Microsoft PowerPoint - Wood Strength and Mechanics by housework


									                                                       Mechanical Properties of wood
               WOOD 474


Wood strength and Mechanics

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


                                                   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

    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




    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.
                                               Work to maximum load: Measure of energy
                                                 of member as it is slowly bent

       Mechanical Properties of wood                                               Stress and Strain
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

       ∆L , ε,                                                                                                                6

       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)

       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
                                                   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
                                                     Where P = concentrated centre load (Newtons)
                                                     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

                 Example 1                                       Example 1

50 x 50 x 762 mm red oak stick                  50 mm

Supported at each end of its 711 mm span
Loaded at centre
Gradual increasing load
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

                             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
                                                         Breaking load is 9,400 N
     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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