MODULE 3
CE 73B- Timber Design
1. Title of the Module: Floor Framing and Notching/Dapping
2. Introduction
This module explains the procedures in designing the floor frame
of a timber structure.
3. Learning Outcome
The students should be able to design and investigate the floor
frame members
of a building structure.
4. Learning Content
Topic 1. Parts of Floor Frame
Topic 2. Design of Floor Frames
Topic 3. Notching / dapping for inclined members
5. Teaching and Learning Activities
Topic 1. Parts of Floor Frame
Classification of Wood Framing
Wood framing may be generally classified into two types known
as:
1. Heavy timber framing
2. Light wood framing
Heavy timber framing. It consists of plant, boards or laminates
structural flooring
that spans between beams and in turn supported by girders as
shown in Figure D.1.1.
Light wood framing. It consists of tongue and grooved flooring that
spans between joists and in turn supported by girders or directly
spanning from wall to wall.
Parts of a Floor Framing
The wood framing generally includes:
1. Girder. It is a principal beam extending from wall to wall of a
building. It is a member supporting the floor joists or floor beams.
Girders maybe either solid or built up.
2. Floor Joists. These ate pars of the floor system placed on the
girders where the floor board are fastened. Floor joists are usually
spaced n the girder at a distance from 30 to 40 centimeters.
3. Flooring. It refers to the material used to cover the entire width
and length of the building. In heavy timber framing, the flooringcould be
plant floors or laminated floors. In light wood framing, the flooring usually
refers to the tongue and groove board.
The thickness of this wood board is from 18 to 25 mm with varying
width that ranges from 87.5 mm to 150 mm
4. Joist Bridging. These are wood strips placed transversely
between joist to prevent joist spring sideways under load
Design of a Floor Bay
The floor design shall conform with the Specfications set forth in
the NSCP.
The design of a typical bay of a building maybe explained by the
following example.
Illustrative Problem
The floor framing of a bay in a building is shown in figure D1.2.
The columnms are spaced 4.50 m on centers in one direction and 6 m in
the other. Each girders supports a beam at the center of its 4.5m span.
The beams in turn support flooring boards spanning parallel to the
girders . The flooring consist of 25x150 mm boars laid side by side and
overlaid within 18mn Tongue and Groove finished flooring. The flooring
weighs 600 Pa. Design the beams and girders for the floor bay. Use
PACE Specification. Unit weight of Guijo – 5000 N/m 3 and allowable
deflection = L/270.
Solution:
25mm x 150mm boards overlaid with
18mm T&G
TRIBUTARY AREA
Allowable stresses:
bending stress, fb = 15.85 MPa
shear stress, fv = 0.96 MPa
modulus of elasticity, E = 13.78 x 103 MPa
δ = L/270 of the span
a. Design the beams for the floor bay
Step 1
Calculate the loads to be carried by the most critical beam. The
floor boards has a span of 2.25 m center-to-center of beams. The floor
area to be supported by the intermediate beam (most critical) is 2.25x
6m = 13.5m2. This is often termed as tributary area and is indicated by
diagonal lines in the Figure D.2
The loads to be carried by the most critical beam are”
Step 2
Determine the maximum external moments. The maximum
bending moment for this loading is
Step 3
Determine the beam sized based on flexure
The trial size may be obtained using flexure formula as
Step 4
Check the beam size
Re-compute the maximum bending moment. Refer to Figure
D.1.3 for the given loading.
6.0m
R=28462N R=28462N
Figure D.1.3
Hence, the total bending moment is
The actual flexure stress is,
Since the maximum flexure stress is lesser than the tabulated
value, adopt 150 mm x 350 mm for flexure.
Step 5
Check the beam for shear
The maximum vertical shear force is equal to the maximum
reaction
56925N
w=350N/m (Weight of girder)
R 2.25m 2.25m R
Using the computed maximum reaction = 29,070 N as the
maximum shear force, the maximum shear stress is.
