Sources and References

Primary textbook — R.C. Hibbeler, Mechanics of Materials, 11th ed., Pearson, 2014.

Supplementary — R.C. Hibbeler, Engineering Mechanics: Statics, 14th ed., Pearson, 2015.

Online resources — MIT OpenCourseWare 2.001 Mechanics and Materials I (Fall 2006); Stanford University ME 180 Mechanics of Solids lecture notes; David J. Morin, Introduction to Classical Mechanics, Cambridge University Press (statics chapters).

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Chapter 1: Statics Review — Forces, Moments, and Equilibrium

Statics is the branch of mechanics that studies bodies at rest or moving with constant velocity — bodies for which the net force and net moment are both zero. Before any material can be analyzed for internal stress and deformation, the external forces acting on a structure must be fully resolved. This chapter reviews the essential tools: vector decomposition, equilibrium conditions, moments, couples, distributed loads, and free-body diagrams.

1.1 Force Vectors in Two and Three Dimensions

A force is a vector quantity possessing both magnitude and direction. In two dimensions, a force \( \mathbf{F} \) of magnitude \( F \) acting at an angle \( \theta \) measured counterclockwise from the positive \( x \)-axis has Cartesian components:

\[ F_x = F \cos\theta, \qquad F_y = F \sin\theta \]

The resultant of a system of concurrent forces is obtained by summing each Cartesian component independently:

\[ R_x = \sum F_x, \qquad R_y = \sum F_y, \qquad R = \sqrt{R_x^2 + R_y^2} \]

In three dimensions, a force is expressed in Cartesian form as:

\[ \mathbf{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} \]

with magnitude \( F = \sqrt{F_x^2 + F_y^2 + F_z^2} \). The direction cosines \( \cos\alpha, \cos\beta, \cos\gamma \) are the direction angles that \( \mathbf{F} \) makes with the \( x \)-, \( y \)-, and \( z \)-axes respectively:

\[ \cos\alpha = \frac{F_x}{F}, \qquad \cos\beta = \frac{F_y}{F}, \qquad \cos\gamma = \frac{F_z}{F} \]

and they satisfy \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \).

1.1.1 Position Vectors and Force Along a Line

When a force acts along the line connecting two points \( A(x_A, y_A, z_A) \) and \( B(x_B, y_B, z_B) \), the unit vector along \( AB \) is:

\[ \hat{u}_{AB} = \frac{\mathbf{r}_{AB}}{|\mathbf{r}_{AB}|}, \qquad \mathbf{r}_{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} + (z_B - z_A)\hat{k} \]

so the force is simply \( \mathbf{F} = F\hat{u}_{AB} \). This is the standard technique for cable and truss member forces.

1.2 Equilibrium of a Particle

A particle is in equilibrium when the vector sum of all forces acting on it is zero. In 2D and 3D respectively:

\[ \sum F_x = 0, \quad \sum F_y = 0 \qquad \text{(2D)} \]\[ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0 \qquad \text{(3D)} \]

The procedure is always: (1) isolate the particle, (2) draw the free-body diagram (FBD) with all external forces, (3) resolve into components, (4) solve the resulting algebraic equations.

1.2.1 Trusses

A truss is an assemblage of two-force members connected at frictionless pins. Each member carries only axial force — tension or compression. Two classical methods of analysis are the method of joints and the method of sections.

In the method of joints, equilibrium is written for each joint (treated as a particle) in turn, yielding two scalar equations per joint in 2D. In the method of sections, a cut is passed through at most three members whose forces are unknown, and moment equilibrium about a clever point can isolate a single unknown.

1.3 Moment of a Force

The tendency of a force to rotate a body about a point or axis is measured by its moment (torque). For a 2D planar problem, the moment of force \( \mathbf{F} \) about point \( O \) has magnitude:

\[ M_O = F \cdot d \]

where \( d \) is the perpendicular distance (moment arm) from \( O \) to the line of action of \( \mathbf{F} \). The sign is positive counterclockwise by convention.

In three dimensions, the cross product gives:

\[ \mathbf{M}_O = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} = (r_y F_z - r_z F_y)\hat{i} - (r_x F_z - r_z F_x)\hat{j} + (r_x F_y - r_y F_x)\hat{k} \]

The scalar moment about a specific axis \( a \) (say, the \( z \)-axis) is the component of \( \mathbf{M}_O \) along that axis.

