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3.2 Statics, Dynamics, and Mechanics of Materials Core
FE Industrial and Systems Study Guide Exam
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1 Chapter 1: Exam Overview, CBT Strategy & Handbook Navigation
1.1 FE Industrial Exam Structure, CBT Logistics, and Pacing 1.2 Using the NCEES FE Reference Handbook and Approved Calculators
2 Chapter 2: Mathematics for Industrial Engineers
2.1 Analytic Geometry, Coordinate Systems, and Trigonometry 2.2 Calculus: Differential, Integral, and Series Approximations 2.3 Linear Algebra: Matrix Operations, Systems of Equations, and Vectors
3 Chapter 3: Engineering Sciences Core
3.1 Thermodynamics Fundamentals and Fluid Flow Principles 3.2 Statics, Dynamics, and Mechanics of Materials Core 3.3 Electrical Circuits, DC/AC Analysis, and Power in Industrial Systems
4 Chapter 4: Ethics and Professional Practice
4.1 NCEES Model Rules of Professional Conduct and Technical Society Ethics 4.2 Engineering Licensure, Public Welfare, and Professional Liability 4.3 Engineering Contracts, Intellectual Property, and Conflicts of Interest
5 Chapter 5: Engineering Economics
5.1 Time Value of Money, Discount Factors, and Equivalence 5.2 Evaluating Alternatives: Present Worth, Equivalent Annual Cost, and IRR 5.3 Incremental Rate of Return, Replacement Analysis, and Break-Even 5.4 Depreciation Methods, MACRS, Income Taxes, and Inflation
6 Chapter 6: Probability Foundations & Distributions
6.1 Basic Probability Concepts, Conditional Probability, and Bayes' Theorem 6.2 Discrete Probability Distributions 6.3 Continuous Probability Distributions
7 Chapter 7: Statistical Inference, Regression & Design of Experiments
7.1 Sampling Distributions, Central Limit Theorem, and Confidence Intervals 7.2 Hypothesis Testing: Means, Proportions, and Variances 7.3 Linear Regression, Correlation Analysis, and Design of Experiments (ANOVA)
8 Chapter 8: Data, Logic, Modeling & Linear Programming
8.1 Optimization Modeling: Decision Variables, Objectives, and Constraints 8.2 Linear Programming: Graphical Solutions and Simplex Fundamentals 8.3 Duality, Sensitivity Analysis, and Integer Linear Programming 8.4 Data, Logic Development, and Analytics
9 Chapter 9: Stochastic Models & Simulation
9.1 Queuing Theory: Little's Law, M/M/1, and Multi-Server Systems 9.2 Discrete-Time Markov Chains: Transition Matrices and Steady-State Probabilities 9.3 Discrete-Event Simulation Principles and Monte Carlo Modeling
10 Chapter 10: Engineering Management, Project Planning & Performance Measurement
10.1 Organizational Structure, Leadership, and Worker Motivation Theories 10.2 Project Management: WBS, Critical Path Method (CPM), PERT, and Crashing 10.3 Earned Value Management (EVM), Decision Trees, and Risk Preference 10.4 Performance Measurement: Productivity, KPIs, Wage Systems, and the Balanced Scorecard
11 Chapter 11: Manufacturing & Production Systems
11.1 Manufacturing Processes, Material Processing, and Tolerancing 11.2 Assembly Line Balancing, Takt Time, Cycle Time, and Throughput 11.3 Sustainable Manufacturing, Waste Minimization, and Life-Cycle Assessment
12 Chapter 12: Production Planning, Inventory Control & Sequencing
12.1 Demand Forecasting Methods and Forecast Error Tracking 12.2 Aggregate Planning, Master Production Scheduling (MPS), and MRP 12.3 Deterministic Inventory Models: EOQ, EPQ, and Safety Stock 12.4 Job Sequencing, Dispatching Rules, and Flow Shop Scheduling
13 Chapter 13: Lean Systems & Theory of Constraints
13.1 Lean Foundations: Value Stream Mapping, Seven Wastes, and 5S 13.2 Pull Systems, Kanban, Just-In-Time (JIT), and SMED Setup Reduction 13.3 Theory of Constraints: Bottleneck Identification and Drum-Buffer-Rope
