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CELE Strength of Materials — Simple Stresses and StrainsDetailed Explanation

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Detailed explanation of Simple Stresses and Strains for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Strength of Materials subtest.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Strength of Materials section sits under a "Core" weighting, and Simple Stresses and Strains is the 1st chapter in the 8-chapter CELE Strength of Materials rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Strength of Materials.

Simple Stresses and Strains - Detailed Explanation

Simple Stresses and Strains is the foundational chapter of Strength of Materials (also called Mechanics of Deformable Bodies or PSAD — Properties of Materials and Structural Analysis and Design in the PRC CE Licensure Examination syllabus). Every subsequent topic — torsion, bending, columns, connections, and combined stresses — is built on the three core relationships introduced here: stress equals force over area, strain equals deformation over length, and stress equals modulus times strain (Hooke's Law). Board examination problems in this chapter test whether you can (1) compute average normal, shear, and bearing stresses; (2) find axial deformations using δ = PL/AE; (3) handle multi-segment bars; (4) solve thermal stress and deformation problems; and (5) analyze statically indeterminate axial members using a compatibility equation. Master these five skill sets and you have the engine that drives the entire PSAD subject. All numerical work in this chapter is in SI units: forces in Newtons (N) or kilonewtons (kN), lengths in millimetres (mm), areas in mm², and stresses in megapascals (MPa = N/mm²).

Concepts

Normal (Axial) Stress

Normal stress σ (sigma) is the intensity of an internal force that acts perpendicular to (normal to) a cross-sectional area. When an external axial load P passes through the centroid of the cross-section, the stress is assumed uniformly distributed and is computed as: σ = P / A where P is the axial force (N) and A is the cross-sectional area (mm²). The resulting unit is N/mm² = MPa. Sign convention: Tensile stress is positive (+); compressive stress is negative (−). In board problems, if a member is being pulled (like a hanger rod or a tie bar), it is in tension. If it is being pushed (like a column or strut), it is in compression. St. Venant's Principle: The formula σ = P/A gives average stress that is essentially uniform only at sections remote from points of load application, holes, notches, or abrupt changes in cross-section. At those local regions, stress concentrations exist and the peak stress can be several times the average. For PRC board purposes, σ = P/A is used unless a stress concentration factor K is explicitly given. Allowable Stress and Factor of Safety (F.S.): Design requires that actual stress ≤ allowable stress. σ_allow = σ_yield / F.S. (for ductile materials) σ_allow = σ_ultimate / F.S. (often used for brittle materials or connections) RA 544 (Civil Engineering Law of the Philippines) and the National Structural Code of the Philippines 2015 (NSCP 2015) mandate that structural members be designed with adequate safety factors. NSCP 2015 Section 502 specifies allowable stresses for various structural materials under the Allowable Stress Design (ASD) method.

Examples

Convert 60 kN to 60,000 N first. Divide by the circular area. The result in N/mm² is directly MPa. Since the force is tensile, the stress is positive.

Scenario

A 25-mm diameter solid steel rod carries an axial tensile load of 60 kN. Find the average normal stress.

Solution

A = π(25)²/4 = π(625)/4 = 490.87 mm² σ = P/A = 60,000 N / 490.87 mm² = 122.2 MPa (tension)

Compressive stress is reported as 6.0 MPa in compression. Since 6.0 < 10 MPa allowable, the pier is adequate. This type of check appears in NSCP 2015 ASD problems.

Scenario

A short concrete pier 300 mm × 300 mm carries a column load of 540 kN. Find the compressive stress. Is it within the allowable of 10 MPa?

Solution

A = 300 × 300 = 90,000 mm² σ = 540,000 / 90,000 = 6.0 MPa (compression) 6.0 MPa < 10 MPa ✓ — acceptable

At the hole, the effective resisting area is reduced by the hole diameter. This net-section concept is fundamental to NSCP 2015 and AISC 360 tension member design.

Scenario

A steel eyebar 50 mm wide × 10 mm thick has a 15-mm diameter hole at its end. It carries a tensile load of 30 kN. Find the stress at the net section (through the hole).

Solution

Net width = 50 − 15 = 35 mm A_net = 35 × 10 = 350 mm² σ = 30,000 / 350 = 85.7 MPa

Applications

Tension member and hanger rod design (NSCP 2015 Section 504, AISC 360 Chapter D)
Column and pier capacity under direct compression (NSCP 2015 Section 505)
Net-section check at bolt holes in tension members
Checking wire rope or cable stresses in suspension structures
Determining required cross-sectional area for a given allowable stress and load

Misconceptions

Using kN directly with mm² — always convert: 1 kN = 1,000 N so that σ comes out in MPa
Applying σ = P/A when the load is eccentric (does not pass through centroid) — that produces combined normal stress and bending, not simple axial stress
Ignoring holes when computing area for tension members — use net area at the hole location
Treating compressive and tensile allowable stresses as always equal — for concrete, allowable tension is near zero

Related Concepts

Shear stress (parallel force on area)
Bearing stress (contact pressure)
Axial deformation δ = PL/AE
Net section in tension members (AISC 360 Chapter D)
Combined axial and bending stress

Common Exam Questions

Example

A 20-mm diameter rod carries 25 kN. σ = 25,000 / [π(20)²/4] = 25,000 / 314.16 = 79.6 MPa

Approach

Convert load to N, compute area in mm², apply σ = P/A to get MPa directly

Question Type

Compute normal stress from given load and dimensions

Example

P = 90 kN, σ_allow = 120 MPa → A = 90,000/120 = 750 mm² → d = √(4×750/π) = 30.9 mm

Approach

Set σ_allow = P/A, solve for A = P/σ_allow, then d = √(4A/π)

Question Type

Find required diameter given allowable stress and load

Example

Rod fails at 200 MPa, carries 80 MPa → F.S. = 200/80 = 2.5

Approach

F.S. = σ_failure / σ_actual or P_failure / P_actual

Question Type

Factor of safety computation

Key Points To Remember

σ = P/A — force perpendicular to the area; force must pass through the centroid for uniform distribution
Units: P in N, A in mm², σ in MPa (N/mm²). Never mix kN with mm² without converting first.
Tensile stress is positive; compressive stress is negative
σ = P/A gives average stress — valid away from load application points (St. Venant's Principle)
Allowable stress = yield stress / F.S. or ultimate stress / F.S. depending on design method
For circular sections: A = π d² / 4; for hollow circular: A = π(d_o² − d_i²) / 4
For a stepped bar, each segment has its own P and A — compute σ separately for each

Shear Stress

Shear stress τ (tau) results when the applied force acts parallel (tangential) to the resisting cross-sectional area: τ = V / A where V is the shear force (N) and A is the area resisting shear (mm²). SINGLE SHEAR: The fastener (bolt, rivet, pin) is cut on only one plane. The entire force P is resisted by one cross-section of the fastener: τ = P / A where A = π d² / 4 DOUBLE SHEAR: The fastener is cut on two planes (classic double-lap joint). Each plane carries half the force: τ = P / (2A) where A = π d² / 4 for one bolt cross-section This concept extends to punching shear (NSCP 2015 / ACI 318-19 Section 22.6), shear in welds, and shear keys in composite beams. The punching shear perimeter in a flat slab is taken at d/2 from the column face, giving a shear area = perimeter × d (effective depth). Allowable shear stress in bolts and rivets is typically specified as a fraction of the tensile yield stress; for example, AISC 360 Table J3.2 gives the nominal shear stress of A325 bolts as 372 MPa. Philippine practice under NSCP 2015 Section 510 provides similar allowable shear values for structural bolts and rivets.

