ngth cfi >M. = U.R5 x 753 k1p-ft = 640 kip-ft. In comparison with the 677 kipft design flexural trength m Prob. 10.14. the 640 k1p-ft determmed herein represent s a mere 5 percent reduction. The inability of the 2 in of concrete within the deck to participate in composite action is not very sign ificant.

10.16. Assume the concrcw-encased W8x40 in Proh. 10.2 and Fig. 10-4 is a beam. Determine the design flexural strength

cp,,M,, and

M,. for bending about the major and minor axes.

A indicated m thl'> chapter. m the section ent itled Concrete-Encased Beams, concrete encasement satisfymg the tated 1111111111um requ1rements prc,cnh hoth local and lateral-torsional buckling of the beam. Of the two method gi,en in the A ISC LRFD Specification for determimng tfJ >M. for concrete-enca\ed beam'>. the s1mpler one i based on the plastic stress distribution on the steel section alone. For x-axis bending (ro:gardless of L )

cp,M., = cp,MI' = rp,Z,F,. 0.90 x 39.8 in ' x 36 kips/in 1 12 in/ ft

= 107 klp-ft

147

COMPOSITE.:. MEMBERS

CHAP. 10]

For .1-ax1., bending (regarc.lh! of L,.)

,.\-1., = ,.Jip = • Z, f 0.90 x ll:l 5 m ' x 36 kips/in:

12 in /ft = 50 kipft

10.17. Assume the concrete-enc, !',,) < 0.3. M,. is detcrmmed by lin ear i nterpolat ion b twccn M. (calculated a., just described) at ( P,./¢>,. P,,) = 0.3. and M, for a composite beam at P,, = 0. A formula is gl\en 111 the Commentary on the AISC LRFD Spec1ticauon (p. 6-175 in the A ISC LRFD Manual) for the determmauon of M.for c:omp. P.) < 1.0: .\1 -L

.. f ., +-IUr:-:!c,)A,F,, ("-:'; ·1.F . , +

-)

--

3

t\,.1· ,

l7f,1r

-

( C-/..J-1)

1

wh en: A,. =web area ot t he enca cd tecl shape , in ': () for concrete-li llcd tubes L = plastic section modulu' of the M el 'hape. in:

c, =average d1-.tance from the ten 1on and compres5ion faces

to

the nearest longitudinal

reinforcmg bar... m

h =width of the compthitc eros secuon perpendicular to the plane of bending. in " = w1dth of the composite cross ecuon parallel to the plane of bending. in From Prob. 10.2 an d Fig. 10-4: lr, - Jr .• = 16 in ,hM,., = 0.1;!5 X 157 kip-ft = 134 ktpf't

l48

COMPOSITE MEMBERS

For

[CH AP. 10

P.. = 0, M.. = ZF, (as in Prob. 10.16 for thearne concrete-enca ed member)

= ()

·852'

q>hMn,

F. = 0.85 X 39.!1 an ' ., /f -In

1

X

36 kip\/an '

t

= 101 kip-ft

rphM.,

=

gcz F. = 0.85 x 18.5 an' x 36 kips/in, 0. -' , , I" . If -In I

= -t7 kip-

ft The results are ploued in Fig. 10-16.

188 kipft

134 kip-fi

':£.

• ':£

...;

101 klp-ft

47 k.ip-fi

0

03 t P.

1.0

0

0.3

o.P.l

1.0

( PJo, P.l

Fig. 10-16

10.18. The built-up beam in Prob. 6.1 and Fig. 6-3 acts compositely with a solid normal-weight concrete floor slab (f; = 5.0 ksi . effective width= 100 in. thickness= 4 in). Assuming full compo itc action and shored construction. determine the design flexural strength 1/Jt>Mn• · A review of the solution to Prob. 6.1 indicates that the web of the beam

IS

noncompact.