14 Chapter 14: Facilities Planning & Material Handling
14.1 Facility Location Analysis: Qualitative Factor Rating and Centroid Method 14.2 Facility Layout Design and Flow Analysis 14.3 Material Handling Equipment, Warehousing, and AS/RS
15 Chapter 15: Supply Chain Design & Logistics
15.1 Supply Chain Network Architecture and Transportation Modes 15.2 Distribution Strategies, Cross-Docking, and Vehicle Routing 15.3 The Bullwhip Effect, Risk Pooling, and Supply Chain Coordination
16 Chapter 16: Human Factors & Cognitive Ergonomics
16.1 Cognitive Engineering, Human-Machine Interface, Displays, and Controls 16.2 Anthropometric Principles, Percentile Design, and Biomechanical Posture 16.3 Environmental Ergonomics: Industrial Noise, Lighting, and Heat Stress
17 Chapter 17: Physical Ergonomics & Workplace Safety
17.1 Physical Ergonomics: The NIOSH Manual Lifting Equation 17.2 Cumulative Trauma Disorders and Postural Assessment 17.3 Workplace Safety: OSHA Standards, Hazard Analysis, and Control Hierarchy
18 Chapter 18: Work Design & Measurement
18.1 Methods Engineering, Process Charting, and Principles of Motion Economy 18.2 Direct Stopwatch Time Study: Observed Time, Performance Rating, and Allowances 18.3 Predetermined Time Systems, Work Sampling, and Learning Curves
19 Chapter 19: Quality Management, Six Sigma & Analytical Tools
19.1 Quality Philosophies, Total Quality Management, and Cost of Quality 19.2 Six Sigma DMAIC Methodology and Quality Function Deployment 19.3 The Seven Basic Quality Tools, Root Cause Analysis, and Taguchi Loss Function
20 Chapter 20: Statistical Process Control
20.1 Statistical Process Control Principles, Variation, and Run Rules 20.2 Variable Control Charts: X-bar & R, X-bar & S, and Individual-Moving Range 20.3 Attribute Control Charts: Defectives (p, np) and Defects (c, u)
21 Chapter 21: Process Capability & Acceptance Sampling
21.1 Process Capability Indices and Defect Levels 21.2 Acceptance Sampling Concepts and Operating Characteristic Curves 21.3 Sampling Risks, AQL, LTPD, and Standard Sampling Plans
22 Chapter 22: Systems Engineering & Life-Cycle Management
22.1 The Systems Engineering V-Model, Stakeholder Requirements, and Functional Architecture 22.2 Configuration Management, Interface Control, and Verification & Validation 22.3 Life-Cycle Costing, Maintainability, and System Availability
23 Chapter 23: Reliability Engineering & Risk Analysis
23.1 Reliability Mathematics, Hazard Rates, and the Bathtub Curve 23.2 System Reliability Configurations: Series, Parallel, and k-out-of-n Systems 23.3 Failure Modes and Effects Analysis and Fault Tree Analysis
3.2 Statics, Dynamics, and Mechanics of Materials Core
Key Takeaways Explain This Topic AI
Rigid-body static equilibrium in two dimensions requires satisfying three independent scalar equations: Σ F_x = 0, Σ F_y = 0, and Σ M_O = 0; support reaction types dictate the unknown constraints (roller = 1 normal force, pin = 2 orthogonal forces, fixed = 2 forces plus 1 moment).
Truss analysis models pin-connected members in pure axial tension or compression; zero-force members can be identified by inspection at unloaded joints with two non-collinear members or three members with two collinear.
Planar dynamics separates rectilinear kinematics (v = v₀ + a t, s = s₀ + v₀ t + ½ a t², v² = v₀² + 2 a Δs) from curvilinear motion, where total acceleration resolves into a tangential component (a_t = dv/dt, changing speed) and a normal centripetal component (a_n = v²/ρ, directed toward the center of curvature).
