226919874-Example-on-Design-of-Timber-Structure-part2.pdf

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 and nailed on series of timber timber joist. The spacing spacing between joists is 500 mm. The timber joists supported by main beam 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 accommodate all the loads loads if the timber used used is from SG3 (standard, (standard, dry). Design data given are as below: Live load Dead load Selfweight of floor

2

= 3.5 kN/m 2 = 0.4 kN/m 2 = 0.5 kN/m

Solution: 1.0 Main Beam Try beam size (named size) 100mm×275mm Step 1: Geometrical Properties

Beam dimension Named size, b (breadth) h (depth) Dressed size, b (breadth) h (depth)

= 100mm = 275mm = 90mm = 265mm

)  (90�265 ℎ 12 = 135�57�10 Second moment of area,  = 12 =

REF

CALCULATION

OUTPUT

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 uniforml y distributed load w= beam distance × (dead load+imposed load) 2 = 2.5m×((0.4+0.5)+3.5)kN/m 2.5m×((0.4+0.5)+3.5)kN/m = 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 depth to breadth ratio 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 3

2A= (3×22×10 N)/(2×90×265) = 1.38 N/mm Maximum shear stress, qs= 3V / 2A

2

Shear modification factor for flexural member (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.1 (load sharing system) s ystem) Notch at end, K 4= 1.0 (no notch at end) 2

Grade stress for shear, qg= 1.61 N/mm (Shear parallel to grain) Permissible shear stress, q p

= qg× K 1× K 2× K 4 2 = 1.61 N/mm ×1.0×1.1×1.0 2 = 1.77 N/mm 2

2

Shear stress, qs=1.38 N/mm < permissible shear stress, q p= 1.77 N/mm . Therefore shear stress is adequate. Step 5: Bending stress, f s 2

2

Maximum bending moment, M=wL /8 = (11×4 )/8 = 22 kNm 2

2

6

3

Section modulus, Z =bh /6= (90×265 )/6= 1.05×10 mm 6

6

3

Bending stress, f s= M/Z = 22×10 Nmm/1.05×10 mm = 20.95 N/mm Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.1 (load sharing system) s ystem) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 6= 1.0 (depth <300mm)

2

REF

CALCULATION

OUTPUT

Grade stress for bending, f g= 15.9 N/mm (Bending parallel to grain) Permissible bending stress, f p = f g× K 1× K 2× K 5× K 6 2 = 15.9 N/mm ×1.0×1.1×1.0×1.0 2 = 17.49 N/mm 2

2

Bending stress, f s=20.95 N/mm > Permissible bending stress, f p= 17.49 N/mm . 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 E mean= 14300 N/mm

2

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

  � 4)(4000) 5  5(11�10   = 384 = 384(14300)(135�57�10) = 18�9  Deflection due to shear

  1�2�22�10   =  = 14300 (90�265) = 1�2  16 where F = shape factor depending on cross section (1.2 for solid rectangular or square) M o= bending moment at mid span (in this case M o= M max 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, δ p= 0.003 Le= 0.003(4000) = 12 mm Total deflection, δt = 20.1 mm > permissible deflection,

δ p=

12 mm. Deflection is not ok.

Step 7: Bearing stress

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

REF

CALCULATION

OUTPUT

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

45mm

100mm

Try joist size (named size) 50mm×125mm Step 1: Geometrical Properties


= 50mm = 125mm = 45mm = 115mm

)  (45�115 ℎ 12 = 5�7�10 Second moment of area,  = 12 = 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 uniforml y distributed load w= joist distance × (dead load+imposed load) 2 = 0.5m×((0.4+0.5)+3.5)kN/m = 2.2 kN/m

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

REF

CALCULATION

OUTPUT

Step 3: Check for lateral stability

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. Step 4: Shear Stress

Maximum shear force, V = wL/ 2 = (2.2×2.5)/2 = 2.75 kN 3

Maximum shear stress, qs= 3V / 2A= (3×2.75×10 N)/(2×45×115) = 0.80 N/mm

2

Shear modification factor for flexural member (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.1 (load sharing system) Shear at notched end, K 4= 1.0 (no notch at end) 2