Examples

In double shear, two cross-sections resist the force, so divide by 2A. If the student mistakenly uses single shear (P/A), the answer doubles to 159.2 MPa — a classic error.

Scenario

A 20-mm diameter bolt connects two plates in DOUBLE SHEAR and carries a force of 50 kN. Find the average shear stress in the bolt.

Solution

A = π(20)²/4 = 314.16 mm² τ = P/(2A) = 50,000 / (2 × 314.16) = 50,000 / 628.32 = 79.6 MPa

The material shears along the cylindrical surface defined by the punch diameter and plate thickness. This is punching shear — same concept as slab punching in RC design.

Scenario

A punch presses a 30-mm diameter hole through a 6-mm thick steel plate. The shear strength of the plate is 200 MPa. Find the required punching force.

Solution

Shear area = perimeter × thickness = π(30)(6) = 565.5 mm² P = τ × A = 200 × 565.5 = 113,097 N ≈ 113.1 kN

Divide total force equally among bolts (assuming equal spacing and same diameter), then apply single-shear formula.

Scenario

Three 16-mm diameter bolts in a lap joint carry a total force of 48 kN in single shear. Find the shear stress per bolt.

Solution

Force per bolt = 48,000 / 3 = 16,000 N A = π(16)²/4 = 201.06 mm² τ = 16,000 / 201.06 = 79.6 MPa

Applications

Bolt, rivet, and pin design in steel connections (NSCP 2015 Section 510, AISC 360 Chapter J)
Punching shear check in RC flat slabs (ACI 318-19 Section 22.6, NSCP 2015 Section 406)
Shear key design in composite beams
Shear stress in welds (throat area of fillet weld)
Shear in mechanical keys and couplings

Misconceptions

Using A = πd (circumferential area) instead of A = πd²/4 (cross-sectional area) for bolt shear
Applying single-shear formula to a double-shear connection — this overstates stress by 100%
Confusing the shear area (bolt cross-section) with the bearing area (projected rectangle d×t)
Assuming shear stress distribution across a bolt cross-section is uniform — it is actually parabolic for beams; for pins/bolts, average shear is used

Related Concepts

Bearing stress (companion check to shear stress in any connection)
Shear flow in built-up sections
Horizontal shear in beams (VQ/Ib formula — a later chapter)
Torsional shear stress (τ = Tc/J)
Punching shear in RC slabs (ACI 318 / NSCP 2015)

Common Exam Questions

Example

A clevis pin passes through a single plate: single shear. A clevis bracket with two outer plates and one inner plate: double shear.

Approach

Count the number of shear planes. One plate sandwiched between two = double shear. Two plates bolted together side-by-side = single shear.

Question Type

Identify single vs. double shear from a connection diagram

Example

P = 120 kN, 3 bolts in double shear, τ_allow = 80 MPa → A = 120,000/(3×2×80) = 250 mm² → d = 17.8 mm, use 20 mm

Approach

τ_allow = P/(nA) where n = number of bolts × shear planes. Solve for A, then d.

Question Type

Find bolt diameter for a given allowable shear stress and load

Example

300×300 column, d = 150 mm → perimeter = 4(300+150) = 1,800 mm → A = 1,800×150 = 270,000 mm²

Approach

Compute critical perimeter at d/2 from column face; shear area = perimeter × d; check V_u ≤ φV_c

Question Type

Punching shear in RC slab

Key Points To Remember

τ = V/A — force parallel to the area
Single shear: τ = P/A (one shear plane); Double shear: τ = P/(2A) (two shear planes)
Forgetting the factor of 2 in double shear is the most common board-exam mistake in this topic
Shear area for a circular fastener is the bolt's cross-sectional area A = πd²/4, NOT the bearing area
Punching shear perimeter = 4(c + d) for a square column of side c; shear area = perimeter × d
Shear stress is uniform on the resisting plane only for pins and fasteners (average shear stress)

Bearing Stress

Bearing stress σ_b is the compressive contact pressure between two bodies — for instance, between a bolt shank and the wall of the hole it presses against, or between a beam end and its support plate. It is computed from the projected area: σ_b = P / A_b where A_b = d × t Here, d is the bolt (or pin) diameter and t is the plate thickness (or the thickness of the thinner connected part). The projected area d×t is a rectangle, not the actual curved contact surface — this simplification is standard in both AISC 360 and NSCP 2015. For a bolt bearing on multiple plates, use the thinnest plate (the critical case). NSCP 2015 Section 510.3 and AISC 360 Section J3.10 give the nominal bearing strength as: R_n = 1.2 F_u t d (for standard holes, deformation is a design consideration) R_n = 1.5 F_u t d (for long-slot perpendicular to load direction) In allowable stress design (ASD), bearing stress is compared to F_p (allowable bearing stress) which varies by bolt type and hole type. Bearing stress governs hole tear-out and plate crushing — two of the three failure modes for bolted connections (the third being net-section fracture of the plate in tension).

Examples

Note: this is the same bolt from the double-shear example. The shear stress was 79.6 MPa but the bearing stress is 208.3 MPa — a much higher value. If the plate's allowable bearing stress is, say, 200 MPa, then the connection fails in bearing despite being safe in shear. Always check both.

Scenario

A 20-mm bolt bears on a 12-mm plate under a force of 50 kN. Find the bearing stress.

Solution

A_b = d × t = 20 × 12 = 240 mm² σ_b = P/A_b = 50,000 / 240 = 208.3 MPa

The clevis is in double shear, so each arm carries half the total load. Despite the thinner arms, the bracket has higher bearing stress because it carries the full load.

Scenario

A pin 25 mm in diameter connects a bracket plate (15 mm thick) to a clevis (two arms each 10 mm thick). The load is 40 kN. Find the bearing stress on the bracket plate and on the clevis arms.

Solution

Bracket: A_b = 25 × 15 = 375 mm²; σ_b = 40,000/375 = 106.7 MPa Each clevis arm carries P/2 = 20,000 N: A_b = 25 × 10 = 250 mm²; σ_b = 20,000/250 = 80.0 MPa Critical (higher): bracket plate at 106.7 MPa

Applications

Bolt bearing check in steel connections (AISC 360 Section J3.10, NSCP 2015 Section 510)
Pin and lug design in bridge hanger bars
Baseplate bearing on concrete pedestal (column baseplate design)
Key and keyway design in shafts
Soil bearing pressure under footings (conceptually the same — contact pressure)

Misconceptions

Using A = πd²/4 (circular bolt area) for bearing — bearing uses the projected area d×t
Applying bearing stress to the bolt itself (bolt stress is shear) — bearing stress is in the PLATE
Using total plate area instead of d×t — only the contact area under the bolt matters
Ignoring bearing check in favor of only shear check — board exams frequently test whether examinees remember to check bearing

Related Concepts

Shear stress in bolts (the companion calculation to bearing)
Net-section tension stress (the third connection failure mode)
Baseplate design and concrete bearing (NSCP 2015 Section 505.3, AISC 360 Chapter J8)
Soil bearing capacity (foundation design)

Common Exam Questions

Example

d = 16 mm, t = 10 mm, P = 30 kN → A_b = 160 mm² → σ_b = 187.5 MPa

Approach

A_b = d×t; σ_b = P/A_b

Question Type

Find the bearing stress given bolt diameter and plate thickness

Example

A common board question gives a lap joint and asks: which failure mode controls? Compute all three and identify the minimum capacity.

