--- title: "Number of bolts for load inclined to grain" subject: "Design of Timber and Masonry Structures" chapter: 2 key: "semester_5/design_of_timber_and_masonry_structures:bolt-load-at-angle-to-grain" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/bolt-load-at-angle-to-grain site: BCT Engineering --- # Number of bolts for load inclined to grain Design of Timber and Masonry Structures · Chapter 2: Joints in Timber Structures - Asked in: year not given, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/bolt-load-at-angle-to-grain - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-2 ## Question A timber tie member is connected to a gusset by 16 mm diameter bolts in single shear. For this member and bolt size, the safe load per bolt is 3.6 kN when the load acts parallel to the grain and 2.1 kN when it acts perpendicular to the grain. The member is loaded by a force of 24 kN acting at 30° to the grain. Using Hankinson's formula, find the safe load per bolt and the number of bolts required. Suggest the spacing of the bolts if bolt spacing along the grain is $7d$ and across the grain is $4d$. ## Short answer Hankinson's formula gives safe load per bolt at 30° = (3.6×2.1)/(3.6 sin²30 +2.1 cos²30) =3.05 kN. Number =24/3.05 ≈7.86, use 8 bolts (2 rows of4). Bolt spacing along grain 115 mm (7d), across 65 mm (4d); minimum member width 200 mm. ## Answer #### 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. ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives the safe load per bolt at angle theta to the grain (P parallel, Q perpendicular)? A. N = (P + Q)/2 B. N = PQ/(P cos^2 theta + Q sin^2 theta) C. N = PQ/(P sin^2 theta + Q cos^2 theta) D. N = P cos theta + Q sin theta Answer: C. Hankinson's formula is N = PQ/(P sin^2 theta + Q cos^2 theta); it gives P at 0 degrees and Q at 90 degrees. (Key point: Hankinson's formula) 2. For P = 3.6 kN, Q = 2.1 kN and theta = 30 degrees, the safe load per bolt is about: A. 3.05 kN B. 2.85 kN C. 3.60 kN D. 2.10 kN Answer: A. N = 7.56/(3.6 x 0.25 + 2.1 x 0.75) = 7.56/2.475 = 3.05 kN, between Q and P. (Key point: Safe load per bolt) 3. For a 24 kN force and 3.05 kN per bolt, how many bolts are provided? A. 7, rounding to the nearest integer B. 8, since 7.86 is rounded up C. 10, to get a factor of safety D. 24, one per kN Answer: B. 24/3.05 = 7.86, which must be rounded up to 8 bolts, arranged as 2 rows of 4. (Key point: Number of bolts) 4. For d = 16 mm, how are the spacing and member width obtained? A. Along grain 4d, across grain 7d, width 150 mm B. Along grain 7d = 112 mm, across grain 4d = 64 mm, width at least 2 x 64 + 65 = 193 mm (take 200 mm) C. Along grain 10d, across grain 5d, width 250 mm D. Along grain 7d, across grain 7d, width 100 mm Answer: B. Spacing is about 115 mm along grain and 65 mm across it, with edge distance 4d = 64 mm each side, so the width is at least 193 mm. 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