Grade stress for shear, qg= 1.61 N/mm (Shear parallel to grain) Permissible bending stress, q p = qg× K 1× K 2× K 4 2 = 1.61 N/mm ×1.0×1.1×1.0 2 = 1.77 N/mm 2

2

Shear stress, qs=0.8 N/mm < permissible shear stress, q p= 1.77 N/mm . Therefore shear stress is adequate. Step 5: Bending stress, f s 2

2

Maximum bending moment, M=wL /8 = (2.2×2.5 )/8 = 1.72 kNm 2

2

4

3


4

3

Bending stress, f s= M/Z = 1.72×10 Nmm/9.9×10 mm = 17.38 N/mm

2

Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.1 (load sharing system) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 = 1.0 (depth <300mm) 2

Grade stress for bending, f g= 15.9 N/mm (bending parallel to grain) Permissible bending stress, f p = f g× K 1× K 2× K 5× K 6 2 = 15.9 N/mm ×1.0×1.1×1.0×1.0 2 = 17.49 N/mm 2

2

Bending stress, f s=17.38 N/mm < permissible bending stress, f p= 17.49 N/mm . Therefore bending stress is adequate.

REF

CALCULATION

OUTPUT

Step 6: Deflection

For load sharing system, use E mean= 14300 N/mm

2

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

5  5(2.2�10 � 2.5)(2500)  � � � 13.�3  3�4 3�4(14300)(5.� � 10 ) Deflection due to shear

 1.2 � 1.�2 � 10  � � 14300 � 0.45    16(45 � 115) where F = shape factor depending on cross section (1.2 for solid rectangular or square) M o= bending moment at mid span (in this case M o= M max as the loading is UDL) G= modulus of rigidity ( E/16 ) A= Cross section area


δt = δb+δs=

13.73mm+0.45mm = 14.18 mm

Permissible deflection, δ p= 0.003 Le= 0.003(2500) = 7.5 mm Total deflection, δt = 14.18 mm > permissible deflection,

δ p=

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 = 4500mm 3

2

Bearing stress, C ts= V/Aa= 2.75×10 /4500 = 0.61 N/mm

2

Bearing modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non load sharing system) Bearing length and position, K 3= 1.0 (bearing length 100mm but <75mm for length from end member) 2

Bearing grade stress, C tg= 2.09 N/mm (compression perpendicular to grain) Permissible bearing stress, C tp= C tg× K 1× K 2× K 3 2 = 2.09 N/mm ×1.0×1.1×1.0 2 = 2.30 N/mm 2

2

Bearing stress, C ts= 0.61 N/mm < permissible bearing stress, C tp= 2.30 N/mm . Therefore bearing stress is adequate.

REF

CALCULATION 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

107.5

Figure 2: Underside notch at beam support

OUTPUT

REF

CALCULATION

OUTPUT

Step 1: Geometrical Properties


=75mm = 225mm = 65mm = 215mm

 Second moment of area,  � ℎ 12 �

(65�215 )12 = 53�83�10

Design span, Le= 3000+[(125/2)×2] = 3125mm Step 2: Check for lateral stability

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. Step 3: Shear Stress

Maximum shear force, V = wL/ 2 = (3.9×3.125)/2 = 6.09 kN 3

Maximum shear stress, qs= 3V / 2A= (3×6.09×10 N)/(2×65×215) = 0.65 N/mm

2

Shear modification factor for flexural member (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (no load sharing system) Shear at notched end factor (underside notch), K 4=

ℎℎ = 107�5215 = 0�5 2

Grade stress for shear, qg= 2.24 N/mm (Shear parallel to grain) Permissible bending stress, q p = qg× K 1× K 2× K 4 2 = 2.24 N/mm ×1.0×1.0×0.5 2 = 1.12 N/mm 2

2

Shear stress, qs=0.65 N/mm < permissible shear stress, q p= 1.12 N/mm . Therefore shear stress is adequate.

