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(c) D.J.Dunn www.freestudy.co.uk
1
CITY AND GUILDS 9210
Unit 130 MECHANICS OF MACHINES AND STRENGTH OF MATERIALS
OUTCOME 1
TUTORIAL 1 - BASIC STRESS AND STRAIN
Outcome 1 Explain static equilibrium, Newton's laws, and calculation of reaction and
internal forces and the basic concepts of different stresses and strains and
determine stresses and strains of components under loading conditions.
The learner can:
1. Explain the concept of stresses and strains: Compressive and shear stresses and linear, lateral, shear,
thermal and volumetric strains.
2. Explain static equilibrium, Newton’s Laws and Calculation of reaction and internal forces.
3. Explain Hooke’s law, Poisson’s ratio, Modulus of Elasticity and Modulus of Rigidity, Bulk Modulus.
4. Explain stress-strain diagrams for ductile and brittle materials and for various strengths of material-
yield strength, ultimate tensile strength etc.
5. Explain the interrelation between elastic constants, proof stress and true stress and strain, axial force
diagrams, stresses and strains in determinate, homogeneous and composite bars under concentrated
loads and self weight.
6. Explain temperature stresses in simple and composite members.
7. Explain strain energy due to axial load (gradual, sudden and impact) and strain energy due to self
weight.
This tutorial should be seen as providing basic knowledge of stress and strain that many
degree students should already know. In this case use it as revision or skip it. On completion
of this tutorial you should be able to do the following.
Define direct stress and strain.
Define shear stress and strain.
Define the modulus of elasticity and rigidity.
Solve basic problems involving stress, strain and modulus.
It is assumed that the student is already familiar with the concepts of FORCE.
(c) D.J.Dunn www.freestudy.co.uk
2
1. STATIC FORCES and STRUCTURES
STATIC FORCES will deform a body in one or more of the following manners.
The force may stretch the body, in which case it is
called a TENSILE FORCE.
The force may squeeze the body in which case it is
called a COMPRESSIVE FORCE.
A rod or rope used in a frame to take a tensile load is
called a TIE. If it takes a compressive force it is called a STRUT.
A strut that is thick compared to its length is called a COLUMN.
The force may bend the body in
which case both tensile and
compressive forces may occur. A
structure used to support a bending
load is called a BEAM or JOIST.
The force may try to shear the body in which case the force
is called a SHEAR FORCE. A scissors or guillotine
produces shear forces.
The force may twist the body in which case SHEAR FORCES occur. A structure that transmits
rotation is called a SHAFT and it experiences TORSION.
(c) D.J.Dunn www.freestudy.co.uk
3
FRAMES
Struts and ties make up the members of lattice frames such as the simple one shown. In this example the
sloping members are compressed and so are struts but the bottom one is stretched and could be a chain so it
is a tie.
BEAMS
A beam is a structure, which is loaded transversely (sideways). The loads may be point loads or uniformly
distributed loads (udl). The diagrams shows examples.
Transverse loading causes bending and bending is a very severe form of stressing a structure. You might
note how easy it is to snap a stick by bending but not so easy by stretching or twisting. The bent beam goes
into tension (stretched) on one side and compression on the other. Failure or fracture will always occur on
the tension side and spread across the section.
(c) D.J.Dunn www.freestudy.co.uk
4
2. SAFETY FACTORS
In the following sections you will learn how to calculate the stress in some basic engineering structures and
components. If we want to be confident that the structure or component does not fail by being overstressed,
we design it so that working stress is smaller than the stress at which it fails. If we want to define what
failure means, we would need to study stress and strain and material science to a much greater depth than
covered by this unit. Failure might mean that the material has yielded or that it has broken. Also calculation
of maximum stress in a component is much more complicated than the cases studied here. If a material has
direct stress and shear stress at the same time, the maximum stress may well exceed either of them on their
own. Also the stress may be raised by things like sharp corners and undercuts. Other factors such as fatigue
and creep also affect the stress. Don't think that this tutorial is anything but the start of your studies in that
area.
If we decide what the fail stress is for a given material, then we design the component so that the working
stress is less. We define the safety factor as the ratio such that:
stress workingMaximum
Failureat StressFactorSafety
3. DIRECT STRESS
When a force is applied to an elastic body, the body deforms. The way in which the body deforms depends
upon the type of force applied to it. A compression force makes the body shorter. A tensile force makes the
body longer.
Tensile and compressive forces are called DIRECT FORCES.
Stress is the force per unit area upon which it acts and the symbol is (called SIGMA).
The units are N/m2 or Pascals.
NOTE ON UNITS The fundamental unit of stress is 1 N/m2 and this is called a Pascal. This is a
small quantity in most fields of engineering so we use the multiples kPa, MPa and GPa.
