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CHAPTER 4
DESIGN AND FABRICATION OF CASTELLATED BEAM
4.1 DESIGN REQUIREMENTS
In the design of beam, two aspects are primary consideration, (1)
Strength requirements, that is the beam has adequate strength to resist the
applied bending moments and accompanying shear forces and (2) Stability
consideration, that is the member is safe against buckling (Duggal 2005).
Because of the opening in the web various failure modes are
expected to happen, which need to be checked and designed for. The strength
of a beam with various web openings shall be determined based on the
interaction of flexure and shear at the web opening. Design constraints
include the displacement limitations, overall beam flexural capacity, beam
shear capacity, overall beam buckling strength, web post buckling and
Vierendeel bending of upper and lower tees.
The design procedure given here are consistent with SP: 6(2) Steel
beams and plate girders (Hand book for Structural Engineers 1998).
4.2 DESIGN PROCEDURE OF CASTELLATED BEAM
Castellated beam is designed for a span length of 3.0 meter. ISMB
150 is selected and depth of the beam is increased to 1.5 times the original
depth as IC 225. The properties of ISMB 150 (Steel table also available in IS
35
800:2007 Annex H, Page No.139) are, A= 19.0 cm2 , Bf = 80mm, tf =7.6 mm,
tw= 4.8mm, Ixx = 726.4 cm4 , Iyy = 52.6 cm4, Zxx = 96.9 cm3, Zyy = 13.1 cm3 .
where,
A - Area of the section
Bf - Breadth of the flange
tf - Thickness of flange
tw - Thickness of web
Ixx & Iyy - Moment of inertia about ‘x’ and ‘y’ axis
Zxx & Zyy - Section modulus about ‘x’ and ‘y’ axis
The basic geometry and notations used for castellated beams are
shown in Figure 4.1.
Figure 4.1 Geometry of castellated beam
These are the steps followed for designing the castellated beam.
1. Calculate moment of resistance (MR )
2. Load applied over the section (W)
36
3. Beam is checked for shear ( va )
4. Beam is checked for deflection ( )
5. Check for combined local bending and direct stresses
6. Design of Stiffeners
For calculating moment of resistance, we have to make out the total
area of the Tee section (A) and the distance from centre of gravity in y
direction (y). Total area of the Tee section is calculated by,
Calculation of properties of the section
Properties of Tee at open throat as shown in Figure 4.2,
Area of flange = 80 X 7.6 = 608 mm2
Area of web = 29.9 X 4.8 = 143.52mm2
Total Area of T section (A) = 751.52 mm2
Figure 4.2 Properties of Tee section
Position of centroid of castellated T-section
= (4.1)
80mm
29.9
mm 4.8 mm
7.6mm
37
= 608 7.62
+ 143.52 1507.62
+ 7.6 751.52
= 29.54 .
Moment of Inertia of T- section about Neutral axis
The moment of inertia of T-section is calculated by,
=
) (4.2)
80 7.612
+ 608 29.547.62
4.812
(37.5 7.6)
+ 4.8 (37.5 7.6) (29.54 22.55)
= 423460.201
1. Moment of Resistance (MR )
Plastic moment capacity is calculated by,
MR = A at d (4.3)
where,
MR = Moment of resistance
A = Area of the T- section at open throat
at = Average stress based on moment capacity
d = Depth of the section
MR = 751.52 X 150 X (225 - 2 X 29.54)
MR = 18703.83 kN.mm
38
2. Load applied over the section (W)
Let W be the load that can be applied over the section
Figure 4.3 Bending moment diagram for simple beam
= 3 (4.4)
W = 3 18703.83 /3000
W = 18.70 kN
Self weight of the beam has to be deducted; hence self weight is
calculated for ISMB 150 section for a span of 3m,
Weight per meter for ISMB 150 is 14.9Kg (IS 800:2007)
Self weight of beam = 146.11 = 0.43833
Load (W) = 18.70 – 0.43833
W = 18.26 kN
39
3. Beam is checked for shear
The average shear stress ( ) at the end is calculated by
= (4.5)
where, P=Maximum end shear =W= 18.26 kN
= Depth of the stem of T- Section
t = Thickness of stem
Figure 4.4 Depth of the stem of T- section
= =18.26 2 102 37.5 7.6
64.07 N/mm2 < 0.40 fy = 0.4 X 250 =100 N/mm2
Hence the section is safe.
