Number of bolts for load inclined to grain

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.