--- title: "Design of a nailed timber lap joint" subject: "Design of Timber and Masonry Structures" chapter: 2 key: "semester_5/design_of_timber_and_masonry_structures:nailed-lap-joint-design" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/nailed-lap-joint-design site: BCT Engineering --- # Design of a nailed timber lap joint 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/nailed-lap-joint-design - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-2 ## Question A tension member 50 mm × 150 mm carries a working tensile load of 9 kN. It is connected by a single lap joint to a similar member using 4 mm diameter round wire nails driven in single shear. The safe lateral load per nail is 0.42 kN and the permissible tensile stress parallel to grain is 7.5 N/mm². Find the number of nails, show a suitable layout (use spacing $10d$ along grain, $5d$ across grain, end distance $15d$, edge distance $5d$) and check the net tension in the member. ## Answer #### 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. ```text <-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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For d = 4 mm, what end distance is used along the grain? A. 15d = 60 mm B. 10d = 40 mm C. 5d = 20 mm D. 7d = 28 mm Answer: A. End distance is 15d = 60 mm; spacing along grain is 10d = 40 mm and across grain and edge distance is 5d = 20 mm. (Key point: Minimum distances for 4 mm nails) 2. With 4 columns at 40 mm pitch and 60 mm end distances, the minimum lap length is: A. 160 mm B. 200 mm C. 240 mm D. 360 mm Answer: C. Lap = 60 + 3 x 40 + 60 = 240 mm, so 250 mm is provided. (Key point: Lap length) 3. What net area is used for the tension check, and what is the stress? A. 7500 mm^2, giving 1.20 N/mm^2 B. 6300 mm^2 after deducting 6 holes of 4 x 50 mm, giving 1.43 N/mm^2 C. 6900 mm^2 after deducting one hole, giving 1.30 N/mm^2 D. 1200 mm^2, giving 7.5 N/mm^2 Answer: B. A_net = 7500 - 6 x 200 = 6300 mm^2, so stress = 9000/6300 = 1.43 N/mm^2, below 7.5 N/mm^2. 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