CELE Strength of Materials — Simple Stresses and StrainsStudy Notes

Normal stress, denoted σ (sigma), arises when an external force P acts perpendicular to and through the centroid of a cross-sectional area A. The average normal stress is given by the fundamental definition: σ = P / A where: - σ = normal stress (Pa, kPa, or MPa; in practice we use N/mm² = MPa) - P = axial force (N) - A = cross-sectional area (mm²) CRITICAL DISTINCTION: • Tensile stress (positive): the force pulls the member apart, creating elongation. In design equations, we often write σₜ or simply σ when the context is clear. • Compressive stress (negative): the force pushes the member together, creating shortening. Written as σc or −σ. St. Venant's Principle states that stress concentrations from applied loads or geometric discontinuities (holes, notches, sudden diameter changes) are local phenomena. They dissipate over a distance roughly equal to the largest dimension of the cross-section. Therefore, the simple formula σ = P/A yields the average stress accurately only away from the point of load application. At holes and notches, local stress concentration factors apply—a topic central to AISC 360 and connection design. Example context for Philippine practice: A reinforced concrete column designed per ACI 318 carries axial load through both concrete and steel reinforcement. Each material sees stress computed separately (σc = Pc/Ac, σs = Ps/As), but they deform together—this is statically indeterminate and requires compatibility equations (Section 7). Units are critical. ALWAYS work in consistent units: - Force in N (newtons) or kN (kilonewtons) - Area in mm² (square millimeters) - Stress emerges in MPa (N/mm²) or kPa (1 MPa = 1 N/mm²) Mixing meters and kilonewtons is the classic mistake: 1 kN over 1 m² = 1 kN/m² = 0.001 MPa, easily missed under exam pressure.

Shear stress τ (tau) develops when a force acts parallel (tangent) to a resisting area. The definition is analogous to normal stress: τ = V / A where: - τ = shear stress (MPa) - V = shear force acting parallel to area (N) - A = area resisting shear (mm²) CRITICAL: You must distinguish between single shear and double shear, a distinction that directly affects many connection problems in steel design (AISC 360) and is a frequent examination pitfall. SINGLE SHEAR: The connector (bolt, rivet, pin, or weld) is cut on one plane only. Imagine a bolt connecting two plates lap-to-lap—the bolt shank fails in one shear plane. The resisting area is one cross-section: τ = P / A For a bolt of diameter d: A = (π/4) d² τ = P / [(π/4) d²] = 4P / (πd²) DOUBLE SHEAR: The connector is cut on two planes. Imagine a bolt in a clevis connection: the bolt passes through a central plate and then through two outer plates. Shear must be resisted on two planes—one on each side of the central plate. The same force P is now resisted by two areas: τ = P / (2A) For double-shear bolts, always check: Do the external members overlap symmetrically? If yes → divide by 2. REAL-WORLD CONTEXT (AISC 360-16): In practice, bolt shear capacity is given as: Φ × Rn = Φ × fnv × Ab where fnv is the nominal shear strength (ksi or MPa), Ab is the bolt's area, and Φ is the resistance factor (≈ 0.75 for bolts). If a bolt is in double shear, the capacity is: Φ × Rn = Φ × fnv × 2Ab (double the area, double the capacity). Note that average shear stress is uniform only in thin webs or where the connector is short relative to its diameter. In thick plates, shear is non-uniform, and the failure surface may be cone-shaped rather than cylindrical. KEY APPLICATION: Riveted and bolted lap joints, welded splices, shear pins, and keys all rely on shear-stress analysis. In the Philippines, NSCP 2015 and PD 60 (timber design) specify allowable shear stresses for wood connections; AISC 360 governs steel bolts and welds.

