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4.2 Engineering Mechanics & Strength of Materials
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1 Chapter 1: Introduction & Board Exam Strategy
1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy
2 Chapter 2: Mathematics Foundations
2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications
3 Chapter 3: Advanced Mathematics for ECE
3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability
4 Chapter 4: GEAS - Engineering Sciences & Economics
4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming
5 Chapter 5: GEAS - Management, Materials, Environment & CAD
5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics
6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts
6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)
7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes
7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators
8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits
8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design
9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems
9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory
10 Chapter 10: EST - Communications Fundamentals & Transmission Lines
10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart
11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems
11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations
4.2 Engineering Mechanics & Strength of Materials
Key Takeaways Explain This Topic AI
Equilibrium conditions for 2D rigid bodies require ∑ F x = 0 \sum F_x = 0 ∑ F x = 0 , ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 , and ∑ M z = 0 \sum M_z = 0 ∑ M z = 0 at any arbitrary point.
Axial stress σ = P A \sigma = \frac{P}{A} σ = A P and normal strain ϵ = δ L \epsilon = \frac{\delta}{L} ϵ = L δ , linked by Hooke's Law σ = E ϵ \sigma = E\epsilon σ = E ϵ , yield axial deformation δ = P L A E \delta = \frac{PL}{AE} δ = A E P L .
Torsional shear stress in circular shafts is τ = T r J \tau = \frac{T r}{J} τ = J T r , where polar moment of inertia J = π d 4 32 J = \frac{\pi d^4}{32} J = 32 π d 4 for solid shafts.
Flexural stress in beams under bending is governed by the flexure formula σ = − M y I \sigma = -\frac{M y}{I} σ = − I M y , reaching maximum value σ m a x = M S \sigma_{max} = \frac{M}{S} σ ma x = S M at extreme fibers.
Centroid of composite areas is computed via x ˉ = ∑ A i x ˉ i ∑ A i \bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i} x ˉ = ∑ A i ∑ A i x ˉ i , and moment of inertia is shifted using the Parallel Axis Theorem I = I c + A d 2 I = I_c + A d^2 I = I c + A d 2 .
Last updated: July 2026
- 4.2 Engineering Mechanics & Strength of Materials
- 1. Statics of Rigid Bodies
Statics investigates bodies at rest or moving at constant velocity under the action of balanced force systems.
Conditions for Static Equilibrium
For a two-dimensional coplanar force system acting on a rigid body, static equilibrium requires three scalar equations: ∑ F x = 0 , ∑ F y = 0 , ∑ M z = 0 \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_z = 0 ∑ F x = 0 , ∑ F y = 0 , ∑ M z = 0
where ∑ M z \sum M_z ∑ M z is the summation of moments about any arbitrary axis perpendicular to the plane.
Moment of a Force & Couples
The moment M ⃗ O \vec{M}_O M O of a force F ⃗ \vec{F} F about a point O O O is a vector cross product: M ⃗ O = r ⃗ × F ⃗ = F ⋅ d \vec{M}_O = \vec{r} \times \vec{F} = F \cdot d M O = r × F = F ⋅ d
where r ⃗ \vec{r} r is the position vector from point O O O to any point on the line of action of F ⃗ \vec{F} F , and d d d is the perpendicular distance (moment arm).
Friction (Dry / Coulomb Friction)
When two contacting surfaces tend to slide relative to each other, a tangential friction force f f f develops:
Static friction: f s ≤ μ s N f_s \le \mu_s N f s ≤ μ s N (Maximum static friction f s , max = μ s N f_{s,\text{max}} = \mu_s N f s , max = μ s N occurs at impending motion)
- Kinetic friction: f k = μ k N f_k = \mu_k N f k = μ k N (where μ k < μ s \mu_k < \mu_s μ k < μ s )
- Angle of static friction: tan ϕ s = μ s \tan\phi_s = \mu_s tan ϕ s = μ s
- Centroids & Moment of Inertia
The centroid ( x ˉ , y ˉ ) (\bar{x}, \bar{y}) ( x ˉ , y ˉ ) of a composite area is determined by dividing it into simple geometric shapes: x ˉ = ∑ A i x ˉ i ∑ A i , y ˉ = ∑ A i y ˉ i ∑ A i \bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i}, \quad \bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i} x ˉ = ∑ A i ∑ A i x ˉ i , y ˉ = ∑ A i ∑ A i y ˉ i
The area moment of inertia I x I_x I x quantifies the resistance of an area to bending about an axis: I x = ∫ y 2 d A I_x = \int y^2 dA I x = ∫ y 2 d A
