How to Calculate and Solve for Modulus of Rupture | Fracture Mechanics

The image above represents modulus of rupture.

To compute for modulus of rupture, three essential parameters are needed and these parameters are Bending Moment (M), Distance (c) and Moment of Inertia (I).

The formula for calculating modulus of rapture:

σ = Mc / I

Where:

σ = Modulus of Rupture
M = Bending Moment
c = Distance
I = Moment of Inertia

Let’s solve an example;
Find the modulus of rupture when the bending moment is 22, the distance is 10 and the moment of inertia is 14.

This implies that;

M = Bending Moment = 22
c = Distance = 10
I = Moment of Inertia = 14

σ = Mc / I
σ = 22(10) / 14
σ = 220 / 14
σ = 15.71

Therefore, the modulus of rupture is 15.71 Pa.

Calculating the Bending Moment when the Modulus of Rupture, the Distance and the Moment of Inertia is Given.

M = σ x I / c

Where;

M = Bending Moment
σ = Modulus of Rupture
c = Distance
I = Moment of Inertia

Let’s solve an example;
Find the bending moment when the modulus of rupture is 22, the distance is 8 and the moment of inertia is 2.

This implies that;

σ = Modulus of Rupture = 22
c = Distance = 8
I = Moment of Inertia = 2

M = σ x I / c
M = 22 x 2 / 8
M = 44 / 8
M = 5.5

Therefore, the bending moment is 5.5.

How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image above represents the maximum velocity to avoid overturning of a vehicle moving along a level circular path.

To compute for the maximum velocity, four essential parameters are needed and these parameters are Acceleration due to Gravity (g), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the maximum velocity:

vmax = √(gra / h)

Where:
vmax = Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path
g = Acceleration due to Gravity
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the maximum velocity when the Acceleration due to Gravity (g) is 10.2, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 14, Radius of Circular Path (r) is 22 and Half of the Distance between the Centre Lines of the Wheel (a) is 32.

This implies that;
g = Acceleration due to Gravity = 10.2
h = Height of Centre of Gravity of the Vehicle from Ground Level = 14
r = Radius of Circular Path = 22
a = Half of the Distance between the Centre Lines of the Wheel = 32

vmax = √(gra / h)
vmax = √((10.2)(22)(32)/14)
vmax = √((7180.79)/14)
vmax = √(512.91)
vmax = 22.647

Therefore, the maximum velocity to avoid Overturning of a Vehicle moving along a Level Circular Path is 22.647 m/s.

How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image represents reaction at the inner wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the inner wheel of a vehicle moving along a level circular path:

RA = mg / 2[1 – v²h / gra]

Where:
RA = Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 13, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 11, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 5, Radius of Circular Path (r) is 7 and Half of the Distance between the Centre Lines of the Wheel (a) is 3.

This implies that;
m = Mass of the Vechicle = 13
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 11
h = Height of Centre of Gravity of the Vehicle from Ground Level = 5
r = Radius of Circular Path = 7
a = Half of the Distance between the Centre Lines of the Wheel = 3

RA = mg / 2[1 – v²h / gra]
RA = 13(9.8) / 2[1 – (11)²(5) / (9.8)(7)(3)]
RA = 127.4 / 2[1 – (121)(5) / 205.8]
RA = 63.7[1 – 605 / 205.8]
RA = 63.7[1 – 2.939]
RA = 63.7[-1.939]
RA = -123.56

Therefore, the reaction at the inner wheel of a vehicle moving along a level of circular path is -123.56 N.

How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

The image represents reaction at the outer wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the outer wheel of a vehicle moving along a level circular path:

RB = mg / 2[1 + v²h/gra]

Where;
RB = Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 12, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 28, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 16, Radius of Circular Path (r) is 8 and Half of the Distance between the Centre Lines of the Wheel (a) is 4.

This implies that;
m = Mass of the Vechicle = 12
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 28
h = Height of Centre of Gravity of the Vehicle from Ground Level = 16
r = Radius of Circular Path = 8
a = Half of the Distance between the Centre Lines of the Wheel = 4

RB = mg / 2[1 + v²h/gra]
RB = 12 x 9.8 / 2[1 + 28² x 16/9.8 x 8 x 4]
RB = 117.60 / 2[1 + 784 x 16/313.6]
RB = 58.80[1 + 40]
RB = 58.80[41]
RB = 58.80[41]
RB = 2410.8

Therefore, the reaction of the outer wheel of a vehicle moving along a level circular path is 2410.8 N.