Formulas for Stress Used in Machine Design

Stress analysis is a fundamental aspect of machine design. Understanding the various formulas used to calculate stress allows engineers to predict the behavior of components under load, ensuring structural integrity and preventing failures. This article provides a comprehensive overview of stress formulas commonly used in machine design, covering axial stress, bending stress, shear stress, torsional stress, combined stresses, and thermal stresses. We'll delve into their derivations, applications, and limitations, equipping you with the knowledge to apply them effectively in real-world scenarios.

Axial Stress

Axial stress, denoted by σ (sigma), arises from forces acting perpendicular to the cross-sectional area of a component. It's a measure of the force distributed over that area and is either tensile (pulling) or compressive (pushing).

The fundamental formula for axial stress is:

σ = F/A

Where: σ = Axial stress (typically in Pascals (Pa) or pounds per square inch (psi))

F = Axial force (typically in Newtons (N) or pounds (lb))

A = Cross-sectional area perpendicular to the force (typically in square meters (m²) or square inches (in²))

This formula assumes that the force is uniformly distributed across the cross-sectional area. This is a critical assumption, and deviations from uniformity (e.g., due to stress concentrations) can significantly affect the actual stress distribution.

Example

Consider a steel rod with a diameter of 20 mm subjected to a tensile force of 50 k N. Calculate the axial stress in the rod.

1.Calculate the cross-sectional area:

A = πr² = π(10 mm)² = π(0.01 m)² ≈

3.14 x 10⁻⁴ m²

2.Calculate the axial stress:

σ = F/A = (50 x 10³ N) / (3.14 x 10⁻⁴ m²) ≈

159.15 MPa

Therefore, the axial tensile stress in the steel rod is approximately 159.15 MPa.

How does axial stress differ from normal stress?

Axial stress is a specific type of normal stress. Normal stress is a more general term that refers to the force acting perpendicular to a surface. Axial stress specifically refers to that normal stress acting along the axis of a structural member (like a rod or column). So, all axial stresses are normal stresses, but not all normal stresses are axial stresses. For example, pressure acting on the surface of a sphere creates normal stress but isn't typically considered "axial" in the same way.

Bending Stress

Bending stress occurs when a moment is applied to a structural member, causing it to bend. This type of stress is common in beams and shafts. The maximum bending stress occurs at the points furthest from the neutral axis.

The bending stress formula, derived from beam bending theory, is:

σ = My/I

Where: σ = Bending stress (typically in Pa or psi)

M = Bending moment (typically in Nm or lb-in)

y = Distance from the neutral axis to the point where stress is being calculated (typically in m or in)

I = Area moment of inertia of the cross-section about the neutral axis (typically in m⁴ or in⁴)

The area moment of inertia (I) depends on the shape of the cross-section. For a rectangular cross-section with widthband heighth, I = (bh³)/12. For a circular cross-section with radiusr, I = (πr⁴)/4.Example:

A cantilever beam with a rectangular cross-section (50 mm wide and 100 mm high) is subjected to a bending moment of 2 k N·m at its fixed end. Calculate the maximum bending stress.

1.Calculate the area moment of inertia:

I = (bh³)/12 = (0.05 m)(0.1 m)³/12 ≈

4.17 x 10⁻⁶ m⁴

2.Determine the maximum distance from the neutral axis:

y = h/2 = 0.1 m / 2 =

0.05 m

3.Calculate the maximum bending stress:

σ = My/I = (2000 Nm)(0.05 m) / (4.17 x 10⁻⁶ m⁴) ≈

23.98 MPa

Therefore, the maximum bending stress in the cantilever beam is approximately 23.98 MPa.

How do you calculate the stress in a curved beam?

Calculating stress in a curved beam is more complex than in a straight beam. The neutral axis shifts towards the center of curvature, and the stress distribution is no longer linear. The bending stress formula for a curved beam is:

σ = M(R - r) / (Ae(r))

Where:

M = Bending Moment

R = Radius of curvature of the neutral axis

r = Distance from the center of curvature to the point where stress is calculated

A = Cross-sectional Area

e = Distance from the centroidal axis to the neutral axis.

Calculating 'e' often requires more advanced calculations using integrals or referencing specialized tables, depending on the beam's cross-sectional shape.

Shear Stress

Shear stress, denoted by τ (tau), arises from forces acting parallel to the cross-sectional area. It's a measure of the force distributed over that area, causing the material to deform by sliding or shearing.

The basic formula for shear stress is:

τ = V/A

Where: τ = Shear stress (typically in Pa or psi)

V = Shear force (typically in N or lb)

A = Area parallel to the force (typically in m² or in²)

However, this formula assumes uniform shear stress distribution, which is often not the case, especially in beams. A more accurate formula for shear stress in beams is:

τ = VQ / (Ib)

Where: τ = Shear stress at a specific point in the cross-section

V = Shear force

Q = First moment of area of the portion of the cross-section above (or below) the point of interest about the neutral axis

I = Area moment of inertia of the entire cross-section

b = Width of the cross-section at the point of interest

Example

A bolt with a diameter of 12 mm is used to connect two plates subjected to a shear force of 8 k N. Calculate the shear stress in the bolt.

1.Calculate the shear area:

A = πr² = π(6 mm)² = π(0.006 m)² ≈

1.13 x 10⁻⁴ m²

2.Calculate the shear stress:

τ = V/A = (8 x 10³ N) / (1.13 x 10⁻⁴ m²) ≈

70.79 MPa

Therefore, the shear stress in the bolt is approximately 70.79 MPa.

Torsional Stress

Torsional stress occurs in shafts and other components subjected to a twisting moment (torque). The stress is highest at the outer surface of the shaft.