Since the maximum shear stress is lesser than the permissive
value, then 150mm x 350 is adequate for shear.
step 6
Investigating for deflection, for this loading the maximum
deflection is,
Adopt 150 mm x 350mm since the maximum deflection is lesser
than the permissible value.
Hence, the beam is 150mm x 350 in cross-section can
satisfactorily carry the loads applied.
a. Design the girders for the floor bay
Step 1
Determine the loads to be carried by the most critical girders.
The critical beam will be framed as shown into the girder at both
sides, the concentrated load at the midspan of the girder is 28,462.50
Step 2
The maximum size of girder maybe obtained using the flexure
formulas,
Step 4
Re-compute the maximum moment. Consider the weight of the
girder.
Hence, the maximum flexure stress is,
Since the actual fiber stress is slightly greater than the permissible
value, the beam will be slightly over stressed:
Step 5
The maximum shear force is computed as, (refer to figure)
Hence, the actual shear stress may be obtained as,
The girder is adequate for shear because the actual shear stress
is lesser than the tabulated value.
Step 6
Check the girder for deflection.
Since the maximum deflection is lesser than the allowable value,
then the girder is adequate to sustain the load.
Illustrative Problem
Figure D1.4 is a floor framing plan of a building . Design the floor
planks, floor joist and girders. The live load is 2000 Pa. The deflection is
limited to 6 mm for floor planks, 8 mm for floor joist and 18mm for
girders. Use NSCP Specification and 67% stress grade of Molave. Use
100 mm T & G floor planks.
6@0.60m=3.60m
3.00m
Figure D1.4
Solution:
Allowable stresses:
bending stress, fb = 19.30 MPa
shear stress, fv = 1.83 MPa
modulus of elasticity, E = 8.14 x 103 MPa
A. Design of Floor Planks
Step 1
Determine the loads to be carried by one floor plank. The floor
planks has a span of 0.60m center-to-center of joists. The floor area
supported by the floor plank is
0.60m
The floor plank will carry the live load, w= 2,000 Pa/ The intensity
of this live load is, 2000 x 0.100 = 200N/m
Step 2
Determine the maximum moment
The maximum moment for this type of loading, simply supported
and uniformly loaded is,
The thickness of the floor plank will be computed using the flexure
formula.
Step 3
Check for flexure.
Re-compute the maximum moment.
Hence, the actual maximum fiber stress is,
Step 4
Check the planks for shear
w =200 + 3 = 203 N/m
0.60m
R R
Step 5
Check for deflection
B. Design of Floor Joists
Determine the loads to be carried by the floor joist. The tributary
area for the floor joists is,
Tributary Area
0.60m
3.0m
Step 2
Check the joist for shear
w =3690 + 75 = 3765 N/m
L=3.0m
R R
Step 3
Check the joist for deflection
C. Design of Girders
Step 1
0.60m
941.25N 1882.5N 1882.5N 1882.5N 1882.5N 1882.5N 941.25N
6@0.60m=3.60m
R 3.60m R
Determine the loads to be carried by the girder.
The exterior joist will only transfer to the girder one-half of the load to be
transferred by the interior joist.
Step 2
For this type of loading (see figure C.16), the maximum moment occurs at the
center.
Hence, the moment may now be computed by taking moment about the
center of the girder.
Step 3
Determine the girder size from flexure
Step 4
Check the girder size
Consider the weight of girder
Step 5
Check the girder size for shear
w =75N/m (weight of girder)
3.60m
R R
0.60m
941.25N 1882.5N 1882.5N 1882.5N 1882.5N 1882.5N 941.25N
3.60m
R
R
The maximum shear is equal shear is equal to the maximum reaction
at the support.
