Floor Systems & Floor Systems & Framing of FloorsFraming of Floors

Residential Architectural Drafting

Floor System Types —Floor System Types —Conventional Dimensional Lumber FramingConventional Dimensional Lumber Framing

Floor System Types —Floor System Types —Open Web Floor JoistOpen Web Floor Joist

Floor System Types —Floor System Types —Truss Joist Floor FramingTruss Joist Floor Framing

Floor System Types —Floor System Types —Post and Beam Floor FramingPost and Beam Floor Framing

Design Criteria for Design Criteria for Structural LoadingStructural Loading

Load Types Dead loads Live loads Dynamic loads

Dead LoadsDead Loads

Definition: (DL) loads that make up the actual weight of the structure, such as walls, floors, roofs and any permanently fixed loads such as furnace, air conditioner or other service equipment. Materials that make up the walls, such as, studs, plywood, insulation, sheet rock, nails, glue, etc. are DL.

Building codes specify a minimum of 10#/sq ft for floors and ceilings, DL = 10#/sqft

Live LoadsLive Loads

Definition: (LL) loads that are fluctuating and changing through the use of the building. These loads include: people, furniture, and exterior weather related items such as, ice, snow, rain, etc.

Building codes specify the amount of live load upon type of use or occupancy. Codes differ, common residential LL = 40#/sq ft

Dynamic LoadsDynamic Loads

Definition: loads imposed on the structure by outside natural forces, such as wind and earthquake.

Wind loads Shear wall design used to resist wind pressure Uplift forces placed upon the roof

Earthquake loads Seismic loads causing lateral forces on entire

structure

Typical Loads for Residential Typical Loads for Residential ConstructionConstruction

See Text

Framing Spacing PracticesFraming Spacing Practices

Code based, acceptance variesSpacing Options

12” OC 16” OC 19.2” OC 24” OC

Common Spacing

Load Consideration for one JoistLoad Consideration for one JoistConsidering 12” OC joist spacing

Considering 16” OC joist spacing

Considering 19.2” OC joist spacing

Considering 24” OC joist spacing

16” 12’-0” SPAN

1’ X 15’ = 15 SQFT X 50 = 750#

12” 15’-0” SPAN

19.2” 18’-0” SPAN

24” 10’-0” SPAN

1.33’ X 12’ = 15.96 SQFT X 50 = 798#

1.6’ X 18’ = 28.8 SQFT X 50 = 1440#

2’ X 10’ = 20 SQFT X 50 = 1000#

Sizing Joist Using Span TablesSizing Joist Using Span TablesLoading reactions of wood members

For every action there is an equal and opposite reaction, creates a “state of rest”

Two types of actions or stresses Fiber Bending Stress (Fb)--a bending stress Modulus of Elasticity (E)--stiffness of structure

• considered as the deflection or amount of sag when structural members are given a load.

Deflection Allowances (Stiffness)

Floor = 1/360

Roof = 1/240

Table ValuesTable ValuesLook up values for lumber type & grade

Normal Duration for fiber stress (Fb)• Typical consideration for floor loads

Modulus of Elasticity (E)See Text

Fb = fiber stress in bending

E = Modulusof Elasticity(Stiffness)

Construction Lumber Considerations —Construction Lumber Considerations —Wood Type & QualityWood Type & Quality

Wood Type Available in Area Douglas Fir-Larch (North) Hemlock-Fir (North) Spruce-Pine-Fir (North) Southern Pine

Wood Quality or Grade Value Select Structural (Best) No. 1/No. 2 (Normally specified) No. 3 (Worst)

EFbDOUGLAS FIR-LARCH

Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2x8 Douglas fir-Larch for grade No.1/No.2 (see text)

Hem-Fir Fb E

Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2x10 Hemlock-Fir for grade No.1/No.2

See Text

Problem #1:Span = 11’-8”Hem-Fir1st Step--find E:E = 1.6

1

2

4

Joist Sizing & Spacing

Value is under1,380 so itworks!

2nd Step--Use E and find size to fit span3rd Step--find Fb value (2x8):Fb = 1,3804th Step--determineif Fb works with E

*

See Text

1

4

Problem #2:Span = 14’-9”Douglas Fir1st Step--find E:E = 1.6

Joist Sizing & Spacing

2nd Step--Use E and find sizes to fit span3rd Step--find Fb value (2x10):Fb = 1,0454th Step-determineif Fb works with E

Value

is over

1045 it 4

doesn’t

work!

5th Step--using Fbfind working column

55

2

Span Table Span Table (not in text)(not in text)Loads 40 LL 10 DL

1--Lumber type

2--Lumber grade

3-Spacing4--Span

Solution: DF #2-2x10 @ 16” OCwill span 14’-11”

Handout on Structural Handout on Structural Analysis #1Analysis #1Use both charts in textRemember that if all values in the “E”

column apply and work then the modulus of Elasticity is the tendency of failure

If values are adjusted in the Fb row then the Fiber stress in bending is the tendency of failure

Anything wrong here?Anything wrong here?

