Timber Truss Design Procedure 2f22i
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Structural Timber Design Course – IEM Dec 2003
TIMBER TRUSS DESIGN PROCEDURE
1. Determine the dead and live loads acting on the truss 2. Compute the stresses 3. Determine the required sizes 4. Design the ts
Example : Standard Truss
7m
Slope 22.5º Spacing of truss 600 mm c/c SG 5, Std Grade, Dry Timber
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
1
Structural Timber Design Course – IEM Dec 2003
Load Determination
Dead Load - Long Term On rafter
: 0.7 kN/m2 on slope
On plan
: 0.7 / cos 22.5º = 0.76 kN/m 2
On ceiling tie : 0.25 kN/m2
Live Load - BS6399 On rafter - 0.75 kN/m2 on plan ( medium term ) On ceiling tie :
1. 0.25 kN/m2 ( consider as long term ) 2. 0.9 kN point load ( short term )
# Assume wind load on rafter as less severe than live load in the design of the .
Wind Load ( very short term ) Taking design wind speed , V = 33 m/s For conservative approach , i = 0.2 and e = 0.9
3 Chap. V
- Rafter Wind Load = 0.613 x 10-3 x 332 x 0.9 = 0.6 kN/m2 ( -ve
)
- Ceiling Tie Wind Load = 0.613 x 10-3 x 332 x 0.2 = 0.134 kN/m2 ( -ve
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
)
2
Structural Timber Design Course – IEM Dec 2003
Stress Computation 3 conditions of loading are required to calculate the member stresses : 1. Long Term ( only long term loads ) 2. Medium Term ( long term + medium term loads ) 3. Short Term ( all loads )
LONG TERM LOADING
On rafter
= 0.76 kN/m2 x 0.6m x 7m 4 bays = 0.798 kN
On ceiling = ( 0.25 + 0.25 ) x 0.6 x 7 3 bays = 0.7 kN LONG TERM
0.798 ( 0.76 + 0.75 ) x 0.6 x 7 / 4 bays
4.4 0.798
0.798 1.7 0.749
4.9
4.6 2.646
0.749
0.8
3.0 0.7
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
0.7
2.646
3
Structural Timber Design Course – IEM Dec 2003
MEDIUM TERM
1.59 ( 0.25 + 0.25 ) x 0.6 x 7 / 3 bays
7.0 1.59
1.59 2.4 1.145
8.0
1.145
1.5
7.4
5.0
4.25
0.7
0.7
4.25
VERY SHORT TERM Rafter
= ( 0.76 + 0.75 – 0.36 ) x 0.6 x 0.7 = 1.21 kN 4
Ceiling Joist = ( 0.25 + 0.25 – 0.134 ) x 0.6 x 0.7 = 0.51 kN 3 1.21 1.21 1.21 0.86
0.86 0.51+ 0.9 0.51
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
4
Structural Timber Design Course – IEM Dec 2003
SHORT TERM 1.59
7.0
7.9 1.59
1.59 3.6 1.145
2.4
8.0
8.9
5.7
4.83
1.145
8.2
0.7 + 0.9 = 1.6
Grade Stresses (SG 5) .
8.9
1.6
1.5
0.7
4.53
Normally (critical) only check for :
σ m,g = 9.5 N/mm2 σ t,g
= 5.7 N/mm2
σ c,g
2
Medium term ( DL + IL )
= 8.5 N/mm
Emean = 9100 N/mm
Short term 2
Emin = 6300 N/mm
( DL + IL + PL)
2
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
5
Structural Timber Design Course – IEM Dec 2003
Example : Assume member size 38 x 100 Finished Size 35 x 97 From table of Properties : Zxx = 54900 mm3 ίxx = 28 mm ίyy = 10.1 mm A
= 3400 mm 2
RAFTER DESIGN Consider medium term load Check for combine bending and axial force.