REF

CALCULATION

OUTPUT

Step 4: Bending stress, f s 2

2

Maximum bending moment, M=wL /8 = (3.9×3.125 )/8 = 4.76 kNm 2

2

4

3


4

3

Bending stress, f s= M/Z = 4.76×10 Nmm/50.1×10 mm = 9.5 N/mm

2

Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (no load sharing system) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 = 1.0 (depth <300mm) 2


2

Bending stress, f s=9.5 N/mm < permissible bending stress, f p= 20.7 N/mm . Therefore bending stress is adequate. Step 5: Deflection 2

For load sharing system, use E min= 11700 N/mm (non load sharing system) Deflection due to bending where w is total of dead load and imposed load in Newton (N) unit,

 5(3�9�10 � 3�125)(3125) 5    = 384 = 384(11700)(53�83�10) = 7�68  Deflection due to shear

  1�2�4�76�10   =  = 11700 (65�215) = 0�56  16 where F = shape factor depending on cross section (1.2 for solid rectangular or square) M o= bending moment at mid span (in this case M o= M max as the loading is UDL) G= modulus of rigidity ( E/16 ) A= Cross section area


δt = δb+δs=

7.68mm+0.56mm = 8.24 mm

Permissible deflection, δ p= 0.003 Le= 0.003(3125) = 9.4 mm Total deflection, δt = 8.24 mm < permissible deflection,

δ p=

9.4 mm. Deflection is ok.

REF

CALCULATION

OUTPUT


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) = 8125mm 3

Bearing stress, C ts= V/Aa= 6.09×10 /8125 = 0.75 N/mm

2

2

Bearing modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non load sharing system) Bearing length and position (bearing length 125mm and >75mm for length from end member) K 3= 1.05 2

Bearing grade stress, C tg= 2.97 N/mm (compression perpendicular to grain) Permissible bearing stress, C tp= C tg× K 1× K 2× K 3 2 = 2.97 N/mm ×1.0×1.0×1.05 2 = 3.12 N/mm 2

2

Bearing stress, C ts= 0.75 N/mm < permissible bearing stress, C tp= 3.12 N/mm . Therefore bearing stress is adequate

REF

CALCULATION 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: Step 1: Geometrical Properties

Beam dimension No need to taking account reaping effect as the question already stated the dimension is dressed size. Dressed size, b (breadth) = 150mm h (depth) = 150mm Column length, L= 5.0m

OUTPUT

REF

CALCULATION

OUTPUT

Step 2: Determine effective length and slenderness ratio

Effective column length, Le= 1.0 L (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, K 8 2

Compression grade stress, C sg= 9.5 N/mm (Compression parallel to grain) 2 Minimum modulus of elasticity, E min= 7400 N/mm For Table 10 E/ σc .ll= E min /C sg= 7400/9.5 = 778.9 λ= 115.5

Λ

E min /C sg 700 778.9 800

100

115.5

120

0.341

x K 8 y

0.254

0.371

0.280

( − 0.341) (115.5 − 100) � � � 0.2�4 (0.254 − 0.341) (120 − 100) ( − 0.3�1) (115.5 − 100) � � � 0.300 (0.2�0 − 0.3�1) (120 − 100) ( − 0.2�4) (��.� − �00) � �  � 0.2�5 (0.3 − 0.2�4) (�00 − �00)  Step 4: Calculate applied and permissible compressive stress 3

2

Applied compressive stress, C sa= F c /A = 80×10 N/(150×150)mm = 3.56 N/mm

2

Bearing modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non load sharing system) Compression member without bending, K 8 = 0.295 Permissible compressive stress, C sp= C sg� K 1×K 2×K 8 2 = 9.5 N/mm ×1.0×1.0×0.295 2 = 2.8 N/mm 2

Applied compressive stress, C sa= 3.55 N/mm > Permissible compressive stress, C sp= 2.8 2 N/mm . Increase column size or provide lateral support to reduce the slenderness of the column.

