Design - Part 1: Beam Design Notes

BEAM DESIGN

• Beam design considerations:
  1. Design for flexure
  2. Design for shear
  3. Design for bearing
  4. Design for deflection

1. Design for Flexure (Clause 3.2.4)

• Requirement: $M* ≤ φ Mn$

  • $M*$: Imposed bending moment as per AS/NZS 1170
  • $φ$: 0.8, strength reduction factor
  • $Mn$: Nominal flexural resistance

• Nominal flexural resistance is calculated as: $Mn = k1 k4 k5 k8 fb Z$

  • $k1$: Load duration factor
  • $k4$: Load sharing factor, if applicable
  • $k5$: Grid system factor, if applicable
  • $k8$: Stability factor
  • $fb$: Characteristic strength in bending
  • $Z$: Section modulus, $bd^2/6$ for a rectangular cross-section
    • $b$: width of the cross section
    • $d$: height of the cross section

• Table 2.7 - Parallel support factor

  • $k<em>4$ or $k</em>6$: load sharing factor

• $K_5$ grid system factor

  • $K<em>5 = 1 + (k</em>4 − 1)(1– \frac{S}{L_B})$
  • $K_5$ but not less than 1.00
  • where
    • $S$ = the value obtained from 2.9.1 that would be applicable if the main beams were fastened together to act as a parallel support system
    • $S$ = the centre-to-centre spacing of the supporting members
    • $L_B$ = the span of the supporting members.

• $k_5$ grid system factor

• $k_8$ stability factor - Dry conditions

• $k_8$ stability factor – Wet conditions

• $L_{ay}$: Is the distance between points of restraint against lateral movement of the compression edge (depends on the structural configuration and on the loading considered)

  • Example: joist with plywood flooring

• Z section modulus, $bd^2/6$ for a rectangular cross-section

  • b width of the cross section
  • d height of the cross section

Design for Flexure – Example 1

• Problem: For the loading shown, what would be adequate cross- sections and grades of sawn timbers?

  • Details:
    • $w = 3.24 kN/m$;
    • $L= 5m$,
    • Medium load duration.
    • Beam laterally restrained at top edge.
    • Dry service conditions

Solution

• Calculations of $M*$
  • $M* = \frac{wL^2}{8} = \frac{3.24 kN/m \times 5^2 m^2}{8}$
  • $M* = 10.13 kN \cdot m$
• $10.13 kN \cdot m ≤ ФMn = Ф k<em>1 k</em>4 k<em>5 k</em>8 f_b Z$
  • $Ф = 0.8$
  • $k_1$ = load duration factor, 0.8 for medium duration
  • $k_4$ = load sharing factor, depends on the number of pieces
  • $k_5$ = grid system factor, 1 assuming not applicable
  • $k_8$ = stability factor, 1 since continuously laterally supported
  • $f_b$ = characteristic strength in bending
  • $Z$ = section modulus, $bd^2/6$ for a rectangular cross-section
  • $15.83 kN \cdot m ≤ k<em>4 f</em>b Z$
• Solution: $15.83 kN \cdot m ≤ k<em>4 f</em>b Z$
  • 1st try:
    • 50x300 MSG8
    • thus $f_b = 14 MPa$
    • and $Z = 0.631 \times 10^6 mm^3$
    • $k<em>4 f</em>b Z = k<em>4 \cdot 14 \cdot 0.631 \times 10^6 = k</em>4 \cdot 8.83 kN \cdot m$
    • Need 2 pieces, fastened together: $2 \times 8.83 kN \cdot m \cdot k_4 = 17.66 kN \cdot m \cdot 1.14 = 20.13 kN \cdot m$
  • 2nd try:
    • 50x250 MSG8
    • $Z = 0.431 \times 10^6 mm^3$ thus
    • $k<em>4 f</em>b Z = 13.75 kN \cdot m$
    • NO GOOD
• Possible solutions for $15.83 kN \cdot m ≤ k<em>4f</em>bZ$
  • Several combinations of dimensions, number of pieces ($k<em>4$), and timber grade ($f</em>b$) are possible to satisfy the flexural design requirement.

2. Design for Shear (Clause 3.2.3)

• Requirement: $V* ≤ φ Vn$
  • $V*$: Imposed shear force as per AS/NZS 1170
  • $φ$: 0.8, strength reduction factor
  • $Vn$: Nominal shear resistance

2.1 Un-notched beams… (Clause 3.2.3.1)

• $Vn = k<em>1 k</em>4 k<em>5 f</em>s A_s$
  • $k_1$: Load duration factor
  • $k_4$: Load sharing factor, if applicable, as for flexure
  • $k_5$: Grid system factor, if applicable, as for flexure
  • $f_s$: Characteristic strength in shear
  • $A_s$: Shear plane area, $2bd/3$ for a rectangular x-section
• For un-notched beams only, one can disregard the loading within a distance “d” of the inside of the supports.
  • $V* = \frac{w (L-2d)}{2}$