apex
3.8
Rafter analysis :
22.5o 3.5 w kN /m
Heel
apex L = 1.9 m
L = 1.9 m
0.125 wL2
0.75L by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
0.0703 wL2 6
Structural Timber Design Course – IEM Dec 2003
Consider lower portion of rafter :
w = ( 0.76 + 0.75 ) x 0.6 = 0.906 kN/m L = 1.9 m M = 0.0703 x 0.906 x 1.9 2 = 0.23 kNm
Applied bending stress,
σ m,a
=
M Z
= 0.23 x 106 54900
= 4.19 N/mm
Under medium term , axial compressive force
= 8.0 kN
Applied compressive stress,
σ c,a
=
P A
= 8000 3400
= 2.35 N/mm2
Effective length = 3 x 1.75 4 cos 22.5 o
= 1.42 m
Rafter is fully restrained by tiling battens in the less stiff direction.
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
7
Structural Timber Design Course – IEM Dec 2003
Slenderress ratio , λ =
σ c//
Le Ίxx
= 1420 28
= 50.7
= 8.5 x 1.25 ( medium term ) = 10.625 N/mm2
E min σ c//
= 6300 10.625
= 592.94
From table 10 ( MS 544 )
K8
= 0.682
σ c,
= 8.5 x 1.25 x 1.1 x 0.682 = 7.97 N/mm2
σ m,
= 9.5 x 1.25 x 1.1
σe
=
2 E
=
2
= 13.0 N/mm2
2 (6300)
= 24.19
(50.7)2
λ
Combine Compression and Bending ( Clause 12.6 )
σ m,a σ m,
+
1 - 1.5 σ c,a K8 σe
3.55 13
σ c,a
< 1
σ c,
+
1 - 1.5 X 2.35 X 0.682
2.35
= 0.598
< 1
7.97
24.19
Therefore it is satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
8
Structural Timber Design Course – IEM Dec 2003
Consider portion over node point. M = 0.125 wL2 = 0.125 x 0.906 x 1.75 2 = 0.347 kNm
Applied bending stress,
σ m,a
=
M Z
= 0.347 x 106 54900
= 6.32 N/mm
Axial Compressive force ( Average lower and upper chord )
8 + 7 = 7.5 kN 2
Applied compressive stress,
σ c,a
=
P A
= 7500 3400
= 2.21 N/mm2
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
9
Structural Timber Design Course – IEM Dec 2003
At node point , λ < 5.0 , rafter is designed as short column. = 8.5 x 1.25 x 1.1 = 11.69 N/mm2
σ c,
Combine Stress calculation for short column
σ m,a
+
σ m,
= 6.32
+
13.0
=
0.68
σ c,a σ c,
2.21 11.69
< 0.9
The upper chord need not be checked because axial compressive force is 7kN < 8 kN for lower chord.
Whole rafter is satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
10
Structural Timber Design Course – IEM Dec 2003
DESIGN OF CEILING TIE
Ceiling tie
– combined bending and tension.