Approach

Compute all three stresses and compare each to its allowable; the governing (smallest capacity) controls

Question Type

Three-check connection problem (shear, bearing, net tension)

Key Points To Remember

σ_b = P/(d×t) — projected area, NOT the circular bolt area and NOT the curved contact surface
d is the bolt diameter; t is the PLATE thickness (not the bolt length)
For multiple plates, check the thinnest plate — it has the highest bearing stress
Bearing area A_b = d×t for bolts; for pins in clevises, use the smaller of the two areas (clevis or pin hole)
Three failure modes in a bolted connection: (1) bolt shear, (2) bearing/crushing of plate, (3) tension fracture at net section
Always run three checks on any connection: shear stress in bolt, bearing stress in plate, and tension stress in net section

Strain and Axial Deformation

NORMAL STRAIN ε (epsilon): When a bar of original length L is axially loaded and deforms by δ (delta), the average normal strain is: ε = δ / L Strain is dimensionless (mm/mm or m/m). Tensile strain is positive; compressive strain is negative. AXIAL DEFORMATION (the workhorse formula): Combine σ = P/A, ε = δ/L, and Hooke's Law σ = Eε: δ = PL / (AE) where: - P = axial force (N) — use the actual force in each segment - L = length of segment (mm) - A = cross-sectional area (mm²) - E = modulus of elasticity (MPa = N/mm²) - δ = axial deformation (mm) — positive = elongation The product AE is the AXIAL RIGIDITY. A stiffer bar (larger AE) deforms less under the same load — a critical insight for composite members. FOR MULTI-SEGMENT BARS: Sum the deformations segment by segment: δ_total = Σ (P_i × L_i) / (A_i × E_i) For each segment, determine P_i by cutting a free body diagram and applying ΣF = 0 to find the internal force. FOR CONTINUOUSLY VARYING LOAD OR AREA (e.g., a bar hanging under its own weight or a tapered bar), integrate: δ = ∫₀ᴸ P(x) / [A(x) × E] dx For a uniform bar hanging vertically under its own weight: the internal force at distance x from the bottom is P(x) = γ × A × x (where γ is unit weight in N/mm³). Integrating gives δ = γL²/(2E) = WL/(2AE) where W is total weight. Note the factor of ½ — the bar's own weight produces half the elongation of an equivalent tip load. Common values of E: - Structural steel: 200 GPa = 200,000 MPa - Aluminum: 70 GPa - Copper/Bronze: 110–120 GPa - Concrete: 20–30 GPa (NSCP 2015: Ec = 4700√f'c MPa for normal weight concrete, ACI 318-19 Section 19.2.2)

Examples

Convert: P = 60,000 N; L = 3,000 mm; E = 200,000 MPa. The rod elongates 1.83 mm — about 0.06% of its length, which is typical for structural steel at working stress.

Scenario

A steel rod 25 mm in diameter, 3 m long, carries a tensile load of 60 kN. E = 200 GPa. Find the elongation.

Solution

A = π(25)²/4 = 490.87 mm² δ = PL/(AE) = (60,000)(3,000) / (490.87 × 200,000) δ = 180,000,000 / 98,174,000 = 1.83 mm

The sign of P determines whether each segment elongates or shortens. Always start from the free end and check equilibrium to find internal forces. The net deformation at D is 0.35 mm elongation.

Scenario

A 3-segment steel bar (E = 200 GPa) is loaded as follows: Segment AB is 500 mm long, 400 mm², carries P_AB; segment BC is 800 mm, 200 mm²; segment CD is 600 mm, 300 mm². External loads: 20 kN applied at B (rightward), 50 kN at C (rightward), 30 kN at D (leftward). Bar is fixed at A. Find total deformation at D.

Solution

FBD at A: R_A = 20 + 50 − 30 = 40 kN (leftward reaction) P_AB (cut between A & B): internal force = 40 kN (tension? Check: A pulled right, so AB is in tension) → P_AB = +40 kN P_BC (cut between B & C): = 40 − 20 = 20 kN tension → P_BC = +20 kN P_CD (cut between C & D): = 40 − 20 − 50 = −30 kN → P_CD = −30 kN (compression) δ_total = Σ PL/(AE) = [40,000×500/(400×200,000)] + [20,000×800/(200×200,000)] + [−30,000×600/(300×200,000)] = [20,000,000/80,000,000] + [16,000,000/40,000,000] + [−18,000,000/60,000,000] = 0.250 + 0.400 − 0.300 = +0.350 mm (elongation)

Self-weight causes only half the deformation of an equivalent tip load. The factor of ½ comes from the linear variation of internal force from zero at the free end to W at the support.

Scenario

An aluminum bar (E = 70 GPa, γ = 26.5 kN/m³) is 4 m long, 40 mm × 40 mm square cross-section, hanging vertically under its own weight. Find the elongation due to self-weight.

Solution

γ = 26,500 N/m³ = 26,500 × 10⁻⁹ N/mm³ A = 40 × 40 = 1,600 mm² L = 4,000 mm W = γAL = 26,500 × 10⁻⁹ × 1,600 × 4,000 = 26,500 × 10⁻⁹ × 6,400,000 = 169.6 N δ = WL/(2AE) = 169.6 × 4,000 / (2 × 1,600 × 70,000) δ = 678,400 / 224,000,000 = 0.00303 mm ≈ 3.03 × 10⁻³ mm

Applications

Checking that member elongations are within serviceability limits (NSCP 2015 Table 405.1, deflection limits)
Settlement analysis of pile groups (axial shortening of pile shaft)
Truss joint displacement by virtual work (unit-load method uses δ = PL/AE for each member)
Compatibility equations for indeterminate structures
Cable elongation in prestressed systems and suspension bridges

Misconceptions

Using L in metres and A in mm² — always keep consistent units; easiest is all mm for L and A, all N for P, giving MPa for E and mm for δ
Using the same P for all segments in a multi-segment bar — internal force changes at each point of external load application
Ignoring the sign of internal force — a segment in compression shortens (negative δ) and this must be accounted for in the total
For self-weight problems, using WL/(AE) instead of WL/(2AE) — the factor of ½ is always required

Related Concepts

Hooke's Law (σ = Eε, the basis of δ = PL/AE)
Thermal deformation δ_T = αLΔT (adds or subtracts from mechanical deformation)
Virtual work / unit-load method for truss deflections
Statically indeterminate axial members (compatibility condition uses δ = PL/AE)
Poisson effect (lateral strain accompanies axial strain)

Common Exam Questions

Example

15-mm diameter steel rod, 2-m long, 45-kN load: δ = 45,000×2,000/(176.71×200,000) = 2.55 mm

Approach

Direct substitution into δ = PL/(AE) after unit conversion

Question Type

Find elongation of a single prismatic bar

Example

Step 1: Reactions. Step 2: Section each segment. Step 3: Sum with signs.