1.3.1 Moment of a Couple

A couple is a pair of forces equal in magnitude, opposite in direction, and separated by a distance \( d \). The resultant force of a couple is zero, but the net moment is non-zero:

\[ M = F \cdot d \]

1.4 Equivalent Force Systems and Resultants

Any system of forces and couples can be reduced to a single resultant force \( \mathbf{F}_R \) plus a resultant moment \( \mathbf{M}_{R,O} \) about a chosen point \( O \):

\[ \mathbf{F}_R = \sum \mathbf{F}_i, \qquad \mathbf{M}_{R,O} = \sum \mathbf{M}_O + \sum \mathbf{M}_{\text{couple},i} \]

If the resultant force is non-zero, the system can often be further reduced to a single force acting at a shifted point — the location where \( \mathbf{M}_{R} = 0 \).

1.4.1 Distributed Loads

A distributed load (load per unit length, \( w(x) \) in N/m) acting along a beam is statically equivalent to a single concentrated force equal to the area under the load diagram, acting at the centroid of that area.

For a uniform distributed load of intensity \( w_0 \) over length \( L \):

\[ F_R = w_0 L, \qquad \bar{x} = \frac{L}{2} \]

For a triangular load varying from 0 at one end to \( w_0 \) at the other end over length \( L \):

\[ F_R = \frac{1}{2} w_0 L, \qquad \bar{x} = \frac{2L}{3} \text{ from the zero end} \]

1.5 Equilibrium of a Rigid Body

A rigid body in static equilibrium satisfies:

\[ \sum \mathbf{F} = \mathbf{0}, \qquad \sum \mathbf{M}_O = \mathbf{0} \]

for any point \( O \). In 2D this gives three independent scalar equations: \( \sum F_x = 0 \), \( \sum F_y = 0 \), \( \sum M_O = 0 \). With three equations, at most three unknowns can be determined — a structure with exactly three unknowns is statically determinate.

Supports introduce reaction forces and moments. A roller provides one force (normal to surface), a pin provides two force components, and a fixed (cantilever) support provides two force components plus one moment.

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Chapter 2: Internal Forces and Cross-Section Properties

Once external reactions are known, internal forces at any cross-section can be determined by the method of sections. Cross-section geometric properties — centroids and moments of inertia — govern how internal forces produce stress.

2.1 Internal Forces and Free-Body Diagrams

To find the internal forces at a cross-section, make an imaginary cut, retain one portion, and apply equilibrium. The cut exposes an internal normal force \( N \), shear force \( V \), and bending moment \( M \) (in a 2D beam).

The procedure: (1) compute support reactions, (2) cut at the desired location \( x \), (3) draw FBD of the left (or right) portion, (4) write \( \sum F_y = 0 \) and \( \sum M_{\text{cut}} = 0 \) to find \( V \) and \( M \).

2.2 Shear Force and Bending Moment Diagrams

The distributions \( V(x) \) and \( M(x) \) along the beam length are the shear and moment diagrams. Two differential relations connect the load, shear, and moment:

Key features to remember:

• Where a concentrated force \( P \) acts, \( V \) jumps by \( P \).
• Where a concentrated couple \( M_0 \) acts, \( M \) jumps by \( M_0 \).
• \( M \) is maximum where \( V = 0 \) (provided no point couples exist there).
• Under a uniform load, \( V \) is linear and \( M \) is parabolic.

2.3 Centroids of Cross-Sections

The centroid of a plane area is the geometric center, defined by:

\[ \bar{x} = \frac{\int x\,dA}{A}, \qquad \bar{y} = \frac{\int y\,dA}{A} \]

For composite areas (built from simple shapes), the centroid is found by a weighted average:

\[ \bar{x} = \frac{\sum \tilde{x}_i A_i}{\sum A_i}, \qquad \bar{y} = \frac{\sum \tilde{y}_i A_i}{\sum A_i} \]

where \( \tilde{x}_i, \tilde{y}_i \) are the centroidal coordinates of each sub-area \( A_i \). Holes are treated as negative areas.

2.4 Moment of Inertia and the Parallel-Axis Theorem

The second moment of area (moment of inertia) about the centroidal \( x \)- and \( y \)-axes measures the distribution of area relative to those axes:

\[ \bar{I}_x = \int y^2\,dA, \qquad \bar{I}_y = \int x^2\,dA \]

Standard results for simple shapes (centroidal axes):

• Rectangle \( b \times h \): \( \bar{I}_x = \frac{bh^3}{12} \), \( \bar{I}_y = \frac{hb^3}{12} \)
• Circle radius \( r \): \( \bar{I}_x = \bar{I}_y = \frac{\pi r^4}{4} \)
• Triangle (base \( b \), height \( h \)): \( \bar{I}_x = \frac{bh^3}{36} \)

For composite sections, the total moment of inertia is:

\[ I_{\text{total}} = \sum \left(\bar{I}_i + A_i d_i^2\right) \]

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Chapter 3: Stress and Strain

With the internal forces known and the cross-section geometry characterized, we can define the fundamental quantities that describe the mechanical state at a point in a solid body: stress and strain.

