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Architecture 314Structures I

Load Tracing

Load PathsLoad DiagramsFloor Systems

Gatti Wool Mill, Rome (Pier Luigi Nervi, 1951)

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Load Combinations

Load Types• Dead Load - D• Roof Live Load - Lr• Floor Live Load - L• Snow Load - S• Wind Load - W• Earthquake - E

Load Combinations

Allowable Stress Design (ASD)• D • D + L• D + (Lr or S)• D + 0.75 L + 0.75 (Lr or S)• D + (W or 0.7 E)

Strength Design (LRFD)• 1.4 D• 1.2 D + 1.6 Lr + 0.5(Lr or S)• 1.2 D + 1.6(Lr or S) + (L or 0.8W)• 1.2 D + 1.6W + L + 0.5(Lr or S)• 1.2 D + 1.6E + L + 0.2S

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Load Paths

Gravity loads trace from top down to their resolution at the foundation.

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Load Paths

Roof loads can be applied as projectedloads (e.g. snow or live loads)

or loads on the surface (e.g. dead or wind)

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Load Paths

Member Hierarchy

Flooring spans between joists

Joists span between beams

Beams span between girders

Girders span between columns

Columns carry load to ground

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Load Paths

Concrete slabs span in the direction of the steel reinforcement.

One-way slabs should span the shortest direction.

Two way slabs span in both directions. Aspect ratios should be between 1:1 and 2:1. The load path divides at 45° from corner.

2-way waffle slab2-way slab tributary area of beam B

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Load Paths

Ideal load paths following the isoclines of maximum tension and compression (principal stress patterns). These give the design with least material, but more complex form.

Cassa di Risparmio, Veniceby Pier Luigi Nervi

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Tributary Area

The tributary area is an area that corresponds to the load on a member.

If geometry and loading is symmetric, then load paths and reactions are also symmetric.

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Tributary Area

The tributary area is an area that corresponds to the load on a member.

Each member has a tributary area that can be used to find the total load on that member.

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Load Transfer example 1

Construct the load diagram and find end reactions for Beam B-1 and Girder G-1

For Load on B1:

Floor live load = 40 PSF40 PSF x (2.5’ + 2.5’) = 200 PLF

Floor dead load = slab + terrazzo + beamDL slab = 150 PCF x (4/12)FT/SF = 50 PSF50 PSF x (2.5’ + 2.5’) = 250 PLF

DL terr. = 13 LB/IN/SF (5/8)IN = 8.13 PSF8.13 PSF x (2.5’ + 2.5’) = 41 PLF

DL beam = 31 PLF

TOTAL DL = 250+41+31 = 322 PLF

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Load Transfer example 1

Construct load diagram and findend reactions for beam B-1

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Load Transfer example 1

Construct load diagram and findend reactions for girder G-1

W14x22

20 ft

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Load Transfer example 2

Find Load Diagrams for:

Joists

Beam

Dead Load

Roof = 8 psf (on real surface)

Rafter = 4 plf (on real length)

Beam = 16 plf (on real length)

Snow Load

Snow Load = 40 psf (on projected area)

24” o.c.

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Load Transfer example 2

Load on Rafter

PLF = PSF x o.c.”/12

SL on projected length

• 40 psf x 24”/12 = 80 plf

DL on real length converted to projected

• (roof DL) + (rafter DL) = total

• 8 psf x 24”/12 + 4 plf = 20 plf on beam length

• Projected DL = 13/12 x 20plf = 21.7 plf

Total Projected Load on the Rafter

• DL + SL = 80 + 21.7 = 101.7 plf

Find Rafter End Reaction

• W = 101.7 plf x 16’ = 1627.2 lbs

• RA = RB = W / 2 = 813.6 lbs

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Load Transfer example 2

Load on Beam:

1. Determine total roof + rafter load in psf

• Snow: 40 psf (projected)

• Roof DL: 8 psf 13/12 = 8.67 psf (projected)

• Rafter DL: 4 plf 12/24”) 13/12 = 2.17 psf (projected)

• TOTAL = 40 + 8.67 + 2.17 = 50.83 psf (projected)

2. Find load per 1’ strip on beam

• 8’ x 50.83 psf = 406.67 plf

3. Include beam self load for total plf

• 16 plf (beam) + 406.67 (roof + SL) = 422.67 plf

Beam Reactions on Columns

• R = wL/2 = 422.67 plf x 24’ / 2 = 5072 lbs

Projected plan

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Floor System example 3

Find Load Diagrams for:B1B2G1

Dead Loadwall 800 PLFfloor slab 70 PSF

Live Loadfloor 90 PSF

Notice the order:B1, then B2, then G1

Concrete slab floor system spanning in directions shown

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Floor System example 3 cont.

Find a beam not loaded by other beams, e.g. a joist or a simple beam.

Sketch the tributary area – ½ span to the next member.

Sketch a load diagram

Calculate the distributed loads in PLF

Wall:

Floor:

(total DL+LL on floor = 160 psf)70 + 90 = 160

Load diagram

20 ft

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Floor System example 3 cont.

From the PLF loading, calculate a total W load.

Locate W at the centroid of the distributed loading.

Solve the end reactions by summingmoments or by proportions.

Wall:

Floor:

Load diagram

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Floor System example 3 cont.

Continue with the next beam supporting a previously solved beam.

Sketch the tributary areas – 2-way slabs divide at 45° from each corner. Areas associated with reactions of other beams are proportional to the load distribution.

Sketch a load diagram

Calculate the distributed loads in PLF, finding peak values of varying loads.

(total DL+LL on floor = 160 psf)

Peak Load on 2-way Slab:

Load diagram

20 ft

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Floor System example 3 cont.

From the PLF loading, calculate a total W load.

Locate W at the centroid of the distributed loading.

Solve the end reactions by summingmoments or by proportions.

Wall:

Floor:

Load diagram

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Floor System example 3 cont.

Continue with the next beam supporting a previously solved beam.

Sketch the tributary areas – 2-way slabs divide at 45° from each corner. Areas associated with reactions of other beams are proportional to the load distribution.

Sketch a load diagram

Calculate the distributed loads in PLF, finding peak values of varying loads.

Peak Load on 2-way Slab:

Floor on one side of G1:

Load diagram

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Floor System example 3 cont.

From the PLF loading, calculate a total W load.

Locate W at the centroid of the distributed loading.

Solve the end reactions by summingmoments or by proportions.

Floor on G1:

2-way Slab:

Load diagram