( = 12s.o) > ( = 106.1) According ly. the design flexural strength is determined as follow!>: (/> o = 0.90. and M. is calculated from the superposition of clastic stresses. considering the effects of sh oring. Correspond ing clast ic stress diagrams for the shored and unshored cases arc shown in Fig. 10·17, wh ere

S, =section modulus of the bare steel beam. in'

s.,_,. . s,_, =section moduli of the transformed section . in ' 11

= modular ratio=

£/ Ec

M,. 1 =required flexural strength due to the (factored) loads applied before the concrete has auained 75 percent off;

M ..! = required flexural strength due to the (factored) loads applied after the concre te has ach1eved 75 percent off' M.=total required flexural strength= (M.., + M,. 1) If con\truction is bored. all loads are resisted by compo lle actton. The two limitations on flexural trength are the maximum stresses in the steel and concrete From Fig. 10-17. the limiting conditions are

M.. srp,.M. = rp,.S,. ,.F.. M.< rp,.M. = rf! onS,. ,(0.85/;)

COMPOSITE MEMBI:.R S

CHAP. 10)

149

-

Effective wtdth - h

'

1

j4 hln I

1

Y/

-' IJ

E._.A (eiJ'ttC neutral

-

---

.,.,"' trJn,lorrned >ection)

+ .If. _, F - c.,. ' S '

M,, S,

'

Shor u wn>t rucuon

Unshorcd conMrucuon

Fig. 10-17 The modular ratto 11 = £! £,

£, = w 1 "Vi:= 1451 'vs.o = 3904 ksi , 19.000 II=

=7

kst

3904 ksi

'
ed W14X 120 in Prob 10.:!0 b a beam (1.c.. P,, = 0. ). dct crmme rp,M.,, and tf>,.•w.. An1 . tf>,. w•. = 572 k1p-ft. 1/>h.w.. = 275 kip-ft

Chapter 11 Connections NOTATION A 1111 = cross-sectional area of the bac."\'f'·p:nt:tr-..tum thrtlat ,tze

>'A.

:

Cilc st rength. as given in Table ll-5. For bolts loaded in shear only, the design shear strength is equal to¢ mult iplied by the nominal shear strength, given in Table 11-5. I f a combination of tension and shear acts on a bolt. the maximum tensile stress is dcterrmned from Table 11-6 and the maximum shear tre !>. lrom Table J J-5. In all cases, tre scs (in h1) arc converted to forces by multiplying by the nominal cross-sectional area of the bolt (ig.nonng the threads). A special category of sftp-cruica/ JOints is recognized by the A I SC' LRFD Specification. Where joint slippage b undesirable (e.g.. if there are frequent load reversab. leading to the possibility of fatigue), the des1gner rna) specify ..slip-critical" high-strength bolts. Because this is a serviceability criterion. the (unfactored) service loads are used in conjunction w1th Table 11-7. If the load combination includes either \\ ind or seismic load together with hve load. the total !>ervice load may be multiplied by 0.75. To determine tlte design shear strength. the nominal values in Table 11-7 are multiplied by ¢ = 1.0 (except ¢ = O.!S5 for long-slotted hob if the load t!> parallel to the slot). If a bolt in a slip-critical connection IS subjected to a service tensile force T. the nominal shear strength

158

CO NECTlONS

Table ll-5

(CHAP. II

Design Strength of Fasteners Shear Strength in Bearing-Type Connections

Tensile Stren gth

Descnpuon of Fa tener A307 bolt>

Resi!>tancc Factor tp 0.75

Nominal Strength. k ·

Resistance Factor tp

Nominal Strength k i

45.0

0.60

27.0

A325 bolts. when threads are nor excluded from hear planes

90.0

0.65

54.0

A325 bolt . when thread are excluded from 'hear planes

90.0

72.0

A490 bolts. when threads are 1101 excluded from hear plane>

112.5

67.5

A490 bolts. when thread , arc excluded from the \hear planes

112.5

90.0

Threaded parts meeung the requirements of Sec. A3. when thread; are 1101 excluded from the shear plane

0.75F,,

0.45F,,

Threaded pam meeting thl' requirements of Sec. A3. when thread are excluded from the shear plane,

0.75F.

Mor:

45.0

36.0

60.0

48.0

A502. Grade I, hot-dnven rivets A502. Grade'> 2 and 3. hot-driven rivets

Table Ll -6 Tension Stress Limit (F,), ksi, for Fasteners in Bearing-Type Connections

Description of Fasten). v.here f, • th e minimumpretension force for th at bolt in Table 11-4. Table tl-7

Nomi na l Slip-Critical Shear Strength of H ij!h-Strcn th Botts• Nommal !>hear Strength. \.. t