3. Beam is checked for deflection
The maximum deflection of the T- section is at the mid span and is
due to the net load carrying capacity and local effects which are calculated by,
d’=37.5mm
d’=37.5mm
40
< (4.6)
where
= Maximum deflection at mid span
1 = Deflection due to net load carrying capacity
2 = Deflection due to local effects
L = Length
where
= (4.7)
To calculate the deflection, we have to recognize the value of
Moment of Inertia (I),
Moment of Inertia (I) = Moment of Inertia of perforated section (Ip)
+ Moment of Inertia of solid section (Is)
Moment of Inertia of perforated section (Ip),
= 2 + IT (4.8)
IT = Moment of Inertia of T- section about N.A
= 2 751.52 29.54 + 2 X 423460.201
= 1119.14 X 104 mm4
Moment of Inertia of solid section (Is)
= 11191385.18 + 4.8 150
12
41
45°
n / 2
n n
m
Stiffener
= 1254.138 X 104 mm4
Average moment of Inertia (I)
I = 11191385.18+12541385.18 /2
I = 11866385.18 mm4
Substituting the value of I in equation (4.6),
=23 648
= 16.45 mm
= ) /24 (4.9)
where,
= Average shear
= Number of perforation panels in half span
m, n = Distance as shown in Figure 4.4
I = Moment of Inertia of T – Section
Figure 4.5 Dimensions of perforations
42
Shear at ends = 18.26kN,
Shear at centre = 0,
Average Shear = 18.26 +
= 9.13 kN.
Let us provide 12 perforations in the entire span. The dimensions of
the perforations will be, p= 6, m=75mm, n= 75 mm, I = 423460.201 mm4
Substituting the values in equation (4.9) we get,
= 0.091 .
Hence total deflection ( ),
+ = 16.54 .
= /325
= 3000 / 325
= 9.23 mm < 29.54 mm.
The deflection is more than the permissible deflection, to make the
beam safe in deflection, stiffeners are introduced.
4. Check for combined local bending and direct stresses
The beam is checked for combined local bending and direct stresses
at various locations as shown in Figure 4.6. The maximum moment and shear
effects are combined to give the worst combination at this point.
43
Figure 4.6 Stresses at various locations
Max end shear (P) =W= 18.26 kN,
Effective length (l) = 3m
= 3 X 103 /2 = 1500 mm
Shear force(vh) = (18.26 X 8 X 225) / 1500 = 21.91 kN
Taking moment about A,
Shear force X distance = vh X (D/2 – y) (4.10)
2252
29.54 = 21.91 225/2
=2464.87
82.96
= 29.71 kN
Shear Stress = Shear force (vh ) /Area
= 29.71 X 103 / (4.8X 75) = 82.53 N/mm2
< 0.40 fy = 0.4 X 250 =100 N/mm2
Maximum combined local bending and direct stress in Tee segment
near ( l / 4) point, that is at about 0.75 m, perforations fall 0.9 m,
° A Vh /2 Vh /2
75 mm
°B
44
Shear = 18.26 X 0.9 / (3/2) = 10.96 kN
There are two flanges, so shear on each flange
= 10.96/2
= 5.48 kN
Maximum bending moment at centre = P X L /3 (4.11)
= 18.26 X 3 X 103 /3
= 18.260 X 103 N.m
Moment = 18.260 X 103 X ( )
= 13.69 X 103 N.m
Stress due to bending taking the T-section as booms,
= M/(area of T X distance between centroids)
Direct stress due to moment =
)
= 110.01 N/mm2
Bending stress due t o shear taking bending on the T-sections
= (shear) X (width across top of opening) / IT (4.12)
= (5.48 X 103) X {75 X (37.5 -29.54)}/ 423460.201
= 7.72 N/mm2
Combined stress = 110.01 + 7.72
= 117.73 N/mm2 < 165 N/mm2
Hence the section is safe.
45
5. Design of Stiffeners
Max width of vertical stiffener = (Space between the two openings –
Thickness of web)/2
= (75 – 4.8) /2
= 35.1 mm.
Hence provide 20 mm wide stiffener.
Thickness of stiffener plate < L /12 = 20/12= 1.66 mm,
Hence let us provide approximately 5 mm thick plate.
Provide a vertical stiffener plate 20 mm X 5mm on both sides of the
web of the castellated beam as shown in Figure 4.7.
