chapter 6- beams and girders-prefinal

8/2/2019 Chapter 6- BEAMS and Girders-Prefinal

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CHAPTER 6

BEAMSANDGIRDERS

6.1 INTRODUCTION

6.2 DESIGN OF BEAMS

6.3 ALLOWABLE BENDING STRESS

6.4 LIMITATIONS OF DEFLECTION

6.5 DEPTH TO SPAN RATIOS

6.6 DESIGN OF PURLINS

6.7

CRANE GIRDERS


2/29

Chapter 6: Beams and Girders

By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-1

6 BEAMS AND GIRDERS

6.1 Introduction:

A structural member is termed a beam when the loading it carries is resisted by bending

action. From the elementary theory of bending the following expression is obtained:

R

E

y

f

I

M== (6-1)

which presupposes that the beam is bent into a circular deflected shape due to a uniformlyapplied bending moment. In practice this is rarely the case. However; since the ratio of

span to depth is usually large the above expression may regarded as quite reliable.

The design of a beam involves checking stress levels from various effects and ensuring that

deflection is within some prescribed limit.

6.2 Design of Beams:

From Equation (6-1), the bending stress can be given as:

Z

My

I

Mf == (6-2)

where Z is the section modulus. Knowing the maximum moment in the beam and the

allowable bending stress, the required section modulus is obtained and the steel section is

selected.

When lateral deflection of the compression flange of the beam is prevented by providing

lateral support, the beam is said to be laterally supported. In this case no reduction of theallowable bending stress is considered. When lateral support is inadequate lateral buckling

of the compression flange occurs and the allowable bending stress must be reduced

accordingly.

6.3 Allowable Bending Stress:

6.3.1 Compact SectionsTension and compression due to bending on extreme fibers of compact sections

symmetric about the plane of their minor axis and bent about their major axis can be


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obtained from the following Equation:

ybFF 64.0= (6-3)

Table (6-1) Allowable bending stresses for compact Sections

Grade of

Steel

)/( 2cmtonFb

mmt 40 mmtmm 10040 <

St 37 1.54 1.38

St 44 1.76 1.63

St 52 2.30 2.14

In order to qualify under this section:

i. The member must meet the compact section requirements of Table (2-1) of the

E.C.P 2008- Clause 2.6.1. which can be summarized for the common sections as

follows;

1.For box section, the ratios of flange and webs should be as follows;

yf Ft

b 58 ===> for flange &

yw

w

Ft

d 127 ===> for web

2.For other sections such as I, [ or T-sec.;

Hot rolled-yf Ft

C 9.16 ===> for flange &

yw

w

Ft

d 127 ===> for web

Welded -yf Ft

C 3.15 ===> for flange &

yw

w

Ft

d 127 ===> for web

(a) Box section

(b) Other sections

Figure 6. 1: Dimensional ratios of compact sections


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Chapter 6: Beams and Gird

By Prof. Kamel Kandil

ii. The laterally un

smaller of:

For box sections:

y

f

uF

bL

84db

Lt

f

uf, for any value of

T

uT

r

L= , the lateral torsional

buckling stress is governed by the torsional strength given by:

yb

u

ff

ltb FCdL

tbF 58.0

800

1

= (6-9)

Forb

C = 1.0, the allowable bending stress is given by:

y

u

ff

ltb F

dL

tbF 58.0

800

1

=


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(b) For deep thin flange sections, where approximately )40.0( , then:

( )yb

Tu

ltb FCrL

F 58.012000

22= (6-12)

For deep thin flange sections made of Steel 37 and considering bC = 1.0 (as in mostcases the moment through the span is greater than the end moments) the allowable

bending stress is given by:

When 54)(


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Figure 6. 2: Allowable Compression Stress in Non-Compact Sections with Slenderness

Ratio of Compression Flange

Alternatively, the lateral torsional buckling stress can be computed more accurately as the

resultant of the above mentioned two components as:

yltbltbltb FFFF 58.022

21+= (6-18)

In the above Equations:

uL = Effective laterally unsupported length of compression flange.

= K x (distance between cross-sections braced against twist or lateral

displacement of the compression flange in cm.

K = Effective length factor.

Tr = Radius of gyration about the minor axis of a section comprising

the compression flange plus one sixth of the web area (cm).

fA = Area of compression flange (cm2).

yF = Yield stress (t /cm2).

