CHAPTER 2THERMAL STRESSES AND

COMPOSITE BARS

Learning OutcomesAt the end of this lecture, the student should

be able to;

1)Understand thermal stresses and composite bars.

2)Understand series and parallel bars.3)Solve problem regarding thermal stresses

and composite bars.

Composite BarsThe composite bar consists of 2 or more

different material and size which bonded together (series or parallel)

When subjected to temperature and load, the composite bar can extent or compress together.

Example:

Composite Bars Subjected to Common Extension (without temperature)

i) Series Bar

Concept of Load, PP1 = P2 (Sum of external load distributed evenly

to all bars)

Concept of Extension/Compression, ∆L∆L = ∆L1 + ∆L2 (Sum of extension/compression of all bars are added)

Bar 1 Bar 2 p p

∆L = PL AE

Law of Equilibrium ForcesWhen all the forces that act upon an object

are balanced, then the object is said to be in a state of equilibrium.

The forces are considered to be balanced if the rightward forces are balanced by the leftward forces and the upward forces are balanced by the downward forces.

This however does not necessarily mean that all the forces are equal to each other.

∑ P Right = ∑ P Left @ ∑ P up = ∑ P down

Example:Consider the two objects pictured in the force

diagram shown below. Note that the two objects are at equilibrium because the forces that act upon them are balanced; however, the individual forces are not equal to each other. The 50 N force is not equal to the 30 N force.

Exercise 1A rod AB is fixed to a wall with a supporter.

The area and length for bar A is 300mm2 and 5m while bar B is 150 mm2 and 3m.

i)By using the law of equilibrium force, calculate the P2.

ii)Calculate the sum of extension in the bar. (Given EA = 210 GPa, EB = 90 GPa)

P2

5m 3m

B A

P3 P1

ii) Parallel Bar

Concept of Load, PP = P1 + P2 (Sum of external load distributed to

each bars)Concept of extension/compression, ∆L∆L1 = ∆L2 (Sum of extension/compression are

same) @

P1L1 = P2L2 @ σ1L1 = σ2L2A1E1 A2E2 E1 E2

Bar 1

Bar 2

P = Load

∆L = PL = σL

AE E

Exercise 2An external load of 10KN were hang on two

wires made from brass and steel. The area of brass is 150mm2 and steel is 80mm2 and length for both wires is considered same. Determine each load for both wires. Given ES= 200 GPa , EB = 100 GPa.

P = 10 KN

Composite Bars Subjected to Temperature Changes

When a material is subjected to a change in temperature its length will change by an amount:

ΔL = αLΔT

Where ;ΔL = amount of expansion/compression (m)α = linear coefficient of expansion (°C-1 )ΔT = temperature changes (°C)L = length (m)

Composite Bars Subjected to Temperature Changes

Ordinary materials expand when heated and contract when cooled, hence , an increase in temperature produce a positive thermal strain.

Thermal Strain, ε = ΔL/L = αΔT

Thermal strains usually are reversible in a sense that the material returns to its original shape when the temperature return to its original value. (movie)

However, if the bar were not constrained, so that it can expand freely, there will be no thermal stress.

If, however, the bar were prevented from expanding then there would be a compressive stress in the

bar.

Thermal stress, σ = E αΔT

Constrained bar

Assumed to expand freely

Steel Brass

BrassSteel P

Example: Parallel barConsider now a composite bar constructed

from two different materials rigidly joined together .

Let us consider that the materials in this case are steel and brass and both materials are applied with temperature and stress.

In general, the coefficients of expansion of the two materials forming the composite bar will be different.

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

Figure below shows the positions to which the individual materials will expand if they are completely free to expand (i.e not joined rigidly together as a compound bar).

The extension of any Length, L is given by;

ΔL = αLΔT

• The difference of free expansion lengths is termed ‘free length’ and equals to:

Free length = αB LB ΔT - αS LS ΔT

= (αB - αS ) L ΔT

where αB > αS

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 effective extension of the steel from its free length position.

Therefore, from the diagrams, it may conclude as follow;

Conclusion 1Extension of steel + compression brass = difference in

“free length”

ΔLS + ΔLB (due to load) = (αB - αS ) L ΔT (due to temp.)

σS LS / Es + σB LB / EB = (αB - αS ) L ΔT

Because L = LB = LS therefore,

σS / Es + σS / Es = (αB - αS )ΔT

Conclusion 2

The tensile force applied to the short member by the long member is equal in magnitude to the compressive force applied to long member by the short member.

Tensile force in steel = Compressive force in brass

σS AS = σB AB

These are two equations with two unknowns which can be solved simultaneously to obtain σSteel and

σBrass

Comparison with Series and Parallel Composite BarSubjected to Temperature Changes

i) Series Bar

ΔL (due to load) = ΔL (due to temp)

σ1 L1 / E1 + σ2 L2 / E2 + …. = α1L1ΔT + α2L2ΔT + ….

ii) Parallel Bar

ΔL (due to load) = ΔL (due to temp)

σS / Es + σS / Es = (αB - αS )ΔT