Approach

Cut FBDs at each segment to find internal forces P_i, then sum δ = Σ P_i L_i/(A_i E_i)

Question Type

Find total deformation of a multi-segment bar

Example

A rod elongates 2 mm under 100 kN; A = 500 mm², L = 3,000 mm → E = 100,000×3,000/(500×2) = 300,000 MPa = 300 GPa (this would be an unusual material — the question tests algebra)

Approach

Rearrange δ = PL/(AE): E = PL/(Aδ) or L = δAE/P

Question Type

Find E or L given δ and other parameters

Key Points To Remember

δ = PL/(AE) — THE single most important formula in Strength of Materials; learn it cold
Units check: P in N, L in mm, A in mm², E in MPa → δ in mm ✓
For multi-segment bars, draw FBDs to find P in each segment before applying δ = PL/(AE)
Elongation is positive (+); shortening is negative (−)
For self-weight hanging bar: δ = WL/(2AE) — half of what a tip load would cause
E_steel = 200 GPa; E_aluminum = 70 GPa; Ec = 4700√f'c MPa (NSCP 2015/ACI 318)
Axial rigidity AE: larger AE = stiffer member = less deformation = attracts more load in composite systems

Hooke's Law, Elastic Constants, and Poisson's Ratio

HOOKE'S LAW for normal stress: Within the proportional limit, stress and strain are linearly proportional: σ = E ε For shear: τ = G γ where E is the Modulus of Elasticity (Young's Modulus) and G is the Shear Modulus (Modulus of Rigidity). Both have units of MPa (or GPa). THE STRESS–STRAIN DIAGRAM for a ductile metal (mild steel) has five landmarks examiners love: 1. Proportional limit (σ_pl): upper end of the straight-line region where σ = Eε holds 2. Elastic limit: the highest stress with full elastic recovery on unloading (very close to the proportional limit for steel) 3. Yield point / Yield strength (σ_y or F_y): stress at onset of plastic (permanent) deformation; steel has a distinct yield plateau 4. Ultimate strength (σ_u or F_u): maximum stress on the engineering stress–strain curve; corresponds to the onset of necking 5. Fracture (rupture) strength: stress at the point of final fracture (lower than σ_u on engineering curve due to area reduction) For brittle materials (cast iron, concrete), there is no yield plateau — fracture follows the elastic zone directly. POISSON'S RATIO ν (nu): An axial tensile strain ε_axial causes a lateral compressive strain ε_lateral in the perpendicular directions: ν = − ε_lateral / ε_axial Typical values: steel ≈ 0.25–0.30; concrete ≈ 0.15–0.20; rubber ≈ 0.50 (incompressible). RELATIONSHIP BETWEEN E, G, AND ν: G = E / [2(1 + ν)] This relation is tested directly in some board questions: given E and ν, find G; or given E and G, find ν. BULK MODULUS K relates volumetric strain to hydrostatic pressure: K = E / [3(1 − 2ν)] When ν = 0.5, K → ∞ (incompressible). This explains why rubber (ν ≈ 0.5) is incompressible. GENERALIZED HOOKE'S LAW (Biaxial/Triaxial): When a point is subjected to stresses in multiple directions, Poisson coupling means each stress contributes to strains in the perpendicular directions: 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)] For biaxial (plane stress, σ_z = 0): ε_x = (1/E)(σ_x − ν σ_y) ε_y = (1/E)(σ_y − ν σ_x) ε_z = −ν/E (σ_x + σ_y) [Poisson contraction/expansion in z-direction even though σ_z = 0] Volumetric (dilatation): e = ε_x + ε_y + ε_z = [(1 − 2ν)/E](σ_x + σ_y + σ_z)

Examples

G ≈ 0.385E for steel (approximately 77 GPa). K > E for steel, meaning volume changes are resisted more strongly than shear. These values appear directly in torsion (using G) and pressure-vessel problems.

Scenario

For steel with E = 200 GPa and ν = 0.30, find G and K.

Solution

G = E/[2(1+ν)] = 200,000/[2(1.30)] = 200,000/2.60 = 76,923 MPa ≈ 77.0 GPa K = E/[3(1−2ν)] = 200,000/[3(0.40)] = 200,000/1.20 = 166,667 MPa ≈ 167 GPa

The negative lateral strain means the bar contracts laterally (becomes thinner) as it elongates axially. The diametral decrease is tiny — 0.00382 mm — but this Poisson effect matters in pressure vessels and multiaxial stress states.

Scenario

A 25-mm diameter, 200-mm long steel bar (E = 200 GPa, ν = 0.30) is subjected to an axial tensile load of 50 kN. Find (a) the axial strain, (b) the lateral strain, and (c) the change in diameter.

Solution

A = π(25)²/4 = 490.87 mm² σ = P/A = 50,000/490.87 = 101.9 MPa (a) ε_axial = σ/E = 101.9/200,000 = 5.09 × 10⁻⁴ (b) ε_lateral = −ν × ε_axial = −0.30 × 5.09×10⁻⁴ = −1.527 × 10⁻⁴ (c) Δd = ε_lateral × d = −1.527×10⁻⁴ × 25 = −3.82 × 10⁻³ mm (decrease)

Applications

Material characterization via tensile testing (determining E, σ_y, σ_u, ε_fracture)
Connecting shear and elastic deformation via G = E/[2(1+ν)]
Pressure vessel wall stress–strain analysis (biaxial state — hoop and longitudinal stresses)
Strain gauge rosette analysis (uses generalized Hooke's Law to find principal stresses from measured strains)
Design of rubber bearings and seismic isolators (ν ≈ 0.5, incompressibility governs)

Misconceptions

Applying Hooke's Law beyond the proportional limit — once yielding occurs, the linear relationship no longer holds
Confusing yield strength with ultimate strength — the board exam sometimes defines 'failure' as either, depending on context (ductile vs. brittle mode)
Ignoring the negative sign in Poisson's ratio — lateral strain is opposite in sign to axial strain
Thinking σ_z = 0 means ε_z = 0 in biaxial stress — ε_z = −ν(σ_x + σ_y)/E ≠ 0 even if no stress acts in z-direction

Related Concepts

Axial deformation δ = PL/(AE) — direct consequence of Hooke's Law
Torsion formula τ = Tc/J and angle of twist φ = TL/(GJ) — uses G
Mohr's Circle and principal stresses (generalized Hooke's Law for combined stresses)
Strain energy and resilience (U = σ²/(2E) × volume)
Thermal strain (additive to mechanical strain in compatibility equations)

Common Exam Questions

Example

E = 70 GPa, ν = 0.33 (aluminum) → G = 70,000/[2(1.33)] = 26,316 MPa ≈ 26.3 GPa

Approach

Directly apply G = E/[2(1+ν)]

Question Type

Find G given E and ν

Example

E = 200 GPa, G = 80 GPa → ν = 200,000/(2×80,000) − 1 = 1.25 − 1 = 0.25

Approach

Rearrange: ν = E/(2G) − 1

Question Type

Find ν given E and G

Example

σ_x = 100 MPa, σ_y = 60 MPa, E = 200 GPa, ν = 0.25 → ε_x = (100 − 0.25×60)/200,000 = 85/200,000 = 4.25×10⁻⁴