Compute the actual maximum shear stress using the following expression,
Step 6
Check the girder for deflection
The maximum deflection occur at the center of the girder.
a. Deflection due to uniform load
w =75N/m (weight of girder)
3.60m
R R
b. Determine the deflection due to the concentrated load. Use Area-Moment
method
0.60m
941.25N 1882.5N 1882.5N 1882.5N 1882.5N 1882.5N 941.25N
A B C
3.60m
R R
56
Topic 3. Notched Beams
Shearing strength of Notched beams
a. Beams Notched on Tension side
Beam should be notched or tapered on the tension side
When a beam is notched on the tension side. When a beam is
notched on the tension side at the supports, the strength of the
beam is decreased by an amount depending on the shape of the
notch. And the ratio of the depth of the notch to the depth of the
beam.
AISC (1974:4-181) limits the ratio of the depth of he noch
to the depth of the beam to 1:10
For square-cornered end notch
The shear stress should be checked by the equation
b. Beams Notched on Compression Side (refer to Figure E.1.2)
When beam is notched or beveled on its upper side at the
ends and such notch is square-cornered, the shear should be
checked by the formula.
End- Notched Beam
If “e” exceeds de, the formula is not used; instead the shear
strength is computed by using the depth of the beam below the
notch, de. According to AISC (1974:4-180) notch on the upper
side of a beam shall nor exceed 40% of the total depth of the
beam.
Bending Strength of notched beams
Notching on the tension side of simple beams on the center of the
span is not recommended. However, if it is necessary to notch a beam at
or neat the midpoint in the open length of a beam, as shown in Figure
E1.3, the net depth should be used in computing the section modulus.
de
Figure E.1.3
Illustrative Problem
A water pipe has to b e installed at the ceiling with its lower side,
of the same level as the bottom of the 150 x 300 mm. The beam has to
carry a uniform load of 5 kN (including its weight) and a concentrated
load of 2.5 kN applied at the misdpan of the beam. Using nominal size, is
the beam safe for flexure and shear? The allowable fiber stress s 12
MPa and the allowable shear stress parallel to the grain s 0.76 MPa.
P = 2.5KN
150mm
150mm
w = 5KN/m 250mm
300mm
Cross section of beam
R 3m R
2m 4m
Figure E1.4
Solution:
The beam shall be investigated at the section where shear and
bending moment are maximum; and at notched section.
Step 1
a. Investigation at the section without Notch
a. The maximum shear for this loading occurs at the supports
and computed as,
b. For this loading, the maximum moment occurs at the
midspan and is equal to.
c. Investigating the beam for flexure,
Since the maximum flexure is lesser than the permissible value,
the beam is safe for flexure.
d. Investigating the beam for shear
The permissible value is greater than the maximum shear stress.
Therefore , the beam is safe for shear.
Investigation at the Notched Section
a. The shear at section 2m from left support is,
Because the actual shear stress at this section s lesser than the
permissible value, then the beam is sage for shear at the notched
section.
b. the bending moment at section 2m from left support is,
In the computation of flexure stress developed at the notched
section, use the effective depth;
Hence, The beam is not safe because the actual flexure stress
developed at the notched section is greater than the permissible
value.
The beam will fail by flexure at the notched section
Illustrative Problem
If simple beam 100mm x 250mm in cross-section be
notched as shown in Figure E.1.5 what is the greatest distance should
the notch be made such that the permissible fiber stress will not be
exceed. The beam will be loaded with a uniform load over its entire span.
50mm
100mm
100mm
250mm 200mm
R=wL/2 R
x Cross section of beam
5.0m
Solution:
Determine the moment that will be developed at the notched section and
equate with the beam resistance to flexure at same section.
a. The uniform load is computed as,
For beam loaded uniformly and simply supported, the maximum moment is equal
to,
b. The moment at section “x” is computed as (refer to Figure E.1.6)
w = 3.67KN/m
R R
5.0 m
Figure E1.6
c. The resisting moment at section “x” can be determined using the
effective depth
Therefore, so that the beam will be safe for flexure, the notch shall be
made at a distance not more than 1m from left support.
HANKINSON’S FORMULA
Derivation of formula
p = allowable compressive parallel to the grain q = allowable compressive stress
perpendicular to the grain r = allowable compressive stress perpendicular to the
inclined section
𝜃 = angle between inclined section end and the line perpendicular to the grain.
