Module 3: Timber Beams - Examples & Applications

Bending Stress (Flexure) and Beam Stability

Basic: Evaluating a Simply Supported Beam

A simply supported rectangular timber beam (width b=100 mmb = 100 \text{ mm}, depth d=200 mmd = 200 \text{ mm}) spans 3 meters3 \text{ meters} and carries a uniform load.
Structural analysis reveals the following maximum internal forces:
  • Maximum bending moment (MM) = 5kNm5 \text{kN}\cdot\text{m}
  • Maximum shear force (VV) = 10kN10 \text{kN}

Given Adjusted Allowable Stresses:

  • Bending (FbF_b'): 12MPa12 \text{MPa}
  • Shear (FvF_v'): 1.0MPa1.0 \text{MPa}

Verify if the beam is safe for both bending and horizontal shear.

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Intermediate: Beam Stability Factor ($C_L$)

Calculate the Beam Stability Factor (CLC_L) for an unbraced roof beam spanning 5.0 m5.0 \text{ m}. The dimensions are 100 mm×300 mm100 \text{ mm} \times 300 \text{ mm}.

Given Parameters:

  • Reference bending design value (FbF_b^*): 14.0 MPa14.0 \text{ MPa}
  • Minimum Modulus of Elasticity (EminE_{min}'): 6,000 MPa6,000 \text{ MPa}
  • Effective unbraced length (lel_e): 5,000 mm5,000 \text{ mm}

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Advanced: Beam with an Overhang

A 150 mm×300 mm150 \text{ mm} \times 300 \text{ mm} timber beam has a main span of 4.0 m4.0 \text{ m} and an overhang of 1.5 m1.5 \text{ m}. It carries a uniform load of 5 kN/m5 \text{ kN/m} over its entire length.
Given Parameters:
  • Main span (L1L_1): 4.0 m4.0 \text{ m}
  • Overhang (L2L_2): 1.5 m1.5 \text{ m}
  • Load (ww): 5 kN/m5 \text{ kN/m}
  • Adjusted Allowable Bending Stress (FbF_b'): 10 MPa10 \text{ MPa}

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Deflection Criteria

Basic: Calculating Total Deflection with Creep

A uniformly loaded simply supported timber floor joist spans 4.0 m4.0 \text{ m}. The immediate (instantaneous) deflections due to the applied loads have been calculated as follows:
  • Immediate Dead Load Deflection (ΔDL\Delta_{DL}): 8 mm8 \text{ mm}
  • Immediate Live Load Deflection (ΔLL\Delta_{LL}): 10 mm10 \text{ mm}

The joist must support unseasoned lumber in wet conditions, requiring a long-term creep factor of 2.02.0. The code requires the total long-term deflection not to exceed L/240L/240.

Determine if the joist is safe for serviceability.

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Intermediate: Deflection with Partial Load

A 100 mm×200 mm100 \text{ mm} \times 200 \text{ mm} timber joist spans 3.0 m3.0 \text{ m}. It carries a live load of 2 kN/m2 \text{ kN/m} applied only to the first half of the span (1.5 m1.5 \text{ m}). E=10,000 MPaE = 10,000 \text{ MPa}. Determine the maximum deflection.

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Advanced: Cambering a Timber Beam

A long-span timber beam (6.0 m6.0 \text{ m}) is expected to deflect 15 mm15 \text{ mm} under dead load and 10 mm10 \text{ mm} under live load. The architect wants the floor to be perfectly flat under dead load. Determine the required initial camber.

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Special Design Considerations

Intermediate: Calculating Required Bearing Length

A 150 mm×300 mm150 \text{ mm} \times 300 \text{ mm} timber beam is simply supported on a concrete wall. The maximum reaction force at the support due to the factored loads is R=45 kNR = 45 \text{ kN}. The adjusted allowable compression stress perpendicular to the grain is Fc=2.5 MPaF_{c\perp}' = 2.5 \text{ MPa}.

Determine the minimum required length of bearing (lbl_b) on the concrete wall to prevent the timber from crushing.

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Advanced: Capacity of a Notched Beam

A 100 mm×250 mm100 \text{ mm} \times 250 \text{ mm} timber joist has an allowable shear stress of Fv=1.2 MPaF_v' = 1.2 \text{ MPa}. However, it is notched at the support on its tension (bottom) face to fit over a ledger board. The notch depth is 50 mm50 \text{ mm}, leaving a reduced net depth (dnd_n) of 200 mm200 \text{ mm} at the support.
Calculate the maximum allowable shear force (VV) the notched end can support.

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Basic: Bearing Area for a Point Load

A 20 kN20 \text{ kN} point load is applied to the top edge of a timber beam via a steel bearing plate. The beam is 100 mm100 \text{ mm} wide. Fc=3.0 MPaF_{c\perp}' = 3.0 \text{ MPa}. Find the minimum required length of the steel plate along the beam axis.

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