design of beams · beams can be further classified by the function that they serve. a girder is a...

Steel Design Misan University Fourth Year Engineering College Dr.Abbas Oda Dawood Civil Department

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DESIGN OF Beams

1. INTRODUCTION

Beams are the most common members found in a typical steel structure. Beams are

structural members that support transverse loads and are therefore subjected primarily to

flexure, or bending. Beams with axial loads are called beam-columns. Beams are usually

thought of as being oriented horizontally and subjected to vertical loads, but that is not

necessarily the case. A structural member is considered to be a beam if it is loaded so as to

cause bending.

Beams can be further classified by the function that they serve. A girder is a member that is

generally larger in section and supports other beams or framing members. Joists are the

closely spaced beams supporting the floors and roofs of buildings and it typically a lighter

section than a beam—such as an open-web steel joist. Stringer are the beams in bridge

floors running parallel to the roadway, whereas floor beams are the larger beams used to

transfer the floor loads from the stringers to the supporting girders or trusses, also stringer

in building is a diagonal member that is the main support beam for a stair. Lintels (or loose

lintels) are beams over openings in masonry walls, such as windows and doors and it is

usually a smaller section. A girt is a horizontal member that supports exterior cladding or

siding for lateral wind loads. Spandrel beams supports the exterior walls of building and

perhaps part of the floor loads.


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2. SECTION SHAPES USED FOR BEAMS AND BEAMS TYPES

The hot-rolled I shape is the one most commonly used for beams. Welded shapes usually fall

into the category classified as plate girders. Commonly used cross-sectional shapes include

the W, S, and M shapes. Doubly symmetric shapes such as the standard rolled W, M, and S

shapes are the most efficient. The beams cross section types are:

2.1 Standard Hot-Rolled Sections

Standard hot-rolled sections in general and wide-flange (or W) sections in particular are the

most popular types of beam sections. The W shapes proved to be the most economical

beam section, and they have largely replaced channels and S sections. The W shapes have

more steel concentrated in their flanges than do S beams and thus have larger moments of

inertia and resisting moments for the same weight and they are relatively wide and have

appreciable lateral stiffness.

The use of S beams has decreased from former years. Today they are used primarily for

special situations, such as when narrow flange width are desirable, where shearing forces

are very high, or for crane rails where larger flange thickness near the web may be

advantageous for lateral bending.

Channels are sometimes used for beams subjected to light loads, such as purlins, and in

places where clearances available require narrow flanges. They have little resistance to

lateral forces, thus channels should be avoided where lateral loading is present or need to

be braced by using sag rods.

.2.2 Plate-Covered Section

2.4 Sheet Beams

These beams are made of thin cold-formed steel sheets in a

variety of shapes. They are useful for very light loads (for example,

light roofs).

2.3 Composite Beams

If the steel beam is covered with a concrete deck, the designer

may take advantage of the additional strength of the concrete,

provided that the concrete deck is properly fastened to the steel

beam (for example, by shear studs).

The bending capacity of available rolled sections may be increased

by adding plate(s) to the section.


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2.3 Plate and Box girders

3- CONSIDERATIONS OF BEAM DESIGN

In the design of beams, the following considerations are necessary:

1.Bending stresses

2.Shearing stresses

3.Local buckling

4.Lateral torsional buckling

5.Web crippling

6.Deflection

Initial selection of the beam is usually made on the basis of the maximum bending stresses.

The other design requirements are checked subsequently. For flexure, the required and

available strengths are moments. For load and resistance factor design (LRFD):

)1(MM unb

Where: Mu= required moment strength = maximum moment caused by the external loads

b = resistance factor for bending (flexure) = 0.90

Mn = nominal moment strength

4- BENDING STRESS AND FLEXURAL BEHAVIOR STAGES OF STEEL BEAMS

To be able to determine the nominal moment strength Mn, firstly the behavior of beams throughout the

full range of loading should be examined, from very small loads to the point of collapse.

Consider the beam shown in Figure, which is oriented so that bending is about the major principal axis

(for an I shape, it will be the x–x axis). The stress at any point of the beam subjected to bending

moment may be computed with usual flexural formula fb=My/I, this formula is applicable only when the

maximum computed stress in the beam is below the elastic limit namely linear elastic range.

For girders covering large spans and/or carrying heavy loads, the available standard rolled

sections may be inadequate. In this case, the designer may use a plate or box girder made of steel

plates. These girders may be homogeneous, made of a single grade of steel, or hybrid, made of

high-strength flanges.


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The distribution of bending stress is linear distribution and the flexure formula about major x-axis as

shown in figure above is:

)2(I

yMf

xb

For a homogeneous material, the neutral axis coincides with the centroidal axis. The maximum stress

will occur at the extreme fiber, where y is maximum. Thus there are two maxima: maximum

compressive stress in the top fiber and maximum tensile stress in the bottom fiber. If the neutral axis is

an axis of symmetry, these two stresses will be equal in magnitude. The value of I/c is known as the

section modulus S. For any cross-sectional shape, the section modulus will be a constant. Thus for

maximum stress:

)3(S

M

c/I

M

I

cMf

xxxmax

For an unsymmetrical cross section, S will have two values: one for the top extreme fiber and one for

the bottom. Values of S for standard rolled shapes are tabulated in the dimensions and properties

tables in the Manual.

Equations above is valid as long as the loads are small enough that the material remains within its

linear elastic range. For structural steel, this means that the stress fmax must not exceed Fy and that the

bending moment must not exceed yield moment My:

)4(SFM xyy

where My is yield moment defined as the moment that will just produce the yield stress in the

outermost fiber of the section.

Initially, when the moment is applied to the beam, the stress will vary linearly from the neutral axis to

the extreme fibers as shown in part b of figure below. If the moment is increased, there will continue to

be a linear variation of stress until the yield stress is reached in the outermost fibers as shown in part

c.

If the moment in a ductile steel beam is increased beyond the yield moment, the outermost fibers that

had previously been stressed to their yield stress will continue to have the same stress, but will yield,

and the duty of providing the necessary additional resisting moment will fall on the fibers nearer to the

neutral axis. This process will continue with more and more parts of the beam cross section stressed

to the yield stress as shown in parts d and e of the figure above, until finally a full plastic distribution is

approached as shown in part f.


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Note that the variation of strain from the neutral axis to the outer fibers remain linear for all of these

cases. When the stress distribution has reached this stage a plastic hinge is said to have form,

because no additional moment can be resisted at the section. Any additional moment applied at the

section will cause the beam to rotate, with little increase in stress.

If the steel stress-strain curve is approximated as a bilinear elasto-plastic curve with yield stress equal

to σy , then the section Moment - Curvature (M-ø) response for monotonically increasing moment is

given by Figure below, where curvature ø= 2ε/d.

