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 5.0 kN and moments of 2.0 kN.m and 1.0 kN.m
about its major and minor axes respectively. Design the member for the ultimate limit
state.
Given
Material properties:
- Timber Strength Class: C24
- Service classes: Class 2, moisture content <=20%
- Load duration: 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 = 14454 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] | |
Fmyd = Kmod·fmyk/γm
= (0.80x24.00)/1.30 = 14.77N/mm2 | |
Fmzd = Kmod·fmzk/γm
= (0.80x24.00)/1.30 = 14.77N/mm2 | |
Cross section loads:
Fx = 5.000 kN
Mz = 2.000 kN·m
My = 1.000 kN·m
Check for Slenderness:
Slenderness ratios:
λrel,z = λz/π·(fc0k/E0,05)1/2 = 17.54/π(21.00/7370)1/2 = 0.298 | |
λrel,y = λy/π·(fc0k/E0,05)1/2 = 47.62/π(21.00/7370)1/2 = 0.809 | |
Since, λrel,y is greater than 0.3, following conditions should be satisfied [Cl 6.3.2.3]:
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.
Sc0d = (1000·Fx/A) = (1000·5.000)/14454 = 0.35 N/mm2 | |
Smzd = (106·Mz)/Wz = (106·2.000)/(4.770x105) = 4.19 N/mm2 | |
Smyd = (106·My)/Wy = (106·1.000)/(1.759x105) = 5.69 N/mm2 | |
Combined stress ratio:
Sc0d/(Kcz·Fc0d) + (Smzd/Fmzd) + Km·(Smyd/Fmyd) = 0.35/(1.008·12.92) + 4.19/14.77 + 0.70(5.69/14.77) = 0.027 + 0.283 + 0.269 = 0.266 | |
Sc0d/(Kcy·Fc0d) + Km·(Smzd/Fmzd) + (Smyd/Fmyd) = 0.35 /(0.820·12.92) + 0.70(4.19/14.77) + 5.69/14.77 = 0.033 + 0.385 + 0.198 = 0.616
| |
Hence the critical ratio is 0.616 < 1.0 and the section is safe.
Comparison
Table 1. EC 5: Part 1-1 Verification Example 2
Criteria |
Reference |
STAAD.Pro
|
Difference |
Critical Ratio (Cl. 6.3.2) |
0.616 |
0.616 |
none |
Output
STAAD.Pro CODE CHECKING - (EC5 )
***********************
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 0.616 1
5.00 C 1.00 -2.00 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 = 5.686 fbz = 4.193 |
| fc = 0.346 |
|--------------------------------------------------------------------------|