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Floor Systems & Framing of Floors

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Presentation on theme: "Floor Systems & Framing of Floors"— Presentation transcript:

1 Floor Systems & Framing of Floors
Residential Architectural Drafting

2 Floor System Types — Conventional Dimensional Lumber Framing

3 Floor System Types — Open Web Floor Joist

4 Floor System Types — Truss Joist Floor Framing

5 Floor System Types — Post and Beam Floor Framing

6 Design Criteria for Structural Loading
Load Types Dead loads Live loads Dynamic loads

7 Dead 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

8 Live 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

9 Dynamic 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

10 Typical Loads for Residential Construction
See Text

11 Framing Spacing Practices
Code based, acceptance varies Spacing Options 12” OC 16” OC 19.2” OC 24” OC Common Spacing

12 Load Consideration for one Joist
Considering 12” OC joist spacing Considering 16” OC joist spacing Considering 19.2” OC joist spacing Considering 24” OC joist spacing 12” 15’-0” SPAN 1’ X 15’ = 15 SQFT X 50 = 750# 16” 12’-0” SPAN 1.33’ X 12’ = SQFT X 50 = 798# 19.2” 18’-0” SPAN 1.6’ X 18’ = 28.8 SQFT X 50 = 1440# 24” 10’-0” SPAN 2’ X 10’ = 20 SQFT X 50 = 1000#

13 Sizing Joist Using Span Tables
Loading 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

14 Table Values Look up values for lumber type & grade See Text
Normal Duration for fiber stress (Fb) Typical consideration for floor loads Modulus of Elasticity (E) See Text Fb = fiber stress in bending E = Modulus of Elasticity (Stiffness)

15 Construction Lumber Considerations — Wood 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)

16 E Fb 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) E Fb DOUGLAS FIR-LARCH

17 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 Hem-Fir Fb E

18 Joist Sizing & Spacing Problem #1: Span = 11’-8” Hem-Fir
See Text Problem #1: Span = 11’-8” Hem-Fir 1st Step--find E: E = 1.6 Joist Sizing & Spacing 1 2nd Step--Use E and find size to fit span 2 * 3rd Step--find Fb value (2x8): Fb = 1,380 Value is under 1,380 so it works! 4th Step--determine if Fb works with E Solution = (E controls failure--all fits) 4

19 Joist Sizing & Spacing See Text Problem #2: Span = 14’-9” Douglas Fir
1st Step--find E: E = 1.6 1 2nd Step--Use E and find sizes to fit span 3rd Step--find Fb value (2x10): Fb = 1,045 2 4th Step-determine if Fb works with E Value is over 1045 it 4 doesn’t work! 5th Step--using Fb find working column Solution: 16 OC(Fb is tendency to failure) 4 5

20 Span Table (not in text)
Loads 40 LL 10 DL 1--Lumber type 2--Lumber grade 3-Spacing 4--Span Solution: DF 16” OC will span 14’-11”

21 Handout on Structural Analysis #1
Use both charts in text Remember 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

22 Beam Design

23 Anything wrong here?

24 Beam Types 1--Solid timber beam 2--Built-up dimensional lumber beam
3--Glued Laminated beam 4--Parallel strand lumber beam (PSL) 5--Laminated veneer lumber beam (LVL) 6--Truss I-Joist beam 7--Box or Plywood beam 8--Flitch beam (wood and steel) 9--Steel beams

25 Beam Type—Solid Lumber Beam

26 Beam Type—Built-up Dimensional Lumber Beam
Dimensional lumber (2x6, 2x8, 2x10, 2x12) nailed and/or glued together Vertical placement

27 Beam and Joist Attached with joist hangers
Joist are attached to beams with metal joist hangers What type of beam is shown?

28 Beam Type — Glued Laminated
Dimensional lumber placed horizontally and glued together

29 Beam Type — Parallel Strand Lumber Beam
See classroom example

30 Beam Type — Laminated Veneer 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

31 Beam Type — Truss I-Joist Beam
Laminated or Solid wood (top and bottom chords) OSB or Plywood web

32 Beam Type — Box or Plywood Beam
12” or 16” structure with plywood skin Designed by architect or engineer

33 Beam Type — Flitch Beam A sandwich of wood and steel

34 Beam Type — Steel Beams S shape (American Standard shape)
Often called an I-beam W & M shapes Wide flange design C shape Channel shape S-- I Shape W or M Shape C--Channel

35 Beam Type — Steel Beams Drawing Callouts:
Shape, Nominal height x Weight/foot Example: W10x25

36 Reaction Reaction 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

37 Calculating the Reactions of a Beam
Total 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# Reaction formula R = wl 2 W = uniform load l = length of span W = 900 #/ linear foot Span = 10’-0” Span = 15’-0” R1 R3 R2

38 Simple Beam Design Simple 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

39 Simple 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 Tributary area of beam 15’-0” Beam span

40 Simple Beam Design Tributary area Total Load on Beam
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# Tributary area of beam 15’-0” Beam span

41 Determine the size of a Solid Wood Beam using Span Table
1)Determine the tributary area and calculate the total load (W) for the beam x 12 x 63 = 7560 TLD Select beam size from table LL = 50# DL = 13# 12’-0” 10’-0” BEAM 20’-0”

42 7560 TLD w/ span of 12’ Solution = 4 x 14 Beam

43 Reading the Steel Table
Table values of load are given in kips 1 kip = 1000 lbs Shape and nominal size across the top Weight per foot is given below designation Span is located along the left side of table

44 Example of Using Steel Table
30’-0” 18’-0” BEAM Calculate load: 18 x 30 x 60 = TLD Select Beam W18 x 40

45 Steel I-Beam Table


47 Glued-Laminated Beam Table

48 Columns and Post

49 Steel Column Table

50 Wood Post Table

51 Load Considerations First floor loads (DL + LL) = 50#/sq ft
First floor partitions (DL) = 10#/sq ft Second floor loads (DL + LL) = 50#/sq ft Second floor partitions (DL) = 10#/sq ft If Truss design no loads on interior structure(DL) If rafter/ceiling joist design (DL) = 20#/sq ft Roof load regionally varies (LL) = 20-50#/sq ft

52 Beam Sizing and Post Spacing
Trial & Error Method 1--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

53 Crawl Space Floor Joist, Beam/Post

54 Handout on Structural Analysis #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

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