Example on Design of Timber Structure part2

REF CALCULATION OUTPUT

Example 1: Design of Flexural/Bending Member

Floor of a building comprises series of timber flooring system with thickness 20mm

arranged and nailed on series of timber joist. The spacing between joists is 500 mm. The

timber joists supported by main beam with distance 2500mm to each other. The main

beam is bolted to column with span 4000mm. Determine the suitable size of beam and

joist used to accommodate all the loads if the timber used is from SG3 (standard, dry).

Design data given are as below:

Live load = 3.5 kN/m2

Dead load = 0.4 kN/m2

Selfweight of floor = 0.5 kN/m2

Solution:

1.0 Main Beam

Try beam size (named size) 100mm×275mm

Step 1: Geometrical Properties

Beam dimension

Named size, b (breadth) = 100mm

h (depth) = 275mm

Dressed size, b (breadth) = 90mm

h (depth) = 265mm

Second moment of area, �� = �ℎ� 12 = (90 × 265�) 12 = 135.57 × 10��


Step 2: Loading

Loadings acted on the main beam are reaction for the joists, which are points load at 0.5m

spacing. The spacing can be consider as closely spaced thus to simplified the design

calculation, the points load can equally presented as uniformly distributed load

w= beam distance × (dead load+imposed load)

= 2.5m×((0.4+0.5)+3.5)kN/m2 = 11 kN/m

Design Span, Le= 4000mm (center-to-center of the column and simply supported

at the column)

Step 3: Check for lateral stability

Both beam-ends are held in position using bolts and the top of the beam is hold by the

joists. From Table 7 the permissible maximum depth to breadth ratio is 5

depth/breadth = 265/90 = 2.94<5

Therefore lateral stability is adequate.

Step 4: Shear Stress

Maximum shear force, V= wL/2 = (11×4)/2 = 22 kN

Maximum shear stress, qs= 3V/2A= (3×22×103 N)/(2×90×265) = 1.38 N/mm

2

Shear modification factor for flexural member (K-Factors)

Duration of loading, K1= 1.0 (long term)

Load sharing system, K2= 1.1 (load sharing system)

Notch at end, K4= 1.0 (no notch at end)

Grade stress for shear, qg= 1.61 N/mm2 (Shear parallel to grain)

Permissible shear stress, qp = qg× K1× K2× K4

= 1.61 N/mm2×1.0×1.1×1.0

= 1.77 N/mm2

Shear stress, qs=1.38 N/mm2

< permissible shear stress, qp= 1.77 N/mm2. Therefore shear

stress is adequate.

Step 5: Bending stress, fs

Maximum bending moment, M=wL2/8 = (11×4

2)/8 = 22 kNm

Section modulus, Z=bh2/6= (90×265

2)/6= 1.05×10

6mm

3

Bending stress, fs=M/Z= 22×106 Nmm/1.05×10

6mm

3 = 20.95 N/mm

2

Bending modification factors (K-Factors)



Form factor, K5= 1.0 (rectangular section)

Depth factor, K6= 1.0 (depth <300mm)


Grade stress for bending, fg= 15.9 N/mm2 (Bending parallel to grain)

Permissible bending stress, fp = fg× K1× K2× K5× K6

= 15.9 N/mm2×1.0×1.1×1.0×1.0

= 17.49 N/mm2

Bending stress, fs=20.95 N/mm2> Permissible bending stress, fp= 17.49 N/mm

2. Therefore

bending stress is not adequate. Increase the timber size or use timber from SG1 or SG2.

