Upload
andrew-yu
View
224
Download
2
Embed Size (px)
Citation preview
1
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Composite Floor and Roof Slab Systems
Professor Thomas A. SabolDepartment of Civil and Environmental Engineering
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Welding shear connectors
(headed studs)
Headed shear stud prior to
weldingMetal deck spanning
perpendicular to beam
Almost all steel framed floors and topped roofs (i.e. those with concrete vs. those with only metal deck) are composite systems
Deck flutes – the deck spans
parallel to the flutesHeaded shear stud welded
through deck to the steel beam
Metal deck spanning
perpendicular to beam
2
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Thickness of concrete slab a function of:•Structural requirements (often not a governing requirement)•Fire-resistive requirements (often governs)•Type of concrete (lightweight vs. normal weight)•Vibration requirements (benefit of added mass)•Seismic and foundation demands (need to reduce weight)
Metal deck thickness and typical spans
Concrete topping
thickness
Deck thickness
3” most common for floors. 2”
sometimes used. 1-1/2” usually
limited to non-composite roofs
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Thickness of metal deck a function of:•Structural requirements (spans between beams)•Vibration requirements•Fire rating
Flute
3
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Fire rating of a metal deck and concrete fill:
• Hourly fire ratings are used as a measure of the ability of the composite deck and slab to contain a fire and keep it from spreading from floor to floor.
• The “fire” is defined in ASTM E119 – it is a laboratory standard, not a fire in a real building
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Fire rating of a metal deck and concrete fill:
• For the duration of the fire test, the floor must carry the design load, not allow a 250o
temperature rise through the slab, and not permit flames or hot gasses to penetrate the assembly.
• The building code controls the number of hours required (see IBC Table 601)
• Concrete cover is often controlled by fire rating required rather than structural requirements.
4
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
COMPOSITE CONSTRUCTION
Fire rating of a metal deck and concrete fill:
Usually, deck fireproofing
isn’t considered economical
110 to 115
pcf
145 to 150 pcf
Very common system for Type I-A
(the most fire-resistive
construction) or I-B construction
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Section I.3 of the AISC Specification addresses composite flexural members
Advantages of Composite Construction: Uses material efficiently: concrete in compression
& steel in tension (web & flange)
Reduces weight and increases stiffness compared to non-composite: deflection only 20-30% of non-composite for the same size beam
COMPOSITE CONSTRUCTION
5
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Advantages of Composite Construction: Reduces structural depth
Efficiently supports high live loads or high “post-composite” dead loads because these are applied after the steel beam has become composite with the concrete – takes advantage of the added strength of the composite section
Provides greater reliability because stability limit states are usually not critical
COMPOSITE CONSTRUCTION
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
“Phases” of Composite Beams
• Pre-composite: Bare steel beam supports uncured concrete. Design of beam would be just as you learned in undergraduate steel class.
• Composite:
Beam and concrete slab work together. A shear transfer mechanism must be present.
Concrete must be sufficiently cured to resist the loads applied after composite action is available.
Composite Construction
6
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
“Types” of Loads
• Pre-composite - loads present before composite action can be developed
Dead (e.g., weight of uncured concrete, beam and deck self-weight)
Live (e.g., temporary construction storage)
• Composite – after composite action is available:
Dead (e.g., superimposed loads such as ceilings, floor finishes, landscaping, etc.)
Live (e.g., the building code prescribed live-load).
Composite Construction
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Prior to development of composite action, total weight of uncured (i.e. wet) concrete must be carried by steel beam
Sometimes, the beam is not strong or stiff enough and it must be shored so it does not fail or it doesn’t deflect too much
SHORED CONSTRUCTION
Shoring – exact
placement varies
Beam
Metal deck and
concrete fill
7
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Shoring supports the steel beam (i.e., shored beam carries “no” load) because shoring supports weight of wet concrete until composite action is available. This usually results in a smaller beam.
SHORED CONSTRUCTION
Shoring post
Note that the supporting beam
must be strong enough to
support weight of shored beam
Beam being
shored
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
If unshored, steel beam alone must have adequate strength (i.e., Mn) to resist all pre-composite loads (i.e. Mu) applied prior to concrete reaching 75% of its design strength, f’c
UNSHORED CONSTRUCTION
Beam
Metal deck and
concrete fill
Beam must carry uncured concrete
as a non-composite steel member
until concrete reaches 75% of f’c
8
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Reasons not to shore:
UNSHORED CONSTRUCTION
High expense and difficult construction logistics make shoring almost always more expensive than potential savings from reduced beam weight in shored construction
Creep - Concrete slab always in compression which can lead to greater long-term deflections
Therefore, shored steel frame construction is almost never used. Unshored construction is usually preferred.
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
In many cases, the size of beam is governed by the “pre-composite” condition, but it may not be stiff enough (i.e., too much deflection).
