D5.E.2 Analysis Methodology

Table 1. EC5 Nomenclature
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 zz-axis.
F_myd	Design bending strength - about yy-axis.
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 z-axis.
I_y	Second moment of area about the weak y-axis.
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 Annex-A 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
SI No.	Property	Symbol	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 Table-1 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 Table-1 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	=	design value of strength property
X_k	=	characteristic value of strength property
γ_m	=	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.Pro 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 EC-5 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 EC-5 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 EC-5 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 EC-5 2004.

Note: In STAAD z-z axis is the strong axis.

(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 EC-5 2004:

S_vd/F_vd ≤ RATIO

Check for Torsional Stresses

Members subjected to torsional stress should satisfy Equation 6.14 of EC-5 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 EC-5 2004:

Note: In STAAD z-z axis is the strong axis.

(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

While evaluating the lateral-torsional stability of a member against the strong axis moment as per Cl.6.3.3.(3) [EN 1995-1-1 (2004)], the applicable bending stress is reduced for the case of axial tension combined with bending. Equation 6.33 [EN 1995-1-1 (2004)] is modified as:

σ_m,d – σ_t,0,d ≤ k_critf_m,d

where

σ_m,d, σ_t,0,d	=	design bending stress and tensile stress, respectively
k_crit	=	a factor which takes into account the reduced bending strength due to lateral buckling
f_m,d	=	the design bending strength

This approach is adopted from Cl.5.4.2 of Manual for the design of timber building structures to Eurocode 5. The Institution of Structural Engineers / TRADA. 2007. London, UK.

Check for Combined Bending and Axial Compression

If members are subjected to bending and axial compression stress, Equations 6.19 and 6.20 of EC-5 2004 should be satisfied:

Note: In STAAD z-z axis is the strong axis.

(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 EC-5 2004.

Note: In STAAD z-z axis is the strong axis.

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 EC-5 2004 should be satisfied.

Note: In STAAD z-z axis is the strong axis.

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 EC-5 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 EC-5 2004:

S_mzd/(Kcrit·F_mzd) ≤ RATIO

Where a combination of moment about the strong z-axis and compressive force exists, the stresses should satisfy Equation 6.35 of EC-5 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 EC-5 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 EC-5 2004:

S_m,crit = 0.78·b²·E_0,05/(h·l_ef)