Type of Bolt

Standard-Size Ifole\

A325

17

A490

21

Over tzcd and Short- Slotted llolt:\

L.ong-Siollcd llolcst

15 IX

12

15

·Cia" A (\lip cocfilcien L 0.33). Clean mill scal e and biH\t ck;mcd 'urfacc wi1h cln;., A COllll ll!l,\. For trengt h , mmimum spacing. and minimum edge distance depend on the dimensions of the bolt h ole!>. Nominal dtmens10ns for standard. mer'>ite. hort- lotted. and long-'>lotted holes are given tn Table ll-8 (Table J3.5 in the AISC LRFD Specification • ). Unli ke '>tandard holes, use of the other t) pes of holes requtres approval of the designer and b subJeCt to the restrictions in Sec. J3.7 of the A I SC LRFO Specification. Table 11-8

Nominal Hol e Dimension ,, i n Hole D1mcn,ion' tn Bolt Diameter. m

Standard (Dia.)

Q,ersile (Dia.)

"

I

l

.
are required. they should be spec1fied t ogether with their required flexural strengths.

z '

I

i

• 11 •

• II •

r-

I

I

Ia)

Fig. 11-4 Be P,, H 0.6(0.85VA2/bjFa,1 = 0.90F, = 0.90 X 36 ksi = 32.4 ksi In kips 3·1n 0 .75 t·n cf> P,= cf>FIHIAIIM ='·' .4 X -k:i-p;"s X Ill"

= 72.9 kips (Note: A' indicated tn Table 11-1. matching E70 weld must be used with A36 steel in thi' case.)

(b) The minimum effect throat thickness of partial-penetration groove weld (as given tn Table 11-2) is ,:; tn f or ,-t.n pates. According to Table Il-l. for tension normal to the effective area of a partial penetration groove weld. the de,1gn Mrength os the lower 'alue of t/>Fn 11 A 11 11 = cpF,A 11 11 = 0.90 X 36 kip/In. X :\on X in= 72.9 kips ct>f:A. = cf>(0.60f.n·lA .. =0.80(0.60 x 70 k1ps/in ) x 3 in x lin = 25.2 kips if an E70 el ectrode i' used. For an E70 dectrode, the design tensile strength is 25.2 kips. As indicated in Table 11-1. an E60 electrode (with strength Frx:o: = 60 ksi, le'>s than the matching E70 weld metal) may also be used. H the weld IS E60, the design strength of the weld cpF,.A. agatn controls: cf>F.A. = 0.80(0.60 F, 11 ) x 3 in x in. where F, 11 = 60 l-..\1. 1/>f.A. =

21.6 kip\ 1f an E60 electrode

1\

used.

11.2. Repeat Prob . 11.1 for plates of unequal thickness:

1 in and n

in.

The effective throat thid..ness for a compl et e-penetrat ion groove weld is the thickn es., of the thinner plate joined , or 1 in . For tension normal w the effective area of a complete penetration groove weld, a matching E70 elect rode mu t be used. The design ten i lc trength is cf>fm 1A 1111 = cf>f;A 8 11 = 0.90

X

36 k1p/ in ,

X

3 in

X

1

in

= 18.2 kips

11.3. A ve rtical complete-penetration groove weld is used to JOtn the two halves of a W24xl76 beam (A36 steel ). Determine the design shear strength of the web splice. According to Table Il-l. for hear on the effect ive area of a complete-penetration groove weld, the design strength is the Iowa value of

r ,-

1=

o.90(0. 60F.) = u. 9 x 0.6 x 36 k si = 19.4 ksi

cf>f;, = 0.80(0.60F,x.d = {0.8 x 0.6 x 70 ksi = 33.6 k i for E70 0.8 x 0.6 x 60 hi = :!8.8 ks1 for E60

166

CONNECTIONS

(CHAP. 11

Regarc.lle of whether an E60 or E70 electrode is U!>ed. the Mrength of the ba c material in the web of the W::!.Jx 176 beam (cpFn" = 19.4 ksi) go\erns.

V. = 19.4 ksi x dr.

= 19.4 ki s x 25.24 m x 0.750 m = 36llk•P' m ·

The tabulated de ign hear strength of a W2-tx76 beam (on p . 3-31 of the A1SC LRFD Manual) is. in fact. 36ll k1p .

