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2017-2018

BASIC LECTURES

ON

STRUCTURAL STEEL

FOR

FOURTH STAGE

IN CIVIL ENGINEERING COLLEGE

Asst. Prof. Dr. Saad Khalaf Mohaisen

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FLEXURAL MEMBERS

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Flexural Members/Beams: Chapter F Part 16.1 page 44

This chapter applies to members subject to simple bending about one principal axis. For simple bending, the member is loaded in a plane parallel to a principal axis that passes through the shear center or is restrained against twisting at load points and supports.

Introduction:

They are defined as members acted upon primarily by transverse loading, often gravity dead and live load effects. Thus, flexural members in a structure may also be referred to as:

Girders

Joists

Purlins

Stringers

Girts

Lintels

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Type of beams for both horizontal & inclined members:

Horizontal members → B.M & V

Inclined members → B.M, V & P

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Typical open joist system Some type of flexural members

Section used for beams:

Among the steel shapes that are used as beam include:

W – Shapes, which normally prove to be the most economical beam sections, and they have largely replaced channels and S – Sections for beam usage.

Channels are sometimes used for beams subjected to light loads, such as purlins, and in places where clearances available require narrow

flanges. Another common type of beam section is the open web joist or bar joist. This type of section, which commonly used to support floor and roof slabs, is actually a light shop – fabricated parallel chord truss.

It is particularly economical for long spans and light loads.

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Subjects to be considered:

The basic design & analysis considers for beams include:- Bending, Shear, Deflection.

Structural beam behavior:

From the relation of un-braced length to moment, we may get the

followings:

Plastic behavior, full plastic moment.

Inelastic lateral torsional buckling behavior.

Elastic lateral torsional buckling behavior.

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Where:

Lb: is the lateral un-braced length.

Lr: is the value of un-braced length that the beam starts to behave

elastically.

Lp: is the value of un-braced length that plastic hinge should be

developed & the value of bending moment reach plastic moment (Mp).

If Lr < Lb → the beam will buckle elastically before the yield stress is

reached.

The beam is still stable up to fully plastic condition → Mn = Mp

If Mr < Mp → stability is not achieved due to lateral torsional buckling,

or flang & web local buckling.

The loading conditions and beam configuration will dictate which of the preceding design parameters controls the size of the beam.

When a beam is subjected to bending loads, the bending stress in the

extreme fiber is define as:-

Elastic Bending:

Mn = Fy Sx where Fy = specified minimum yield stress of the type of steel being used, ksi

(MPa). Sx = elastic section modulus about the x-axis, in3. (mm3).

Where:

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fb=Maximum bending stress, My=Yield moment,

Fy=Yield stress, M=Bending moment, c =Distance from the neutral axis to the extreme fiber, I =Moment of inertia, and S =Section modulus. The above formulation is based on the elastic behavior of the beam.

Stress Distribution for Bending Members

Plastic Moment: A plastic hinge occurs when the entire cross section of the beam is at its yield point, not just the extreme fiber. The moment at which a plastic hinge is developed in a beam is called the plastic moment and is defined as:-

Mp = Fy . Zx ……..F2-1 where Mp =plastic moment, which must be ≤ 1.5 My for homogenous cross-sections, and Zx =plastic section modulus.

Classification of Shapes

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AISC classified cross-sectional shapes as: Compact,

Non-compact, and Slender.

Depending on the values for the width-thickness ratios of the individual elements that form the shape, there are also two types of elements that are defined in the AISC Specification:

Stiffened elements, and Unstiffened elements. For I-shape, the ratio for the projection flange (an unstiffened

element) is ( /2 ), and the ratio for the web (a stiffened element) is

(ℎ/ ). The classification of shapes is found in Section B4 of the specification, “Local Buckling”, in Table B4-1. it can be summarized as follows:

=width-thickness ratio (b/t).

=upper limit for compact category.

=upper limit for non-compact category.

If ≤ and the flange is continuously connected to the web, the shape

is compact;

If < ≤ , the shape is non-compact; and

If > , the shape is slender.

It is important to note that:

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

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 λ

≤ λr then elastic local buckling of the individual plate element

occurs. 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).

