Stress Formula Derivations and Worked Examples
Stress, a fundamental concept in mechanical engineering and materials science, quantifies the internal forces that molecules within a continuous material exert on each other. It arises when external forces act on a deformable body, causing internal resistance to deformation. Understanding stress and its associated formulas is crucial for designing safe and reliable structures and machines. This article provides a comprehensive overview of stress formulas, their derivations, and practical applications with worked examples, targeting engineering students, practicing engineers, and researchers.
Normal Stress: Axial Loading
Normal stress, often denoted by σ (sigma), is a measure of the force acting perpendicular to a surface area. It is commonly associated with axial loading, where a force is applied along the longitudinal axis of a member, like a rod or a column.
The fundamental formula for normal stress is:
σ = F/A
Where: σ = Normal stress (typically in Pascals (Pa) or pounds per square inch (psi))
F = Applied force (in Newtons (N) or pounds (lb))
A = Cross-sectional area (in square meters (m²) or square inches (in²))
Derivation
Imagine a slender rod subjected to a tensile force F. Internally, the material resists this force by generating an equal and opposite force distributed across the cross-sectional area. The stress is simply the intensity of this internal resistive force per unit area. Therefore, if we divide the applied force by the area it acts upon, we get the average normal stress. This assumes the force is uniformly distributed over the area.
Worked Example 1: Tension in a Steel Rod
A steel rod with a diameter of 20 mm is subjected to a tensile force of 50 k N. Calculate the normal stress in the rod.
Given: Diameter, d = 20 mm = 0.02 m
Force, F = 50 k N = 50,000 N
Solution: Area, A = πr² = π(d/2)² = π(0.02/2)² = π(0.01)² =
3.1416 x 10⁻⁴ m²
Stress, σ = F/A = 50,000 N / (3.1416 x 10⁻⁴ m²) =
159.15 x 10⁶ Pa =
159.15 MPa
Worked Example 2: Compression in a Concrete Column
A concrete column with a square cross-section of 300 mm x 300 mm is subjected to a compressive load of 500 k N. Calculate the normal stress in the column.
Given: Side of square, s = 300 mm = 0.3 m
Force, F = 500 k N = 500,000 N
Solution: Area, A = s² = (0.3 m)² =
0.09 m²
Stress, σ = F/A = 500,000 N / 0.09 m² =
5.56 x 10⁶ Pa =
5.56 MPa
Important Note: The normal stress calculated here is an average stress. Local stress concentrations can occur near holes, corners, or other geometric discontinuities, which significantly increase the actual stress at those locations. Stress concentration factors are used to account for these increases.
Shear Stress: Tangential Loading
Shear stress, denoted by τ (tau), is a measure of the force acting parallel or tangential to a surface area. It arises from forces that tend to cause one part of a material to slide or shear relative to another part. Common examples include forces on rivets, bolts, or adhesives.
The basic formula for shear stress is:
τ = V/A
Where: τ = Shear stress (typically in Pascals (Pa) or pounds per square inch (psi))
V = Shear force (in Newtons (N) or pounds (lb))
A = Area parallel to the force (in square meters (m²) or square inches (in²))
Derivation
Similar to normal stress, shear stress represents the intensity of the internal resistive force. However, in this case, the force is parallel to the area. Imagine two plates glued together, subjected to a force trying to slide them apart. The adhesive resists this force by generating an equal and opposite shear force distributed over the bonded area. The shear stress is this force divided by the area.
Worked Example 1: Shear Stress in a Rivet
Two steel plates are joined by a single rivet with a diameter of 10 mm. If the plates are subjected to a tensile force of 15 k N, calculate the shear stress in the rivet. Assume the rivet is in single shear.
