Equilibrium of Forces and Laws of Motion

Equilibrium of Forces

In mechanics, a system is in equilibrium when there is no net force or net moment acting upon it. This means that the object remains at rest if it was initially at rest, or it continues to move with a constant velocity if it was initially in motion. For a rigid body, equilibrium is achieved when two conditions are met:

• The vector sum of all external forces acting on the body is zero.
• The sum of all external moments about any point is zero.

The first condition ensures that there is no translational acceleration, while the second condition ensures that there is no rotational acceleration.

Types of Equilibrium

Equilibrium can be classified into three types based on the object's response to a slight disturbance:

• Stable Equilibrium: If an object, when slightly displaced from its equilibrium position, experiences a restoring force or moment that tends to bring it back to its original position. Think of a pendulum hanging vertically.
• Unstable Equilibrium: If an object, when slightly displaced, experiences a force or moment that tends to move it further away from its equilibrium position. An example is a cone balanced on its tip.
• Neutral Equilibrium: If an object, when slightly displaced, remains in its new position without any tendency to return or move further away. A ball on a flat horizontal surface exhibits neutral equilibrium.

Free Body Diagrams (FBD)

To analyze forces and determine equilibrium, the concept of a Free Body Diagram is crucial. An FBD is a diagram that shows an isolated object and all the external forces acting upon it. By drawing an FBD, we can systematically apply the conditions of equilibrium.

To draw an FBD:

1. Isolate the object of interest from its surroundings.
2. Represent the object as a simple geometric shape (e.g., a point or a block).
3. Identify all external forces acting on the object. These can include:
   • Weight (gravitational force)
   • Applied forces (pushed or pulled)
   • Reaction forces from supports
   • Friction forces
   • Tension in ropes or cables
4. Draw each force as a vector originating from the object, indicating its direction and magnitude (if known).
5. Label each force clearly.

Laws of Motion

Sir Isaac Newton's three laws of motion are fundamental principles that describe the relationship between an object and the forces acting upon it.

Newton's First Law of Motion (Law of Inertia)

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced external force. Inertia is the tendency of an object to resist changes in its state of motion. The mass of an object is a measure of its inertia.

Newton's Second Law of Motion

The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This law is mathematically expressed as:

$$ \sum \vec{F} = m\vec{a} $$

Where:

• $ \sum \vec{F} $ is the vector sum of all external forces acting on the object (net force).
• $ m $ is the mass of the object.
• $ \vec{a} $ is the acceleration of the object.

This is the most important law for analyzing the motion of objects under the influence of forces. It implies that if the net force is zero, the acceleration is zero, which is consistent with the first law.

Newton's Third Law of Motion

For every action, there is an equal and opposite reaction. This means that if object A exerts a force on object B, then object B simultaneously exerts a force on object A that is equal in magnitude and opposite in direction. These forces act on different objects. For example, when you push on a wall, the wall pushes back on you with the same force.

Friction

Friction is a force that opposes the relative motion or tendency of motion between two surfaces in contact. It arises from the microscopic irregularities on the surfaces and adhesive forces between them. Friction is a force that can be both beneficial (e.g., allowing us to walk) and detrimental (e.g., causing wear and tear in machinery).

Types of Friction

Friction can be broadly categorized into several types:

1. Static Friction ($f_s$)

Static friction is the force that opposes the initiation of motion between two surfaces that are at rest relative to each other. It acts when an external force is applied to an object, but the object does not move. The magnitude of static friction is variable and adjusts itself to be equal and opposite to the applied force, up to a certain maximum value.

The maximum static friction ($f_{s,max}$) is given by:

$$ f_{s,max} = \mu_s N $$

Where:

• $ \mu_s $ is the coefficient of static friction, a dimensionless quantity that depends on the nature of the two surfaces in contact.
• $ N $ is the normal force, the force perpendicular to the surfaces in contact.