Under long term – Loads 0.25 + 0.25
The BMD for UDL : W / unit length
L= 2.33
L
L
0.1WL2 ,
+
0.08wL2
+
+
0.025w L2
Check Outer Bay
W
= ( 0.25 + 0.25 ) x 0.6 = 0.3 kN/m
L
= 7/3 = 2.33
M
= 0.08wL 2 = 0.08 x 0.3 x ( 2.33 ) 2 = 0.13 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
11
Structural Timber Design Course – IEM Dec 2003
σ m, a
= M/Z = 0.13 x 10 6 54900 = 2.39 N / mm2
Axial tensile force ( long term stress )
σ t, a
= 4.6 kN
= 4600 3400 = 1.355 N / mm2 = 9.5 x 1 x 1.1 = 10.45 N /mm 2
σ m, σ t,
= 5.7 x 1 x 1.1 = 6.27 N / mm2
Combination
:
= 2.39
+ 1.355
10.45
6.27
σ m,a + σ t,a σ m, σ t,
<1
= 0.45 < 1.0
*Satisfactory
At ,
M
= 0.1wL2 = 0.1 x 0.3 x 2.33 2 = 0.163 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
12
Structural Timber Design Course – IEM Dec 2003
Axial tensile force
= 4.6 + 3.0
= 3.8 kN
2
σ m, a
= 0.163 x 10 6 54900 = 2.97 N / mm2
σ t, m
= 3800
= 1.12 N / mm 2
3400
Combination :
2.97
+
10.45
1.12
= 0.46 < 1
6.27
* Satisfactory
Under short term - Loads = point load 0.9 kN + UDL
P
0.075 PL 0.175PL 2.33 M
2.33
2.33
at center of ceiling tie due to UDL , M = 0.025 wL 2 = 0.025 x 0.3 x 2.33 2 = 0.041 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
13
Structural Timber Design Course – IEM Dec 2003
M due to point load 0.9 kN ,
M = 0.175 PL = 0.175 x 0.9 x 2.33 = 0.367 kNm
*∑M
= 0.408 kNm
σ m, a
= 0.408 x 10 6 = 7.43 N / mm 2 54900
Axial tensile force , ( max ) = 8.9 kN
σ t, a
= 8900 = 2.62 N / mm 2 3400
Permissible stresses :
σ m,
= 9.5 x 1.5 x 1.1 = 15.68 N / mm2
σ t,
= 5.7 x 1.5 x 1.1 = 9.41 N / mm2
Combination : 7.43 15.68
+ 2.62
= 0.75 < 1
9.41
Satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
14
Structural Timber Design Course – IEM Dec 2003
At , M for UDL , M = 0.1wL2 = 0.1 x 0.3 x 2.332 = 0.163 kNm
M for point load , M = 0.075 PL = 0.075 x 0.9 x 2.33 = 0.157 kNm
*∑ M
= 0.321 kNm
σ m, a
= 0.321 x 106 = 5.85 N / mm2 54900
Axial tensile force =
8.9 + 5.7
= 7.3 kN
2 σ t, a
= 7300 = 2.15 N / mm2 3400
Combination , 5.85 15.68
+
2.15
= 0.60
< 1
9.41
* Satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
15
TIMBER TRUSS DESIGN PROCEDURE
1. Determine the dead and live loads acting on the truss 2. Compute the stresses 3. Determine the required sizes 4. Design the ts
Example : Standard Truss
7m
Slope 22.5º Spacing of truss 600 mm c/c SG 5, Std Grade, Dry Timber
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
1
Structural Timber Design Course – IEM Dec 2003
Load Determination
Dead Load - Long Term On rafter
: 0.7 kN/m2 on slope
On plan
: 0.7 / cos 22.5º = 0.76 kN/m 2
On ceiling tie : 0.25 kN/m2
Live Load - BS6399 On rafter - 0.75 kN/m2 on plan ( medium term ) On ceiling tie :
1. 0.25 kN/m2 ( consider as long term ) 2. 0.9 kN point load ( short term )
# Assume wind load on rafter as less severe than live load in the design of the .
Wind Load ( very short term ) Taking design wind speed , V = 33 m/s For conservative approach , i = 0.2 and e = 0.9
3 Chap. V
- Rafter Wind Load = 0.613 x 10-3 x 332 x 0.9 = 0.6 kN/m2 ( -ve
)
- Ceiling Tie Wind Load = 0.613 x 10-3 x 332 x 0.2 = 0.134 kN/m2 ( -ve
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
)
2
Structural Timber Design Course – IEM Dec 2003
Stress Computation 3 conditions of loading are required to calculate the member stresses : 1. Long Term ( only long term loads ) 2. Medium Term ( long term + medium term loads ) 3. Short Term ( all loads )
LONG TERM LOADING
On rafter
= 0.76 kN/m2 x 0.6m x 7m 4 bays = 0.798 kN
On ceiling = ( 0.25 + 0.25 ) x 0.6 x 7 3 bays = 0.7 kN LONG TERM
0.798 ( 0.76 + 0.75 ) x 0.6 x 7 / 4 bays
4.4 0.798
0.798 1.7 0.749
4.9
4.6 2.646
0.749
0.8
3.0 0.7
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
0.7
2.646
3
Structural Timber Design Course – IEM Dec 2003
MEDIUM TERM
1.59 ( 0.25 + 0.25 ) x 0.6 x 7 / 3 bays
7.0 1.59
1.59 2.4 1.145
8.0
1.145
1.5
7.4
5.0
4.25
0.7
0.7
4.25
VERY SHORT TERM Rafter
= ( 0.76 + 0.75 – 0.36 ) x 0.6 x 0.7 = 1.21 kN 4
Ceiling Joist = ( 0.25 + 0.25 – 0.134 ) x 0.6 x 0.7 = 0.51 kN 3 1.21 1.21 1.21 0.86
0.86 0.51+ 0.9 0.51
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
4
Structural Timber Design Course – IEM Dec 2003
SHORT TERM 1.59
7.0
7.9 1.59
1.59 3.6 1.145
2.4
8.0
8.9
5.7
4.83
1.145
8.2
0.7 + 0.9 = 1.6
Grade Stresses (SG 5) .