Given: Diameter, d = 10 mm = 0.01 m
Force, V = 15 k N = 15,000 N
Solution: Area, A = πr² = π(d/2)² = π(0.01/2)² = π(0.005)² =
7.854 x 10⁻⁵ m²
Shear Stress, τ = V/A = 15,000 N / (7.854 x 10⁻⁵ m²) =
191.0 x 10⁶ Pa =
191.0 MPa
Worked Example 2: Shear Stress in a Double Shear Connection
Now, consider the same two steel plates, but joined by a pin indouble shear. This means the shear force is distributed acrosstwocross-sections of the pin.
Given: Diameter, d = 10 mm = 0.01 m
Force, V = 15 k N = 15,000 N
Solution: Area per shear plane, A = πr² = π(d/2)² = π(0.01/2)² = π(0.005)² =
7.854 x 10⁻⁵ m²
Total shear area = 2A = 2 7.854 x 10⁻⁵ m² =
1.5708 x 10⁻⁴ m²
Shear Stress, τ = V / (2A) = 15,000 N / (1.5708 x 10⁻⁴ m²) =
95.49 x 10⁶ Pa =
95.49 MPa
Notice that the shear stress in the double shear connection is half of that in the single shear connection for the same applied force and pin diameter.
Common Pitfall: Forgetting to account for the number of shear planes in bolted or pinned connections. Always carefully examine the connection to determine how many areas are resisting the shear force.
Bending Stress in Beams
Beams are structural elements designed to resist bending moments. Bending stress, a form of normal stress, arises within a beam due to the applied bending moment. It varies linearly across the beam's cross-section, with maximum tensile stress on one side and maximum compressive stress on the other.
The bending stress formula is:
σ = My/I
Where: σ = Bending stress (typically in Pascals (Pa) or pounds per square inch (psi))
M = Bending moment at the section (in Newton-meters (N·m) or pound-inches (lb·in))
y = Distance from the neutral axis to the point where stress is calculated (in meters (m) or inches (in))
I = Second moment of area (area moment of inertia) about the neutral axis (in meters to the fourth power (m⁴) or inches to the fourth power (in⁴))
Derivation
The derivation of this formula relies on several assumptions: the beam is linearly elastic, the material is homogeneous and isotropic, and plane sections remain plane during bending (Bernoulli-Euler beam theory). The derivation involves relating the curvature of the beam to the applied moment and then relating the strain to the curvature. Finally, using Hooke's Law (σ = Eε), where E is the Young's modulus, the bending stress is related to the bending moment.
Worked Example: Bending Stress in a Rectangular Beam
A rectangular beam with a width of 50 mm and a height of 100 mm is subjected to a bending moment of 2 k N·m. Calculate the maximum bending stress in the beam.
Given: Width, b = 50 mm = 0.05 m
Height, h = 100 mm = 0.1 m
Moment, M = 2 k N·m = 2000 N·m
Solution: Second moment of area (for a rectangle about its neutral axis), I = (bh³)/12 = (0.05 m (0.1 m)³)/12 =
4.167 x 10⁻⁶ m⁴
Maximum distance from the neutral axis, y = h/2 = 0.1 m / 2 =
0.05 m
Stress, σ = My/I = (2000 N·m 0.05 m) / (4.167 x 10⁻⁶ m⁴) =
24.0 x 10⁶ Pa =
24.0 MPa
Important Note: The maximum bending stress occurs at the outermost fibers of the beam (farthest from the neutral axis).
Torsional Shear Stress in Shafts
Torsion refers to the twisting of an object due to an applied torque. Shafts, often used in rotating machinery to transmit power, are particularly susceptible to torsional shear stress.
The torsional shear stress formula is:
τ = Tρ/J
Where: τ = Torsional shear stress (typically in Pascals (Pa) or pounds per square inch (psi))
T = Applied torque (in Newton-meters (N·m) or pound-inches (lb·in)) ρ (rho) = Radial distance from the center of the shaft to the point where stress is calculated (in meters (m) or inches (in))
J = Polar moment of inertia (in meters to the fourth power (m⁴) or inches to the fourth power (in⁴))
For a solid circular shaft: J = (πd⁴)/32, where d is the diameter.