Bearing stress σb is not truly a 'simple' stress in the Mohr-circle sense, but it is treated as a distinct entity in connection design. It represents the **contact pressure between two surfaces in contact**, typically between a bolt shank and the wall of a hole in a plate, or between a rivet and a plate. The bearing stress is defined as: σb = P / Ab where: - P = force pressing the surfaces together (N) - Ab = projected bearing area (mm²) The **projected bearing area** is not the circular contact area but rather the rectangular projection: for a bolt of diameter d bearing on a plate of thickness t: Ab = d × t This is the most common formula. It assumes bearing failure occurs when the contact pressure is high enough to crush or shear the material around the hole. Why the projected area, not the circular area? Empirically, bearing failure (hole tear-out, plate crushing, or local yield around the hole) correlates best with the rectangular 'shadow' cast by the bolt onto the plate. This is reflected in AISC 360 Appendix D (Bolted Connections), where: Φ × Rn = Φ × (2.4 × Fu × d × t) or (1.2 × Fu × d × t) depending on whether the bolt is or is not at the edge of the connection. The factor of 2.4 or 1.2 is multiplied by Fu (ultimate tensile strength of plate material) and by d × t (the projected area). GEOMETRIC CONTEXT: Bearing stress governs: 1. **Hole tear-out**: if the hole is too close to the edge of the plate, it tears out in a cone-like pattern. 2. **Local crushing**: if the plate material is weak relative to the bolt, the area immediately around the hole yields or crushes. 3. **Prying action**: in bolted connections with thin plates, external moments can induce additional tension (prying) that affects effective bearing area. In design, bearing area sometimes is reduced by an edge distance factor or a spacing factor if holes are too close to edges or to each other (AISC 360, Section J3). PRACTICAL THRESHOLD: Bearing stress is typically higher than shear stress in the same connection. For a bolt connecting structural steel plates: - Shear stress ~150–300 MPa (depending on bolt grade) - Bearing stress ~600–1200 MPa or higher (depends on plate thickness and Fu) If bearing stress is excessive, the remedy is to: - Increase plate thickness t - Use a larger diameter bolt d - Use a plate material with higher Fu - Space bolts farther apart (only if edge distance and center-to-center distance are controlled) CRITICAL FOR EXAMS: A bolt may be safe in shear but fail in bearing, or vice versa. Always check both.

Strain ε (epsilon) quantifies deformation normalized by the original dimension. It is **dimensionless** and is the complement to stress: stress is what the material experiences; strain is how the material responds. NORMAL (AXIAL) STRAIN: For a member of original length L that undergoes axial deformation (elongation or shortening) Δ (delta) or δ: ε = δ / L where: - ε = normal strain (dimensionless, often expressed in micro-strain μ-strain, where 1 μ = 1×10⁻⁶) - δ = axial deformation: positive for elongation, negative for shortening (mm) - L = original length (mm) Example: A steel cable 10 m long stretches by 5 mm under load: ε = 5 / 10,000 = 0.0005 = 500 μ SHEAR STRAIN: Shear strain γ (gamma) is the change in angle (in radians) of an originally right-angled corner of the element due to shear deformation. For small deformations: γ ≈ tan(γ) = Δy / h where Δy is the relative displacement of one face and h is the perpendicular distance. Shear strain is also dimensionless and typically very small (a few hundredths of a radian or smaller). STRAIN-STRESS RELATIONSHIP (HOOKE'S LAW): Within the elastic (linear) region of the material's stress–strain curve, stress and strain are proportional: σ = E × ε (for normal stress and strain) τ = G × γ (for shear stress and strain) where: - E = modulus of elasticity (Young's modulus), in MPa or GPa - G = shear modulus (modulus of rigidity), in MPa or GPa These are material properties, independent of size or shape. For structural steel: - E ≈ 200 GPa = 200,000 MPa - G ≈ 80 GPa (related to E by Poisson's ratio; see Section 5) For concrete: - E ≈ 20–40 GPa (depends on f'c; per ACI 318, E = 4700√f'c in psi-units, or roughly 14,000–42,000 MPa depending on concrete strength) For aluminum: - E ≈ 70 GPa KEY INSIGHT: A stiffer material (higher E) deforms less for the same stress. This is why steel is preferred over wood for precision and why composite columns (steel + concrete) require careful load-sharing analysis. ELASTIC vs. PLASTIC STRAIN: - Elastic strain: fully recovers when load is removed. It is the strain you must predict to ensure serviceability. - Plastic strain: permanent, does not recover. Occurs when stress exceeds the yield stress σy. In design, we typically restrict stresses to the elastic range (below σy) to avoid permanent set. POISSON'S EFFECT: When a material is stretched axially (positive ε_axial), it contracts laterally (negative ε_lateral). This lateral contraction is governed by **Poisson's ratio**, detailed in Section 5.