If the applied force is less than $f_{s,max}$, the object remains at rest, and the static friction force equals the applied force. Motion begins only when the applied force exceeds $f_{s,max}$.

2. Kinetic Friction ($f_k$)

Kinetic friction (or sliding friction) is the force that opposes the relative motion between two surfaces that are sliding past each other. Once motion has started, the friction force reduces to the kinetic friction.

The magnitude of kinetic friction is generally constant and is given by:

$$ f_k = \mu_k N $$

Where:

• $ \mu_k $ is the coefficient of kinetic friction. For most materials, $ \mu_k < \mu_s $.
• $ N $ is the normal force.

3. Rolling Friction

Rolling friction occurs when a round object (like a wheel or ball) rolls over a surface. It is generally much smaller than sliding friction and is caused by the deformation of the rolling object and the surface.

4. Fluid Friction

Fluid friction (also known as drag) is the resistance force experienced by an object moving through a fluid (liquid or gas). Its magnitude depends on the speed of the object, the properties of the fluid, and the shape and size of the object.

Factors Affecting Friction

The main factors influencing friction are:

• Nature of the surfaces: Rougher surfaces generally have higher coefficients of friction.
• Normal force: The greater the normal force pressing the surfaces together, the greater the friction force.
• Area of contact: For dry friction, the area of contact has a negligible effect on the friction force, as long as the normal force and surface properties remain the same. This is a counter-intuitive but experimentally verified fact for many materials.

Applications and Examples

• Walking: We walk because the friction between our shoes and the ground allows us to push backward on the ground, and the ground pushes us forward.
• Brakes: Brake pads create friction with the brake discs or drums to slow down a vehicle.
• Tires: The friction between tires and the road surface provides the grip needed for acceleration, braking, and steering.
• Machinery: Lubricants are used to reduce friction between moving parts in engines and other machines, minimizing wear and energy loss.

Problem-Solving Tip for Friction:

When dealing with friction problems, always draw a Free Body Diagram.

1. Determine if the object is at rest or in motion.
2. If at rest, assume static friction and check if the applied force exceeds the maximum static friction.
3. If in motion, use kinetic friction.
4. Resolve forces into components parallel and perpendicular to the surface of contact.
5. Apply Newton's laws of motion and the friction equations.

Stress and Strain

When a material is subjected to external forces, internal forces develop within the material to resist deformation. Stress and strain are fundamental concepts used to quantify these internal forces and the resulting deformation. They are crucial for understanding the mechanical behavior of materials and designing safe structures and components.

Stress ($ \sigma $)

Stress is defined as the internal resisting force acting per unit area within a deformable body. It represents the intensity of the internal forces.

Mathematically, stress is given by:

$$ \sigma = \frac{P}{A} $$

Where:

• $ \sigma $ (sigma) is the stress.
• $ P $ is the internal resisting force (or the applied load).
• $ A $ is the cross-sectional area over which the force is distributed.

The unit of stress is typically Pascals (Pa) in the SI system, where 1 Pa = 1 N/m². Other common units include megapascals (MPa) and gigapascals (GPa). In the imperial system, it is often measured in pounds per square inch (psi).

Types of Stress

Stress can be categorized based on the direction of the force relative to the area:

• Normal Stress ($ \sigma $): Occurs when the force is perpendicular (normal) to the cross-sectional area. It can be further divided into:
  • Tensile Stress: Caused by a pulling force that tends to elongate the material.
  • Compressive Stress: Caused by a pushing force that tends to shorten the material.
• Shear Stress ($ \tau $): Occurs when the force is parallel (tangential) to the cross-sectional area. It tends to cause one part of the material to slide relative to another. Mathematically, $ \tau = \frac{V}{A} $, where $ V $ is the shear force and $ A $ is the area.

Strain ($ \epsilon $)

Strain is a measure of the deformation of a material relative to its original size. It is a dimensionless quantity.

