CITY AND GUILDS 9210 Unit 130 MECHANICS OF · PDF fileAll engineering materials expand when heated and this ... F = 30 000 = aAa + bAb 30 000 ... x = b L = 3.248 x 10-6 x 1.5 = 4.872

(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 3 - THERMAL STRAIN AND SUDDEN LOADING

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.

Calculate stress due to dimensional changes brought about by temperature changes.

Calculate stresses in composite bars.

Calculate stresses induced by thermal expansion in compound bars.

Explain basic strain energy theory.

Calculate the deflection of simple structures due to gradual, sudden and impact

loading.


1. TEMPERATURE STRESSES

Metals expand when heated. This can be put to good use. For example a ring may be expanded by warming

it and then fitted onto a shaft and on cooling grips the shaft very tightly.

Thermal expansion can also produce unwanted stresses in structures. For example, suddenly allowing hot

fluid into a badly designed pipe could cause it to fracture as it tries to get longer but is prevented from doing

so.

COEFFICIENT OF LINEAR EXPANSION

All engineering materials expand when heated and this expansion is usually equal in all directions. If a bar

of material of length L has its temperature increased by degrees, the increase of length L is directly

proportional to the original length L and to the temperature change . Hence

L =constant x L

The constant of proportionality is called the coefficient of linear expansion ().

L = L

INDUCED STRESS IN A CONSTRAINED BAR

When a material is heated and not allowed to expand freely, stresses are induced which are known as

"temperature stresses." Suppose the bar was allowed to expand freely by distance L and then compressed

back to its original length. The compressive strain is then

=L /L = L /L =

Since stress/strain = modulus of elasticity (E) then the induced stress is

= E = E

If the material is cooled and then stretched back to the original length the stress and strain are tensile.


WORKED EXAMPLE No. 1

A thin steel band 850 mm diameter must be expanded to fit around a disc 851 mm diameter. Calculate

the temperature change needed and the stress produced in the ring. The coefficient of linear expansion is

15 x 10-6

per oC and the modulus of elasticity E is 200 GPa.

SOLUTION

Initial circumference of ring = D = x 850 = 2670.35 mm

Required circumference = x 851 = 2673.50 mm

L = 2673.50 - 2670.35 = 3.15 mm L = L

3.15 = 15 x 10-6

x 2670.35 x

= 3.15/(15 x 10-6

x 2670.35) = 78.6 Kelvin

= E = 200 x 109 x 15 x 10

-6 x 78.6 = 235.8 MPa.

Alternatively

strain = L/ L = 3.15/2670.35 = 0.011796

stress = E = 200 x 109 x 0.011796 = 235.8 MPa


A brass bar is 600 mm long and it is turned on a centre lathe to 100 mm diameter. It is held between the

chuck jaws and a running tail stock so that it is not free to expand. During the turning process it has

become heated from 20 o

C to 95oC. Calculate the thermal stress induced in the bar and the resulting

thrust on the chuck and tail stock. E for brass is 90 Gpa and is 18 x 10-6

per oC.

SOLUTION

Stress induced = = E = 90 x 109 x 18 x 10

-6 x (95 – 20) = 121.5 MPa

Force = stress x cross sectional area = 121.5 x 106 x x 0.1

2/4 = 954.3 kN


Determine the induced stress and thrust if the centre lathe flexed so that the bar changed length.

SOLUTION

Check your solution here. The free expansion of the bar is

L = L = 18 x 10-6

x 600 x (95 – 20) = 0.81 mm Actual change in length is 0.6 mm.

Strain induced = change in length/original length = (0.81 – 0.6)/600 = 0.00035

Stress induced = E = 90 x 109 x 0.00035 = 31.5 MPa

Force = stress x cross sectional area = 31.5 x 106 x x 0.01

2/4 = 247.4 kN


SELF ASSESSMENT EXERCISE No. 1

1. A steel ring is 50 mm diameter and 2 mm thick. It must be fitted onto a shaft 50.1 mm diameter.

Calculate the temperature to which it must be heated in order to fit on the shaft. The initial temperature

is 20 oC and the coefficient of linear expansion is 15 x 10

-6 per

oC.

