D5.E.2 Analysis Methodology
Symbol  Description 

S_{t0d}  Design tensile stress parallel (at zero degree) to grain alignment. 
S_{t90d}  Design tensile stress perpendicular (at 90 degrees) to grain alignment. 
S_{c0d}  Design compressive stress parallel to grain alignment. 
S_{c90d}  Design compressive stress perpendicular to grain alignment. 
S_{mzd}  Design bending stress about zz axis. 
S_{myd}  Design bending stress about yy axis. 
S_{vd}  Design shear stress. 
S_{tor_d}  Design torsional stress. 
F_{t0d}  Design tensile strength  parallel to the grain alignment. 
F_{t90d}  Design tensile strength  perpendicular to the grain alignment. 
F_{c0d}  Design compressive strength  parallel to the grain alignment. 
F_{c90d}  Design compressive strength  perpendicular to the grain alignment. 
F_{mzd}  Design bending strength  about zzaxis. 
F_{myd}  Design bending strength  about yyaxis. 
F_{vd}  Design shear strength about yy axis. 
RATIO 
Permissible ratio of stresses as input using the RATIO parameter. The default value is 1. 
l_{z} ,l_{rel,z}  Slenderness ratios corresponding to bending about zz axis. 
l_{y},l_{rel,y}  Slenderness ratios corresponding to bending about yy axis. 
E_{0,05}  Fifth percentile value of modulus of elasticity parallel to grain. 
G_{0,05}  Fifth percentile value of shear modulus parallel to grain. 
I_{z}  Second moment of area about the strong zaxis. 
I_{y}  Second moment of area about the weak yaxis. 
I_{tor}  Torsional moment of inertia. 
f_{mk}  Characteristic bending strength. 
b, h  Width and depth of beam. 
Equations for Characteristic Values of Timber Species as per AnnexA of EN 338:2003
The following equations were used to determine the characteristic values:
For a particular Timber Strength Class (TSC), the following characteristic strength values are required to compute the other related characteristic values.

Bending Strength – f_{m,k}

Mean Modulus of Elasticity in bending – E_{0, mean}

Density  ρ_{k}
SI No.  Property  Symbol  Wood Type  

Softwood (C)  Hardwood (D)  
1.  Tensile Strength parallel to grain  f_{t,0,k}  0.6 * f_{m,k}  
2.  Tensile Strength perpendicular to grain  f_{t,90,k}  Minimum of {0.6 and (0.0015*r_{k})}  
3.  Compressive Strength parallel to grain  f_{c,0,k}  5 * (f_{m,k })^{ 0.45}  
4.  Compressive Strength perpendicular to grain  f_{c,90,k}  0.007*r_{k}  0.0015*r_{k} 
5.  Shear Strength  f_{v,k}  Minimum of {3.8 and (0.2*f_{m,k} ^{ 0.8})}  
6.  Modulus of Elasticity parallel to grain  E_{0,05}  0.67* E_{0,mean}  0.84* E_{0,mean} 
7.  Mean Modulus of Elasticity perpendicular to grain  E_{90,mean}  E_{0,mean} /30  E_{0,mean} /15 
8.  Mean Shear Modulus  G_{mean}  E_{0,mean} /16  
9.  Shear Modulus  G_{0,05}  E_{0,05} /16 
The values of the characteristic strengths computed using the above equations, may differ with the tabulated values in Table1 of EN 338:2003. However, in all such cases, the values obtained from the provided equations are treated as actual and is used by the program, as the values of Table1 are based on these equations.
Design values of Characteristic Strength
As per clause 2.4.1, Design values of a strength property shall be calculated as:
X_{d} = K mod·(X_{k}/γ_{m})
Where:
 X_{d }is design value of strength property
 X_{k} characteristic value of strength property
 γ_{m} is partial factor for material properties.
The member resistance in timber structure is calculated in STAAD according to the procedures outlined in EC5. This depends on several factors such as cross sectional properties, different load and material factors, timber strength class, load duration class, service class and so on. The methodology adopted in STAAD for calculating the member resistance is explained here.
Check for Tension stresses
If the direction of applied axial tension is parallel to the direction of timber grain alignment, the following formula should be checked per Equation 6.1 of EC5 2004:
S_{t0d}/F_{t0d} ≤ RATIO
If the direction of applied axial tension is perpendicular to the direction of timber grain alignment, the following formula should be checked:
S_{t90d}/F_{t90d} ≤ RATIO
Check for Compression stresses
If the direction of applied axial compression is parallel to the direction of timber grain alignment, the following formula should be checked per Equation 6.2 of EC5 2004:
S_{c0d}/F_{c0d} ≤ RATIO
If the direction of applied axial compression is perpendicular to the direction of timber grain alignment, the following formula should be checked per Equation 6.3 of EC5 2004:
S_{t0d}/(F_{t0d}·Kc90) ≤ RATIO
Check for Bending stresses
If members are under bending stresses, the following conditions should be satisfied per Equations 6.11 and 6.12 of EC5 2004.
(S_{mzd}/F_{mzd}) + Km·(S_{myd}/F_{myd}) ≤ RATIO 
Km·(S_{mzd}/F_{mzd}) + (S_{myd}/F_{myd}) ≤ RATIO 
Check for Shear stresses
Horizontal stresses are calculated and checked against allowable values per Equation 6.13 of EC5 2004:
S_{vd}/F_{vd} ≤ RATIO
Check for Torsional stresses
Members subjected to torsional stress should satisfy Equation 6.14 of EC5 2004:
S_{tor_d}/(Kshape·F_{tor_d}) ≤ RATIO
Check for combined Bending and Axial tension
Members subjected to combined action of bending and axial tension stress should satisfy Equations 6.17 and 6.18 of EC5 2004:
(S_{t0d}/F_{t0d}) + (S_{mzd}/F_{mzd}) + Km·(S_{myd}/F_{myd}) ≤ RATIO 
(S_{t0d}/F_{t0d}) + Km·(S_{mzd}/F_{mzd}) + (S_{myd}/F_{myd}) ≤ RATIO 
Check for combined Bending and axial Compression
If members are subjected to bending and axial compression stress, Equations 6.19 and 6.20 of EC5 2004 should be satisfied:
(S_{c0d}/F_{c0d})^{2} + (S_{mzd}/F_{mzd}) + Km·(S_{myd}/F_{myd}) ≤ RATIO 
(S_{c0d}/F_{c0d})^{2} + Km·(S_{mzd}/F_{mzd}) + (S_{myd}/F_{myd}) ≤ RATIO 
Stability check

