Design Principles Outline Design Process · Design Values PS 20 Section 01 – TPI 1-2007 Selection...
Transcript of Design Principles Outline Design Process · Design Values PS 20 Section 01 – TPI 1-2007 Selection...
1Level 1: Section 06 – Version 3© 2014 SBCA
1
Section 06: Design Principles
2
Section Downloads
Download & Print TTT I Sec 06 Slides
TTT I Sec 06 Handout
Section 05 – Truss Materials Design Values
PS 20
Section 01 – TPI 1-2007 Selection 6.4.2 Repetitive Member Increase
Version 2.1
3
Design Principles Outline
Design Process
Basic Design Principles Statics Forces
Moments
Static Equilibrium
Mechanics of Materials
4
Design Process
The two functions of the structural design process: Load
Resistance
LOAD ≤ RESISTANCE
or in other terms
ACTUAL ≤ ALLOWABLE
1000# ≤ 2000#
Factor of Safety = 2
5
Design Process
Section 06 - Resistance
Section 07 - Load
6
Basic Design Principles
3 basic principles in the design of metal plate connected wood trusses: Statics
Mechanics of Materials
Triangulation
Section 06
Section 01
2Level 1: Section 06 – Version 3© 2014 SBCA
7
Statics
Statics – bodies at rest Equilibrium of bodies subjected to the action of
forces
Dynamics – bodies in motion
The following 3 concepts are used to statically analyze a structure: Forces
Moments
Static Equilibrium
8
Forces
Externally applied loads become internal forces
9
Axial Forces
Act through the length of the truss member Inward or pushing force is axial compression
Outward or pulling force is axial tension
10
Transverse Forces
Act perpendicular to the length of the truss member Cause bending & shear stresses
11
Internal Forces
Internal Tension & Compression Forces
Beam with Point Load(Exaggerated Bending)
Each Beam Element isDistorting Under Internal Forces
Unloaded Beam
Compression
Tension
12
Neutral Axis
Occurs at some depth in the beam where there is neither tension nor compression
3Level 1: Section 06 – Version 3© 2014 SBCA
13
Parallel Truss Chord Axial Forces
14
Pitched Truss Chord Axial Forces
Howe TrussAxial ForcesTension (T) and Compression (C)
acting at each joint
T
T T
T
T T T T T T
C C
C
C
C
C
C C
CCC
C
C C
T
T
T T
C C
15
Stress Reversals
16
Truss Action Under Gravity Load
Image Courtesy of Alpine Engineered Products
Movement
17
Compression Buckling Stabilize the column to increasing
compression load capacity by 4x
18
Incorrect Bracing Top chord buckling under its own weight Lateral Bracing to Prevent Compression Instability
Section 08 – Truss Design, Manufacture & Installation Overview
4Level 1: Section 06 – Version 3© 2014 SBCA
19
Incorrect Bracing
Permanent Web Bracing Broken webs that buckled too far out of plane.
7-1
0-2
7-1
0-2
0-1
0-0
1-4
-0
0-1
0-0
14-0-05-5-12
5-4-0 5-4-04-2-4 4-2-44-5-12 4-5-1228-0-0
2500 LB
12 126
1
2
3
4
5
6
7
8
9
10
6
4-5
4-5 4 -5
4 -5
6 -8
10-1410-14 3-73-7
4-4-0 4-2-4 4-2-4 4-4-014-0-0
5-5-12
10-106-8 8-128-121112131415
1-4
-0Section 03 – Design Responsibilities
20
Permanent Web Bracing
Truss Design Drawing shows if permanent CLR is required on a web
All lateral braces require diagonal braces for stability
21
Permanent Web Bracing
Diagonal Bracing
22
Purlins
4 ft. on-center
Top Chord Continuous Lateral
Brace
Diagonal Bracing
23
Moments
Force that produces a rotation of a member &
subsequent bending stresses
P
L
aa
24
Moment Equation
P
L
Force (lbs)
aa
Length (ft)
Torque /Bending Moment = Forces that act a distance away to produce rotation about a point
5Level 1: Section 06 – Version 3© 2014 SBCA
25
Moment Equilibrium
P
L
Mwall = Mend of board
26
Truss Bending Moments
Rotation Point Rotation Point
27
Static Equilibrium
L/2 L/2
28
Static Equilibrium
+ positive direction of force
negative direction of force
Sign Convention can be reversed...just need to be consistent
To achieve static equilibrium: the sum of all forces = zero
ΣForces = 0
Sigma is “the sum of”
29
Static Equilibrium
2000 - R1 - R2 = 0
R1 + R2 = 2000 lbs.