(Answer 133.3 K)

2. A stub shaft 85.2 mm diameter must be shrunk to 85 mm diameter in order to insert it into a housing. By

how much must the temperature be reduced? Take the coefficient of linear expansion is 12 x 10-6

per oC.

(Answer -195.6 K)

3. A steam pipe is 120 mm outer diameter and 100 mm inner diameter. It has a length of 30 m and passes

through a wall at both ends where it is rigidly constrained. Steam at 200oC is suddenly released into the

pipe. The initial temperature of the pipe is 15 oC and the coefficient of linear expansion is 15 x 10

-6 per

oC. E is 200 Gpa.

Calculate

i. the thermal stress produced. (555 MPa)

ii. the force exerted by the pipe against the walls. (1.918 MN)


2. COMPOSITE BARS

A typical composite bar might be as shown. Consider the whole bar being compressed by a force applied to

flanges at each end.

Figure 1

The two parts are the same length throughout and both parts are hence strained the same amount. Let the

original length be L and the change in length be x.

Strain =

A = x/L For material A

B= x/L For material B

The cross areas are AA and AB

Stress in A = A = F/AA F1 = AAA

Stress in B = B = F/AB F2 = BAB

F = F1 + F2

F = AAA + BAB .........(1)

The strain in a is A = x/L

The strain in b is B= x/L

The modulus of elasticity for a is EA = A/A A = A/EA

The modulus of elasticity for b is EB = B/B B = B/EB

Since the strains are equal A/EA = B/EB

A = (EA /EB)B .........(2)

Combining equations (1) and (2)

F = B [(EA /EB)AA +AB]

or F = A [(EB /EA)AB +AA]



A steel cylinder is 0.5 m outer diameter and 0. 4 m inner diameter and 1.5 m long. It is filled with

concrete and used as a vertical column to support a weight of 30 kN. Determine the compression and

the stresses in both the steel and concrete.

E is 205 GPa for the steel and 10 GPa for concrete.

SOLUTION

Aa = (0.52 - 0.45 2)/4 = 0.0373 m2. (steel)

Ab = ( 0.45 2)/4 = 0.159 m2. (concrete)

a/Ea = b/Eb

a = Ea (b/Eb)

F = 30 000 = aAa + bAb

30 000 = Ea (b/Eb)Aa + bAb

30 000 = (Ea /Eb)bAa + bAb

30 000 = (205 x 109 /10 x 109)0.0373b + 0.159 b

30 000 = 20.5 x 0.0373b+ 0.159 b

30 000 = 0.7646b+ 0.159 b

30 000 = 0.9236b

b = 30 000/0.9236

b = 32.48 kPa

a = (Ea /Eb)b

a = 20.5 x 32.38 = 665.8 kPa

b = b/Eb

b = 32480/10 x 109 = 3.248 x 10-6

x = b L = 3.248 x 10-6 x 1.5 = 4.872 10-6m

Check for material a.

a = a/Ea

a = 665800/205 x 109 =3.248 x 10-6

x = a L = 3.248 x 10-6 x 1.5 = 4.872 10-6m



1. Reinforced concrete column is rectangular in section and measures 0.4 m x 0.6 m. The column contains

40 steel rods 10 mm diameter. Calculate the maximum load which can be supported given that the stress

in the steel must not exceed 400 MPa.


(6.94 MN)

2. A cast iron cylinder is filled with concrete and used as a pillar to support a weight of 40 kN. The

cylinder is 0.5 m outer diameter and 0.49 m inner diameter. Determine the stress in the iron and

concrete.