The composite section usually possesses more than enough strength to support the superimposed composite loads.
So, to improve efficiency, we “camber” the beam (pre-deflect the beam upwards)
CAMBER
Beam
Camber – deform beam in opposite direction
of anticipated dead load deflection
9
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Camber allows us to use a more efficient section
Generally, it is recommended that beams be cambered if 80% of pre-composite deflection, based on pre-composite dead load, is > 1/2" to 3/4"
Camber in 0.25 in. increments and limit to about 2.5 in. max. for 30 to 40 ft. long beams (longer beams may have more camber)
Beam
Camber – deform beam in opposite direction
of anticipated dead load deflection
CAMBER
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Goal of using camber: obtain a level or nearly level slab once the pre-composite dead load has been applied and the composite action develops
CAMBER
Camber in pre-composite beam
Level slab after the concrete has been placed
10
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Use a consistent topping thickness rather than trying to obtain a level beam during placement. “Level” beams run the risk of excessive deflection due to “ponding” of concrete
CAMBER
Camber in pre-composite beam
Level slab after the concrete has been placed
“Level” pre-composite beam
Composite beam with excessive concrete to maintain level slab
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Do not over-camber
Sometimes the camber “doesn’t come out” because of fixity and over-estimation of pre-composite loads
CAMBER
Camber in pre-composite beam
Level slab after the concrete has been placed
In most situations, highly restrained
connections should not be cambered
Pinned connections allow the rotation
required to flatten-out a cambered beam
11
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Problem locations: interior columns adjacent to longer spans, roof beams with large camber
Area around interior column won’t deflect. If camber in adjacent beam doesn’t come out, an unintended low spot at the column may result
If camber doesn’t come out of roof beams, roof drainage can be impacted unfavorably.
CAMBER
Long-span beam with excessive
camber (potential high spot)
Area around column
(potential low spot)
Roof beam with
excessive camber
(potential high
spot that
interferes with
roof drainage)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
If camber doesn’t come out:
Shear connector can protrude above slab
Concrete cover may be unequal or insufficient to obtain required fire rating
CAMBER
Camber in pre-composite beam
Level slab after the concrete has been placed
Over-cambered beam with protruding shear connectors
Over-cambered beam with uneven concrete thickness
12
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
General recommendations:
Camber ≤ 0.8Δpre-composite (considers restraint)
Δpre-composite should only include those loads present before composite condition (e.g., beam self-weight, concrete slab)
When evaluating camber for cantilevered beams with and without backspan, consider actual conditions
CAMBER
Does this beam rotate?
Is this a column with significant stiffness?
Relative length of cantilever and backspan
will change apparent deformed shape
Cantilever is unlikely to exhibit
significant composite action
(concrete is in tension)
Based on “pre-composite” dead loads or
other loads you are sure will be present
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Cambering uses hydraulic rams or heat. Operator has to be careful not to buckle the beam. Cambering is a trial and error process
Beam
Beam is pushed beyond elastic limit to achieve a
permanent set. When hydraulic rams are released,
the beam springs back by the elastic deformation.
The residual set is the camber.
Hydraulic ramHydraulic ram
Reference line
(a string)
CAMBER
Cambering with
hydraulic rams
13
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
CAMBER
Heat is applied to create camber on
the opposite site of the desired
crown (i.e., upward camber).
Heat marks go on bottom side (beam’s
crown (camber) is to the top)
Cambering with heat
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Pre-Composite Condition (Positive Moment)
If decking is perpendicular to the beam, it is usually assumed that the unbraced length is 0 ft.
If decking is parallel to the beam (often applies to girders) the deck isn’t oriented in its strongest direction. The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).
Composite Condition (Positive Moment)
Concrete slab braces beam; therefore, unbraced length is 0 ft.
UNBRACED LENGTH OF BEAM
14
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Pre-Composite Condition (Negative Moment)
Bottom flange is in compression (often applies to cantilevers). The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).
Composite Condition (Negative Moment)
Concrete is in tension and bottom flange is in compression (often applies to cantilevers). The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).
UNBRACED LENGTH OF BEAM
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
How much slab acts as part of the composite beam? (See Section I3.1a)
Concrete stress decreases with perpendicular distance from beam. Concrete over the beam flange is under greatest stress. Nevertheless, AISC Specification assumes constant stress over assumed (effective) width with reasonable accuracy.
Concept of effective width is similar to that used for “t-beams” by ACI 318.