Parallel Axis Theorem : The moment of inertia I x I_x I x about any axis parallel to a centroidal axis I x ˉ I_{\bar{x}} I x ˉ at distance d d d is: I x = I x ˉ + A d 2 I_x = I_{\bar{x}} + A d^2 I x = I x ˉ + A d 2
Shape Area A A A Centroidal Moment of Inertia I x ˉ I_{\bar{x}} I x ˉ Polar Moment J c J_c J c
Rectangle ( b × h b \times h b × h ) b h b h bh b h 3 12 \frac{b h^3}{12} 12 b h 3 b h ( b 2 + h 2 ) 12 \frac{b h(b^2 + h^2)}{12} 12 bh ( b 2 + h 2 )
Circle (diameter d d d ) π d 2 4 \frac{\pi d^2}{4} 4 π d 2 π d 4 64 \frac{\pi d^4}{64} 64 π d 4 π d 4 32 \frac{\pi d^4}{32} 32 π d 4
- Triangle (base b b b , height h h h ) b h 2 \frac{b h}{2} 2 bh b h 3 36 \frac{b h^3}{36} 36 b h 3 N/A
- 2. Dynamics of Moving Bodies
- Kinematics of Rectilinear Motion (Constant Acceleration a a a )
- v = v 0 + a t v = v_0 + at v = v 0 + a t
- s = s 0 + v 0 t + 1 2 a t 2 s = s_0 + v_0 t + \frac{1}{2} a t^2 s = s 0 + v 0 t + 2 1 a t 2
- v 2 = v 0 2 + 2 a ( s − s 0 ) v^2 = v_0^2 + 2a(s - s_0) v 2 = v 0 2 + 2 a ( s − s 0 )
- s − s 0 = ( v 0 + v 2 ) t s - s_0 = \left(\frac{v_0 + v}{2}\right) t s − s 0 = ( 2 v 0 + v ) t
- Kinetics & Newton's Second Law
When an unbalanced force system acts on a particle of mass m m m , it accelerates in the direction of the resultant force: ∑ F ⃗ = m a ⃗ \sum \vec{F} = m \vec{a} ∑ F = m a
Work-Energy Principle
The total work U 1 − 2 U_{1-2} U 1 − 2 performed by all forces on a body equals the change in kinetic energy Δ T \Delta T Δ T : U 1 − 2 = T 2 − T 1 = 1 2 m v 2 2 − 1 2 m v 1 2 U_{1-2} = T_2 - T_1 = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 U 1 − 2 = T 2 − T 1 = 2 1 m v 2 2 − 2 1 m v 1 2
Impulse-Momentum Principle
The linear impulse J ⃗ \vec{J} J of a force acting over time interval Δ t \Delta t Δ t equals the change in linear momentum Δ p ⃗ \Delta \vec{p} Δ p : J ⃗ = ∫ t 1 t 2 F ⃗ d t = m v ⃗ 2 − m v ⃗ 1 \vec{J} = \int_{t_1}^{t_2} \vec{F} dt = m\vec{v}_2 - m\vec{v}_1 J = ∫ t 1 t 2 F d t = m v 2 − m v 1
3. Strength of Materials & Stress-Strain Analysis
Strength of Materials analyzes internal stress distributions and structural deformations under external loading.
Normal Stress & Strain
Normal Stress σ \sigma σ : Force per unit area acting perpendicular to the cross section: σ = P A \sigma = \frac{P}{A} σ = A P
Normal Strain ϵ \epsilon ϵ : Fractional change in length: ϵ = δ L 0 \epsilon = \frac{\delta}{L_0} ϵ = L 0 δ
Hooke's Law : Within the elastic limit of a material: σ = E ϵ \sigma = E \epsilon σ = E ϵ where E E E is Young's Modulus of Elasticity (e.g., E steel ≈ 200 GPa E_{\text{steel}} \approx 200 \text{ GPa} E steel ≈ 200 GPa ).
Axial Deformation Formula
Combining σ = P A \sigma = \frac{P}{A} σ = A P , ϵ = δ L \epsilon = \frac{\delta}{L} ϵ = L δ , and σ = E ϵ \sigma = E\epsilon σ = E ϵ yields the total elongation δ \delta δ of an axially loaded bar: δ = P L A E \delta = \frac{P L}{A E} δ = A E P L
Thermal Stress & Strain
When a constrained bar experiences a temperature change Δ T \Delta T Δ T , internal thermal stress arises: δ thermal = α L Δ T ⟹ σ thermal = E α Δ T \delta_{\text{thermal}} = \alpha L \Delta T \implies \sigma_{\text{thermal}} = E \alpha \Delta T δ thermal = αL Δ T ⟹ σ thermal = E α Δ T
where α \alpha α is the coefficient of thermal expansion.
Poisson's Ratio & Shear Modulus
Poisson's Ratio ν \nu ν : Ratio of lateral strain to axial strain under uniaxial loading: ν = − ϵ lateral ϵ axial \nu = -\frac{\epsilon_{\text{lateral}}}{\epsilon_{\text{axial}}} ν = − ϵ axial ϵ lateral
Shear Modulus G G G (Modulus of Rigidity): Relates shear stress τ \tau τ to shear strain γ \gamma γ : τ = G γ , G = E 2 ( 1 + ν ) \tau = G \gamma, \quad G = \frac{E}{2(1 + \nu)} τ = G γ , G = 2 ( 1 + ν ) E
- 4. Shaft Torsion & Beam Flexure
- Torsion of Circular Shafts
When a torque T T T is applied to a circular shaft of radius r = d / 2 r = d/2 r = d /2 and length L L L , the maximum shear stress τ max \tau_{\text{max}} τ max occurs at the outer boundary: τ max = T r J = 16 T π d 3 \tau_{\text{max}} = \frac{T r}{J} = \frac{16 T}{\pi d^3} τ max = J T r = π d 3 16 T
where J = π d 4 32 J = \frac{\pi d^4}{32} J = 32 π d 4 is the polar moment of inertia for a solid circular cross section.
The angle of twist ϕ \phi ϕ (in radians) over length L L L is: ϕ = T L G J \phi = \frac{T L}{G J} ϕ = G J T L
Flexural Stress in Beams (Flexure Formula)
When a beam is subjected to a bending moment M M M , internal normal stresses σ \sigma σ vary linearly with distance y y y from the neutral axis: σ = − M y I \sigma = -\frac{M y}{I} σ = − I M y
The maximum flexural stress σ max \sigma_{\text{max}} σ max at the extreme fiber ( y = c y = c y = c ) is: σ max = M c I = M S \sigma_{\text{max}} = \frac{M c}{I} = \frac{M}{S} σ max = I M c = S M
where S = I c S = \frac{I}{c} S = c I is the Section Modulus of the beam.