Approach

Apply generalized Hooke's Law: ε_x = (σ_x − ν σ_y)/E, ε_y = (σ_y − ν σ_x)/E

Question Type

Biaxial stress state — find strains

Key Points To Remember

E = σ/ε = slope of the initial linear portion of the stress–strain diagram
G = τ/γ = shear modulus; G = E/[2(1+ν)] links the three elastic constants
For structural steel: E = 200 GPa, G ≈ 77 GPa, ν ≈ 0.30
Yield strength F_y is where permanent deformation begins; ultimate strength F_u is the peak stress
The stress–strain curve for brittle materials (concrete, cast iron) has NO yield plateau
Poisson: lateral strain opposes axial strain in sign — lateral contraction accompanies axial tension
ν = 0.5 → incompressible (volume unchanged under load); this is the rubber/biological tissue limit
Generalized Hooke's Law: each strain depends on all three stresses via Poisson coupling

Thermal Stress and Deformation

A temperature change ΔT causes a material to expand or contract freely (with no stress) by: δ_T = α L ΔT where: - α = coefficient of thermal expansion (°C⁻¹ or /°C) - L = original length (mm) - ΔT = change in temperature (°C), positive for increase - δ_T = free thermal deformation (mm), positive for expansion For structural steel, α ≈ 11.7 × 10⁻⁶ /°C (NSCP 2015 uses 11.7×10⁻⁶; some references use 12×10⁻⁶ — use whatever the problem states). KEY PRINCIPLE: Free expansion (or contraction) produces NO STRESS. Thermal stress arises ONLY when deformation is restrained. FULLY RESTRAINED BAR: If both ends are fixed so no movement is possible, the thermal deformation must be entirely resisted by an internal mechanical force. Setting the mechanical strain equal and opposite to the thermal strain: σ_T = E α ΔT This is compressive when temperature rises (bar wants to expand but cannot), tensile when temperature drops. PARTIAL RESTRAINT / GAP PROBLEM: If there is a gap g between the bar and one support: - If δ_T = αLΔT ≤ g: bar expands but does not touch the wall — no stress develops - If δ_T > g: bar closes the gap (δ_T − g) of expansion is restrained → stress develops from the excess: Compatibility: αLΔT − PL/(AE) = g Solve: P = AE(αΔT − g/L) Then: σ = P/A = E(αΔT − g/L) FOR INDETERMINATE STRUCTURES WITH THERMAL LOADING: Use equilibrium + compatibility + force-deformation, with δ_T included in the compatibility equation. For example, a bar with a support spring of stiffness k at one end: δ_mechanical + δ_spring = δ_T PL/(AE) + P/k = αLΔT Solve for P, then σ = P/A.

Examples

The bar wants to expand 40°C × 11.7×10⁻⁶ × L = 4.68×10⁻⁴ L. The rigid walls prevent this expansion, inducing compressive stress. Note: length L cancels out — the thermal stress formula has no L.

Scenario

A steel bar is fixed at both ends with no initial stress at 20°C. Temperature rises to 60°C. Find the thermal stress. α = 11.7 × 10⁻⁶ /°C, E = 200 GPa.

Solution

ΔT = 60 − 20 = 40°C σ_T = EαΔT = 200,000 × 11.7×10⁻⁶ × 40 σ_T = 200,000 × 4.68×10⁻⁴ = 93.6 MPa (compression)

Step 1: Check if free expansion exceeds the gap. Step 2: If yes, only the excess deformation is mechanically restrained. Step 3: Find the force and stress from the excess. If the bar had expanded less than 1.5 mm, no stress would develop.

Scenario

A 3-m long steel bar (A = 600 mm², E = 200 GPa, α = 11.7 × 10⁻⁶ /°C) has a gap of 1.5 mm between its right end and a rigid wall at 20°C. Temperature rises to 70°C. Find the stress developed.

Solution

ΔT = 50°C; L = 3,000 mm Free expansion δ_T = αLΔT = 11.7×10⁻⁶ × 3,000 × 50 = 1.755 mm Gap = 1.5 mm < δ_T = 1.755 mm → wall is contacted; excess = 1.755 − 1.500 = 0.255 mm is restrained Compatibility: PL/(AE) = 0.255 mm P = 0.255 × AE/L = 0.255 × (600 × 200,000)/3,000 = 0.255 × 40,000 = 10,200 N σ = P/A = 10,200/600 = 17.0 MPa (compression)

Copper has a higher α, so it wants to expand more. Both are restrained by the walls. Series arrangement: equal forces, different stresses (different areas). The stiffer steel carries higher stress per unit area.

Scenario

A copper rod (α_c = 17 × 10⁻⁶ /°C, E_c = 120 GPa, A_c = 300 mm²) and a steel rod (α_s = 11.7 × 10⁻⁶ /°C, E_s = 200 GPa, A_s = 200 mm²) are placed end-to-end between rigid walls with no initial stress. Temperature rises 30°C. Find the stress in each.

Solution

Free expansion: δ_cu = 17×10⁻⁶ × L_cu × 30; δ_st = 11.7×10⁻⁶ × L_st × 30 Assuming equal lengths L: δ_cu_free = 5.1×10⁻⁴ L; δ_st_free = 3.51×10⁻⁴ L Compatibility (total deformation = 0, since both ends fixed): δ_cu + δ_st = δ_T_cu + δ_T_st −P_cu × L/(A_cu E_cu) − P_st × L/(A_st E_st) + δ_T_total = 0 Equilibrium (both carry same compressive force P since in series): P_cu = P_st = P P × L / (300 × 120,000) + P × L / (200 × 200,000) = (5.1 + 3.51) × 10⁻⁴ × L P × L [1/36,000,000 + 1/40,000,000] = 8.61 × 10⁻⁴ L P [2.778×10⁻⁸ + 2.500×10⁻⁸] = 8.61×10⁻⁴ P × 5.278×10⁻⁸ = 8.61×10⁻⁴ P = 16,313 N σ_cu = 16,313/300 = 54.4 MPa (compression); σ_st = 16,313/200 = 81.6 MPa (compression)

Applications

Rail gap design (expansion joints in railway tracks to prevent thermal buckling — a practical Philippine LRT/MRT engineering concern)
Expansion joints in concrete pavements, bridges (NSCP 2015 Section 307 on bridge expansion joints)
Prestressed concrete — thermal effects modify effective prestress
Pipe stress analysis in hot-fluid distribution systems
Bimetallic strip thermostats (differential thermal expansion)

Misconceptions

Assuming thermal stress exists even when the member is free to expand — ZERO stress unless restrained
Using the wrong sign: a restrained bar that heats up is in COMPRESSION (not tension)
Applying σ = EαΔT to a gap problem without checking whether the gap is closed first
Forgetting that in the gap problem, L in PL/(AE) is the original bar length, not the gap size

Related Concepts

Statically indeterminate axial members (thermal problems are almost always indeterminate)
Compatibility equations (thermal deformation adds to or subtracts from mechanical deformation)
Expansion joints in structural design (NSCP 2015)
Composite members with differential thermal expansion (bimetallic beam analogy)

Common Exam Questions

Example

Steel bar, ΔT = +50°C → σ_T = 200,000 × 11.7×10⁻⁶ × 50 = 117 MPa compression

Approach

σ_T = EαΔT; compression if temperature rises, tension if drops

Question Type

Fully restrained bar — find thermal stress

Example

Gap g = 2 mm, L = 4,000 mm, α = 11.7×10⁻⁶ → ΔT_contact = g/(αL) = 2/(11.7×10⁻⁶ × 4,000) = 42.7°C

Approach

Set δ_T = αLΔT = gap for contact temperature. For stress: excess deformation = PL/(AE).