When a material is stretched in one direction, it simultaneously contracts in the perpendicular directions—a phenomenon rooted in the conservation of volume and the material's molecular structure. **Poisson's ratio** ν (nu) quantifies this lateral contraction relative to the axial extension. DEFINITION: ν = − (lateral strain) / (axial strain) = − ε_lateral / ε_axial The negative sign ensures ν is positive: when ε_axial is positive (elongation), ε_lateral is negative (contraction), so the ratio is positive. TYPICAL VALUES: - Structural steel: ν ≈ 0.27–0.30 (often taken as 0.3) - Aluminum: ν ≈ 0.33 - Concrete: ν ≈ 0.15–0.20 (lower than steel due to aggregate restraint) - Rubber: ν ≈ 0.5 (nearly incompressible; volume changes minimally under hydrostatic pressure) RELATIONSHIP TO E AND G: The three elastic constants E, G, and ν are not independent. They are linked by the fundamental relation: G = E / [2(1 + ν)] where G is the shear modulus. This relation arises from elasticity theory and holds for isotropic, homogeneous materials (materials with the same properties in all directions). Verification for steel: G = 200 GPa / [2(1 + 0.3)] = 200 / 2.6 ≈ 77 GPa (close to the typical 80 GPa) If you know any two of E, G, and ν, you can find the third. In practice, E and ν are tabulated; G is rarely given directly but computed from this formula. BULK MODULUS: The bulk modulus K relates volumetric strain (dilatation) to hydrostatic pressure. It represents the resistance of a material to uniform compression: K = E / [3(1 − 2ν)] Note the denominator 3(1 − 2ν): this explains why ν = 0.5 (rubber) gives infinite K (incompressible) and why most materials have ν < 0.5 (compressible under high pressure). For concrete with ν ≈ 0.2: K = E / [3(1 − 0.4)] = E / 1.8 If E_concrete = 25 GPa, then K ≈ 13.9 GPa (significant stiffness against uniform compression). GENERALIZED HOOKE'S LAW (MULTIDIMENSIONAL STRESS): When a point in a material experiences stress in more than one direction (e.g., biaxial or triaxial), the strain in each direction is affected by stresses in all directions. The generalized Hooke's law states: ε_x = [1/E] [σ_x − ν(σ_y + σ_z)] ε_y = [1/E] [σ_y − ν(σ_z + σ_x)] ε_z = [1/E] [σ_z − ν(σ_x + σ_y)] Each normal strain is the sum of two effects: 1. **Direct effect**: the stress in that direction stretched or compressed by 1/E. 2. **Poisson contraction**: the perpendicular stresses cause lateral contraction, subtracted from the direct effect. Example: Biaxial stress state with σ_x = 100 MPa, σ_y = 50 MPa, σ_z = 0, E = 200 GPa, ν = 0.3: ε_x = [1/200,000] [100 − 0.3(50 + 0)] = [1/200,000] [100 − 15] = 85 / 200,000 = 0.000425 ε_y = [1/200,000] [50 − 0.3(0 + 100)] = [1/200,000] [50 − 30] = 20 / 200,000 = 0.0001 ε_z = [1/200,000] [0 − 0.3(100 + 50)] = [1/200,000] [−45] = −0.000225 Notice: ε_z is negative (contraction) even though σ_z = 0, because the lateral stresses σ_x and σ_y cause lateral contraction via Poisson's effect. VOLUMETRIC STRAIN (DILATATION): The total change in volume per unit original volume is: ε_vol = ε_x + ε_y + ε_z = [(1 − 2ν) / E] (σ_x + σ_y + σ_z) For hydrostatic pressure σ_x = σ_y = σ_z = −p (all equal, all compressive): ε_vol = [(1 − 2ν) / E] (−3p) = −3p / K where K = E / [3(1 − 2ν)] is the bulk modulus. This shows the link between bulk modulus and volumetric compression. PRACTICAL IMPLICATIONS: - **Pressure vessels**: circumferential and longitudinal stresses cause hoop strain and axial strain; radial strain (wall thickness reduction) is computed from the generalized Hooke's law. - **Reinforced concrete under combined loading**: concrete in compression is accompanied by lateral Poisson contraction; if this contraction is restrained (e.g., by transverse reinforcement or confining pressure), the concrete becomes stronger—the basis for tied column design in ACI 318. - **Composite members**: different materials have different ν; in a steel-concrete composite, the lateral strains may not match, causing internal stresses. COMPRESSIBILITY: A material with ν close to 0.5 is nearly incompressible (bulk modulus approaches infinity). This is why rubber (ν ≈ 0.5) is incompressible and used for seals and bearings. Most metals and concrete are compressible (ν ≈ 0.3–0.2), so they do change volume under hydrostatic pressure, but not drastically.

