Structural calculations

Simple calculation

Not in accordance with the latest building regulations. The calculation is intended to show which factors influence the result.

Step 1

You can get a feel for strength of materials by calculating a beam projecting from a wall at A (see figure). To prevent the beam from failing under load, we check the shear and bending stresses caused by the load (see below) at the most heavily loaded cross-section at A, where the bending moment is greatest. This moment equals the force(s) multiplied by the distance perpendicular to the direction of the force.

Example figure for the calculation. 

M = F * a

• M = bending moment at the fixed support
• F = force (in newtons)
• a = shortest distance from the line of action of the force to the fixed support at A

A fixed support is a connection at the wall that can resist horizontal and vertical reactions and a bending moment without rotation. If a 2 m long fixed-ended (cantilever) beam is loaded at its end by a downward force P = 2.0 kN, the vertical reaction V_AP at A (see figure) is, by equilibrium, 2.0 kN upwards.

The self-weight (a line load q in kN/m) of a beam with a chosen depth h = 225 mm and a chosen width b = 70 mm is

q = 0.225 m * 0.07 m * 5 kN/m³ (unit weight)
q = 0.08 kN/m

From the equilibrium of q it follows that
V_Aq = ql = 0.08 * 2 = 0.16 kN

The total vertical reaction at A is then
V_A = V_AP + V_Aq = 2.16 kN

As a result of the vertical reaction V_A a vertical shear stress occurs in section A of
τ_A = V_A/bh
τ_A = 2,160 N/(225 mm * 70 mm)
τ_A = 0.14 N/mm² < 0.8 N/mm² = 2,0 (shear strength of the common timber grade C18) * 0,7 (factor for outdoor climate and load duration) / {1,35 (load factor) *1,3 (material factor)}

Because the shear stress is lower than the adjusted shear strength, the structure is adequate in shear.

Step 2

The bending moment at the most heavily loaded section A due to P is:
M_AP = 2.0 kN * 2 m = 4.0 kNm

The moment due to self-weight gives
M_Aq = ql (= force due to q) * 1/2 (= distance to the centroid)
M_Aq = ql²/ 2 = 0,08 * 2²/ 2 = 0.16 kNm
M_A = M_AP + M_Aq = 4.16 kNm

The section modulus W³⁾ determines the stresses σ in a cross-section due to a bending moment M according to
σ = M/W

If the beam has the same rectangular cross-section over its whole length, this rectangular beam is prismatic and
W = bh²/ 6 = 70 * 225²/ 6 = 590,625 mm³

At the bottom and top edges of the beam (see figure) there is then a horizontal bending stress
σ_A = 4.16 kNm/590,625 mm³ = 7.0 N/mm²
7.2 N/mm² = 18 (bending strength C18) * 0.7 (factor for outdoor climate and load duration)/{1.3 (material factor) * 1.35 (load factor)}

Because the bending stress is lower than the adjusted bending strength, the structure is adequate in bending.

Step 3

We also check the deflection of the beam.
For a cantilever beam, the immediate elastic deformation
u_BP;elastisch at B: u_BP;elastisch = Pl³/ {3 E¹⁾ I²⁾} =
2000 N * 2000³ mm³/ { 3 * 9000 N/mm² * (1/12) * 70 mm * 225³ mm³ } = 8.9 mm

For a cantilever beam, q causes the immediate elastic deformation
u_BP;elastisch = ql⁴/ { 8 EI } =
0.08 N/mm * 2000⁴ mm⁴/ { 8 * 9000 N/mm² * (1/12) * 70 mm * 225³ mm³ } = 0.3 mm
u_B;elastic = 8.9 + 0.3 = 9.2 mm

The total deflection of the beam equals the immediate elastic deflection u_B;elastic plus the time-dependent deflection u_B;creep

u_B;creep = elastic deformation due to the permanent load + 0.6 (creep factor) * 0.4 (if the variable load were present 40% of the time) * the elastic deformation due to the variable load.
u_B;creep = 0.3 + 0.6 * 0.4 * 8.9 = 2.4 mm

u_B;total = 9.2 + 2.4 = 11.6 mm

u_B;inst = the immediate deflection at B due to the permanent load = u_Bq;elastic = 0.3 mm

u_B;add = u_B;total – u_B;inst

Requirement A) u_B;add = 11.6 – 0.3 = 11.3 mm < 12 mm = 0.006 * 2,000 mm = 0.006L = the additional deflection limit for a cantilever, i.e. fixed-ended, beam

Requirement B) u_B;total = 11.6 mm < 16 mm = 0.008 * 2000 = 0.008L

Because here u_B;total = 11.6 mm < 0.006 * 2,000 mm, requirement A) is also met at the same time, so u_B;add did not need to be calculated here.

Because the deformations are smaller than the maximum permissible deformation, the structure is adequate in deformation.

Conclusion: The material strength is used efficiently, which makes the structure economical.

¹⁾
E = modulus of elasticity = stress / [strain = extension per unit length]. For timber grade C18, E = 9,000 N/mm²

²⁾
I = second moment of area = {∫∫z² dy dz with z = distance to the centre of the beam, integration limits h/2 and -h/2 for z and the side edges for y } = bh³/ 12 for a rectangular cross-section, because then ∫dy = b

³⁾
W = I / (h/2). For a rectangular cross-section this gives
W = bh³/ 12/ (h/ 2) = bh²/ 6.