Areas may be calculated in mm2 and units of stress in N/mm
2 are quite acceptable. Since 1 N/mm
2
converts to 1 000 000 N/m2 then it follows that the N/mm
2 is the same as a MPa
(c) D.J.Dunn www.freestudy.co.uk
5
4. DIRECT STRAIN
In each case, a force F produces a deformation x. In engineering we usually change this force into stress and the deformation into strain and we define these as follows.
Strain is the deformation per unit of the original length. The symbol is (called EPSILON).
Strain has no units since it is a ratio of length to length. Most engineering materials do not stretch very much
before they become damaged so strain values are very small figures. It is quite normal to change small
numbers into the exponent form of 10-6
. Engineers use the abbreviation (micro strain) to denote this
multiple. For example a strain of 0.000068 could be written as 68 x 10-6
but engineers would write 68 .
WORKED EXAMPLE No.1
A metal wire is 2.5 mm diameter and 2 m long. A force of 1200 N is applied to it and it stretches 0.3
mm. Assume the material is elastic. Determine the following.
i. The stress in the wire .
ii. The strain in the wire .
If the stress that produces failure is 300 MPa, calculate the safety factor.
SOLUTION
2222
N/mm 2444.909
1200
A
Fσ mm 4.909
4
2.5 x π
4
πdA
Answer (i) is hence 244 MPa
150or 0.000152000
mm 0.3
L
x
Safety Factor = 300/244 = 1.23
SELF ASSESSMENT EXERCISE No.1
1. A steel bar is 10 mm diameter and 2 m long. It is stretched with a force of 20 kN and extends by
0.2 mm. Calculate the stress and strain. If the stress at failure is 300 MPa calculate the safety factor.
(Answers 254.6 MPa, 100 and 1.18)
2. A rod is 0.5 m long and 5 mm diameter. It is stretched 0.06 mm by a force of 3 kN. Calculate the stress
and strain. If the stress at failure is 250 MPa calculate the safety factor.
(Answers 152.8 MPa, 120 and 1.64)
(c) D.J.Dunn www.freestudy.co.uk
6
5. STRESS AND STRAIN RELATIONSHIPS FOR MATERIALS
The relationship between direct stress and strain is determined with a tensile test. A typical testing machine
is shown. These measure and plot the force against extension of the test piece and converts them into stress
and strain so that the relationship between them may be determined. The details of the test are not needed
here. The tensile test is conducted in order to find the following properties of a material.
The yield stress.
The ultimate tensile stress.
The elastic range.
The ductile range.
The modulus of elasticity. Ductility or lack of ductility is indicated by two results from
the test as follows.
The % elongation.
The % area reduction.
Pictures courtesy of Lloyd Instruments www.lloyd-instruments.co.uk/
The following applies to materials tested in tension (direct stress). Similar tests and descriptions apply to
materials in torsion (the torsion test). The diagram below shows typical Force – extension graphs for some
materials.
Graph A is typical for a brittle material.
The force is directly proportional to extension right up to the point of fracture
and hence obeys Hooke's Law. The
material fractures without warning. Cast
Iron is a typical example.
Graph B shoes the result for a material
that has some ductility such as steel. The
material obeys Hooke's Law up to a point
where the structure of the material starts
to give and slip; a phenomenon known as
yielding. After yielding the material
continues to extend with little or no increase in the force and this is known as plasticity.
Graph C is typical for a very ductile material such as copper with a very large plastic range.
In the elastic range the material will spring back when
the force is removed. In the plastic range the material
is permanently damaged and remains stretched when
the force is removed. The permanent damage starts at
the yield point. Usually this is the same point as the
limit of proportionality. At the necking point the
material starts to narrow at the point where it is going
to break and a pronounced neck forms. Because the
cross sectional area is reduced, the force needed to
stretch it further becomes smaller but the stress in the
material remains the same. Eventually at some point
the specimen will break.
(c) D.J.Dunn www.freestudy.co.uk
7
DIRECT STRESS AND STRAIN
Stress = Force/Area
Strain = Extension/Gauge length
Yield stress y = Yield Load/Original Area.
Ultimate Tensile Stress u = Maximum Load/Original Area
Normally the ultimate tensile stress is based on the maximum load as the reduction in area causes the load
(but not the stress) to fall.
6. MODULUS of ELASTICITY
Elastic materials always spring back into shape when released. They usually obey HOOKE'S LAW. This is
the law of a spring which states that deformation is directly proportional to the force. As long as the material
is within the elastic range and the graph is a straight line, the ratio of F/x and hence / is constant. The ratio
/ is called the modulus of Elasticity and has a symbol E. This property determines how elastic the material
is.