The axial deformation formula synthesizes the three foundational concepts: stress, strain, and elasticity. It is arguably the most-used single formula in Strength of Materials and recurs in nearly every topic and exam problem. DERIVATION: Start with the three relationships: 1. **Stress definition**: σ = P / A 2. **Strain definition**: ε = δ / L 3. **Hooke's law**: σ = E ε Combine (1) and (2) into (3): E = σ / ε = (P/A) / (δ/L) = (P/A) × (L/δ) = PL / (Aδ) Rearrange for δ: δ = PL / (AE) where: - δ = axial deformation (elongation or shortening), mm - P = axial force, N - L = original length, mm - A = cross-sectional area, mm² - E = modulus of elasticity, MPa (or N/mm²) MEMORIZE THIS FORMULA. It is as fundamental as F = ma in physics. COMPOSITE BARS (MULTIPLE SEGMENTS): When a bar is composed of several segments with different materials or cross-sections, or when the load changes along the length, the total deformation is the sum of individual segment deformations: δ_total = Σ (P_i L_i) / (A_i E_i) where subscript i denotes each segment. Example: A composite tension member with: - Segment 1: Steel, P₁ = 100 kN, L₁ = 1 m, A₁ = 500 mm², E₁ = 200 GPa - Segment 2: Aluminum, P₂ = 100 kN, L₂ = 0.5 m, A₂ = 500 mm², E₂ = 70 GPa δ_total = (100,000 × 1000) / (500 × 200,000) + (100,000 × 500) / (500 × 70,000) = 100,000,000 / 100,000,000 + 50,000,000 / 35,000,000 = 1.0 + 1.43 = 2.43 mm The aluminum segment contributes more to total deformation (1.43 mm vs. 1.0 mm) because its modulus is lower. VARYING LOAD OR AREA: If load or area varies continuously along the length (e.g., a bar hanging under its own weight, or a tapered bar), integrate: δ = ∫₀ᴸ [P(x) / (A(x)E)] dx For a bar hanging under its own weight: - Weight at distance x from free end: W(x) = γ A (L − x), where γ is unit weight - P(x) = γ A (L − x) - δ = ∫₀ᴸ [γ A(L−x)] / (AE) dx = (γ/E) ∫₀ᴸ (L−x) dx = (γ/E) [Lx − x²/2]₀ᴸ = (γ/E) [L² − L²/2] = γL² / (2E) The elongation due to self-weight is δ_self = γL² / (2E), independent of cross-section (as long as E and γ are uniform). RIGIDITY: The product AE is called the **axial rigidity** or **axial stiffness**. A higher rigidity means less deformation for a given load. In composite members, the member's overall rigidity is the sum of the component rigidities in series (i.e., the compliances 1/(AE) add). STRUCTURAL IMPLICATIONS: 1. **Serviceability**: Deformation limits (deflection, settlement, differential movement) are often the governing design criterion, not stress. 2. **Composite members**: The stiffer material (higher AE) attracts proportionally more load in a statically indeterminate system. 3. **Temperature-dependent**: The same load produces different deformations if temperature (hence E) changes. COMPARISON: STRESS vs. DEFORMATION LIMITS In design, we must satisfy both: - **Strength**: σ ≤ σ_allow (or per LRFD, Φσ_n ≥ σ_demand) - **Serviceability**: δ ≤ δ_allow (often L/300 or L/500 for buildings per NSCP) Sometimes deformation governs before stress. For example, a very long, lightly-loaded cable may satisfy stress but violate sag limits.

