Transcript Document 7324580

Floor Systems &
Framing of Floors
Residential Architectural
Drafting
Floor System Types —
Conventional Dimensional Lumber Framing
Floor System Types —
Open Web Floor Joist
Floor System Types —
Truss Joist Floor Framing
Floor System Types —
Post and Beam Floor Framing
Design Criteria for
Structural Loading
 Load Types



Dead loads
Live loads
Dynamic loads
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
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
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
Typical Loads for Residential
Construction
See Text
Framing Spacing Practices
 Code based, acceptance varies
 Spacing Options




12” OC
16” OC
Common Spacing
19.2” OC
24” OC
Load Consideration for one Joist
 Considering 12” OC joist spacing
12”
1’ X 15’ = 15 SQFT X 50 =
750#
15’-0” SPAN
 Considering 16” OC joist spacing
16”
12’-0” SPAN
1.33’ X 12’ = 15.96 SQFT X 50 =
798#
 Considering 19.2” OC joist spacing
19.2”
18’-0” SPAN
1.6’ X 18’ = 28.8 SQFT X 50
= 1440#
 Considering 24” OC joist spacing
24”
10’-0” SPAN
2’ X 10’ = 20 SQFT X 50 =
1000#
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”
Deflection Allowances
(Stiffness)
Floor = 1/360
Two types of actions or stresses


Roof = 1/240
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.
Table Values
 Look 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 = Modulus
of Elasticity
(Stiffness)
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)
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)
DOUGLAS FIR-LARCH
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
See Text
Joist Sizing & Spacin
Problem #1:
Span = 11’-8”
1
Hem-Fir
1st Step--find E:
E = 1.6
2nd Step--Use E
and find size to fit
span
3rd Step--find Fb
value (2x8):
Fb = 1,380
4th Step--determine
if Fb works with E
Soution =
2x8
@19.2OC
2
See Text
Problem #2:
Span = 14’-9”
Douglas Fir
1st Step--find E:
E = 1.6
2nd Step--Use E
and find sizes to fit
span
3rd Step--find Fb
value (2x10):
Joist Sizing & Spacing
1
2
Fb = 1,045
4th Step-determine
if Fb works with E
5th Step--using Fb
find working column
Soution:
2x10 @ 16
OC(Fb is
5
5
 Loads


40 LL
10 DL
 1--Lumber
type
 2--Lumber
grade
 3-Spacing
 4--Span
Solution: DF
#2-2x10 @ 16”
OC
will span 14’-
Span Table (not in text)
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
Anything wrong here?
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
Beam Type—Solid Lumber Beam
Beam Type—Built-up
Dimensional Lumber Beam
 Dimensional lumber (2x6, 2x8, 2x10, 2x12)
nailed and/or glued together
 Vertical placement
Beam and Joist Attached with
joist hangers
 Joist are
attached
to beams
with
metal
joist
hangers
 What type
of beam
is shown?
Beam Type — Glued Laminated
 Dimensional lumber placed horizontally
and glued together
Beam Type — Parallel Strand
Lumber Beam
 See classroom example
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
Beam Type — Truss I-Joist Beam
 Laminated or Solid wood (top and bottom
chords)
 OSB or Plywood web
Beam Type — Box or Plywood Beam
 2x @ 12” or 16” structure with plywood skin
 Designed by architect or engineer
Beam Type — Flitch Beam
 A sandwich of wood and steel
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 C--Channel
Shape
Shape
Beam Type — Steel Beams
 Drawing Callouts:


Shape, Nominal height x Weight/foot
Example: W10x25
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
Calculating the Reactions of a Beam
 Total load on beam should equal reaction
Reaction
loads:
formula
 25 x 900 = 22500#
R = wl
 R1 = 15/2 x 900# = 6750#
2
W = uniform load
 R2 = 10/2 x 900# = 4500# l = length of span
 R3 = (15/2 + 10/2) x 900 =11250#
W = 900 #/ linear foot
R1
Span = 15’-0”
Span = 10’-0”
R3
R2
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
 Terminology
Tributary area
 Joist/Rafter
of beam
 Beam/Girder
 Post/Column
 Span
 Tributary area
 Conditions of Design
 Uniform load over length of beam
 Beam supported at each end
15’-0”
Simple Beam Design
Simple Beam Design

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”
 Tributary area
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 LL =
10 x 12 x 63 = 7560 TLD 50# DL
20’-0”
 Select beam size from table = 13#
BEAM
10’-0”
7560 TLD w/ span of 12’
 Solution = 4 x 14 Beam
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
Example of
Using Steel
Table
BEAM
18’-0”
 Calculate load:
18 x 30 x 60 =
32400 TLD
 Select Beam
W18 x 40
Steel I-Beam Table
Glued-Laminated Beam Table
Columns and Post
Steel Column Table
Wood Post Table
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
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
Crawl Space
Floor Joist, Beam/Post
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