In the realm of mechanical engineering, understanding stress within materials is paramount for designing safe and efficient structures and components. Strength of materials, also known as mechanics of materials, provides the foundational principles for analyzing stresses and strains in solid objects subjected to various loads. This article serves as a practical reference guide to stress formulas, covering fundamental concepts and applications across diverse engineering scenarios.
Stress: Definition and Fundamental Types
Stress, denoted by the Greek letter sigma (σ), is defined as the force (F) acting per unit area (A) within a material. Mathematically, it is expressed as:
σ = F/A
Stress is typically measured in Pascals (Pa) or pounds per square inch (psi). It is crucial to distinguish between two main types of stress: normal stress and shear stress. Normal stress acts perpendicular to the surface, while shear stress acts parallel to the surface.
Normal Stress
Normal stress arises from forces that are perpendicular to the surface area. It can be either tensile (pulling) or compressive (pushing). Tensile stress is considered positive, while compressive stress is considered negative. The formula remains the same, σ = F/A, but the interpretation of the sign changes.
Example: Consider a steel rod with a cross-sectional area of 0.01 m² subjected to a tensile force of 10,000 N. The tensile stress in the rod is σ = 10,000 N /
0.01 m² = 1,000,000 Pa = 1 MPa.
Shear Stress
Shear stress, denoted by the Greek letter tau (τ), is caused by forces that are parallel to the surface area. It is often associated with bolted or riveted joints, where the force is transmitted tangentially. The shear stress formula is:
τ = F/A
where F is the shear force and A is the area resisting the shear.
Example: Imagine two plates bolted together, with a bolt diameter of 10 mm. If the plates are subjected to a shear force of 5,000 N, the shear stress on the bolt is τ = 5,000 N / (π (0.005 m)²) =
63.66 MPa.
Stress Due to Axial Loading
Axial loading refers to forces applied along the longitudinal axis of a structural member. These forces can cause either tension or compression, as previously discussed.
Axial Stress Formula
The axial stress formula, σ = P/A, is a direct application of the normal stress definition. Here, P represents the axial force and A represents the cross-sectional area perpendicular to the force.
Deformation Due to Axial Loading
The deformation (elongation or contraction) of a member under axial loading is given by:
δ = (PL) / (AE)
where: δ is the deformation.
P is the axial force.
L is the original length of the member.
A is the cross-sectional area.
E is the modulus of elasticity (Young's modulus) of the material.
Example: A 2-meter-long aluminum bar with a cross-sectional area of 0.001 m² is subjected to a tensile force of 20,000 N. The modulus of elasticity for aluminum is approximately 70 GPa. The elongation of the bar is δ = (20,000 N 2 m) / (0.001 m² 70 x 10⁹ Pa) =
0.000571 m =
0.571 mm.
Bending Stress in Beams
Beams are structural members designed to resist bending loads. Bending stress varies across the cross-section of the beam, with the maximum stress occurring at the points farthest from the neutral axis.
Bending Stress Formula
The bending stress formula is:
σ = (My) / I
where: σ is the bending stress.
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 area moment of inertia of the cross-section about the neutral axis.
The maximum bending stress occurs at the outermost fibers of the beam, where y is the largest. This is often denoted as c, the distance from the neutral axis to the outermost fiber. The formula then becomes σ_max = (Mc) / I.
Shear Stress in Beams
While bending stress dominates in beam analysis, shear stress also plays a role, especially in short, heavily loaded beams. The shear stress formula for beams is:
τ = (VQ) / (Ib)
where: τ is the shear stress.
V is the shear force at the section of interest.
Q is the first moment of area of the portion of the cross-section above (or below) the point where stress is being calculated, taken about the neutral axis.
I is the area moment of inertia of the entire cross-section about the neutral axis.
b is the width of the cross-section at the point where stress is being calculated.
Example: A simply supported rectangular beam has a width of 100 mm, a height of 200 mm, and a length of 3 meters. It is subjected to a uniformly distributed load of 10 k N/m. Calculate the maximum bending stress and maximum shear stress.
Maximum bending moment: M = (w L²) / 8 = (10,000 N/m (3 m)²) / 8 = 11,250 Nm
Area moment of inertia: I = (bh³) / 12 = (0.1 m (0.2 m)³) / 12 =
6.67 x 10⁻⁵ m⁴
Maximum bending stress: σ_max = (Mc) / I = (11,250 Nm 0.1 m) / (6.67 x 10⁻⁵ m⁴) =
16.87 MPa
Maximum shear force: V = (w L) / 2 = (10,000 N/m 3 m) / 2 = 15,000 N
Maximum shear stress (at the neutral axis): τ_max = (3V) / (2A) = (3 15,000 N) / (2 0.1 m
0.2 m) =
1.125 MPa
Torsional Stress in Shafts
Torsion occurs when a twisting moment, or torque (T), is applied to a structural member. This results in shear stress within the member.
Torsional Shear Stress Formula
The torsional shear stress formula is:
τ = (Tr) / J
where: τ is the torsional shear stress.
T is the applied torque.
r is the distance from the center of the shaft to the point where stress is being calculated (maximum at the outer radius).
J is the polar moment of inertia of the cross-section.
For a solid circular shaft, J = (πd⁴) / 32, where d is the diameter. For a hollow circular shaft, J = (π(d_o⁴ - d_i⁴)) / 32, where d_o is the outer diameter and d_i is the inner diameter.
Angle of Twist
The angle of twist (θ) due to torsion is given by:
θ = (TL) / (GJ)
where: θ is the angle of twist in radians.
T is the applied torque.
L is the length of the shaft.
G is the shear modulus of the material.
J is the polar moment of inertia.