5- PLASTIC HINGES AND THE PLASTIC MOMENT

The plastic moment is the maximum moment, or nominal bending strength of a beam with full lateral

stability.The plastic moment is the moment that will produce full plasticity in a member cross section

and create a plastic hinge. The ratio of the plastic moment Mp to the yield moment My is called the

shape factor. The shape factor equals 1.5 for rectangular sections and varies from about 1.10 to 1.2

for standard rolled beam sections.


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A plastic hinge is said to have formed at the center of the beam, and this hinge along with the actual

hinges at the ends of the beam constitute an unstable mechanism. During plastic collapse, the

mechanism motion will be as shown in Figure. Structural analysis based on a consideration of collapse

mechanisms is called plastic analysis.

The plastic moment capacity, which is the moment required to form the plastic hinge, can easily be

computed from a consideration of the corresponding stress distribution. In Figure below, the

compressive and tensile stress resultants are shown, where Ac is the cross-sectional area subjected to

compression, and At is the area in tension. These are the areas above and below the plastic neutral

axis, which is not necessarily the same as the elastic neutral axis. From equilibrium of forces,

Consider a simply supported beam with a

concentrated load at midspan is shown at

successive stages of loading. Once yielding

begins, the distribution of stress on the cross

section will no longer be linear, and yielding will

progress from the extreme fiber toward the neutral

axis. At the same time, the yielded region will

extend longitudinally from the center of the beam

as the bending moment reaches M at more

locations. These yielded regions are indicated by

the dark areas in Figure below.

In Figure b, yielding has just begun. In Figure c, the

yielding has progressed into the web, and in Figure

d the entire cross section has yielded. The

additional moment required to bring the beam from

stage b to stage d is 10 to 20% of the yield

moment, My , for W shapes. When stage d has

been reached, any further increase in the load will

cause collapse, since all elements of the cross

section have reached the yield plateau of the

stress–strain curve and unrestricted plastic flow will

occur.


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tc

ytyc

AA

FAFA

TC

Thus the plastic neutral axis divides the cross section into two equal areas. For shapes that are

symmetrical about the axis of bending, the elastic and plastic neutral axes are the same. The plastic

moment, M , is the resisting couple formed by the two equal and opposite forces, or

ZFa)2

A(Fa)A(Fa)A(FM yytycyp

where A = total cross-sectional area

)yy(a ct =distance between the centroids of the two half-areas

a)2

A(Z = plastic section modulus

Values for the plastic section modulus of the cross-section.Z are tabulated for various cross-sections

in the properties section of the LRFD manual.

The plastic centroid for a general cross-section corresponds to the axis about which the total

area is equally divided, i.e., Ac = At = A/2. The plastic centroid is not the same as the elastic centroid

or center of gravity (c.g.) of the cross-section. As shown below, the c.g. is defined as the axis about

which A1y1 = A2y2.

For a cross-section with at-least one axis of symmetry, the neutral axis corresponds to the centroidal

axis in the elastic range. However, at Mp, the neutral axis will correspond to the plastic centroidal axis.

For a doubly symmetric cross-section, the elastic and the plastic centroid lie at the same point.

The design plastic strength

)5(M5.1ZF9.0M yypb

Mp= plastic moment, which must be ≤ 1.5 M for homogenous cross-sections

My = moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution

= Fy S for homogenous cross-sections

Z = plastic section modulus from the Properties section of the AISC manual.

S = elastic section modulus, also from the Properties section of the AISC manual.


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Example 1: Determine the elastic section modulus, S, plastic section modulus, Z, yield moment, My , and the plastic moment Mp , of the cross-section shown below. What is the design moment for the beam cross-section. Assume 50 ksi steel.

Solution

21f in975.0*12A

22f in150.1*15A

2w in125.7)0.175.016(5.0A

2

g in125.31125.7159A

1- Distance of elastic centroid from top = y

in38.9125.31

5.15*15875.7*125.72/75.0*9y

433

23

23

x

in14303

5.0*62.5

3

5.0*63.8

12.6*1512

15*1005.9*9

12

12*75.0I

3xx in45.152

38.9

1430

c

IS

ft.k2.635in.k5.762245.152*50SFM xyxy

2- Distance of plastic centroid from top = py

in875.13y

5625.152

125.31)75.0y(*5.075.0*12

p

p

y1 = centroid of top half area about plastic centroid

y2 = centroid of bottom half area about plastic centroid

in5746.105625.15

5625.6*5.0*125.135.13*75.0*12y1

in5866.15625.15

5625.0*5.0*125.1625.1*1*15y2

321x in26.189)5866.15746.10(*5625.15)yy(

2

AZ

ft.k58.788in.k93.946250*26.189FZM yxxp

Design strength according to AISC

yypb M5.1ZF9.0M

OK

ft.k725.94915.635*5.1ft.k72.70958.788*9.0Mpb

Example 2:Compute the plastic moment, Mp

, for a W10×60 of A992 steel.

Solution

From AISC Manual, for W10 x 60:

2g in7.17A

2in85.82

7.17

2

A

The centroid of the half-area can be found in the tables for WT shapes, which are cut from W shapes. The shape here is the WT5 × 30, and the distance from the outside face of the flange to the centroid is 0.884 inch, as shown in Figure.

in216.4884.02

dyy 21

321x in62.74216.4*2*85.8)yy(

2

AZ

This result, is the same as the value given in the dimensions and properties tables.

ft.k311in.k373150*62.74FZM yxxp


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6- LOCAL BUCKLING OF BEAM SECTION

AISC classifies cross-sectional shapes as compact, noncompact, or slender, depending on the

values of the width-to-thickness ratios.

r

rp

p

hastionseccrosstheofelementanyIftionsecSlender

hastionseccrosstheofelementoneanyIftionsseccompactNon

havetionseccrossofelementsallIfSectionCompact

The classification of a beam is necessary since the design strength of the beam is a function

of its classification. The classification of shapes is found in Section B4 of the Specification,

"Member Properties,” in Table B4.1.

Note that compression members have different criteria than flexural members, so a shape

could be compact for flexure but slender for compression.

Local buckling of flange

due to compressive stress

The hot-rolled steel sections are thin-walled sections

consisting of a number of thin plates. When normal stresses

due to bending and/or direct axial forces are large, each plate

(for example, flange plate or web plate) may buckle locally in

a plane perpendicular to its plane. In order to prevent this

undesirable phenomenon, the width-to-thickness ratios of

the thin flange and the web plates are limited by the code.


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Thus, slender sections cannot develop Mp due to elastic local buckling. Non-compact

sections can develop My but not Mp before local buckling occurs. Only compact sections can

develop the plastic moment Mp.

All rolled wide-flange shapes are compact with the following exceptions, which are non-

compact : W40x174, W14x99, W14x90, W12x65, W10x12, W8x10, W6x15 (made from

A992).

7- LATERAL TORSIONAL BUCKLING

The compression flange of a beam behaves like an axially loaded column. Thus, in beams

covering long spans the compression flange may tend to buckle. Unlike a column, however,

the compression portion of the cross section is restrained by the tension portion, and the

outward deflection (flexural buckling) is accompanied by twisting (torsion). This form of

instability is called lateral-torsional buckling (LTB).