Column Stability check
The relative slenderness ratios should be calculated per Equations 6.21 and 6.22 of EC5 2004.
λ_{rel,z} = λ_{z}/π·(S_{c0k}/E_{0,05})^{1/2} λ_{rel,y} = λ_{y}/π·(S_{c0k}/E_{0,05})^{1/2} If both λ_{rel,z} and λ_{rel,y} are less than or equal to 0.3 the following conditions should be satisfied:
(S_{c0d}/F_{c0d})^{2} + (S_{mzd}/F_{mzd}) + Km·(S_{myd}/F_{myd}) ≤ RATIO (S_{c0d}/F_{c0d})^{2} + Km·(S_{mzd}/F_{mzd}) + (S_{myd}/F_{myd}) ≤ RATIO In other cases, the conditions in Equations 6.23 and 6.24 of EC5 2004 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 (Equations 6.25 through 6.28 of EC5 2004):
Kcz = 1/{K_{z} + [(K_{z})^{2}  (λ_{rel,z})^{2}]^{1/2}} Kcy = 1/{K_{y} + [(K_{zy})^{2}  (λ_{rel,y})^{2}]^{1/2}} Kz = 0.5·[1 + β_{c}·(λ_{rel,z}  0.3) + (λ_{rel,z})^{2}] Ky = 0.5·[1 + β_{c}·(λ_{rel,y}  0.3) + (λ_{rel,y})^{2}] The value of β_{c} incorporated in the software is the one for solid timber (i.e., 0.2).

Beam Stability check
If members are subjected to only a moment about the strong axis z, the stresses should satisfy Equation 6.33 of EC5 2004:
S_{mzd}/(Kcrit·F_{mzd}) ≤ RATIO
Where a combination of moment about the strong zaxis and compressive force exists, the stresses should satisfy Equation 6.35 of EC5 2004 (ref. to Equations 6.32 and 6.34 of the same):
[S_{mzd}/(Kcrit·F_{mzd})]^{2} + S_{c0d}/(Kcz·F_{c0d}) ≤ RATIO
Where:
 Kcrit = 1.0 when λ_{rel,m} ≤ 0.75
 Kcrit = 1.56  0.75·λ_{rel,m} when 0.75 < λ_{rel,m} ≤ 1.4
 Kcrit = 1/( λ_{rel,m})^{2} when 1.4 < λ_{rel,m}
 λ_{rel,m} = (f_{mk}/S_{m,crit})^{1/2}
For hardwood, use Equation 6.30 of EC5 2004:
S_{m,crit} = π·(E_{0,05}·I_{y}·G_{0,05}·I_{tor})^{1/2}/(l_{ef}·W_{z})
For softwood, use Equation 6.31 of EC5 2004:
S_{m,crit} = 0.78·b^{2}·E_{0,05}/(h·l_{ef})