How to Calculate and Solve for Flexural Strength for Rectangular Cross-section in Defects | Ceramics

The image above represents flexural strength for rectangular cross-section in defects.

To compute for flexural strength for rectangular cross-section in defect, four essential parameters are needed and these parameters are Load at fracture (Ff), Length of the cross-section (b), Width of the cross-section (d) and Distance between support points (L).

The formula for calculating flexural strength for rectangular cross-section in defects:

σfs = 3FfL / 2bd²

Where:

σfs = Flexural Strength
L = Distance between Support Points
b = Length of Cross-section
d = Width of Cross-section

Let’s solve an example;
Find the flexural strength when the load at fracture is 10, the length of cross-section is 7, width of cross-section is 9 and the distance between support points is 21.

This implies that;

Ff = Load at Fracture = 10
L = Distance between Support Points = 21
b = Length of Cross-section = 7
d = Width of Cross-section = 9

σfs = 3FfL / 2bd²
σfr = 3(10)(21) / 2(7)(9)²
σfr = (630) / 2(7)(81)
σfr = (630) / (1134)
σfr = 0.55

Therefore, the flexural strength for rectangular cross-section in defect is 0.55 Pa.

Calculating the Distance Between Support Points when the Flexural Strength for Rectangular Cross-Section in Defects, Load at Fracture, length of the cross-section and width of cross-section is Given.

L = σfs x 2bd2 / 3Ff

Where;

L = Distance between Support Points
σfs = Flexural Strength
b = Length of Cross-section
d = Width of Cross-section

Let’s solve an example;
Find the distance between support points when the flexural strength is 20, the load at fracture is 12, the length of cross-section is 7 and the width of cross-section is 9.

This implies that;

σfs = Flexural Strength = 20
Ff = Load at Fracture = 12
b = Length of Cross-section = 7
d = Width of Cross-section = 9

L = σfs x 2bd2 / 3Ff
L = 20 x 2 x 7 x 92 / 3 x 12
L = 280 x 81 / 36
L = 22680 / 36
L = 630

Therefore, the distance between support points is 630.

How to Calculate and Solve for Flexural Strength for Circular Cross-section in Defects | Ceramics

The image above represents flexural strength for circular cross-section in defects.

To compute for flexural strength for circular cross-section in defects, three essential parameters are needed and these parameters are Load at fracture (Ff), Specimen radius (R) and Distance between support Points (L).

The formula for calculating flexural strength for circular cross-section in defects:

σfs = FfL / πR³

Where:

σfs = Flexural Strength
L = Distance between Support Points

Lets’s solve an example;
Find the flexural strength when the distance between support points is 30, load at fracture is 21 and the specimen radius is 11.

This implies that;

L = Distance between Support Points = 30
Ff = Load at Fracture = 21
R = Specimen Radius = 11

σfs = FfL / πR³
σfs = (21)(30) / π(11)³
σfs = (630) / π(1331)
σfs = (630) / (4181.4)
σfs = 0.150

Therefore, the flexural strength for circular cross-section is 0.150 Pa.

Calculating the Distance Between Support Points when the Flexural Strength for Circular Cross-section, the Specimen Radius and the Load at Fracture is Given.

L = σfs x πR³ / Ff

Where;

L = Distance between Support Points
σfs = Flexural Strength

Let’s solve an example;
Find the distance between support points when the flexural strength is 44, load at fracture is 3 and specimen radius is 20.

This implies that;

σfs = Flexural Strength = 44
Ff = Load at Fracture = 3
R = Specimen Radius = 20

L = σfs x πR³ / Ff
L = 44 x π x 3³ / 20
L = 44 x π x 27 / 20
L = 3732.2 / 20
L = 186.61

Therefore, the distance between support point is 186.61.

How to Calculate and Solve for Flexural Strength with Relation to Volume Fraction | Ceramics

The image above represents Flexural strength with relation to volume.

To compute for flexural strength with relation to volume, three essential parameters are needed and these parameters are Initial Stress (σo), Constant (n) and Volume fraction porosity (P).

The formula for calculating flexural strength with relation to volume:

σfs = σo exp (-nP)

Where:

σfs = Flexural Strength
σo = Initial Stress
n = Constant
P = Volume Fraction Porosity

Let’s solve an example;
Find the flexural strength when the initial stress is 11, the constant is 8 and the volume fraction porosity is 22.

This implies that;

σo = Initial Stress = 11
n = Constant = 8
P = Volume Fraction Porosity = 22

σfs = σo exp (-nP)
σfr = (11)exp(-(8)(22))
σfr = (11)exp(-176)
σfr = (11)(3.665e-77)
σfr = 4.032e-76

Therefore, the flexural strength is 4.032e-76 Pa.

Calculating the Initial Stress when the Flexural Strength, the Constant and the Volume Fraction Porosity is Given.

σo = σfr / exp (-nP)

Where;

σo = Initial Stress
σfs = Flexural Strength
n = Constant
P = Volume Fraction Porosity

Let’s solve an example;
Find the initial stress when the flexural strength is 20, the constant is 10 and the volume fraction porosity is 8.

