 # V. EC5 - Timber Column with Bending

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.

## Details

Material properties:

• Timber Strength Class:  C24
• Service classes: Class 2, moisture content <=20%

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

## Validation

Characteristic material properties for timber:

Modification factor Kmod = 0.80 …from table 3.1

Material factors γm = 1.30 … from table 2.3

 fc0k= 21.00 N/mm2

 E0,05 = 7370 N/mm2

 Fc0d= (Kmod·fc0k)/γm = (0.80·21.00)/1.30 = 12.92 N/mm2 [Cl 2.4.1(1)P]

 fmyk = 24.00 N/mm2

 Fmyd = Kmod·fmyk/γm = (0.80x24.00)/1.30 = 14.77N/mm2

 fmzk = 24.00 N/mm2

 Fmzd = Kmod·fmzk/γm = (0.80x24.00)/1.30 = 14.77N/mm2

Fx = 5.000 kN

Mz = 2.000 kN·m

My = 1.000 kN·m

Check for Slenderness:

Slenderness ratios:

 λz = (1000/57) = 17.54

 λy = (1000/21) = 47.62

 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.

## Results

Table 1. EC 5: Part 1-1 Verification Example 2
Critical Ratio (Cl. 6.3.2) 0.616 0.616 none

The following file is included as C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ .

``````STAAD SPACE
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 1 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC WOOD
E 1.10316e+007
POISSON 0.15
DENSITY 0.00231749
ALPHA 5.5e-006
END DEFINE MATERIAL
CONSTANTS
MATERIAL WOOD MEMB 1
MEMBER PROPERTY
1 PRIS YD 0.198 ZD 0.073
SUPPORTS
1 FIXED
2 FY -5.0 MX 1.0 MZ 2.0
PERFORM ANALYSIS
PARAMETER
CODE TIMBER EC5
ALPHA 0 ALL
LDC 3 ALL
SCL 2 ALL
TSC 6 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH
``````

```                         STAAD.Pro CODE CHECKING - (EC5 )
***********************
ALL UNITS ARE - KN   METE (UNLESS OTHERWISE Noted)
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                                 |
|--------------------------------------------------------------------------|
```