Newton's second law (Σ F = m a) directly couples applied kinetics to particle acceleration, while the work-energy principle (T₁ + Σ U₁₋₂ = T₂) and impulse-momentum equation (I = ∫ F dt = m v₂ - m v₁) solve velocity and contact force problems without time-dependent integration.
Mechanics of materials evaluates structural safety: normal stress σ = P / A, direct shear stress τ = V / A (or V / [2 A_pin] in double shear), axial elastic elongation δ = P L / (A E), constrained thermal stress σ_T = E α ΔT, and factor of safety FS = σ_fail / σ_allow.
Last updated: September 2026
3.2 Statics, Dynamics, and Mechanics of Materials Core
Quick Answer: Statics establishes rigid-body equilibrium by enforcing ∑ F x = 0 \sum F_x = 0 ∑ F x = 0 , ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 , and ∑ M O = 0 \sum M_O = 0 ∑ M O = 0 . Planar truss members are evaluated via the Method of Joints (local node equilibrium) or Method of Sections (cutting through up to three unknown members). Dynamics governs particle motion via constant-acceleration rectilinear formulas and curvilinear normal/tangential accelerations ( a t = d v / d t a_t = dv/dt a t = d v / d t , a n = v 2 / ρ a_n = v^2/\rho a n = v 2 / ρ ). In mechanics of materials, direct stresses ( σ = P / A \sigma = P/A σ = P / A , τ = V / A \tau = V/A τ = V / A ), elastic elongation ( δ = P L A E \delta = \frac{PL}{AE} δ = A E P L ), and constrained thermal stress ( σ T = E α Δ T \sigma_T = E \alpha \Delta T σ T = E α Δ T ) ensure structural integrity through the Factor of Safety ( F S = σ f a i l / σ a l l o w FS = \sigma_{fail} / \sigma_{allow} FS = σ f ai l / σ a ll o w ).
Industrial engineers evaluate static structures (storage racks, mezzanines, crane supports), dynamic material transport systems (conveyors, robotic arms, Automated Guided Vehicles [AGVs]), and mechanical components (shafts, pins, bolted joints) to ensure safe, cost-effective manufacturing operations. The FE exam rigorously tests these foundational engineering science competencies.
1. Statics and Rigid-Body Equilibrium
A body is in static equilibrium when it remains at rest under the combined action of all applied external forces and moments.
Free-Body Diagrams (FBDs) and Support Reactions
Constructing an accurate FBD requires isolating the body from its surroundings and drawing all applied external loads, body weights, and support constraints:
Support Type Diagram Representation Restrained Degrees of Freedom Reactive Unknowns
Roller / Rocker / Smooth Surface Wheel on ground or single triangle Motion perpendicular to supporting surface 1 reaction: Normal force R n R_n R n perpendicular to surface
Pin / Hinge / Frictionless Pivot Pin through clevis Translation in x x x and y y y planes 2 reactions: Orthogonal forces R x R_x R x and R y R_y R y
Fixed (Built-in / Cantilever) Embedded beam end Translation in x , y x, y x , y and rotation 3 reactions: R x , R y R_x, R_y R x , R y , and reaction moment M z M_z M z
Planar Equilibrium Equations
For two-dimensional coplanar systems, static equilibrium requires three independent scalar equations: ∑ F x = 0 , ∑ F y = 0 , ∑ M O = 0 \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0 ∑ F x = 0 , ∑ F y = 0 , ∑ M O = 0
Where the moment of a force about reference point O O O is M O = F ⋅ d ⊥ = r × F M_O = F \cdot d_\perp = \mathbf{r} \times \mathbf{F} M O = F ⋅ d ⊥ = r × F , using counterclockwise as the standard positive convention.
2. Planar Truss Analysis and Zero-Force Members
A truss is a structural framework composed of slender members connected at their ends by frictionless pins. In ideal trusses:
All loadings are applied strictly at the joints.