𝐹1 = 𝐴1𝑞1 𝐹 = 𝐴𝑟
𝐹2 = 𝐴2p
A = 𝐴1 sin 𝜃 + 𝐴2 cos 𝜃
F = 𝐹1 sin 𝜃 + 𝐹2
cos 𝜃
q p
𝐹1 = 𝐹𝑠𝑖𝑛𝜃
𝐹2 = 𝐹𝑐𝑜𝑠𝜃
F = 𝐹𝑟𝑠𝑖𝑛2𝜃 + 𝐹𝑟𝑐𝑜𝑠2𝜃
𝑞 𝑝
pq = pr 𝑠𝑖𝑛2𝜃+ qr 𝑐𝑜𝑠2𝜃
r(p 𝑠𝑖𝑛2𝜃 + q 𝑐𝑜𝑠2𝜃) = pq
𝑝𝑞
r = p 𝑠𝑖𝑛2𝜃 + q 𝑐𝑜𝑠2𝜃
PROBLEM:
In the given figure; M1 is 100mm x 100mm, F = 25 kN and = 36° . The allowable compressive
stress are p = 8.3 MPa & q = 2.5 MPa. The projection of M1 and M2 is restricted to a vertical distance of
62mm. Design a suitable notch.
Solution:
The most feasible notch is to let AC & BC bisecs the
angle between the intersecting edges
100 Sin 30 = 𝐴𝐵
AB = 200 mm
𝐴𝐶
Sin 15 = ---
𝐴𝐵
AC = Sin 15 (200)
AC = 52 mm
𝐵𝐶
Cos 15 =
𝐴𝐵
BC = 200 cos 15°
BC = 193.2 mm
𝐵𝐸
Sin 15 = ---
𝐵𝐶
BE = 193.2 (sin 15)
BE = 50 mm < 62mm (𝑠𝑎𝑓𝑒)
𝐹2
Cos(15°) =
𝐹
F2 = 25000Cos(15) = 24148 N
F1 = 25000sin(15) = 6470 N
𝐴𝐴𝐶 = 100(52) = 5200 𝑚𝑚2
𝐴𝐵𝐶 = 100(193.2) = 19320 𝑚𝑚2
𝑝𝑞
r= 𝑃 𝑠𝑖𝑛2𝜃+𝑞𝑐𝑜𝑠2𝜃
𝑟𝐴𝐶 = = 𝟕. 𝟏𝟖𝟒 𝑴𝑷𝒂
8.
𝑝𝑞 8.3(2.5)
= ---------------------
𝑟𝐵𝐶 =
𝑃 𝑠𝑖𝑛2𝜃+𝑞𝑐𝑜𝑠2𝜃 8.3𝑠𝑖𝑛275+2.5𝑐𝑜𝑠275
= 2.623 MPa
F2
𝑓𝐴𝐶 = ---- =
AAC
𝑓𝐴𝐶 = 4.64 𝑀𝑃𝑎 < 7.184 (𝑠𝑎𝑓𝑒)
𝐹1 6470
𝑓𝐵𝐶 = = = 0.33 𝑀𝑃𝑎 < 2.623 (𝑠𝑎𝑓𝑒)
𝐴𝐵𝐶 19320
Remark: Plates/problems are given to apply the design principles for this topic
6. Recommended learning materials and resources for supplementary reading.
Students may refer to NSCP as their guide in the different specification in designing
7. Flexible Teaching Learning Modality (FTLM) adopted
module, exercises, plates, messenger etc…
8. Assessment Task
Problems/ plates in every topics will be given to students to assess their
knowledge and skills. Plates, quizzes and periodic exams will be given for
evaluation
9. References
1. Association of Structural Engineers of the Philippines. National Structural Code of
the Philippines. 5th edition, 2001
2. Venancio, Timber Design and Construction Method.
3. Padilla, Perfecto B., Jr. Structural Design and Construction, Padilla Civil
Engineering Review School and Publishing, copyright 2003
