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Deformation of a Component of Two Materials

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Prequisite Knowledge
Deformation Due To Axial Loading

Problem
An axial force of 110 kN is applied to the component by means of rigid end plates as shown. Determine (a) the normal stress in the steel shell and (b) the corresponding deformation of the assembly.



Solution
(This is known as a statically indeterminate problem)

Part (a), stress

The force applied to the plate is being distributed to the steel outer shell and the brass inner core. Therefore it can be written:

(Eq1)     P = P₁ + P₂

where P is the force shown in the diagram, P₁ is the force experienced by the steel outer shell and P₂ is the force experienced by the brass inner core.

(note: for the duration of this problem subscript 1 will refer to the steel outer shell and the subscript 2 will refer to the brass inner core.)

Given the way the plate is pushing on both the steel and brass, it can be said that the deformation in the steel and brass would be the same. That is, the plate will squeeze the steel shell and the brass core by equal amounts.

This means that:

δ₁ = δ₂

and because

δ =

PL AE

(Eq2)

P1L1 A1E1 = P2L2 A2E2

Re-writing Eq2 for P₂ and substituting into Eq1,

P = P1 +

P1A2E2 A1E1

A1 =

π(0.06 m)2 4

–

π(0.035 m)2 4

= 0.0018653 m2

A2 =

π(0.035 m)2 4

= 0.000962113 m2

then substituting the known values:

110 kN = P1 +

P1(0.000962113 m2)*(105 GPa) (0.0018653 m2)*(200 GPa)

110 kN = P₁ + P₁(0.2708)

110 kN = (1.2708)P₁

P₁ = 86.56 kN

the stress σ₁ is then:

σ1 =

P1 A1

=

86.56 kN 0.0018653 m2

σ = 46.405 MPa

Part (b), elongation

recall that the elongation is the same for both materials and also the elongation is:

δ =

PL AE

then:

δ =

PL AE

and

δ =

P1L A1E1

=

P2L A2E2

so:

δ =	(86560 N)*(2 m) (0.0018653 m2)*(200*109 Pa)

δ = 0.000464 m