Temperature changes cause materials to expand or contract. Understanding thermal effects is crucial for design in the Philippines, where tropical climates and seasonal variations are significant, and for infrastructure like bridges, pipelines, and buildings that experience wide temperature swings. THERMAL DEFORMATION (FREE EXPANSION): When a material is **free to expand** (no external constraints), a temperature change ΔT produces a thermal strain that is independent of stress: ε_T = α ΔT where α is the **coefficient of linear thermal expansion** (per °C or per K). The resulting thermal deformation is: δ_T = ε_T × L = α L ΔT where L is the original length. TYPICAL THERMAL EXPANSION COEFFICIENTS: - Structural steel: α ≈ 11.7 × 10⁻⁶ /°C - Aluminum: α ≈ 23.1 × 10⁻⁶ /°C (roughly twice steel) - Concrete: α ≈ 10–14 × 10⁻⁶ /°C (depends on aggregates; quartzite aggregate → higher α) - Glass: α ≈ 9 × 10⁻⁶ /°C KEY INSIGHT: **Free thermal expansion produces deformation but NO STRESS.** If a steel rail has room to expand, temperature rise causes elongation but zero internal stress. The member is deformation-free in the sense that it carries no thermal load. THERMAL STRESS (RESTRAINED EXPANSION): When thermal deformation is **prevented or restricted** by external constraints (e.g., a rail fixed at both ends, or a member restrained by adjacent structures), internal stress develops. For a member that is **fully restrained** (cannot move at all): σ_T = E α ΔT where: - σ_T = thermal stress (MPa), tensile if ΔT > 0 (heating, restrained expansion) or compressive if ΔT < 0 (cooling, restrained contraction) - E = modulus of elasticity (MPa) - α = thermal expansion coefficient (/°C) - ΔT = temperature change (°C) Derivation: If a bar of length L is fully restrained, the free thermal deformation δ_T = α L ΔT is prevented. An internal force P must counteract this deformation: δ_internal = P L / (AE) = −δ_T = −α L ΔT P = −α E A ΔT σ_T = P / A = −α E ΔT The negative sign indicates compression under heating (counterintuitive at first: the bar wants to expand but is held; this creates compressive stress). Conversely, cooling creates tensile stress. PARTIAL RESTRAINT (INITIAL GAP): If a member is not initially fully restrained but has a gap g (or clearance) available for expansion before hitting a constraint, thermal stress develops only **after the gap closes**. The net deformation is: δ_free − δ_internal = g α L ΔT − (σ_T L / E) = g Solving for σ_T: α L ΔT − g = σ_T L / E σ_T = [E / L] (α L ΔT − g) = E(α ΔT − g/L) If ΔT is small such that α L ΔT < g, then σ_T ≤ 0 (no stress; the bar expands into the gap). PRACTICAL EXAMPLES: 1. **Concrete bridge deck** on a sunny day: The top surface heats faster than the bottom, creating a temperature gradient. The deck wants to arch upward. Expansion joints allow the movement; without them, large compressive stresses develop, potentially causing buckling or spalling (ACI 318, NSCP 2015). 2. **Steel rail in the Philippines**: A rail laid on a hot day (say, 45°C) without expansion joints will contract when cooled to winter temperature (say, 25°C). ΔT = −20°C. The thermal contraction is: δ_T = 11.7 × 10⁻⁶ × 15,000 mm × (−20) = −3.51 mm per rail. If the rail is fully restrained, σ_T = 200,000 × 11.7 × 10⁻⁶ × (−20) = −46.8 MPa (compressive). Over a 100 m rail: δ_T = 11.7 × 10⁻⁶ × 100,000 × (−20) = −23.4 mm (significant, may buckle if not relieved). 