REF

CALCULATION

OUTPUT

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: Step 1: Geometrical Properties

Beam dimension Taking account of reaping effect at four side of the timber section Named size, b (breadth) = 75mm h (depth) = 150mm Dressed size, b (breadth) h (depth)

= 65mm = 140mm

Column length, L= 2.5m Step 2: Determine effective length and slenderness ratio

Effective column length about x-x axis Lex= 0.7 L (restrained at both ends in position and in direction) = 0.7(2.5m) = 1.75m Radius of gyration about x-x axis, r x= h / √12 = 140/ √12 = 40.4 Slenderness ratio about x-x axis, λ x= Lex / r x = 1750/40.4 = 43.3<180, ok. Effective column length about y-y axis Ley= 1.0 L (restrained at both ends in position and in direction) = 1.0(2.5m) = 2.5m Radius of gyration about y-y axis, r y= b / √12 = 65/ √12 = 18.8 Slenderness ratio about y-y axis, λ y= Ley / r y = 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 K 8 Step 3: Determine modification factor for compression member without bending, K 8 2

Compression grade stress, C sg= 18.8 N/mm (Compression parallel to grain) 2 Minimum modulus of elasticity, E min= 11700 N/mm For Table 10 E/ σc .ll= E min /C sg= 11700/18.8 = 622.3 λ= 133

Λ

E min /C sg 600 622.3 700

120

133

140

0.226

x K 8 y

0.172

0.254

0.195

REF

CALCULATION

OUTPUT

( − 0.226) (133 − 120) � � � 0.1�1 (0.1�2 − 0.226) (140 − 120) ( − 0.254) (133 − 120) � � � 0.216 (0.1�5 − 0.254) (140 − 120) ( − 0.1�1) (622.3 − 600) � �  � 0.1�� (0.216 − 0.1�1) (�00 − 600)  Step 4: Calculate applied and permissible compressive stress

Bearing modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non load sharing system) Compression member without bending, K 8 = 0.197 Permissible compressive stress, C sp= C sg� K 1×K 2×K 8 2 = 18.8 N/mm ×1.0×1.0×0.197 2 = 3.7 N/mm 2

Ultimate axial load, F c= C sp×A = 3.7 N/mm2×(65×140)mm = 33670 N= 33.7 kN

REF

CALCULATION

OUTPUT

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.

h

b

Timber cross section

Figure 1: Example of metal nail plate Solution: Step 1: Geometrical Properties


=50mm =150mm = 45mm = 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= 6300mm 3

2

2

/ t sa= T An = 35×10 N/6300mm 2 = 5.56 N/mm

REF

CALCULATION

OUTPUT

Step 3: Calculate the permissible tension stress

Modification factors: Duration of loading, K 1= 1.0 (long term loading) Load sharing system, K 2= 1.0 (non load sharing) Tension parallel to grain grade stress, t sg= 10.9 N/mm

2

Permissible tension stress, t sp= t sg×K 1× K 2 2 = 10.9 N/mm ×1.0×1.0 2 = 10.9 N/mm 2

2

Applied tension stress, t sa= 5.56 N/mm < permissible tension stress, t sp= 10.9 N/mm . Therefore, tension capacity of the timber section is adequate.

REF

CALCULATION

OUTPUT

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: Step 1: Geometrical Properties

Beam dimension Named size, b (breadth) h (depth)

=50mm =150mm

Dressed size, b (breadth) h (depth)

= 45mm = 140mm

Step 2: Calculate applied bending stress and permissible bending stress Applied bending stress 2 2 Maximum bending moment, M=wL /8 = (0.5×2 )/8 = 0.25 kNm 2

2

3

3

Section modulus, Z =bh /6= (45×140 )/6= 147×10 mm 6

3

3

Applied Bending stress, f a= M/Z = 0.25×10 Nmm/147×10 mm = 1.7 N/mm

2

Permissible bending stress Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non load sharing system) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 = 1.0 (depth <300mm) 2


2

Bending stress, f a=1.7 N/mm < permissible bending stress, f p= 8.8 N/mm . Therefore bending stress is adequate.

