10. Beam 3 - UB

BEAMS

By.Ir.Sugeng P Budio,MSc 1

INTRODUCTION

Beams are structural members that support transverse loads and are therefore subjected primarily to flexure, or bending. If a substantial amount of axial load is also present, the member is referred to as a beam-column.This figure (1) shows both a hotrolled shape and a built-up shape


This figure (1) shows both a hotrolled shape and a built-up shape along with the dimensions to be used for the width-thickness ratios.

yw F

970

t

h ≤If

the member is to be treated as a beam, regardless of whether it is a rolled shape or built-up.

Figure (1)


If yw F

970

t

h ≥

the member is considered to be a plate girder

All of the standart hot-rolled shapes found in the Manual are in the first category (beams). Most built-up shapes will be classified as plate girders, but some will be beams by the AISC definition.

For beams, the basic relationship between load effects and strength For beams, the basic relationship between load effects and strength can be written as

nbu MM ∅≤where : Mu = controlling combination of factored load moments

Øb = resistance factro for beams = 0.90Mn = nominal moment strength

The design strength Øb Mn is sometimes called the design moment.

BENDING STRESS AND THE PLASTIC MOMENT


Figure (2)


From elementary mechanics of materials, the stress at given point can be found from the flexure formula :

(1)

Where M is the bending moment at the cross section under consideration, y is the perpendicular distance from the neutral plane to the point of interest, and Ix is the moment of inertia of the area of the cross section with respect to the neutral axis.

xb I

Myf =

the cross section with respect to the neutral axis.

For maximum stress, Equation (1) takes the following form :

xxxmax S

M

c/I

M

I

Mcf ===

(2)

where c is the perpendicular distance from the neutral axis to the extreme fiber, and Sx is the elastic section modulus of the cross section.

The stress fmaxx must not exceed Fy, and the bending moment must not exceed

xy S FM =

where My is the bending moment that brings the beam to the point of yielding.

In figure above, a simply supporte beam with a concentrated load at midspan is shown at succesive stages of loading. Once yielding begins, the distribution of stress on the cross section will no longer be begins, the distribution of stress on the cross section will no longer be linier, and yielding will progress from the extreme fiber toward the neutral axis.

Figure (3)


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 axis are the same. The plastic moment Mp is the resisting couple formed by the two equal and opposite forces :

ZFa2

AFa)A(Fa)A(FM yytycyp =

===

where : A = total cross-sectional areaa = distance between the centroids of the two half-areasa = distance between the centroids of the two half-areasZ = (A/2)a = plastic section modulus

Figure (5)

Compute the plastic moment Mp for a W10 x 60 of A36 steell

SolutionFrom the Dimensions and properties tables :


From the Dimensions and properties tables :

2

2

in 8.82

17.6

2

A

in 17.6A

==

=

The centroid of half of this are can be found in the tables for WT-shapes,which are cut from W-shapes. the relevant shape here is the WT5x30, and the distance from the outside faceof the flange to the centroid is 0.884 inches (figure above)

3in 74.388.8(8.452)a2

AZ

in 8.4522(0.844)10.222(0.884)da

==

=

=−=−=

This compares favorablly with the value of 74.6 given in the Dimensions and properties tables (the difference resulut from roundoff in the tabular values)

ANSWERMp = FyZ = 36(74.38) =2678 in-kips = 223 ft-kipsMp = FyZ = 36(74.38) =2678 in-kips = 223 ft-kips

Classification Of Shapes

AISC classifies cross-sectional shapes as compact, noncompact, orslender. depending on the values of the width-thickness ratios. For I-and H- shapes, the ratio for the projecting flange (an unstiffenedelement) is bf/2tf, and the ratio for the web ( a stiffened element) ish/t . It can be summarized in a general way as follows :


h/tw. It can be summarized in a general way as follows :

Let

λ = width : thickness ratioλp = upper limit for the compact categoryλr = upper limit for noncompact category

Width – Thickness parameters ( for hot-rolled I- and H-shapes

Element λ λp λr

Flangef

t2

b

F

6510F

141

−


Web

ft2 yF 10Fy −

wt

hyF

640

yF

970

Then

if λ � λp and the flange is countinuously to the web, the shape is compactif λp < λ � λr , the shape is noncompact ; and


if λp < λ � λr , the shape is noncompact ; andif λ > λr , the shapes is slender

The category is based on the worst width-thickness ratio of the cross section. For example, if the web is compact and the flange is noncompact, the shape is classified as noncompact. Table has been extracted from AISC Table and contains width-thickness raios for hot-rolled I- and H- shapes cross sections.

