Transcript Composite Beams (cont`d)

Composite Beams (cont’d)
Floor beam
Girder
S
L
b
tc
h
Effective concrete-steel T-Beam
• The composite beam can be designed as an
effective T-Beam, where width of the slab on
either side is limited to:
– 1/8 of the beam span
– ½ distance to centerline of adjacent beam
– The distance to the end of the slab
Shoring
• Temporary shores (supports) during
construction are optional.
• If temporary shores are NOT used, the steel
section must have adequate strength to
support all loads prior to concrete attaining
75% of f’c
Shear Strength
• Design shear strength and allowable shear
strength of composite beams are based on
just the steel section!
Flexural Strength
• Positive Flexural strength fbMn (or Mn/Wb) are
determined as follows:
– fb = 0.90 (LRFD) and/or Wb = 1.67 (ASD)
• Mn depends on h/tw as follows:
• If h tw  3.76 E yf determine Mn for yield from plastic
stress distribution on composite section (flange yield)
• Else, determine Mn from yielding from superposition of
elastic stresses, considering shoring

b
0.85 f’c
tc
a
h
y
Negative moment
• The design Negative moment can be based on
the steel section alone.
• Could be based on plastic stress distribution
through composite section provided
– Steel beam is adequately braced compact section
– Shear connectors in the negative moment area
– Slab reinforcement parallel to steel is properly
developed
Shear Connectors
Steel section
Concrete Slab
Ribbed steel deck
Effective width
b
tc
Yc
hr
tw
d
tf
bf
Composite beam
with formed steel deck
• Nominal rib height is limited to 3 inches.
• Width of rib or haunch must be at least 2 inch.
For calculations, never more than minimum clear
width
• Must be connected with shear connectors ¾” or
less in diameter. Can be welded through deck or
to steel cross-section.
• Connectors must not extend more than 1.5”
above the top of the deck.
• Must be at least ½” cover
Composite beam with formed steel
deck (cont)
• Slab thickness must be at least 2”
• Deck must be anchored to all supporting
members at max spacing of 18”.
• Stud connectors, or a combination of stud
connectors and arc spot (puddle) welds may be
used
• If ribs are perpendicular to steel, concrete below
the steel deck must be neglected for calculation
section properties and concrete area
Composite beam with formed steel
deck (cont)
• For deck ribs parallel to steel beam, concrete
below top of steel deck may be included in
determining composite section properties and
area of concrete.
• Deck ribs over beams may be split and separated
to form concrete haunch.
• When depth of deck is 1.5” or greater, average
width of supported haunch or rib must be at least
2” for the first stud plus four stud diameters for
each additional stud.
Shear Connectors
• Shear force is transferred by the connectors
• The total horizontal shear force, V’, between
max positive moment and zero moment is the
smallest of
– Concrete crushing: V’ = 0.85 f’c Ac
– Steel yielding: V’ = As y
– Connectors fail: V’ = ∑Qn
Shear Connectors
• For negative moments, concrete cannot
withstand tension. Rebar yields
– Tensile yielding: V’ = Ar yr
– Shear connectors: V’ = ∑ Qn
Number of shear connectors
• Number of shear connectors = V’/Qn
• Strength of one shear connector
Qn  0.5Asc f c' E c  Rg Rp Ascu
Asc = x-sectional area of 1 connector,
Rg and Rp on next pages

u = tensile strength of connector
Rg
• Rg = 1 for
– One stud welded in steel deck rib with deck perpendicular to
steel shape;
– Any number of studs welded in a row through steel deck with
deck parallel to steel shape and ratio of rib width to depth ≥ 1.5
• Rg = 0.85 for
– Two studs welded through steel deck rib with deck
perpendicular;
– One stud welded through deck parallel to steel and rib width to
depth < 1.5
• Rg = 0.7 for
– Three or more studs welded in the deck rib, perpendicular to
steel
Rp
• Rp = 1.0 for
– Studs welded directly to steel shape (not through steel deck) and
having a haunch detail with not more than 50% of the top flange
covered by deck or sheet steel closures.
• Rp = 0.75 for
– Studs welded in composite slab, deck perpendicular to steel, emid-ht ≥ 2
inch
– Studs welded through deck, deck parallel to steel
• Rp = 0.6 for
– Studs welded in composite slab, deck perpendicular to steel and emid-ht
< 2 inch
• emid-ht = distance from edge of stud shank to steel deck web
measured at mid height of deck rib in the load bearing direction of
the stud (direction of maximum moment)
Channels
• Channels welded to steel beam may be used as
shear connectors.
• Welds must develop the shear resistance Qn
• Effects of eccentricity must be considered