EFFECTIVE FLANGE WIDTHRefers to AISC 360 – Specification
for Structural Steel Buildings
Actual stress distribution
in concrete slab
Effective width with uniform
stress distribution
15
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
How much slab acts as part of the composite beam?
be is taken on each side of the beam center line. Effective width is sum of the be values.
t is concrete topping thickness
EFFECTIVE FLANGE WIDTH
t
be1
Effective Flange Width
be2
Slab edge
(where occurs)
Bottom of
steel deck
Top of
steel deckbe2
be1
be1
be2 Slab edge
Effective
flange width
Effective
flange width
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EFFECTIVE FLANGE WIDTH
1 span of beam
81
distance from centerline of beam to beam Use smallest value2
of beam to slab edge
eb
Centerline
Effective flange width = be1 + be2
t
be1
Effective Flange Width
be2
Slab edge
(where occurs)
Bottom of metal
deck
Top of
steel deck
16
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR TRANSFER
Headed shear stud
prior to welding
For the metal deck/concrete and the steel beam to work together effectively, adequate shear transfer must be provided
Shear connectors – studs, headed (“Nelson”) studs –most common
3/4" most common diameter
Length > 4 stud
Beam flange
Beam web
Shear
connectorShear
demand Q
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR TRANSFER
Although not directly related to composite strength calculations, the deck must also be attached to the steel framing
Decking usually attached to steel beam using puddle welds or power-actuated fasteners
Puddle welds used to attach
decking to steel framing
Powder-actuated fasteners
fired through deck into steel
attach decking
17
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR TRANSFER
Most failures occur with slab crushing so we usually assume plastic behavior in both steel and concrete (i.e., we can fully develop Mp in the beam)
Required strength of shear connectors (V’), between point of M+
max and M = 0, is the least of the following (Section I3.2d):
V’ = 0.85 f'cAc Compressive strength of concreteV’ = AsFy Tensile strength of steel beamV’ = Qn Total nominal strength of shear studs
where Ac = area of concrete within effective slab width (in2)
As = area of steel cross section (in2)
Ac = Effective width x slab
topping thickness
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR TRANSFER
Beam
Uniform load
M+maxn* = required number
of shear studs
from M+max to M0
*For most beams (e.g. those
with symmetric uniform
loads), total number of studs
required for a beam is 2n and
they are evenly distributed
along the length of the beam.
M0
If two or more studs per
flute are required, the
additional studs are added
starting at the supports
and working toward the
middle
Usually install one
stud per flute (i.e., 12”
on center)
18
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR TRANSFER
Beam (Girder)
Point loads
M0
M+max
n = required number
of shear studs
from M+max to M0
If the shear diagram shows no (or very little) shear
along a portion of beam, theoretically, no shear
connectors are required. Nevertheless, minimum
spacing of shear connectors governs in this region
Shear diagram
Example of Beam with Point Loads
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
No -factor for strength of studs.
Nominal strength of shear connector is:
where
Asa = Area of shear connector shank (in.2)
Ec = Young's modulus of concrete (ksi)
Fu = Specified minimum tensile strength (ksi)
Rg = Varies between 1.0 and 0.75 depending on number of studs and direction of deck flutes
Rp = Varies between 1.0 to 0.6 depending on how many studs are welded in a given flute
SHEAR CONNECTOR STRENGTH (Qn)
0 5n sa c c g p sa uQ . A f ' E R R A F (Eq. I8-1)
19
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
“Parallel” usually
refers to girders
“Perpendicular”
usually refers to
beams
SHEAR CONNECTOR STRENGTH (Qn)
Condition Rg RP
No decking 1.0 0.75
Decking oriented parallel to the steel shape
wr/hr≥ 1.5
wr/hr < 1.5
1.0
0.85
0.75
0.75
Decking oriented perpendicular to the steel shape
Number of studs occupying the same decking rib
1
2
3 or more
1.0
0.85
0.7
0.6
0.6
0.6
hr = nominal rib height, in.wr = average width of concrete rib or haunch
(as defined in Section I3.2c), in.
Decking oriented perpendicular
to steel shape
(See User Note on page 16.1-98 of the Steel Manual for common values of Rg and Rp)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
SHEAR CONNECTOR STRENGTH (Qn)
f'c Qn
(ksi) (pcf) (kips)3.0 110 17.7 3/4" A36 studs
145 17.7
4.0 110 17.7145 17.7
Most common stud diameter and concrete strength
SHEAR CONNECTOR STRENGTH (Qn) for one stud per rib in the “weak” position (see next slide)
Table I3-21 of the
Specification gives values
for other conditions
20
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Stiffening rib in
deck flute
Weak and Strong Stud Positions
Direction of
shear
Direction of
shear
Stud in the
“strong” position
Stud in the
“weak” position
Deck flutes usually have a stiffening rib that requires stud be located to one side or the other
Stud strength can be a function of position“Weak” position typically assumed unless otherwise
indicated (since specific placement not usually specified)
“More” concrete
to support stud
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
2” min
3” max
1.5” min
Ignore concrete in the flute (rib) unless
the rib is parallel to beam (e.g. a girder
condition -- typically about 50% of the
concrete is available in this case)
Metal deck flutes (ribs)
perpendicular to beams
0.5” min
Metal deck flutes (ribs)
parallel to beam (girder)
Partial Framing Plan
Concrete Cover and Thickness
Metal deck flutes (ribs)
perpendicular to beams
21
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
For uniform loads, a uniform spacing of studs shows similar performance compared with a spacing that follows the statical shear distribution (i.e. VQ ÷ Ib) …therefore we use uniform spacing for uniform loads
For Point Loads - Number of connectors between load and nearest point of zero moment must be sufficient to develop M+
max at point load.
