In mechanical engineering, understanding stress is paramount for designing safe and reliable structures and machines. Stress, defined as force per unit area, quantifies the internal forces that molecules within a continuous material exert on each other. Correctly formulating and calculating stress allows engineers to predict how materials will behave under load, preventing failures and optimizing designs. This article delves into various stress formulations, covering normal stress, shear stress, bending stress, torsional stress, and combined stress scenarios. We will explore each concept with clear explanations, relevant formulas, and practical engineering examples.
Understanding Normal Stress
Normal stress, often denoted by σ (sigma), is the force acting perpendicular to a surface area. It's crucial for analyzing tension and compression in structural elements. The basic formula for normal stress is:
σ = F/A
Where: σ is the normal stress (typically in Pascals (Pa) or pounds per square inch (psi))
F is the normal force acting on the area (in Newtons (N) or pounds (lb))
A is the area over which the force is distributed (in square meters (m²) or square inches (in²))
It’s vital to distinguish between tensile stress (positive σ, pulling force) and compressive stress (negative σ, pushing force). In tensile stress, the material is being stretched, while in compressive stress, the material is being squeezed.
Axial Loading Example
Consider a steel rod with a diameter of 20 mm subjected to a tensile force of 50 k N. Calculate the tensile stress in the rod.
1.Calculate the cross-sectional area (A):
A = πr² = π(d/2)² = π(0.02 m / 2)² ≈
3.1416 x 10⁻⁴ m²
2.Calculate the tensile stress (σ):
σ = F/A = (50 x 10³ N) / (3.1416 x 10⁻⁴ m²) ≈
159.15 x 10⁶ Pa =
159.15 MPa
Therefore, the tensile stress in the steel rod is approximately 159.15 MPa.
Compressive Loading Example
A concrete column with a square cross-section of 300 mm x 300 mm supports a compressive load of 500 k N. Determine the compressive stress in the column.
1.Calculate the cross-sectional area (A):
A = (0.3 m) x (0.3 m) =
0.09 m²
2.Calculate the compressive stress (σ):
σ = F/A = -(500 x 10³ N) / (0.09 m²) ≈ -5.56 x 10⁶ Pa = -5.56 MPa
The compressive stress in the concrete column is approximately -5.56 MPa (negative sign indicates compression).
Shear Stress Formulation
Shear stress, denoted by τ (tau), arises when a force acts parallel to a surface area, causing the material to deform by sliding or shearing. The formula for shear stress is:
τ = V/A
Where: τ is the shear stress (typically in Pascals (Pa) or pounds per square inch (psi))
V is the shear force acting on the area (in Newtons (N) or pounds (lb))
A is the area over which the shear force is distributed (in square meters (m²) or square inches (in²))
Common examples of shear stress include the stress in a bolt connecting two plates or the stress in a keyway of a rotating shaft.
Single Shear Example
Two steel plates are connected by a bolt with a diameter of 12 mm. If the plates are subjected to a shear force of 10 k N, calculate the shear stress in the bolt.
1.Calculate the cross-sectional area of the bolt (A):
A = πr² = π(d/2)² = π(0.012 m / 2)² ≈
1.131 x 10⁻⁴ m²
2.Calculate the shear stress (τ):
τ = V/A = (10 x 10³ N) / (1.131 x 10⁻⁴ m²) ≈
88.42 x 10⁶ Pa =
88.42 MPa
The shear stress in the bolt is approximately 88.42 MPa.
Double Shear Example
If the same two steel plates are connected using a clevis joint such that the bolt is in double shear, how does the shear stress change? The shear force is still 10 k N and the bolt diameter remains 12 mm.
In double shear, the shear force is distributed overtwocross-sectional areas of the bolt.
1.Calculate the total shear area (A_total):
A_total = 2 A = 2 (1.131 x 10⁻⁴ m²) ≈
2.262 x 10⁻⁴ m²
2.Calculate the shear stress (τ):
τ = V / A_total = (10 x 10³ N) / (2.262 x 10⁻⁴ m²) ≈
44.21 x 10⁶ Pa =
44.21 MPa
The shear stress in the bolt under double shear is approximately 44.21 MPa, which is half the value of the single shear case, illustrating the benefit of using double shear connections.