Lateral–torsional buckling occurs when the distance between lateral brace points is large

enough that the beam fails by lateral, outward movement in combination with a twisting

action (Δ and θ, respectively, in Figure below).

It is important to note that:

- If λ ≤ λp, then the individual plate element can develop and

sustain σy (Fy) for large values of ε before local buckling

occurs.

- If λp ≤ λ ≤ λr , then the individual plate element can develop

σy but cannot sustain it before local buckling occurs.

- If λr ≤ λ, then elastic local buckling of the individual plate

element occurs.


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Lateral torsional buckling may be prevented through the following provisions:

1.Lateral supports at intermediate points in addition to lateral supports at the vertical

supports

2.Using torsionally strong sections (for example, box sections)

3. I-sections with relatively wide flanges: Beams with wider flanges are less susceptible to

lateral–torsional buckling because the wider flanges provide more resistance to lateral

displacement

In general, adequate restraint against lateral–torsional buckling is accomplished by the

addition of a brace or similar restraint somewhere between the centroid of the member and

the compression flange. For simple-span beams supporting normal gravity loads, the top

flange is the compression flange, but the bottom flange could be in compression for

continuous beams or beams in moment frames.

Lateral–torsional buckling can be controlled in several ways, but it is usually dependant on

the actual construction details used. Thus Lateral-torsional buckling can be prevented by

bracing the beam against twisting at sufficiently close intervals. This can be accomplished

with either of two types of stability bracing: lateral bracing, and torsional bracing.

Torsional bracing prevents twist directly; it can be either nodal or continuous, and it can

take the form of either cross frames or diaphragms.

Lateral bracing, which prevents lateral translation, should

be applied as close to the compression flange as possible.

Like beams with a metal deck oriented perpendicular to

the beam span are considered fully braced


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Lateral-torsional buckling is fundamentally similar to the flexural buckling or flexural-

torsional buckling of a column subjected to axial loading. The differences are that lateral-

torsional buckling is caused by flexural loading (M), and the buckling deformations are

coupled in the lateral and torsional directions. Also there is one very important difference.

For a column, the axial load causing buckling remains constant along the length. But, for a

beam, usually the lateral-torsional buckling causing bending moment M(x) varies along the

unbraced length.

If a beam can be counted on to remain stable up to the fully plastic condition, the nominal

moment strength can be taken as the plastic moment capacity; that is,

Mn= Mp

Otherwise due to local or lateral bucking, Mn will be less than Mp. Thus the nominal bending

strength, Mn , is a function of the following:

1. Lateral–torsional buckling (LTB),

2. Flange local buckling (FLB), and

3. Web local buckling (WLB).

8- BENDING DESIGN STRENGTH OF COMPACT SHAPES

A beam can fail by reaching Mp and becoming fully plastic, or it can fail by:

1. lateral-torsional buckling (LTB), either elastically or inelastically;

2. flange local buckling (FLB), elastically or inelastically; or

3. web local buckling (WLB), elastically or inelastically.

If the maximum bending stress is less than the proportional limit when buckling occurs, the

failure is said to be elastic. Otherwise, it is inelastic.

For convenience, first categorize beams as compact, noncompact, or slender, and then

determine the moment resistance based on the degree of lateral support.

The compact shapes, defined as those whose webs are continuously connected to the

flanges and that satisfy the following width-to-thickness ratio requirements for the flange

and the web: ywyf

f

F

E76.3

t

hand

F

E38.0

t2

b

The web criterion is met by all standard I and C shapes listed in the Manual for Fy≤ 65 ksi;

therefore, in most cases only the flange ratio needs to be checked (note that built-up welded


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I shapes can have noncompact or slender webs). Most shapes will also satisfy the flange

requirement and will therefore be classified as compact. The noncompact shapes are

identified in the dimensions and properties table with a footnote (footnote f).

The moment strength of compact shapes is a function of the unbraced length, Lb, defined as

the distance between points of lateral support, or bracing indicated as “×”. The relationship

between the nominal strength, Mn , and the unbraced length is shown in Figure. If the

unbraced length is no greater than Lp , the beam is considered to have full lateral support,

and Mn = Mp . If Lb is greater than Lp but less than or equal to Lr, the strength is based on

inelastic LTB. If Lb is greater than Lr, the strength is based on elastic LTB.

y

ypF

Er76.1L

2

oxy

oxy

tsrcJ

hS

E

F7.076.611

hS

cJ

F7.0

Er95.1L

2/1

x

wy

tsS

CIr

Lb= Laterally unsupported length

Lp = Plastic length: maximum unbraced length at

which the nominal bending strength equals the

plastic moment capacity and at which inelastic

lateral–torsional buckling occurs.

Lr = unbraced length at which elastic lateral–

torsional buckling occurs.


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ShapesChannelFor

C

I

2

h

ShapesIsymmerticdoublyFor0.1

c

w

yo

where

J = Torsional constant,

Cw = Warping constant, and

ho = Distance between flange centroids = d -tf

In the above equation, the term 0.7Fy Sx is also referred to as Mr, which corresponds to the

limiting buckling moment and is the transition point between inelastic and elastic lateral–

torsional buckling, namely the moment corresponding to first yield is:

xyr SF7.0M

9- DETERMINATION OF NOMINAL FLEXURAL STRENGTH FOR COMPACT SHAPES

The nominal bending strength for compact I and C-shaped sections can be summarized

as follows:

pnpb MMLLFor1

p

pr

pb

xyppbnrbp MLL

LL)SF7.0M(MCMLLLFor2

2

ts

b

ox

2

tsb

2

bcrpxcrnrb

r

L

hS

cJ078.01

)r/L(

ECFwhereMSFMLLFor3

where : Cb = Moment gradient factor

If the moment within the unbraced length Lb is uniform (constant), there is no moment

gradient and Cb = 1.0. If there is a moment gradient, the value of Cb is given by:

b

CBAmax

maxb

bb

b

LwithmomentorminfuNonForM3M4M3M5.2

M5.12C

Lwithinmoment)ttancons(uniformFor0.1C

C

where

Mmax= absolute value of the maximum moment within the unbraced length (including the

end points of the unbraced length)


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MA = absolute value of the moment at the quarter point of the unbraced length

MB = absolute value of the moment at the midpoint of the unbraced length

MC = absolute value of the moment at the three-quarter point of the unbraced

length

Cb for several common cases of loading and lateral support can be found in Part 3 of the AISC

Manual (example is shown in figure below). A value of 1.0 is always conservative, regardless

of beam configuration or loading, but in some cases it may be excessively conservative.


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Example 3: The beam shown in Figure is a W16*31 of A992 steel. It supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange. The service dead load is 450 lb/ft. This load is superimposed on the beam; it does not include the weight of the beam itself. The service live load is 550 lb/ft. Does this beam have adequate moment strength?