Step 6: Deflection

For load sharing system, use Emean= 14300 N/mm2

Deflection due to bending where w is total of dead load and imposed load in Newton (N)

unit,

�� =5��

384!� = 5(11 × 10� × 4)(4000)�384(14300)(135.57 × 10�) = 18.9��

Deflection due to shear

�" =#$%&' = 1.2 × 22 × 10�

(14300 16 )(90 × 265) = 1.2��

where

F= shape factor depending on cross section (1.2 for solid rectangular or square)

Mo= bending moment at mid span (in this case Mo= Mmax as the loading is UDL)

G= modulus of rigidity (E/16)

A= Cross section area

Total deflection due to bending and shear, δt= δb+δs= 18.9mm+1.2mm = 20.1 mm

Permissible deflection, δp= 0.003Le= 0.003(4000) = 12 mm

Total deflection, δt= 20.1 mm > permissible deflection, δp= 12 mm. Deflection is not ok.

Step 7: Bearing stress

No need to check as the beam connected to the column using bolts.


2.0 Timber joist

The joists are fulfilling the requirement for load sharing system where the number of

member is more than 4 and the distance between joist is 500 mm less than 610 mm. The

flooring system is assumed distributing the loading to sideways on the joist.Assume the

joist is simply supported on the main beam

Try joist size (named size) 50mm×125mm


Beam dimension


h (depth) = 125mm


h (depth) = 115mm

Second moment of area, �� = �ℎ� 12 = (45 × 115�) 12 = 5.7 × 10��

Step 2: Loading

Loadings acted on the main beam are reaction for the joists, which are points load at 0.5m

spacing. The spacing can be consider as closely spaced thus to simplified the design

calculation, the points load can equally presented as uniformly distributed load

w= joist distance × (dead load+imposed load)

= 0.5m×((0.4+0.5)+3.5)kN/m2 = 2.2 kN/m

Design Span, Le= 2500mm (from the surface of the beam to another beam)

100mm

45mm



Both beam-ends are held in position using steel bracket and the top of the joist is hold by

timber flooring system. From Table 7 the permissible maximum depth to breadth ratio is 5

depth/breadth = 115/45 = 2.56<5

Therefore lateral stability is adequate.


Maximum shear force, V= wL/2 = (2.2×2.5)/2 = 2.75 kN

Maximum shear stress, qs= 3V/2A= (3×2.75×103 N)/(2×45×115) = 0.80 N/mm

2




Shear at notched end, K4= 1.0 (no notch at end)


Permissible bending stress, qp = qg× K1× K2× K4

= 1.61 N/mm2×1.0×1.1×1.0

= 1.77 N/mm2

Shear stress, qs=0.8 N/mm2 < permissible shear stress, qp= 1.77 N/mm

2. Therefore shear

stress is adequate.


Maximum bending moment, M=wL2/8 = (2.2×2.5

2)/8 = 1.72 kNm


2)/6= 9.9×10

4mm

3

Bending stress, fs=M/Z= 1.72×106 Nmm/9.9×10

4mm

3 = 17.38 N/mm

2






Grade stress for bending, fg= 15.9 N/mm2 (bending parallel to grain)


= 15.9 N/mm2×1.0×1.1×1.0×1.0

= 17.49 N/mm2

Bending stress, fs=17.38 N/mm2 < permissible bending stress, fp= 17.49 N/mm

2. Therefore

bending stress is adequate.


Step 6: Deflection

For load sharing system, use Emean= 14300 N/mm2


unit,

�� = 5��

384!� =

5(2.2 × 10� × 2.5)(2500)�

384(14300)(5.7 × 10�)= 13.73 ��


�" = #$%

&' =

1.2 × 1.72 × 10�

(1430016 )(45 × 115)

= 0.45 ��

where






Permissible deflection, δp= 0.003Le= 0.003(2500) = 7.5 mm

Total deflection, δt= 14.18 mm > permissible deflection, δp= 7.5 mm. Deflection is not ok.


