Chapter 1 · 2 hours

Introduction

Practice questions

Practice questions and answers

2 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

Practice · 6 marks

Define normal stress, shear stress and bearing stress, giving the formula and SI unit of each. Differentiate between ultimate stress, allowable (working) stress and factor of safety.

Answer

Types of stress

Stress is the internal resisting force per unit area of a cross-section. Its SI unit is the pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$ (in practice MPa $= \text{N/mm}^2$ ).

Stress	Cause	Formula	Acts
Normal stress	Axial force $P$ perpendicular to area	$\sigma = P/A$	Perpendicular to the section
Shear stress	Force $V$ parallel to the area (cutting action)	$\tau = V/A_s$	In the plane of the section
Bearing stress	Compressive contact force between two bodies, e.g. a pin and the hole in a plate	$\sigma_b = P/(d\,t)$	On the projected contact area

Normal stress is tensile (pull, positive) or compressive (push, negative).
For a rivet or pin of diameter $d$ in single shear $A_s = \pi d^2/4$ ; in double shear $A_s = 2\times\pi d^2/4$ .
The bearing area is the projected area (diameter $\times$ plate thickness), not the curved surface.

Ultimate stress, allowable stress and factor of safety

Term	Meaning
Ultimate stress $\sigma_u$	Maximum stress the material can carry before fracture (from the test, load at failure / original area)
Allowable stress $\sigma_{all}$	Maximum stress permitted in design; always below the elastic or yield limit
Factor of safety (FOS)	Ratio of failure stress to allowable stress, $\text{FOS} = \sigma_u/\sigma_{all}$ (or $\sigma_y/\sigma_{all}$ for ductile materials)

\sigma_{all} = \frac{\sigma_u}{\text{FOS}}

FOS is always greater than 1. It covers uncertainty in loads, material defects, workmanship and the consequences of failure. Typical values are about 1.5 to 2 for steel structures under static load and larger for brittle materials or for impact loads.

Practice · 2+2+2 marks

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.

Answer

Given data

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

 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).

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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