Question Type

Gap problem — find temperature for first contact or stress after contact

Example

Common in board exams with steel and aluminum or steel and copper combinations

Approach

Compatibility: sum of mechanical deformations = sum of free thermal deformations. Equilibrium: sum of forces = 0.

Question Type

Composite bar with different α values — find internal forces

Key Points To Remember

Free expansion/contraction → δ_T = αLΔT, ZERO thermal stress
Fully restrained → σ_T = EαΔT (compression when temperature rises)
α_steel ≈ 11.7 × 10⁻⁶ /°C; α_aluminum ≈ 23 × 10⁻⁶ /°C; α_concrete ≈ 10 × 10⁻⁶ /°C
Gap problem: check if αLΔT > gap first; if yes, excess deformation is restrained and causes stress
Thermal stress is independent of member length (σ_T = EαΔT has no L) — but deformation depends on L
In composite bars with different α, each material wants to expand differently; compatibility forces them to a common deformation, generating internal forces
Temperature RISE in a restrained bar → COMPRESSION; temperature DROP → TENSION

Statically Indeterminate Axial Members

A structure is STATICALLY INDETERMINATE (or hyperstatic) when the equations of equilibrium alone (ΣF = 0, ΣM = 0) are insufficient to determine all unknown forces — the number of unknowns exceeds the number of equilibrium equations. The degree of indeterminacy = number of unknowns − number of equilibrium equations. For axial members, the extra equations needed are COMPATIBILITY EQUATIONS — conditions that specify how the deformations of different members must be geometrically consistent (they must fit together after loading). THREE-STEP METHOD (the universal approach for indeterminate axial problems): STEP 1 — EQUILIBRIUM: Write ΣF = 0 (or ΣFx = 0) in terms of the unknown forces. For a bar fixed at both ends, there are two unknown reactions; ΣF = 0 gives one equation relating them. STEP 2 — COMPATIBILITY: State a geometric condition on the deformations. Examples: - Bar fixed at both ends: total deformation from A to B = 0 (δ_AB = 0) - Two members connected at a rigid plate: both deform the same amount (δ_1 = δ_2) - Bar fixed at one end with a support spring at the other: δ_bar + δ_spring = 0 (or = some specified value) STEP 3 — FORCE–DEFORMATION: Substitute δ = PL/(AE) (plus αLΔT for thermal effects) into the compatibility equation and solve simultaneously with the equilibrium equation. COMMON CONFIGURATIONS tested in the PRC board: 1. BAR FIXED AT BOTH ENDS under a midpoint load P: R_A + R_B = P (equilibrium) Compatibility: segment AB shortens by R_A×L_AB/(AE) = R_B×L_BC/(AE) ... [set equal because total = 0] Result: R_A = P × L_BC/L; R_B = P × L_AB/L (forces share load inversely proportional to their segment lengths, if AE is constant) 2. COMPOSITE SHORT COLUMN (steel pipe filled with concrete, or steel post with concrete collar): P_s + P_c = P (equilibrium) δ_s = δ_c (compatibility — same shortening since materials are bonded and same length) P_s L/(A_s E_s) = P_c L/(A_c E_c) → P_s/P_c = (A_s E_s)/(A_c E_c) = axial rigidity ratio Result: the stiffer material (higher AE) carries more load. 3. TWO WIRES SUPPORTING A RIGID BAR: ΣFy = 0 (equilibrium); ΣM = 0 about one attachment point; compatibility: elongations of the two wires satisfy the geometry of the rigid bar's rotation (linear variation — similar triangles). 4. BAR WITH THERMAL LOADING: Add δ_T = αLΔT as an additional deformation term in the compatibility equation.

Examples

Axial rigidities: AE_s = 3,000×200,000 = 6×10⁸ N; AE_c = 60,000×20,000 = 12×10⁸ N. The concrete's axial rigidity is twice the steel's, so it carries twice the load. This is the fundamental load-sharing rule for composite systems.

Scenario

A composite short column (E_s = 200 GPa, A_s = 3,000 mm²; E_c = 20 GPa, A_c = 60,000 mm²) carries P = 600 kN. Both materials are the same length. Find the load in each.

Solution

Equilibrium: P_s + P_c = 600 kN ... (1) Compatibility: δ_s = δ_c → P_s L/(A_s E_s) = P_c L/(A_c E_c) → P_s/(3,000×200,000) = P_c/(60,000×20,000) → P_s/600,000,000 = P_c/1,200,000,000 → P_c = 2P_s ... (2) Substitute (2) into (1): P_s + 2P_s = 600 → P_s = 200 kN; P_c = 400 kN

The reaction closer to the applied load is larger. R_A at the near end = 60 kN (3/4 of total); R_B at the far end = 20 kN (1/4 of total). This makes physical sense: the stiff short segment resists more.

Scenario

A steel bar (A = 500 mm², E = 200 GPa, L = 2,000 mm) is fixed at both ends. A load P = 80 kN is applied at 500 mm from the left end. Find the reactions.

Solution

Let R_A = left reaction, R_B = right reaction. Equilibrium: R_A + R_B = 80,000 N ... (1) Compatibility: total deformation from A to B = 0 (rigid walls) δ_AC + δ_CB = 0 (but note AC is being 'pushed' by R_A while CB is 'pulled' — set up signs carefully) Using P_AC = R_A (tension positive) and P_CB = R_A − 80,000 = −R_B (compression): δ_total = R_A×500/(500×200,000) + (−R_B)×1,500/(500×200,000) = 0 R_A×500 = R_B×1,500 R_A = 3R_B ... (2) From (1): 3R_B + R_B = 80,000 → R_B = 20,000 N = 20 kN; R_A = 60 kN

Applications

Composite columns (steel-encased concrete, steel-filled tube — NSCP 2015 Section 506 composite compression members)
Prestressed concrete: the tendon and the concrete are a composite indeterminate system
Bolted flange connections where bolt preload interacts with the clamped member
Portal frames with fixed supports (indeterminate in multiple ways)
Bridge continuous spans (horizontal thermal force with fixed abutments)

Misconceptions

Assuming the load distributes equally between materials in a composite — it distributes proportional to AE (axial rigidity), NOT equally and NOT proportional to area alone
Forgetting the compatibility equation entirely and trying to solve indeterminate problems with equilibrium alone — you will always be one equation short
Using δ_1 = δ_2 for members in series (end-to-end) — compatibility for series members is that internal forces are equal, not deformations
For a bar fixed at both ends, thinking both reactions are P/2 — reactions depend on the position of the load and the AE values of each segment

Related Concepts

Virtual work / energy methods for more complex indeterminate structures
Thermal stress in restrained bars (thermal loading in indeterminate systems)
Composite beams in bending (same load-sharing principle, but for flexural rigidity EI)
Statically indeterminate beams (additional topic using moment-area and three-moment theorem)

Common Exam Questions

Example

3-material column: ΣF = 0 gives one equation; two compatibility equations (δ_1=δ_2, δ_2=δ_3); 3 equations, 3 unknowns.