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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). The definition of λ and the values for λp and λr for the individual elements of various cross-sections are given in Table B5.1 and shown graphically on page 16.1-183. For example,

The compact shapes defined as those whose webs are continuously connected to the flanges and that satisfy the following width-thickness ratio requirements for the flange and web:

&

If the beam is compact and has continuous lateral support, or if the unbraced length is very short (Lb≤Lp), the nominal moment strength, Mn, is the full plastic moment capacity of the shape, Mp.

Mn=Mp=Fy . Zx ………………………………. (AISC Equ. F2-1)

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Local buckling of beam section – Compact and Non-compact:

Mp, the plastic moment capacity for the steel shape, is calculated by

assuming a plastic stress distribution (+ or - σy) over the cross-

section.

The development of a plastic stress distribution over the cross-section

can be hindered by two different length effects:

(1) Local buckling of the individual plates (flanges and webs) of the

cross-section before they develop the compressive yield stress fy (бy).

(2) Lateral-torsional buckling of the unsupported length of the beam

member before the cross-section develops the plastic moment Mp.

Stress-strain response of plates subjected to axial compression and local buckling.

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Zones for beam design:

Figure shown below shows that beams have three distinct ranges, or zones, of behavior, depending on their lateral bracing situation.

More type of beams may be found in many structures as in:

Bridges

Girder bridges

Suspension bridges

Cable-stayed bridges

Truss bridges

Arch bridges

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Rigid-frame (Ramen) bridges

Composite deck slab bridges

I-shaped steel grating deck slabs

Monorails and new transportation systems Bending Strength of Compact Shapes:

The basic design strength equation for beams in bending is:

LRFD Ø = 0.9 Mu ≤ Øb Mn

Where:

= Ultimate Moment.

= Nominal bending strength.

∅ = Ultimate design bending strength.

The nominal bending strength, , is a function of the following:

1) Lateral – torsional buckling (LTB),

2) Flange local buckling (FLB), and

3) Web local buckling (WLB).

Flange local buckling and web local buckling are localized failure modes and are only of concern with shapes that have noncompact webs or flanges.

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 ), as shown in figure shown below:

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Local buckling of individual plate

Lateral torsional buckling behavior

Lateral-Torsional Buckling:

The laterally unsupported length of a beam-member can undergo lateral-torsional buckling due to the applied flexural loading (bending moment).

Lateral-torsional buckling is fundamentally similar to the flexural buckling or flexural-torsional buckling of a column subjected to axial loading.

- The similarity is that it is also a bifurcation-buckling type phenomenon. - 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.

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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.

- The worst situation is for beams subjected to uniform bending moment along the unbraced length. Why?

Lateral-torsional buckling – Uniform bending moment: Consider a beam that is simply-supported at the ends and subjected to four-point loading as shown below. The beam center-span is subjected to uniform bending moment M. Assume that lateral supports are provided at the load points.

- Lb = Laterally unsupported length. - If the laterally unbraced length Lb is less than or equal to a plastic

length Lp then lateral torsional buckling is not a problem and the beam will develop its plastic strength Mp.

- Lp = 1.76 ry

[For I members & channels (See Pg. 16.1-33)]

If Lb is greater than Lp then lateral torsional buckling will occur and the moment capacity of the beam will be reduced below the plastic strength Mp as shown below.

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Unbraced length

Moment capacity (Mn) versus unsupported length (Lb).

If Lb ≤ Lp, no LTB:

Mn = Mp = Fy Zx ≤ 1.5My .………………………………...AISC Eq. (F1-1)

If Lp < Lb ≤ Lr. Inelastic LTB.

…………AISC Eq. (F1-2)

Note: that Mn is a linear function of Lb.

If Lb > Lr, (slender member), elastic LTB.

………………………………………………AISC Eq. (F1-12)

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As shown in the above figure, the lateral-torsional buckling moment (Mn

=Mcr) is a function of the laterally unbraced length L

b and can be calculated

using the equation:

AISC Eq. (F1-13)

Where: M

n = moment capacity

Lb = laterally unsupported length.

Mcr

= critical lateral-torsional buckling moment.

E = 29000 ksi; G = 11,200 ksi

Iy = moment of inertia about minor or y-axis (in

4

)

J = torsional constant (in4

) from the AISC manual pages of tables 1.