11.4. Two vertical partial-penetration groove welds. each with an effective throat thickness of

1in,

arc used to join the two halves of a W2 x 176 beam Determine the design shear strength of the web splice. According to Table 11-1, for ;hear parallel to the axes of partial-penetration groove welds the following l im 1t ;tatcs hould be con!.idercd: Shear fracture of the base material (Eq. (14-1)]

Rn = 0. 75(0.6A...,F,,) = 0. 75 x 0.6(25.24 in x 0. 750 in)

x 58 m

i s

= 494 kips Shear y1elding of the base material (Eq. (15-J)]

cpR.= 0.90(0.6A ,.•F.) =0.9 X0.6(25.:!41n = 368 kips

X

0.750m)

X

361..i hear area for the two partial-penetration groove weld-." d x 2 x! in; 1.e. A.= 25.24 in X 2 X 0.25 in= 12.62

31.5 -k-:i--p7"' x 12.62.m'· = 398 k1.ps

In

for E7Cl

m•

kips , 27.0 --:--T" X 12.62m·-34 1 !..1psfor E60

m·

In condu,ion. the des1gn shear strength at the !>plice b 368 kip'> (based on the limit state of shear yielding of the b F..A. the design tensile strengths of the plates are not critical in this case.

11.6.

In Prob. 11.5, determine the design tensile strength for the maximum size of fillet weld : (a) E70. (b) E60. According to Sec. J2.2(b) of the AISC LRFD Specification. the maximum size of fillet weld s Equals plate thickness. if

l.umed in this exercise.

Fig. 11-8

169

CONNECfiON S

CHA P. II]

The suffix es N and X refer to a beari n g-type (i.e.. non-slip-critical) connection, where N designates threads included in the shear plane . X designates threads excluded from the shear plane . The design shear strength of a bolt

where ¢ and F,." are as listed in Table 11-5. 2 The nomi nal cross-sectio nal area of a -in-diameter bolt is A= 0.60 in (as calculated in Prob . 11.7). (a)

For a -in-diameter A325-N bolt, the design shear strength kips . , . cp V,, = 0.65 X 54.0 X 0.60 = 21.1 S --:'2 m· kip m

(b)

For a -in-diameter A325-X bolt. the design shear strength kips . , . cpV,, = 0.65 X 72.0-:-:;- X 0.60 in·= 28.1 kipS

m ·

(c)

For a -in-diameter A490-N bolt, the design shear strength

cpv" = 0.65 X

kips . , . x 0.60 = 26.4 kips 67.5-:-:;m·

m·

(d)

For a ij-i n-diameter A490-X boll, the design shear strength

cpv. = 0.65 x 90.0 (e)

k'

sx 0.60 in 2 = 35.2 kips

m ·

For a -in-diameter A307 bolt, the design shear strength

cp V,, = 0.60 X

kips X

27.0-:-:;-

m·

11.9.

. , 0.60

. = 9.7

m·

S kip

A -in-diameter A325 bolt is subjected to combined shear and tension. Determine the design tensile force assuming the required shear force is 10 kips. The nominal cross-sectional area of a -in-diameter bolt is 0.60 in

2

•

2

The shear stress J,, = 10 kips/0.60 in = 16.6 kips/in (a)

2

According to Table 11-6. for A325-N bolts (thread s included in the shear plane), the design tensile stress F, = (85 - l.8[,, :s68) ksi = (85- 1.8 x 16.6 < 68) ksi = 55.1 ksi The design tensile force kips . , . F,A =55. I -:-:;- X 0.60 m·= _,3.0 kipS

m·

(b)

For A325-X bolts {threads excluded from the shear plane), the design tensile stress

F, = (85 - 1.4[.. < 68) ksi = (85- 1.4 X 16.6 < 68) ksi =61.7ksi

170

CONNECT I ONS

(CHAP.

11

The design tensile force kip . F,A = 61.7 . , 0.6010 = 37.1 ktp X

10"

11.10. Determine the shear strength of a -in-diameter A325 bolt in a slip-critical connection. (Please nore : The strengths of slip-critical connection are expressed a unfactored force in Table 11-

7.) As uming tandard-size holes. f,. = 17 ksi for A325 bolts. Shear strength kips

. •

.

f,.A = 17 -:-;- x 0.60 10·= 10.2 ktps

m·

Maximum service load shear on the bolt is 10.2 kips. As noted in Table 1 1-7, j;. = 17 ksi and the other hear strengths tabu lated therein are for class A surfaces (with slip coefficie n t 0.33). Higher shear strength for high-strength bolts in slip-critical connections are available for cl:t s B (sl ip coefticient 0.50) and cia; C (slip coefficient 0.40) surfaces. l11e higher values are given m the Specification for Structuml Joint\ Using ASTM A325 or A490 Bolts, which appears in Part 6 of the A ISC LRFD Manual.