Bending Srength Of Compact Shapes

A beam can fail by reaching Mp and becoming fully plastic, or it can fail by buckling in one of the following ways :

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


1. Lateral-torsional buckling (LTB), either elastically or inelastically2. Flange local buckling (FLB), elastically or inelastically3. Web local buckling (WLB), elastically or inelastically

We begin with compact shapes, defined as those whose webs are continuously connected to the flanges and that satisfy the following width-thickness ratio requirement for the flange and the web :

ywyf

f

F

640

t

h and

F

65

2t

b ≤≤

The first category, laterally supported compact beams, is quite common and is the simple case. AISC gives the nominal strength as :

Mn = Mp

where : Mp = FyZ<1.5My

The limit of 1.5 M for M is to prevent excessive working load


The limit of 1.5 My for Mp is to prevent excessive working load deformations and is satisfied when :

1.5S

Zor S1.5FZF yy ≤≤

The beam shown in Figure above is a W16 x 31 of A36 steel. It supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange. The service dead load, including the weight of the beam, is 500 lb/ft. The service live load os 550 lb/ft. Does this beam have adequate moment strength?


beam have adequate moment strength?

SolutionFor simply supported, uniformly loaded beam, the maximum bending moment occurs at midspan and is equal to

2max wL

8

1M =

where w is the load in units of force per units length, and L is the span length.

kipsft 61.888

0.550(30)M

kipsft 56.258

0.500(30)Lw

8

1M

2

L

22

DD

−==

−===

Since the dead load is less than 8 times the live load, load combination


Since the dead load is less than 8 times the live load, load combination controls :

Mu = 1.2MD + 1.6ML = 1.2(56.25)+1.6(61.88)=166 ft-kips

Alternatively, the loads can be factored at the outset :

kipsft 1668

1.480(30)Lw

8

1M

kips/ft 1.4801.6(0.550)1.2(0.500)1.6w1.2ww2

2uu

LDu

−===

=+=+=

Check for compactness:

compact is flange the 6.3 10.836

65

F

65

3.6t2

b

y

f

f

∴⟩==

=


steel A36for compact is W16x31a

shapes) all(for F

640

t

h

yw

∴

⟨

Since the beam is compact and laterally supported,

Mn = Mp = FyZx = 36(54) = 1944 in-kips = 162 ft-kips

Check for Mp < 1.5My

kipsft 166 kipsft 1460.90(162)M

5.114.12.47

54

S

Z

nb

x

x

−⟨−==∅

⟨== OK

N.G


kipsft 166 kipsft 1460.90(162)Mnb −⟨−==∅ N.G

Answer : Since the design moment is less than

the factored load moment,

the W16x31 is unsatisfactory


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. In this book, points of lateral support will be indicated by an “x”, as shown in Figure (10), The relationship between nominal strength Mnand unbraced length is shown in next Figure (11).

If the unbraced length is nograter than Lp, to be defined presently, the beam is considered to have full lateral support, and Mn = Mp.

If Lb is greater than Lp but less than or equal to the


If Lb is greater than Lp but less than or equal to the parameter Lr, the strength is based on inelastic LTB.

If Lb is graterthan Lr, the strength is based on elastic LTB.The equation for the theoretocal elastic latera-torsional buckling strength can be found in the Theory of Elastic stability and, with some notational changes, is as follows :

wy

2

by

bn CI

L

EGEI

LM

π+π=

Figure (11)


Figure (11)

where :Lb = unbraced length (in.)G = shear modulus = 11,200 ksi for structural steelJ = torsional constant (in.4)Cw = warping constant (in.6)

If the moment is greater than that corresponding to first yield, the strength is based on inelastic behaviour. The moment corresponding to first yield is taken as :

Mr = FLSx


Mr = FLSx

where FL is the smaller of (Fyf – Fr) or Fyw. In this expression, the flange yield stress is reduced by Fr, the residual stress. For a nonhybrid member, Fyf = Fyw = Fy, and FL will always vbe equal to Fy –Fr . For example, AISC Equation above will be written as