8.9
1.6
1.5
0.7
4.53
Normally (critical) only check for :
σ m,g = 9.5 N/mm2 σ t,g
= 5.7 N/mm2
σ c,g
2
Medium term ( DL + IL )
= 8.5 N/mm
Emean = 9100 N/mm
Short term 2
Emin = 6300 N/mm
( DL + IL + PL)
2
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
5
Structural Timber Design Course – IEM Dec 2003
Example : Assume member size 38 x 100 Finished Size 35 x 97 From table of Properties : Zxx = 54900 mm3 ίxx = 28 mm ίyy = 10.1 mm A
= 3400 mm 2
RAFTER DESIGN Consider medium term load Check for combine bending and axial force.
apex
3.8
Rafter analysis :
22.5o 3.5 w kN /m
Heel
apex L = 1.9 m
L = 1.9 m
0.125 wL2
0.75L by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
0.0703 wL2 6
Structural Timber Design Course – IEM Dec 2003
Consider lower portion of rafter :
w = ( 0.76 + 0.75 ) x 0.6 = 0.906 kN/m L = 1.9 m M = 0.0703 x 0.906 x 1.9 2 = 0.23 kNm
Applied bending stress,
σ m,a
=
M Z
= 0.23 x 106 54900
= 4.19 N/mm
Under medium term , axial compressive force
= 8.0 kN
Applied compressive stress,
σ c,a
=
P A
= 8000 3400
= 2.35 N/mm2
Effective length = 3 x 1.75 4 cos 22.5 o
= 1.42 m
Rafter is fully restrained by tiling battens in the less stiff direction.
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
7
Structural Timber Design Course – IEM Dec 2003
Slenderress ratio , λ =
σ c//
Le Ίxx
= 1420 28
= 50.7
= 8.5 x 1.25 ( medium term ) = 10.625 N/mm2
E min σ c//
= 6300 10.625
= 592.94
From table 10 ( MS 544 )
K8
= 0.682
σ c,
= 8.5 x 1.25 x 1.1 x 0.682 = 7.97 N/mm2
σ m,
= 9.5 x 1.25 x 1.1
σe
=
2 E
=
2
= 13.0 N/mm2
2 (6300)
= 24.19
(50.7)2
λ
Combine Compression and Bending ( Clause 12.6 )
σ m,a σ m,
+
1 - 1.5 σ c,a K8 σe
3.55 13
σ c,a
< 1
σ c,
+
1 - 1.5 X 2.35 X 0.682
2.35
= 0.598
< 1
7.97
24.19
Therefore it is satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
8
Structural Timber Design Course – IEM Dec 2003
Consider portion over node point. M = 0.125 wL2 = 0.125 x 0.906 x 1.75 2 = 0.347 kNm
Applied bending stress,
σ m,a
=
M Z
= 0.347 x 106 54900
= 6.32 N/mm
Axial Compressive force ( Average lower and upper chord )
8 + 7 = 7.5 kN 2
Applied compressive stress,
σ c,a
=
P A
= 7500 3400
= 2.21 N/mm2
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
9
Structural Timber Design Course – IEM Dec 2003
At node point , λ < 5.0 , rafter is designed as short column. = 8.5 x 1.25 x 1.1 = 11.69 N/mm2
σ c,
Combine Stress calculation for short column
σ m,a
+
σ m,
= 6.32
+
13.0
=
0.68
σ c,a σ c,
2.21 11.69
< 0.9
The upper chord need not be checked because axial compressive force is 7kN < 8 kN for lower chord.
