Understanding Stress Formulas in Strength of Materials
Stress is a fundamental concept in strength of materials, representing the internal forces that molecules within a continuous material exert on each other. Understanding stress and its related formulas is crucial for engineers to design safe and reliable structures and components. This article provides a comprehensive overview of stress formulas, their applications, and key considerations for their use.
Stress is defined as force per unit area. Mathematically, it's expressed as:
σ = F/A
where: σ (sigma) represents stress (typically in Pascals (Pa) or pounds per square inch (psi)).
F is the force acting on the area.
A is the area over which the force is distributed.
It’s important to note that stress is a tensor quantity, meaning it has both magnitude and direction. In simpler terms, stress describes both how much force is being applied and the orientation of that force relative to the surface.
Types of Stress
Stress can be categorized into different types based on the direction of the applied force relative to the surface area. The most common types are: Normal Stress:This stress acts perpendicular to the surface. It can be either tensile (pulling) or compressive (pushing). Tensile stress is considered positive, while compressive stress is considered negative.
Formula: σ = F/A (same as the general stress formula, but specifically applied to normal forces) Shear Stress:This stress acts parallel to the surface. It is also called tangential stress.
Formula: τ = V/A
where: τ (tau) represents shear stress (typically in Pascals (Pa) or pounds per square inch (psi)).
V is the shear force acting on the area.
A is the area over which the shear force is distributed. Bearing Stress:This stress is a specific type of compressive stress that occurs when one object presses against another, like a bolt pressing against a plate.
Formula: σb = F/Ab
where: σb is the bearing stress.
F is the applied force.
Ab is the projected area of contact. For example, the area of a bolt hole in a plate is typically calculated as the bolt diameter multiplied by the plate thickness.
Stress due to Axial Loading
Axial loading occurs when a force is applied along the longitudinal axis of a member. The stress produced by an axial load is a normal stress. The key assumption here is that the force is applied uniformly across the cross-sectional area.
Formula: σ = P/A
where: P is the axial force.
A is the cross-sectional area of the member.
Example: A steel rod with a diameter of 20 mm is subjected to a tensile force of 50 k N. Calculate the stress in the rod.
Area, A = π (d/2)^2 = π (0.02 m / 2)^2 =
3.1416 x 10^-4 m^2
Stress, σ = P/A = (50,000 N) / (3.1416 x 10^-4 m^2) =
159.15 MPa
Bending Stress in Beams
Beams are structural elements designed to resist bending moments. Bending stress varies linearly across the beam's cross-section, with maximum tensile stress at the outer fiber on one side and maximum compressive stress at the outer fiber on the opposite side. The neutral axis experiences zero stress.
Formula: σ = My/I
where: M is the bending moment at the section of interest.
y is the distance from the neutral axis to the point where stress is being calculated.
I is the second moment of area (also known as the area moment of inertia) of the beam's cross-section about the neutral axis.
For maximum bending stress: σmax = Mc/I = M/S, where 'c' is the distance from the neutral axis to the outermost fiber and S = I/c is the section modulus.
Example: A rectangular beam with a width of 50 mm and a height of 100 mm is subjected to a bending moment of 5 k N-m. Calculate the maximum bending stress.
Second moment of area, I = (bh^3)/12 = (0.05 m (0.1 m)^3)/12 =
4.1667 x 10^-6 m^4
Distance to outermost fiber, c = h/2 = 0.1 m / 2 =
0.05 m
Maximum bending stress, σmax = Mc/I = (5000 N-m 0.05 m) / (4.1667 x 10^-6 m^4) = 60 MPa
Torsional Shear Stress in Shafts
Torsion occurs when a twisting moment (torque) is applied to a shaft. This induces shear stress in the shaft. The shear stress is maximum at the outer surface of the shaft and zero at the center.
Formula: τ = Tr/J
where: T is the applied torque.
r is the radial distance from the center of the shaft to the point where stress is being calculated.
J is the polar moment of inertia of the shaft's cross-section.
For a solid circular shaft, J = (π*d^4)/32, where d is the diameter of the shaft.
For a hollow circular shaft, J = (π(do^4 - di^4))/32, where do is the outer diameter and di is the inner diameter.
Example:A solid steel shaft with a diameter of 40 mm is subjected to a torque of 500 N-m. Calculate the maximum shear stress.
Polar moment of inertia, J = (πd^4)/32 = (π (0.04 m)^4)/32 =
2.513 x 10^-7 m^4
Maximum shear stress, τmax = Tr/J = (500 N-m 0.02 m) / (2.513 x 10^-7 m^4) =
39.79 MPa
Stress in Thin-Walled Pressure Vessels
Pressure vessels are containers designed to hold fluids or gases under pressure. Thin-walled pressure vessels are defined as those where the wall thickness is significantly smaller than the vessel's radius (typically, thickness < radius/10). Two primary stresses are developed in thin-walled cylinders:hoop stress (circumferential stress) and longitudinal stress.