Mr = (Fy – Fr)Sx

As shown in Figure, the boundary between elastic and inelasticbehavior will be for an unbraced length of Lr, which is the value ofLb obtained from Equation when Mn is set equal to Mr. Thefollowing equation is obtained :

( ) ( )

1

2ry2

ry

1yr

2

EGJA

SX

FFX11FF

XrL

π=

−++−

=where


2

x

y

w2

x1

GJ

S

I

C4X

2SX

=

=

As with columns, inelastic behavior of beams is more complicated than elastic behavior, and empirical formulas are often used. With one minor modification, the following equation is used by AISC :

( )

y

yp

pr

pbrppn

F

r300L

LL

LLMMMM

=

−−

−−=

where

The nominal bending strength of compact beams is completely described by Equation subject to an upper limit of Mp for inelastic


described by Equation subject to an upper limit of Mp for inelastic beams, provided the applied moment is uniform over the unbraced length Lb. Otherwise there is a moment gradient, and Equation must be modified by a factro Cb. this factro is given in AISC as

CBAmax

maxb M3M4M3M5.2

M5.12C

+++=

where :

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

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 unbracedlength


B

unbracedlengthMC =absolute value of the moment at the three-quarter point of

the unbraced length.

when the bending moment is uniform, the value of Cb is

0.1M3M4M3M5.2

M5.12Cb =

+++=

Determine Cb for a uniformly loaded, simply supported beam with lateral support at its ends only.

Because of symmetry, the maximum moment is at midspan :


Because of symmetry, the maximum moment is at midspan :

2Bmax wL

8

1MM ==

Also because of symmetry, the moment at the quarter point is equal to the moment at three-quarter point. From Figure ;

M3M4M3M5.2

M5.12C

wL32

3

32

wL

8

wL

8

L

4

wL

4

L

2

wLMM

maxb

222

CA

+++=

=−=

−

==

Answer : Cb = 1.14


14.1

323

381

4323

381

5.2

81

5.12

M3M4M3M5.2C

CBAmaxb

=

+

+

+

=

+++=

Figure (12)


Figure (13)next shows the value of Cb for several common cases of loading and lateral support.

For unbraced cantilever beams, AISC specifies a value of C of 1.0 . A value of 1.0 is always conservative,


of Cb of 1.0 . A value of 1.0 is always conservative, regardless of beam configuration or loading, but in some cases it may be excessively conservative.

The complete specification of nominal moment strength for compact shapes can be summarized.

Figure (13)


( )

2

5.1

pcrn

pbr

pbrppbn

pn

MMM

MLL

LLMMMCM

MMM

≤=

≤

−−

−−=

≤= For Lb ≤ Lp

For Lp<Lb≤Lr

For Lb > Lr

22

211

2

)/(21

/

2

ybyb

xb

wyb

yb

bcr

rL

XX

rL

XSC

CIL

EGJEI

LCM

+=

+= ππ

Figure (14)


The ons ants X1 and X2 have been previously defined and are tabulated in the Dimensions and properties Tables

The effect of Cb on the nominal strength is illustrated in


The effect of Cb on the nominal strength is illustrated in Figure (14). Although the strength is directly proportional to Cb, this graph clearly shows the importance of observing the upper limit of Mp , regardless of which equation is used for Mn

Example

Determine the design strength ØbMn for a W14x68 of A242 steel suject to the following conditions.

a) Continous lateral supportb) Unbraced length=20ft;Cb=1.0c) Unbraced length=20ft;Cb=1.75

Solution


a) From Part 1 of the Manual, aW14 x 68 is seen to be in shape group 2 and is therefore available with a yield stress Fy of 50 ksi

Determine whether shape is compact,noncompact,or slender “

<∴ this shape is compact and

kipsft 479.2

kipsin 5750

50(115)ZFMM xypn

−=−=

===

Check for Mp ≤ 1.5My

5.112.1103

115⟨==x

S

Z


103xS

Answer : ∅nMn = 0.90(479.2)=431 ft-kips

b) Lb = 20 ft, and Cb =1.0, Compute Lp and Lr :

ft 8.700in 104.450

300(2.46)

F

300L

y

p ====

From the torsion Properties Tables,

J = 3.02 in.4 Cw = 5380 in.6


Although X1 and X2 are tabulated in the Dimensions and Properties Tables in Manual, they will be computed here for illustration.