For a hollow circular shaft: J = (π(do⁴ - di⁴))/32, where do is the outer diameter and di is the inner diameter.
Derivation
The derivation of this formula is similar to that of bending stress, relying on assumptions of linear elasticity, homogeneity, and isotropy. It relates the angle of twist to the applied torque and then relates the shear strain to the angle of twist. Finally, using the shear modulus (G) and the relationship between shear stress and shear strain (τ = Gγ), the torsional shear stress is related to the applied torque.
Worked Example: Torsional Shear Stress in a Solid Shaft
A solid steel shaft with a diameter of 50 mm is subjected to a torque of 1 k N·m. Calculate the maximum shear stress in the shaft.
Given: Diameter, d = 50 mm = 0.05 m
Torque, T = 1 k N·m = 1000 N·m
Solution: Polar moment of inertia, J = (πd⁴)/32 = (π(0.05 m)⁴)/32 =
6.136 x 10⁻⁷ m⁴
Maximum radial distance, ρ = d/2 = 0.05 m / 2 =
0.025 m
Shear Stress, τ = Tρ/J = (1000 N·m 0.025 m) / (6.136 x 10⁻⁷ m⁴) =
40.75 x 10⁶ Pa =
40.75 MPa
Important Note: The maximum torsional shear stress occurs at the outer surface of the shaft (farthest from the center).
Hoop Stress in Thin-Walled Cylinders
Thin-walled cylinders, such as pressure vessels and pipes, experience stresses due to internal pressure. Hoop stress (also called circumferential stress) acts tangentially to the circumference of the cylinder.
The hoop stress formula for a thin-walled cylinder is:
σh = (Pr)/t
Where: σh = Hoop stress (typically in Pascals (Pa) or pounds per square inch (psi))
P = Internal pressure (typically in Pascals (Pa) or pounds per square inch (psi))
r = Radius of the cylinder (in meters (m) or inches (in))
t = Wall thickness of the cylinder (in meters (m) or inches (in))
Derivation
The derivation involves considering a free-body diagram of a section of the cylinder. The internal pressure acts on the curved surface, creating a force that is balanced by the hoop stress acting over the cross-sectional area of the cylinder wall. Forthin-walledcylinders, the assumption is that the stress is uniformly distributed across the wall thickness. The hoop stress is twice the longitudinal stress in a closed-end cylinder.
Worked Example: Hoop Stress in a Pressure Vessel
A thin-walled cylindrical pressure vessel has a radius of 1 meter and a wall thickness of 10 mm. It is subjected to an internal pressure of 2 MPa. Calculate the hoop stress in the vessel.
Given: Radius, r = 1 m
Thickness, t = 10 mm = 0.01 m
Pressure, P = 2 MPa = 2 x 10⁶ Pa
Solution: Hoop Stress, σh = (Pr)/t = (2 x 10⁶ Pa 1 m) / 0.01 m = 200 x 10⁶ Pa = 200 MPa
How do you calculate hoop stress in thin-walled cylinders?
The hoop stress is calculated using the formula σh = (Pr)/t, where P is the internal pressure, r is the radius of the cylinder, and t is the wall thickness. This formula is valid for thin-walled cylinders, where the ratio of radius to thickness (r/t) is greater than 10.
What is the difference between true stress and engineering stress?
Engineering stress is calculated using the original cross-sectional area of the material, while true stress is calculated using the instantaneous cross-sectional area, which decreases during tensile deformation. True stress is a more accurate representation of the stress experienced by the material at any given instant, especially at large strains.
When should principal stress formulas be applied in design?
Principal stress formulas should be applied in design when the material is subjected to complex stress states, involving multiple normal and shear stresses acting simultaneously. Principal stresses represent the maximum and minimum normal stresses at a point and are crucial for determining the failure criteria of the material. The maximum shear stress theory and von Mises criterion utilize principal stresses to predict yielding.