In order to consider the sections shown in Table (6-3) as compact the following conditions

must be satisfied:

i.

The limits of width to thickness ratio are according to Table (2-1) (See E.C.P. 2008Clause 2.6.1).

0

200

400

600

800

1000

1200

1400

1600

0 40 80 120 160 200

SLENDERNESS RATIO LU/rT

ALLAWABLEBENDIN

STRESSKg/Cm2


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ii. The section is sy

iii. The lateral un-b

value of Eq. (6-7

Table 6- 3: Summary reg

ITEMTYPE OF

SECTI

1

DOUB

SYMMET

I-SHA

2

BOX SEC

3

CANN

SECTI

4SOLID SE

ers

Dr. Maher Elabd (2011/2012)

mmetrical about its minor axis.

raced length of compression flange uL m

).

rding the allowable bending stresses

ROSS

ONCOMPACTNESS

STR

AC

LY

RICAL

PE

COMPACT

NON-COMPACT

TION COMPACT

NON-COMPACT

EL

ON

NON-COMPACT

TION

6-7

st satisfy the smaller

INING

IONSb

F

X yF64.0

Y yF72.0

X yF58.0

Y yF58.0

X yF64.0

Y yF64.0

X yF58.0

Y yF58.0

X

yF58.0

Y yF58.0

X yF72.0

Y yF72.0


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6.4 Limitations of deflection:

The calculated deflection due to live load only without dynamic effect of any beam shall

not be greater than the values shown in Table (6-3).

]Table 6- 4: Maximum allowable deflections

MEMBER MAX.DEFLECTION

Beams in building carrying plaster or other brittle finish. L/300

All other beams L/200

Cantilevers L/180

Crane track girders L/800

6.5 Depth to span ratios:

i. The depth of rolled beams in floors shall preferably be not less than 1/24 of their

span.

ii. The depth of beams and girders subjected to shocks or vibrations shall preferablybe not less than 1/20 of their span.

iii. The depth of simply supported roof purlins shall preferably be not less than 1/40 of

their span.


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6.6 Design of Purli

6.6.1 General Layout:

6.6.2 Cross-section of pHot rolled section: [ , SI

For relatively long span

ers


s:

urlin:

B, or BFIB.

a trussed purlin may be used.

6-9


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Cold formed section ([

6.6.3 Structural systemPurlins may be designed

i. Simply supporte

ii. Continuous bea

iii. Continuous bea

iv. Compound bea

v. Simple beam wi

ers


r Z sections )

:

as:

beam.

over two spans (No saving in design).

over three spans.

th knee

6-10


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6.6.4 Loads and straining actions:Purlins are generally subjected to the following loads:

i. Own weight. This is to be assumed.

ii. Weight of covering material.

iii. Imposed live load or a concentrated live load of 100 Kg.

iv. Wind load.

CASES OF LOADING:

The purlin should be designed for the following cases of loading:

Case (I) #1: Dead load + Superimposed load on roof (D.L. + L.L.)

Case (I) #2: Dead load + Concentrated load of 100 Kgs. (D.L. +100 Kg)

Case (II): Dead load + Superimposed load + Wind pressure. (D.L. + L.L. + W.L.).

The total load acting on the purlin in each case is analyzed in the directions of the principal

axes of the cross-section namely Wy, and Wx where :

Wy is the component of the total load in the direction of the y-y axis of the cross-

section.

Wx is the component of the total load in the direction of the x-x axis of the cross-

section..

Knowing the values of Wy and Wx the straining action on the beam (Mx and My) areobtained.


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6.6.5 Finding cross-sec

.

.......ZHence

secttrytablesFrom

=reqOr

sectiongivena

x

=+=

=

+

+=

+=

y

y

x

xact

all

x

x

x

y

x

x

y

y

x

x

Z

M

Z

Mf

c

F

MZ

Z

CM

Z

Mf

For

Z

M

Z

Mf

The chosen cross sectio

The shear stress in the

the shear stress in purlin

N.B:

For relevant secti

for Channel sectio

6.6.6 Effect of roof typeThe type of roof coveri

roof covering (e.g. corr

the principal axes of its

case of rigid roof coveri

deflection in the directi

and the load component

6.6.7 Depth of purlin:To avoid excessive defl

L is the span of the purli

ers


ion of purlin:

O.K..........