Cw

= warping constant (in6

) from the AISC manual pages.

For more explanations:

This equation is valid for ELASTIC lateral torsional buckling only (like the Euler equation). That is it will work only as long as the cross-section is elastic and no portion of the cross-section has yielded.

As soon as any portion of the cross-section reaches the yield stress Fy,

the elastic lateral torsional buckling equation cannot be used. - L

r is the unbraced length that corresponds to a lateral-torsional

buckling moment Mr = S

x (F

y –10).

- Mr will cause yielding of the cross-section due to residual stresses.

When the unbraced length is less than Lr, then the elastic lateral

torsional buckling equation cannot be used.

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When the unbraced length (Lb) is less than Lr but more than the plastic length Lp, then the lateral-torsional buckling Mn is given by the

equation below:

If Lp ≤ Lb ≤ Lr, then

…AISC Eq. (F1-2)

- This is linear interpolation between (Lp, Mp) and (Lr, Mr),see the figure above.

Moment Capacity of beams subjected to non-uniform bending moments:

As mentioned previously, the case with uniform bending moment is worst for lateral torsional buckling.

For cases with non-uniform bending moment, the lateral torsional buckling moment is greater than that for the case with uniform

moment. The AISC specification says that:

- The lateral torsional buckling moment for non-uniform bending moment case = C

b x lateral torsional buckling moment for uniform moment case.

Cb is always greater than 1.0 for non-uniform bending moment.

Cb is equal to 1.0 for uniform bending moment.

- Sometimes, if you cannot calculate or figure out Cb, then it can be

conservatively assumed as 1.0.

…………..AISC Eq. (F1-3)

Where: Cb is a factor that takes into account the non-uniform bending moment distribution over an unbraced length Lb. M

max = Magnitude of maximum bending moment in L

b

Ma = Magnitude of bending moment at quarter point of L

b

Mb = Magnitude of bending moment at half point of L

b

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Mc = Magnitude of bending moment at three-quarter point of L

b

Rm = Section symmetry factor

Rm = 1.0 , for doubly symmetric members ( I-shapes), Rm = 1.0 , for singly symmetric sections in single-curvature bending,

, for single symmetric shapes subjected to reverse

curvature bending, and Iyc =Moment of inertia of the compression flange about the y-axis. Cb = 1.0, for cantilevers or overhangs where the free end is unbraced.

Some of values are given in Table 3-1 of the AISC (shown in the next

page).

If the bending moment is uniform, all moment values are the same giving . This is also true for a conservative design. Cb=1.0. The moment capacity Mn for the case of non-uniform bending moment

- Mn = C

b x M

n for the case of uniform bending moment ≤ M

p

- Important to note that the increased moment capacity for the non-uniform moment case cannot possibly be more than M

p.

- Therefore, if the calculated (Mn) value is greater than (Mp), then you

have to reduce it to (Mp)

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Moment capacity versus Lb for non-uniform moment case.

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Shear requirements of Beams: chapter G, 16.1 –page 67

Shear stresses are usually not a controlling factor in the design of beams, except for the following cases: 1) The beam is very short. 2) There are holes in the web of the beam. 3) The beam is subjected to a very heavy concentrated load near one of the supports. 4) The beam is coped. From the elementary mechanics of materials, the shear stress at any point can be found:

fv = shear stress at the point of interest V = vertical shear force at the section 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.

Shear in steel beam section Two methods of calculating shear strength are presented below. The method presented in Section G2 does not utilize the post buckling strength of the member (tension field action). The method presented in Section G3 utilizes tension field action. The design shear strength, ØvVn, and the allowable shear strength, Vn/Ωv ,

shall be determined as follows.

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LRFD

Фb = 0.9

1- Nominal Shear Strength: This section applies to webs of singly or doubly symmetric members and channels subject to shear in the plane of the web.