NUMBER, SPACING AND COVER
Moment Diagram M+max
Provide required number of
connectors over this distance
P Key concept
for girder
design
M = 0
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
In areas of zero (low) shear, the maximum spacing of studs would usually govern
NUMBER, SPACING AND COVER
M = 0
M+max
Provide required number of
connectors over this distance
P
Key concept
for girder
design
Moment Diagram
Shear Diagram
P
Provide shear connectors no
further apart than 8 x slab
thickness
22
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Minimum center to center spacing along beam is 6(where is stud diameter) and minimum center to center transverse spacing along beam is 4. If “formed slab” (i.e., no steel deck): 4 either direction.
Studs require min. 1" lateral concrete cover (except where deck is used)
of stud < 2.5 tf if not over webs
NUMBER, SPACING AND COVER
6min
4min
Studs
Top of beam flange
May stagger studs if beam is
too narrow – this can squeeze
in a few more studs over a
given distance.Multiple Studs in a Row (or Flute)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
When composite action is small or a beam has a point of “zero” shear (e.g., the middle segment of a girder loaded at its third points), provide shear connectors spacing governed by maximum spacing
Maximum spacing < 8 x total slab thickness (36 in. max.) to prevent vertical separation between slab and beam flange when slab goes into compression
NUMBER, SPACING AND COVER
6 (min) -
4min
Studs
Top of beam flange8 x toal slab
thickness
(max)
23
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
When decking runs perpendicular to the beam, studs are installed based on the flute spacing in the deck, typically 12 in. on center
When decking runs parallel to the beam (e.g., a girder), the stud spacing is independent of the deck flute spacing and 6 x stud diameter would govern
NUMBER, SPACING AND COVER
If flutes are perpendicular
to beam, flute spacing
establishes stud spacing
If flutes are parallel to beam,
stud spacing is independent of
flute spacing
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
If fully composite section has moment strength Mn*, but you only need Mn* where < 1.0, you canprovide enough studs just to reach Mn*
A partially composite beam will have fewer than the maximum number of studs required to develop the full composite strength of the beam and slab system: same beam and slab but fewer studs = lower moment strength
Partially composite beams are the most common in actual building design
It is recommended that > 25%
PARTIALLY COMPOSITE BEAMS
24
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Many designers will often specify beam studs at 12 in. on center, even if design requires fewer, because:
• Reduces construction errors by keeping things consistent
• Provides added flexural strength at a nominal cost (i.e. studs are less expensive than strengthening the beam later)
PARTIALLY COMPOSITE BEAMS
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Flexural capacity of composite section may be governed by:• Tensile strength of beam, • Compressive strength of concrete slab or • Strength of shear connectors.
The least of these values establishes the limit state that governs the design
Composite section cannot transfer more shear than can be developed by the shear connectors – if the shear connectors have less strength than the concrete slab or the beam, then this will establish the limit on the flexural state of the composite section.
FLEXURAL CAPACITY Key
concept
Usually, when it is a
partially composite
beam
25
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Plastic stress
distribution signifies
section can achieve Mp
Wide flange sections
with Fy ≤ 50 ksi satisfy
this, so you may assume
entire section yields
hc = distance between toes of fillet d-2k
yfw
c
FE
th
76.3
FLEXURAL CAPACITY
If web is slender and in compression, it may buckle.
Use plastic stress distribution in the composite section if:
h c
The flange and the web do not
meet at a right angle due to the
rolling process that creates the
wide flange shape. There is a
small radius at the intersection
called a “fillet.”k
k
d
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Use elastic stress distribution if:
Superimpose elastic stresses and effect of shoring must be considered.
The elastic stress distribution requirement applies most frequently when built-up members are used (e.g. plate girders) and relatively thin plates are used.
3 76 where 0 9c
yfw
h E. ; .Ft
FLEXURAL CAPACITY Not a common condition
-- you must calculate
stress distribution
based on level of strain
26
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
0.85f’c
Fy
Calculating the moment capacity of a composite section is a function of plastic neutral axis (PNA) location.
Possible plastic neutral axis locations:•Neutral axis in the concrete slab •Neutral axis in top flange of beam•Neutral axis in web of steel beam
T (from steel beam)
C (from concrete)
b = effective
widthConcrete in tension
(doesn’t contribute
to flexural strength
but does influence
internal moment arm
dimension)
PNA
FLEXURAL CAPACITY
PNA in slab
Steel beam
weaker than
concrete slab
Stress
Distribution
t = total slab
thickness
Internal
moment arm
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Plastic Neutral Axis in Concrete Slab:
(Note: If a > t, then PNA is not in slab and you need to revise your assumptions and recalculate the flexural capacity.)
