Lecture 2-thermal Strain

Thermal Strain

When the temperature of a component is increased or decreased the material respectively expands or contracts.

Thermal Expansion, (or the increase in length)

for

most

materials,

results

from an increase in temperature. The extent of the expansion (∆ L) depends on the temperature change (∆ T), the length of the part (L0), and the coefficient of thermal expansion of the material ∆ L = α Linvolved. (linear 0 ∆ T

expansion)

Typical values of coefficient of linear expansion Carbon Steel 12 x 10-6 /0C Aluminium

24 x 10-6 /0C

Copper

17 x 10-6 /0C

Cast Iron

10 x 10-6 /0C

Brass

16 x 10-6 /0C

Bronze

18 x 10-6 /0C

Thermal Strain The thermal strain (εT) is given by: εT = ∆ L / L0

εT = α ∆ T If this expansion or contraction is not resisted in any way then the processes take place free of stress.

If the changes in dimensions are restricted, then stresses termed temperature stresses will

be

material.

set

up

within

the

Thermal Stress The thermal stress (FT) is given by:

FT = εE where E – modulus

Elastic (a)Bar of initial length L; (b)Elongation ∆ L due to heat; (c)Thermal stress in

For both

Large

temperature

α

and

E

changes,

vary

with

temperature. We

shall

temperature

assume

that

changes

the are

sufficiently small for α and E to be considered as constants.

Both the thermal strain α ∆ T and the elastic strain F/E may exist together.

The total strain is the

sum of the two:

α ∆ T + F/E

Thermal Stress In (c), when the bar is subjected to change in temperature, the total strain is zero, i.e. the elastic strain must be equal and opposite to the thermal strain, thus

(a)Bar of initial length L; (b)Elongation ∆ L due to heat; (c)Thermal stress in

Differential thermal expansion Results

whenever

there

is

a

temperature difference or gradient from point to point in metals. The differential occurs because most metals expand with

increasing

temperature.

If

the

increase (or decrease) in temperature is different material,

in

different

the

sections

sections

will

expanded to a different extent.

of

a

have

Differential thermal expansion In this case, the compressive and tensile stresses can result in the bending of the part, as shown below:

Compound Bars

In certain applications it is necessary to use a combination of elements or bars made from different materials, each material

performing

a

different

function, such as electric cables.

Example: In overhead electric cables, for example, it is often convenient to carry the current in a set of copper wires surrounding steel wires,

the

latter

being

designed

to

support the weight of the cable over large spans.

When an external load W is applied to such

a

between

compound the

bar

individual

it

is

shared

component

materials in proportions depending on

When two or more members are rigidly fixed together so that they share the same load and extend or compress the same amount, the members form a compound bar. The stresses in each member are calculated using the following: 1.The total load is the sum of the loads taken by each member. 2.The total load taken by each member is given by the product of its stress and its area. 3.The extension or contraction is the same

Each member carries a portion of the total load W proportional to its EA/L value. Force in member 1 :

If all members are of equal length the stress of one member (F1) is given by:

Stress in member 1:

If the compound bar is subjected to a temperature rise each material will attempt to expand by different amount.

Hea t

If the compound bar is subjected to a temperature rise each material will attempt to expand by different amount.

The two materials are now rigidly joined as a compound bar and subjected to the same temperature rise, each material will attempt to expand to its free length position but each will be affected by the movement of the other. The higher coefficient of expansion material (brass) will therefore seek to pull the steel up to its free length position and conversely the lower coefficient of expansion material (steel) will try to hold the brass back to its free length position. The result is an effective compression of the brass from its free length position and an

The tensile force applied to the short member by the long member is equal in magnitude to the compressive force applied to the long member by the short member. Tensile force in steel = compressive force in brass σsteel Asteel = σbrass Abrass These are two equations with two unknowns which can be solved simultaneously to obtain σsteel and σbrass .