Members are modeled as weightless two-force members , carrying only axial tensile ( T T T ) or compressive ( C C C ) forces.
Zero-Force Member Inspection Rules
Zero-force members carry no load under the given loading condition and can be identified immediately by inspection:
Rule 1 (Two non-collinear members): If two non-collinear members meet at an unloaded joint with no external forces or support reactions, both members are zero-force members .
Rule 2 (Three members with two collinear): If three members meet at an unloaded joint where two members are collinear, the third non-collinear member is a zero-force member .
Method of Joints vs. Method of Sections
Method of Joints: Isolates individual pin joints as concurrent force particles. Because forces are concurrent, ∑ M = 0 \sum M = 0 ∑ M = 0 provides no information, leaving two scalar equations per joint ( ∑ F x = 0 \sum F_x = 0 ∑ F x = 0 , ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 ). Useful when solving for forces in all truss members, starting at a joint with at most two unknown member forces.
Method of Sections: Cuts an imaginary line through the entire truss, dividing it into two separate rigid bodies. The cut must pass through the target member and at most three unknown members. Taking ∑ M = 0 \sum M = 0 ∑ M = 0 about the intersection point of two unknown members solves for the third unknown directly in a single algebraic step.
3. Dynamics: Kinematics of Particles
Kinematics describes motion without considering the forces causing it.
Rectilinear Motion with Constant Acceleration ( a = const a = \text{const} a = const )
For one-dimensional straight-line motion under uniform acceleration: v = v 0 + a t v = v_0 + a t 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 + 2 a (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
Curvilinear Motion in Normal-Tangential ( n - t n\text{-}t n - t ) Coordinates
When a particle moves along a curved path with local radius of curvature ρ \rho ρ , velocity is tangent to the path ( v = v u t \mathbf{v} = v \mathbf{u}_t v = v u t ). Acceleration resolves into two orthogonal components: a = a t u t + a n u n \mathbf{a} = a_t \mathbf{u}_t + a_n \mathbf{u}_n a = a t u t + a n u n
Tangential Acceleration ( a t a_t a t ): Measures the rate of change of speed along the path: a t = d v d t = v ˙ = v d v d s a_t = \frac{dv}{dt} = \dot{v} = v \frac{dv}{ds} a t = d t d v = v ˙ = v d s d v
Normal (Centripetal) Acceleration ( a n a_n a n ): Measures the rate of change of velocity direction, always directed inward toward the center of curvature: a n = v 2 ρ a_n = \frac{v^2}{\rho} a n = ρ v 2
Total Acceleration Magnitude: a = a t 2 + a n 2 a = \sqrt{a_t^2 + a_n^2} a = a t 2 + a n 2
If speed is constant ( v = const v = \text{const} v = const ), a t = 0 a_t = 0 a t = 0 , but normal acceleration remains non-zero ( a n = v 2 / ρ a_n = v^2/\rho a n = v 2 / ρ ).
4. Kinetics: Newton's Second Law, Work-Energy, and Impulse-Momentum
Kinetics relates forces to the resulting particle motion.
