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© University of Western Australia
School of Civil and Resource Engineering 2006
16. Beam-and-Slab Design
• Beam-and-Slab System
• How does the slab work?
• L- beams and T- beams
• Holding beam and slab together
ENDP3112 Structural Concrete Design
BEAM - AND - SLAB SYSTEM
Here is a conventional
beam:
Here is a wide beam:
Here is a T-beam:
and an L-beam:
We can save all this
dead weight - provided
that the reduced web
can resist the induced
shear actions.
So we can design a floor
system, using beam and
slab design as follows . . .
One-way continuous slab spanning
One-way continuous slab spanning
Simply
supported T-
beams,
spanning
Simply
supported
L-beam
spanning:
Beams supported on columns or walls
Edge beam
(L-beam)
Interior beams
(T-beams)
Span direction
of beams
Span direction
of slab
Support
columns
Model for slab design:
Model for beam design:
These reactions per metre . . .
. . . provide
this UDL on
the beams.
First,
consider the
slab . . .
HOW DOES THE SLAB WORK ?
• Like a one-way continuous slab, supported by walls,
which in this case are in fact, beams,
• Could proceed to use our methods for continuous
beams (slabs in this case) adopting linear-elastic
methods (moment distribution, stiffness
methods, etc.), or using moment re-distribution.
However, in most cases a simplified ‘coefficient’ method is
easier. The method applies when :
• ratio of adjacent span lengths <= 1.2
• loads are essentially UDL - g and q
• q <= 2g
• slab is of uniform section
• rebar layout complies with Code arrangement
• actions at supports are solely due to slab loads
Let’s consider this method . . .
Cl. 7.2
Ln Ln
Ln Ln Ln
TWO SPAN MULTI-SPAN
AT EDGE
BEAM
Design moments:
Fd = 1.2 g + 1.5 q
Ln =clear distance between faces of supports
- Fd Ln
2
9
+ Fd Ln
2
11
+ Fd Ln
2
11
- Fd Ln
2
10
- Fd Ln
2
11
+ Fd Ln
2
11
+ Fd Ln
2
16
+ Fd Ln
2
16
- Fd Ln
2
24
Design shear forces :
Ln Ln
Shear force at RH end: 1.15 Fd Ln
2
Fd Ln
2
Shear force at LH end:
Fd Ln
2
Fd Ln
2
Shear force at mid span:
Fd Ln
7
Fd Ln
8
T- BEAMS AND L - BEAMS
b
bef
bef
C
T
C
C
T
T
This is the
‘flange’
..and this is
the ‘web’
Since T-beams and
L-beams have a
greater effective
width of compression
flange (bef > b), a
greater lever arm is
available.
lever arm
lever arm
lever arm
So T- and L-beams
will have a higher
ultimate bending
capacity.
(Likewise, at working
moment, the section
will be stiffer.)
Conventional
beam
T-beam
L-beam
Simply supported T-beams and L-beams
Usually, the neutral axis at ultimate moment is
within the flange. So for bending, we proceed
as for a rectangular beam, using bef .
(Sometimes the approximation LA = d - t / 2 is
used for preliminary calculations.)
So a T-beam or an L-beam can be designed just as for a rectangular
beam, except for the following considerations:
1. Two-dimensional stresses at slab/beam interfaces.
2. An appropriate bef is selected for the flange.
3. The beam and slab are securely held together.
First, consider (1) . . .
Bending Resistance
bef
neutral axis
Two dimensional stresses
At mid-span of the beam, the top of the slab is subjected to both
compression in the direction of the beam span, and tension caused by the
hogging of the slab across the beam here:
The concrete is subjected to a two-dimensional
stress state. At first glance, this may appear to be
a problem. However, this is not the case.
The negative moment in the slab is resisted by the
tensile rebar. In the beam direction, the ability of
the concrete to carry the compressive stress due
to positive bending remains relatively unaffected.
Now consider bef . . .
Effective width of beam flange
The width of flange clearly cannot exceed
half the distance to the next adjacent beam
(otherwise we’d be double-counting).
But the width is also limited by the ability of
the section to distribute actions from the
web to the flange, and this will be governed
by the span of the beam.