Solution

ft.k1648

30*457.1

8

LwM

ft/k457.155.0*6.1481.0*2.1w6.1w2.1w

ft/k55.0ft/lb550w

ft/k481.0ft/lb48131450w

selfweightincludingnotft/lb450w

ksi50F

22u

u

LDu

L

D

D

y

From AISC Manual, for W16*31: bf =5.53 in tf =0.44 in , bf /2 tf = 6.28, Zx =54 in3

- Determine whether the shape is compact, noncompact, or slender:

)Manualinshapesallforcompactisweb(

compactiswebTheF

E76.3

t

h

compactisFlange28.615.950

2900038.0

F

E38.0

28.6Manualfromdirectlyor28.644.0*2

53.5

t2

b

yw

y

f

f

Thus the section is classified as Compact Section

continuous lateral support 0Lb

pnpb MMLLandshapeCompactFor

ft.kips225in.kips27000.54*50ZFMM xypn

Design strength = ft.kips203225*9.0Mnb

OKft.k164Mft.kips203M unb

Thus W16*31 is satisfactory

Example 4: Determine the flexural strength of a W14* 68 of A992 steel subject to: a. Continuous lateral support. b. An unbraced length of 20 ft with Cb= 1.0. c. An unbraced length of 30 ft with Cb= 1.0.

Solution

From AISC Manual, for W14*68: bf /2 tf = 6.97, Zx =115 in3

compactiswebTheF

E76.3

t

h

compactisFlange28.615.9F

E38.097.6

t2

b

yw

yf

f

Thus the section is classified as Compact Section

a- Continuous lateral support



ft.kips2.479in.kips5750115*50ZFMM xypn

Design strength = ft.kips4312.479*9.0Mnb

b- An unbraced length of 20 ft with Cb= 1.0. ft20Lb , 0.1Cb

From AISC Manual, for W14*68: ry = 2.46 in, Sx =103 in3, Iy = 121 in4, Cw=5380 in6 , rts = 2.8 in tf = 0.72 in, d = 14.0 in, ho = 13.3 in, J=3.01 in4

ft692.8in3.10450

2900046.2*76.1

F

Er76.1L

yyp

2

oxy

oxytsr

cJ

hS

E

F7.076.611

hS

cJ

F7.0

Er95.1L

2

2/12/1

x

wy

ts in833.7103

5380*121

S

CIr

in28.1372.00.14tdh fo

ShapesIsymmerticdoublyFor0.1c

002201.028.13*103

0.1*01.3

hS

cJ

ox

571.82850*7.0

29000

F7.0

E

y


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Example 4: cont.

2

oxy

oxytsr

cJ

hS

E

F7.076.611

hS

cJ

F7.0

Er95.1L

ft28.29in3.351002201.0

1

571.828

176.611

*002201.0*571.828*799.2*95.1L

2

r

ppr

pb

xyppbn

rbp

MLL

LL)SF7.0M(MCM

28.29L20L692.8L

OK2.479Mft.k0.381in.k4572

692.828.29

692.820)103*50*7.05750(57500.1M

p

n

Design strength = ft.kips343381*9.0Mnb

c- An unbraced length of 30 ft with Cb= 1.0. ft30Lb , 0.1Cb

ft28.29Lr

2

ts

b

ox2

tsb

2b

cr

pxcrnrb

r

L

hS

cJ078.01

)r/L(

ECFwhere

MSFMLL

ksi9.33

799.2

12*30

28.13*103

0.1*01.3078.01

)799.2/12*30(

29000**0.1F

2

2

2

cr

OKft.k2.479Mft.k0.291

ft.k0.291in.k3492103*90.33SFM

p

xcrn


Example 5: Determine Cb for a uniformly loaded, simply supported W shape with lateral support at its ends only.

Solution

For non-uniform moment within Lb

CBAmax

maxb

M3M4M3M5.2

M5.12C

8

LwMM

2

maxB

Also because of symmetry, the moment at the quarter point equals the moment at the three-

quarter point. MA= MC

2CA Lw

32

3

8

L*

4

Lw

4

L*

2

LwMM

14.1

32

3*3

8

1*4

32

3*3

8

1*5.2

8

1*5.12

Cb

Example 6: Determine Cb for the beam below.

L

Cb=1.67 Cb=1.0 Cb=1.67


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10. BENDING STRENGTH OF NONCOMPACT SHAPES

As previously noted, most standard W, M, S, and C shapes are compact. A few are

noncompact because of the flange width-to-thickness ratio, but none are slender.

In general, a noncompact beam may fail by lateral-torsional buckling, flange local buckling,

or web local buckling. Any of these types of failure can be in either the elastic range or the

inelastic range. The strength corresponding to each of these three limit states must be

computed, and the smallest value will control.

The webs of all hot-rolled shapes in the Manual are compact, so the noncompact shapes are

subject only to the limit states of lateral-torsional buckling LTB and flange local buckling FLB.

Built-up welded shapes, however, can have noncompact or slender webs as well as

noncompact or slender flanges.

From AISC, , the flange is noncompact ifrp and buckling will be inelastic, where

f

f

t2

b

y

pF

E38.0

y

rF

E0.1

11. Determination of Nominal Flexural Strength

If the shape is noncompact because of the flange, the nominal strength will be the smaller of

the strengths corresponding to flange local buckling FLB and lateral-torsional buckling LTB.

a- Flange Local Buckling FLB:

pr

p

xyppnrp

pnp

)SF7.0M(MMcompactnonisflangetheIf

MMFLBnoisthereIf

b. Lateral-Torsional Buckling LTB:

pnpb MMLTBnoisthereLLFor1


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4

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4

p

pr

pb

xyppbnrbp MLL

LL)SF7.0M(MCMLTBInelasticLLLFor2

2

ts

b

ox

2

tsb

2

bcrpxcrnrb

r

L

hS

cJ078.01

)r/L(

ECFwhereMSFMLTBelasticLLFor3

c- Use smaller value from FLB and LTB

Note: For non-compact shapes, the value of Lp, must be computed from the equation :

y

ypF

Er76.1L

Example 7: A simply supported beam with a span length of 45 feet is laterally supported at its ends and is subjected to the following service loads:

Dead load = 400 lb/ft (including the weight of the beam)

Live load = 1000 lb/ft

If Fy = 50 ksi, is a W14 × 90 adequate?