For normal stress, the corresponding strain is normal strain (or linear strain):

$$ \epsilon = \frac{\delta L}{L_0} $$

Where:

• $ \epsilon $ (epsilon) is the normal strain.
• $ \delta L $ is the change in length of the object.
• $ L_0 $ is the original length of the object.

Tensile strain is positive (elongation), and compressive strain is negative (shortening).

For shear stress, the corresponding strain is shear strain ($ \gamma $):

$$ \gamma = \tan(\theta) \approx \theta \text{ (for small angles)} $$

Where $ \theta $ is the change in angle (in radians) between two lines that were originally perpendicular. Shear strain is also dimensionless.

Stress-Strain Relationship (Material Behavior)

The relationship between stress and strain is unique to each material and is often represented by a stress-strain curve obtained from tensile or compression tests.

Elastic Region

In the initial part of the stress-strain curve, stress is directly proportional to strain. This region is called the elastic region. If the load is removed within this region, the material returns to its original shape. The constant of proportionality in this region is the modulus of elasticity.

Hooke's Law

Within the elastic limit, stress is directly proportional to strain. This principle is known as Hooke's Law.

For normal stress and strain:

$$ \sigma = E \epsilon $$

Where $ E $ is the Modulus of Elasticity (Young's Modulus).

For shear stress and strain:

$$ \tau = G \gamma $$

Where $ G $ is the Shear Modulus (Modulus of Rigidity).

Plastic Region

Beyond the elastic limit, if the stress is increased, the material undergoes permanent deformation. This is the plastic region. Even if the load is removed, the material will not return to its original shape.

Ultimate Tensile Strength (UTS)

This is the maximum stress a material can withstand while being stretched or pulled before necking begins.

Fracture Strength

This is the stress at which the material breaks.

Elastic Constants and Poisson's Ratio

Elastic constants are material properties that describe how a material deforms under stress within its elastic limit. They are fundamental to predicting a material's mechanical response and designing structures. The primary elastic constants are Young's Modulus, Shear Modulus, and Bulk Modulus. Poisson's ratio relates the lateral strain to the axial strain.

Young's Modulus (Modulus of Elasticity) ($ E $)

Young's Modulus measures a material's stiffness or resistance to elongation or compression under tensile or compressive stress. It is defined as the ratio of normal stress to normal strain in the elastic region.

$$ E = \frac{\text{Normal Stress}}{\text{Normal Strain}} = \frac{\sigma}{\epsilon} $$

A higher value of $ E $ indicates a stiffer material, meaning it deforms less under a given load. Steel has a high $ E $, while rubber has a very low $ E $.

Shear Modulus (Modulus of Rigidity) ($ G $)

The Shear Modulus measures a material's resistance to shear deformation. It is defined as the ratio of shear stress to shear strain in the elastic region.

$$ G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{\tau}{\gamma} $$

A higher value of $ G $ indicates that the material is more resistant to twisting or shearing.

Bulk Modulus ($ K $)

The Bulk Modulus measures a material's resistance to uniform compression or expansion under hydrostatic pressure. It is defined as the ratio of hydrostatic stress (pressure) to the resulting volumetric strain.

$$ K = -\frac{P}{\Delta V / V_0} $$

Where:

• $ P $ is the hydrostatic pressure (compressive stress).
• $ \Delta V $ is the change in volume.
• $ V_0 $ is the original volume.

The negative sign is included because an increase in pressure ($ P > 0 $) causes a decrease in volume ($ \Delta V < 0 $), resulting in a positive Bulk Modulus. It quantifies how much a substance will compress volumetrically when subjected to external pressure. Liquids and solids have high bulk moduli, meaning they are difficult to compress.

