load tracing 19.ppt - structures1.tcaup.umich.edu2-way slab tributary area of beam b 2-way waffle...
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University of Michigan, TCAUP Structures I Slide 1 of 22
Architecture 314Structures I
Load Tracing
Load PathsLoad DiagramsFloor Systems
Gatti Wool Mill, Rome (Pier Luigi Nervi, 1951)
University of Michigan, TCAUP Structures I Slide 2 of 22
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
University of Michigan, TCAUP Structures I Slide 3 of 22
Load Paths
Gravity loads trace from top down to their resolution at the foundation.
University of Michigan, TCAUP Structures I Slide 4 of 22
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)
University of Michigan, TCAUP Structures I Slide 5 of 22
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
University of Michigan, TCAUP Structures I Slide 6 of 22
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
University of Michigan, TCAUP Structures I Slide 7 of 22
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
University of Michigan, TCAUP Structures I Slide 8 of 22
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.
University of Michigan, TCAUP Structures I Slide 9 of 22
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.
University of Michigan, TCAUP Structures I Slide 10 of 22
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
University of Michigan, TCAUP Structures I Slide 11 of 22
Load Transfer example 1
Construct load diagram and findend reactions for beam B-1
University of Michigan, TCAUP Structures I Slide 12 of 22
Load Transfer example 1
Construct load diagram and findend reactions for girder G-1
W14x22
20 ft
University of Michigan, TCAUP Structures I Slide 13 of 22
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.
University of Michigan, TCAUP Structures I Slide 14 of 22
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
University of Michigan, TCAUP Structures I Slide 15 of 22
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
University of Michigan, TCAUP Structures I Slide 17 of 22
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
University of Michigan, TCAUP Structures I Slide 18 of 22
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
University of Michigan, TCAUP Structures I Slide 19 of 22
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
University of Michigan, TCAUP Structures I Slide 20 of 22
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
University of Michigan, TCAUP Structures I Slide 21 of 22
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
University of Michigan, TCAUP Structures I Slide 22 of 22
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