Solution

ft.k5278

45*080.2

8

LwM

ft/k080.2000.1*6.1400.0*2.1w6.1w2.1w

22u

u

LDu

- Determine whether the shape is compact, noncompact, or slender:

From AISC Manual, for W14*90:

bf /2 tf = 10.2, Zx =157 in3 , Sx =143 in3

noncompactFlange2.1015.950

2900038.0

F

E38.0

y

2.10t2

b

f

f

15.9F

E38.0

yp

1.2450

290000.1

F

E0.1

yr

compactnonisshapetherp

a- Flange Local Buckling Capacity, FLB:

in.kips7850157*50ZFM xyp

pr

p

xyppn )SF7.0M(MM

ft.kips5.637in.k7650

15.91.24

15.92.10)143*507.07850(7850Mn

b. Lateral-Torsional Buckling Capacity LTB:

ft45Lb

From AISC Zx Table for W14*68: Lp = 15.1 ft Lr = 42.5 ft

LTBElasticLLL rbp

pxcrn MSFM

From AISC Manual, for W14*68: Iy = 362 in4, Cw=16000 in6 , rts = 4.11 in, ho = 13.3 in, J=4.06 in4

2

ts

b

ox2

tsb

2b

crr

L

hS

cJ078.01

)r/L(

ECF

ShapesIsymmerticdoublyFor0.1c

For simply supported beam with uniformly distributed load 14.1Cb

ksi2.37

11.4

12*45

3.13*143

0.1*06.4078.01

)11.4/12*45(

29000**14.1F

2

2

2

cr

OK7850Min.k5320

ft.k33.443in.k5320143*2.37SFM

p

xcrn

33.443)LTB(M5.637)FLB(M nn

ft.kips33.443Mn


NG527M399M unb

Thus the beam does not have adequate moment strength.


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12. SHEAR STRENGTH

In the design process for steel beams, shear rarely controls the design; therefore, most

beams need to be designed only for bending and deflection. Special loading conditions, such

as heavy concentrated loads or heavy loads on a short span beam, might cause shear to

control the design of beams.

From mechanics of materials, the general formula for shear stress in a beam is

where

bI

QVfv

f v = shear stress at the point under consideration,

V = vertical shear at a point along the beam under consideration,

Q = first moment, about the neutral axis, of the area of the cross section between the point

of interest and the top or bottom of the cross section,

I = moment of inertia about the neutral axis, and

b = thickness of the section at the point under consideration.

the AISC specification allows the design for shear to be based on an approximate or average

shear stress distribution as shown in Figure, where the shear stress is concentrated only in

the vertical section of the beam, for which the aspect ratio between the beam depth, d, and

the web thickness, tw, is generally high. In the AISC specification, the shear yield stress is

taken as 60% of the yield stress, Fy. The design shear strength is defined as

yw

v F6.0A

Vf

where Aw = area of the web.


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13. AISC Specification Requirements for Shear

For LRFD, the relationship between required and available strength is

nvu VV

where

Vu = maximum shear based on the controlling combination of factored loads

v = resistance factor for shear

The design shear strength is defined as

vwyn CAF6.0V

where

Aw = area of the web = d * t

d = overall depth of the beam

Cv= ratio of critical web stress to shear yield stress

The value of Cv depends on whether the limit state is web yielding, web inelastic buckling, or

web elastic buckling.

Case 1: For Hot-rolled I shapes, the limit state is shear yielding if

0.1and0.1CF

E24.2

t

hvv

yw

Most W shapes with F y ≤ 50 ksi fall into this category.

Case 1: For all other doubly or singly symmetric shapes

9.01 v

:foolowsasedminerdetisC2 v

0.1C

yinstabilitwebNoF

Ek10.1

t

hIfa

v

y

v

w

w

y

v

v

y

v

wy

v

t/h

F

Ek10.1

C

bucklinngwebinelasticisstateLimiteF

Ek37.1

t

h

F

Ek10.1Ifb


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4

y

2w

vv

y

v

w

Ft/h

Ek51.1C

bucklinngwebelasticisstateLimiteF

Ek37.1

t

hIfc

5kWhere v

Shear is rarely a problem in rolled steel beams; the usual practice is to design a beam for

flexure and then to check it for shear.

Values of nv V are given in several tables in Part 3 of the Manual, including the Zx table, so

computation of shear strength is unnecessary for hot-rolled shapes.

Example 8: A simply supported beam with a span length of 45 feet is laterally supported at its ends and is subjected to the following service loads:

Dead load = 400 lb/ft (including the weight of the beam)


If Fy = 50 ksi, check the W14 × 90 for shear?

Solution

ft/k080.2000.1*6.1400.0*2.1

w6.1w2.1w LDu

From AISC Manual, for W14*90: h / tw = 25.9, tw=0.44

d=14

Aw = d * tw = 14 * 0.44 = 6.160

0.5450

2900024.2

0.1and0.1CF

E24.2

t

hvv

yw

kips8.1840.1*160.6*50*6.0

CAF6.0V vwyn

kips1858.184*0.1Vnv

OK

kips185kips8.462

45*080.2

2

LwV u

u


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14. DEFLECTION

In addition to designing for bending and shear, beams also need to be checked for

serviceability. There are two main serviceability requirements: deflection and floor

vibrations.

For a beam, being serviceable usually means that the deformations, primarily the vertical

sag, or deflection, must be limited. For beams, deflections must be limited such that the

occupants of the structure perceive that the structure is safe.

Excessive deflection is usually an indication of a very flexible beam, which can lead to

problems with vibrations. The deflection itself can cause problems if elements attached to

the beam can be damaged by small distortions. In addition, users of the structure may view

large deflections negatively and wrongly assume that the structure is unsafe.

For the common case of a simply supported, uniformly loaded beam such as that in Figure

below, the maximum vertical deflection is:

IE

Lw

384

5 4

Deflection formulas for a variety of beams and loading conditions can be found in Part 3, of

the AISC Manual. For more unusual situations, standard analytical methods such as the

method of virtual work may be used.

The basic deflection limits are

summarized in Table below. Note that

only service level loads are used for

serviceability considerations. The limits

shown due to dead load do not apply to

steel beams, because the dead load

deflection is usually compensated for by

some means, such as cambering.

Example 9: Is the beam W18*35 shown below satisfactory with respect to deflection requirements.

wD = 500 lb/ft and wL = 550 lb/ft

Solution

From AISC Manual, for W18*35: Ix= 510 in4

in616.0510*29000

)12*30(*12/5.0

384

5

IE

Lw

384

544

DD

in678.0510*29000

)12*30(*12/55.0

384

5

IE

Lw

384

544

LL

in294.1678.0616.0LD

The maximum permissible deflection is

NGIN294.1in0.1360

12*30

360

L

Thus the beam not satisfies the deflection criterion.


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15.DESIGN TABLES OF BEAMS

1- AISCM, Tables 3-2 through 3-5 can be used to select the most economical beam based on

section properties. AISCM, Table 3-2 lists the plastic section modulus, Zx, for a given series of

shapes, with the most economical in one series at the top of the list in bold font. The most

economical shapes for Ix, , Zy, , and Iy are provided in AISCM, Tables 3-3, 3-4, and 3-5,

respectively.