3.1 Normal Stress

The assumption of uniform normal stress is valid when the load is applied at the centroid of the section and the cross-section is remote from any stress concentrations (Saint-Venant’s principle, discussed in Chapter 4).

3.2 Shear Stress

Shear stress has two subscripts: the first indicates the face normal direction, the second indicates the direction of the stress component. By the complementary property of shear: \( \tau_{xy} = \tau_{yx} \), \( \tau_{xz} = \tau_{zx} \), \( \tau_{yz} = \tau_{zy} \).

3.2.1 Bearing Stress

Bearing stress acts on the contact area between two bodies:

\[ \sigma_b = \frac{P}{A_b} \]

where \( A_b = t \cdot d \) is the projected area (thickness times pin diameter) for a pin-in-hole connection.

3.3 Factor of Safety

3.4 Normal Strain

3.5 Shear Strain

3.6 Mechanical Properties of Materials

3.6.1 The Stress–Strain Diagram

The stress–strain diagram is obtained by loading a standard tensile specimen and recording \( \sigma \) versus \( \varepsilon \). For a typical mild steel, the diagram exhibits:

1. Proportional limit: up to this stress, \( \sigma \) and \( \varepsilon \) are linearly proportional.
2. Elastic limit: the maximum stress for which the material is fully elastic (no permanent deformation).
3. Yield point: stress at which significant permanent deformation begins (often accompanied by a visible yield plateau in mild steel).
4. Ultimate stress \( \sigma_u \): the maximum engineering stress.
5. Fracture stress: the stress at which the specimen breaks (lower than \( \sigma_u \) in engineering stress due to necking).

3.6.2 Ductility and Toughness

3.6.3 True Stress and True Strain

Engineering stress and strain are computed using original dimensions. True (Cauchy) stress and logarithmic strain use the current (deformed) dimensions:

\[ \sigma_{\text{true}} = \frac{P}{A_{\text{current}}}, \qquad \varepsilon_{\text{true}} = \ln\!\left(\frac{L_f}{L_0}\right) \]

True stress is always higher than engineering stress beyond the yield point because the cross-sectional area decreases. True strain is related to engineering strain by \( \varepsilon_{\text{true}} = \ln(1 + \varepsilon_{\text{eng}}) \).

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Chapter 4: Axial Loading

Axial loading is the simplest case of internal force application: a prismatic bar subjected to forces along its centroidal axis. This chapter derives the deformation formula, treats statically indeterminate bars, introduces thermal effects, and discusses Saint-Venant’s principle.

4.1 Deformation of an Axially Loaded Bar

Consider a bar of original length \( L \), cross-sectional area \( A \), and elastic modulus \( E \), subjected to axial force \( N \). From Hooke’s Law:

\[ \sigma = \frac{N}{A} = E\varepsilon = E\frac{\delta}{L} \]

Solving for deformation:

\[ \delta = \frac{NL}{AE} \]

The quantity \( AE \) is called the axial stiffness of the member.

4.2 Statically Indeterminate Axial Structures

A structure is statically indeterminate when the number of unknown reactions exceeds the number of independent equilibrium equations. Additional equations come from compatibility (geometric) conditions.

The three-step approach for statically indeterminate problems:

1. Equilibrium: write \( \sum F = 0 \) involving the unknowns.
2. Compatibility: express the geometric constraint (e.g., total deformation is zero for a fixed–fixed bar).
3. Force–deformation: substitute \( \delta = NL/(AE) \) to obtain additional equations.

4.3 Thermal Strain

A temperature change \( \Delta T \) causes a material to expand or contract. The thermal strain is:

\[ \varepsilon_T = \alpha \Delta T \]

where \( \alpha \) is the coefficient of thermal expansion (units: 1/°C or 1/K). For steel, \( \alpha \approx 12 \times 10^{-6} \) /°C; for aluminum, \( \alpha \approx 23 \times 10^{-6} \) /°C.

