Timber Design
Choice of Timber on strength basis
a.) For beam, rafters & floor joints, high bending strength & adequate stiffness
b.) For column & posts; high compressive strength & adequate stiffness
c.) For poles : high bending strength
Rules of Thumb on size of joints to Span
Depth of Floor Joist (inches)
=
Dressed Dimension ( :
a.) For dimension 4 inches or less = - 3/8 “
b.) For dimension greater than 4 inches = - ½”
Where = nominal dimension
Example:
1.) Floor joist span = 10ft.
Recommended depth =
6”
2.) Determine dressed dimensions
a.) 2 x 6 S2S lumber
- 5½“
Dressed dimension: 2 x 5 ½ “
6”
2”
b.) 2 x 4 S4S lumber
-
6” -
Dressed dimension=
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
2”
Flexure Formula Shearing stress
Q= statistically moment of axes above
or below the point of question about its N.A.
Unit measurement : board foot
=
=
A.) Rectangular Section
d/2 d
d/2
B.) Circular Section
r
NA
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
C.) Triangular Section
2y/3 y
Fvmax
d NA
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
PROBLEM:
Two carpenters are to carry a 6m wooden plank (500mm x 200 mm. At the midspan of the plank or cement
weighing 980 N. , , . Check the adequacy of the
plank .
980N
W = self wt. 6m
a.) Flexure
200
50
b.) Shear
50=d
a) Flexure
b=200
b.)
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
Design of Floor Joist
T&G Flooring
Floor joist
s
s
girder
Laterally supported timber beams:
If
If
If
Where;
Fb’= reduced fiber stress
Fb= allowable bending stress
k = capacity reduction factor
b= width of beam
L= laterally unsupported span of beam
Values of k
1.
P P
or
L/3 L/3 L/3
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
2.
3.
P P P
L/4 L/4 L/4 L/4
4.
SAMPLE PROBLEM:
Design the floor joist using Guijo
Given:
Residential
Span of joist =3.6 m
Spacing of joist = 0.3 m
Allo. Deflection = L/300
Properties:
Solution:
Loadings;
DEADLOAD
a.) Flooring (T&G) & floor joists. . . . . 10psf
b.) Ceiling & ceiling joists. . . . . . . . . . . 5psf
c.) Partitions ( ¼ “ plywood). . . . . . . . . 10psf
LIVELOAD
Residential . . . . . . . . . . . . . . . . . . . . . . . 40psf
_________________
Total : 65 psf
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
W=0.93 KN/m
a.) Flexure
Check against shear:
Check against deflection:
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
Try 50 x 150 mm
Safe on shear
Deflection
Lateral stability
Provide bridging @ midspan
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
Timber Columns
Effective column length = Le = keL
Buckling Factor ke:
0.65 0.8 1.2 1.0 2.10 2.4
Allowable unit stress in Mpa of cross- sectional area of square or rectangular simple solid columns.
1. Short columns
When
2. Intermediate columns
When
3. Long Column
When
Where:
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
For round section
D= diameter of the column
d = equivalent square section
Sample Problem:
A timber column of rectangular cross- section ( 150 x 200mm) has a length of 3.6m. It is fixed at both ends.
Given:
; k = 0.65
Calculate the max. safe axial load that the column could carry.
P=?
Solution:
3.6m 200
150
Lecture Notes of Eng’r. Edgardo S. Cruz, MSCE
Mapua Institute of Technology
School of CE-EnSE-CEM