R1 = R2
R1 = 2000 lbs./2 = 1000 lbs.
R2 = 1000 lbs.
+ positive direction of force
negative direction of force
ΣForces = 0
ΣV = 0Set unknown values = known values
30
Equations of Equilibrium
Vertical Forces:
Horizontal Forces:
Moments:
ΣV = 0
ΣH = 0
ΣM = 0
6Level 1: Section 06 – Version 3© 2014 SBCA
31
Vertical Forces: ΣV = 0
ΣV = 0: P + w – R1 – R2 = 0
+ positive direction of force
negative direction of force
32
Horizontal Forces: ΣH = 0
ΣH = 0: W – R1 – R2 = 0
Environmental Loads
Generated by Truss-to-Bearing Connection
33
Moments: ΣM = 0
ΣMRP = 0: (P x 2L) – (2P x L) = 0
2PL – 2PL = 0
0 = 0+ clockwise rotation
- counter clockwise rotation
Fulcrum
[Rotation Point (RP)]
3434
Sign Convention Review
ΣV = 0
ΣH = 0
ΣM = 0
+ positive direction of force
negative direction of force
+ clockwise rotation
- counter clockwise rotation
35
Free Body Diagrams
Account for all forces acting on a structure
36
Quiz 1
7Level 1: Section 06 – Version 3© 2014 SBCA
37
Mechanics of Materials
Strength of material
Focused on lumber
Lumber is anisotropic
Section 05 – Truss Materials38
Wood Cells
Like drinking straws Stronger lengthwise than crosswise
39
Crushing
(see design values download page 3
Example 2x4 No. 1 SP)Compression Perpendicular
to Grain
Compression Parallel to Grain
565 psi 1650 psi
40
Six Lumber Design Values
Fb – Bending Stress
Ft – Tension Parallel to Grain
Fv – Shear Stress
Fc┴ – Compression Perpendicular to Grain
Fc – Compression Parallel to Grain
E – Modulus of Elasticity (MOE)
41
Lumber Design Values
Section 02 – Terminology42
Using the Lumber Guides
Southern Pine Use Guide Southern Pine Council
Western Lumber Product Use Manual Western Wood Products Association
The U.S. Span Book for Major Lumber Species Canadian Wood Council
Design Values Handout
8Level 1: Section 06 – Version 3© 2014 SBCA
43
Bending Stress (Fb)
44
Tension Stress (Ft)
45
Shear Stress (Fv)
Horizontal shear
46
Compression Stress Perpendicular to Grain (Fc┴
)
47
Compression Stress Parallel to Grain (Fc)
48
Stress
Stress = Force/Area
ℓ in.
w in.
Area (A) = ℓ" x w" = ℓw in2
P lbs.
Stress (F) = Force (P) / Area (A) psi
9Level 1: Section 06 – Version 3© 2014 SBCA
49
Stress
Stress =Axial Compression Force (lbs.)
(1.5 in. x 3.5 in.)Fc =
50
TTT 1 Sec 06 Handout
51
Stress Example
P = 4000 lbs. axial tension force in a 2x4 chord
A = cross-sectional area of the chord: A = (1.5 in.) x (3.5 in.) = 5.25 in.2
Stress (F) in chord: F = P/A
F = (4000 lbs.)/(5.25 in.2)
F = 761.9 lbs./in.2
F = 762 psi
52
Stress Example
Ft = 762 psi (actual stress)
Fc =P
A
Actual ≤ Allowable
Stress Example
53
Select Structural
762 ≤ 900
54
Combined Stress Index (CSI)
Summation of axial & bending stresses divided by their respective allowable stress for a specific truss member.
No transverse loads
Section 02 – Terminology
=762 psi
900 psi= 0.85CSI =
Ft (actual)
Ft (allowable)
Fb (actual)
Fb (allowable)+
0
10Level 1: Section 06 – Version 3© 2014 SBCA
55
Stress
Pressure and Stress
56
Repetitive Member Factor
57
Repetitive Member Factor
15% increase for bending stress
Selections from ANSI/TPI 1-2007
12
3
58
Modulus of ElasticityE = Measure of material stiffness or how it will deform under load
Section 05 – Truss Materials
The higher the E value...the stiffer the material.
E value does not correlate to strength.
59
Quiz 2
60
Feedback