GJ

S

I

CX

EGJA

SX

xw

x

4

ksi30212

)20)(01.3)(200,11(000,29

103

2

2

2

1

=

=

=

=

π

π


ftin

FFXFF

XrL

ksi

x

GJIX

ryry

yr

y

40.268.316)1050(001649.011)1050(

)3021(46.2

)(11)(

)(001649.0

02.3200,11

103

121

53804

4

2

22

1

2

2

2

==−++−

=

−++−

=

=

=

=

−

Since Lp < Lb < Lr , the strength is based on inelastic LTB.

xryr

kipsft

SFFM

−=

−=

−=

3.34312

)103)(1050(

)(


p

pr

pbrppbn

Mkipsft

LL

LLMMMCM

⟨−=

−

−−−=

−−

−−=

4.392

700.840.26

700.820)3.3432.479(2.4790.1

)(

Answer : ØbMn = 0.90(392.4) = 353 ft-kips

c) Lb = 20 ft, and Cb = 1.75. The design strength for Cb = 1.75 is 1.75 times the design strength for Cb

= 1.0

� Mn = 1.75(392.4) = 686.7 ft-kips > Mp = 479.2 ft-kips


� Mn = 1.75(392.4) = 686.7 ft-kips > Mp = 479.2 ft-kips

The nominal strength cannot exceed Mp, therefore, use a nominal strength of Mn = 479.2 ft-kips :

Answer : Ø Mn = 0.90(479.2) = 431 ft-kips

BENDING STRENGTH OF NONCOMPACT SHAPES


Where :

141

65

=

=y

pF

λ

λ

The strength corresponding to both limit states must be computed and the smaller value will control. From AISC ,


shapes rolledfor ksi 10stress residualF

)(

r ==

−=

−=

xryr

ry

r

SFFM

FFλ

These terms have been specialized for nonhybrid beams

A simply supported beam with a span length of 40 feet islaterally supported at its end and is subjected to thefollowing service loads.

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


Dead load = 400 lb/ft (including the weight of the beam)Live load = 1000/ft

If ASTM A572 Grade 50 steel is used, is a W14 x 90 adequate ?

Factored load moment :

ftkipswww =+=+= /080.2)000.1(6.1)400.0(2.16.12.1


kipsftLwM

ftkipswww

uu

LDu

−===

=+=+=

0.4168

)40(080.2

8

1

/080.2)000.1(6.1)400.0(2.16.12.12

2

Determine whether shape is compact, noncompact,or slender

19.950

6565

2.102

===

==

y

p

f

f

F

t

b

λ

λ


3.221050

141141 =−

=−

=ry

rFF

λ

Since λp < λ < λr , this is a noncompact shape

Check the capacity based on the limit state of flange local buckling.

MMMM

kipsftSFFM

kipsftZFM

pr

prppn

xryr

xyp

−

−−

−−=

−=−=−=

−===

)(

7.47612

)143)(1050()(

2.65412

)157(50

λλλλ


kipsft −=

−−−−= 5.640

19.93.22

19.92.10)7.4762.654(2.654

The design strength based on FLB is therefore

ØbMn = 0.90(640.5) = 576 ft-kips

Check the capacity based on the limit state of lateral-torsional buckling.From the Load Factor Design Selection Tables,

Lp = 15, Lr = 38.4 ftLb = 40 ft > Lr, � failure is by elastic LTB

From part 1 of manual

Iy = 362 in.4


Iy = 362 in.J = 4.06 in.4

Cw = 16,000 in.6

For a uniformly loaded, simply supported beam with lateral support at the ends,

Cb = 1.14

AISC Equation :

x

MCIL

EGJEI

LCM pwy

by

bbn

+=

≤

+=

)000,16)(362(000,29

)06.4)(200,11)(362(000,2914.12

2

ππ

ππ


kipsftkipsin

x

−=−==

+=

0.5150.618)5421(14.1

)000,16)(362(1240

)06.4)(200,11)(362(000,29)12(40

14.1

Mp = 654.2 ft-kips > 515.0 ft-kips (OK)

Since 515.0 < 640.5, LTB controls :