Newton's Second Law ∑ F = m a \sum \mathbf{F} = m \mathbf{a} ∑ F = m a
In n - t n\text{-}t n - t coordinates: ∑ F t = m a t = m d v d t \sum F_t = m a_t = m \frac{dv}{dt} ∑ F t = m a t = m d t d v ∑ F n = m a n = m v 2 ρ \sum F_n = m a_n = m \frac{v^2}{\rho} ∑ F n = m a n = m ρ v 2
Work-Energy Principle
Work done by external forces equals the change in kinetic energy: T 1 + ∑ U 1 → 2 = T 2 T_1 + \sum U_{1\to 2} = T_2 T 1 + ∑ U 1 → 2 = T 2
Where:
- Kinetic energy of a particle: T = 1 2 m v 2 T = \frac{1}{2} m v^2 T = 2 1 m v 2
- Work of a constant force: U = F d cos θ U = F d \cos\theta U = F d cos θ
Work of gravity: U g = − m g Δ y = − m g ( y 2 − y 1 ) U_g = -m g \Delta y = -m g (y_2 - y_1) U g = − m g Δ y = − m g ( y 2 − y 1 )
Work of an elastic spring: U s = − 1 2 k ( x 2 2 − x 1 2 ) U_s = -\frac{1}{2} k (x_2^2 - x_1^2) U s = − 2 1 k ( x 2 2 − x 1 2 ) , where x x x is stretch or compression from unstretched length
For conservative systems with no non-conservative work (such as friction): T 1 + V g 1 + V e 1 = T 2 + V g 2 + V e 2 T_1 + V_{g1} + V_{e1} = T_2 + V_{g2} + V_{e2} T 1 + V g 1 + V e 1 = T 2 + V g 2 + V e 2
Principle of Linear Impulse and Momentum
Linear impulse of a resultant force acting over time interval Δ t = t 2 − t 1 \Delta t = t_2 - t_1 Δ t = t 2 − t 1 equals the change in linear momentum: m v 1 + ∑ ∫ t 1 t 2 F d t = m v 2 m \mathbf{v}_1 + \sum \int_{t_1}^{t_2} \mathbf{F} dt = m \mathbf{v}_2 m v 1 + ∑ ∫ t 1 t 2 F d t = m v 2 I = F a v g Δ t = m v 2 − m v 1 \mathbf{I} = \mathbf{F}_{avg} \Delta t = m \mathbf{v}_2 - m \mathbf{v}_1 I = F a vg Δ t = m v 2 − m v 1
If external resultant forces are zero ( ∑ F e x t = 0 \sum \mathbf{F}_{ext} = \mathbf{0} ∑ F e x t = 0 ), total linear momentum is conserved: m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f m_1 \mathbf{v}_{1i} + m_2 \mathbf{v}_{2i} = m_1 \mathbf{v}_{1f} + m_2 \mathbf{v}_{2f} m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f . In impacts, the coefficient of restitution ( e e e ) along the line of impact is: ≠ = v 2 f − v 1 f v 1 i − v 2 i , 0 ≤ e ≤ 1 \ne = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}, \quad 0 \le e \le 1 = = v 1 i − v 2 i v 2 f − v 1 f , 0 ≤ e ≤ 1
5. Mechanics of Materials: Stress, Strain, and Deformation
Mechanics of materials determines the internal stresses and deformations within structural members subjected to external loads.
Direct Normal and Shear Stresses
Direct Normal Stress ( σ \sigma σ ): Axial force P P P perpendicular to cross-sectional area A A A : σ = P A \sigma = \frac{P}{A} σ = A P Tension is positive ( + + + , member elongates); compression is negative ( − - − , member shortens).
Direct Shear Stress ( τ \tau τ ): Internal shear force V V V acting parallel to cross-sectional area A A A : τ s i n g l e = V A \tau_{single} = \frac{V}{A} τ s in g l e = A V τ d o u b l e = V 2 A p i n \tau_{double} = \frac{V}{2 A_{pin}} τ d o u b l e = 2 A p in V In a double-shear pin connection, the total transmitted force V V V is distributed equally across two shear planes, cutting the required pin area in half.