The thermal deformation of a free bar:

\[ \delta_T = \alpha \Delta T L \]

If the bar is constrained, the thermal deformation is suppressed and a thermal stress develops. Compatibility requires the mechanical deformation to equal the negative of the thermal deformation:

\[ \delta_{\text{mech}} + \delta_T = 0 \implies \frac{\sigma L}{E} + \alpha \Delta T L = 0 \implies \sigma = -E\alpha\Delta T \]

A temperature increase in a constrained bar produces compressive stress.

4.4 Saint-Venant’s Principle

This principle justifies the use of \( \sigma = N/A \) for prismatic bars away from connections, holes, or abrupt changes in geometry. At stress concentrations, the local peak stress exceeds the average by a stress concentration factor \( K \):

\[ \sigma_{\max} = K \sigma_{\text{avg}} \]

Values of \( K \) depend on the geometry and are tabulated in reference handbooks. For a circular hole in a wide plate under uniaxial tension, \( K = 3 \) at the edges of the hole.

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Chapter 5: Torsion

Torsion arises when a structural member is twisted about its longitudinal axis by applied torques. The analysis of circular shafts is exact within linear elasticity; non-circular sections require more advanced methods (not covered in ME 219).

5.1 Shear Stress in Circular Shafts

Consider a solid circular shaft of radius \( c \) and length \( L \), subjected to an internal torque \( T \). The fundamental assumptions are: plane cross-sections remain plane and circular after twisting; radii remain straight; material is linear elastic and homogeneous; the shaft is in pure torsion (no bending or axial load).

5.2 Angle of Twist

From the derivation above, the angle of twist \( \phi \) over a length \( L \) under torque \( T \) is:

\[ \phi = \frac{TL}{GJ} \]

For a shaft with multiple segments or a varying torque distribution:

\[ \phi = \sum_{i=1}^n \frac{T_i L_i}{G_i J_i}, \qquad \text{or} \qquad \phi = \int_0^L \frac{T(x)}{GJ(x)}\,dx \]

The sign convention for torque follows the right-hand rule: positive torque on a cross-section face whose outward normal points in the positive \( x \)-direction means the torque vector points in the positive \( x \)-direction.

5.3 Statically Indeterminate Torsion

When a shaft is fixed at both ends, the torque reactions cannot be determined from equilibrium alone. The procedure parallels the axial case:

1. Equilibrium: \( T_A + T_B = T_{\text{applied}} \)
2. Compatibility: total angle of twist of one end relative to the other is zero.
3. Force–deformation: \( \phi = TL/(GJ) \).

5.4 Power Transmission

Rotating shafts transmit mechanical power. The relationship between power \( P \), torque \( T \), and angular velocity \( \omega \) (in rad/s) is:

\[ P = T\omega = T(2\pi f) = T(2\pi N/60) \]

where \( f \) is frequency in Hz and \( N \) is rotation speed in rpm. Solving for the required torque for a given power:

\[ T = \frac{P}{\omega} \]

This torque is then used in the torsion formula to check that \( \tau_{\max} \le \tau_{\text{allow}} \) and that the angle of twist does not exceed the design limit.

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Chapter 6: Bending

Bending is the most common loading mode in structural engineering. When a beam carries transverse loads, internal bending moments develop that produce normal stresses varying across the cross-section. This chapter derives the flexure formula and extends it to composite sections and eccentric axial loading.

6.1 Assumptions in Pure Bending

The Euler–Bernoulli beam theory rests on several assumptions:

1. The beam is initially straight and has a uniform cross-section.
2. The material is linear elastic, homogeneous, and isotropic.
3. The cross-section is symmetric about the plane of bending.
4. Plane cross-sections remain plane after bending (the kinematic hypothesis).
5. Deformations are small.

6.2 The Flexure Formula

The section modulus \( S \) is a single cross-section property that encapsulates both \( I \) and the extreme-fiber distance \( c \). Maximizing \( S \) for a given cross-sectional area is the design goal: I-beams and wide-flange sections concentrate material far from the neutral axis to maximize \( S \) and hence minimize bending stress.

6.3 Composite Beams

A composite beam is made of two or more materials bonded together. The analysis uses the transformed-section method: convert all materials to an equivalent cross-section in a single reference material (usually the stiffer one).

The key insight: both materials have the same strain at the same \( y \)-location (perfect bonding, plane sections remain plane), but the stress differs by the factor \( n \) due to the different stiffnesses.