2- AISCM, Table 3-6 provides a useful summary of the beam design parameters for W-

shapes. The lower part of the table provides values for pM , rM nV , Lp , and Lr for

any given shape. The upper portion of the table provides the maximum possible load that a

beam may support based on either shear or bending strength. When the unbraced length is

between Lp and Lr, the design bending strength is

)LL(BFMM pbpbnb


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3- AISCM, Tables 3-10 and 3-11 gives the design bending strength nb M for sections

normally used as beams are given in AISC (W-shapes and C-shapes) and plotted for wide

range of unbraced length in the plastic and inelastic and elastic ranges. They are cover

almost all of the unbraced lengths encountered in practice and they are plotted for a

moment gradient factor of Cb = 1.0, which is conservative for all cases, and yield strengths of

Fy= 50 ksi for W-shapes and Fy= 36 ksi for C-shapes.

For beams with Cb greater than 1.0, multiply the moment capacity calculated using these

tables by the Cb value to obtain the actual design moment capacity of the beam for design

moment. Note that pb MC must always be less than pM .

b

uu

nbbnbb

C

MMeffective

M*CMeffective0.1CIfchartfrom


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For each of the shapes, Lp is indicated with solid circle (●), while Lr is shown with a hollow

circle (⃝).

In the beam design tables, the sections that appear in bold font are the lightest and,

therefore, the most economical sections available for a given group of section shapes; these

sections should be used especially for small unbraced lengths where possible.

First proceed up from the bottom of the chart for an Lb until intersect a horizontal line from

the nb M column. Any section to the right and above this intersection point (↗) will have a

greater unbraced length and a greater design moment capacity. When moving up and to the

right (↗), if dash lines faced, the dashed lines indicate that the sections will provide the

necessary moment capacities, but are in an uneconomical range, thus proceed further up

and to the right (↗), the first solid line encountered will represent the lightest satisfactory

section.


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Example 10: Solve example 3 by AISCM Tables

The beam shown in Figure is a W16*31 of A992 steel. It supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange. The service dead load is 450 lb/ft. This load is superimposed on the beam; it does not include the weight of the beam itself. The service live load is 550 lb/ft. Does this beam have adequate moment strength, span =30ft?

Solution

ft.k1648

30*457.1

8

LwM

ft/k457.155.0*6.1)031.045.0(*2.1w6.1w2.1w

22u

u

LDu


- From Table 3-10 AISC Manual for W16*31 and Lb=0, Fy=50 ksi:

OKft.k164Mft.kips203M unb

Thus W16*31 is satisfactory

Example 11: Solve Example 4 by AISCM Tables Determine the flexural strength of a W14* 68 of A992 steel subject to: a. Continuous lateral support. b. An unbraced length of 20 ft with Cb= 1.0. c. An unbraced length of 30 ft with Cb= 1.0. Solution

- For W14*68 and Fy=50 ksi:

a- Lb=0

From Table 3-10 AISC Manual : ft.kips431Mnb

Or from Table 3-6 or 3.2 for W14*68 : ft.kips431MM pbnb

b- Lb=20 ft with Cb= 1.0. From Table 3-10 AISC Manual : ft.kips343Mnb

Or from Table 3-6 or 3.2 for W14*68 : 431Mpb

270Mrb 3.29L,69.8L rp 81.7BF

43134369.83.29

69.820)270431(4310.1

MLL

LL)MM(MCM

LLL

ppr

pb

rbpbpbbnb

rbp

343)69.820(81.7431

)LL(BFMMOR pbpbnb

c- Lb=30 ft with Cb= 1.0. From Table 3-10 AISC Manual : ft.kips9.261Mnb

Example12: Solve example7 by AISCM Tables

A simply supported beam with a span length of 45 feet is laterally supported at its ends and is subjected to the following service loads:

Dead load = 400 lb/ft (including the weight of the

beam). Live load = 1000 lb/ft

If Fy = 50 ksi, is a W14 × 90 adequate?

Solution

ft.k5278

45*080.2

8

LwM

ft/k080.2000.1*6.1400.0*2.1w

22u

u

u

- From Table 3-10 AISC Manual for W14*90 and Lb=45, Fy=50 ksi:

Maximum unbraced length for W14*90 is 33 ft

thus for Lb=45 ---- NG

Or from Table 3-6 maximum span for W14*90

=35ft thus for Lb=45 ---- NG

Example13: Solve example8 by AISCM Tables

A simply supported beam with a span length of 45 feet is laterally supported at its ends and is subjected to the following service loads:Dead load = 400 lb/ft (including the weight of the beam)


If Fy = 50 ksi, check the W14 × 90 for shear?

Solution

ft/k080.2000.1*6.1400.0*2.1

From Table 3-6 or 3.2 for W14*90: kips185Vnv

OKkips185kips8.462

45*080.2

2

LwV u

u


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16. DESIGN OF BEAMS

Beam design entails the selection of a cross-sectional shape that will have enough strength

in bending and adequate stiffness for serviceability. Shear typically does not control, but it

should be checked as well.

The design process can be outlined as follows.

1. Determine the service and factored loads on the beam. Service loads are used for

deflection calculations and factored loads are used for strength design. The weight of the

beam would be unknown at this stage, but the self-weight can be initially estimated and is

usually comparatively small enough not to affect the design.

2. Determine the factored shear and moments on the beam.

3. Select a shape that satisfies this strength requirement. This can be done in one of two

ways:

a. For shapes listed in the AISC beam design tables, select the most economical beam to

support the factored moment. Then check deflection and shear for the selected shape.

b. For shapes not listed in the AISC beam design tables, assume a shape, compute the

available strength, and compare it with the required strength. Revise if necessary.

4. Check the deflection, if deflection exceed limits redesign the section and select new

section based on Ix Tables. Firstly find required Ix from deflection formula, then enter to Ix

Tables and select Ix > Ix required.

5. Check the shear strength.


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Example 13: Select a standard hot-rolled shape of A992 steel for a 30ft simply supported beam. The beam has continuous lateral support and must support a uniform service live load of 4.5 kips/ft. The maximum permissible live load deflection is L=240.

Solution

Assume selfweight of beam = 0.1 kips/ft

ft.k5.8238

30*32.7

8

LwM

ft/k32.75.4*6.11.0*2.1w6.1w2.1w

22u

u

LDu


Assume compact section

yxpnpb FZMMLLandshapeCompactFor

3

yb

u

x

uyxbunb

in6.21950*9.0

12*5.823

F

MZ

MFZMM

Enter to Zx table and select section W 24*84 of Zx=224 > 219.5 OK Selfweight of beam 0.084 < 0.1 assumed OK, no need recalculation of Zx. - Check live load deflection

in19.12370*29000

)12*30(*12/5.4

384

5

IE

Lw

384

544

LL


OKin19.1in5.1240

12*30

240

L

-Check for local buckling:

compactiswebTheF

E76.3

t

h

compactisFlange15.9F

E38.086.5

t2

b

yw

yf

f

Thus the section is compact as assumed ----OK - Check Shear

From Zx Table for W24*84: kips340Vnv

OKkips340kips8.1092

30*32.7

2

LwV u

u

Use W24*84

Example 14: Design a 30 ft span simply supported beam subjected to uniformly distributed dead load of 450 lbs/ft. and a uniformly distributed live load of 550 lbs/ft. The dead load does not include the self-weight of the beam. The beam has continuous lateral support. The maximum permissible load deflection is L=360. Fy = 50 ksi.