Bearing occur as the joist is supported by a steel bracket and the bearing area as follows,

Bearing area, Aa= 45mm×100mm = 4500mm2

Bearing stress, Cts= V/Aa= 2.75×103/4500 = 0.61 N/mm

2

Bearing modification factors (K-Factors)


Load sharing system, K2= 1.0 (non load sharing system)

Bearing length and position, K3= 1.0 (bearing length 100mm but <75mm for

length from end member)

Bearing grade stress, Ctg= 2.09 N/mm2 (compression perpendicular to grain)

Permissible bearing stress, Ctp= Ctg× K1× K2× K3

= 2.09 N/mm2×1.0×1.1×1.0

= 2.30 N/mm2

Bearing stress, Cts= 0.61 N/mm2 < permissible bearing stress, Ctp= 2.30 N/mm

2. Therefore

bearing stress is adequate.


Example 2: Design of flexural member

Figure 1 shows main beams of 3250 mm length span over an opening 3000 mm wide and

support a flooring system that exerts a long-duration loading of 3.9 kN/m including its

own self-weight over it span. The beam is supported by 125 mm wide side-walls on either

side and have underside notched as in Figure 2. Carry out design checks to show that a 75

mm×225 mm (named size) full swan SG2 (select, wet) timber is suitable to carry the load.

Figure 1: Timber beams supported by side walls

Figure 2: Underside notch at beam support

107.5



Beam dimension

Named size, b (breadth) =75mm

h (depth) = 225mm


h (depth) = 215mm

Second moment of area, �� = �ℎ�

12 = (65 × 215�) 12 = 53.83 × 10��

Design span, Le= 3000+[(125/2)×2] = 3125mm


Assume the ends of member held in position. From Table 7 the permissible maximum

depth to breadth ratio is 3

depth/breadth = 215/65 = 3.3>3

Therefore lateral stability is not adequate. Increase the beam dimension or provide lateral

support to the member.


Maximum shear force, V= wL/2 = (3.9×3.125)/2 = 6.09 kN

Maximum shear stress, qs= 3V/2A= (3×6.09×103 N)/(2×65×215) = 0.65 N/mm

2



Load sharing system, K2= 1.0 (no load sharing system)

Shear at notched end factor (underside notch),

K4= ℎ* ℎ = 107.5 215 = 0.5


Permissible bending stress, qp = qg× K1× K2× K4

= 2.24 N/mm2×1.0×1.0×0.5

= 1.12 N/mm2

Shear stress, qs=0.65 N/mm2 < permissible shear stress, qp= 1.12 N/mm

2. Therefore shear

stress is adequate.



Maximum bending moment, M=wL2/8 = (3.9×3.125

2)/8 = 4.76 kNm


2)/6= 50.1×10

4mm

3

Bending stress, fs=M/Z= 4.76×106 Nmm/50.1×10

4mm

3 = 9.5 N/mm

2



Load sharing system, K2= 1.0 (no load sharing system)





= 20.7 N/mm2×1.0×1.0×1.0×1.0

= 20.7 N/mm2

Bending stress, fs=9.5 N/mm2 < permissible bending stress, fp= 20.7 N/mm

2. Therefore


Step 5: Deflection

For load sharing system, use Emin= 11700 N/mm2 (non load sharing system)


unit,

�� =5��

384!� = 5(3.9 × 10� × 3.125)(3125)�384(11700)(53.83 × 10�) = 7.68��


�" =#$%&' = 1.2 × 4.76 × 10�

(11700 16 )(65 × 215) = 0.56��

where






Permissible deflection, δp= 0.003Le= 0.003(3125) = 9.4 mm

Total deflection, δt= 8.24 mm < permissible deflection, δp= 9.4 mm. Deflection is ok.