This implies that;

σfs = Flexural Strength = 20
n = Constant = 10
P = Volume Fraction Porosity = 8

σo = σfr / exp (-nP)
σo = 20 / exp (-10 x 8)
σo = 20 / exp (-80)
σo = 20 / – 5.54e+34
σo = – 3.61e-34

Therefore, the initial stress is 3.61e-34.

How to Calculate and Solve for Relationship between Modulus of Elasticity on Volume Fraction Porosity, E | Ceramics

The image above represents modulus of elasticity.

To compute for the relationship between modulus of elasticity on volume fraction porosity, two essential parameters are needed and these are Modulus of Elasticity of Non Porous Material (Eo) and Volume Fraction Porosity (P).

The formula for calculating the modulus of elasticity:

E = Eo(1 – 1.9P + 0.9P²)

Where:

E = Modulus of Elasticity
Eo = Modulus of Elasticity of Non Porous Material.
P = Volume Fraction Porosity

Let’s solve an example;
Find the modulus of elasticity when the modulus of elasticity of non porous material is 12 and volume fraction porosity is 22.

This implies that;

Eo = Modulus of Elasticity of Non Porous Material = 12
P = Volume Fraction Porosity = 22

E = Eo(1 – 1.9P + 0.9P²)
E = (12)(1 – 1.9(22) + 0.9(22)²)
E = (12)(1 – (41.8) + 0.9(484))
E = (12)(1 – (41.8) + (435.6))
E = (12)(394.8)
E = 4737.6

Therefore, the modulus of elasticity is 4737.6 Pa.

How to Calculate and Solve for Viscosity | Ceramics

The image above represents viscosity.

To compute for viscosity, three essential parameters are needed and these parameters are Force applied (F), Area (A) and Derivation Ratio of Velocity to Distance of Fluid Flow (dv/dy).

The formula for calculating viscosity:

η = F/A / dv/dy

Where:

η = Viscosity
F = Force Applied
A = Area
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow

Let’s solve an example;
Find the viscosity when the force applied is 21, area is 14 and derivation ratio of velocity to distance of fluid flow is 19.

This implies that;

F = Force Applied = 21
A = Area = 14
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow = 19

η = F/A / dv/dy
η = (21/14) / (19)
η = (1.5) / (19)
η = 0.0789

Therefore, the viscosity is 0.0789 Pa s.

Calculating Force Applied when the Viscosity, the Area and the Derivation ratio of velocity to distance of fluid flow is Given.

F = (η x dv/dy) A

Where;

F = Force Applied
η = Viscosity
A = Area
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow

Let’s solve an example;
Find the force applied when the viscosity is 20, the area is 30 and the derivation is 8.

This implies that;

η = Viscosity = 20
A = Area = 30
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow = 8

F = (η x dv/dy) A
F = (20 x 8) 30
F = (160) 30
F = 4800

Therefore, the force applied is 4800.

How to Calculate and Solve for Normally Occupied Positions | Ceramics

The image above represents normally occupied positions.

To compute for normally occupied positions, four essential parameters are needed and these parameters are Avogadro’s number (NA), Density (ρ), Atomic weight (AK) and Atomic weights (AG).

The formula for calculating normally occupied positions:

N = NAρ / (AK + AG)

N = Normally Occupied Positions
AK, AG = Atomic Weights
ρ = Density

Let’s solve an example;
Find the normally occupied positions when the avogadro’s number is 6.022e+23, atomic weight is 12, atomic weight is 16 and the density is 10.

This implies that;

NA = Avogadro’s Number = 6.022e+23
AK, AG = Atomic Weights = 12, 16
ρ = Density = 10

N = NAρ / (AK + AG)
N = (6.022e+23)(10)/(12 + 16)
N = (6.022e+24)/(28)
N = 2.15e+23

Therefore, the normally occupied positions is 2.15e+23.

Calculating the Density when the Normally Occupied Positions, the Avogadro’s Number, the Atomic Weights is Given.

ρ = N (AK + AG) / NA

Where;

ρ = Density
N = Normally Occupied Positions
AK, AG = Atomic Weights

Let’s solve an example;
Given that normally occupied positions is 20, the avogadro’s number is 6.022e+23, atomic weights is 18, 6.

This implies that;

N = Normally Occupied Positions =20
NA = Avogadro’s Number = 6.022e+23
AK, AG = Atomic Weights = 18, 6

ρ = N (AK + AG) / NA
ρ = 20 (18 + 6) / 6.022e+23
ρ = 20 (24) / 6.022e+23
ρ = 480 / 6.022e+23
ρ = 7.970e-22

Therefore, the density is 7.970e-22.

How to Calculate and Solve for Schottky Defect | Ceramics

The image above represents schottky defect.

To compute for schottky defect, four essential parameters are needed and these parameters are N, Activation energy (Qs), Boltzmann’s Constant (K) and Temperature (T).