Solution


ft.k25.1738

30*54.1

8

LwM

ft/k54.155.0*6.1)45.01.0(*2.1w6.1w2.1w

22u

u

LDu


Assume compact section


ubb MM

Enter to Zx table with 25.173Mbb and select

section W 14*30 of OK25.173177Mbb

Selfweight of beam 0.030 < 0.1 assumed OK, no need recalculation of bb M .

- Check service load deflection

in296.2301*29000

)12*30(*12/1.1

384

5

IE

Lw

384

544

LLD


NGin19.1in0.1360

12*30

360

L

Redesign with service-load deflection as design

criteria

3.691Irequired0.1I*29000

)12*30(*12/1.1

384

5

360

L

IE

Lw

384

5

xx

4

4L

Enter to Ix table with 3.691Ix and select section

W 21*44 of OK3.691843Ix

Ok360

Lin82.0

843*29000

)12*30(*12/1.1

384

5

IE

Lw

384

5 44L

LD

-Check for local buckling: compact as assumed - Check Shear:From Zx Table for W21*44:

kips217Vnv

OKkips217kips1.232

30*54.1

2

LwV u

u

Use W24*84


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Example 13: Design the beam shown below. The concentrated live loads acting on the beam are shown in the Figure. The beam is laterally supported at the load and reaction points. Do not check deflection.

Solution


ft/k12.00*6.11.0*2.1w6.1w2.1w

kips4830*6.10*2.1P

LDu

u

If the weight of the beam is neglected, The bending moment diagram: From AISC Table 3-23

'28L,'10b,'12a

6.44)baL(L

PR1

4.51)baL(L

PR2

bRM,aRMM 221max1

Span Lb,ft Cb Mu, k.ft Mueffective = Mu/Cb

AB 12 1.67 535.2 320.48

BC 8 1.0 535.2 535.2

CD 10 1.67 514 307.78

Assume that span BC is the controlling span because it has the largest Mu/Cb although the corresponding Lb is the smallest. Enter AISC Table 3-10 with Lb = 8' and Mu=535.2 and Select W21*68 which is first solid line with ɸbMn =570. Check the selected section for spans AB, BC, and CD

Span Lb

,ft

ɸbMn from chart

Cb Cb*ɸbMn Limit ɸbMn

AB 12 495 1.67 826.7 600

BC 8 570 1.0 570

CD 10 531 1.67 886.8 600

ft.k7.68

28*068.0MwtodueMoment

2

uu

Thus, for span AB, ɸbMn=600 >Mu+6.7 → OK for span BC, ɸbMn=570> Mu +6.7 → OK For span CD, ɸbMn=600> Mu+6.7 → OK

-Check for local buckling: Section is compact (there is no footnote in the dimensions and properties indicate that it isnon-compact) - Check Shear: From Table for W21*68: kips273Vnv

OKkips273kips4.51RV 2u

Use W21*68

Example 14: Design the simply-supported beam shown below. The uniformly distributed dead load is equal to 1 kips/ft. and the uniformly distributed live load is equal to 2 kips/ft. A concentrated live load equal to 10 kips acts at the mid-span. Lateral supports are provided at the end reactions and at the mid-span.

Solution Assume selfweight of beam = 0.1 kips/ft

kips1610*6.1P

ft/k52.42*6.1)11.0(*2.1w6.1w2.1w

u

LDu

Since this is a symmetric problem, need to consider only span AB, Lb=12 ft.

- Calculate Cb : From AISC Table 3-23

2

uu

x26.2x24.622

x16)x24(

2

x52.4

2

xP)xL(

2

xw)x(Mu

'12Lalongmoment.max44.421)ft12x(MuM

'12Lalongintpoquarterthree1.377)ft9x(MuM

'12Lalongintpohalf08.292)ft6x(MuM

'12Lalongintpoquarterft.k38.166)ft3x(MuM

bmax

bA

bB

bA

37.1M3M4M3M5.2

M5.12C

CBAmax

maxb

Span Lb,ft Cb Mu, k.ft Mueffective = Mu/Cb

AB 12 1.37 421.44 421.44/1.37=307.62

Enter AISC Table 3-10 with Lb = 12' and Mu=307.62 and Select W21*48 which is first solid line with ɸbMn =311

Check the selected section

Span Lb

,ft

ɸbMn from chart

Cb Cb*ɸbMn Limit ɸbMb

AB 12 311 1.37 426.07 398

Thus, for span AB, ɸbMn=398 <Mu=421.44 → NG

Redesign the section - Select the next section with greater capacity than W21 x 48. Select W18*55 with ɸbMn =342 k.ft Check the selected section

Span Lb

,ft

ɸbMn from chart


AB 12 342 1.37 468.54 420

30 kips 30 kips

Cb =1.0 assume

d Cb =1.67

Cb =1.67


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Thus, for span AB, ɸbMn=420 <Mu=421.44 → NG

Redesign the section - Select the next section W21*55 with ɸbMn =376 k.ft Check the selected section

Span Lb

,ft

ɸbMn from chart


AB 12 376 1.37 515.12 473

Thus, for span AB, ɸbMn=473 >Mu=421.44 → OK -Check for local buckling: Section is compact (there is no footnote in the dimensions and properties indicate that it is non-compact). - Check Shear: From Table for W21*55: kips234Vnv

OKkips234kips24.622

16

2

24*52.4Vu

Use W21*55

Example 14: The beam shown in Figure must support

two concentrated live loads of 20 kips each at the quarter points. The maximum live load deflection must not exceed L/240. Lateral support is provided at the ends of the beam. Use A992 steel and select a W shape. Solution Assume selfweight of beam = 0.1 kips/ft

ft/k12.01.0*2.1w

kips3220*6.1P

u

u

If the weight of the beam is neglected, The bending moment diagram: From AISC Table 3-23

'6a

ft.kipd1926*32aPMmax

0.1CMMMM bmaxCBA

Thus the charts to be used without modification.

Enter AISC Table 3-10 with Lb = 24' and Mu=192 and Select W12*53 which is first solid line with ɸbMn =208.5.

ft.k82.38

24*053.0MwtodueMoment

2

uu

ɸbMn =208.5 > Mu=192+3.82 ------OK

-Check for local buckling: Section is compact (there is no footnote in the dimensions and properties indicate that it is non-compact).

- Check live load deflection

in11.1])12*6(4)12*24(*3[425*29000*24

)12*6(*20

)a4L3(IE24

aP

22

22L


OKin11.1in2.1240

12*24

240

L

- Check Shear: From Table for W12*53:

kips125Vnv

OKkips125kips8.32322

24*)053.0(2.1Vu

Use W12*53


Lecture 4 .......