2 2
0 9
n p s y
d aM M A F t
.
t
d
a
d/2 + t - a/2
a/2
d/2
T = AsFy
C = 0.85f’cab
b = effective widthConcrete in
tension (doesn’t
contribute to
flexural strength
but does influence
internal moment
arm dimension)
PNA
FLEXURAL CAPACITY
T must equal
C to satisfy
equilibrium
0.85f’c
Fy
Internal
moment arm
27
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
If deck is used, the methodology to calculate the flexural capacity is the same except the slab is elevated above the beam flange by the height of the deck (hr). Adjust geometry.
FLEXURAL CAPACITY
(Note: If a > hc, then PNA is not in slab and you need to limit the compression strength to C = 0.85f’cbhc. If PNA is no lower than t, you can assume PNA is “in the slab” to calculate Mn.)
td
a
d/2 + t - a/2
a/2
d/2
T = AsFy
C = 0.85f’cabPNA
0.85f’c
Fy
hc
hr
Concrete in
compression
Concrete in tension
(below PNA)
Deck
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Fy
0.85f’c
Fy
Neutral Axis in Top Flange of Beam:
As a first guess, we assume neutral axis is at base of flange:
FLEXURAL CAPACITY
0 85 c f y
s f y
C . f' b t AF
T A A F
t C
T
0.85f’cbt
b
y
d-y
FybfybfN.A.
Fy(As - bfy)
y is depth of
steel in
compression
Not all of
the steel is
in tension
Cconc
Csteel
Tsteel
Adjust “t” as required
for concrete topping
over steel deck
28
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Fy
0.85f’c
FyTsteel
Plastic Neutral Axis in Top Flange of Beam:
C = Cconc + Csteel = Tsteel = T
But if C ≠ T with assumed PNA at base of flange, then our assumption is wrong and PNA is not at base of flange.
FLEXURAL CAPACITY
t C
T
0.85f’cbt
b
y
d-y
FybfybfP.N.A.
Fy(As - bfy)
Cconc Csteel
Note: T = C to
satisfy equilibrium
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Fy
0.85f’c
FyTsteel
Plastic Neutral Axis in Top Flange of Beam:
Setting C = T and solving for Mn:
FLEXURAL CAPACITY
0 85
0 85
2
0 9 0 85 22 2 2
c y f y s y f
y s c
y f
n p c y f y s
. f ' b t F b y F A F b y
F A . f' bty
F b
t y dM M . . f' bt y F b y F A y
C
T
t C
T
0.85f’cbt
b
y
d-y
FybfybfP.N.A.
Fy(As - bfy)
Cconc
Csteel
29
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Neutral Axis in Web of Steel Beam
Where Cs = Fy[(bftf) + tw(d-2tf-x)]
Using equilibrium and summing moments about the PNA, one can solve for Mn (derivation not shown).
FLEXURAL CAPACITY
0.85f’c
Fy
Fy
t
C
T
b
bf
0.85f’cbt = Cconc
Fybftf = Cflange
Fy(PNA - tf)tw = Tweb
Fytw(d-2tf-x) = Cweb
d
x
tw
tf
PNAFybftf = Tflange
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
0.85f’c
Fy
Fy
Neutral Axis in Web of Steel Beam
Note that ΣQn = Cconc since this is the force that must be transferred across the slab(deck)/beam interface
Note: PNA = 0.5(AsFy+Cconc) = Fy[As/2 +tw(PNA-d/2)]
FLEXURAL CAPACITY
t
C
T
b
bf
0.85f’cbt = Cconc
Fybftf = Cflange
Fy(PNA - tf)tw = Tweb
Fytw(d-2tf-x) = Cweb
d
x
tw
tf
PNAFybftf = Tflange
30
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
“Inadequate slab” means that
strength of concrete slab is less
than tensile strength of steel
Limit States of Composite Beams Based on Relative Strength of Components
FLEXURAL CAPACITY
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
(Assuming Beams Without Shoring)
The bare steel section must have enough strength to resist weight of uncured concrete and other pre-composite loads (i.e., pre-composite dead and live loads).
If partially cured composite section will support incidental construction loads, it is recommended that an allowance for 10 – 20 psf be included as a “pre-composite” live load.
FLEXURAL CAPACITY
31
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
If the decking is judged not provide bracing for the compression flange, Lb must be considered in designing the beam. Otherwise, Lb = 0.