Bearing occur as the beam is supported directly on side wall and the bearing area as

follows,

Bearing area, Aa= 65mm (beam breadth) ×125mm (wall breadth) = 8125mm2

Bearing stress, Cts= V/Aa= 6.09×103/8125 = 0.75 N/mm

2




Bearing length and position

(bearing length 125mm and >75mm for length from end member)

K3= 1.05

Bearing grade stress, Ctg= 2.97 N/mm2 (compression perpendicular to grain)

Permissible bearing stress, Ctp= Ctg× K1× K2× K3

= 2.97 N/mm2×1.0×1.0×1.05

= 3.12 N/mm2

Bearing stress, Cts= 0.75 N/mm2 < permissible bearing stress, Ctp= 3.12 N/mm

2. Therefore

bearing stress is adequate


Example 2: Design of Compression Member

Check the adequacy of 5.0 m long of timber column for long term loading if a column

section of 150 mm x 150 mm (dressed size) is subjected to an axial load of 80 kN

(including selfweight of column). The timber used is in SG4 (standard, wet) and the

column is not restrained about both axes but restrained at both ends in position

Solution:


Beam dimension

No need to taking account reaping effect as the question already stated the dimension is

dressed size.


h (depth) = 150mm

Column length, L= 5.0m


Step 2: Determine effective length and slenderness ratio

Effective column length, Le= 1.0L (restrained at both ends in position but not in direction)

= 1.0(5m) = 5m

Radius of gyration, r= b/√12 = 150/√12 = 43.3

Slenderness ratio, λ= Le/r = 5000/43.3 = 115.5 <180, slenderness ok. (for compression

member carrying dead and imposed load other than loads resulting from wind)

Step 3: Determine modification factor for compression member without bending, K8

Compression grade stress, Csg= 9.5 N/mm2

(Compression parallel to grain)

Minimum modulus of elasticity, Emin= 7400 N/mm2

For Table 10

E/σc.ll= Emin/Csg= 7400/9.5 = 778.9

λ= 115.5

Λ

Emin/Csg 100 115.5 120

700 0.341 x 0.254

778.9 K8

800 0.371 y 0.280

(+ − 0.341)

(0.254 − 0.341)=

(115.5 − 100)(120 − 100)

; + = 0.274

(. − 0.371)

(0.280 − 0.371)=

(115.5 − 100)(120 − 100)

; . = 0.300

(/0 − 0.274)(0.3 − 0.274)

=(778.9 − 700)(800 − 700)

; /0 = 0.295

Step 4: Calculate applied and permissible compressive stress

Applied compressive stress, Csa= Fc/A = 80×103 N/(150×150)mm

2= 3.56 N/mm

2




Compression member without bending, K8= 0.295

Permissible compressive stress, Csp= Csg×K1×K2×K8

= 9.5 N/mm2×1.0×1.0×0.295

= 2.8 N/mm2

Applied compressive stress, Csa= 3.55 N/mm2 > Permissible compressive stress, Csp= 2.8

N/mm2. Increase column size or provide lateral support to reduce the slenderness of the

column.


Example 3: Design of Compression Member

A 2.5 meter compression member with named sized 75mm×150mm is restrained at both

end in position. However, the column is restrained about major axis and unrestrained

about minor axis at top and column base. Determine the ultimate axial load capacity of the

column section for long term loading if the column section in use is SG2 (select, wet).

Solution:


Beam dimension

Taking account of reaping effect at four side of the timber section


h (depth) = 150mm


h (depth) = 140mm

Column length, L= 2.5m

Step 2: Determine effective length and slenderness ratio

Effective column length about x-x axis

Lex= 0.7L (restrained at both ends in position and in direction)

= 0.7(2.5m) = 1.75m

Radius of gyration about x-x axis, rx= h/√12 = 140/√12 = 40.4

Slenderness ratio about x-x axis, λx= Lex/rx = 1750/40.4 = 43.3<180, ok.

Effective column length about y-y axis

Ley= 1.0L (restrained at both ends in position and in direction)

= 1.0(2.5m) = 2.5m

Radius of gyration about y-y axis, ry= b/√12 = 65/√12 = 18.8

Slenderness ratio about y-y axis, λy= Ley/ry = 2500/18.8 = 133<180, ok.