11.11. Repeat Problem ll.IO for a service tensile force of 20 kips acting in combination with the shear. If tens• on " pre;ent. the shear values in Table 11-7 are to be muluphed b} ( 1 - TIT,). ' here T i the '>ervice ten;ile force and T,is the minimum pretension load for the bolt 10 Table Il-4. 10.2 kips

x

( I- ) = l0.21o.ps (1-20 ki.ps ) =50 kip maxtmum scrvtce toad 'hear T- x T,

39 ktps

11.12. Check t he bearing strengths of the -in-diameter bolts 10 Probs. 11.8 and 11.10. Assume two or more bolts in the line of force connecting two -in plates of A36 steel; standard holes; center-to-center distance of 3 in; and edge distance of l! in. Edge distance (L

= 1.5 in)> (1.5d = 1.5 x

in= 1.31in).

(3.0d = 3.0 X in= 2.63 in). Equation (JJ-Ja) is appljcable and t h e desi gn bea ring strength is R.. = 2.4 dt F,,.

(f>Rn = 0.75 X 2.4 X

X

Spacing (C = 3.0 in) >

cpR...

where cp = 0.75 and

in X 58 k ips/in

= 34.3 kip per bolt

In Prob. li.R. the only bolt governed by bearing strength is the A490-X in part (d). for which (rpR. = 34.3 1-. ips) < (rf>V. = 35.2 kips). All the other bolts are governed by shear strength, because rpV. < (rpR. = 3-t.3 kips) . Regarding Prob . 11.10. where the bolt is in a :.lip-critical connecuon, the limiting service load shear of 10.2 kip!> obviously go,erns over tbe limiting factored load bearing value of 34.3 kips .

11.13. The end of a WJ2x87 beam (A36 steel) has been prepared as shown in Fig. 11-9 for connection to a support ing member. The three holes are 10 diameter for -m-diameter bolts. Determine the design shear strength of the beam web . The applicable limit states are shear yielding. shear fracture. and block !>hear rupture. For shear yielding

CONNEC..IIONS

CHAP. II]

171

2 in cope

---+d

-' 7

I:!S.l1n

l

1 in

I

0 0

2 R.. = 0. 75 X 0.6A..,F,, A,..= (d-copc-3d.)t = (12.53 in- 2 in- 3 x : in)

x 0.515 in= 3.97 in'

1/>R. = 0.75 X 0.6 X 3.97 in' X 58 kM = 10-l kip; For block shear rupture (of ;ection (2) in Fig. 11-9] q> = 0. 75 and R. = the greater value of

0.61\,.• F. + AJ. 0.6A.,..f.

where

(C-14·1) (C-14-2)

+ A.F.

A, • = gro; area of the vcrttca1 part of (2) A ... = net area of the vertical part of (2)

A. = gro;s area of the horizontal part of (2) A .. = net area of the horizontal part of (2) +2 x 3m) x0.515 in =3.86m' 1 - ._, ,X 16) X o.::>1c-' A,,-- ( ,tn + 2 X ·tn--2.66 3 m'

A..= (I

A8 = I

10

tn''

in x 0.515 in = 0.77 in!

A .. = (I in - x

1

in) X 0.515 in = 0.53 in'

R.is the greater of . ,

kips . • • ktp . 0.6 x 3.86 m·x 36-;-;- + 0.53 m· x :>8-;-;- = 114 ktps tn· •n· . , _ kip . , kips . 0.6 X 2.66 sn· X :>8 77 + 0.77 1n· X 36-;--; = 120 kipS 1n· 1n· R. = 120 k1p\ 1/>R. = 0.75 X 120 kips= 90 k1p; The design ;hear strength is 90 kips. based on the gove rning limit state of block shear rupture.

11.14. Design a base plate for a Wl4x90 column with a factored axial load of 700 kipl>. All '>tccl is A36. The base plate is on a footing 2ft 0 in X 2ft 0 in;[;= 4 ksi .

•

172

CONNECTIONS

[CHAP. 1l

The design beanng strength for steel bearing on concrete is determined from Eq. (11.5) or (11.6); the former for bearing on the full area of concrete. and the latter for bearing on less than the full area. The dimensions of the W 14X90 column d b1 = 14.02 in x 14.52 in . Try a 16in x 16 in base pl ate and use

Eq. (11.7).