Poisson's Ratio ($ \nu $)

When a material is stretched in one direction (axially), it tends to contract in the perpendicular directions (laterally). Poisson's ratio ($ \nu $, nu) is the ratio of lateral strain to axial strain in the elastic region.

$$ \nu = -\frac{\text{Lateral Strain}}{\text{Axial Strain}} = -\frac{\epsilon_{lat}}{\epsilon_{axial}} $$

The negative sign is used because when a material is stretched axially (positive axial strain), it contracts laterally (negative lateral strain), resulting in a positive value for $ \nu $. For most engineering materials, $ \nu $ ranges from 0.25 to 0.35. For example, steel has a $ \nu $ of about 0.3.

A material with $ \nu = 0.5 $ is called incompressible, as its volume does not change under elastic deformation (e.g., rubber).

Relationships Between Elastic Constants

For an isotropic and homogeneous material (properties are the same in all directions and at all points), the three elastic constants ($ E $, $ G $, $ K $) and Poisson's ratio ($ \nu $) are related. The most common relationships are:

• $$ E = 2G(1 + \nu) $$
• $$ E = 3K(1 - 2\nu) $$
• $$ G = \frac{E}{2(1 + \nu)} $$
• $$ K = \frac{E}{3(1 - 2\nu)} $$

These relationships are very useful. If any two of the four constants are known, the other two can be calculated.

Example:

A steel rod is subjected to a tensile stress of 100 MPa. If Young's Modulus for steel is 200 GPa and Poisson's ratio is 0.3, calculate the strain.

Using Hooke's Law: $ \sigma = E \epsilon $

$ \epsilon = \sigma / E $

$ \epsilon = \frac{100 \text{ MPa}}{200 \text{ GPa}} = \frac{100 \times 10^6 \text{ Pa}}{200 \times 10^9 \text{ Pa}} = 0.0005 $

The axial strain is 0.0005.

The lateral strain would be $ \epsilon_{lat} = -\nu \epsilon_{axial} = -0.3 \times 0.0005 = -0.00015 $. The rod contracts laterally.

Bending Moments and Shear Force Diagrams

Beams are structural elements that primarily resist loads applied laterally to their longitudinal axis. When a beam is subjected to loads, internal forces develop within it, which vary along its length. The Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) are graphical representations of these internal forces, essential for analyzing the strength and stability of beams.

Shear Force ($ V $)

Shear force at any section of a beam is the algebraic sum of the vertical forces acting on either side of the section. It represents the tendency of one part of the beam to slide vertically relative to the adjacent part.

• Conventionally, upward forces on the left of the section and downward forces on the right are considered positive.
• Downward forces on the left and upward forces on the right are considered negative.

Bending Moment ($ M $)

Bending moment at any section of a beam is the algebraic sum of the moments of all the forces acting on either side of the section. It represents the tendency of the forces to bend the beam.

• Conventionally, moments that cause the beam to sag (concave upwards, like a smile) are considered positive (sagging moment).
• Moments that cause the beam to hog (concave downwards, like a frown) are considered negative (hogging moment).

Relationship Between Load, Shear Force, and Bending Moment

There are fundamental relationships between the distributed load ($ w $), shear force ($ V $), and bending moment ($ M $) along the beam's length ($ x $):

• $$ \frac{dV}{dx} = -w(x) $$ (The rate of change of shear force is equal to the negative of the distributed load intensity).
• $$ \frac{dM}{dx} = V(x) $$ (The rate of change of bending moment is equal to the shear force).

These relationships are crucial for constructing SFDs and BMDs:

• If $ w(x) = 0 $ (no distributed load), $ V $ is constant and $ M $ varies linearly.
• If $ V = 0 $, $ M $ is constant (a point of maximum or minimum bending moment).
• If $ V $ changes sign, $ M $ has a local maximum or minimum.
• The area under the SFD curve between two points equals the change in bending moment between those points.