3. **Composite members**: Steel and concrete have different α values. In a reinforced concrete column, differential thermal deformation can induce internal stresses and potentially spalling or cracking. THERMAL STRESS COMBINED WITH MECHANICAL LOAD: If a member is subjected to both an applied load P and a temperature change, the total stress is: σ_total = σ_mechanical + σ_thermal = P/A + E α ΔT For instance, a steel cable in tension under load P, heated: σ_total = P/A + E α ΔT If the cable is cooled (ΔT < 0), thermal stress is compressive and reduces the total tension. If heated, it increases the tension. DESIGN IMPLICATIONS (NSCP 2015, AISC 360): - **Expansion joints**: Required in long structures (buildings, bridges, pavements) to accommodate thermal movement. Typical spacing is 20–40 m depending on expected ΔT and material. - **Stress relief**: Thermal stress is often the governing criterion in tall chimneys, long pipelines, or members under sustained high temperature. - **Load factor**: In LRFD design, thermal loads are factored (typically 1.2 × Dead + 0.5 × Live + 1.0 × Thermal, or similar per code). - **Anchorages**: Thermal movement must be accommodated by connections; stiff, fully-restrained connections can fail if thermal stresses are not managed. COMPREHENSIVE EXAMPLE: THERMAL STRESS WITH PARTIAL CONSTRAINT A steel member initially 5 m long is installed with no gap (g = 0) between two rigid supports at 25°C. The temperature rises to 55°C. α = 11.7 × 10⁻⁶ /°C, E = 200 GPa. (a) If the member is **free to expand**: δ_T = ? (b) If the member is **fully restrained**: σ_T = ? (c) If the member had an initial **1 mm gap** (now closed after expanding 1 mm), what happens when it is heated further to 75°C? Solution (a): ΔT = 55 − 25 = 30°C δ_T = α L ΔT = 11.7 × 10⁻⁶ × 5000 mm × 30 = 1.755 mm (elongation, no stress) Solution (b): Fully restrained from 25°C to 55°C (the same ΔT = 30°C): σ_T = E α ΔT = 200,000 × 11.7 × 10⁻⁶ × 30 = 70.2 MPa (compressive, since the member is held and cannot expand) Solution (c): Initially, the member has a 1 mm gap. Heated from 25°C to 55°C: δ_T = 1.755 mm > g = 1 mm → gap closes, and compressive stress develops. σ_T,55 = E [α ΔT − g/L] = 200,000 [11.7 × 10⁻⁶ × 30 − 1/5000] = 200,000 [0.000351 − 0.0002] = 200,000 × 0.000151 = 30.2 MPa (compressive) Further heating to 75°C (ΔT = 50°C from the baseline 25°C): Now the gap is closed, and the member is fully restrained: σ_T,75 = E α ΔT = 200,000 × 11.7 × 10⁻⁶ × 50 = 117 MPa (compressive)

A system is **statically determinate** when the number of unknown forces or moments equals the number of equilibrium equations available. A system is **statically indeterminate** when there are more unknowns than equations; additional equations come from **compatibility conditions** (constraints on deformation). Common indeterminate axial problems include: 1. A bar fixed at both ends, carrying an axial load 2. Composite columns or tension members (e.g., steel tube filled with concrete) 3. Multiple bars in parallel sharing a load 4. Thermal restraint problems SOLUTION PROCEDURE: **Step 1: Draw free-body diagram and write equilibrium equations.** **Step 2: Identify compatibility condition(s) relating deformations.** **Step 3: Write force–deformation relations (δ = PL/AE for each segment).** **Step 4: Solve simultaneously.