---
title: "Pin in double shear and bearing"
subject: "Strength of Materials"
chapter: 1
key: "semester_4/strength_of_materials_mechanical:pin-double-shear-bearing-check"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/pin-double-shear-bearing-check
site: BCT Engineering
---

# Pin in double shear and bearing

Strength of Materials · Chapter 1: Introduction

- Asked in: year not given, 2+2+2 marks
- Video: [Mechanics of Materials: Lesson 4 - Shear Stress, Single & Double Shear!](https://www.youtube.com/watch?v=hAS42U0x40w) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/pin-double-shear-bearing-check
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-1

## Question

A 25 mm diameter pin joins a central plate 12 mm thick to two outer plates. The central plate is pulled to the right by a tensile load of 60 kN and each outer plate pulls to the left with 30 kN, so the pin is in double shear. Find (a) the average shear stress in the pin and the bearing stress between the pin and the central plate, (b) the factor of safety against shear failure if the ultimate shear strength of the pin material is 300 MPa, (c) the largest load the joint may carry if a factor of safety of 3 against shear failure is required.

## Short answer

For double shear, shear stress τ = P/(2A) = 61.1 MPa; bearing stress σb = P/(d t) = 200 MPa. Factor of safety = τu/τ = 4.91. With required FOS = 3, allowable load = 98.2 kN.

## Answer

#### Given data
$P = 60\ \text{kN}$, $d = 25\ \text{mm}$, $t = 12\ \text{mm}$, double shear (two planes), $\tau_u = 300\ \text{MPa}$.

```text
 outer plate  <-- 30 kN ===[ pin ]
 centre plate      60 kN --->[ pin ]
 outer plate  <-- 30 kN ===[ pin ]
   (two shear planes in the pin)
```

#### (a) Shear and bearing stress
Area of pin in one plane:

$$
A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(25)^2 = 490.9\ \text{mm}^2
$$

The pin has two shear planes, each carrying $P/2$:

$$
\tau = \frac{P}{2A} = \frac{60\times10^3}{2\times 490.9} = 61.1\ \text{MPa}
$$

Bearing on the central plate (it carries the whole 60 kN):

$$
\sigma_b = \frac{P}{d\,t} = \frac{60\times10^3}{25\times 12} = 200\ \text{MPa}
$$

#### (b) Factor of safety
$$
\text{FOS} = \frac{\tau_u}{\tau} = \frac{300}{61.1} = 4.91
$$

#### (c) Allowable load for FOS = 3
$$
\tau_{all} = \frac{300}{3} = 100\ \text{MPa}, \qquad P_{all} = \tau_{all}\,(2A) = 100\times 2\times 490.9 = 98.2\times10^3\ \text{N}
$$

**Answer: $\tau = 61.1$ MPa, $\sigma_b = 200$ MPa; FOS $= 4.91$; $P_{all} = 98.2$ kN (bearing and plate tension should also be checked).**

## Check yourself

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

1. In the pin joint, how is the average shear stress in the pin found?
   A. tau = P/A, using the full 60 kN on one plane
   B. tau = P/(2A), each of two planes carries P/2
   C. tau = P/(3A), using three planes
   D. tau = 2P/A, because two plates pull
   Answer: B. The pin has two shear planes, each carrying half the 60 kN, so tau = 60000/(2 x 490.9) = 61.1 MPa. (Key point: Shear planes in double shear)

2. Which load and area give the bearing stress between the pin and the central plate?
   A. 30 kN on d x t
   B. 60 kN on d x t
   C. 60 kN on pi d^2/4
   D. 30 kN on 2 x pi d^2/4
   Answer: B. The central plate carries the whole 60 kN over the projected area 25 x 12, giving 200 MPa. (Key point: Bearing stress on central plate)

3. How is the factor of safety against shear failure calculated here?
   A. Applied shear stress divided by ultimate shear strength
   B. Ultimate shear strength divided by the applied shear stress
   C. Ultimate shear strength divided by the bearing stress
   D. Ultimate shear strength times the applied shear stress
   Answer: B. FOS = tau_u / tau = 300/61.1 = 4.91. (Key point: FOS against shear)

4. To find the largest load with FOS = 3, what is done?
   A. Use tau_all = 300/3 = 100 MPa and P = tau_all x (2A)
   B. Use tau_all = 300 x 3 and P = tau_all x A
   C. Use tau_all = 300/3 and P = tau_all x A
   D. Multiply the 60 kN load by 3
   Answer: A. The allowable stress is 100 MPa and it acts on two shear planes, so P = 100 x 2 x 490.9 = 98.2 kN. (Key point: Largest load for FOS 3)

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