Joints in Timber Structures

Mechanical fasteners (nails, screws, bolts) transfer load between timber members mainly by bearing of the fastener on the wood together with bending of the fastener.

Behaviour of bolted joints

• A bolt in a loaded joint bears against the wood. Early load–slip behaviour is nearly linear, then wood crushes under the bolt and the bolt bends, giving a ductile yielding response.
• A bolt placed in a hole 1–1.5 mm larger than its diameter first slips before taking load, so tight holes and washers (about $3d$ in size) under the head and nut are needed.
• Capacity depends on the bolt diameter $d$, member thickness $t$ (ratio $t/d$), the angle between load and grain (Hankinson's formula) and whether the joint is in single or double shear.

Behaviour of nailed joints

• Nails carry lateral load by bearing and bending; under axial load they resist withdrawal by friction.
• Joints with many small nails are ductile. Nails should not be used to carry permanent withdrawal loads.
• Hard timber is pre-drilled to avoid splitting.

Failure modes

1. Wood crushing under the fastener (bearing failure).
2. Bending / yielding of the fastener.
3. Splitting along the grain when end distance or spacing is too small.
4. Shear-out of a block (plug) at the loaded end.
5. Tension failure of the net section through the holes.
6. Withdrawal of nails or screws.

Detailing rules (typical code values; check the relevant code)

Other rules: stagger the fasteners in adjacent rows, avoid placing all bolts on one grain line, provide at least two bolts per connection, and use nail penetration in the second member of not less than $10d$ (clinched where the nail passes through).

Given data

Load $P = 9$ kN, $d = 4$ mm, safe load per nail $= 0.42$ kN, member $50 \times 150$ mm, $\sigma_{t} = 7.5$ N/mm².

Number of nails

$$n = \frac{9}{0.42} = 21.4 \;\Rightarrow\; \text{provide } 24 \text{ nails (6 rows} \times 4 \text{ columns)}$$

24 nails give a symmetrical arrangement.

Layout

Minimum distances: along grain $10d = 40$ mm; across grain $5d = 20$ mm; end $15d = 60$ mm; edge $5d = 20$ mm.

• Across the 150 mm width: 6 rows with edge distance 20 mm each side, so spacing $= (150 - 40)/5 = 22$ mm $> 20$ mm. OK.
• Along the grain: 4 columns at 40 mm pitch, so lap length $\geq 60 + 3 \times 40 + 60 = 240$ mm. Provide a lap of 250 mm.

  <-60->|<-40->|<-40->|<-40->|<-60->
  +--------------------------------+  ^
  |   o      o      o      o       | 20
  |   o      o      o      o       | 22
  |   o      o      o      o       | 22     50x150
  |   o      o      o      o       | 22     member
  |   o      o      o      o       | 22
  |   o      o      o      o       | 22
  +--------------------------------+ 20
      Lap length = 250 mm

Check of net tension

Net section is reduced by one row of nail holes (6 holes). Conservatively the holes are deducted fully:

$$\begin{aligned} A_{net} &= 50 \times 150 - 6 \times (4 \times 50) = 7500 - 1200 = 6300\ \text{mm}^2 \\ \sigma_t &= \frac{9000}{6300} = 1.43\ \text{N/mm}^2 < 7.5\ \text{N/mm}^2 \end{aligned}$$

The member is safe in tension. The joint is also checked so that the nail length gives penetration of at least $10d = 40$ mm into the second member.

Answer: 24 nails of 4 mm diameter (6 rows × 4 columns), lap length 250 mm; net tension 1.43 N/mm² < 7.5 N/mm², safe.

Hankinson's formula

When the load makes an angle $\theta$ with the grain, the safe load per bolt is

$$N = \frac{P\,Q}{P\sin^2\theta + Q\cos^2\theta}$$

where $P$ is the safe load parallel to grain and $Q$ is the safe load perpendicular to grain.

Calculation

$P = 3.6$ kN, $Q = 2.1$ kN, $\theta = 30^\circ$: $\sin^2 30^\circ = 0.25$, $\cos^2 30^\circ = 0.75$.

$$\begin{aligned} N &= \frac{3.6 \times 2.1}{3.6 \times 0.25 + 2.1 \times 0.75} \\ &= \frac{7.56}{0.90 + 1.575} = \frac{7.56}{2.475} = 3.05\ \text{kN} \end{aligned}$$

Number of bolts

$$n = \frac{24}{3.05} = 7.86 \;\Rightarrow\; \text{provide } 8 \text{ bolts}$$

Arrange as 2 rows of 4 bolts (symmetric about the axis of the member).

Spacing for $d = 16$ mm

• Along the grain: $7d = 112$ mm, provide 115 mm. Between 4 bolts in a row the length is $3 \times 115 = 345$ mm.
• Across the grain: $4d = 64$ mm, provide 65 mm between the rows.
• End distance: $7d = 112$ mm, provide 115 mm. Edge distance $4d = 64$ mm, so the member must be at least $2 \times 64 + 65 = 193$ mm wide, take 200 mm.

   115  115  115  115          bolts in two rows
 +--------------------------------------
 |  o    o    o    o     <- 64 edge
 |                        65 between rows
 |  o    o    o    o     <- 64 edge
 +--------------------------------------

Answer: Safe load per bolt = 3.05 kN; number of bolts = 8 (two rows of four), member width at least 200 mm.