Hooke's Law and Axial Elastic Elongation
Within the linear elastic region, stress is proportional to engineering strain ( ϵ = δ / L 0 \epsilon = \delta / L_0 ϵ = δ / L 0 ): σ = E ϵ \sigma = E \epsilon σ = E ϵ
Where E E E is Young's modulus (Modulus of Elasticity, e.g., structural steel E ≈ 200 GPa = 29 × 10 6 psi E \approx 200\text{ GPa} = 29 \times 10^6\text{ psi} E ≈ 200 GPa = 29 × 1 0 6 psi ; aluminum E ≈ 70 GPa = 10 × 10 6 psi E \approx 70\text{ GPa} = 10 \times 10^6\text{ psi} E ≈ 70 GPa = 10 × 1 0 6 psi ). Combining σ = P / A \sigma = P/A σ = P / A and ϵ = δ / L \epsilon = \delta/L ϵ = δ / L gives total axial deformation: δ = P L A E \delta = \frac{P L}{A E} δ = A E P L
Poisson's Ratio and Shear Modulus
When a bar is stretched axially, it contracts laterally. Poisson's ratio ( ν \nu ν ) is the ratio of lateral strain to longitudinal axial strain: ν = − ϵ l a t e r a l ϵ a x i a l \nu = -\frac{\epsilon_{lateral}}{\epsilon_{axial}} ν = − ϵ a x ia l ϵ l a t er a l
For metals, ν \nu ν typically ranges from 0.25 0.25 0.25 to 0.35 0.35 0.35 . Shear modulus G G G relates to E E E and ν \nu ν via: G = E 2 ( 1 + ν ) G = \frac{E}{2(1 + \nu)} G = 2 ( 1 + ν ) E
- 6. Thermal Stress and Factor of Safety
- Thermal Deformation and Constrained Thermal Stress
An unrestrained member subjected to temperature change Δ T \Delta T Δ T undergoes thermal expansion without developing internal stress: δ T = α L Δ T , σ u n c o n s t r a i n e d = 0 \delta_T = \alpha L \Delta T, \quad \sigma_{unconstrained} = 0 δ T = αL Δ T , σ u n co n s t r ain e d = 0
Where α \alpha α is the linear coefficient of thermal expansion ( 1 / ∘ C 1/^\circ\text{C} 1 / ∘ C or 1 / ∘ F 1/^\circ\text{F} 1 / ∘ F ). For structural steel, α ≈ 12 × 10 − 6 / ∘ C ≈ 6.5 × 10 − 6 / ∘ F \alpha \approx 12 \times 10^{-6}\text{ /}^\circ\text{C} \approx 6.5 \times 10^{-6}\text{ /}^\circ\text{F} α ≈ 12 × 1 0 − 6 / ∘ C ≈ 6.5 × 1 0 − 6 / ∘ F .
If the member is constrained between unyielding, rigid walls, the total deformation must be zero ( δ t o t a l = δ T − δ P = 0 \delta_{total} = \delta_T - \delta_P = 0 δ t o t a l = δ T − δ P = 0 ): α L Δ T − P L A E = 0 ⟹ σ T = P A = E α Δ T \alpha L \Delta T - \frac{P L}{A E} = 0 \implies \sigma_T = \frac{P}{A} = E \alpha \Delta T αL Δ T − A E P L = 0 ⟹ σ T = A P = E α Δ T
CRITICAL SIGN CONVENTION: When temperature increases ( Δ T > 0 \Delta T > 0 Δ T > 0 ), a constrained member attempts to expand, and the unyielding supports exert inward compressive forces. Thus, heating induces compressive thermal stress , while cooling ( Δ T < 0 \Delta T < 0 Δ T < 0 ) induces tensile thermal stress .
Factor of Safety ( F S FS FS )
Structural designs incorporate a safety margin to account for material variability, dynamic shock loads, and manufacturing tolerances: F S = σ f a i l σ a l l o w = σ y i e l d σ w o r k i n g or σ u l t i m a t e σ a l l o w FS = \frac{\sigma_{fail}}{\sigma_{allow}} = \frac{\sigma_{yield}}{\sigma_{working}} \quad \text{or} \quad \frac{\sigma_{ultimate}}{\sigma_{allow}} FS = σ a ll o w σ f ai l = σ w or kin g σ y i e l d or σ a ll o w σ u lt ima t e
Allowable working stress is therefore: σ a l l o w = σ f a i l F S \sigma_{allow} = \frac{\sigma_{fail}}{FS} σ a ll o w = FS σ f ai l .