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6.6.8 Design of purlinLAYOUT OF TIE RODS

Using tie rod will affe

ers


using tie rods:

:

t the value of My only. My can be calcu

6-13

lated as follows:


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For one tie rod:

For two tie rods:

The ties should be desig

effect ofWx.

In case of one tie rod th

and T2 are calculated as

WT x 25.11

where N is the number

the tie rod as 0.7 of the

grosA

It should be mentioned

no need for tie rods as

Example:Design the intermediate

ers


8

2

2

=

SW

Mx

y

12

3

2

=

SW

Mx

y (Give reaso

ned for the maximum tension force acting o

e design force in the tie rod is the maximum

follows:

NS

2or

( )

sin2

12

25.1

2x

NS

W

T

x +

of supported purlins by the system. Assumi

ross area hence, the net area of the tie rod ca

pt

sF

T

7.0

max=

that in case of cast-in-place reinforced conc

y = 0

purlin of the shown roof truss (flexible roo

6-14

s) !

the system due to the

ofT1 or T2 where T1

g the effective area of

n be estimated as:

rete covering, there is

) as hot rolled [ cross-


16/29



section considering the

Weight of covering

Superimposed load

Wind load (pressur

Spacing of trusses

For a span of 6.0 the hei

160 the own weight 1

CONSIDERING CASE (

Wtotal = 19 + 201.9 + 5

Sin = 0.394

Wx = 56.92 Kg/m

1.2568

692.56M

.5978

6132.82=M

2

y

2

x

==

=

CONSIDERING CASE (

ers


ollowing:

sheets = 20 Kg/m2

= 50 Kg/m2

) = 15 Kg/m2

= 6.0 m

ght of the cross section = 600 / 40 =15 cm (i.

Kg/m.

I)#1:(D.L + L.L.)

1.75 = 144.5 Kg/m

, Cos = 0.919

, Wy = 132.82 Kg/m

Kg.m.

Kg.m.

I)#2:(D.L + Concentrated load)

6-15

e. [ #160 ). For a [ #


17/29



Kg.m.3.1018

65.22M

Kg.m.235.8=8

64.52M

Kg/m'5.22W,kg/m'4.52W

'Kg/m579.12019

2

DL-y

2

D.L-x

xy

==

=

==

=+=DLW

For a concentrated load of P = 100 Kg at the center of purlin:

Py = 91.9 Kg and Px = 39.4 Kg

Kg.m.60.411.593.101)(M

Kg.m.373.7=9.1378.235)(M

Kg.m.1.59

4

64.39M

Kg.m.9.1374

69.91

y

x

y

=+=

+=

==

==

total

total

Mx

From the above two cases of loading the critical case is the case of D.L.+ L.L.

Mx = 597.7 Kg.m and My = 256.1 Kg.m

CONSIDERING CASE (II):(D.L + LL + WL)

The effect of wind load will be in the y-y direction (i.e. it will affect Mx Only).

Wy = 15x1.9 = 28.5 Kg/m/, Wx = 0.0

Kg.m.25.1288

628.5=M

2

x =

Mx (total) = 597.7 + 128.25 = 725.95 Kg.m

My = 256.10 Kg.m

Stress equation:

The channel is always non-compact section:

2/4.158.0 cmtFFFF ybbybx ====


18/29



127

1.256

191

7.597

191Z200#[

1400

2577.597(Z

7=reqZ

7

7Zsection Z[

x

x

x

yx

+=

=

+=

+

=+

+=

act

all

yx

x

y

x

x

y

y

x

x

f

cTry

req

F

MMOr

M

Z

M

Z

M

f

For

Z

M

Z

Mf

Check for case II:

/54.1326

1

725

cKg

Z

M

Z

Mf

y

y

x

xact

=

=+=

DESIGN THE PURLIN USI

Redesign the previous exa

6.6.8.A

From previous example:

Mx maximum = 597.7 Kg.