Figure 4.7 Stiffeners
As per the I.S specifications the bearing stiffener should be
designed as column, with a distance of web = 40 X t, as a part of the stiffener.
Therefore length of web as stiffener = 40 X 5 = 200 mm.
Effective area of stiffener = 2 X 20 X 5 + 200 X 5
= 1200 mm2
Stiffener 20mm X 5mm 7.6mm 29.9mm
7.6mm
150mm
29.9mm
46
Let each stiffener plate be splayed by 5 mm to fit on the weld.
Bearing area of stiffener = 2 X (20 – 5) X 5
= 150 mm2
Stress = 18.26 10 /150
= 121.73 N/mm2 < 0.75 f y (185 N/mm2 )
Moment of Inertia of stiffener
I = b d3 /12 (4.13)
= 5 ( ) + (200 5)
= 40000
Radius of gyration,
= (4.14)
where
I = Moment of Inertia
A = Effective area of stiffener
Substituting the values in (4.10)
= 40000/1200 = 5.77 mm
Effective length of stiffener (l) = 0.7 X height of stiffener
l = 0.7 X (225- (2 X 7.6)) = 146.86 mm
l / r = 146.56 / 5.77 = 25.40
For l / r and fy = 250N/mm2 , cd = 223 N/mm2
cd = Design compressive stress for column buckling) (IS 800:2007)
47
Load carrying capacity = Design compressive stress ( cd ) X Area
= 223 X 1200 = 267600 N
267.600 kN > 20.65 kN Which is sufficient.
Connection of Stiffeners
Let us provide the fillet weld of 5 mm for its full length of the
stiffener.
4.3 FABRICATION OF CASTELLATED BEAM
Castellated beams are structural members, which are made by
flame cutting a rolled beam along its web in a definite pattern and then
rejoining the two halves by welding so that the overall beam depth is
increased by 50% without any increase in weight for improved structural
performance against bending. Figure 4.8 explains the castellation process and
Figure 4.9 gives the mathematical formulation of castellated beam.
Figure 4.8 Fabrication of castellated beam
48
Figure 4.9 Mathematical formulation
Two types of Rolled steel wide flange beam is considered
ISMB 150
ISMB 200
Castellated beam is fabricated for a span of 3m. Two sections
ISMB 150 and ISMB 200 are selected, the beam is cut along its web in a zig
zag manner and then rejoined together to get a 50% increased depth of 225
mm from 150mm and 300 mm from 200mm respectively. The Increased
depth of Castellated beam is indicated as IC 225 and IC300.
Table 4.1 Detail of the beams
I section beam
Increased depth of
Castellated beam
No of openings
Span length(m)
Depth of opening
(mm)
Overall height (mm)
ISMB 150 IC 225 11 3 150 225
ISMB 200 IC 300 8 3 200 300
49
PROPERTIES OF ISMB 150 SECTION
Figure 4.10 shows the marking on ISMB 150 Rolled steel wide
flange beam section which is of following dimensions. The section properties
are assigned from steel table. tf =7.5 mm, tw =5mm, bf =80mm, Hw =150
mm and length of the beam is taken as L=3.2 m, with angle of cut 45°. Where
‘tf ’ is the thickness of the flange, ‘tw’ is the thickness of the web, ‘bf’
breadth of the flange and ‘Hw’ height of the web. Figure 4.11 shows
increased depth of castellated section IC 225 without any increase in weight
from ISMB 150.
Figure 4.10 Marking on ISMB 150
Figure 4.11 Increased depth (225mm) of Castellated beam from ISMB 150 (IC 225)
50
PROPERTIES OF ISMB 200 SECTION
Figure 4.12 shows the marking on ISMB 200 with Fe250 grade
Rolled steel wide flange beam section which is of following dimensions. The
section properties are assigned from steel table. tf =10 mm, tw = 6mm, bf
=100mm, Hw =200 mm and length of the beam is taken as L=3.2 m with
angle of cut 45°. Figure 4.13 gives the Increased depth of Castellated beam(IC
225) from ISMB 150. Table 4.2 gives the properties of steel.