Bending Stress Formulation
Bending stress occurs in beams subjected to bending moments. These stresses vary linearly across the cross-section, with maximum tensile stress at one extreme fiber and maximum compressive stress at the opposite extreme fiber. The bending stress formula, derived from beam bending theory, is:
σ = My/I
Where: σ is the bending stress (typically in Pascals (Pa) or pounds per square inch (psi))
M is the bending moment at the section of interest (in Newton-meters (N·m) or pound-inches (lb·in))
y is the distance from the neutral axis to the point where the stress is being calculated (in meters (m) or inches (in))
I is the second moment of area (moment of inertia) of the cross-section about the neutral axis (in meters to the fourth power (m⁴) or inches to the fourth power (in⁴))
The neutral axis is the axis within the beam's cross-section where the bending stress is zero.
Rectangular Beam Example
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.
1.Calculate the second moment of area (I):
For a rectangular section, I = (bh³) / 12 = (0.05 m (0.1 m)³) / 12 ≈
4.167 x 10⁻⁶ m⁴
2.Determine the maximum distance from the neutral axis (y):
y = h/2 = 0.1 m / 2 =
0.05 m
3.Calculate the maximum bending stress (σ_max):
σ_max = (M y) / I = (2 x 10³ N·m 0.05 m) / (4.167 x 10⁻⁶ m⁴) ≈ 24 x 10⁶ Pa = 24 MPa
The maximum bending stress in the beam is 24 MPa.
Circular Beam Example
Consider a circular beam with a diameter of 80 mm subjected to a bending moment of 1.5 k N·m. Determine the maximum bending stress in the beam.
1.Calculate the second moment of area (I):
For a circular section, I = (πd⁴) / 64 = (π (0.08 m)⁴) / 64 ≈
2.01 x 10⁻⁶ m⁴
2.Determine the maximum distance from the neutral axis (y):
y = d/2 = 0.08 m / 2 =
0.04 m
3.Calculate the maximum bending stress (σ_max):
σ_max = (M y) / I = (1.5 x 10³ N·m
0.04 m) / (2.01 x 10⁻⁶ m⁴) ≈
29.85 x 10⁶ Pa =
29.85 MPa
The maximum bending stress in the circular beam is approximately 29.85 MPa.
Torsional Stress Formulation
Torsional stress occurs in shafts subjected to torque or twisting moments. The stress distribution is zero at the center of the shaft and maximum at the outer surface. The torsional stress formula is:
τ = Tr/J
Where: τ is the torsional shear stress (typically in Pascals (Pa) or pounds per square inch (psi))
T is the applied torque (in Newton-meters (N·m) or pound-inches (lb·in))
r is the radial distance from the center of the shaft to the point where the stress is being calculated (in meters (m) or inches (in))
J is the polar moment of inertia of the cross-section (in meters to the fourth power (m⁴) or inches to the fourth power (in⁴))
For a solid circular shaft, J = (πd⁴) / 32, and for a hollow circular shaft, J = (π/32) (D⁴ - d⁴), where D is the outer diameter and d is the inner diameter.
Solid Shaft 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 in the shaft.
1.Calculate the polar moment of inertia (J):
J = (πd⁴) / 32 = (π (0.05 m)⁴) / 32 ≈
6.136 x 10⁻⁸ m⁴
2.Determine the maximum radius (r):
r = d/2 = 0.05 m / 2 =
0.025 m
3.Calculate the maximum torsional shear stress (τ_max):
τ_max = (T r) / J = (1 x 10³ N·m 0.025 m) / (6.136 x 10⁻⁸ m⁴) ≈
40.75 x 10⁶ Pa =
40.75 MPa
The maximum torsional shear stress in the shaft is approximately 40.75 MPa.
Hollow Shaft Example
A hollow shaft has an outer diameter of 100 mm and an inner diameter of 60 mm. If it is subjected to a torque of 5 k N·m, what is the maximum shear stress?
1.Calculate the polar moment of inertia (J):
J = (π/32) (D⁴ - d⁴) = (π/32) ((0.1 m)⁴ - (0.06 m)⁴) ≈
8.59 x 10⁻⁷ m⁴
2.Determine the maximum radius (r):
r = D/2 = 0.1 m / 2 =
0.05 m
3.Calculate the maximum torsional shear stress (τ_max):
τ_max = (T r) / J = (5 x 10³ N·m 0.05 m) / (8.59 x 10⁻⁷ m⁴) ≈
291.04 x 10⁶ Pa =
291.04 MPa
The maximum torsional shear stress in the hollow shaft is approximately 291.04 MPa.
Combined Stress States
In many practical engineering situations, structural components are subjected to a combination of normal and shear stresses. Analyzing these combined stress states requires techniques like Mohr's circle and the concept of principal stresses. Principal stresses are the maximum and minimum normal stresses at a point, acting on planes where the shear stress is zero.