** EXAMPLE 1: BAR FIXED AT BOTH ENDS A steel bar of length L and area A is fixed at both ends and carries a concentrated load P at distance a from the left end. Equilibrium: R_L + R_R = P (R_L and R_R are support reactions) Compatibility: The total deformation is zero (the bar cannot move): δ_L + δ_R = 0 Where: δ_L = deformation of left segment (length a): δ_L = (R_L × a) / (A E) (positive means elongation toward the right) δ_R = deformation of right segment (length L−a): δ_R = (−R_R × (L−a)) / (A E) (negative because R_R points leftward) Compatibility: (R_L × a) / (A E) − (R_R × (L−a)) / (A E) = 0 R_L × a = R_R × (L−a) R_L / R_R = (L−a) / a From equilibrium: R_L = P − R_R Substitute: (P − R_R) × a = R_R × (L−a) P a = R_R [a + L − a] = R_R × L R_R = P a / L R_L = P − P a / L = P(L−a) / L Conclusion: The load is distributed inversely to the distance; the reaction closer to the load is larger. EXAMPLE 2: COMPOSITE SHORT COLUMN (STEEL + CONCRETE) A composite column consists of a steel H-section (area A_s, E_s) and concrete fill (area A_c, E_c). They share an axial load P and must compress equally. Equilibrium: P_s + P_c = P Compatibility (equal compression): δ_s = δ_c (same length L, same compression) (P_s L) / (A_s E_s) = (P_c L) / (A_c E_c) P_s / (A_s E_s) = P_c / (A_c E_c) Define the "stiffness" k_i = A_i E_i. Then: P_s / k_s = P_c / k_c P_s = P × (k_s) / (k_s + k_c) = P × (A_s E_s) / (A_s E_s + A_c E_c) P_c = P × (k_c) / (k_s + k_c) = P × (A_c E_c) / (A_s E_s + A_c E_c) Interpretation: Load shares proportionally to axial rigidity (AE). The stiffer material carries proportionally more load. This principle extends to any composite or parallel system. NUMERICAL EXAMPLE: Steel tube: A_s = 3000 mm², E_s = 200 GPa → k_s = 600,000 Concrete: A_c = 60,000 mm², E_c = 20 GPa → k_c = 1,200,000 Total k = 1,800,000 For P = 600 kN: P_s = 600 × (600,000 / 1,800,000) = 600 × 1/3 = 200 kN P_c = 600 × (1,200,000 / 1,800,000) = 600 × 2/3 = 400 kN Steel carries 33% despite being a minority area, because its modulus is 10× higher. EXAMPLE 3: THERMAL RESTRAINT (INDETERMINATE) A steel bar is fixed at both ends with zero stress at 20°C. Temperature rises to 80°C. α = 11.7 × 10⁻⁶ /°C, E = 200 GPa, L = 2 m. Free thermal expansion: δ_T,free = 11.7 × 10⁻⁶ × 2000 × 60 = 1.404 mm (elongation wanted) Compatibility: The bar cannot move, so internal compression must balance the thermal expansion: δ_thermal + δ_internal = 0 α L ΔT − (σ L) / E = 0 σ = E α ΔT = 200,000 × 11.7 × 10⁻⁶ × 60 = 140.4 MPa (compressive) This internal compression is the **thermal stress**—already covered in Section 7, but notice it emerges naturally from the compatibility condition of a statically indeterminate (fixed) system. MULTIPLE SEGMENTS WITH DIFFERENT LOADS: Consider a composite bar with segment 1 (steel, length L_1, area A_1) and segment 2 (aluminum, length L_2, area A_2) loaded by forces P_1 and P_2 at junctions. Equilibrium: Internal force in segment 1 = internal force in segment 2 (for the system to be in equilibrium internally). But if segment 1 is load **and** the system is over-constrained (fixed at both ends), compatibility demands that deformations relate to the internal forces. The key insight: **Always add compatibility equations until you have as many equations as unknowns.** COMMON MISTAKES: 1. **Forgetting compatibility**: Assuming all materials carry equal stress (wrong). They carry equal strain if in parallel, or equal deformation if in series. 2. **Misidentifying statically indeterminate systems**: If a problem gives more constraints (e.g., fixed at two ends plus a middle support) than necessary to prevent motion, it's indeterminate. 3. **Ignoring the physical meaning**: Indeterminacy arises from **constraints**. Remove or relax a constraint, and you can solve using equilibrium alone. APPLICATION TO DESIGN: In AISC 360, composite columns and connections often involve load-sharing between materials. ACI 318 specifies that steel reinforcement and concrete must be analyzed together—a form of statically indeterminate analysis. In both, the relationship **P_i = P × (A_i E_i) / Σ(A_j E_j)** recurs constantly.