Table 4.2 Properties of steel
Material Grade
Young’s modulus (N/mm2 )
Yield stress (MPa)
Ultimate stress (MPa)
Fe 250 2 X105 295 426
Figure 4.12 Marking on ISMB 200
Figure 4.13 Increased depth (300mm) of Castellated beam from ISMB 200 (IC 300)
51
4.3.1 Distribution of Stresses in I section
i. The nature of shear distribution in an I – beam is shown in
Figure 4.14. In an I- Section the value of Q which is zero at
the extreme fiber increases to a high value at top flange- web
interactions and attain maximum value at the neutral axis.
ii. The bending stress distribution diagram shown in Figure 4.15,
it can be seen that flanges carries maximum bending stress.
Figure 4.14 Shear stress distributions
Figure 4.15 Bending stress distribution
iii. From the comparison of the shear and flexural stress
distributions, it is observed that the flanges carry a major
portion of the flexural load, whereas the web carries most of
the shear load.
52
iv. The primary modes of failure of the beam are the local
buckling of compression flange and shear buckling of web.
v. Castellated beam depends on web post buckling. That is at
high shear locations, normally near the supports and neutral
axis; the principle planes would be inclined to the longitudinal
axis of the member.
vi. Along the principle planes, the principle stresses would be
diagonal tension and diagonal compression causes the web to
buckle in a direction perpendicular to its action.
vii. This problem can be solved by reducing the depth to thickness
ratio of the web and we can also provide web stiffeners that
would develop tension field action to resist diagonal
compression.
4.3.2 Tension Field Action
For shear, the design of sections for strength is usually governed by
the web plate subjected to shear force and undergoing shear buckling, or
yielding in shear or a combination of the two. For webs, with relatively high
depth-to-thickness ratios, the shear stress distribution in the web after
buckling changes and significant post-buckling strength may occur as a result
of development of a diagonal tension which is called “Tension Field Action”
(TFA). This Tension Field Action involves the effects of out of plane forces
on the transverse stiffeners due to the shear post-buckling response of the web
panels. When the girder is loaded beyond its limit, first it will deflect
significantly because it has lost its stiffness, but after significant deflection,
the profile of girder will be too curved so the web in the region between the
stiffeners will experience tension which will be directed towards the support.
This will cause some truss kind of action with stiffeners as vertical member of
truss, compression flange as top chord, tension flange as tie member and our
53
web as a diagonal member. Now this girder will again carry good amount of
load just because of this action. This is tension field action of plate girder
where web plays an important role. Also the plate girders with a high ratio of
tension field to elastic buckling contribution are those containing a web with a
high slenderness ratio. When the contribution of tension field action is
considered, the hybrid girders are able to obtain the predicted shear strength
(Atorod Azizinamini et al 2007).
For beam subjected to shear if flange cannot increase the shear
resistance, rearrangement of stresses favorable for the web is possible which
might utilize an element other than web most likely, the element of paramount
importance in the shear case is the transverse stiffener upon which a tension
field might be supported. Figure 4.16 shows the tension field action. As the
web begins to buckle, the web loses its ability and no longer support to increase
in compressive principal stresses. However, the tensile stresses will continue to
increase, with the stiffeners of the panel carrying the balance of the compressive
forces. Under additional loading, the panel behaves similarly to Pratt truss, with
the web carrying the additional diagonal tension and the stiffeners supporting
additional compression (Basler 1961). The web resists only the diagonal
tension and this behavior of the web is called tension field action
Figure 4.16 Tension field actions
54
4.3.3 Stiffeners
The webs, when they are inadequate to carry the load are made
strong and stable by the provision of a wide variety of stiffeners.
1. Intermediate Transverse Web Stiffener
i. Transverse intermediate stiffeners shall be provided when the
average shear stress is larger than the allowable buckling shear
stress (qb). The stiffeners may be pairs one on either side of
the web or single. When the single stiffeners are used, these
should be placed alternately on the opposite sides of the web.
The stiffener shall extend from flange to flange.
ii. The intermediate stiffeners shall be designed as a compression
member (with effective length = 0.80 the length of stiffener)
to resist a force and to improve the buckling strength of
slender web due to shear.
2. Load Carrying Stiffener
i. The function of these stiffeners is to distribute the load to the
web. These are provided at the supports and at the points of
concentrated loads.
ii. The bearing stiffeners shall be designed as a column assuming
the section to consist of the pair of stiffeners together with a
length of web equal to 12, 25 times the web thickness for end
and intermediate stiffeners respectively
55
3. Torsion Stiffener
These types of stiffeners are used to provide torsional restraint to
beams and girders at supports.
