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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
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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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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-2
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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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-5
(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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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-6
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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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-8
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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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
s:
urlin:
B, or BFIB.
a trussed purlin may be used.
6-9
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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
r Z sections )
:
as:
beam.
over two spans (No saving in design).
over three spans.
th knee
6-10
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-11
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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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
ion of purlin:
O.K..........
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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
6.6.8 Design of purlinLAYOUT OF TIE RODS
Using tie rod will affe
ers
Dr. Maher Elabd (2011/2012)
using tie rods:
:
t the value of My only. My can be calcu
6-13
lated as follows:
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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
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-
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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
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 [ #
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-16
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 ====
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Chapter 6: Beams and Gird
By Prof. Kamel Kandil
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
Dr. Maher Elabd (2011/2012)
O.K.1400
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-18
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
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-19
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.
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Chapter 6: Beams and Girders
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
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil -
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
r. Maher Elabd (2011/2012)
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.
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil -
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
r. Maher Elabd (2011/2012)
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
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil -
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
r. Maher Elabd (2011/2012)
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.
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-24
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 ===
==
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-25
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
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil -
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
r. Maher Elabd (2011/2012)
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,
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Chapter 6: Beams and Girders
By Prof. Kamel Kandil - Dr. Maher Elabd (2011/2012) 6-27
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
8/2/2019 Chapter 6- BEAMS and Girders-Prefinal
29/29
Chapter 6: Beams and Girders
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)(