The nominal shear strength, Vn, of unstiffened or stiffened webs, according to the limit states of shear yielding and shear buckling, is

Vn = 0.6Fy Aw Cv ……………..(G2-1)

For LRFD Ø=1.0

Vu ≤ Øv Vn

(a) For webs of rolled I-shaped members with h/tw ≤ 2.24 : so

Cv = 1.0 ………………………..(G2-2) (b) For webs of all other doubly symmetric shapes and singly symmetric shapes and channels, except round HSS, the web shear coefficient, Cv, is determined as follows:

(i) For h/tw ≤ 1.10

Cv = 1.0 …………………….(G2-3)

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(ii) For 1.10 < h/tw 1.37

…………………………(G2-4)

(iii) For h/tw > 1.37

………………….(G2-5)

Where: Aw = the overall depth times the web thickness, dtw , in2 (mm2). The web plate buckling coefficient, kv, is determined as follows: (i) For unstiffened webs with h/tw < 260, kv = 5, except for the stem of tee shapes where kv = 1.2. (ii) For stiffened webs,

= 5 when a/h > 3.0 or a/h >

Where: a = clear distance between transverse stiffeners, in. (mm). h = for rolled shapes, the clear distance between flanges less the fillet or corner radii, in. (mm). = for built-up welded sections, the clear distance between flanges, in.

(mm). = for built-up bolted sections, the distance between fastener lines, in. (mm). = for tees, the overall depth, in. (mm).

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2- Transverse Stiffeners

Transverse stiffeners are not required where h/tw ≤ 2.46 , or where

the required shear strength is less than or equal to the available shear strength provided in accordance with Section G2.1 for kv = 5. Transverse stiffeners used to develop the available web shear strength, as provided in Section G2.1, shall have a moment of inertia about an axis in the web center for stiffener pairs or about the face in contact with the web plate for single stiffeners, which shall not be less than at3wj. Where j is:

j=

Transverse stiffeners are permitted to be stopped short of the tension flange, provided bearing is not needed to transmit a concentrated load or reaction. The weld by which transverse stiffeners are attached to the web shall be terminated not less than four times nor more than six times the web thickness from the near toe to the web-to-flange weld. When single stiffeners are used, they shall be attached to the compression flange, if it consists of a rectangular plate, to resist any uplift tendency due to torsion

in the flange. When lateral bracing is attached to a stiffener, or a pair of stiffeners, these, in turn, shall be connected to the compression flange to transmit 1 percent of the total flange force, unless the flange is composed only of angles. Bolts connecting stiffeners to the girder web shall be spaced not more than 12 in. (305 mm) on center. If intermittent fillet welds are used, the clear distance between welds shall not be more than (16tw) nor more than 10 in.(250 mm).

Tension field action: 1. Limits on the Use of Tension Field Action Consideration of tension field action is permitted for flanged members when the web plate is supported on all four sides by flanges or stiffeners. Consideration of tension field action is not permitted for: (a) End panels in all members with transverse stiffeners; (b) Members when a/h exceeds 3.0 or [260/(h/tw )]2;

(c) 2Aw /(Afc + Aft) > 2.5; or

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(d) h/bfc or h/bft > 6.0 Specification for

where Afc = area of compression flange, in2 (mm2) Aft = area of tension flange, in2 (mm2) bfc = width of compression flange, in. (mm) bft = width of tension flange, in. (mm) In these cases, the nominal shear strength, Vn, shall be determined according to the provisions of Section G2.

2- Nominal Shear Strength with Tension Field Action When tension field action is permitted according to Section G3.1, the nominal shear strength, Vn, with tension field action, according to the limit state of tension field yielding, shall be

(a) For h/tw ≤ 1.10

Vn = 0.6 Fy Aw …………………………………………………………………(G3-1)

(b) For h/tw > 1.10

Vn = 0.6Fy Aw

……………………………………(G3-2)

Where: kv and Cv are as defined in Section G2.1.

3- Transverse Stiffeners: Transverse stiffeners subject to tension field action shall meet the requirements of section G2.2 and the following limitations:

(1) (b/t)st ≤ 0.56

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(2) Ast >

ℎ

………………..(G3-3)

where (b/t)st = the width-thickness ratio of the stiffener (Fy)st = specified minimum yield stress of the stiffener material, ksi (MPa) Cv = coefficient defined in Section G2.1 Ds = 1.0 for stiffeners in pairs = 1.8 for single angle stiffeners = 2.4 for single plate stiffeners Vr = required shear strength at the location of the stiffener, kips (N) Vc = available shear strength; ØvVn (LRFD) or Vn/Ωv (ASD) with Vn as defined in Section G3.2, kips (N) NOTE: SEE THE TALES 3-23 IN PAGE 3-211 FOR USING MANY EXPRESSIONS TO FIND THE MOMENTS, SHEARS & DEFLECTIONS, WHERE AISC STABULATE SOME LOADING CASES FOR BEAMS.