KgMy 0.64

8

392.56

2

==

ers


O.K.1400


19/29



37.741001400

6477.597.)(Re cmx

xqZx =

+=

From table try [ # 140

33 8.14,4.86 cmZcmZ yx ==

..1400/11241008.14

64

4.86

7.597 2 KOcmKgfact


20/29



6.7 6.7 Crane Girders:

6.7.1 Design Loads of Crane Girders:Crane girders are members used as runways for overhead cranes serving shops and other

industrial building. The main feature characterizing the behavior of crane girders are:

i.The withstanding of a vertical live load of the crane, which has a dynamic action on

the girder.

ii.The action of comparatively large concentrated loads applied by the wheels of the

crane and transmitted through the flange connection to the web of the girder.

iii.The presence of lateral braking forces that induce bending of the top beam flange in ahorizontal plane.

An overhead crane consists of one (or two) main girders (Crane bridge ), along which the

crane trolley with its load runs.


21/29


By Prof. Kamel Kandil -

The load being handled, as

the crane girders through tcrane wheel load may have

In view of possibility of sh

crane runway and other re

between 1.2 1.9 and is g

Owing to braking of the t

This is taken as 10% of the

Also, due to braking of the

r. Maher Elabd (2011/2012)

well as the weight of the crane and the troll

e crane wheels. Depending upon the locatioa maximum or minimum value.

arp changes in the speed of hoisting the load

sons, the crane load is multiplied by a dyn

nerally taken as 1.25.

olley, along the crane bridge, lateral horiz

wheel loads (without dynamic effect) at the

trolley, along the crane girder, a horizontal

6-20

y, is transmitted to

n of the trolley, the

, unevenness of the

mic factor ranging

ntal force appears.

op of rail level.

force along the rail


22/29



is considered. This is taken

6.7.1 CROSS SECTIONS OF

Cross sections (a), (b) and

capacity up to 35 tons. C

to 30 tons. Section (e) ma

members are provided to w

6.7.2 Analysis of crane giThe design moments and

using the influence lines pl

finding the maximum ben

that the middle of the gird

and from the nearest load.

determine the maximum

support and the remaining

It should be noted that the

by the vertical and the hori

Arrangement


as1

7(or 15%) of the wheel loads without i

CRANE GIRDERS:

(c) for cranes having a span of up to 6 m

ross section (d) is used for cranes with a spa

y be used for heavy cases, in which special

ithstand the lateral forces.

ders:

hear forces originated by the crane load m

otted according to the structural system of t

ing moment in a simple beam, the loads sh

er will be at equal distances from the result

Under the latter the maximum moment w

hear force it is necessary to place one of

nes as near as possible to it.

location of the crane loads for determining t

ontal forces should be identical.

of crane loads for maximum bending m

6-21

pact.

eters with a lifting

n of six meters in 5

horizontal bracing

ay be computed by

e crane girder. For

uld be so arranged

ant of all the loads

ill be observed. To

the loads above a

he stresses induced

ment.


23/29



Arrangeme

The required section mod

design strength reduced by

girder, which is simultane

induced by these forces ap

The strength of a solid cran

For the top fiber of

++=

yt

y

xt

x

tZ

M

Z

M

A

Nf

For the bottom fibe

FZ

M=f b

xb

x

where:

At = Area

Zxt = Net

Zyt = Sec

the cr

Zxb = Net s

6.7.3 Deflection of craneThe deflection of crane gir

10384

5 24=

EI

LM

EI

wL s

where:

L = Span of cra

Ms = Moment p


t of crane loads for maximum shearing f

lus (Zx) of the crane girder is determined

150250 Kg/cm2. This is done because in t

usly subjected to horizontal braking forces,

ear.

e girder is checked by means of the followin

the girder:

BCaseFb

r of the girder:

ACase

of top flange.

ection modulus for top fiber of girder.

ion modulus of the top flange (or of the

ne girder with respect to the vertical axis

ection modulus for the bottom fiber of th

irder:

er can be checked by means of the equation.

800

L

e girder.

oduced by the vertical service load witho

6-22

rce.

on the basis of the

he top flange of the

additional stresses

expressions:

bracing beam) of

y-y.

girder.

ut introducing the


24/29



dynamic eff

Example (1):

Design a crane girder with

Pmax = 6 tons. The arrange

Impact Coeff. = 25%,

Lateral shock = 10% ,

Braking force = 1/7.