AS3600-2001 provides direct guidance:
For T-beams bef = bw + 0.2 a
For L-beams bef = bw + 0.1 a
Where for a simply supported beam,
a = span length L of the beam.
bef
bef
bw
bw
So it is important to check bef before proceeding too far with the
design. Remember not to encroach on the territory of any
adjacent beams.
Cl. 8.8
HOLDING BEAM AND SLAB TOGETHER
Vertical shear action is carried by the web,
which includes the common web/flange area.
This is bv.do, where bv and do are as shown.
bv
do
To ensure that the common web/flange is
properly engaged in its role of carrying
some of the shear force, stirrups are
ALWAYS used, and are carried as high as
possible.
Note how this explains the use of bv in all shear resistance
formulas considered so far.
Note too that bv = bw for reinforced concrete beams.
(Footnote: This is not the case for prestressed concrete
beams, hence the different terms.)
But there is also a longitudinal (horizontal) shear action to
concern us . . .
Resistance to Longitudinal Shear Forces
x δx
C
C + δC
T T + δT
C C + δC
δx
δCV V - δV
Over distance δx, a horizontal
shear force δC must be provided
to ensure integrity of the beam.
From the equilibrium of the
element,
δC / δx = V / (Lever Arm)
So the rate of change of C is
proportional to V, or V* at
ultimate load.
So how do we make provision for these stresses?
Does the concrete carry the actions, or do we need
special rebar arrangements? . . .
LA
δx
SHEAR ON WEB SHEAR PLANE:
Critical section
Shear force to be resisted = V* stirrups area Asv at s spacing
Critical section
Shear force to be resisted = V*.A1/A2
A2 = total area of
flange
A1 A1
SHEAR ON FLANGE SHEAR PLANE:
Vuf = β4 As fsy d / s
+ β5 bf d f ’ct
< 0.2 f ’c bf d
For monolithic construction:
β4 = 0.9 and β5 = 0.5
Check φ Vuf > Longitudinal
shear force
Steel contribution
Concrete
contribution
Cl. 8.4
s
δx = Lever Arm
Lower Ductility Check
For any concrete beam, we must ensure that,
when the first crack occurs, the rebar is
adequate to carry the moment which caused
the first crack, with an appropriate margin.
This requirement is stated:
Muo >= (Muo)min = 1.2 Mcr
For T- and L- beams, there is no simple
formula for doing this. We must calculate the
uncracked second moment of area of the
section Ig, the section modulus with respect to
the tension fibre Z, and proceed from there.
Note that the neutral axis
of the uncracked section
may be below (as shown)
or within the flange.
Then Mcr = f ’cf Z and we check that Muo >= 1.2 Mcr
If not, then increase Muo until it is.
Cl. 8.1.4.1
Upper Ductility Check
We must also check that the section is not prone to compressive concrete
fracture before the rebar has yielded sufficiently to warn the user of a
problem.
The check is (as for rectangular beams): ku <= 0.4
This is seldom a problem, since we have a wide compression flange to help
us:
kud
But we should always check that this is satisfied.
Effective Inertia for Deflection Ief
This can be worked out by Branson’s formula, but can be time
consuming.
Best to use an approximation to start with, and then use the more
complicated method only if deflection appears to be a problem.
A very simple approximation for Ief was provided in AS3600 - 1994:
Ief = 0.045 bef d3 ( 0.7 + 0.3 bw / bef )3
Using EcIef, proceed to compute deflections as for a rectangular beam.
If calculated deflection is much less than the allowable deflection, then
further calculation is not required.
So what about detailing ? . . .
Detailing
The slab is designed as a one-way,
continuous slab, spanning across
the beam supports.
…and the beam is designed as a
simply supported T- (or L-) beam,
spanning across its supports.
Primary slab rebar, top
and bottom:
Secondary
slab rebar:
Stirrups extended
into slab:
Main beam rebar:
Termination of bars to AS3600
Figure 9.1.3.2
SUMMARY
• A beam-and-slab system, with a one-way slab, and beams
cast compositely with the slab, is a highly efficient
floor system.
• The slab is designed as a continuous slab, using theory of
continuous beams (slabs), or the simplified ‘coefficient’
method (where applicable i.e. most of the time).
• Edge beams act as L-beams, and interior beams as
T-beams.
• L- and T-beams save weight, and provide a greater lever arm
for flexural strength and stiffness.