To avoid yielding the beam, it is recommended that for “pre-composite” condition:
Mu < FyZ
where:
Mu = Required strength (i.e., factored moment) due to loads from concrete + weight of other pre-composite loads
= 0.9
FLEXURAL CAPACITY
This is Z for the
steel beam alone
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
“Transformed moment of
inertia” substitutes steel for
concrete at the ratio of n =
(Es/Ec) ≈ 10 and then calculates
Itr using parallel axis theorem
Lower Bound Moment Of Inertia
Moment of inertia will vary with the applied moment and location of the neutral axis because of the amount of uncracked concrete
Use of a transformed moment of inertia (Itr)using elastic theory will underestimate deflections by 15% to 30%
DEFLECTION ESTIMATE
32
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Lower bound moment of inertia (ILB) is moment of inertia at required strength level (i.e., it uses only the provided shear transfer (Qn) and enough of the slab to balance Qn)
AISC recommends that effective moment of inertia (Ieff) be about 0.75Itr
DEFLECTION ESTIMATE
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Transformed Moment Of Inertia
DEFLECTION ESTIMATE
Effective width of concrete is
transformed (reduced) to equivalent
width of steel based on ratio of n = Es/Ec
(i.e., beff/n)
Transformed moment of inertia (Itr) of the composite
section can be calculated using parallel axis theorem
using the transformed properties
If some of the concrete in the decking
is in compression, an effective width
of 50% may be used when doing the
transformation (only applies when
flutes are parallel to the beam (e.g.,
girders)
beff
33
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Lower Bound Moment Of Inertia
Full elastic analysis involves many assumptions, so, as an alternative, AISC offers the use of Lower Bound Moment of Inertia
DEFLECTION ESTIMATE
2 23 3 12n
LB s s ENA ENAy
QI I A Y d d d Y
F
d
d1
Equivalent concrete area =Qn/Fy
YENA
ENA
a/2
Location of effective
Concrete flange force Qn
C
T d3
Distance from resultant
tension force (for full
tension yield) to top of
steel (usually, d/2)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
YENA = distance from bottom of beam to elastic neutral axis (ENA)
=
DEFLECTION ESTIMATE
3 3 12ns
y
ns
y
QA d d d
F
QA
F
d
d1
Equivalent concrete area =Qn/Fy
YENA
ENA
a/2
Location of effective
Concrete flange force Qn
C
T d3
34
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Design Method Summary
1. Determine pre-composite and composite loads for strength evaluation (factored loads) and serviceability evaluation (unfactored loads)
2. Determine required strength of beam supporting pre-composite loads (using factored loads)
3. Check deflection of pre-composite beam using unfactored dead loads and establish initial camber recommendation (at most, 80% of DL deflection)
4. Check beam under composite loads (factored) and determine required number of studs
5. Check deflection under composite loads (unfactored)
6. Summarize design recommendations
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Center of beam spacing
is 10’/2 = 5’
Design a composite beam to satisfy the following requirements.
Assume beams at 10' center to center, L = 36’, 4" lightweight concrete + 3" deck (t = 7”). Steel is ASTM A992 (i.e. Fy = 50 ksi)
DL = 0.78klf, LL = 1.2klf , Fy = 50 ksi, f'c = 3 ksi
EXAMPLE 1
klfuw 862216178021 .).(.).(.
kftklf
uM 74628
36862 2
.)(.
Factored Loads:
Moment:
Governs2 36 12
1088
2 5 12 120
( )b "
b ( ) "
Effective Flange Width: L = 36’
35
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 1
Try a W18 x 35 As = 10.3 in2
Where is plastic neutral axis? Assume that PNA is in slab.
.in87.1
.)in108()ksi3(85.0)ksi50(in.3.10
'85.02
bf
FAa
c
ys
t
d
ad/2 + t - a/2
a/2
d/2
T = AsFy
C = 0.85f’cba
fy
b = effective
widthConcrete in
tension
N.A
.
Since a < t, the PNA is in concrete slab
Note: really
it is because
a < t-3”h
ch
r
Deck
Concrete in
compression
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Composite section may be oversized because Mn is much greater than Mu. We may be able to reduce the steel section, but we still have to check its ability to support the pre-composite loads (i.e. the loads prior to development of composite action).
What is available strength assuming full composite action?
EXAMPLE 1
nM2 2
17 7 1 87 10 9 10 3 50 7
2 2 12
576 1 462 7
p s y
ft k ft k
d aM A F t
. .( . ) . ( )
. . O.K .
36
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Governs (assuming you
use only the minimum
number of studs – no
longer assuming full
composite action)
Design of Studs
Use Qn = 17.7k for f'c = 3 ksi (weak position) lightweight concrete and =3/4"
What is magnitude of force used to establish the number of studs for shear transfer? Select minimum of:
(a) 0.85 f'cab = 0.85 (3) (1.87) (108) = 515k
or
(b) AsFy = 10.3 (50) = 515k
or
EXAMPLE 1
kx
Qinin
ftinkft
n 3372
2871
72717
127462.
..
. /
(c)22a
td
Mu
This equation only
applies if the PNA is
above the top of the
beam flange!