Slenderness ratio about y-y axis is larger than slenderness ratio about x-x axis. This

indicates the column tends to buckle about y-y axis. Therefore, use slenderness ratio about

y-y axis to determine K8

Step 3: Determine modification factor for compression member without bending, K8

Compression grade stress, Csg= 18.8 N/mm2 (Compression parallel to grain)

Minimum modulus of elasticity, Emin= 11700 N/mm2

For Table 10

E/σc.ll= Emin/Csg= 11700/18.8 = 622.3

λ= 133

Λ

Emin/Csg 120 133 140

600 0.226 x 0.172

622.3 K8

700 0.254 y 0.195


(+ − 0.226)(0.172 − 0.226)

=(133 − 120)(140 − 120)

; + = 0.191

(. − 0.254)

(0.195 − 0.254)=

(133 − 120)(140 − 120)

; . = 0.216

(/0 − 0.191)

(0.216 − 0.191)=

(622.3 − 600)(700 − 600)

; /0 = 0.197

Step 4: Calculate applied and permissible compressive stress




Compression member without bending, K8= 0.197

Permissible compressive stress, Csp= Csg×K1×K2×K8

= 18.8 N/mm2×1.0×1.0×0.197

= 3.7 N/mm2

Ultimate axial load, Fc= Csp×A = 3.7 N/mm2×(65×140)mm2= 33670 N= 33.7 kN


Example 4: Design of Tension Member

A 2.5m internal timber truss member is subjected with a long-term tension loading of

35kN. The member is connected using metal nail plate at the connection as shown in

example in Figure 1.Check the adequacy of 50×150mm timber cross section to be used as

the internal truss member if the timber is SG3 (select, wet) group.

Timber cross section

Figure 1: Example of metal nail plate

Solution:


Beam dimension


h (depth) =150mm


h (depth) = 140mm

Step 2: Calculate the net tension area and applied tension stress

The member is connected using metal nail plate. It can be assume that the reduction of the

cross section area due to nail holes is too small. Therefore, in author opinion and

judgment it is not significantly affect the tension capacity of the timber section. Thus, the

net area of the timber is equal to the gross cross section of the timber.

Ag=An= 45×140mm= 6300mm2

tsa= T/An = 35×103

N/6300mm2

= 5.56 N/mm2

b

h


Step 3: Calculate the permissible tension stress

Modification factors:

Duration of loading, K1= 1.0 (long term loading)

Load sharing system, K2= 1.0 (non load sharing)

Tension parallel to grain grade stress, tsg= 10.9 N/mm2

Permissible tension stress, tsp= tsg×K1× K2

= 10.9 N/mm2×1.0×1.0

= 10.9 N/mm2

Applied tension stress, tsa= 5.56 N/mm2 < permissible tension stress, tsp= 10.9 N/mm

2.

Therefore, tension capacity of the timber section is adequate.


Example 4: Design of Combined Tension and Bending Stress

Figure shows a portion of a ceiling tie in a roof truss subjected to a UDL of 0.5 kN/m and

axial tension of 22kN. The loading is long term and no load sharing. Check if a

SG4(common, wet) 50×150mm (named size) timber section is adequate to sustain the

load.

Solution:

Solution:


Beam dimension


h (depth) =150mm


h (depth) = 140mm

Step 2: Calculate applied bending stress and permissible bending stress

Applied bending stress

Maximum bending moment, M=wL2/8 = (0.5×2

2)/8 = 0.25 kNm


2)/6= 147×10

3mm

3

Applied Bending stress, fa=M/Z= 0.25×106 Nmm/147×10

3mm

3 = 1.7 N/mm

2

Permissible bending stress








= 8.8 N/mm2×1.0×1.0×1.0×1.0

= 8.8 N/mm2

Bending stress, fa=1.7 N/mm2 < permissible bending stress, fp= 8.8 N/mm

2. Therefore



Step 3: Calculate the applied tension stress and permissible tension stress

Applied tension stress

Assume no reduction of area due unavailability of detailing at connection.