A = 24 in

'_

x 24 in = 576 inz

kips ! ,.-4 . .,

A, = 16in

x 16 in =256in 2

4> = 0.60

Ill"

The design beanng strength

1/>rPp = 0 ·8)-J.r•;A, -

A,

= 0.85 X

2

lops . , v576 in 4-. -, X 256m· . , X 256 m·

m·

= 1306 kips > 700 kips required

o.k.

Referring to Fig. 11·6

N= 16.0in,

m =0.5(N-0.95d)

d=l4.0in

= 0.5( 16m-0.95 x 14 in)= 1.35 in

b1 = 14.52 in

B= 16.0in,

n

= 0.5(8- 0.80b1) = 0.5(16 10-0.80 x 14.52 in)= 2.1910

To determme c, solve Eqs. [11.8] to I11.10)

P,, = Nbrt P,,=

8 = 556 kips

700 kips. x 14.021nX . . 14.52m

16 m .

X I

6m

556 kips -::-0--:. 6-;:(0::-. 8:-:5;:-Vr.5;::7;:=6 in"1"J""( t47.'='52:=1,;.n,;,X=I4=. O:=i:=n:=-)4:-:k-:-i p-s-/-:;-;in2

> -0.6

x

556 kips (1.7 x 4 kips/in:)

= 162 in 2 > 136 in 2 =

162 in

c = !f(d + b1 - t1) - V(d + b1 -

td-

4(A" -t1b1))

(d +b1 -t1 )= (14.02+ 14.52-0.71)in=27.1Bin

c = ![27.83 in- Y(27 .83m)2 -4( 162 in - 0.71 in x 14.52 in)[

c = 4.26 in Referring to Eq. [11.8) m = 1.35 in,

v

2. 19 in

c=

4.26 in

2 x 700 kips • X I 6"111 X J6 in =0.41 0.9 X 36 kips ;· 111·

= =

11 =

/

2 x 556 kips

Y Yo.9 x 36 kips/i n

2

=

0 46

xl62 in 1 •

CONNECTIONS

CH A P. II]

173

Base plate thickness rP ts the largest of ( 1.35 in x 0.41 = 0.55 tn). (2.19 in x 0.41 = 0.90 in), and (4.26 in x 0.46 = 1.96 in). Use a base plate 16 in >: 2 to v 16 tn .

Supplementary Problems 11.15. Complete penetration groove welds are used to join the flange of the two halve of the W24x 176 beam (A36 steel) in Prob . 11.3. Determine (a) the design flexural strength at the splice and (b) the appropriate electrode . An.f .

(a) cphM, = 1115 kip-ft. (b) matching E70.

11.16. The flange of the two halves of the same W24X 176 beam are joined by Hn partial-penetration groove welds. Determine (a) the design flexural strength at the splice and (b) the appropriate electrode. AilS .

cp.M , = 446 kip-ft for E70; cpbM• = 383 kip-ft for E60.

11.17. In Fig . 11-l(d), the plates are 3 in wide and in thick . the base material is A36 steel. Determine the design tensile strength of the splice for the minimum size fillet weld : (a) E70. (b) E60. AilS . (a) 33.4 kips , (b) 28.6 kips .

11.18. In Prob. 11.17. determ10e the design tensile strength for the maximum size fillet weld : (a) E70. {b) E60. AilS. (a) 72.9 kips , (b) 72.9 kips.

11.19. Repeat Prob . ll. for a -in-diameter bolt. Arts . (a) 15.5 kips , (b) 20.7 kips. (c) 19.4 kips. (d) 25.8 kips. (e) 7.2 kips .

11.20. Repeat Prob . 11.9 for a -in-diameter bolt. Arts.

(a) 19.6 kips. (b) 23.6 kips.

11.21. Repea t Probs. 11.10 and 11.11 for i-in-diameter bolts. Arts . 7.5 kips, 2.1 kips.

11.22. Determtne the bearing strength of -in-diameter bolts connecting -in plates of A36 steel: standard holes: center-to-cente r distance of 2! in; and edge distance oft!in . A ItS.

19.6 ktps .

11.23. Determtne the destgn shear strength of the web of the W21X44 beam (A36 steel) in Fig. 11-10. The five holes are 1 1 -in-diameter for l-in-diameter bolts. Arts . 121 ktp .