Constructing SFD and BMD

The process typically involves:

1. Determine Support Reactions: Calculate the vertical and moment reactions at the supports using the equations of static equilibrium ($ \sum F_y = 0 $, $ \sum M = 0 $).
2. Choose a Section: Consider a section at a distance $ x $ from one end of the beam.
3. Calculate Shear Force ($ V(x) $): Sum the vertical forces to the left (or right) of the section.
4. Calculate Bending Moment ($ M(x) $): Sum the moments of forces to the left (or right) of the section about the section itself.
5. Plot the Diagrams: Plot $ V(x) $ and $ M(x) $ against $ x $ along the length of the beam.

Examples of SFD and BMD for Simple Beams:

1. Cantilever Beam with a Point Load at the Free End:

• Load: $ P $ at $ x = L $ (free end). Support at $ x = 0 $.
• Reactions: $ R_A = P $, $ M_A = PL $ (hogging moment).
• SFD: Constant value of $ -P $ from $ x=0 $ to $ x=L $.
• BMD: Varies linearly from $ -PL $ at $ x=0 $ to $ 0 $ at $ x=L $. Maximum moment is at the support.

2. Simply Supported Beam with a Uniformly Distributed Load (UDL):

• Load: $ w $ per unit length over the entire span $ L $.
• Reactions: $ R_A = R_B = wL/2 $.
• SFD: Varies linearly from $ +wL/2 $ at the left support to $ -wL/2 $ at the right support. Zero shear occurs at the center ($ x = L/2 $).
• BMD: Parabolic curve. Maximum positive moment occurs at the center ($ x = L/2 $) where shear is zero. $ M_{max} = wL^2/8 $. Moment is zero at the supports.

Significance:

The maximum shear force and maximum bending moment values obtained from the SFD and BMD are critical for designing the beam. Engineers use these values to ensure that the stresses induced in the beam do not exceed the material's allowable limits, preventing failure. The location of the maximum bending moment is particularly important as it often dictates the beam's cross-sectional requirements.

Torsion

Torsion is a type of deformation that occurs when a structural element is subjected to twisting forces, resulting in a twisting moment or torque. This is commonly experienced by shafts, axles, and screws. The analysis of torsion is crucial for designing components that transmit rotational power.

Torsion in Circular Shafts

The analysis of torsion is simplest for circular cross-sections. When a circular shaft is subjected to a torque ($ T $), it twists, and points on the cross-section move tangentially.

Torsion Formula for Circular Shafts

For a circular shaft of radius $ r $ subjected to a torque $ T $, the shear stress ($ \tau $) at a distance $ \rho $ from the center is given by:

$$ \tau = \frac{T \rho}{J} $$

Where:

• $ T $ is the applied torque.
• $ \rho $ (rho) is the radial distance from the center of the shaft.
• $ J $ is the polar moment of inertia of the cross-section.

The shear stress is maximum at the outer surface ($ \rho = r $) and zero at the center.

$$ \tau_{max} = \frac{T r}{J} $$

Polar Moment of Inertia ($ J $)

The polar moment of inertia is a geometric property that represents the shaft's resistance to torsion.

• For a solid circular shaft of radius $ r $ (or diameter $ d = 2r $): $$ J = \frac{\pi r^4}{2} = \frac{\pi d^4}{32} $$
• For a hollow circular shaft with outer radius $ r_o $ and inner radius $ r_i $: $$ J = \frac{\pi}{2} (r_o^4 - r_i^4) = \frac{\pi}{32} (d_o^4 - d_i^4) $$

Angle of Twist ($ \phi $)

The angle of twist ($ \phi $) is the total angle through which one end of the shaft rotates relative to the other end when subjected to a torque $ T $ over a length $ L $.

$$ \phi = \frac{TL}{JG} $$

Where:

• $ T $ is the applied torque.
• $ L $ is the length of the shaft.
• $ J $ is the polar moment of inertia.
• $ G $ is the shear modulus of the material.

The angle of twist is usually measured in radians.

