A Timber Column of length 1.0 meter, having c/s dimension of 73 mm × 198 mm, is subjected to an axial compressive force of 50.0 kN. Design the member for the ultimate limit state.
Details
Material properties:
- Timber class: C24
- Service classes: Class 2, moisture content ≤ 20%
- Load duration classes: Medium-term
Cross section properties:
- Length of the member is 1 m.
- Rectangular cross section, b = 73 mm, h = 198 mm,
- Effective cross sectional area A = 14,454 mm2,
- Radius of gyration of cross section about y-axis ry = 21 mm,
- Radius of gyration of cross section about z-axis rz = 57 mm,
- Section modulus of cross section about z-axis: Wz =
4.770x105 mm3
- Section modulus of cross section about y-axis: Wy =
1.759x105 mm3
Characteristic material properties for timber:
Modification factor Kmod = 0.80 …from table 3.1
Material factors γm = 1.30 … from table 2.3
Fc0d= (Kmod·fc0k)/γm
= (0.80·21.00)/1.30 = 12.92 N/mm2 [Cl 2.4.1(1)P] | |
Cross section loads:
Compression parallel to the grain:
Sc0d = (1000xFx)/A = (1000x50.000)/14454 = 3.46N/mm2 < 12.92N/mm2 (Fc0d) | |
The ratio of actual compressive stress to allowable compressive strength:
Sc0d /Fc0d = 3.46 / 12.92 = 0.268 < 1.0 [Cl. 6.1.4.(1)P] | |
Check for Slenderness:
Slenderness ratios:
As timber grade is C24 (i.e., Soft Wood)
E0,05 = 0.67·E0,mean = 0.739 kN/m2
| |
[Annex A,EN 338:2003]
λrel,z = λz/π·(fc0k/E0,05)1/2 = 17.54/π(21.00/0.739)1/2 = 0.298 | |
λrel,y = λy/π·(fc0k/E0,05)1/2 = 47.62/π(21.00/0.739)1/2 = 0.809 | |
Since, λrel,y is greater than 0.3, following conditions should be satisfied:
Sc0d/(Kcz·Fc0d) + (Smzd/Fmzd) + Km·(Smyd/Fmyd) ≤ RATIO
| |
Sc0d/(Kcy·Fc0d) + Km·(Smzd/Fmzd) + (Smyd/Fmyd) ≤ RATIO
| |
Where:
Kz = 0.5·[1 + βc·(λrel,z - 0.3) + (λrel,z)2] = 0.50·[1 + 0.2(0.298 - 0.3) + (0.298)2] = 0.541 | |
Ky = 0.5·[1 + βc·(λrel,y - 0.3) + (λrel,y)2] = 0.50·[1 + 0.2(0.809 - 0.3) + (0.809)2] = 0.878 | |
Kcz = 1/{Kz + [(Kz)2 - (λrel,z)2]1/2} = 1/{0.541 + [(0.541)2 - (0.298)2]1/2}= 1.008 | |
Kcy = 1/{Ky + [(Kzy)2 - (λrel,y)2]1/2} = 1/{0.878 + [(0.878)2 - (0.809)2]1/2} = 0.820 | |
For Rectangular cross section Km = 0.70. The member is subjected to Compression only, so actual bending stress is zero.
Sc0d/(Kcz·Fc0d) + (Smzd/Fmzd) + Km·(Smyd/Fmyd) = 3.46/(1.008·12.92) + 0.0 + 0.0 = 0.268 + 0.0 + 0.0 = 0.266 | |
Sc0d/(Kcy·Fc0d) + Km·(Smzd/Fmzd) + (Smyd/Fmyd) = 3.46 /(0.820·12.92) + 0.0 + 0.0 = 0.326 + + 0.0 + 0.0 = 0.326
| |
Hence the critical ratio is 0.326 < 1.0 and the section is safe.
Results
Table 1. EC 5: Part 1-1 Verification Example 1
Criteria |
Reference |
STAAD.Pro
|
Difference |
Comments |
Critical Ratio (Cl. 6.3.2) |
0.326 |
0.327 |
negligible |
|
STAAD.Pro Output
STAAD.Pro CODE CHECKING - (EC5) v1.0
***********************
ALL UNITS ARE - KN METE (UNLESS OTHERWISE Noted)
MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/
FX MY MZ LOCATION
=======================================================================
1 PRIS ZD = 0.073 YD = 0.198
PASS CL.6.3.2 0.327 1
50.00 C 0.00 0.00 0.0000
|--------------------------------------------------------------------------|
| AX = 0.01 IY = 0.00 IZ = 0.00 |
| LEZ = 1.00 LEY = 1.00 |
| |
| ALLOWABLE STRESSES: (NEW MMS) |
| FBY = 14.769 FBZ = 14.769 |
| FC = 12.859 |
| ACTUAL STRESSES : (NEW MMS) |
| fby = 0.000 fbz = 0.000 |
| fc = 3.459 |
|--------------------------------------------------------------------------|