Beam TypesBeam Types1--Solid timber beam2--Built-up dimensional lumber beam3--Glued Laminated beam4--Parallel strand lumber beam (PSL)5--Laminated veneer lumber beam (LVL)6--Truss I-Joist beam7--Box or Plywood beam8--Flitch beam (wood and steel)9--Steel beams

Beam Type—Solid Lumber BeamBeam Type—Solid Lumber Beam

Beam Type—Built-up Beam Type—Built-up Dimensional Lumber BeamDimensional Lumber Beam

Dimensional lumber (2x6, 2x8, 2x10, 2x12) nailed and/or glued together

Vertical placement

Beam and Joist Attached with Beam and Joist Attached with joist hangersjoist hangers

Joist are attached to beams with metal joist hangers

What type of beam is shown?

Beam Type — Glued LaminatedBeam Type — Glued LaminatedDimensional lumber placed horizontally

and glued together

Beam Type — Parallel Strand Beam Type — Parallel Strand Lumber BeamLumber Beam

See classroom example

Beam Type — Laminated Beam Type — Laminated Veneer Lumber BeamVeneer Lumber Beam

Laminated Veneer Lumber (LVL)

Made of ultrasonically graded douglas fir veneers with exterior adhesives under heat and pressure

1 3/4” wide x (5 1/2 to 18”) depth

Beam Type — Truss I-Joist BeamBeam Type — Truss I-Joist BeamLaminated or Solid wood (top and bottom

chords)OSB or Plywood web

Beam Type — Box or Plywood BeamBeam Type — Box or Plywood Beam2x @ 12” or 16” structure with plywood skinDesigned by architect or engineer

Beam Type — Flitch BeamBeam Type — Flitch Beam

A sandwich of wood and steel

Beam Type — Steel BeamsBeam Type — Steel BeamsS shape (American

Standard shape) Often called an

I-beamW & M shapes

Wide flange design

C shapeChannel shape S--

I ShapeW or MShape

C--ChannelShape

Beam Type — Steel BeamsBeam Type — Steel BeamsDrawing Callouts:

Shape, Nominal height x Weight/foot Example: W10x25

ReactionReactionReaction is the

portion of the load that is transferred to the bearing points of the beam

A simple beam reaction to a load would be at the end supports. Each end would support or be required to carry half the total load

Calculating the Reactions of a BeamCalculating the Reactions of a BeamTotal load on beam should equal

reaction loads: 25 x 900 = 22500#

R1 = 15/2 x 900# = 6750#R2 = 10/2 x 900# = 4500#R3 = (15/2 + 10/2) x 900 =11250#

R2R3R1

W = 900 #/ linear foot

Span = 15’-0”

Span = 10’-0”

Reaction formula R = wl 2

W = uniform load l = length of span

Simple Beam DesignSimple Beam DesignSimple beam has a uniform load evenly

distributed over the entire length of the beam and is supported at each end.

Uniform load = equal weight applied to each foot of beam

Simple Beam DesignSimple Beam Design

Terminology Joist/Rafter Beam/Girder Post/Column Span Tributary area

Conditions of Design Uniform load over length of beam Beam supported at each end

Bea

m sp

anTributary area of beam 15

’-0”

Simple Beam DesignSimple Beam Design

Tributary area 16’ x 15’ = 240 sq ft

Total Load on Beam 240 x 50#/sq ft =

12,000#Load at each

supporting end 12,000/2 = 6000#

Bea

m sp

anTributary area of beam 15

’-0”

Determine the size of a Solid Determine the size of a Solid Wood Beam using Span TableWood Beam using Span Table

1)Determine the tributary area and calculate the total load (W) for the beam

10 x 12 x 63 = 7560 TLD

Select beam size from table

12’- 0

”10’-0”

BEAM20’-0”

LL = 50# DL = 13#

7560 TLD w/ span of 12’

Solution = 4 x 14 Beam

Reading the Steel Table Reading the Steel Table

Table values of load are given in kips 1 kip = 1000 lbs

Shape and nominal size across the topWeight per foot is given below designationSpan is located along the left side of table

Example of Example of Using Steel Using Steel TableTable

Calculate load: 18 x 30 x 60 = 32400 TLD

Select Beam W18 x 40

30’-0

”

18’-0”

BEAM

Steel I-Beam TableSteel I-Beam Table

Glued-Laminated Beam TableGlued-Laminated Beam Table

Columns and PostColumns and Post

Steel Column TableSteel Column Table

Wood Post TableWood Post Table

Load ConsiderationsLoad Considerations

First floor loads (DL + LL) = 50#/sq ftFirst floor partitions (DL) = 10#/sq ftSecond floor loads (DL + LL) = 50#/sq ftSecond floor partitions (DL) = 10#/sq ftIf Truss design no loads on interior structure(DL)If rafter/ceiling joist design (DL) = 20#/sq ftRoof load regionally varies (LL) = 20-50#/sq ft

Beam Sizing Beam Sizing and Post and Post SpacingSpacing

Trial & Error Method1--Locate tributary area

2--Determine various conditions placing post

to shorten the beam span

3--Go to tables & choose beam

4--Smaller beams are less expensive and usually better

Crawl SpaceCrawl SpaceFloor Joist, Beam/PostFloor Joist, Beam/Post

Handout on Structural Handout on Structural Analysis #2Analysis #2

Before doing calculations sketch problem to visualize conditions

Calculate the tributary loads for beams and columns conditions

Use Handout charts and tables and select beams and columns for conditions