Hoop Stress (σh): This stress acts in the circumferential direction. It is twice as large as the longitudinal stress.
Formula: σh = (pr)/t
where: p is the internal pressure.
r is the internal radius of the cylinder.
t is the wall thickness. Longitudinal Stress (σl):This stress acts along the longitudinal axis of the cylinder.
Formula: σl = (pr)/(2t)
For spherical pressure vessels, the stress is uniform in all directions.
Formula: σ = (pr)/(2t)
Example: A thin-walled cylindrical pressure vessel has an internal diameter of 1 meter and a wall thickness of 10 mm. It is subjected to an internal pressure of 2 MPa. Calculate the hoop stress and the longitudinal stress.
Internal radius, r = 1 m / 2 = 0.5 m
Hoop stress, σh = (pr)/t = (2 MPa 0.5 m) / (0.01 m) = 100 MPa
Longitudinal stress, σl = (pr)/(2t) = (2 MPa 0.5 m) / (2
0.01 m) = 50 MPa
Principal Stresses and Maximum Shear Stress
At any point in a stressed material, it is possible to find a coordinate system where the shear stresses are zero. The normal stresses in this coordinate system are called principal stresses (σ1, σ2, and σ3), where σ1 is the maximum principal stress, σ3 is the minimum principal stress, and σ2 is the intermediate principal stress. Principal stresses are critical for determining the failure criteria of materials.
In 2D stress situations (plane stress), the principal stresses can be calculated using the following formulas: σ1,2 = (σx + σy)/2 ± √[((σx - σy)/2)^2 + τxy^2]
where: σx is the normal stress in the x-direction.
σy is the normal stress in the y-direction.
τxy is the shear stress in the xy-plane.
The maximum shear stress (τmax) is related to the principal stresses: τmax = (σ1 - σ3)/2
When should principal stress formulas be applied in design?
Principal stress formulas are essential when dealing with combined loading scenarios where multiple stress components (normal and shear) are present. They are crucial for predicting material failure under complex stress states because they represent the maximum and minimum normal stresses the material experiences, regardless of the coordinate system. Comparing these values to material strength properties helps determine if a design is safe.
How do you determine the orientation of the principal planes?
The angle (θp) of the principal planes (the planes on which principal stresses act) relative to the original coordinate system can be found using:
tan(2θp) = (2τxy)/(σx - σy)
Thermal Stress
Thermal stress arises when a material is subjected to temperature changes while constrained from expanding or contracting freely.
Formula: σ = αEΔT
where: α is the coefficient of thermal expansion of the material.
E is the Young's modulus of the material.
ΔT is the change in temperature.
Example: A steel bar is fixed between two rigid supports. The temperature of the bar increases by 50°C. Given that the coefficient of thermal expansion of steel is 12 x 10^-6 /°C and the Young's modulus is 200 GPa, calculate the thermal stress in the bar.
Thermal stress, σ = αEΔT = (12 x 10^-6 /°C) (200 x 10^9 Pa) (50 °C) = 120 MPa (compressive)
Common Pitfalls and Misconceptions
Confusing Stress and Force: Stress is forceper unit area, not just force. A larger force applied over a larger area may result in the same stress as a smaller force applied over a smaller area. Ignoring Stress Concentrations: Stress concentrations occur at sharp corners, holes, and other geometric discontinuities. The stress at these locations can be significantly higher than the average stress. Failure often initiates at these points. Assuming Uniform Stress Distribution: Stress is not always uniformly distributed, especially in complex geometries or under complex loading conditions. Bending stress, for example, varies linearly across the cross-section of a beam. Using Engineering Stress Instead of True Stress: Engineering stress is calculated using the original cross-sectional area of the material. True stress is calculated using the instantaneous cross-sectional area, which decreases as the material deforms under tensile loading. For large deformations, true stress is a more accurate representation of the stress state. Engineering stress is sufficient for most design applications where deformations are small. What is the difference between true stress and engineering stress?Engineering stress (also known as nominal stress) is calculated by dividing the applied force by the original cross-sectional area of the material. True stress, on the other hand, is calculated by dividing the applied force by the instantaneous cross-sectional area of the material, which changes during deformation. True stress provides a more accurate representation of the stress experienced by the material, especially at large strains, as it accounts for the reduction in cross-sectional area during deformation.
Understanding stress formulas is paramount for engineers in various fields. Whether designing pressure vessels, analyzing beam structures, or optimizing rotating machinery, applying these formulas correctly ensures structural integrity and safety. By carefully considering the types of stress, loading conditions, and material properties, engineers can make informed decisions and create robust and reliable designs. Always remember to account for stress concentrations and consider using appropriate safety factors to compensate for uncertainties and variations in material properties.