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
This beam is an example of a partially composite section. A fully composite section would require sufficient studs to transfer at least 515k but only 372k is required to transfer shear due to design loads. Amount of composite action is about 72% (i.e. 372/515 = 0.72)
Minimum number of required studs on each side of beam centerline is:
Say, 22 studs each side of beam centerline or a total of 44 studs
Note that actual Qn is 22(17.7k) = 378.4kThis value is used later in the deflection calculations.
EXAMPLE 1
372 3
21 6517 7min
. kn .
. k
37
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Compare strength of W18x35 against loads present before composite section is developed (i.e. pre-composite condition) :
2
52 10 0 52
0 78 0 52 0 26
20 101 2 0 52 1 6
1000
0 94
0 94 36152
8249 152
psf klfconc
klf
psfklf
u
klf
klfft k
u
ft k ft kp
w ' .
DL' . . .
x 'w . ( . ) .
.
. ( ')M
M
OK
A W18x35 is OK for non-composite loads
Assumed construction
live load
EXAMPLE 1
Weight of concrete slab
Superimposed
dead load (not
used in this check
because it is
applied when the
composite
strength is
available)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Typically, anything
below L/240 for DL
only is considered too
flexible
Service load deflection before composite section forms:
4
0.52
5(0.52)(36) 17281.29"
384(29000)(510)
36' 12325 240
1.29 325
klfw
x LOK
A W18x35 has significant pre-composite dead load deflection, but we can camber out most of this (or we could upsize the beam).
EXAMPLE 1 Camber is based
on DL only in
this case
38
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 1
Beam
Camber 1.0”
How much camber is required?Normally, we camber a section if the pre-composite deflection is greater than about 0.5 to 0.75 in. And we camber in 0.25in. increments.
Therefore, camber in this case is at most 1.25 in. (Δpre-composite =1.29 in. initial deflection).
AISC recommends camber be based on 0.8 Δpre-composite = 1.03”, so use 1 in. of camber
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
2 2
3 3 1
3 3 1
2
2
17 7 378 4 1 8710 3 17 7 7
2 50 2378 4
10 350
15 17
nLB s s ENA ENA
y
ns
y
ENA
ns
y
QI I A Y d d d Y
F
QA d d d
FY
QA
F
( . ) . .. .
..
. "
d3
= d
/2
d1
Equivalent concrete
area =Qn/Fy
YE
NA
ENA
a/2
Service Live Load Deflection After Composite Section
Lower Bound I:
EXAMPLE 1
As = 10.3 in2
Is = 510 in4
d = 17.7 in.
ΣQn = 378.4 k
d1 = tslab – a/2
tslab = 7 in.
a = 1.87 in.
a
t sla
b
39
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
22
4
4
17 7 378 4 1 87510 10 3 15 17 17 7 7 15 17
2 50 2
1481
5 1 2 36 1728384 29000 1481
36 121 06 1 2
360 36036 12
408 3601 06
LB
LL
. . .I . . . .
in.
( . )( )( )( )
L x. " Max.Deflection . "
xO.K .
.
Substitute YENA into equation for Lower Bound Moment of Intertia:
EXAMPLE 1
Building code
typically limits LL
deflection to
L/360
AISC recommends
limiting deflection
to less than 1” for
typical spans
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Final design recommendations are:
W18x35 with 44 - 3/4”diameter studs. Camber beam 1.0”
Note: For a complete design, you must list the beam size, the total number of studs, and the amount of camber.
EXAMPLE 1
This is
important!
40
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Design composite beam for the following loads using AISC Manual
L = 45' Beams spaced 10’o.c.Fy = 50 ksi f'c = 4.0 ksi3" deck + 4.5" hardrock ( = 145 pcf)
DL slab= 0.075ksf
DL beam weight = 0.008ksf (assumed)DL all other = 0.010ksf (ceiling, floor, MEP, etc.)
LL = 0.1ksf (unreduced)
Line loads on beamTotal dead load = 0.093ksf (10 ft) = 0.93klf
Total live load = 0.10ksf (10 ft) = 1.0klf
EXAMPLE 2
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Construction loadsConstruction dead load = 0.083ksf (10 ft) = 0.83klf
Construction live load = 0.020ksf (10 ft) = 0.20klf
Determine required flexural strength
EXAMPLE 2
2
1.2(0.93 ) 1.6(1.0 ) 2.72
2.72 (45 )688
8
klf klf klfu
klf ftft k
u
w
M
Determine beffThe effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline
Concrete slab
plus beam
weight
41
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Note use of “2” to
account for beff on
both sides of beam
EXAMPLE 2
Determine beff(1) One-eighth of the beam span
(2) One-half the distance to centerline of adjacent beam
(3) One-half the distance to edge of slab (N.A.)