- 7. Step-by-Step Worked Engineering Examples
- Example 1: Truss Member Force via Method of Sections
Problem: A symmetrical Warren truss carries an overhead crane hoist in a distribution warehouse. The truss has lower chord joints A , B , C , D A, B, C, D A , B , C , D and upper chord joints E , F , G E, F, G E , F , G , with bay widths of 3 m 3\text{ m} 3 m and height 4 m 4\text{ m} 4 m . A vertical downward payload of P = 60 kN P = 60\text{ kN} P = 60 kN is suspended at joint B B B , which sits 3 m 3\text{ m} 3 m from support A A A on a 9 m 9\text{ m} 9 m span. Support reactions follow from global equilibrium: R A y = 60 ( 9 − 3 ) / 9 = 40 kN R_{Ay} = 60(9 - 3)/9 = 40\text{ kN} R A y = 60 ( 9 − 3 ) /9 = 40 kN (pin at A A A ) and R D y = 60 ( 3 ) / 9 = 20 kN R_{Dy} = 60(3)/9 = 20\text{ kN} R Dy = 60 ( 3 ) /9 = 20 kN (roller at D D D ). Determine the internal axial force in the top horizontal chord member E F EF EF .
Step-by-Step Solution:
Select the cut section: Pass an imaginary section cut through upper chord member E F EF EF , diagonal member E B EB EB , and lower chord member A B AB A B . Retain the left-hand portion of the truss containing joint A A A .
Identify exposed internal forces: The cut exposes internal axial forces F E F F_{EF} F EF (top chord), F E B F_{EB} F EB (diagonal), and F A B F_{AB} F A B (bottom chord). Assume all unknown forces are in tension (pointing away from the cut face).
Select optimal moment center: Notice that both unknown diagonal force F E B F_{EB} F EB and unknown lower chord force F A B F_{AB} F A B pass directly through joint B B B . Taking moments about joint B B B eliminates both F E B F_{EB} F EB and F A B F_{AB} F A B in a single equation: ∑ M B = 0 \sum M_B = 0 ∑ M B = 0
Formulate moment equilibrium on left section:
Upward support reaction at A A A ( R A y = 40 kN R_{Ay} = 40\text{ kN} R A y = 40 kN ) acts at horizontal distance 3 m 3\text{ m} 3 m from B B B , creating a clockwise (negative) moment: − 40 kN × 3 m = − 120 kN ⋅ m -40\text{ kN} \times 3\text{ m} = -120\text{ kN}\cdot\text{m} − 40 kN × 3 m = − 120 kN ⋅ m .
Force F E F F_{EF} F EF acts along the top chord at height 4 m 4\text{ m} 4 m above joint B B B . Assuming tension (pointing rightward toward F F F ), it creates a clockwise (negative) moment about B B B : − F E F × 4 m -F_{EF} \times 4\text{ m} − F EF × 4 m . ∑ M B = − R A y ( 3 m ) − F E F ( 4 m ) = 0 \sum M_B = -R_{Ay}(3\text{ m}) - F_{EF}(4\text{ m}) = 0 ∑ M B = − R A y ( 3 m ) − F EF ( 4 m ) = 0 − 120 − 4 F E F = 0 ⟹ F E F = − 120 4 = − 30.0 kN -120 - 4 F_{EF} = 0 \implies F_{EF} = -\frac{120}{4} = -30.0\text{ kN} − 120 − 4 F EF = 0 ⟹ F EF = − 4 120 = − 30.0 kN
Interpret the sign: The negative sign confirms that member E F EF EF is in compression ( 30.0 kN (C) 30.0\text{ kN (C)} 30.0 kN (C) ).
Example 2: Dynamic AGV Braking and Normal Acceleration
Problem: An automated guided vehicle (AGV) in a semiconductor cleanroom has a total mass of m = 400 kg m = 400\text{ kg} m = 400 kg . The AGV enters a circular curve of radius ρ = 25 m \rho = 25\text{ m} ρ = 25 m at an initial speed of v 0 = 6.0 m/s v_0 = 6.0\text{ m/s} v 0 = 6.0 m/s . While traversing the curve, its friction brakes apply a constant tangential retarding force of F b r a k e = 600 N F_{brake} = 600\text{ N} F b r ak e = 600 N . Calculate the magnitude of the total acceleration vector of the AGV immediately after braking commences.