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Deflection Requirments of Beams – Serviceability:

Steel beams are designed for the factored design loads. The moment capacity, i.e., the factored moment strength (Ø

bM

n) should be greater

than the moment (Mu) caused by the factored loads.

A serviceable structure is one that performs satisfactorily, not causing discomfort or perceptions of unsafety for the occupants or users of the structure.

- For a beam, being serviceable usually means that the deformations, primarily the vertical slag, or deflection, must be limited.

- The maximum deflection of the designed beam is checked at the

service-level loads. The deflection due to service-level loads must be less than the specified values.

The AISC Specification gives little guidance other than a statement in Chapter L, “Serviceability Design Considerations,” that deflections should be checked. Appropriate limits for deflection can be found from the governing building code for the region.

The following values of deflection are typical maximum allowable total (service dead load plus service live load) deflections.

- Plastered floor construction – = L/360

- Unplastered floor construction – = L/240

- Unplastered roof construction – = L/180

So:

where ℎ

Deflection of steel beam

NOTE: In calculating deflection, the loads should be used without factoring

in methods LRFD.

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EX: CHECK THE STRENGTH ADEQUACY OF OF THE FULL LATERALLY

SUPPORTED BEAM (W21x44) FOR THE LOADING CONDITION SHOWN.

ASSUME (A572G50) STEEL MATERIAL.

Wd= 1 k.ft, Wl= 3 k/ft W21x44

L=21ft

SOLUTION:

1- SPE. DIM., & PROPERTIES:

steel Fy Fu Sec w.t A Zx

44 lb/ft

94.5 in3

2- Strength:

LRFD

Wu = 1.2(1+0.044) + 1.6(3) = 6.05kip/ft

Mu = wu L2 / 8 = 333.5kips.ft Mn = Mpx = Fy Zx = 50 x 94.5 / 12 = 397.5 kips.ft Øb Mn = 0.9 x 397.5 = 358kips.ft > Mu O.K

Design strength is adequate

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EX: SELECT A WF SECTION FOR THE BEAM CASE SHOWN, USE A572G50

STEEL MATERIAL & ASSUME FULL LATERALLY SUPPORTED BEAM where

(Lb=0).

PL = 30K wd = 1.5k/ft

15ft 15ft

L=30ft

SOLUTION:

1- SPEC., DIM.,& PROPERTIES:

Steel Fy Fu PL PD L Lb

0.0

2- LOADING:

LRFD

Wu = 1.2(1.5)=1.8k/ft, Pu=1.6(30)=48kips Mu = wu x L2/8 + Pu x L /4 = 562.5k.ft

3- Select section from AISC Tables 3-2 @ PAGE 3-16 , Where Øb Mpx >

Mu & Mpx/ Ωb > Ma, such as W24x62

LRFD

ØbMpx = 574k.ft

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4- Check section:

LRFD

Wu = 1.2(1.5+0.062) = 1.874k/ft Pu + 1.6(30) = 48 kips Mu =570.8k.ft < ØbMpx = 574k.ft O.K

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EX: Determine the classification (compact or non compact) of a W18x35

and a W21x48 for A572G50 steel material. Check both the flange and the

web.

Solution:

1- Specification, dimensions & properties:

For W18x35

steel Fy Fu sec Bf/2tf h/tw

A572G50 50 65 WW18x35 7.06 53.5

For W21x48

steel Fy Fu sec Bf/2tf h/tw

A572G50 50 65 WW21x48 9.47 53.6

2- Check flange width thickness limitations, from Table 4-1B:

for W18x35:

λf=b/t= bf/2tf=7.06 <

, so flange of W18x35 is compact.

For W21x48:

Since λf=b/t =bf/tf = 9.47 >

, so

λf = 9.47 <

, so flange of W21x48 is non compact.

