V. EC5 - Timber Column

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 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

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

Cross section loads:

F_x = 50.000 kN

Compression parallel to the grain:

S_c0d= (1000xF_x)/A = (1000x50.000)/14454 = 3.46N/mm² < 12.92N/mm² (F_c0d)

The ratio of actual compressive stress to allowable compressive strength:

S_c0d/F_c0d= 3.46 / 12.92 = 0.268 < 1.0 [Cl. 6.1.4.(1)P]

Check for Slenderness:

Slenderness ratios:

λ_z= (1000/57) = 17.54

λ_y = (1000/21) = 47.62

E_{0,mean =}1.1031 kN/m²

As timber grade is C24 (i.e., Soft Wood)

E_0,05 = 0.67·E_0,mean = 0.739 kN/m²

[Annex A,EN 338:2003]

λ_rel,z = λ_z/π·(f_c0k/E_0,05)^1/2 = 17.54/π(21.00/0.739)^1/2 = 0.298

λ_rel,y = λ_y/π·(f_c0k/E_0,05)^1/2 = 47.62/π(21.00/0.739)^1/2 = 0.809

Since, λ_rel,yis greater than 0.3, following conditions should be satisfied:

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. The member is subjected to Compression only, so actual bending stress is zero.

S_c0d/(Kcz·F_c0d) + (S_mzd/F_mzd) + Km·(S_myd/F_myd) = 3.46/(1.008·12.92) + 0.0 + 0.0 = 0.268 + 0.0 + 0.0 = 0.266

S_c0d/(Kcy·F_c0d) + Km·(S_mzd/F_mzd) + (S_myd/F_myd) = 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 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 1.0 0 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 FX -50
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            0.327         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                                 |
  |--------------------------------------------------------------------------|