Approach

Step 1: ΣF = 0. Step 2: δ_1 = δ_2 → P_1/(A_1E_1) = P_2/(A_2E_2). Step 3: Solve simultaneously.

Question Type

Composite column — find load in each material

Example

Bar fixed at both ends with load at 1/3 point: R_near = 2/3 P, R_far = 1/3 P (inverse proportion to distances)

Approach

ΣF = 0 (1 equation); compatibility δ_total = 0 (1 equation); 2 equations, 2 unknowns R_A and R_B.

Question Type

Bar fixed at both ends — find reactions

Example

Two wires of different L, A, E: wire forces are NOT simply proportional to their positions; axial rigidities must be included.

Approach

ΣF = 0; ΣM = 0 (gives one compatibility equation via geometry); solve for wire forces.

Question Type

Two wires supporting a rigid horizontal bar

Key Points To Remember

Indeterminate axial: always need Equilibrium + Compatibility + Force-Deformation (three steps, never skip one)
Compatibility tells you how the deformations of different elements must relate geometrically
For composite columns (bonded materials): δ_1 = δ_2 (same deformation), so P_1/P_2 = (A_1 E_1)/(A_2 E_2)
The stiffer material in a composite system attracts MORE load — a key board concept
For a bar fixed at both ends: total elongation = 0; reactions are inversely proportional to segment lengths (if AE = constant)
For a rigid bar on two springs/wires: use similar triangles (linear geometry) for compatibility
Thermal effects: add δ_T = αLΔT to the mechanical deformation δ = PL/(AE) in the compatibility equation

Practice Problems

Unit check: P = 80,000 N, L = 2,500 mm, A = 706.86 mm², E = 70,000 MPa → δ in mm ✓. The rod is stressed to 113.2 MPa, which is 42% of yield (F.S. = 2.39). This is a typical working-stress level for aluminum structures. Note: for concrete or brittle members, F.S. would typically be referenced to ultimate strength.

Problem

PROBLEM 1 (Normal Stress + Deformation): An aluminum rod (E = 70 GPa) has a diameter of 30 mm and a length of 2.5 m. It carries an axial tensile load of 80 kN. Find: (a) the normal stress, (b) the elongation, and (c) the factor of safety if the yield strength is 270 MPa.

Solution

A = π(30)²/4 = π(900)/4 = 706.86 mm² (a) σ = P/A = 80,000/706.86 = 113.2 MPa (b) δ = PL/(AE) = (80,000)(2,500)/(706.86 × 70,000) = 200,000,000/49,480,200 = 4.04 mm (c) F.S. = σ_yield/σ_actual = 270/113.2 = 2.39

Three checks, three different stresses: bolt shear (149.2 MPa), plate bearing (187.5 MPa), net tension (93.75 MPa). The governing failure mode is bearing at 187.5 MPa — the highest stress. If the allowable bearing stress is 200 MPa, the connection is just acceptable. This three-check procedure is fundamental to NSCP 2015 Section 510 connection design and appears frequently in board examinations.

Problem

PROBLEM 2 (Shear + Bearing + Net Tension): A steel plate 10 mm thick is connected to a gusset plate using two 16-mm diameter bolts in single shear. The applied tensile force is 60 kN. The plate is 80 mm wide. Find: (a) shear stress in bolts, (b) bearing stress on the main plate, (c) net-section tensile stress in the plate.

Solution

(a) Bolt shear (single shear, 2 bolts): A_bolt = π(16)²/4 = 201.06 mm² per bolt Total shear area = 2 × 201.06 = 402.12 mm² τ = P/(n×A) = 60,000/402.12 = 149.2 MPa per bolt... OR: τ per bolt = (60,000/2)/201.06 = 30,000/201.06 = 149.2 MPa (b) Bearing stress (on main plate, each bolt carries 30,000 N): A_b = d × t = 16 × 10 = 160 mm² per bolt σ_b = 30,000/160 = 187.5 MPa (c) Net-section tension (at the bolt hole): Net width = 80 − 16 = 64 mm A_net = 64 × 10 = 640 mm² σ_net = P/A_net = 60,000/640 = 93.75 MPa

Three-part logic: (1) Check if bar touches the wall — it does when ΔT > 42.7°C. Since 50°C > 42.7°C, it touches. (2) Only the EXCESS deformation beyond the gap is resisted. (3) Find P from δ_excess = PL/(AE). Note: had the temperature only risen to 55°C (ΔT = 40°C < 42.7°C), the bar would not touch the wall and σ = 0.

Problem

PROBLEM 3 (Thermal Stress with Gap): A 4-m long steel bar (A = 900 mm², E = 200 GPa, α = 11.7 × 10⁻⁶ /°C) is fixed at one end. The other end has a 2-mm gap to a rigid wall at 15°C. (a) At what temperature will the bar first touch the wall? (b) If the temperature rises to 65°C, what is the compressive stress in the bar?

Solution

(a) Contact temperature: δ_T = αLΔT = gap 11.7×10⁻⁶ × 4,000 × ΔT = 2 ΔT = 2/(11.7×10⁻⁶ × 4,000) = 2/0.0468 = 42.7°C T_contact = 15 + 42.7 = 57.7°C (b) Stress at 65°C (ΔT = 65 − 15 = 50°C): Free expansion = 11.7×10⁻⁶ × 4,000 × 50 = 2.34 mm Excess (restrained) deformation = 2.34 − 2.00 = 0.34 mm Compatibility: PL/(AE) = 0.34 mm P = 0.34 × AE/L = 0.34 × (900 × 200,000)/4,000 P = 0.34 × 180,000,000/4,000 = 0.34 × 45,000 = 15,300 N σ = P/A = 15,300/900 = 17.0 MPa (compression)

AE_s = 800 × 10⁶ N; AE_c = 2,000 × 10⁶ N. Concrete is 2.5× stiffer (higher AE), so it carries 2.5× more load. Despite the much larger area of concrete, its lower E means the stress ratio is not directly obvious — the load-sharing ratio is the axial rigidity ratio. This composite column arrangement is common in Philippine construction (steel pipe columns filled with concrete).

Problem

PROBLEM 4 (Statically Indeterminate Composite Column): A 600-mm long composite column consists of a steel core (A_s = 4,000 mm², E_s = 200 GPa) surrounded by a concrete jacket (A_c = 80,000 mm², E_c = 25 GPa). Both ends are covered by rigid plates, and a compressive load of 1,200 kN is applied. Find: (a) load carried by steel, (b) load carried by concrete, (c) stress in each material.