Web for both sections:

λh = h/tw=53.5 & 53.6 respectively <

, so web of both

sections are compact.

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Ex: Design a 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. Assume A992G50 steel

material.

WD=450lb/ft, WL=550lb/ft

30ft

Solution:

1- Spec., Dimensions & Properties:

Steel Fy Fu WD WL L

A992 50 65 0.45 0.55 30

2- Loading: Calculate the factored design loads (without self-weight).

LRFD

wU = 1.2 w

D + 1.6 w

L = 1.42 k/ft.

MU = w

u L

2

/ 8 = 1.42 x 302

/ 8 = 159.75 k.ft.

3- Select the lightest section from the AISC Manual design tables.

From page (3-18) of the AISC manual, select W16 x 26 of Fy=50 ksi with Ø

bM

p = 166.0 k.ft.

4- Check section with self-weight: Add self-weight of designed section and

check design.

Wsw = 0.026 k/ft , WD=0.45+0.026 = 0.476 k/ft.

Wu = 1.2 x 0.476 + 1.6 x 0.55 = 1.4512 k/ft.

Therefore, Mu = 1.4512 x 302 / 8 = 163.26 k.ft. < ØbMp of W16 x 26.

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5- Check deflection at service loads: w = 0.45 + 0.026 + 0.55 k/ft. = 1.026 k/ft. (Note: use loads without factoring).

δ = 5 w L4

/ (384 E Ix) = 5 x (1.026/12) x (30 x 12)

4

/ (384 x 29000 x

301) =2.142" max deflection limitations: For plastered, unplastered floor & unplastered roof construction.

L/360 L/240 L/180

1" 1.5" 2.0"

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δ = 2.142" > 1", 1.5" & 2" (This check is done for all limitations, because the question doesn’t

mention the corresponding supported materials). So the section is not good for all serviceability requirements.

6- Redesign with service-load deflection as design criteria: Let L /360 = 1.0 in. = 5 w L4/(384 E Ix) Therefore, Ix > 644.8 in4 Select the section from the moment of inertia selection tables in the AISC manual. (tables 3-3, See page 3-21):

Section Ix Zx Sx ØbMp

W21 x 44 843 358

Deflection at service load = δ = 0.765". < L/360 =1.0" - OK! Note that the serviceability design criteria controlled the design and the section. H.W: Check shear requirements:

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EX: Design the beam shown below. The concentrated live loads acting on the beam are shown below. The beam is laterally supported at the load

and reaction points.

X in beam means position of lateral support.

Solution: 1- Specification, Dimensions & Properties:

Steel Fy Fu PL1 PL2 L Lb1 Lb2 Lb3

2- Loadings:

LRFD

PL = 30 kips

Pu = 1.6 PL = 48 kips

wu = 1.2 x Wsw = 0.12 kips/ft.

The reactions and bending moment diagram for

the beam are shown below.

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3- Determine Lb, C

b, M

u, and M

u/C

b for all spans:

Span Lb

(ft.)

Cb M

u

(kip-ft.)

Mu/C

b

(kip-ft.)

AB 12 1.67 550.6 329.7

BC 8 1.0 (assume)

550.6 550.6

CD 10 1.67 524.0 313.8

NOTE: It is important to note that it is possible to have different Lb and Cb values for different laterally unsupported spans of the same beam.

4- Design the beam and check all laterally unsupported spans Assume that span BC is the controlling span because it has the largest M

u/C

b although the corresponding L

b is the smallest.

From the AISC-LRFD manual select W21 x 68 made from 50 ksi steel (page 3-16):

Sec ØMpx Lp Lr Ix

W21x68 600 6.36 18.7 1480

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Check the selected section for spans AB, BC, and CD Span

Use curve in table 3-10 page 3-119 (find Lb & section on table then read Ø

bM

n on section curve).

Span Lb

(ft.)

ØbM

n , for C

b

=1.0 From

tabl 3-10 page 3-119

Cb Ø

bM

n

for Cb value

col. 3 x col. 4

ØbM

p

limit

AB 12 495 1.67 826.65 600 kip-ft

BC 8 572 1.0 572.0

CD 10 531 1.67 886.77 600 kip-ft.