(115 kPa 2.36MPa)

3. A compound bar is constructed by riveting a steel strip 5 mm x 20 mm to a brass strip 8 mm x 20 mm to

form a cross section 13 mm x 20 mm. The bar is 300 mm long and is stretched until the stress in the

brass is 5 MPa. Determine

i. the force used to stretch the bar. (1.8 kN)

ii. the extension of the bar. (0.0015 mm)

Take E for steel as 200 GPa and for brass as 100 GPa.


3. THERMAL STRESSES IN COMPOSITE BARS

Consider a composite bar of length L. If the temperature is raised

by θ and material A expands more than material B the change in lengths for each is:

A L and B L respectively.

If the materials are clamped so that they are constrained to the

same length then A is compressed a distance xA and B is stretched

a distance xB.

It follows that: A L - B L = xA + xB

Divide by L and A - B = (xA + xB)/L

(A - B) = (εA + εB)

The force in each part is equal and opposite. F = σA AA = σB AB


A compound bar is 1.5 m long and made from a steel strip riveted along its length to a copper strip on

both sides. All three strips are 50 mm wide and 10 mm thick. Assuming no stress in the strips when they

were riveted, calculate the stress and change in length when the assembly is 60oC hotter.

E is 200 GPa for the steel and 110 GPa for copper.

α is 12 x 10-6

/K for steel and 17 x 10-6

/K for copper.

SOLUTION

(c - s) = 60 (17 x 10-6

- 12 x 10-6

) = 300 x 10-6

(εc + εs) = 300 x 10-6

σc/Ec + σs/Es = 300 x 10-6

Ac = 2 x 50 x 10 = 1000 mm2 As = 50 x 10 = 5000 mm

2

F = 15714 N

Stress in the copper is 15714/1000 = 15.714 N/mm2= 15.714 MPa

Stress in the steel is 15714/500 = 31.429 N/mm2 = 31.429 MPa

Strain in the copper is 15.714 x 106/110 x 10

9 = 142.855 x 10

-6

Strain in the steel is 31.428 x 106/200 x 10

9 = 157.143 x 10

-6

Change in length of the copper is L εc = 1500 x 142.855 x 10-6

= 0.214 mm

Change in length of the steel is L εs = 1500 x 157.14 x 10-6

= 0.236 mm

Final length of the bar is:

L + L c - 0.214 mm = 1500(1 + 60 x 17 x 10-6

) - 0.214 = 1501.32 mm or

L + L s + 0.236mm = 1500(1 + 60 x 12 x 10-6

) + 0.236 = 1501.32 mm


SELF ASSESSMENT No. 3

A compound bar as illustrated is an aluminium tube with a steel draw bar through the middle as shown.

The nuts at each end are tightened just sufficiently to be firm without inducing any stress. The tube is

300 mm long, 30 mm outer diameter and 25 mm inner diameter. The bolt is 20 mm diameter. The

assembly is heated and the temperature rises by 200o C. Find the stress in each material and the

extension of the bar.

For steel, E = 200 GPa and α = 12 x 10-6

/K

For aluminium, E = 70 GPa and α = 23 x 10-6

/K.

Answers Steel, 42.67 MPa (tensile) aluminium 62 MPa (compressive) and 0.424 mm


4 STRAIN ENERGY

When an elastic body such as a spring is deformed, work is done. The energy used up is stored in the body

as strain energy and it may be regained by allowing the body to relax. The best example of this is a

clockwork device which stores strain energy and then gives it up.

Springs take many forms. Any elastic material may be stretched, compressed, twisted, sheared or bent. In all

cases work is done to deform the material so strain energy is stored in it. Strain energy is usually given the

symbol U. Consider the two simple cases shown below of linear springs that obey Hooke’s law.

F = k x

k is the stiffness in N/m and x is the deflection in m. If we start from zero and gradually increase the force to

Fmax, the F – x graph produced is a straight line with a gradient k. The work done is the area under the graph.

W = F x/2 so U = F x/2

F and x are the final or maximum values. If we substitute F = kx then U = kx2/2

The use of strain energy is used in many applications to determine the deflection of a structure. A solid

material will deform elastically and the stiffness may be related to the material properties. Here are some

examples with no further explanation. (Look in the tutorial on strain energy for the full theory).

