---
title: "Riveted lap joint strength and efficiency"
subject: "Workshop Technology"
chapter: 10
key: "semester_2/workshop_technology:riveted-joint-strength-efficiency"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/riveted-joint-strength-efficiency
site: BCT Engineering
---

# Riveted lap joint strength and efficiency

Workshop Technology · Chapter 10: Metal Joining Process

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/riveted-joint-strength-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-10

## Question

Name the types of riveted joints and the possible modes of failure of a riveted lap joint. A single-riveted lap joint connects two plates 10 mm thick with 18 mm diameter rivets at a pitch of 50 mm. The permissible stresses are: tension 100 MPa, shear 60 MPa and crushing 150 MPa. Calculate the strength of the joint per pitch length and its efficiency.

## Short answer

Per pitch: tearing strength = 32 000 N, shear = 15 268 N, crushing = 27 000 N. Joint strength is governed by shear at 15.27 kN. Solid plate strength = 50 000 N. Efficiency = 15 268/50 000 = 30.5 %.

## Answer

#### Types of riveted joints
- Lap joint: plates overlap and are riveted (single, double or triple riveted; chain or zigzag arrangement).
- Butt joint: plates are placed edge to edge and covered by one or two cover plates (single or double strap).

#### Modes of failure
1. Tearing of the plate across the rivet hole line.
2. Shearing of the rivet (single shear in lap joint).
3. Crushing of the rivet or plate where they bear on one another.
4. Tearing of the plate at the edge (avoided by margin of 1.5d).
5. Shearing of the plate in front of the rivet (avoided by margin).

#### Calculation (per pitch length)
Given: t = 10 mm, d = 18 mm, p = 50 mm, $\sigma_t$ = 100 MPa, $\tau$ = 60 MPa, $\sigma_c$ = 150 MPa; one rivet per pitch.

Tearing of plate:

$$
P_t = (p - d)\,t\,\sigma_t = (50-18)\times 10\times 100 = 32{,}000\ \text{N}
$$

Shearing of rivet (single shear):

$$
P_s = \frac{\pi}{4}d^2\tau = \frac{\pi}{4}(18)^2\times 60 = 15{,}268\ \text{N}
$$

Crushing:

$$
P_c = d\,t\,\sigma_c = 18\times 10\times 150 = 27{,}000\ \text{N}
$$

Strength of the joint is the least of the three: $P = 15{,}268$ N (shear failure).

Strength of the solid plate:

$$
P_{solid} = p\,t\,\sigma_t = 50\times 10\times 100 = 50{,}000\ \text{N}
$$

$$
\eta = \frac{P}{P_{solid}} = \frac{15{,}268}{50{,}000} = 0.305
$$

| Mode | Load (N) |
|---|---|
| Tearing | 32,000 |
| Shearing | 15,268 |
| Crushing | 27,000 |

**Answer: joint strength = 15.27 kN per pitch (rivet shearing governs); efficiency = 30.5%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is NOT a mode of failure of a riveted lap joint?
   A. Tearing of plate across the rivet hole
   B. Shearing of rivet
   C. Crushing of rivet or plate
   D. Buckling of the rivet head under axial load
   Answer: D. The listed modes are tearing, shearing, crushing, edge tearing and shearing of plate ahead of the rivet. (Key point: Modes of failure)

2. What is the tearing resistance of the plate per pitch?
   A. p t σt = 50,000 N
   B. d t σt = 18,000 N
   C. (p - d) t σt = 32,000 N
   D. (p + d) t σt = 68,000 N
   Answer: C. The plate section at the hole line is (p - d) t, so P = 32 x 10 x 100 = 32,000 N. (Key point: Tearing strength)

3. A lap-joint rivet is in single shear. What is its shear strength here?
   A. 15,268 N
   B. 30,536 N
   C. 27,000 N
   D. 10,800 N
   Answer: A. P = (π/4)(18)^2 x 60 = 15,268 N. Double shear (30,536 N) does not apply to a lap joint. (Key point: Shear strength of rivet)

4. How is the joint strength decided?
   A. Average of the three values
   B. The highest of the three values
   C. The least of the tearing, shearing and crushing values
   D. Sum of the three values
   Answer: C. The joint fails at the weakest mode, here shearing at 15,268 N. (Key point: Joint strength rule)

5. What is the efficiency of the joint?
   A. 54%
   B. 30.5%
   C. 64%
   D. 85%
   Answer: B. η = 15,268 / (p t σt = 50,000) = 0.305. (Key point: Efficiency)

## Related in Workshop Technology

- [Upsetting diameter and force calculation](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Aupsetting-diameter-force-numerical) — Workshop Technology, chapter 9
- [Resistance spot welding heat and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Aresistance-spot-welding-heat-numerical) — Workshop Technology, chapter 10
- [Blanking force and press capacity for washer](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Ablanking-force-press-capacity) — Workshop Technology, chapter 8
- [Gear manufacturing and simple indexing](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Agear-milling-indexing-numerical) — Workshop Technology, chapter 11
- [Drilling time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Adrilling-time-numerical) — Workshop Technology, chapter 5
- [Soldering process and fluxes](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Asoldering-process-flux) — Workshop Technology, chapter 3

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