The formula for calculating schottky defect:

Ns = N exp (-Qs/2KT)

Where:

Qs = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the schottky defect when the activation energy is 44, N is 22, boltzmann’s constant is 1.38064852E-23 and the temperature is 30.

This implies that;

N = 22
Qs = Activation Energy = 44
K = Boltzmann’s Constant = 1.38054852E-23
T = Temperature = 30

Ns = N exp (-Qs/2KT)
Ns = (22)exp(-(44)/2(1.38064852e-23)(30))
Ns = (22)exp((-44)/(8.283891119e-22))
Ns = (22)exp(-5.3115135583771414e+22)
Ns = (22)(0)
Ns = 0

Therefore, the schottky defect is 0.

Calculating the N when the Schottky Defect, the Activation Energy, the Boltzmann’s Constant and the Temperature is Given.

N = Ns / e (-Qs / 2KT)

Where;

Ns = Schottky Defect
Qs = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the N when the schottky defect is 40, the activation energy is 24, the boltzmann’s constant is 1.38064852E-23 and the temperature is 10.

This implies that;

Ns = Schottky Defect = 40
Qs = Activation Energy = 24
K = Boltzmann’s Constant = 1.38064852E-23
T = Temperature = 10

N = Ns / e (-Qs / 2KT)
N = 40 / e (-24 / 2 x 1.38064852E-23 x 10)
N = 40 / e (-24 / 2.76129704E+23)
N = 40 / e (8.691567e-23)
N = 40 / 8.691567e+23
N = 4.602e-23

Therefore, the is 4.602e-23.

How to Calculate and Solve for Frenkel Defect | Ceramics

The image above represents frenkel defect.

To compute for frenkel defect, four essential parameters are needed and these parameters are N, activation energy (Qfr), Boltzmann’s Constant (K) and temperature (T).

The formula for calculating the frenkel defect:

Nfr = N exp (-Qfr / 2KT)

Where:

Nfr = Frenkel Defect
Qfr = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the frenkel defect when the activation energy is 34, N is 22, Temperature is 12 and the boltzmann’s constant is 1.38064852e-23.

This implies that;

Qfr = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Nfr = N exp (-Qfr / 2KT)
Nfr = (22)exp(-(34) / 2(1.38064852e-23)(14))
Nfr = (22)exp((-34) / (3.865815856e-22))
Nfr = (22)exp(-8.795038684325786e+22)
Nfr = (22)(0)
Nfr = 0

Therefore, the frenkel defect is 0.

Calculating the Activation Energy when the Frenkel Defect, the Boltzmann’s Constant and the Temperature is Given.

Qfr = – (In (Nfr / N) x 2KT)

Where;

Qfr = Activation Energy
Nfr = Frenkel Defect
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Given that the frenkel defect is 20, the boltzmann’s constant is 5, the temperature is 2 and N is 10. Find the activation energy?

This implies that;

Nfr = Frenkel Defect = 20
K = Boltzmann’s Constant = 5
T = Temperature = 2
N = 10

Qfr = – (In (Nfr / N) x 2KT)
Qfr = – (In (20 / 10) x 2 x 5 x 2)
Qfr = – (In 2 x 20)
Qfr = – (In 40)
Qfr = – 3.688

Therefore, the activation energy is – 3.688.

How to Calculate and Solve for Theoretical Density of Ceramics | Ceramics

The image above represents theoretical density of ceramics.

To compute for theoretical density of ceramics, five essential parameters are needed and these parameters are Number of formula units in unit cell (n’), Sum of atomic weights of atoms (ΣAc), Sum of atomic weights of anions (ΣAA), Unit cell volume (Vcand Avogadro’s number (NA).

The formula for calculating theoretical density of ceramics:

ρ = n'(ΣAc + ΣAA) / VcNA

Where:

ρ = Theoretical Density of Ceramics
n’ = Number of Formula Units in Unit Cell
ΣAc = Sum of Atomic Weights of Atoms
ΣAA = Sum of Atomic Weights of Anions
Vc = Unit Cell Volume

Let’s solve an example;
Find the theoretical density of ceramics when the number of formula units in unit cell is 12, the sum of atomic weights of atoms is 8, the sum of atomic weights of anions is 10, the unit cell volume is 9 and the avogadro’s number is 6.02214e+23.

This implies that;

n’ = Number of Formula Units in Unit Cell = 12
ΣAc = Sum of Atomic Weights of Atoms = 8
ΣAA = Sum of Atomic Weights of Anions = 10
Vc = Unit Cell Volume = 9
NA = Avogadro’s Number = 6.02214e+23

ρ = n'(ΣAc + ΣAA) / VcNA
ρ = (12)(8 + 10) / (9)(6.02214e+23)
ρ = (12)(18) / (5.4199e+24)
ρ = (216)/(5.4199e+24)
ρ = 3.98e-23

Therefore, the theoretical density of ceramics is 3.98e-23.