174

CONNECTIONS

[CHAP. 1I

l'-----....,

0 0 0 0 0

W2 1 X 44

1m 2 10

Fig. 11-10

11.24. Design a base plate for a W8x 67 column with a factored axial load of 450 kips. All steel is A36. The base plate will occupy the full area of concrete support; = 3.5 ksi .

r:

Ans.

Base plate 14 in x I in x 14 in.

Chapter 12 Other Design Considerations NOTATION

b1 = flange WH.Ith , in d = depth of the member , in

d, = web depth clear of fillets. in= d2k

F.. = specified minimum yield stress K =effective length factor for columns k = distance from outer face of the flange to web toe of the fillet , in

I =sti ffener height. in N = length of bearing, in P,= nominal axial compressive strength of the column, kips P,, = required axial compressive strength of the column, kips R, =nomin al st rength. kips

R,. =req uired strength, kips R,, = nominal shear strength. kips r1 = flange thickness. in

r.. =web thickness. in X = param eter in Eq s. ( K 1-6) and (K 1-7) Y = parameter in Eqs. (Kl-6) and (K 1-7)

¢ = resist ance factor

¢R, =design mength, kips ¢R., = design !>hear strength , kips INTRODUCfiOI'I Additional provisions for steel structures are given in the final three chapters of the AlSC LRFD Specification. as follows: Chap. K-Strength Design Considera tions Chap. L-Serviceability Design Considerations Chap . M-Fabrication, Erection , and Quality Control The strength and stability provisions relating to concentrated forces are discussed herein .

CONCENTRATED LOADS AND REACTIONS A concentrated force acting on a member introduces high stresses in its vicinity. To prevent failure , the required (or factored) concentrat ed load or reaction R,. (kips) must be checked against the design strength ¢R, (kips). as determined b) the appropriate limit states. For each limit state. ¢ is the resistance factor and R, is the nominal strength. (I) Local Web Yielding. This limit st ate applies to aU concentrated forces (tensile or compressive) in the plane of the web. The design strength of th e web at the toe of the fillet

175

176

OTHER DESIGN CONSIDERATIONS

ICHAP.

12

is R,. where

= 0.75 and Rn depends on whether the concen trated force is a load or a reaction . a.

For a concentrated load (acting along a member at a distance from either end greater than d/2) (Kl-4)

b.

For a concentrated reaction (acting at or near the end of the member)

-

F.tr I ._

(Kl-5)

where dis the depth of the member , in, and t1is flange thickness, in. If the concentrated force exceeds Rn, a pair of stiffeners must be provided in accordance with the Stiffener Requirements section later in this chapter. (3) Sidesway Web Buck ling. This limit state relates to concentra ted compressive force applied to one flan ge in the plane of the web, where no lateral bracing or (h alf-depth ) stiffeners arc provided. The design compressive strength is rJ>Rn, where

+j

8

I I

t

n

I

I

'

---

"

1 II /1

I

\

•

I

'

I "

I

Fig. 12-2

For the concentrated compressive reaction of 100 kip acllng on the bottom flange. the applicable limit tntcs arc (I) local web yielding and (2) web crippling. (It is as umcd that the beam is welded to the base pl ate and hoth arc anchor-bolted to the concrete :.upport. This hould rrovidc adequate lateral bracing to prevent idesway web buck lin g.) Corresponding to the applicable limit states arc Eqs. (K 1-3) and (K 1-5), each of which has N. the length of hearing. a a parameter. Solving for N, we obtain R.. < tf>R. = tf>(2.5k + N)F,t.

in + N) x

100 kips -;; JJJ(2 .5 x I 36 kirs/in'x 0.40 in (K 1-3)

N?::3.5 in (K 1-5)

100 ktp s 0.75 X 68(0.40 in)··[I + 3(

_N ) (00.0460I1S11 )'

36 X

k ps

·:. :. 6:. :. 15:. :. i:.n.:.