Solution:

t714.17

12=N

25.15.121.0M

125.1)(M5.125.2

56

5.212

y

x

=

==

=+

==

=

=

ILLRM

tR

ax

a

Assuming moment due to

Mx total =1.05 15.63


ct.

a span of L = 6 meters. The maximum load

ent of the wheels is shown diagrammaticall

Wheel arrangement

t.m.

t.m.63.155.t.m=

wn weight of the beam = 5% of Mx (LL+I):

=16.41 t.m.

6-23

on a crane wheel is

in the figure.


25/29



t12.5=1.2510.0=I)+Q(LL

t106

466

=+=Q

Assuming the shear force due to own weight of the beam = 3% of Q(LL+I), hence:

t12.95.1203.1Q+I)+Q(LL D.L

Choosing of cross-section:

Assume Fall =1200 kg/cm2

280#BFIBTry

cm1367

1200

1041.16 35

==

all

xx

f

MreqZ

Local buckling of elements (Table (2-1) See E.C.P. 2004 clause 2.6.1):

cmsttbcfw 4.11)0.222.128(05)2(5.0 ===

==


26/29



2/140058.0 cmkgFFyb

==

For BFIB # 280

Area of top flange = 2 x 28.0 = 56.0 cm2

O.K.Kg/cm140020.1/1618

1480

1041.16

261

1025.1

56

10714.1

cm1480

cm2616

282

22

553

3

32

=

=

==

=

=

cmKgf

f

ZZ

Z

t

t

xbxt

yt

Check for deflection:

O.K.800Lcm03.1

20720210010

6001005.12

10

2

22

=

==

=

=

Try

EI

LMs

Check of shear stress for B.F.I.B #320:

( ) O.K.kg/cm84024000.35


27/29



Straining actions on cr

.16.14)(M

5.283.5)(

83.56

5.214

x =+

==

==

xILL

xLLM

tox

R

x

a

Assuming own weight of c

0.8

62.0.).(

2

==x

LDMx

Mx total = 18.25 + 0.9

As the effect of horizonta

hence:

My = 0.1 x 14.6 = 1.46

Shearing force:

x

xQ

.025.167.11Q

.116

470.7

total

max

+=

=+=

Normal force on crane

The maximum normal forc

Choosing of cross-secti

Assume the cross section t


ne girder:

m.t.25.185

m.t.6.4

.

=

ane girder = 200 Kg/m

m.t.9.0

= 19.15 m.t.

l shock is taken as 10% of the vertical lo

m.t.

tonx 2.1532

ton67

=

irder:

= 0.15 x 14 = 2.1 ton.

n:

be as shown in the figure (S.I.B.+ [ )

6-26

ds without impact,


28/29



220#channel5.15

cm1041400

1.46x10

[for thereq

5.15,cm1460Zx,400#S.I.B.

cm13681400

1015.19)(

35

3

35

choosetohavewecmbFor

Z

cmbTry

x

F

MSIBtheforZ

x

pt

xx

=

=

==

==

Properties of built up section:

For a S.I.B. # 400:

Area = 118 cm2

, Ix = 29210 cm4

, b = 15.5 cm

tweb = 1.44 cm , tflange = 2.16 cm.

For a channel # 220:

Area = 37.4 cm2

Ix = 2690 cm4

, Iy = 197 cm4 ,

tweb = 0.9 cm, ec = 2.14 cm

43

y

4

22

x

cm3.336012

16.2)5.15(2690I

channel).flangeupper(for the

cm39401

)51.414.29.020(4.37197)51.4(111829210(I

cm.51.41324.37

)14.29.025.21(4.37

=

+=

+

=

++++=

=+

+=

y

x

y

I

I

e

Atop = 37.4+2.16x15.5 = 70.88 cm2

The un-braced length:

The actual un-braced length = 600 cms.

yrb 4

cmsbLu 35888.70

3.336041313 === (see Eq. 6-4)

cmsLu 600358


29/29


Lateral torsional buckling:

2

.. t/cm25.1)16.2240(600

16.25.15800=

=TBLf

Check of stresses:

OK

Casef

Casef

casef

top

top

bottom

2235

225

225

kg/cm20.11250kg/cm6.130488.70

101.2

3.3360

0.11101.46797B)(

kg/cm1250kg/cm79739401

)51.49.020(1019.16A)(

kg/cm1400kg/cm119139401

)51.420(1019.16A)(

chapter 6- beams and girders-prefinal

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