Thus, for span AB, Ø

bM

n = 600 kip-ft. > M

u - OK!

For span BC, ØbM

n = 572 kip-ft. > M

u - OK!

For span CD, ØbM

n = 600 kip-ft. > M

u - OK!

5- Check for local buckling:

For flange, λ = bf / 2t

f = 6.0; Corresponding λ

p =

= 9.192

Therefore, λ < λp - compact flange

For web, λ = h/tw

= 43.6; Corresponding λp =

= 90.55

Therefore, λ < λp - compact web

Compact section. - OK! Note; This example demonstrates the method for designing beams with several laterally unsupported spans with different L

b and C

b values.

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EX: Check the shear adequacy of the beam (W21x55) for the subjected loading condition shown below. Use A572G50 steel material.

WD=2k/ft, WL=4k/ft

21ft Solution:

1- Specification, dimentions & properties:

Steel Fy Fu Sec A D Tw Kdes

20.8 0.375 1.02

htw = d-2Kdes =20.8-2(1.02)=18.76

2- Loading:

LRFD

Wu=1.2WD+1.6WL=8.8k/ft

Vu=wu x L/2=88k

3- Find shear strength:

h/tw = 18.76/0.375=50.03 <

so Cv =1.0

So: Vn= 0.6Fy Aw = 0.6 x 50 (20.8x0.375) x 1.0 = 234kips

LRFD

So Vu=88 < ØvVn=1.0x234=234k

Or read that ØvVn=234k from

Table 3-2 @ page 3-17

So Shear is adequate

Note: to use the AISC manual for getting the value of ØvVnx & Vnx/Ωv.

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EX: Check the adequacy of beam (W24x55) for deflection requirements. Assume maximum permissible deflection is for L/360

w=3k/ft (incl. s.w) 21ft Solution:

1- Spec. , dimensions & properties:

steel Fy Fu Sec A Ix L

A572G50 W24x55 1350 21

2- Deflection:

δ=

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EX: Determine Cb for the beam shown below, parts (a) and (b). Assume

the beam is a doubly symmetric member.

Solution:

a-

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b-

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EX: Verify the available flexural strength of the (W18x50), ASTM A992 beam assume the beam braced at the ends and third points. Apply the

requirements of the AISC Specification directly. WD=0.45k/ft, WL=0.75k/ft bracing or lateral support 35 ft

Lb=11.67ft Lb=11.67ft Lb=11.67ft Solution:

1- Specification, dimensions & properties: From AISC Manual Table 1-1, the geometric properties are as follows: for Lp & Lr see table

Steel Fy Fu Sec Sx Zx Lbend Lpmid Lp Lr

A992 50 65 W18x50 88.9 101 11.67 11.67 5.83 16.9

2- Loading:

LRFD

wu = 1.2(0.45 kip/ft) +1.6 (0.75 kip/ft)

= 1.74 kip/ft

Mu = wu L2 / 8 = 266 kips.ft

3- Calculate nominal flexural strength, Mn, first find Cb: For the lateral-torsional buckling limit state, the nonuniform moment modification factor can be calculated using AISC Specification Equation F1-1:

Use Rm= 1.0

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For the center segment of the beam, the required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the

maximum midspan moment as: Mmax = 1.00, Ma = 0.972, Mb= 1.0, and Mc = 0.972.

For the end-span beam segments, the required moments for AISC Specification Equation F1-1 can be calculated as a percentage of the

maximum midspan moment as: Mmax = 0.889, Ma = 0.306, Mb = 0.556, and Mc = 0.750.

Thus, the center span, with the higher required strength and lower Cb, will govern.

For a compact beam with an unbraced length of Lp < Lb ≤ Lr, the lesser of either the flexural yielding limit state or the inelastic lateral-torsional buckling limit state controls the nominal strength. 4- Find plastic moment:

Mn = Mp = Fy . Zx = 50 x 101 / 12 = 421 kip-ft

5- From AISC Specification Section F1, the flexural strength is:

LRFD Øb = 0.9

Øb Mn = 0.90(339 kip.ft)= 379

kip.ft > 266 kip-ft ………………….O.K.