• Care is required to ensure that the beam and slab are
properly held together - hence attention to vertical and
horizontal shear force actions is required.
NEXT
14. Member Strength of Columns
(Lecture 15. on ‘Compression rebar for bending’ will be on
Monday next week)

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Beamand slab design

  • 1. © University of Western Australia School of Civil and Resource Engineering 2006 16. Beam-and-Slab Design • Beam-and-Slab System • How does the slab work? • L- beams and T- beams • Holding beam and slab together ENDP3112 Structural Concrete Design
  • 2. BEAM - AND - SLAB SYSTEM Here is a conventional beam: Here is a wide beam: Here is a T-beam: and an L-beam: We can save all this dead weight - provided that the reduced web can resist the induced shear actions. So we can design a floor system, using beam and slab design as follows . . .
  • 3. One-way continuous slab spanning One-way continuous slab spanning Simply supported T- beams, spanning Simply supported L-beam spanning: Beams supported on columns or walls
  • 4. Edge beam (L-beam) Interior beams (T-beams) Span direction of beams Span direction of slab Support columns Model for slab design: Model for beam design: These reactions per metre . . . . . . provide this UDL on the beams. First, consider the slab . . .
  • 5. HOW DOES THE SLAB WORK ? • Like a one-way continuous slab, supported by walls, which in this case are in fact, beams, • Could proceed to use our methods for continuous beams (slabs in this case) adopting linear-elastic methods (moment distribution, stiffness methods, etc.), or using moment re-distribution. However, in most cases a simplified ‘coefficient’ method is easier. The method applies when : • ratio of adjacent span lengths <= 1.2 • loads are essentially UDL - g and q • q <= 2g • slab is of uniform section • rebar layout complies with Code arrangement • actions at supports are solely due to slab loads Let’s consider this method . . . Cl. 7.2
  • 6. Ln Ln Ln Ln Ln TWO SPAN MULTI-SPAN AT EDGE BEAM Design moments: Fd = 1.2 g + 1.5 q Ln =clear distance between faces of supports - Fd Ln 2 9 + Fd Ln 2 11 + Fd Ln 2 11 - Fd Ln 2 10 - Fd Ln 2 11 + Fd Ln 2 11 + Fd Ln 2 16 + Fd Ln 2 16 - Fd Ln 2 24 Design shear forces : Ln Ln Shear force at RH end: 1.15 Fd Ln 2 Fd Ln 2 Shear force at LH end: Fd Ln 2 Fd Ln 2 Shear force at mid span: Fd Ln 7 Fd Ln 8
  • 7. T- BEAMS AND L - BEAMS b bef bef C T C C T T This is the ‘flange’ ..and this is the ‘web’ Since T-beams and L-beams have a greater effective width of compression flange (bef > b), a greater lever arm is available. lever arm lever arm lever arm So T- and L-beams will have a higher ultimate bending capacity. (Likewise, at working moment, the section will be stiffer.) Conventional beam T-beam L-beam
  • 8. Simply supported T-beams and L-beams Usually, the neutral axis at ultimate moment is within the flange. So for bending, we proceed as for a rectangular beam, using bef . (Sometimes the approximation LA = d - t / 2 is used for preliminary calculations.) So a T-beam or an L-beam can be designed just as for a rectangular beam, except for the following considerations: 1. Two-dimensional stresses at slab/beam interfaces. 2. An appropriate bef is selected for the flange. 3. The beam and slab are securely held together. First, consider (1) . . . Bending Resistance bef neutral axis
  • 9. Two dimensional stresses At mid-span of the beam, the top of the slab is subjected to both compression in the direction of the beam span, and tension caused by the hogging of the slab across the beam here: The concrete is subjected to a two-dimensional stress state. At first glance, this may appear to be a problem. However, this is not the case. The negative moment in the slab is resisted by the tensile rebar. In the beam direction, the ability of the concrete to carry the compressive stress due to positive bending remains relatively unaffected. Now consider bef . . .