The units of E are the same as the units of stress.
WORKED EXAMPLE No. 2
An aluminium tensile test specimen is 5 mm diameter with a gauge length of 50 mm. The force measure
at the yield point was 982 N and the maximum force was 1.6 kN. Calculate the yield stress and the
ultimate tensile stress.
Cross sectional area A = π d2/4 = π x 5
2/4 = 19.63 mm
2
Yield Stress = 982/19.63 = 50 N/mm2 or 50 MPa (Note 1 MPa = 1 N/mm
2)
Ultimate Tensile Stress = 1600/19.63 = 81.5 N/mm2 or 81.5 MPa
At a point on the proportional section the extension was 0.03 mm and the force 800 N. Calculate the
modulus of Elasticity (Young's Modulus)
GPa 67.9or N/mm 790660.03 x 19.63
50 x 800
Ax
FLE 2o
SELF ASSESSMENT EXERCISE No. 2
A tensile test specimen has a cross sectional area of 100 mm2. The force measure at the yield point was
41 kN and the maximum force was 42 kN. Calculate the following.
The yield stress (410 MPa)
The tensile strength. (420 MPa)
The same tensile test showed that the force at the limit of proportionality was 40 KN and the extension
was 0.095 mm over a gauge length of 50 mm.
Calculate the modulus of elasticity. (210 GPa)
(c) D.J.Dunn www.freestudy.co.uk
8
WORKED EXAMPLE No. 3
A steel tensile test specimen has a cross sectional area of 100 mm2 and a gauge length of 50 mm. In a
tensile test the graph of force - extension produces a gradient in the elastic section of 410 x 103 N/mm.
Determine the modulus of elasticity.
The specimen breaks when the force is 35 kN. What is the ultimate tensile stress based on the original
area?
SOLUTION
The gradient gives the ratio F/A = and this may be used to find E.
GPa 205or MPa 000 205or N/mm 000 205 100
50 x 10 x 410
A
Lx
x
F
ε
σE 23
350MPaor 350N/mm100
35000
Area
Force Breakingσ 2
u
WORKED EXAMPLE No. 4
A Steel column is 3 m long and 0.4 m diameter. It carries a load of 50 MN. Given that the modulus of
elasticity is 200 GPa, calculate the compressive stress and strain and determine how much the column is
compressed.
SOLUTION
mm 5.97 mm 3000 x 0.001989L ε xso L
xε
0.001989 x10200
x10397.9
E
σε so
ε
σE
Pa x10397.90.126
x1050
A
Fσ m 0.126
4
0.4 x π
4
πdA
9
6
66
222
(c) D.J.Dunn www.freestudy.co.uk
9
SELF ASSESSMENT EXERCISE No. 3
1. A bar is 500 mm long and is stretched to 500.45 mm with a force of 15 kN. The bar is 10 mm diameter.
Calculate the stress and strain.
Assuming that the material has remained within the elastic limit, determine the modulus of elasticity.
If the fail stress is 250 MPa calculate the safety factor.
(Answers 191 MPa, 900με , 212.2 GPa and 1.31.)
2. The maximum safe stress in a steel bar is 300 MPa and the modulus of elasticity is 205 GPa. The bar is
80 mm diameter and 240 mm long. If a factor of safety of 2 is to be used, determine the following.
i. The maximum allowable stress. (150 MPa)
ii. The maximum tensile force allowable. (754 kN)
iii. The corresponding strain at this force. (731.7 με)
iv. The change in length. (0.176 mm)
3. A circular metal column is to support a load of 500 Tonne and it must not compress more than 0.1 mm.
The modulus of elasticity is 210 GPa. The column is 2 m long. Calculate the following.
i. The cross sectional area(0.467 m2)
ii. The diameter. (0.771 m)
Note 1 Tonne is 1000 kg.
(c) D.J.Dunn www.freestudy.co.uk
10
7. SHEAR STRESS
Shear force is a force applied sideways on to the material (transversely loaded). Here are some examples:
Shear stress is the force per unit area carrying the load. This means the cross sectional area of the material
being cut and of the beam and pin.
Shear stress = F/A The symbol is called Tau. Note that A is the area being sheared.
As with direct stress, we may define a safety factor as the ratio of the fail shear stress to the actual working
stress.
The sign convention for shear force and stress is based on how it shears the materials and this is shown
below.
8. SHEAR STRAIN
In order to understand the basic theory of shearing, consider a block of material being deformed sideways as
shown. (Note in reality the deformation is very small and the diagram is much exaggerated)
The shear force causes the material to deform as shown. The shear strain is defined as the ratio of the
distance deformed (x) to the height (L).
