# Microsoft PowerPoint - Wood Strength and Mechanics by housework

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```									                                                       Mechanical Properties of wood
WOOD 474

Stresses

Wood strength and Mechanics

Mechanical Properties of wood                     Mechanical Properties of wood
Stresses
Tension
Breaking strength (MOR): determines load a beam
will carry

Compression

Shear
MOR is accepted criterion of strength, although not a true stress
as it is only true to the proportional limit.

Mechanical Properties of wood                     Mechanical Properties of wood
Compression strength // to grain: Important in   Compression strength ⊥ to grain: Important in
piles, and columns                               design strength at connections between wood
members and beam supports

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Mechanical Properties of wood                 Mechanical Properties of wood
Tension strength // to grain: Important for    Shear strength // to grain: Often determines
bottom member (chord) in wood truss             the load capacity of short beams

Bending                 Mechanical Properties of wood
Toughness: measure of amount of
work needed to break small
specimen by impact
Neutral axis

compression

shear

tension

Mechanical Properties of wood                 Mechanical Properties of wood
Tension ⊥ to grain: Important in design of     Resilience: Measured by amount of energy
connections in buildings                        absorbed when a timber is bent in the
elastic range
Side hardness: Resistance to denting e.g.
flooring
Work to maximum load: Measure of energy
of member as it is slowly bent

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Mechanical Properties of wood                                               Stress and Strain
Stress
Modulus of Elasticity (MOE): Measure of resistance             It occurs when a load acts on a solid timber
to bending (i.e. directly related to stiffness of a
beam) also a factor in the strength of a long                 (column or beam). It also occurs internally within a
column                                                           timber.
Modulus of elasticity // to grain (Young’s Modulus):           Force per unit area.
Measure of resistance to elongation or shortening                P , σ
of a specimen under uniform tension or                           A
compression.                                                  Units of stress are force per unit area,
e.g. psi (or lb/in2); Pa or N/m2

Stress and Strain                                           Stress and Strain

Strain                                                                        2”
8000 pound force

The external forces deform the shape and size of                         2”
Stress = 8000 = 2000 psi
the timber. The change in length per unit of                                               2”                          4 in2
length in the direction of the stress is called the
strain.                                                                                                   Strain    = 6.000 - 5.9928
6.0000”

∆L , ε,                                                                                                                6
59928”

L                                                                                                              = 0.0012

Because it is expressed in length/per unit length, it                                                    Young’s Modulus = Stress/Strain
has no units.
= 2000      = 1.67 x 106 psi
2”       0.0012

Mechanical Properties of wood                                               Stress and Strain

Strain results when stress applied to
Stress and strain is simple in uniaxial
wood                                                             tension/compression
There is a linear relationship up to the                         More complex in a beam (bending
proportional limit.
Compression           timber)
When stress is removed strain is           parallel to grain
completely recovered                                             Top half is under compression and
Below the proportional limit the ratio                           bottom half is under tension
is constant - Young’s Modulus                                    Maximum stresses are at the upper and
(Previous example Young’s Modulus is                             lower surfaces
1.67 x 106 psi. or 11,500 MPa)

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Stress and Strain - Bending                              Stress and Strain
Simple beam analysis stresses vary linearly
from top to bottom
No tension/compression at mid-point –
neutral axis
Because upper surface is under compression
it shortens and the lower surface lengthens
The length at the neutral axis remains
unchanged
Amount of bending is called the “deflection”

Stress and Strain                                       MOE

Determined by static bending test
Load a simple beam at the centre

MOE                                        MOE

Measure deflection                                 MOE is calculated using:
Beam size, Beam span, Load, Deflection
Neutral axis                                       MOE = PL3/48ID
Deflection
Where P = concentrated centre load (Newtons)
compression
D = deflection (m) at midspan
L = span (m)
I = moment of inertia, which depends on beam
shear      size - width x depth3/12 so units are m4 or in4
tension

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Example 1                                       Example 1

50 x 50 x 762 mm red oak stick                  50 mm

Supported at each end of its 711 mm span
At a load of 6,680 N deflection at midspan is                     711 mm
6.6 mm
What is MOE?                                                     762 mm

Example 1                                       Example 2

MOE = PL3/48ID                                  0.075 m square piece red oak of same type
MOE = [6,680 N (0.711 m)3]/48ID                 and quality as in example 1
MOE =       [6,680 N (0.711 m)3                 Will be placed between two roof beams 1.27
{48[0.050 m (0.050 m)3/12]x 0.0066 m}   m apart
MOE = 2391/(0.000025 x 0.0066) N/m2             Space heater 1,360 kg (13,350 N) is to be
hung from the center.
MOE = 14600 x 106 N/m2
How much will the specimen deflect?
MOE = 14600 MPa

Example 2                                           MOR

MOE = PL3/48ID                                  Determined by static bending test
Reorganise D = PL3/48 x MOE x I                 Load a simple beam at the centre until it
D=      [ 13350 N (1.27 m)3]                    fails
[48(14.6 x 109 Pa)(0.075 m (0.075 m)3/12]
D = 27346/1.848 x 106 m
D = 0.015 m                                     Bending strength = Mc/I where M =
maximum moment, c = half depth, I is
moment of inertia

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MOR                                           Example 3
Sample of red oak (as in example 1)
This reduces to
Loaded to failure in a testing machine
MOR = 1.5 P x L/width x depth2
Valid only if beam is supported at both ends
and loaded in centre                                Calculate the MOR
Below the Proportional limit the P is the           MOR = 1.5PL/d x h2
maximum stress at the top and bottom                MOR = 1.5 x 9,400 N x 0.711 m) = 10025
surface                                                     0.05 m x (0.05 m)2        1.25 x 10-4
MOR = 80.2 MPa

Example 4                                         Example 5
Sample of red oak (as in example 2)                 Sample of red oak (as in example 2)
What is maximum space heater weight which           Loaded only to 13,350 N (1,360 kg)
can be loaded?                                      What is maximum stress at the centre of
top and bottom surfaces
MOR = 1.5PL/d x h2
MOR = 1.5PL/d x h2
Reorganise P = MOR x d x h2/1.5 x L
MOR = 1.5 x 13350 N x 1.27 m)
P = (80.2 x 106 Pa) x 0.075 m x (0.075 m)2
(0.075 m x (0.075 m)2
(1.5 x 1.27 m)
MOR = 25432 Nm /4.2 x 10-4 m3
P = 0.0338 x 106 Pa /1.905 m = 17,760 N
MOR = 60.3 x 106 Pa = 60.3 MPa
or 1,860 kg (1 kg = 9.807 N)

Practice questions 1                               Practice questions 2

As in Example 2 calculate the deflection but        As in Example 4 calculate the maximum
in place of red oak use:                            space heater weight for:
a)   coastal Douglas-fir of same dimensions         a)   coastal Douglas-fir of same dimensions
and MOE of 13,400 MPa                               and MOR of 72.0 MPa
b)   Western hemlock 75 x 75 mm and 800 mm          b)   Western hemlock 75 x 75 mm and 800 mm
span, with MOE of 11,300 MPa.                       span, with MOR 66.0 MPa.

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