The formula for torsional shear stress is:

τ = Tr/J

Where: τ = Torsional shear stress (typically in Pa or psi)

T = Applied torque (typically in Nm or lb-in)

r = Radius of the shaft (typically in m or in)

J = Polar moment of inertia of the cross-section (typically in m⁴ or in⁴)

For a solid circular shaft, J = (πr⁴)/2. For a hollow circular shaft with outer radius r_o and inner radius r_i, J = (π/2)(r_o⁴ - r_i⁴).

Example

A solid steel shaft with a diameter of 50 mm is subjected to a torque of 1 k N·m. Calculate the maximum torsional shear stress.

1.Calculate the polar moment of inertia:

J = (πr⁴)/2 = (π(0.025 m)⁴)/2 ≈

6.14 x 10⁻⁸ m⁴

2.Calculate the maximum torsional shear stress:

τ = Tr/J = (1000 Nm)(0.025 m) / (6.14 x 10⁻⁸ m⁴) ≈

40.72 MPa

Therefore, the maximum torsional shear stress in the shaft is approximately 40.72 MPa.

What is the difference between polar moment of inertia and area moment of inertia?

Thearea moment of inertia(I) quantifies a beam's resistance to bending about a specific axis. It describes how the cross-sectional area is distributed relative to that bending axis. Thepolar moment of inertia(J), on the other hand, quantifies a shaft's resistance to twisting (torsion). It describes how the area is distributed relative to thecenterof the cross-section. They are both geometric properties that describe the distribution of area, but they relate to different types of loading.

Combined Stresses

In many real-world scenarios, components are subjected to a combination of stresses, such as axial and bending stress, or shear and torsional stress. To analyze these situations, we need to combine the individual stress components.

Principal Stresses

Principal stresses (σ₁ and σ₂) represent the maximum and minimum normal stresses at a point, acting on planes where the shear stress is zero. They are found using the following formulas, derived from Mohr's circle:

σ_1,2 = (σ_x + σ_y)/2 ± √(((σ_x - σ_y)/2)² + τ_xy²)

Where: σ_x = Normal stress in the x-direction σ_y = Normal stress in the y-direction τ_xy = Shear stress in the xy-plane

Maximum Shear Stress

The maximum shear stress (τ_max) is given by:

τ_max = √(((σ_x - σ_y)/2)² + τ_xy²) = (σ₁ - σ₂)/2

Example

A shaft is subjected to a bending stress of 80 MPa and a torsional shear stress of 60 MPa. Calculate the principal stresses and the maximum shear stress. Assume σ_y = 0.

1.Calculate the principal stresses:

σ_1,2 = (80 + 0)/2 ± √(((80 - 0)/2)² + (60)²)

σ_1,2 = 40 ± √(40² + 60²)

σ_1,2 = 40 ± √(1600 + 3600)

σ_1,2 = 40 ± √5200

σ_1,2 = 40 ± 72.11

σ₁ ≈ 112.11 MPa

σ₂ ≈ -32.11 MPa

2.Calculate the maximum shear stress:

τ_max = √(40² + 60²) ≈ 72.11 MPa

Therefore, the principal stresses are approximately 112.11 MPa and -32.11 MPa, and the maximum shear stress is approximately

72.11 MPa.

When should principal stress formulas be applied in design?

Principal stress formulas are essential when analyzing components under combined loading conditions. They are critical for predicting failure based on failure theories like the maximum shear stress theory or the distortion energy theory (von Mises criterion), which rely on principal stresses to determine if a material will yield or fracture under complex stress states. These formulas provide a more accurate representation of the stress state than simply adding individual stress components.

Thermal Stress

Thermal stress arises from the expansion or contraction of materials due to temperature changes. If the material is constrained, this expansion or contraction generates stress.

The formula for thermal stress is:

σ = αEΔT

Where: σ = Thermal stress (typically in Pa or psi) α = Coefficient of thermal expansion (typically in /°C or /°F)

E = Young's modulus of elasticity (typically in Pa or psi) ΔT = Change in temperature (typically in °C or °F)

Example

A steel bar is heated from 20°C to 100°C. The bar is constrained, preventing it from expanding. Calculate the thermal stress developed in the bar. Assume α = 12 x 10⁻⁶ /°C and E = 200 GPa.

1.Calculate the change in temperature:

ΔT = 100°C - 20°C = 80°C

2.Calculate the thermal stress:

σ = αEΔT = (12 x 10⁻⁶ /°C)(200 x 10⁹ Pa)(80°C) ≈ 192 MPa

Therefore, the thermal stress developed in the steel bar is approximately 192 MPa.

What are some practical applications of understanding thermal stress?

Understanding thermal stress is crucial in designing various engineering systems: Pressure Vessels: Accounting for thermal gradients and expansions is crucial for pressure vessel design, especially in high-temperature applications. Piping Systems: Pipelines transporting hot fluids experience significant thermal expansion. Expansion loops are designed to accommodate this expansion and prevent excessive stress. Engine Components: Pistons, cylinders, and engine blocks are subjected to high temperatures and thermal stresses. Material selection and design must consider these factors. Bridges and Buildings: Thermal expansion and contraction of structural elements in bridges and buildings must be accounted for to prevent damage. Expansion joints are incorporated to allow for these movements.

Conclusion

This article has covered the fundamental formulas for calculating stress in machine design, including axial, bending, shear, torsional, combined, and thermal stresses. Understanding these formulas and their applications is crucial for engineers to design safe, reliable, and efficient machines and structures. Remember to carefully consider the assumptions underlying each formula and to account for factors such as stress concentrations and combined loading conditions. Mastery of these concepts will enable you to effectively analyze and design mechanical components to withstand the loads they will encounter in service.