Gross section area Ag= Net cross section, An= 45×140mm= 6300mm2

tsa= T/An = 22×103

N/6300mm2

= 3.49 N/mm2

Permissible tension stress






= 5.3 N/mm2×1.0×1.0

= 5.3 N/mm2

Applied tension stress, tsa= 3.49 N/mm2 < permissible tension stress, tsp= 5.3 N/mm

2.

Therefore, tension capacity of the timber section is adequate.

Step 4: Interaction Equation

12

13+

52

53≤ 1.0

= 3.495.3 + 1.78.8 = 0.85 < 1.0

The timber section is adequate to carries the combined tension and bending load.


Example 4: Design of Combined Tension and Bending Stress

Figure shows a portion of a ceiling tie in a roof truss subjected to a UDL of 0.5 kN/m and

axial tension of 22kN. The loading is long term and no load sharing. Determine a suitable

size of timber for timber in SG3 (wet, common).

Step 1: Calculate applied bending stress and permissible bending stress



2)/8 = 0.25 kNm

As the size is unknown, therefore the section modulus is

Section modulus, Z=bh2/6

Applied Bending stress, fa=M/Z= 0.25×106 Nmm/((bh

2)/6)= 1.7 N/mm

2




Load sharing system, K2= 1.0 (non-load sharing system)





= 11.3 N/mm2×1.0×1.0×1.0×1.0

= 11.3 N/mm2

Step 3: Calculate the applied tension stress and permissible tension stress



b and h are unknown, therefore the net area is

Gross section area Ag= Net cross section, An= bh

tsa= T/An = 22×103

N/bh


Permissible tension stress






= 6.8 N/mm2×1.0×1.0

= 6.8 N/mm2

Step 4: Interaction Equation

1213 +

5253 ≤ 1.0

= 899×:;< �= >

�.0 + ?;.9@×:;A (�=B �⁄ )D E::.� = 1 … Equation 1

The depth to breadth ratio for member with rectangular shape should be limited to an

appropriate value as in Table 7 to ensure no risk of buckling under design load. The

selection of depth/breadth ratio is base on degree of lateral support provided in the timber

structure system to the member. Therefore, this ratio can be used in estimating a suitable

size for a timber member.

Take depth/breadth=2 (no lateral support)

h/b= 2, thus h=2b ... Equation 2

Insert equation 2 into equation 1

= 899×:;< �(9�) >

�.0 + ?;.9@×:;A (�(9�)B �⁄ )D E::.� = 1

Rearranging the equation we will get

=�� − 1.618 × 10�� − 33.201 × 10� = 0 … Equation 3

Solving Equation 3, we will get b=48.1 mm and inserting into Equation 2 h= 96.2 mm.

The b and h obtained are minimum dimensions required to resist the combine tension and

bending loading. Taking account the effect of reaping the proposed suitable timber size

(named size) is 60mm×110mm.

For confirmation, recalculate for combined tension and bending interaction equation

Beam dimension


h (depth) =110mm


h (depth) = 100mm




2)/8 = 0.25 kNm


2)/6= 83.33×10

3mm

3

Applied Bending stress, fa=M/Z= 0.25×106 Nmm/83.33×10

3mm

3 = 3 N/mm

2




Load sharing system, K2= 1.0 (non-load sharing system)





= 11.3 N/mm2×1.0×1.0×1.0×1.0

= 11.3 N/mm2



b and h are unknown, therefore the net area is

Gross section area Ag= Net cross section, An= bh= 50×100=5000mm2

tsa= T/An = 22×103

N/5000 mm2 = 4.4 N/mm

2

Permissible tension stress Modification factors:





= 6.8 N/mm2×1.0×1.0

= 6.8 N/mm2

Interaction Equation

12

13+

52

53≤ 1.0

= 4.46.8 +3

11.3 = 0.91 < 1.0

The timber section is adequate to carries the combined tension and bending load.

Documents

Example on Design of Timber Structure part2