DEFORMATION BY TWISTING

Consider a shaft being twisted an angle θ radian by application of torque T Nm. If T is gradually increased

from zero to a maximum, we find again that the relationship between T and θ is directly proportional and we

get a straight line graph with a gradient kt and this is the torsional stiffness in Nm/radian.

The work done is W=Tθ/2 The strain energy is U =Tθ/2 = ktθ2/2

It can be shown that U = (2/4G) x volume of the bar.

( is the maximum shear stress on the surface)


DEFORMATION BY STRETCHING

U = (2/2E ) x volume of the bar

DEFORMATION DUE TO BENDING (BEAMS)

It can be shown that dxM2EI

1U 2 but of course M may vary with x.

5. APPLICATION TO IMPACT LOADS

Consider a mass M that is dropped a height z onto a spring of stiffness k

as shown. When the mass hits the spring, the spring will deflect a

distance x before the mass stops moving down.

At the moment the spring is compressed to its maximum the force in the

spring is F and the strain energy is:

U = kx2/2

The potential energy given up by the falling mass is:

P.E. = Mg(z + x)

SIMPLIFIED SOLUTION

If x is small compared to the distance z then we may say P.E. = Mgz

Equating the energy lost to the strain energy gained we have

Mgz = kx2/2 Hence

k

2Mgzx

EXACT SOLUTION

Equating P.E and Strain energy we have:

Mg{z + x) = kx2/2

Rearrange into a quadratic equation

kx2 - 2Mgx – 2Mgz = 0

Solving with the quadratic equation we find:

2k

8kMgz2Mg2Mgx

2

There are two solutions and without explanation we take the result as:

2k

8kMgz2Mg2Mgx

2


SUDDENLY APPLIED LOADS

A suddenly applied load occurs when z = 0. This is not the same as a static load. Putting z = 0 yields the

result:

k

2Mg

2k

2Mg2Mgx

2

The static deflection of the spring when the mass just rests on it is xs = Weight/k = Mg/k from which it

follows that x = 2 xs

The deflection is double that of the static load.

It also follows that the instantaneous force in the spring is double the static weight.

This theory also applies to loads dropped on beams.


A mass of 1600 kg rolls onto the end of a simple cantilever bridge as shown. When the mass rests on

the end of the cantilever, the deflection is 50 mm. What is the maximum deflection when the ball first

rolls onto the end?

SOLUTION

This is a suddenly applied load so the deflection is twice the static deflection and is 100 mm.



A mass of 5 kg is slides on a rod suspended from a spring as shown. The spring stiffness is 4000 N/m.

Calculate the maximum deflection of the spring when mass is dropped from a height of 0.3 m onto

collar at the end. Calculate the deflection of the spring when the mass comes to a rest.

SOLUTION

mm 99or m 0.099x

4000 x 2

0.3 x 9.81 x 5 x 4000 x 89.81 x 5 x 29.81 x 5 x 2

2k

8kMgz2Mg2Mgx

22

Static deflection xs = Mg/k = 5 x 9.81/4000 = 0.012 m or 12 mm.



1. A spring loaded platform supports a mass of 150 kg as shown and the platform deflects 20 mm from its

normal position. If the same mass is dropped onto the platform from a height of 200 mm what will be

the maximum deflection of the platform?

(Answers 111 mm)

2. A beam is placed across a span as shown. When a force of 20 kN is applied at the centre, it deflects 3

mm. Calculate the maximum deflection when a mass of 5000 kg is dropped from a height of 10 mm

onto the middle and the deflection when the mass rests on the beam.

(Answers 22 mm and 7.36 mm)

Documents

CITY AND GUILDS 9210 Unit 130 MECHANICS OF · PDF fileAll engineering materials expand when heated and this ... F = 30 000 = aAa + bAb 30 000 ... x = b L = 3.248 x 10-6 x 1.5 = 4.872