45(2) 11.3
8
ftft
10(2) 10.0 controls
2
ftft
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Using the figures on pages 3-30 and 3-31 and the tables starting on page 3-156 of the AISC Steel Manual, we can significantly simplify the effort required to design the beam.
EXAMPLE 2
Different locations for PNA
in beam top flange
Different locations for PNA
in beam web
TFL assumes
compression
force is in
concrete only
Y2 = tslab – a/2
42
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Using the figures on pages 3-14 and 3-15 and the tables starting on page 3-158 of the AISC Steel Manual, we can significantly simplify the effort required to design the beam.
EXAMPLE 2
…
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Calculate moment arm for concrete force measured from top of steel shape (Y2):
Assume a = 1” (this is often a reasonable first guess)
Enter Manual Table 3-19 with the required strength and Y2 = 7.0in. Select a beam and neutral axis location that indicates sufficient available strength
Select a W21x50 as a trial beam.
When PNA Location 5 (BFL), this composite shape has an available strength of
Mn = 769ft-k > 688ft-k OK
EXAMPLE 2
12 7.5" 7.0"
2 2slab
aY t
This is our required
strength for the
composite beam
43
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
From Table 3-19:
EXAMPLE 2
Y2 = 7.0 in.
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Check the beam deflections and available strength
Check the deflection of the beam under pre-composite conditions, considering only the pre-composite dead loads as contributing to the deflection (i.e. no pre-composite live load)
Limit Δpre-composite to a maximum of 2.5 in. to facilitate concrete placement
From Manual Table 3-20, a W21x50 has Ix = 984in4, so the member does not satisfy the deflection criterion under construction loads
3 3
44 4 /5 5(0.83 )(45 ) (1728 )
1060384 384(29000 )2.5
klf ft in ftinDL
req ksi in
wI
E
44
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Check the beam deflections and available strength
Ix of beam
(only)
(Table 3-20)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Using Manual Table 3-20, revise the trial member selection to a W21x55, which has Ix = 1140in4
Check selected member strength as an unshored beam under construction loads assuming adequate lateral bracing through the deck attachment to the beam flange.
Calculate the required strength
2
1.4 1.4(0.83 ) 1.16
1.2 1.6 1.2(0.83 ) 1.6(0.2 ) 1.32
1.32 (45 )331
8
klf klf
klf klf klf
klf ftft k
u
DL
DL LL
M
Governs
45
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
The design strength for a W21x55 is 473ft-k > 331ft-k OK
For a W21x55 with Y2 = 7.0in, the member has sufficient available strength when the PNA is at Location 6 and ΣQn = 292kips
Mn = 767ft-k > 688 ft-k OK
Mp
Y2 = 7.0 in.
ΣQn
(Table 3-19)
(Table 3-19)
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
Check a (were we right to assume a = 1 in?)
0.716in < 1.0in OK (we are close and required concrete compression area is less than we assumed)
Check live load deflectionΔLL= ℓ/360 = [(45ft)12in/ft]/360 = 1.5in
Lower bound moment of inertia for composite beams is tabulated in Manual Table 3-20.
In some situations
you need to go back
and recalculate
strength
EXAMPLE 2
' /
2920.716
0.85 0.85(4 )10 (12 )
kipsinn
ksi ft in ftc
Qa
f b
46
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
For a W21x55 with Y2 = 7.0in and the PNA at Location 6, ILB = 2440in4
1.3in < 1.5in OK
3 3
4
4 4 /5 5(1.0 )(45 ) (1728 )1.3
384 384(29000 )2440
klf ft in ftinLL
LL ksi inLB
wEI
(Table 3-20)
Y2 = 7.0 in.
Note that PNA is in the web
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Determine if beam has sufficient shear strength
Determine the required number of shear stud connectors
Using perpendicular deck with one ¾ in. diameter weak stud per rib (per foot) in normal weight 4ksi concrete, Qn = 17.7k/stud
Total number of shear connectors = 2(17 studs) = 34 studs
45(2.72 ) 61.2
2234 61.2 OK
ftklf k
u
k kn
V
V
/
29217 studs on each side of the beam
17.7
kn
k studn
47
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Check spacing of shear connectors
Since each flute is 12in, use one stud every flute, starting at each support, and proceed for 17 studs on each end of the span (checking to make sure maximum spacing is satisfied).
(Usually, the designer would simply require one stud per foot without trying to save the 4 or 5 extra studs for a 22.5 ft half span.)
6dstud < 12in < 8tslab, therefore shear stud spacing requirements are met
The studs are to be 5in. long, so they will extend a minimum of 1.5in. into the slab
Professor Thomas A. Sabol
CEE 241– Advanced Steel Design
EXAMPLE 2
Final design:
Use W21x55 beam with minimum of 34 studs and camber 1.75 in.
(Note:
(So, use camber of 1.75 in.)
4
4max21 55
10600.8(2.5 ) 1.86
1140
inreq in in
inW x
I
I