Example: A solid steel shaft with a diameter of 50 mm and a length of 1 meter is subjected to a torque of 500 Nm. The shear modulus of steel is 80 GPa. Calculate the maximum shear stress and the angle of twist.
Polar moment of inertia: J = (πd⁴) / 32 = (π (0.05 m)⁴) / 32 =
6.14 x 10⁻⁷ m⁴
Maximum shear stress: τ_max = (Tr) / J = (500 Nm 0.025 m) / (6.14 x 10⁻⁷ m⁴) =
20.36 MPa
Angle of twist: θ = (TL) / (GJ) = (500 Nm 1 m) / (80 x 10⁹ Pa 6.14 x 10⁻⁷ m⁴) =
0.0102 radians =
0.584 degrees
Stress in Thin-Walled Pressure Vessels
Pressure vessels are containers designed to hold fluids or gases at high pressures. Thin-walled pressure vessels are those where the wall thickness is significantly smaller than the radius.
Hoop Stress
Hoop stress (σ_h), also known as circumferential stress, acts tangentially to the surface of the vessel. The formula for hoop stress in a thin-walled cylindrical pressure vessel is:
σ_h = (pr) / t
where: σ_h is the hoop stress.
p is the internal pressure.
r is the inner radius of the cylinder.
t is the wall thickness.
Longitudinal Stress
Longitudinal stress (σ_l) acts along the longitudinal axis of the cylinder. The formula for longitudinal stress in a thin-walled cylindrical pressure vessel is:
σ_l = (pr) / (2t)
Spherical Pressure Vessels
For thin-walled spherical pressure vessels, the stress is uniform in all directions and is given by:
σ = (pr) / (2t)
Example: A cylindrical pressure vessel has an inner 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 longitudinal stress.
Hoop stress: σ_h = (pr) / t = (2 x 10⁶ Pa 0.5 m) /
0.01 m = 100 MPa
Longitudinal stress: σ_l = (pr) / (2t) = (2 x 10⁶ Pa 0.5 m) / (2
0.01 m) = 50 MPa
Combined Stresses and Principal Stresses
In many real-world scenarios, structural members are subjected to multiple types of stress simultaneously. For instance, a shaft may experience both torsional and bending stresses. To analyze such situations, we need to determine the principal stresses.
Principal Stresses
Principal stresses are the maximum and minimum normal stresses at a point, and they occur on planes where the shear stress is zero. They are found by solving for the eigenvalues of the stress tensor. For a 2D stress state, the principal stresses (σ₁ and σ₂) are given by:
σ₁, σ₂ = (σ_x + σ_y)/2 ± √(((σ_x - σ_y)/2)² + τ_xy²)
where: σ_x is the normal stress in the x-direction. σ_y is the normal stress in the y-direction. τ_xy is the 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
Mohr's Circle
Mohr's circle is a graphical representation of stress at a point, used to determine principal stresses, maximum shear stress, and stresses on any arbitrary plane.
Example: A point in a structural member is subjected to normal stresses σ_x = 80 MPa and σ_y = 20 MPa, and a shear stress τ_xy = 30 MPa. Calculate the principal stresses and the maximum shear stress.
σ₁ = (80 + 20)/2 + √(((80 - 20)/2)² + 30²) = 50 + √(900 + 900) = 50 + 42.43 =
92.43 MPa σ₂ = (80 + 20)/2 - √(((80 - 20)/2)² + 30²) = 50 - √(900 + 900) = 50 -
42.43 =
7.57 MPa τ_max = √(((80 - 20)/2)² + 30²) =
42.43 MPa
Thermal Stress
Thermal stress arises when a material is subjected to temperature changes and its expansion or contraction is constrained.
Thermal Stress Formula
The thermal stress is given by:
σ = αEΔT
where: σ is the thermal stress. α is the coefficient of thermal expansion of the material.
E is the modulus of elasticity. ΔT is the change in temperature.
Example: A steel bar is heated from 20°C to 100°C. The coefficient of thermal expansion for steel is 12 x 10⁻⁶ /°C, and the modulus of elasticity is 200 GPa. If the bar is constrained from expanding, calculate the thermal stress.
ΔT = 100°C - 20°C = 80°C σ = (12 x 10⁻⁶ /°C) (200 x 10⁹ Pa) (80°C) = 192 MPa
Common Pitfalls and Considerations
Units: Always ensure consistent units throughout calculations. Converting all values to a common unit system (e.g., SI units) is crucial. Assumptions: Be aware of the assumptions underlying each formula. For instance, the thin-walled pressure vessel formulas are only valid when the wall thickness is significantly smaller than the radius. Stress Concentrations: Stress concentrations occur at geometric discontinuities such as holes, corners, or notches. These areas experience significantly higher stresses than predicted by simple formulas. Stress concentration factors must be considered in such cases. Material Properties: Accurate material properties (e.g., modulus of elasticity, shear modulus, coefficient of thermal expansion) are essential for reliable stress calculations. Use reputable sources for material property data. Sign Conventions:Consistently apply sign conventions for tensile and compressive stresses, as well as shear stress.
How do you calculate hoop stress in thin-walled cylinders?
Hoop stress in thin-walled cylinders is calculated using the formula σ_h = (pr) / t, where 'p' is the internal pressure, 'r' is the inner radius, and 't' is the wall thickness.
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 changes as the material deforms. True stress provides a more accurate representation of the stress experienced by the material during deformation, particularly at large strains.
When should principal stress formulas be applied in design?
Principal stress formulas should be applied in design when components are subjected to multi-axial loading conditions, where stresses act in multiple directions simultaneously. They are critical for determining the maximum normal and shear stresses, which are essential for predicting material failure and ensuring structural integrity.
This reference provides a foundation for understanding and calculating stresses in various engineering applications. By carefully applying these formulas and considering the underlying assumptions, engineers can design safe and reliable structures and components.