Composite Beams 2

Composite Beams (contd)

Girder

Floor beam

L

S

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

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 fc

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 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

tc

h

a

0.85 fc

sy

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

Concrete Slab

Ribbed steel deck

Steel section

Effective width

b

Yc

tw

tf

bf

tc

hr

d

Composite beamwith 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 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 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 fc Ac

Steel yielding: V = As sy

Connectors fail: V = Qn

Shear Connectors

For negative moments, concrete cannot withstand tension. Rebar yields

Tensile yielding: V = Ar syr

Shear connectors: V = Qn

Number of shear connectors

Number of shear connectors = V/Qn

Strength of one shear connector

Asc = x-sectional area of 1 connector,

Rg and Rp on next pages

su = 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

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 fc Ac

Steel yielding: Ct = As sy

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

Eff slab

d

hr

d/2

tf

PNA

0.85fc

a

sy

sy

Cc

e

Case 2

Eff slab

d

hr

d/2

tf

PNA

0.85fc

a

sy

sy

Cc

e

Case 3

Eff slab

d

hr

d/2

tf

PNA

0.85fc

a

sy

Cc

e

Example

Composite framing in typical multi-story building

3.25 lightweight concrete, 2 steel deck.

Concrete: r = 115 lb/ft2; fc = 3 ksi

Additional 30% dead load assumed for equipment during construction

Deck is supported on steel beams with stud connectors.

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

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

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 fc Ac = 0.85 x 3 x 90 x 3.25 = 745.9 kips

Ct = As sy = 6.49 x 50 = 324.5 kips **

Depth of concrete stress block

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

We are expecting 183.4, so this passes