REF

CALCULATION

OUTPUT

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= 6300mm 3

2

2

t sa= T An = 22×10 N/6300mm / 2 = 3.49 N/mm

Permissible tension stress Modification factors: Duration of loading, K 1= 1.0 (long term loading) Load sharing system, K 2= 1.0 (non load sharing)

Tension parallel to grain grade stress, t sg= 5.3 N/mm

2

Permissible tension stress, t sp= t sg×K 1× K 2 2 = 5.3 N/mm ×1.0×1.0 2 = 5.3 N/mm 2

2

Applied tension stress, t sa= 3.49 N/mm < permissible tension stress, t sp= 5.3 N/mm . Therefore, tension capacity of the timber section is adequate. Step 4: Interaction Equation

  + ≤ 1.0   �

3�49 � 1�7 = 0�85 < 1�0 5�3 8�8

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

REF

CALCULATION

OUTPUT

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 Applied bending stress 2 2 Maximum bending moment, M=wL /8 = (0.5×2 )/8 = 0.25 kNm

As the size is unknown, therefore the section modulus is 2 Section modulus, Z =bh /6 6

2

Applied Bending stress, f a= M/Z = 0.25×10 Nmm/((bh )/6)= 1.7 N/mm

2

Permissible bending stress Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non-load sharing system) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 = 1.0 (depth <300mm) 2

Grade stress for bending, f g= 11.3 N/mm (bending parallel to grain) Permissible bending stress, f p = f g× K 1× K 2× K 5× K 6 2 = 11.3 N/mm ×1.0×1.0×1.0×1.0 2 = 11.3 N/mm 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. b and h are unknown, therefore the net area is Gross section area Ag= Net cross section, An= bh 3

t sa= T An = 22×10 N/ bh /

REF

CALCULATION

OUTPUT



2

Permissible tension stress, t sp= t sg×K 1× K 2 2 = 6.8 N/mm ×1.0×1.0 2 = 6.8 N/mm Step 4: Interaction Equation

 �  ≤ 1�0   =

� ��() =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

=

�() ��(()) � =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 t he proposed suitable timber size (named size) is 60mm×110mm. For confirmation, recalculate for combined tension and bending interaction equation Beam dimension Named size, b (breadth) h (depth) Dressed size, b (breadth) h (depth)

=60mm =110mm = 50mm = 100mm

REF

CALCULATION

OUTPUT

Applied bending stress 2 2 Maximum bending moment, M=wL /8 = (0.5×2 )/8 = 0.25 kNm 2

2

3

3


3

3

Applied Bending stress, f a= M/Z = 0.25×10 Nmm/83.33×10 mm = 3 N/mm

2

Permissible bending stress Bending modification factors (K-Factors) Duration of loading, K 1= 1.0 (long term) Load sharing system, K 2= 1.0 (non-load sharing system) Form factor, K 5= 1.0 (rectangular section) Depth factor, K 6 = 1.0 (depth <300mm) 2

Grade stress for bending, f g= 11.3 N/mm (bending parallel to grain) Permissible bending stress, f p = f g× K 1× K 2× K 5× K 6 2 = 11.3 N/mm ×1.0×1.0×1.0×1.0 2 = 11.3 N/mm Applied tension stress Assume no reduction of area due unavailability of detailing at connection. b and h are unknown, therefore the net area is 2 Gross section area Ag= Net cross section, An= bh= 50×100=5000mm 3

2

t sa= T An = 22×10 N/5000 mm = 4.4 N/mm /

2



2

Permissible tension stress, t sp= t sg×K 1× K 2 2 = 6.8 N/mm ×1.0×1.0 2 = 6.8 N/mm Interaction Equation

  + ≤ 1.0   �

4�4 � 3 = 0�91 < 1�0 6�8 11�3

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

226919874-Example-on-Design-of-Timber-Structure-part2.pdf

Recommend Documents