flange beam design · flange beam flanged beams occur when beams are cast integrally with and...
TRANSCRIPT
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Flange Beam Design
-
Semester 1 2016/2017
Department of Structures and Material Engineering
Faculty of Civil and Environmental Engineering
University Tun Hussein Onn Malaysia
Introduction
▪ Main reinforcement
▪ Shear reinforcement
▪ Traverse reinforcement
Key
Different
Design Procedures
Step Task Standard
1 Determine design life EN 1990:2002 Table 2.1
2 Determine beam size EN 1992-1-1: Table 7.4N
EN 1992-1-2: Table 5.5 & 5.6
3 Determine the effective width EN 1992-1-1: Sec.5.3.2.1
4 Determine design actions on beam EN 1991-1-1
5 Durability and characteristic strengths EN 1992-1-1: Sec. 3 & 4
6 Determine nominal cover EN 1991-1-1: Sec.4.4.1
7 Calculate moment and shear force EN 1992-1-1: Sec.5
8 Design of flexural reinforcement EN 1992-1-1: Sec.6.1, 9.2.1.1
9 Design of shear reinforcement EN 1992-1-1: Sec.6.2
10 Design for traverse
11 Check deflection EN 1992-1-1: Sec.7.4
12 Check cracking EN 1992-1-1: Sec.7.3
13 Detailing EN 1992-1-1: Sec.8 & 9.2
Moment
▪ Negative moment - it should be noted that when T-beam is
subjected to negative moment, the slab at top of web will be in
tension while the bottom web is in compression. This usually
occurs at interior support of continuous beam.
Moment
Flange Beam
▪ Flanged beams occur when beams are cast integrally with
and support a continuous floor slab.
▪ Part of the slab adjacent to the beam is counted as acting in
compression to form T- and L-beam.
▪ The effective width of flange, beff is given in Sec. 5.3.2.1 of
EC2 and should be based on the distance lo.
where;
beff = effective flange
width
bw = breadth of the web
of the beam.
hf = thickness of the
flange.
beff beff
bw bw
hf
h
T-Beam L-Beam
Flange Beam
▪ The design procedure of flange beam depends on where the
neutral axis lies.
▪ The neutral axis may lie in the flange or in the web.
▪ There are three cases that should be considered:
- Neutral axis lies in flange (M < Mf)
- Neutral axis lies in web (M > Mf but < Mbal)
- Neutral axis lies in web (M > Mbal)
beff
bw
hf
d
x
beff
bw
hf
d
x
Flange Beam
▪ The effective width of flange, beff is given in Sec. 5.3.2.1 of
EC2.
▪ beff should be based on the distance lo between points of zero
moment as shown in the figure below.
Flange Beam
▪ The effective flange width, beff for T-beam or L-beam may be
derived as:
where:
,eff eff i wb b b b= +
, 0.2 0.1 0.2eff i i o ob b l l= +
,eff i ib b
Example 1
▪ Based on figure below, determine the effective flange width,
beff of beam B/1-3.
3000 4500
25
00
40
00
1 2 3
A
B
C
200 x 500
200 x 500 200 x 500
200 x 500 200 x 500
20
0 x
50
02
00
x 5
00
20
0 x
50
02
00
x 5
00
FS1 (150 thk.)
FS2 (150 thk.) FS3 (150 thk.)
20
0 x
50
0
Example 1
▪ lo (distance between points of zero moment)
▪ Effective flange width, beff
3000 mm 4500 mm
lo = 0.85 x 3000 = 2550 mm
lo = 0.85 x 4500 = 3825 mm
lo = 0.15 x (3000 + 4500) = 1125 mm
1 2 3
,eff eff i wb b b b= +
Example 1
▪ Span 1-2
beff1 = 0.2(1250) + 0.1(2550)
= 505 mm < 0.2lo = 510 mm < b1 = 1250 mm
beff2 = 0.2(2000) + 0.1(2550)
= 655 mm > 0.2lo = 510 mm < b2 = 2000 mm
beff = (505 + 510) + 200 = 1215 mm < 3250 mm
2500 4000
b1 = 2500/2 = 1250 b2 = 4000/2 = 2000
beff,1 beff,2
bw = 200 mm
beff
b = 1250 + 2000 = 3250 mm
A B C
Example 1
▪ Span 2-3:
beff1 = 0.2(1250) + 0.1(3825)
= 632.5 mm < 0.2lo = 765 mm < b1 = 1250 mm
beff2 = 0.2(2000) + 0.1(3825)
= 782.5 mm > 0.2lo = 765 mm < b2 = 2000 mm