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17.BEAM BEARING PLATES

The function of the bearing plate is to distribute a concentrated load to the supporting

material. Two types of beam bearing plates are considered: one that transmits the beam

reaction to a support such as a concrete wall and one that transmits a load to the top flange

of a beam. In the case of bearing on concrete or masonry, the bearing plate is large enough

such that the bearing strength of the concrete or masonry is not exceeded. In the case of

bearing on another steel section, the bearing plate is designed to be large enough such that

local buckling does not occur in the supporting steel section as shown in Figure below.

The design of the bearing plate consists of three steps.

1. Determine dimension N so that web yielding and web crippling are prevented.

2. Determine dimension B so that the area B*N is sufficient to prevent the supporting

material (usually concrete) from being crushed in bearing.

3. Determine the thickness t so that the plate has sufficient bending strength.

For practical purposes, N is usually a minimum of 6 in. and B is usually greater than or equal

to the beam flange width, bf. This allows for reasonable construction tolerances in placing

the bearing plate and beam. Both the plate dimension B and N should be selected in

increments of 1 in., and the plate thickness, t , is usually selected in increments of 1⁄4 in.

The basic design checks for beam bearing are web yielding and web crippling in the beam,

plate bearing and plate bending in the plate, and bearing stress in the concrete or masonry.


Lecture 4 .......

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17.1Web Yielding

At the support and the points of concentrated loads, the web behaves like a short column,

and the beam must transfer compression from the wide flange to the narrow web. When

the magnitude of the reaction or concentrated load is excessive, high compressive stresses

may cause yielding at the junction of the web and the flange.

Web yielding is the crushing of a beam web subjected to compression stress due to a

concentrated load. This concentrated load could be an end reaction from a support , or it

could be a load delivered to the top flange by a column or another beam.

Yielding occurs when the compressive stress on a horizontal section through the web

reaches the yield point. When the load is transmitted through a plate, web yielding is

assumed to take place on the nearest section of width tw. In a rolled shape, this section will

be at the toe of the fillet, a distance k from the outside face of the flange.

The compression stress is assumed to be distributed on a ratio of 1:2.5 through the beam

flange and inner radius as shown below.

a- The area at the support subject to yielding is tw (2.5k + N). Multiplying this area by the

yield stress gives the nominal strength for web yielding at the support:

)Nk5.2(tFR wyn

b- At the interior load, the length of the section subject to yielding is 2(2.5k) + N= 5k + N and

the nominal strength is

)Nk5(tFR wyn

unwy RR where 0.1wy


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17.2 Web Crippling

Web crippling is buckling of the web caused by the compressive force delivered through the

flange.

a- For an interior load, the nominal strength for web crippling is

w

fy5.1

f

w2wn

t

tFE

t

t

d

N31t80.0R

b- For a load at or near the support (no greater than half the beam depth from the end),

the nominal strength is

w

fy5.1

f

w2wn

w

fy5.1

f

w2wn

n

t

tFE

t

t2.0

d

N41t40.0R2.0

d

NIf

t

tFE

t

t

d

N31t40.0R2.0

d

NIf

R

unwc RR where 75.0wc

17.3 Concrete Bearing Strength

The concrete support beam must resist the bearing load applied by the steel plate. The

bearing strength of the supporting concrete in crushing is:

1cc1

21ccpc Af7.1

A

AA)f85.0(P and 65.0c

A1 = Area of steel bearing = BN, and

A2 = Maximum area of the support geometrically similar and concentric with the

loaded area = (B + 2e)(N + 2e). Note that the dimension e is the minimum distance from

the edge of the plate to the edge of the concrete support.


Lecture 4 .......

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17.4 Plate Thickness

Once the length and width of the plate have been determined, the average bearing pressure

is treated as a uniform load on the bottom of the plate. The plate is treated as a cantilever of

span length n = (B – 2k)/2 and a width of N.

2

k2Bnwhere

FNB9.0

nR2t

y

2u

For the limit states of web yielding and web crippling, a pair of transverse stiffeners or a web

doubler plate is added to reinforce the beam section when the design strength is less than

the applied loads.

The design procedure for bearing plates can be summarized as follows:

1. Assume a value for the bearing plate length, N.

2. Check the beam for web yielding and web crippling for the assumed value of N;

adjust the value of N as required.

3. Determine the bearing plate width, B, such that the bearing plate area, A1= BN, is

sufficient to prevent crushing of the concrete or masonry support.

5. Determine the thickness, t, of the beam bearing plate so that the plate has adequate

strength in bending.

N

N


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Example 13: Check web yielding and web crippling for the beam shown in Figure. The steel is ASTM A572, grade 50.

Solution

From AISCM for W 18*50: k=0.972" , tw=0.355"

d=18" , tf=0.57"

N = 6" (From figure)

a- Check Web Yielding

OKkips100kips192

)6972.0*5(355.0*50*0.1

)Nk5(tFR wywynwy

b- Check Web Crippling

OKkips100kips172

355.0

57.0*FE

57.0

355.0

18

631355.0*80.0*75.0

t

tFE

t

t

d

N31t80.0R

y5.1

2

w

fy

5.1

f

w2wwcnwc

The W18 *50 beam is adequate for web yielding and web crippling.

Example 13: A W18 *50 beam is simply supported on 10-in.-thick concrete walls at both ends as shown in Figure. Design a beam bearing plate at the concrete wall supports assuming the following: – Beam span =20 ft. center-to-center of support - concrete strength = 4000 psi –wL = 2 kips/ft , – wD =1.5 kips/ft – ASTM A36 steel for the beam and bearing plate

Solution

kips1502

20*5

2

LwR

ft/k0.52*6.1)5.1(*2.1w6.1w2.1w

uu

LDu

From AISCM for W 18*50: k=0.972" , tw=0.355"

d=18" , tf=0.57"

Assume N=6" (recommended practical value)

a- Check Web Yielding

OKkips50Rkips107

)6972.0*5.2(355.0*36*0.1

)Nk5.2(tFR

u

wywynwy

b- Check Web Crippling

2.033.018

6

d

N

w

fy

5.1

f

w2wwcnwc

t

tFE

t

t2.0

d

N41t40.0R

OKkips50Rkips72

t

tFE

57.0

355.02.0

18

6*41355.0*40.0*75.0R

u

w

fy5.1

2nwc

-Determine dimension B from a consideration of

bearing strength. If conservatively assumed that

the full area of the support is used, the required

plate area A1

"8BSay

495.7BB495.7bB

"78.3N

AB

in62.22A

A)4*85.0(*65.050

RA)f85.0(P

minfmin

1

21

1

u1ccpc

- Determine Thickness

028.32

972.0*28

2

k2Bn

768.036*6*8*9.0

028.3*50*2

FNB9.0

nR2t

2

y

2u

Say t = 1"

Use Bearing Plate 6" * 8" * 1"

design of beams · beams can be further classified by the function that they serve. a girder is a...

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