Whole rafter is satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
10
Structural Timber Design Course – IEM Dec 2003
DESIGN OF CEILING TIE
Ceiling tie
– combined bending and tension.
Under long term – Loads 0.25 + 0.25
The BMD for UDL : W / unit length
L= 2.33
L
L
0.1WL2 ,
+
0.08wL2
+
+
0.025w L2
Check Outer Bay
W
= ( 0.25 + 0.25 ) x 0.6 = 0.3 kN/m
L
= 7/3 = 2.33
M
= 0.08wL 2 = 0.08 x 0.3 x ( 2.33 ) 2 = 0.13 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
11
Structural Timber Design Course – IEM Dec 2003
σ m, a
= M/Z = 0.13 x 10 6 54900 = 2.39 N / mm2
Axial tensile force ( long term stress )
σ t, a
= 4.6 kN
= 4600 3400 = 1.355 N / mm2 = 9.5 x 1 x 1.1 = 10.45 N /mm 2
σ m, σ t,
= 5.7 x 1 x 1.1 = 6.27 N / mm2
Combination
:
= 2.39
+ 1.355
10.45
6.27
σ m,a + σ t,a σ m, σ t,
<1
= 0.45 < 1.0
*Satisfactory
At ,
M
= 0.1wL2 = 0.1 x 0.3 x 2.33 2 = 0.163 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
12
Structural Timber Design Course – IEM Dec 2003
Axial tensile force
= 4.6 + 3.0
= 3.8 kN
2
σ m, a
= 0.163 x 10 6 54900 = 2.97 N / mm2
σ t, m
= 3800
= 1.12 N / mm 2
3400
Combination :
2.97
+
10.45
1.12
= 0.46 < 1
6.27
* Satisfactory
Under short term - Loads = point load 0.9 kN + UDL
P
0.075 PL 0.175PL 2.33 M
2.33
2.33
at center of ceiling tie due to UDL , M = 0.025 wL 2 = 0.025 x 0.3 x 2.33 2 = 0.041 kNm
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
13
Structural Timber Design Course – IEM Dec 2003
M due to point load 0.9 kN ,
M = 0.175 PL = 0.175 x 0.9 x 2.33 = 0.367 kNm
*∑M
= 0.408 kNm
σ m, a
= 0.408 x 10 6 = 7.43 N / mm 2 54900
Axial tensile force , ( max ) = 8.9 kN
σ t, a
= 8900 = 2.62 N / mm 2 3400
Permissible stresses :
σ m,
= 9.5 x 1.5 x 1.1 = 15.68 N / mm2
σ t,
= 5.7 x 1.5 x 1.1 = 9.41 N / mm2
Combination : 7.43 15.68
+ 2.62
= 0.75 < 1
9.41
Satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
14
Structural Timber Design Course – IEM Dec 2003
At , M for UDL , M = 0.1wL2 = 0.1 x 0.3 x 2.332 = 0.163 kNm
M for point load , M = 0.075 PL = 0.075 x 0.9 x 2.33 = 0.157 kNm
*∑ M
= 0.321 kNm
σ m, a
= 0.321 x 106 = 5.85 N / mm2 54900
Axial tensile force =
8.9 + 5.7
= 7.3 kN
2 σ t, a
= 7300 = 2.15 N / mm2 3400
Combination , 5.85 15.68
+
2.15
= 0.60
< 1
9.41
* Satisfactory
by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn
15
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