Power Transmission by Shafts

Shafts are often used to transmit power. The power ($ P $) transmitted by a rotating shaft is related to the torque ($ T $) and angular velocity ($ \omega $).

$$ P = T \omega $$

Where $ \omega $ is in radians per second. If the angular speed is given in revolutions per minute (RPM), $ N $, then $ \omega = \frac{2 \pi N}{60} $.

So, $$ P = T \left( \frac{2 \pi N}{60} \right) $$

If power is in Watts (W) and torque is in Newton-meters (Nm), angular velocity is in rad/s. Common units for power are horsepower (hp) or kilowatts (kW).

Torsion in Non-Circular Shafts

The torsion formula derived for circular shafts ($ \tau = T\rho/J $) is not directly applicable to shafts with non-circular cross-sections (e.g., square, rectangular). For these shapes, the shear stress distribution is non-uniform, and the stress lines are not radial. The analysis requires more advanced techniques, such as finite element methods or empirical formulas derived from experiments.

For a rectangular shaft with sides $ a $ and $ b $ ($ a \ge b $), the maximum shear stress and angle of twist can be approximated using formulas involving a factor that depends on the aspect ratio $ a/b $.

Buckling of Columns

Buckling is a phenomenon where a slender structural member subjected to compressive axial load suddenly deforms laterally. This instability can lead to catastrophic failure even if the compressive stress is below the material's yield strength. Columns are typically vertical members that are susceptible to buckling.

Euler's Formula for Buckling Load

Leonhard Euler derived a formula for the critical buckling load ($ P_{cr} $) of a perfectly straight, slender column with pinned ends (hinged at both ends, allowing rotation but not translation).

$$ P_{cr} = \frac{\pi^2 EI}{(L_e)^2} $$

Where:

• $ E $ is the Young's Modulus of the column material.
• $ I $ is the minimum area moment of inertia of the column's cross-section. This is critical because buckling occurs about the axis with the least resistance to bending.
• $ L_e $ is the effective length of the column, which depends on the end support conditions.

Effective Length ($ L_e $)

The effective length accounts for how the end supports affect the buckling behavior. It is often expressed as $ L_e = K L $, where $ K $ is the effective length factor and $ L $ is the actual length of the column.

Common end conditions and their $ K $ factors:

Slenderness Ratio ($ \lambda $)

The slenderness ratio is a dimensionless quantity that indicates the susceptibility of a column to buckling. It is defined as the ratio of the effective length to the least radius of gyration ($ r $).

$$ \lambda = \frac{L_e}{r} $$

Where $ r = \sqrt{I/A} $, and $ A $ is the cross-sectional area.

Euler's formula can also be expressed in terms of the slenderness ratio:

$$ P_{cr} = \frac{\pi^2 EA}{\lambda^2} $$

This shows that for a given material, the buckling load is inversely proportional to the square of the slenderness ratio.

Limitations of Euler's Formula

Euler's formula is valid only for long, slender columns where the stress at buckling is below the material's proportional limit (i.e., within the elastic range). For shorter, stockier columns, the failure mode is typically yielding due to compressive stress rather than buckling, or a combination of both.

Inelastic Buckling

For intermediate-length columns, buckling may occur after the material has entered the plastic region. This is known as inelastic buckling. Theories like the Tangent Modulus theory (Engesser's theory) are used to analyze inelastic buckling, where the effective modulus $ E $ is replaced by the tangent modulus $ E_t $ at the stress level corresponding to buckling.

Design Considerations

In structural design, columns are often designed to avoid buckling by ensuring they are not too slender or by using materials with high Young's Modulus. Safety factors are applied to the calculated critical buckling load to account for imperfections in columns, material variations, and unexpected load increases.

Thin-Walled Pressure Vessels

Pressure vessels are containers designed to hold fluids (liquids or gases) at a pressure substantially different from the ambient pressure. Thin-walled pressure vessels are those where the ratio of the inner radius ($ r $) to the wall thickness ($ t $) is large (typically $ r/t \ge 10 $). In such vessels, the stress distribution across the thickness is assumed to be uniform.