Qn  0.3 t f  0.5t w Lc
f c' E c
• Where tf = flange thickness of channel connector
• tw = web thickness of channel shear connector
• Lc =length of channel shear connector
Compressive Strength
– Concrete crushing: Cc = 0.85 f’c Ac
– Steel yielding: Ct = As y
– Connectors fail: Cs = ∑Qn
• Similar to shear values
• The location of the plastic neutral axis affects
the failure criteria
Location of Plastic Neutral Axis
• Case 1: PNA is in the web of the steel. Occurs
when concrete compressive force is less than
web force, Cc ≤ Pyw
• Case 2: PNA is in the thickness of the top
flange. Pyw < Cc < Ct
• Case 3: PNA is in the concrete slab. Cc ≥ Ct
– Note: in Case 3, concrete below PNA is neglected!
Case 1
0.85f’c
a
Eff slab
Cc
hr
e
PNA
y
d
d/2
tf
y
Case 2
0.85f’c
a
Eff slab
Cc
hr
PNA
y
d
d/2
tf
y
e
Case 3
0.85f’c
a
Eff slab
Cc
hr
PNA
e
d
d/2
tf
y
Example
• Composite framing in typical multi-story building
• 3.25” lightweight concrete, 2” steel deck.
– Concrete: r = 115 lb/ft2; f’c = 3 ksi
– Additional 30% dead load assumed for equipment during
construction
• Deck is supported on steel beams with stud
connectors.
– ¾” diameter, 3.5” long
• Unshored construction
– Beams must support their own weight, weight of concrete
before it hardens, deck weight and construction loads.
• Check floor for vibration with damping ration of 5%.
Example (p2)
•
•
•
•
Typical beam is 30 ft long.
Distance to adjacent beams is 10 ft.
Ribs are perpendicular to the beam
Uniform dead loads on beam are, 500 lb/ft +
30% for equipment loads
• Superimposed loads are 250 lb/ft
• Live loads (uniform) 500 lb/ft
Example (p3)
• Have to pick a beam. Must handle 1.3*0.5 +
wt of beam.
• Using A992 (50 ksi) steel. Assume 22 lb/ft
starting estimate
• W = 1.3*0.5 + 0.022 kip/ft = 0.672 kip/ft
• Factored load: 1.4*0.672 = 0.941
• Factored moment: 0.941 * L2/8 = 0.941*302/8
= 105.8 kip-ft
Plastic section modulus
105.8  12
Z

 28.2 in 3
f y
0.9  50
Mu

Fortunately, a W14x22 has a Z=33.2 in3, I=199 in4, and w=22
Deflection of the beam
• The deflection of the beam is given as
5wL
5  0.522 30  12


 1.6"
384EI
384  29,000 199
4
4
3
• So camber the beam by 1.6” prior to pouring

the concrete. Probably make it 1.5” in
drawings.
Next step
• We know that a W14x22 will handle the
unshored loads. We need to consider live
loads as well.
• We can apply the load reduction factor
considering our area (30’ x 10’ between
beams and supports)
• R = 0.0008(A-150) = 0.0008(300-150)=0.12
• So our live load is 0.5*(1-0.12) = 0.44 kip/ft
Factored load
• Greater of
– 1.2(0.5+0.25+0.022) + 1.6(0.44) = 1.63 kip/ft
– 1.4(0.5+0.25+0.22) = 1.081 kip/ft
• Factored moment is thus
– Mn = 1.63 * 302/8 = 183.4 kip-ft
Concrete compressive force
• Concrete flange with is lesser of
– B = 10x12 = 120” or
– B = 2 (30 x 12/8) = 90” **
• Compressive force in concrete is smaller of
– Cc = 0.85 f’c Ac = 0.85 x 3 x 90 x 3.25 = 745.9 kips
– Ct = As y = 6.49 x 50 = 324.5 kips **
Depth of concrete stress block
C
324.5
a

 1.414 in
'
0.85 f c b 0.85  3.0  90
Since Cc > Ct, PNA is in the concrete slab.
The distance between the compression and tension forces, e, on the W14x22
e = 0.5d + 5.25 – 0.5a
= 0.5 x 13.7 + 5.25 – 0.5*1.414 = 11.393 in
fM n  0.9Ct e  0.9  324.5  11.393 /12  277.3
We are expecting 183.4, so this passes