4. Diagonal Stiffener
The diagonal stiffeners are designed for the portion of the
combination of applied shear and bearing that exceeds the capacity of the
web.
4.3.4 Stiffeners Along the Shear Zone
The major disadvantage of castellated beam is the introduction of
an opening in the web of the beam alters the stress distribution within the
member and also influences its collapse behavior. These openings decrease
the stiffness of the beams resulting in larger deflection (Chapkhanea et al
2012). Hence to improve the shear strength of castellated beam and to reduce
the deflection stiffeners are introduced on the web.
From the concept of tension field action and Pratt truss, stiffeners
are introduced on the web. Two types of stiffeners are introduced
1. Diagonal stiffeners on the web opening along the shear zone
2. Vertical stiffeners on the solid portion of the web along the
shear zone.
4.3.5 Different Types of Sections Adopted
Stress concentration is more near the opening than in the solid
portion, hence stiffeners are introduced on the web opening and also on the
solid portion of the web. In order to study the effect of stiffeners along shear
zone the following three cases are considered.
56
The following are the different types of cases that are used for the
experimental and analytical work.
a) Case I – Castellated beam Without Stiffeners (WOS).
b) Case II – Castellated beam With Diagonal Stiffeners on
the web opening along the shear zone (WDS).
c) Case III – Castellated beam With Vertical Stiffeners on
the solid portion of the web along the shear zone
(WVS).
Figure 4.17 Shear force and stiffeners details (IC 300)
L
S1
L
FS = P/2
FS = P/2
S0
57
Figure 4.17 shows the shear force and stiffeners detail for all the
three cases. To reduce the stress concentration and deflection stiffeners are
introduced in two forms. One type of stiffeners is provided diagonally in the
hole region and second type of stiffeners are provided vertically on the solid
portion of the web. Figure (a) shows the first case without stiffeners (WOS),
Figure (b) shows the second case with diagonal stiffeners (WDS), Figure(c)
shows the third case with vertical stiffeners (WVS) of castellated beam along
the shear zone. The following are the two sections that are adopted throughout
the analysis. The Table 4.3 and 4.4 shows the dimensions adopted for three
cases of castellated beam. The dimensions of all the three cases are
maintained throughout the work. For each case two sections are adopted
IC225 and IC300.
Table 4.3 Beam type I (IC 225)
Spec
imen
det
ail
Len
gth
(m)
Thi
ckne
ss o
f fla
nge
t f
(mm
)
Thi
ckne
ss o
f web
t w
(mm
)
Bre
ath
of th
e Fl
ange
b f
(mm
)
Hei
ght o
f the
web
op
enin
g H
W (m
m)
Len
gth
of st
iffen
er
(mm
)
Wid
th o
f stif
fene
r (m
m)
Thi
ckne
ss o
f stif
fene
r (m
m)
WOS 225 3.2 7.5 5 80 150 - - -
WDS 225 3.2 7.5 5 80 150 190 15 5
WVS 225 3.2 7.5 5 80 150 200 20 5
58
Table 4.4 Beam type II (IC 300) Sp
ecim
en d
etai
l
Len
gth
(m)
Thi
ckne
ss o
f fla
nge
t f (m
m)
Thi
ckne
ss o
f web
t w
(mm
)
Bre
ath
of th
e fla
nge
b f (m
m)
Hei
ght o
f the
web
op
enin
g H
w (m
m)
Len
gth
of st
iffen
er
(mm
)
Wid
th o
f stif
fene
r (m
m)
Thi
ckne
ss o
f st
iffen
er (m
m)
WOS 300 3.2 10 6 100 200 - - -
WDS 300 3.2 10 6 100 200 260 20 6
WVS 300 3.2 10 6 100 200 260 30 6
In this table thickness of the flange, breath of the flange, thickness
of the web, height of the web, overall height, depth of the hole, length of
stiffeners, width and thickness of the stiffeners and length of the span are
discussed.
Table 4.5 gives the details of the test beam and Figure 4.18 shows
the castellated beam with diagonal and vertical stiffeners.
Table 4.5 Details of Test beams
S.No. Types No. of Specimen Test
No. of Specimens IC 225 IC 300
1. WOS 1 1 2
2. WDS 1 1 2
3. WVS 1 1 2
Total No. of beams 6
59
Figure 4.18 CB with diagonal and vertical stiffeners along the shear zone