Types of Thin-Walled Pressure Vessels

Based on their shape, thin-walled pressure vessels are commonly classified into two types:

• Cylindrical Vessels
• Spherical Vessels

Stresses in Cylindrical Pressure Vessels

For a cylindrical pressure vessel with internal pressure $ p $, inner radius $ r $, and wall thickness $ t $, two primary stresses are developed:

1. Hoop Stress ($ \sigma_h $) (Circumferential Stress)

This stress acts tangential to the circumference of the cylinder and tends to split the cylinder along its length. Consider a longitudinal section through the center of the cylinder. The internal pressure $ p $ acts on the projected area ($ 2r \times L $, where $ L $ is the length), creating an outward force. This force is resisted by the hoop stress acting on the inner surface area of the walls ($ 2t \times L $).

Equating the forces:

$$ p (2rL) = \sigma_h (2tL) $$

Therefore, the hoop stress is:

$$ \sigma_h = \frac{pr}{t} $$

Hoop stress is generally twice the longitudinal stress.

2. Longitudinal Stress ($ \sigma_l $) (Axial Stress)

This stress acts parallel to the axis of the cylinder and tends to split the cylinder across its ends. Consider a transverse section cutting the cylinder into two halves. The internal pressure $ p $ acts on the circular area ($ \pi r^2 $) creating an outward force. This force is resisted by the longitudinal stress acting on the wall's cross-sectional area ($ 2\pi r t $).

Equating the forces:

$$ p (\pi r^2) = \sigma_l (2\pi rt) $$

Therefore, the longitudinal stress is:

$$ \sigma_l = \frac{pr}{2t} $$

Note that $ \sigma_h = 2 \sigma_l $. The higher hoop stress is usually the critical factor in designing cylindrical pressure vessels.

Stresses in Spherical Pressure Vessels

For a spherical pressure vessel with internal pressure $ p $, inner radius $ r $, and wall thickness $ t $, the stress is uniform in all directions along the tangent to the surface. This stress is often called the circumferential stress or meridional stress.

Consider a hemisphere subjected to internal pressure. The pressure acts on the projected area of the circle ($ \pi r^2 $), creating an outward force. This force is resisted by the stress acting on the wall's cross-sectional area ($ 2\pi r t $).

Equating the forces:

$$ p (\pi r^2) = \sigma (2\pi rt) $$

Therefore, the stress in a spherical pressure vessel is:

$$ \sigma = \frac{pr}{2t} $$

In a spherical vessel, the stress is the same in all tangential directions.

Comparison

Comparing the stresses:

• Cylindrical vessel: $ \sigma_h = pr/t $, $ \sigma_l = pr/(2t) $. Maximum stress is $ pr/t $.
• Spherical vessel: $ \sigma = pr/(2t) $.

This means that for the same radius, thickness, and pressure, a cylindrical vessel experiences twice the stress in the hoop direction compared to the stress in a spherical vessel of the same radius and thickness. Therefore, spherical vessels are more efficient in handling pressure.

Design Considerations

When designing pressure vessels, engineers must ensure that the maximum stress (usually the hoop stress in cylindrical vessels) is less than the allowable stress for the material. The allowable stress is typically the yield strength or ultimate tensile strength divided by a factor of safety.

$$ \sigma_{allowable} = \frac{\text{Yield Strength}}{FS} \quad \text{or} \quad \frac{\text{Ultimate Tensile Strength}}{FS} $$

The required thickness $ t $ can then be calculated:

For cylindrical vessels: $ t \ge \frac{pr}{\sigma_{allowable}} $

For spherical vessels: $ t \ge \frac{pr}{2\sigma_{allowable}} $

In practice, a small additional thickness is usually added to account for corrosion, erosion, and manufacturing tolerances.