Mohr's Circle
Mohr's circle is a graphical representation of the stress state at a point. It allows engineers to visualize and calculate principal stresses, maximum shear stress, and the orientation of the principal planes.
Principal Stress Formulas
The principal stresses (σ₁) and (σ₂) can be calculated using the following formulas:
σ₁,₂ = (σₓ + σᵧ)/2 ± √[((σₓ - σᵧ)/2)² + τₓᵧ²]
Where: σₓ is the normal stress in the x-direction σᵧ is the normal stress in the y-direction τₓᵧ is the shear stress in the xy-plane
The maximum shear stress (τ_max) is given by:
τ_max = √[((σₓ - σᵧ)/2)² + τₓᵧ²] = (σ₁ - σ₂)/2
Combined Bending and Torsion Example
A shaft with a diameter of 60 mm is subjected to a bending moment of 800 N·m and a torque of 1200 N·m. Determine the maximum principal stress and maximum shear stress.
1.Calculate the bending stress (σₓ):
I= (πd⁴)/64 = (π (0.06m)⁴)/64 ≈
6.36 x 10⁻⁷ m⁴
y= d/2 = 0.06m / 2 =
0.03m
σₓ = (My) / I = (800 N·m 0.03m) / (6.36 x 10⁻⁷ m⁴) ≈
37.74 x 10⁶ Pa =
37.74 MPa
2.Calculate the torsional shear stress (τₓᵧ):
J= (πd⁴)/32 = (π (0.06m)⁴)/32 ≈
1.27 x 10⁻⁶ m⁴
r= d/2 = 0.06m / 2 =
0.03m
τₓᵧ = (Tr) / J = (1200 N·m 0.03m) / (1.27 x 10⁻⁶ m⁴) ≈
28.35 x 10⁶ Pa =
28.35 MPa
3.Assume σᵧ = 0 (no normal stress in the y-direction):
4.Calculate the principal stresses:
σ₁,₂ = (37.74 MPa + 0)/2 ± √[((37.74 MPa - 0)/2)² + (28.35 MPa)²]
σ₁,₂ = 18.87 MPa ± √[(18.87 MPa)² + (28.35 MPa)²]
σ₁,₂ = 18.87 MPa ± √(356.08 +
803.72) MPa²
σ₁,₂ = 18.87 MPa ± √1159.8 MPa²
σ₁,₂ = 18.87 MPa ±
34.06 MPa
σ₁ = 18.87 MPa +
34.06 MPa =
52.93 MPa
σ₂ = 18.87 MPa -
34.06 MPa = -15.19 MPa
5.Calculate the maximum shear stress:
τ_max = (σ₁ - σ₂) / 2 = (52.93 MPa - (-15.19 MPa)) / 2 = (68.12 MPa) / 2 =
34.06 MPa
Therefore, the maximum principal stress is approximately 52.93 MPa, and the maximum shear stress is approximately
34.06 MPa.
People Also Ask
How do you calculate hoop stress in thin-walled cylinders?
Hoop stress (σ_h), also known as circumferential stress, in a thin-walled cylinder subjected to internal pressure (p) is calculated using the formula: σ_h = (p r) / t, where r is the radius of the cylinder and t is the wall thickness. This formula assumes that the cylinder's wall thickness is significantly smaller than its radius (typically, t < r/10).
What is the difference between true stress and engineering stress?
Engineering stress is calculated by dividing the applied force by theoriginalcross-sectional area of the material, while true stress is calculated by dividing the applied force by theinstantaneouscross-sectional area of the material during deformation. Engineering stress is simpler to calculate but becomes less accurate at large deformations. True stress provides a more accurate representation of the stress experienced by the material as it deforms and its cross-sectional area changes.
When should principal stress formulas be applied in design?
Principal stress formulas are essential when designing components subjected to combined loading conditions, such as simultaneous bending and torsion, or when dealing with complex stress distributions. They help determine the maximum normal and shear stresses experienced by the material, which are critical for assessing the risk of yielding or fracture. Applying principal stress formulas ensures that the design accounts for the most critical stress state within the component, leading to safer and more reliable designs.
Conclusion
Mastering stress formulations is crucial for any mechanical engineer. By understanding the principles behind normal stress, shear stress, bending stress, torsional stress, and combined stress states, and by applying the correct formulas and techniques, engineers can accurately predict material behavior, optimize designs, and ensure the safety and reliability of structures and machines. The worked examples provided in this article offer practical guidance for applying these concepts to real-world engineering problems. Continued practice and application of these principles are key to developing a strong understanding of stress analysis.