6.4 Eccentric Axial Loading

When an axial load is applied eccentrically (not at the centroid of the cross-section), it produces both axial normal stress and bending stress. The total normal stress is the superposition:

\[ \sigma = \frac{N}{A} \pm \frac{M y}{I} \]

where \( M = N \cdot e \) is the bending moment created by the eccentricity \( e \) (distance from centroid to load line).

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Supplementary Topics and Integration

Review: Putting It Together — A Complete Structural Analysis

Modern structural analysis in ME 219 proceeds through the following logical chain, which unifies all six chapters:

1. Statics (Chapters 1–2): Find external reactions using equilibrium of the entire structure. Then cut at the section of interest and draw a free-body diagram to find internal forces \( N \), \( V \), \( M \), \( T \).

2. Cross-section geometry (Chapter 2): Calculate \( A \), \( \bar{y} \), \( I \), \( J \), \( c \) for the cross-section.

3. Stress (Chapters 3–6): Apply the appropriate formula:

   • Axial: \( \sigma = N/A \)
   • Bending: \( \sigma = My/I \)
   • Torsion: \( \tau = T\rho/J \)
   • Shear (transverse): \( \tau = VQ/(It) \) (advanced, beyond ME 219 scope here)

4. Deformation (Chapters 4–5): Calculate elongation \( \delta = NL/(AE) \) or twist \( \phi = TL/(GJ) \) to check serviceability.

5. Design check: Compare computed stresses against allowable values and verify factors of safety.

Sign Convention Summary

Consistency in sign conventions is essential. The following conventions are adopted throughout:

Quantity	Positive
Normal stress \( \sigma \)	Tensile
Axial force \( N \)	Tensile (away from face)
Shear force \( V \)	Left-portion upward on right face
Bending moment \( M \)	Sagging (concave up), bottom in tension
Torque \( T \)	Right-hand rule along positive \( x \)-axis
Normal strain \( \varepsilon \)	Elongation
Shear strain \( \gamma \)	Decrease in right angle
Coordinate \( y \) (bending)	Upward from neutral axis

Common Errors and How to Avoid Them

Key Formulas Reference

The following table collects the central formulas from all chapters:

Topic	Formula	Conditions
Axial stress	\( \sigma = N/A \)	Uniform, centroidal
Axial deformation	\( \delta = NL/(AE) \)	Prismatic, elastic
Thermal deformation	\( \delta_T = \alpha \Delta T L \)	Unconstrained
Bending stress	\( \sigma = My/I \)	Elastic, homogeneous
Max bending stress	\( \sigma_{\max} = Mc/I = M/S \)	At extreme fiber
Torsional shear stress	\( \tau = T\rho/J \)	Circular shaft, elastic
Max torsional stress	\( \tau_{\max} = Tc/J \)	At outer surface
Angle of twist	\( \phi = TL/(GJ) \)	Uniform shaft
Parallel-axis theorem	\( I = \bar{I} + Ad^2 \)	Any parallel axis
Polar MOI (solid)	\( J = \pi c^4/2 \)	Solid circle
Section modulus	\( S = I/c \)	Bending design
Shear modulus	\( G = E/\left[2(1+\nu)\right] \)	Isotropic material
Poisson’s ratio	\( \nu = -\varepsilon_{\text{lat}}/\varepsilon_{\text{long}} \)	Uniaxial load

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Appendix: Physical Intuition and Conceptual Synthesis

Why the Neutral Axis Passes Through the Centroid

This result follows directly from the requirement that net axial force be zero in pure bending. Mathematically, \( \int_A \sigma\,dA = 0 \) with \( \sigma \propto y \) forces \( \int_A y\,dA = 0 \), which defines the centroid. Physically: if the neutral axis were above the centroid, more area would be in tension than compression, creating a net tensile force that would violate equilibrium in the absence of an applied axial load.

Energy Methods Preview

The elastic strain energy stored per unit volume in a member under normal stress is:

\[ u = \frac{\sigma^2}{2E} \]

For a bar under axial load, integrating over the volume:

\[ U = \int_0^L \frac{N^2}{2AE}\,dx \]

For a shaft under torsion:

\[ U = \int_0^L \frac{T^2}{2GJ}\,dx \]

These expressions underlie Castigliano’s theorem and virtual work methods used in ME 321 and beyond.

The Importance of Linearity

All formulas in ME 219 assume linear elastic behavior: the constitutive law is Hooke’s law, deformations are small (so geometry does not change appreciably under load), and superposition is valid. These are excellent approximations for metals at stresses well below yield. The course lays the essential foundation; nonlinear behavior, large deformations, plasticity, and fracture mechanics are treated in upper-year courses.