Example calculations for a timber beam ceiling

Load combinations for verifications of load-bearing capacity

LK 2 is clearly decisive in this case, and this is used for further consideration.

3. Characteristic strength and rigidity properties C 24

Comparison with the results from dimensioning reference table 6.2.2

For dimensioning criterion B:       erf b/h = 12/24 cm:

Dimensioning values for stress resistance

4. Stresses - Internal forces and moments and reactions at support

Internal forces and moments per beam

(e = 62.5 cm)

Characteristic reactions at support for the decisive LK 2:

5. Preliminary estimate

Required section modulus:

W_y,req = M_d / f_m,d = 10.38 · 10³/ 14.8     W_y,req = 701 cm³

6. Verifications for the limit of load-bearing capacity state

Dimensioning value for bending stress:

σ _m,y,d = M_d / W_y = 10.38 / 768 · 10³

σ _m,y,d = 13.5 N/mm²

Verification:

Results from dimensioning reference table 6.1 (Cross-sectional values):

für Md = 10.38 kNm       erf b/h = 8/24 cm

Dimensioning value for shearing stress:

τ _d = 1.5 · V_d / A = 1.5 · 9.23 / 192

τ _d = 0.72 N/mm²

Verification:

7. Verifications for the limit of serviceability state

According to DIN 1052, Section 9.2, the following three cases have to be examined each one is only shown for variable action:

a) Limitation of deflection as a result of variable loads (initial deflection without creep influences):

w_Q,inst = w_Q,1,inst + χ_0,2 · w_Q,2,inst ≤ l/300

b) Limitation of final deflection with creep influences as a result of all loads without consideration of initial deflection:

w_fin – w_G,inst = w_G,inst (1 + k_def) + w_Q,1,inst (1 + y_2,1 · k_def) – w_G,inst ≤ l/200

c) Limitation of deflection in the quasi-permanent dimensioning situation for guaranteeing general usability and appearance:

w_fin – w₀ = w_G,inst (1 + k_def) + w_Q,1,inst · y_2,1 (1 + k_def) – w₀ ≤ l/200

Calculation of deflection

E_0,mean · I_y = 1.100 · 9.216 = 10.14 · 10⁶ kNcm²

w_G,fin = w_G,inst (1 + k_def) = 0.58 · (1 + 0.6) = 0.928 cm

Proof of deflection

Case a) w_Q,inst = 0.92 cm = l/489 < l/300

Case b) wfin – wG,inst = 0.58 (1 + 0.6) + 0.92 (1 + 0.3 · 0.6) – 0.58 = 1.43 cm = l/313 < l/200

Case c) w_fin – w₀ = 0.58 (1 + 0.6) + 0.92 · 0.3 (1 + 0.6) - 0 = 1.37 cm = l/329 < l/200

Simplified vibration verification

(Limitation of deflection as a result of quasi-permanent action to w = 6 mm in accordance with DIN 1052, Section 9.3):

w = w_G,inst + c₂ · w_Q,inst = 0.58 + 0.3 · 0.92 = 0.86 cm

w = 8:6 mm > 6 mm (verification requirement not met)

Option 1:

Increase the moment of inertia / the beam width by 8.6/6 • 100 = 43%.

Option 2

It is recommended that exact vibration verifications be calculated, which should provide more cost-effective results. Suitable verification methods can be found, for example, in "Erläuterungen zu DIN 1052:2004-08" [1] ("Commentary on DIN 1052:2004-08").

A_g,k = 3.94 kN/m

A_q,k = 6.30 kN/m

A_g,k = B_g,k = 1.75 · 4.50 / 2

A_q,k = B_q,k = 2.80 · 4.50 / 2

End support A and B:

M_d = 10.38 kNm

V_d = 9.23 kN

Reference moment for LK 2:

M_d = Σq_d · l² / 8 = 6.56 · 4.50² / 8 · 0.625

Reference lateral force for LK 2:

V_d = Σq_d · l / 2 = 6.56 · 4.50 / 2 · 0.625

f_m,d = 14.8 N/mm²

f_v,d = 1.23 N/mm²

k_mod = 0.80

Ψ_M = 1.3

f_m,d = 0.8 · 24.0/ 1.3

f_v,d = 0.8 · 2.0 / 1.3

Modification factor for solid timber

Partial safety factor for wood

Dimensioning value for flexural strength

Dimensioning value for shear strength

f_m,k = 24.0 N/mm²

f_v,k = 2.0 N/mm²

E_0,mean = 11.000 N/mm²

Characteristic bending stress

Characteristic shearing stress

Modulus of elasticity

Results from dimensioning reference table – Ceiling beams, simply supported beams, e = 62.5 cm, C24

Result B (with simplified vibration verification)

KVH^® C24: 12/24 cm

Alternative: Duobalken^®,Triobalken^® C24: 14/22 cm

Result A (without vibration
verification)

KVH^® C24: 8/24 cm

For: l = 4.50 m

g_k = 1.75 kN/m²

q_k,N = 2.80 kN/m²