1Level 1: Section 06 – Version 3© 2014 SBCA

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Section 06: Design Principles

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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

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Design Principles Outline

Design Process

Basic Design Principles Statics Forces

Moments

Static Equilibrium

Mechanics of Materials

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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

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Design Process

Section 06 - Resistance

Section 07 - Load

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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

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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

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Forces

Externally applied loads become internal forces

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Axial Forces

Act through the length of the truss member Inward or pushing force is axial compression

Outward or pulling force is axial tension

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Transverse Forces

Act perpendicular to the length of the truss member Cause bending & shear stresses

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Internal Forces

Internal Tension & Compression Forces

Beam with Point Load(Exaggerated Bending)

Each Beam Element isDistorting Under Internal Forces

Unloaded Beam

Compression

Tension

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Neutral Axis

Occurs at some depth in the beam where there is neither tension nor compression

3Level 1: Section 06 – Version 3© 2014 SBCA

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Parallel Truss Chord Axial Forces

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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

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Stress Reversals

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Truss Action Under Gravity Load

Image Courtesy of Alpine Engineered Products

Movement

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Compression Buckling Stabilize the column to increasing

compression load capacity by 4x

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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

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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

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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

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Permanent Web Bracing

Truss Design Drawing shows if permanent CLR is required on a web

All lateral braces require diagonal braces for stability

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Permanent Web Bracing

Diagonal Bracing

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Purlins

4 ft. on-center

Top Chord Continuous Lateral

Brace

Diagonal Bracing

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Moments

Force that produces a rotation of a member &

subsequent bending stresses

P

L

aa

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Moment Equation

P

L

Force (lbs)

aa

Length (ft)

Torque /Bending Moment = Forces that act a distance away to produce rotation about a point

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Moment Equilibrium

P

L

Mwall = Mend of board

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Truss Bending Moments

Rotation Point Rotation Point

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Static Equilibrium

L/2 L/2

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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”

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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

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Equations of Equilibrium

Vertical Forces:

Horizontal Forces:

Moments:

ΣV = 0

ΣH = 0

ΣM = 0

6Level 1: Section 06 – Version 3© 2014 SBCA

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Vertical Forces: ΣV = 0

ΣV = 0: P + w – R1 – R2 = 0

+ positive direction of force

negative direction of force

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Horizontal Forces: ΣH = 0

ΣH = 0: W – R1 – R2 = 0

Environmental Loads

Generated by Truss-to-Bearing Connection

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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)]

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Sign Convention Review

ΣV = 0

ΣH = 0

ΣM = 0

+ positive direction of force

negative direction of force

+ clockwise rotation

- counter clockwise rotation

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Free Body Diagrams

Account for all forces acting on a structure

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Quiz 1

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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

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Crushing

(see design values download page 3

Example 2x4 No. 1 SP)Compression Perpendicular

to Grain

Compression Parallel to Grain

565 psi 1650 psi

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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)

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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

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Bending Stress (Fb)

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Tension Stress (Ft)

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Shear Stress (Fv)

Horizontal shear

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Compression Stress Perpendicular to Grain (Fc┴

)

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Compression Stress Parallel to Grain (Fc)

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Stress

Stress = Force/Area

ℓ in.

w in.

Area (A) = ℓ" x w" = ℓw in2

P lbs.

Stress (F) = Force (P) / Area (A) psi

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Stress

Stress =Axial Compression Force (lbs.)

(1.5 in. x 3.5 in.)Fc =

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TTT 1 Sec 06 Handout

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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

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Stress Example

Ft = 762 psi (actual stress)

Fc =P

A

Actual ≤ Allowable

Stress Example

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Select Structural

762 ≤ 900

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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

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Stress

Pressure and Stress

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Repetitive Member Factor

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Repetitive Member Factor

15% increase for bending stress

Selections from ANSI/TPI 1-2007

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3

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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.

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Quiz 2

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