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%
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 r_y = 21 mm,
Radius of gyration of cross section about z-axis r_z = 57 mm,
Section modulus of cross section about z-axis: W_z= 4.770x10⁵ mm³
Section modulus of cross section about y-axis: W_y= 1.759x10⁵ mm³

Validation

Characteristic material properties for timber:

Modification factor Kmod = 0.80 …from table 3.1

Material factors γ_m = 1.30 … from table 2.3

f_c0k= 21.00 N/mm²

E_0,05 = 7370 N/mm²

F_c0d= (Kmod·f_c0k)/γ_m = (0.80·21.00)/1.30 = 12.92 N/mm² [Cl 2.4.1(1)P]

f_myk= 24.00 N/mm²

F_myd= Kmod·f_myk/γ_m = (0.80x24.00)/1.30 = 14.77N/mm²

f_mzk= 24.00 N/mm²

F_mzd= Kmod·f_mzk/γ_m = (0.80x24.00)/1.30 = 14.77N/mm²

Cross section loads:

F_x= 5.000 kN

M_z= 2.000 kN·m

M_y= 1.000 kN·m

Check for Slenderness:

Slenderness ratios:

λ_z= (1000/57) = 17.54

λ_y = (1000/21) = 47.62

λ_rel,z = λ_z/π·(f_c0k/E_0,05)^1/2 = 17.54/π(21.00/7370)^1/2 = 0.298

λ_rel,y = λ_y/π·(f_c0k/E_0,05)^1/2 = 47.62/π(21.00/7370)^1/2 = 0.809

Since, λ_rel,yis greater than 0.3, following conditions should be satisfied [Cl 6.3.2.3]:

S_c0d/(Kcz·F_c0d) + (S_mzd/F_mzd) + Km·(S_myd/F_myd) ≤ RATIO

S_c0d/(Kcy·F_c0d) + Km·(S_mzd/F_mzd) + (S_myd/F_myd) ≤ RATIO

Where:

Kz = 0.5·[1 + β_c·(λ_rel,z - 0.3) + (λ_rel,z)²] = 0.50·[1 + 0.2(0.298 - 0.3) + (0.298)²] = 0.541

Ky = 0.5·[1 + β_c·(λ_rel,y - 0.3) + (λ_rel,y)²] = 0.50·[1 + 0.2(0.809 - 0.3) + (0.809)²] = 0.878

Kcz = 1/{K_z + [(K_z)² - (λ_rel,z)²]^1/2} = 1/{0.541 + [(0.541)² - (0.298)²]^1/2}= 1.008

Kcy = 1/{K_y + [(K_zy)² - (λ_rel,y)²]^1/2} = 1/{0.878 + [(0.878)² - (0.809)²]^1/2} = 0.820

For Rectangular cross section Km = 0.70.

S_c0d= (1000·F_x/A) = (1000·5.000)/14454 = 0.35 N/mm²

S_mzd= (10⁶·M_z)/W_z= (10⁶·2.000)/(4.770x10⁵) = 4.19 N/mm²

S_myd= (10⁶·M_y)/W_y = (10⁶·1.000)/(1.759x10⁵) = 5.69 N/mm²

Combined stress ratio:

S_c0d/(Kcz·F_c0d) + (S_mzd/F_mzd) + Km·(S_myd/F_myd) = 0.35/(1.008·12.92) + 4.19/14.77 + 0.70(5.69/14.77) = 0.027 + 0.283 + 0.269 = 0.266

S_c0d/(Kcy·F_c0d) + Km·(S_mzd/F_mzd) + (S_myd/F_myd) = 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
Criteria	Reference	STAAD.Pro	Difference	Comments
Critical Ratio (Cl. 6.3.2)	0.616	0.616	none

STAAD.Pro Input

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
LOAD 1 LOADTYPE NONE TITLE LOAD CASE 1
JOINT LOAD
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 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            0.616         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                                 |
  |--------------------------------------------------------------------------|