ØbMn = 0.90(515.0)= 464 ft-kips > Mu = 416 ft-kips (OK)


Answer :Since Mu < ØbMn, the beam has adequate moment strength

Shear StrengthThe shear strength of a beam must be sufficient to satisfy the relationship

Vu ≤ ØvVn

Vu = maximum shear based on the controlling combination of factored


combination of factoredØv = resistance factor for shear = 0.90Vn = nominal shear strength

Consider the simple beam of Figure(15). At a distance xfrom the left end and at the neutral axis of the crosssection, the state of stress is as shown in part d. since thiselement is located at the neutral axis, it is not subjected toflexural stress. From elementary mechanics of materials,the shearing stress is given by


It

VQfv =

fv = vertical and horizontal shearing stress at the point of interest

V = vertical shear force at the section under consideration


considerationQ = 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 axist = width of the cross section at the point of

interest

Taking the shear yield stress as 60% of the tensile yield stress, the stress in the web at failure can be written as

yw

nv F

A

Vf 60.0==

where Aw = area of the web. The nominal strength corresponding to this limit state is


Vn = 0.6 FyAw

For h/tw ≤ 418/√Fy , there is no web instability, and

Vn = 0.6FyAw

For 418/√Fy < h/tw ≤ 523/√Fy, inelastic web buckling can occur, and

w

y

wyn th

FAFV

/

/4186.0=

For 523/√Fy < h/tw ≤ 260, the limit state is elastic web buckling :

2)/(

000,132

w

wn th

AV =


Aw = area of the web = dtwd = overall depth of the beam


Check the beam of Example for shear


From example, wu = 2.080 kips/ft, and L = 40 ft. A W14x90 with Fy = 50 ksi,is used. For a simply supported, uniformly loaded beam, the maximum shear occurs at the support and is equal to the reaction :


the support and is equal to the reaction :

kipsLw

V uu 6.41

2

)40(080.2

2===

From the Dimensions and properties Tables in Part1 of the Manual, the web width-thickness ratio of a W14x90 is :

11.5950

418418

9.25

==

=

y

w

F

t

h

Since h/tw is less than 418/√Fy, the strength is governed by shear yielding of the web

Vn = 0.60 FyAw

= 0.6 Fy(dtw) = 0.6(50)(14.02)(0.440) = 185.1 kips


= 185.1 kips

ØvVn = 0.90(185.1) = 167 kips > 41.6 kips (OK)

Answer : Since the shear design strength is greater than the factored load shear, the beam is satisfactory

Figure (18)

Block Shear


AISC “Block Shear Rupture Strength”,gives twoequation for the block shear design strength :

[ ][ ]gtynvun

ntugvyn

AFAFR

AFAFR

+∅=∅

+∅=∅

6.0

6.0

Ø = 0.75A = gross area in shear (in Figure(18),length AB


Agv = gross area in shear (in Figure(18),length AB times the web thickness)

Anv = net area in shearAgt = gross area in tension (in Figure(18),length BC

times the web thickness)Ant = net area in tension

The gorvening equation is the one which has the larger fracture term

Determine the maximum factored load reaction, based on block shear, that can be resisted by the beam shown in Figure (19)


Effective hole diameter = ¾ + ⅛ = ⅞ in

Gross and net tension areas :


Agv = (2+3+3+3)tw = 11(0.300) = 3.300 in.2

Anv = (11-3.5 x ⅞)(0.300) = 2.381 in.2

AISC Equation

ØRn = Ø[0.6FyAgv + FuAnt] = 0.75[0.6(36)(3.300)+58(0.2438)]= 0.75[71.28 + 14.14]=64.1 kips

AISC Equation

ØRn = Ø[0.6 FuAnv + FyAgt]= 0.75[0.6(58)(2.381)+36(0.375)]= 0.75[82.86 + 13.50] = 72.3 kips


= 72.3 kips

The fracture term in AISC Equation is the larger one (that is, 82.86>14.14) so this equation governs

Answer : Maximum factored load reaction = design strength for block shear = 72.3 kips

Deflection

wL45

For the common case of a simply supported, uniformlyloaded beam such as that in Figure, the maximumvertical deflection is given by


EI

wL4

384

5=∆

The following values are typical maximum allowabletotal (service dead load plus service live load)deflections.

