Materials are also subjected to compressive stress which causes deformation in two or three dimension of the material. To help you visualize this concept, imagine a sponge being squeezed from all dimensions. We will see that there is a change in volume.
2. BULK MODULUS
β’ Bulk modulus, a numerical constant that
describes the elastic properties of
a solid or fluid when it is under pressure on all
surfaces. The applied pressure reduces the
volume of a material, which returns to its
original volume when the pressure is removed.
Sometimes referred to as the incompressibility,
the bulk modulus is a measure of the ability of
a substance to withstand changes in volume
when under compression on all sides. It is
equal to the quotient of the applied pressure
divided by the relative deformation.
definition
7. BULK MODULUS
Compressibility is a measure of the relative
volume change of a solid or a fluid in response
to a pressure change.
The inverse of bulk modulus is called
compressibility.
8. BULK MODULUS
π =
1
π΅
= β
βπ
π
π
βπ
= β
1
π
π
β
βπ
βπ
(πππππππ π ππππππ‘π¦)
The inverse of bulk modulus is called
compressibility.
9. BULK MODULUS
A hydraulic press contains 0.25π3
(250L) of oil.
Find the decrease in volume when it is subjected to
an increase in pressure βπ = 1.6 Γ 102
ππ (about
160atm or 2300psi). The bulk modulus of the oil is
π΅ = 5.0 Γ 109
ππ (about 5.0 Γ 104
ππ‘π), and its
compressibility is π =
1
π΅
= 20 Γ 10β6
ππ‘πβ1
.
Problem solving
10. BULK MODULUS
β’ We need to find the decrease in volume βπ .
Identifying the target
11. BULK MODULUS
β’ What do we have in hand?
β’ βπ = 1.6 Γ 102
ππ
β’ π΅ = 5.0 Γ 109
ππ
β’ π =
1
π΅
= 20 Γ 10β6
ππ‘πβ1
.
β’ π
π = 0.25π3
(250L)
data
12. BULK MODULUS β’Execute
β’We derive a formula for βπ.
β’
π΅βπ
ππ
= β
βπ
1
β’βπ = β
βπ
1
ππ
π΅
solution
13. BULK MODULUS
β’ βπ = β
1.6Γ102ππ
1
0.25π3
5.0Γ109ππ
= β8.0 Γ 10β4
π3
ππ β .80πΏ
β’ Alternatively we can use the concept of compressibility
which will give us,
β’ π =
1
π΅
= β
βπ
ππ
βπ
= β
1
ππ
β
βπ
βπ
β’ So βπ = βππ
πβπ
β’ βπ = βππ
πβπ
β’ βπ = β 20 Γ 10β6
ππ‘πβ1
0.25π3
1.6 Γ 102
ππ =
β 8.0 Γ 10β4
π3
ππ β .80πΏ
solution