Step-by-Step Solution:
Determine tangential acceleration ( a t a_t a t ): By Newton's second law along the path tangent: ∑ F t = − F b r a k e = m a t \sum F_t = -F_{brake} = m a_t ∑ F t = − F b r ak e = m a t a t = − 600 N 400 kg = − 1.50 m/s 2 a_t = \frac{-600\text{ N}}{400\text{ kg}} = -1.50\text{ m/s}^2 a t = 400 kg − 600 N = − 1.50 m/s 2
Determine normal (centripetal) acceleration ( a n a_n a n ): At the instant braking begins, speed is v = 6.0 m/s v = 6.0\text{ m/s} v = 6.0 m/s : a n = v 2 ρ = ( 6.0 m/s ) 2 25 m = 36.0 25 = 1.44 m/s 2 a_n = \frac{v^2}{\rho} = \frac{(6.0\text{ m/s})^2}{25\text{ m}} = \frac{36.0}{25} = 1.44\text{ m/s}^2 a n = ρ v 2 = 25 m ( 6.0 m/s ) 2 = 25 36.0 = 1.44 m/s 2
Calculate total acceleration magnitude: The tangential and normal components are mutually perpendicular ( 90 ∘ 90^\circ 9 0 ∘ ): a = a t 2 + a n 2 = ( − 1.50 ) 2 + ( 1.44 ) 2 = 2.25 + 2.0736 = 4.3236 = 2.08 m/s 2 a = \sqrt{a_t^2 + a_n^2} = \sqrt{(-1.50)^2 + (1.44)^2} = \sqrt{2.25 + 2.0736} = \sqrt{4.3236} = 2.08\text{ m/s}^2 a = a t 2 + a n 2 = ( − 1.50 ) 2 + ( 1.44 ) 2 = 2.25 + 2.0736 = 4.3236 = 2.08 m/s 2
8. Common FE Exam Traps in Statics, Dynamics, and Materials
Concept Fatal Exam Pitfall Correct Engineering Methodology
Single vs. Double Shear Using τ = V / A \tau = V / A τ = V / A for clevis or double-lap bolted connections. For double shear, the applied load is divided between two shear planes: τ = V / ( 2 A p i n ) \tau = V / (2 A_{pin}) τ = V / ( 2 A p in ) . Required pin area is halved.
Thermal Stress Sign Assuming thermal expansion results in tensile stress. When constrained against expansion, heating ( Δ T > 0 \Delta T > 0 Δ T > 0 ) causes the rigid supports to push inward, producing compressive stress .
Moment of Inertia Base/Height Swapping base b b b and height h h h in beam bending moment of inertia: I = b h 3 12 I = \frac{b h^3}{12} I = 12 b h 3 . Height h h h is always the dimension perpendicular to the neutral bending axis ( h h h is cubed).
Curvilinear Acceleration Omitting centripetal acceleration ( v 2 / ρ v^2/\rho v 2 / ρ ) when an object moves around a curve at constant speed. Constant speed only implies a t = 0 a_t = 0 a t = 0 . Normal acceleration a n = v 2 / ρ a_n = v^2/\rho a n = v 2 / ρ remains active because velocity direction changes continuously.
Mass vs. Weight in Kinetics Substituting weight in pounds-force ( W W W ) directly for mass m m m in F = m a F = m a F = ma . In US Customary units, mass must be expressed in slugs: m = W / g m = W / g m = W / g (where g = 32.2 ft/s 2 g = 32.2\text{ ft/s}^2 g = 32.2 ft/s 2 ). In SI, mass is kilograms ( k g kg k g ).
Truss Zero-Force Members Misidentifying joints with external loads as zero-force members. Zero-force member inspection rules apply only to joints where no external forces or support reactions are applied.