  • 10. Effective width of beam flange The width of flange clearly cannot exceed half the distance to the next adjacent beam (otherwise we’d be double-counting). But the width is also limited by the ability of the section to distribute actions from the web to the flange, and this will be governed by the span of the beam. AS3600-2001 provides direct guidance: For T-beams bef = bw + 0.2 a For L-beams bef = bw + 0.1 a Where for a simply supported beam, a = span length L of the beam. bef bef bw bw So it is important to check bef before proceeding too far with the design. Remember not to encroach on the territory of any adjacent beams. Cl. 8.8
  • 11. HOLDING BEAM AND SLAB TOGETHER Vertical shear action is carried by the web, which includes the common web/flange area. This is bv.do, where bv and do are as shown. bv do To ensure that the common web/flange is properly engaged in its role of carrying some of the shear force, stirrups are ALWAYS used, and are carried as high as possible. Note how this explains the use of bv in all shear resistance formulas considered so far. Note too that bv = bw for reinforced concrete beams. (Footnote: This is not the case for prestressed concrete beams, hence the different terms.) But there is also a longitudinal (horizontal) shear action to concern us . . .
  • 12. Resistance to Longitudinal Shear Forces x δx C C + δC T T + δT C C + δC δx δCV V - δV Over distance δx, a horizontal shear force δC must be provided to ensure integrity of the beam. From the equilibrium of the element, δC / δx = V / (Lever Arm) So the rate of change of C is proportional to V, or V* at ultimate load. So how do we make provision for these stresses? Does the concrete carry the actions, or do we need special rebar arrangements? . . . LA δx
  • 13. SHEAR ON WEB SHEAR PLANE: Critical section Shear force to be resisted = V* stirrups area Asv at s spacing Critical section Shear force to be resisted = V*.A1/A2 A2 = total area of flange A1 A1 SHEAR ON FLANGE SHEAR PLANE: Vuf = β4 As fsy d / s + β5 bf d f ’ct < 0.2 f ’c bf d For monolithic construction: β4 = 0.9 and β5 = 0.5 Check φ Vuf > Longitudinal shear force Steel contribution Concrete contribution Cl. 8.4 s δx = Lever Arm
  • 14. Lower Ductility Check For any concrete beam, we must ensure that, when the first crack occurs, the rebar is adequate to carry the moment which caused the first crack, with an appropriate margin. This requirement is stated: Muo >= (Muo)min = 1.2 Mcr For T- and L- beams, there is no simple formula for doing this. We must calculate the uncracked second moment of area of the section Ig, the section modulus with respect to the tension fibre Z, and proceed from there. Note that the neutral axis of the uncracked section may be below (as shown) or within the flange. Then Mcr = f ’cf Z and we check that Muo >= 1.2 Mcr If not, then increase Muo until it is. Cl. 8.1.4.1
  • 15. Upper Ductility Check We must also check that the section is not prone to compressive concrete fracture before the rebar has yielded sufficiently to warn the user of a problem. The check is (as for rectangular beams): ku <= 0.4 This is seldom a problem, since we have a wide compression flange to help us: kud But we should always check that this is satisfied.
  • 16. Effective Inertia for Deflection Ief This can be worked out by Branson’s formula, but can be time consuming. Best to use an approximation to start with, and then use the more complicated method only if deflection appears to be a problem. A very simple approximation for Ief was provided in AS3600 - 1994: Ief = 0.045 bef d3 ( 0.7 + 0.3 bw / bef )3 Using EcIef, proceed to compute deflections as for a rectangular beam. If calculated deflection is much less than the allowable deflection, then further calculation is not required. So what about detailing ? . . .
  • 17. Detailing The slab is designed as a one-way, continuous slab, spanning across the beam supports. …and the beam is designed as a simply supported T- (or L-) beam, spanning across its supports. Primary slab rebar, top and bottom: Secondary slab rebar: Stirrups extended into slab: Main beam rebar: Termination of bars to AS3600 Figure 9.1.3.2
  • 18. SUMMARY • A beam-and-slab system, with a one-way slab, and beams cast compositely with the slab, is a highly efficient floor system. • The slab is designed as a continuous slab, using theory of continuous beams (slabs), or the simplified ‘coefficient’ method (where applicable i.e. most of the time). • Edge beams act as L-beams, and interior beams as T-beams. • L- and T-beams save weight, and provide a greater lever arm for flexural strength and stiffness. • Care is required to ensure that the beam and slab are properly held together - hence attention to vertical and horizontal shear force actions is required.
  • 19. NEXT 14. Member Strength of Columns (Lecture 15. on ‘Compression rebar for bending’ will be on Monday next week)