'] t

,

_o

20 99111

111

in-

0.40111

in The m1111mum length of bearing tS N = 8.6 in. Roundtng up to the ne:>.t full inch. let V = 9111. ,\'? 8.6

OTHER DESIGN CO SIDERA TIONS

CHAP. 12)

179

The area of the bearing plate i determined by the bearing >trength of the concrete \upport. U mg Eq. [ J1. 6] from Chap. II. the design bearing strength b

q>.,P,, = cp, x O.K5[,'A 1 ! A , where VA1/A , < 2. Substituting in Ett . Ill . 6]. we obtam . klp 100 kip> = 0.60 X 0.85 X 3 X A IX 2

m·

The area of the bearing plate A 1 = 32.7 in1 • Becau>e the bcanng plate dimensions are . A, 32.7 in' = 3 6 11 8 2: -= 1 N 9 in ·

BN>A,:

However. 8 cannot be le,, than the flange width of the W21 x62 beam. br = 8.24. Rounding up. let 8 = 9 in . A formula for beanng plate thickness i> gl\'Cn on page 3-50 of the AISC LRFD Manual : t=

where R = 100 kip• II

2 .22Rn 1

A ,F.

8 - 2k 9m-2 X I in . 3 13 = =Ill ')

2

-

.

A 1 = BN = 9 in x 9 in = 81 in

F..= 36 ksi t=

2.22 x 100 kips x (3.13 inf . , = 0.86m 81 in·x 36 k>•

Use a bearing plate I in x 9 m X 9 in .

12.2. ln Prob. 12.I. can the bearing plate be eliminated'? For the W21 X62 beam to bear directly on the concrete upport. it bottom flange muM be ,ufficicntly thick to act as a bearing plate. Let t=

2 22 111 · RF = 0.615 in AI

•

the flange thickne•!> of the W21 X62 beam . Because 8 = b1 = i\.24 in ,, . ? I . 8 - 2k 8._.. X • Ill- = 2.75 Ill

n=

t=

'

Ill

-

)

2

2.22 x 100 kips x (2.75 ml ' . ,. = 0.615 Ill m· 'IpS A 1 X 36 k

A ,= 123 in 1 (>32.7 in 1 required for bearing on concrete) A, A , 123 in1 N =-= - = = 15.0in

8

h1

8.24 in

By increasing the length of bearing of the beam on the concrete to 15 in. the bearing plate can be eliminated .

OTHER DESIGN CONSIDERATIONS

180

12.3.

[CHAP. 12

A column with a 12-in-long base plate rests on the top flange of a Wl8xSO beam (A36 steel). 20ft long. Determine the maximum column load if the beam is (a) not stiffen ed or braced along its entire span and (b) not stiffened but braced at the load point. (a)

For a concentrated compressive force acting on the top flange of a beam. the applicable limit states are (I) local web yielding. (2) web cripphng. and (3) sidesway web buckling. Th.: corresponding equations are ( K 1-2), (Kl-·n. and (Kl -7) (assuming no restramt agatn t rotation). ktps P.tresses. 43 Resistance. 8 Reststance factors. 8- J 0 compression. 28. 124 flexure. 4-1, 63. 67. 126, 129 shear. 48, 68. 130 tcnston . 16 tor ion . I 08 Rolled shapes. 5-6, 30. 70

Second order: analyst . 93 effect . 9-1 moments. 92-95 Sectton modulus: ela\ttc. 43 plastic. 43 Sections (see Cross sections) Serviceability, 8 Shape . \tructural: built-up. 5-6. 66. 70 rolled . S-6, 30. 70 Shear center. 41. 106-107 Shear connectors. 126-J 29 Shear 'trength. 48--19. 68. 130 Shoring. 126-127 Slender clement members. 24 Slenderness ratio. 25-28 Slip-critical joints. 157 Stability (see Buckling) Stiffeners, web: details. 69, 178 requirements, 65. 69. 176-178 Stiffnes\, 5 Stiffness reduction factors. 26-27

INDEX

Strength, 8 Stress-strain diagram . 3-4, 41-43 Structural steel , 3-7 Sway forces, 93

Temion field action. 67 Ten ion members. 14-16 Tensile strength. 4. 16 Torsion. 106-110 avo1ding or mimmiZing. I 07 lOX deformation, 110 design criteria. 108 St. Venant, 109 hear center , 41, 106-107 warping. 110

Unbraced frames, 25-27

187

Unbraced length, 26, 44-47

Vibration of beams. 49-50

Web : huckhng. 176-177 cnpphng. 176 paneb. 177 uffener (see Stiffeners. web) y•elding, 175-176 Weld . 153-156 Width-thickness ratios. 24. 41, 63-65

Yield point. 3 Yield wcngth. 4 Y1cld strcs . 4. 123

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