Plastered construction :360

L

Unplastered floor construction : L


Unplastered floor construction :240

Unplastered roof construction :180

L

Where L is the span length

Check the deflection of the beam shown in Figure(21). The maximum permissible total deflection is L/240


Maximum permissible deflection = inL

500.1240

)12(30

240==


Since it is more convenient to express the deflectionin inches rather than feet, use units of inches in thefollowing equation :

Total service load = 500 + 550 = 1050 lb/ft = 1.050 kips/ft

Maximum total deflection :

)1230)(12/050.1(55 44

= xwL


Answer : the beam satisfied the deflection criterion

(OK) 1.500in. 1.294in.

)510(000,29

)1230)(12/050.1(

384

5

384

5

⟨=

= x

EI

wL

Design

Beam design entails the selection of a cross-sectionalshape that will have enough strength and will meetthe serviceability requirement. As far strength isconcerned, flexure is almost always more critical than


concerned, flexure is almost always more critical thanshear, so the usual practice is to design for flexureand the check shear. The design process can beoutlined as follows :

1. Compute the factored load moment Mu. This will be required design strength ØbMn. The weight of the beam is part of the dead load but is unknown at this point. A value may be assumed, or the weight may be ignored initially and checked after a shape has been selected.

2. Select a shape that is satisfies this strength requirement. This can be


2. Select a shape that is satisfies this strength requirement. This can be done on one of two ways :

• Assume a shape ,compute the design strength,and compare it with the factored load moment. Revise if necessary. The trial shape can be easily selected in only a limited number of situations (Example)

• Use the beam design charts in Part 3 of the manual. This is preferred method and will be explained following Example

3. Check the shear strength4. Check the deflection.

Select a standart hot-rolled shape of A36 steel forthe beam shown in Figure (22). The beam hascontinuous lateral support and must support auniform service live load of 5 kips/ft. the maximum


uniform service live load of 5 kips/ft. the maximumpermissible live load deflection is L/360.

Assume a beam weight of 100 lb/ft

wu = 1.2wD + 1.6wL = 1.2(0.100) + 1.6(5.000) = 8.120 kips/ft

2

8

1= LwM uu


nb

2

M required5.9138

)30(120.8

8

∅=−=

=

=

kipsft

LwM uu

Assume a compact shape. For a compact shape and continuous lateral support,

Mn = Mp = ZxFy ≤ 1.5My

From ØbMn ≥ Mu

ØbFyZx ≥ Mu

3.3.338)36(90.0

)12(5.913in

F

MZ

yb

ux ==

∅≥

Try a W30x108

This shape is compact as assumed (noncompact


This shape is compact as assumed (noncompactshapes are marked as such in the table), and Zx/Sx =346/299 = 1.16. this is less than 1.5,therefore Mn =Mp as assumed.Since the weight is slightly larger than assumed, the required strength will be recomputed, although the W30x108 has more capacity than originally required, and it will almost certainly be adequate.

kipsftM

ftkipsw

u

u

−==

−=+=

6.9148

)30(130.8

130.8)000.5(6.1)108.0(2.12

From the Load Factor Design selection Table,

ØbMp = ØbMn = 934 ft-kips > 914.6 ft-kips (OK)


In lieu of basing the search on the required sectionmodulus, the design strength ØbMn could be used, sinceit directly proportional to Zx and is also tabulated.

Check shear : kips 1222

)30(130.8

2=== Lw

V uu

From the Tables of Uniform Load Constants,

ØvVn = 316 kips > 122 kips OK

Check deflection :

The maximum permissible live load deflection isOK


The maximum permissible live load deflection isL/360 = 30(12)/360 =1 in.

in. 1 in. 0.703)4470(000,29

)1230)(12/000.5(

384

5

384

5 44

⟨===∆ x

EI

Lw

x

L

OK

Answer : Use a W30 x 108

TUGAS KELAS STRUKTUR BAJA21 21 21 21 OktoberOktoberOktoberOktober 2011201120112011


Tentukan dimensi profil baja mutu A 50 untuk balokseperti terlihat pada gambar di atas. Lateral support padabalok tersebut hanya pada tumpuan saja dan harusmemikul beban seperti pada gambar. Lendutan yangdiijinkan adalah L/360.

Date post:	17-Apr-2022
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

10. Beam 3 - UB

Documents