beff = (632.5 + 765) + 200 = 1597.5 mm < 3250 mm
2500 4000
b1 = 2500/2 = 1250 b2 = 4000/2 = 2000
beff,1 beff,2
bw = 200 mm
beff
b = 1250 + 2000 = 3250 mm
A B C
Example 1
▪ Dimension of flange beam:
Span 1-2 Span 2-3
beff = 1215mm
bw = 200 mm
hf = 150 mm
beff = 1598 mm
bw = 200 mm
hf = 150 mm
Design For Flexural
▪ Design procedure for flange beam:
1) Calculate Mf,
2) If M ≤ Mf, neutral axis in the flange and hence provide
tensile reinforcement only
( )0.567 0.5f ck f fM f bh d h= −
2
Ed
ck
MK
bd f=
0.5 0.25 0.951.134
Kz d d
= + −
,0.87
Ed
s req
yk
MA
f z=
Design For Flexural
3) If M > Mf, neutral axis in the web
, compression reinforcement is not required
, compression reinforcement is required
2
bal f ckM f bd=
( )
( )
0.1 0.36
0.87 0.5ck w f
s
yk f
M f b d d hA
f d h
+ −=
−
0.167 0.567 1 12
w f w f
f
b h b h
b d b d
= + − −
( )
( )'
'0.87f
s
yk
M MA
f d d
−=
−
( )'
0.167 0.567
0.87ck w ck f w
s s
yk
f b d f h b bA A
f
+ −= +
balM M
balM M
Design For Shear
▪ Design procedure:
1) Determine design shear force, VEd
2) Determine the concrete strut capacity for cot θ =1.0 and
cot θ =2.5 (θ = 22o and θ = 45o respectively)
If VEd > VRd,max cot θ = 1.0 (θ = 45o), redesign the section11
If VEd < VRd,max cot θ = 2.5, use cot θ = 2.5 (θ = 22o), and
calculate the shear reinforcement as follows,
( ),max
0.36 1 / 250
cot tanw ck ck
Rd
b df fV
−=
+
0.78 cotsw Ed
yk
A V
s f d = ( ); cot 2.5 =
Cl.6.2.3 (3)
Design For Shear
If VRd,max cot θ = 2.5 < VEd < VRd,max cot θ = 1.0
11
3) Maximum spacing
4) Calculate the minimum links,
( )10.5sin
0.18 1 / 250Ed
w ck ck
V
b df f − = −
0.78 cotsw Ed
yk
A V
s f d =
0.08sw w ck
yk
A b f
s f= Cl.9.2.2(5)
0.75s dMaximum
spacing
Design for Traverse
▪ Design procedure:
1) Determine distance Δx=0.5(L/2) which is the greatest
longitudinal shear stresses.
2) Change of moment over distance Δx
3) Change in longitudinal force
4) Calculate longitudinal shear stress
23
32
wLM =
0.5 2w
d
f
M b bF
d h b
− =
−
0.27d
ed ctk
f
Fv f
h x
=
Traverse shear reinforcement is required
Design for Traverse
5) Check concrete strut capacity in flange
6) Traverse shear reinforcement
7) Calculate minimum traverse reinforcement area
8) Check additional longitudinal reinforcement
( ),max
0.4 1 / 250
cot tanck ck
ed
f fv
−=
+
0.78 cotsf ed f
f yk
A v h
s f d =
,min 0.26 ctm
s f
yk
fA bh
f
=
,
0.5 cot
0.87Ed
s td
yk
VA
f
=
Deflection
▪ To control deflection to a maximum of span/250.
▪ Procedure:
1) Calculate ρo = √fck 10-3
2) Calculate ρ = As,req / bd
3) Calculate ρ’ = As’,req / bd
4) Determine K and calculate l/d
5) Calculate modification factor MFflange, MFspan and MFarea
6) For adequate deflection control, (l/d)actual < (l/d)allow
3/2
11 1.5 3.2 1 ; o o
ck ck o
lK f f
d
= + + −
'
'
111 1.5 ;
12o
ck ck o
lK f f
d
= + + −
Cl.9.2.2(5)
Deflection
▪ MFflange
- In Cl. 7.4.2(2), bf/bw > 3, Mfflange = 0.8
- Or bw/bf ≤ 0.3, Mfflange = 0.8, and for any, it should be
determined using the following graph.
Deflection
▪ Factor for structural system can be determined from Table
7.4N:
Cracking
▪ Crack control for beam design can be directly referred to
Cl.7.3.
▪ For a convenient, crack control without direct calculation is
preferable, Cl. 7.3.3, Table 7.2N and Table 7.3N.
▪ Procedure:
1) Calculate steel stress for limiting crack width, wk = 0.3mm
2) Determine maximum bar size or bar spacing
3) For adequate crack control,
Sprov .< Smax
+ = +
,
,
0.3435
1.35 1.5s reqk k
s
k k s prov
AG Qf
G Q ATable 7.2N
Table 7.3N
Cracking
▪ Maximum bar spacing for crack control (Table 7.3N)