The end face rotates through an angle . Since this is a very small angle, it is accurate to say the distance x is
the length of an arc of radius L and angle so that = x/L
It follows that is the shear strain. The symbol is called Gamma.
(c) D.J.Dunn www.freestudy.co.uk
11
9. MODULUS OF RIGIDITY G
If we were to conduct an experiment and measure x for various
values of F, we would find that if the material is elastic, it
behave like a spring and so long as we do not damage the
material by using too big a force, the graph of F and x is a
straight line as shown (obeying Hooke's Law).
The gradient of the graph is constant so F/x = constant and this is the shear spring stiffness of the block in
N/m.
If we divide F by the area A and x by the height L, the relationship is still a constant and we get:
But F/A = and x/L = so:
This constant will have a special value for each elastic material and is called the Modulus of Rigidity with
symbol G.
10. ULTIMATE SHEAR STRESS
If a material is sheared beyond a certain limit it becomes permanently distorted and does not spring all the
way back to its original shape. The elastic limit has been exceeded. If the material is stressed to the limit so
that it parts into two (e.g. a guillotine or punch), the ultimate limit has been reached. The ultimate shear
stress is u and this value is used to calculate the force needed by shears and punches.
WORKED EXAMPLE No. 5
Calculate the force needed to guillotine a sheet of metal 5 mm thick and 0.8 m wide given that the
ultimate shear stress is 50 MPa.
SOLUTION
kN 200or N 000 200 4000 x 50A x F so A
F
N/mm 50 is stressshear ultimate The mm 4000 5 x 800 A
mm 5 x mm 800 rectangle a iscut be toarea The
22
(c) D.J.Dunn www.freestudy.co.uk
12
WORKED EXAMPLE No. 6
Calculate the force needed to punch a hole 30 mm diameter in a sheet of metal 3 mm thick given that
the ultimate shear stress is 60 MPa.
SOLUTION
kN 16.965or N 16965 282.7 x 60A x F so A
F
N/mm 60 is stressshear ultimate The mm 282.7 3 x 30 x πA
x tπd ss x thicknencecircumfere theiscut be toarea The
22
WORKED EXAMPLE No. 7
Calculate the force needed to shear a pin 8 mm diameter given that the ultimate shear stress is 60 MPa.
SOLUTION
kN 3.016or N 3016 50.26 x 60A x F so A
F
N/mm 60 is stressshear ultimate The 50.26mm4
8 x A
4
πdA areacircular theis sheared be toarea The
222
SELF ASSESSMENT EXERCISE No. 4
1. A guillotine must shear a sheet of metal 0.6 m wide and 3 mm thick. The ultimate shear stress is 45
MPa. Calculate the force required. (Answer 81 kN)
2. A punch must cut a hole 30 mm diameter in a sheet of steel 2 mm thick. The ultimate shear stress is 55
MPa. Calculate the force required. (Answer10.37 kN)
3. Two strips of metal are pinned together as shown with a rod 10 mm diameter. The ultimate shear stress
for the rod is 60 MPa. Determine the maximum force required to break the pin. (Answer 4.71 kN)
(c) D.J.Dunn www.freestudy.co.uk
13
11. DOUBLE SHEAR
Consider a pin joint with a support on both ends as shown. This is called a
CLEVIS and CLEVIS PIN. If the pin shears it will do so as shown.
By balance of forces, the force in the two supports is F/2 each.
The area sheared is twice the cross section of the pin so it takes twice as much
force to break the pin as for a case of single shear. Double shear arrangements
doubles the maximum force allowed in the pin.
WORKED EXAMPLE No. 8
A pin is used to attach a clevis to a rope. The force in the rope will be a maximum of 60 kN. The
ultimate shear stress allowed in the pin is 120 MPa. Calculate the diameter of a suitable pin if the safety
factor must be 3.0 based on the ultimate value.
SOLUTION
The working stress = 120/3 = 40 MPa
mm 30.9π
750 x 4d
4
d mm 750A m10 x 750
40x10 x 2
60000
2τ
F A
2A
F is sressshear thesoshear doublein ispin The
2226
6
SELF ASSESSMENT EXERCISE No. 5
1. A clevis pin joint as shown is used to joint two steel rods in a structure. The pin is 8 mm diameter. The
ultimate shear stress for the pin material is 80 MPa. Determine the maximum force that can be exerted
using a safety factor of 2.0. (Answer 4.02 kN)
2. The steel tie rods used in question 1 are 20 mm diameter. The steel yield stress is 160 MPa. Based on
the answer to question 1, determine the safety factor for the rods based on yield stress. (Answer 12.5)