BTE 1013ENGINEERING SCIENCES

7. MATERIALS

NAZARIN B. NORDINnazarin@icam.edu.my

What you will learn:

• Strength, elesticity, ductility, malleability, brittleness, toughness, hardness

• Ferrous/ non-ferrous metals, tensile stress, yield stress, shear force, percentage of elongation and percentage of reduction in plain carbon steel, shear force, bending moment and fatigue test

7.1 Strength, elasticity, ductility, malleability, brittleness, toughness, hardness

7.2 Ferrous/non-ferrous metals, tensile stress, yield stress, shear force, percentage of elongation, percentage of reduction in plain carbon steel, shear force, bending moment and fatigue test

• For practical purposes, components are designed to withstand forces and loads that a device is designed for and, so long as the instructions for use and maintenance, such as safe loads and tightening torques, are observed, problems should not be experienced.

Strength of materials

Other terms used in describing materials

• Hardness• Toughness

Hardness• A hard material is one that resists

indentation or abrasion by another material.

Toughness• A material is said to be tough when a

large amount of energy is required to fracture it.

Brittleness• Materials that break without undergoing

local distortion and are unable to withstand sharp blows are said to be brittle. Most types of cast iron are brittle.

Ductility• A material that can be drawn out by

tensile force is said to be ductile. The steel sheet that is used in the construction of motor car panels is of a type known as deep drawing steel and this is a ductile material.

Malleability• Metals that can hammered and bent

without cracking are said to be malleable. Lead is an example of a malleable material.

Non-ferrous metals• These are mainly alloys that contain no iron.

Commonly used non-ferrous alloys are those made from copper, lead, tin, aluminium or magnesium. Non-ferrous alloys are used extensively in automotive engineering.

Stress• Forces that tend to stretch, or pull something

apart, are known as tensile forces and they produce two important effects:

• 1. In trying to pull the bolt apart, internal resisting

• forces are created and these internal forces are

• known as stress.• 2. The length of the bolt will increase, and this

change in the bolt’s dimensions is known as strain.

• Stress is calculated by dividing the applied force

• by the cross-sectional area of the bolt.• Stress = Force/Cross-sectional area

Types of stress• There are three basic forms of stress:

– 1. tensile stress;– 2. compressive stress;– 3. shear stress – torsional stress is a form of shear

stress.

Examples of stress measure• Example 1: A cylinder head bolt with an

effective diameter of 15mm carries a tensile load of 10 kN. Calculate the tensile stress in the bolt.

• Example 2: A connecting rod has a cross-sectional area of 200mm2 and it carries a compressive force of 2.4 tonnes. Calculate the compressive stress in the

• connecting rod.

• Example 3: The hand brake linkage shown in Figure

• carries a tensile force of 600 N. Calculate the shear stress in the clevis pin, which is 12mm in diameter.

• In this case the shearing action is attempting to shear the clevis pin across two cross-sectional areas.

• Example 4: A propeller shaft coupling of a truck is secured by four bolts of 14mmdiameter that are equally spaced at a radius of 50mm from the centre of the propeller shaft. Calculate the shear stress in each bolt when the shaft is transmitting a torque of 500N m.

Strain• When a load is applied to a metal test bar a

change of shape takes place. A tensile load will stretch the bar and a compressive load will shorten it. This change of shape is called strain. The three basic types of strain are shown in Figure

Example strain measure• A steel rod 200mm in length stretches by

0.12mm when it is subjected to a tensile load of 2 tonnes. Determine the strain.

• Solution• Strain = change in length/original length• = 0.12mm/200mm• Tensile strain in the steel rod = 0.0006

• Note: strain does not have any units.

THANK YOU