2.2 Notched beams (Clause 3.2.6)

• $V* + \frac{1.2 M*}{d<em>n} = k</em>1 k<em>4 k</em>5 k<em>7 f</em>s A_{sn}$

  • $d_n$: net depth of member at notch
  • $k_7$: notch coefficient factor
  • $A<em>{sn}$: shear plane area, $2bd</em>n/3$ for a rectangular x-section

• Table for values of $k_7$

  • Notch slope

    • (a and d'expressed in millimetres)

    • a≥0.1d

      • .5 for notch slope 2

      • .333 for notch slope 3

      • .25 for notch slope 4

    • a<0.1d

      • 1 for parallel to grain

• IMPORTANT It must be emphasized that it is not good practice to notch a beam on the tension side. It will lead to splits and potentially, catastrophic failure of the member.

Design for Shear – Example 2

• Problem: For the beam shown, verify if it is adequate to resist the applied shear.

  • Details:
    • $w = 3.24 kN/m$,
    • Medium load duration
    • Beam laterally restrained at top edge
    • Dry service conditions

Solution

• Calculations of $V*$ (the beam is un-notched)
  • $V* = \frac{w (L-2d)}{2} = \frac{3.24 kN/m \times (5 – 2 \cdot 0.290) m}{2}$
  • $V* = 7.16 kN$
• $7.16 kN ≤ φV<em>n = φ k</em>1 k<em>4 k</em>5 f<em>s A</em>s$
  • $φ = 0.8$
  • $k_1$ = load duration factor, 0.8 for medium load duration
  • $k_4$ = load sharing factor, 1.14 for 2 pieces
  • $k_5$ = grid system factor, not applicable thus 1
  • $f_s$ = characteristic strength in shear, 3.8 MPa
  • $A_s$ = shear area, $2bd/3$, equal to $2 \cdot 2 \cdot 45 \cdot 290/3 = 17400 mm^2$
  • φV_n = 48.2 kN > 7.16, thus Adequate

3. Design for Bearing

• Once a beam has been sized for flexure and shear, its adequacy to resist the concentrated forces at the bearing points is verified.

• If it is not adequate, one needs to change the bearing surface dimensions, NOT the beam size.

• (Clause 3.2.9)

  • $Nb* ≤ φ Nnb$
    • $Nb*$: design bearing load as per AS/NZS 1170
    • $φ$: 0.8, strength reduction factor
    • $Nnb$: nominal bearing resistance

3.1 Load applied perpendicular-to-grain

• $N<em>{nbp} = k</em>1 k<em>3 f</em>p A_p$
  • $k_1$: Load duration factor
  • $k_3$: bearing area factor
  • $f_p$: characteristic bearing strength perpendicular-to-grain
  • $A_p$: bearing area for loading perpendicular-to-grain
• $k_3$ bearing area factor is applicable only if away from the end by at least 75 mm

Design for Bearing – Example 3

• Problem:
  • $w = 3.24 kN/m$,
  • Medium Duration (Includes permanent and live loads)
  • Dry service conditions
• Determine the length of bearing required.

Solution

• Calculations of $N_{nbp}*$ (the beam reactions)
  • $N_{nbp}* = \frac{w \cdot L}{2} = \frac{3.24 kN/m \times 5 m}{2}$
  • $N_{nbp}* = 8.1 kN$
• $8.1 kN ≤ φ N<em>{nbp} = φ k</em>1 k<em>3 f</em>p A_p$
  • $φ = 0.8$
  • $k_1$ = load duration factor, 0.8 for normal duration
  • $k_3$ = bearing area factor, 1, as near the end of the member
  • $f_p$ = characteristic strength in compression, 8.9 MPa
  • $A<em>p$ = bearing area, equal to $2 \cdot 45 \cdot l</em>b = 90l_b mm^2$
  • $φ N<em>{nbp} = 0.51l</em>b kN/mm ≥ 8.1 kN$
• Thus $l_b ≥ \frac{8.1}{0.51} mm = 15.9 mm$
• In practice, a bearing length of 25 mm would be specified as minimum

3.2 Load applied parallel-to-grain

• $N<em>{nbl} = k</em>1 f<em>c A</em>l$
  • $f_c$: characteristic bearing strength parallel-to-grain
  • $A_l$: bearing area for loading parallel-to-grain

3.3 Load applied at any angle θ

• $N<em>{nbθ} = \frac{N</em>{nbl} \cdot N<em>{nbp}}{N</em>{nbl} sin^2 θ + N_{nbp} cos^2 θ}$

• The following 3 verifications would be necessary

  • $N<em>{bl}* ≤ φ N</em>{nbl}$
  • $N<em>{bp}* ≤ φ N</em>{nbp}$
  • $N<em>{bθ} * ≤ φ N</em>{nbθ}$