Solution

Step 1 — Equilibrium: P_s + P_c = 1,200,000 N ... (1) Step 2 — Compatibility: δ_s = δ_c (same length, deform equally) P_s × L/(A_s × E_s) = P_c × L/(A_c × E_c) L cancels: P_s/(4,000 × 200,000) = P_c/(80,000 × 25,000) P_s/800,000,000 = P_c/2,000,000,000 P_c = (2,000,000,000/800,000,000) × P_s = 2.5 P_s ... (2) Step 3 — Solve simultaneously: Substitute (2) into (1): P_s + 2.5P_s = 1,200,000 3.5 P_s = 1,200,000 (a) P_s = 342,857 N ≈ 342.9 kN (b) P_c = 2.5 × 342,857 = 857,143 N ≈ 857.1 kN Check: 342.9 + 857.1 = 1,200 kN ✓ (c) Stresses: σ_s = P_s/A_s = 342,857/4,000 = 85.7 MPa (compression) σ_c = P_c/A_c = 857,143/80,000 = 10.7 MPa (compression)

Key steps: (1) Find wall reaction by ΣF = 0 for the whole bar. (2) Cut each segment and find internal force P_i by ΣF = 0 on one portion. (3) Compute δ_i = P_i L_i/(A_i E_i) for each segment. (4) Sum with proper signs. Segment 3 carries zero internal force because equilibrium at D (free end) requires P_3 = 0. The bar's rightward movement at D is 1.312 mm — driven primarily by the elongation of the aluminum segment.

Problem

PROBLEM 5 (Multi-Segment Bar with Different Materials): A bar consists of three segments in series: Segment 1 — bronze (E = 83 GPa, A = 500 mm², L = 500 mm); Segment 2 — aluminum (E = 70 GPa, A = 400 mm², L = 600 mm); Segment 3 — steel (E = 200 GPa, A = 300 mm², L = 400 mm). Forces applied: 30 kN rightward at junction of segments 1 and 2, and 50 kN leftward at junction of segments 2 and 3. Bar is fixed at the left end (through segment 1) and free at the right end. Find the total deformation at the free right end.

Solution

Label: A—(Seg1/Bronze)—B—(Seg2/Al)—C—(Seg3/Steel)—D (free end) External loads: 30 kN rightward at B; 50 kN leftward at C Wall reaction at A (ΣFx = 0): R_A = 30 − 50 = −20 kN → 20 kN leftward (or: R_A points left, meaning it pulls the bar to prevent rightward movement) Actually, let right = positive: At A (fixed): R_A + 30 − 50 = 0 → R_A = +20 kN (rightward reaction — meaning wall pulls bar rightward, or bar is being PULLED to the left — let's check by FBD) FBD segment 1 (A to B), cutting just right of A: P_1 = R_A = +20 kN? Check: sum of forces on left portion = R_A − P_1 = 0 ... no. Correct FBD: Cut Segment 1 between A and B. Left portion has reaction R_A. ΣFx = 0: R_A + P_1 = 0 where P_1 is the internal force (positive = tension). R_A is the reaction at A. Let right = positive for external forces: 30 kN at B (→), 50 kN at C (←= −50 kN) Equilibrium whole bar: R_A + 30 − 50 = 0 → R_A = +20 kN (rightward, so wall pushes bar rightward = bar is in equilibrium pulling left) Internal forces (section method, sum forces left of cut): P_1 (Seg 1 A→B): left portion has R_A = +20 kN (→). For equilibrium of left portion: R_A − P_1 = 0 (if tension positive means right face of cut has P_1 pointing right). → P_1 = +20 kN (tension) ✓ P_2 (Seg 2 B→C): left portion has R_A + 30 kN. ΣF = R_A + 30 − P_2 = 0 → P_2 = 20 + 30 = +50 kN (tension) P_3 (Seg 3 C→D): left portion has R_A + 30 − 50 = 0 kN → P_3 = 0 kN (no internal force in Seg 3!) Deformations: δ_1 = P_1 L_1/(A_1 E_1) = 20,000 × 500/(500 × 83,000) = 10,000,000/41,500,000 = +0.241 mm δ_2 = P_2 L_2/(A_2 E_2) = 50,000 × 600/(400 × 70,000) = 30,000,000/28,000,000 = +1.071 mm δ_3 = P_3 L_3/(A_3 E_3) = 0 δ_total = 0.241 + 1.071 + 0.000 = +1.312 mm (elongation at free end D)

Under compression: the bar shortens axially (negative ε_axial) but expands laterally (positive ε_lateral — Poisson effect causes the bar to 'bulge'). The volume still decreases because the axial shortening dominates. If ν were exactly 0.5, ΔV = 0 (incompressible). The shear modulus G ≈ 76.9 GPa is used directly in the torsion chapter.

Problem

PROBLEM 6 (Poisson Effect + Elastic Constants): A 40-mm diameter, 300-mm long steel bar (E = 200 GPa, ν = 0.30) is subjected to a 150-kN compressive load. Find: (a) axial stress, (b) axial strain, (c) lateral strain, (d) change in diameter, (e) change in volume. Also compute G.

Solution

A = π(40)²/4 = 1,256.6 mm² (a) σ_axial = −150,000/1,256.6 = −119.4 MPa (compression = negative) (b) ε_axial = σ/E = −119.4/200,000 = −5.97 × 10⁻⁴ (c) ε_lateral = −ν × ε_axial = −0.30 × (−5.97×10⁻⁴) = +1.791 × 10⁻⁴ (positive = lateral expansion) (d) Δd = ε_lateral × d = 1.791×10⁻⁴ × 40 = +7.16 × 10⁻³ mm (diameter INCREASES under compression) (e) Volume: V₀ = A × L = 1,256.6 × 300 = 376,991 mm³ ε_vol = ε_axial(1 − 2ν) = (−5.97×10⁻⁴)(1 − 0.60) = (−5.97×10⁻⁴)(0.40) = −2.388×10⁻⁴ ΔV = ε_vol × V₀ = −2.388×10⁻⁴ × 376,991 = −90.1 mm³ (volume DECREASES under compression) G = E/[2(1+ν)] = 200,000/[2(1.30)] = 76,923 MPa ≈ 76.9 GPa

Exam Preparation Tips

MEMORIZE THE FIVE KEY FORMULAS: σ = P/A, τ = V/A (or P/2A for double shear), σ_b = P/(dt), δ = PL/(AE), and σ_T = EαΔT. These five cover 70–80% of board exam problems in this chapter.
UNIT DISCIPLINE: Always work in N and mm so that stress comes out directly in MPa (= N/mm²). Convert kN to N at the start: 1 kN = 1,000 N. Never put kN into δ = PL/(AE) with A in mm².
DOUBLE-SHEAR REFLEX: As soon as you see a pin between two outer plates and one inner plate (or a clevis), immediately write τ = P/(2A). The board frequently tests this specific configuration.
THREE CONNECTION CHECKS: For every bolted or riveted connection problem, check all three: (1) bolt shear τ = P/nA, (2) plate bearing σ_b = P/(dt), (3) net-section tension σ = P/A_net. The problem may ask for all three or just one, but knowing all three prevents omission errors.
THERMAL LOGIC SEQUENCE: (1) Is the bar free to move? If yes → δ_T = αLΔT, σ = 0. (2) Is there a gap? If yes → check if αLΔT > gap; if yes, excess = αLΔT − gap is restrained → stress develops. (3) Fully restrained → σ = EαΔT directly. This three-step mental checklist handles every thermal variation.
INDETERMINATE THREE-STEP DRILL: Equilibrium → Compatibility → Force-Deformation. Never skip the compatibility step. For composite columns, compatibility is δ_1 = δ_2 (same deformation), leading to P_1/P_2 = (A_1 E_1)/(A_2 E_2).
STEEL PROPERTIES TO MEMORIZE: E = 200 GPa, G ≈ 77 GPa, ν ≈ 0.30, α = 11.7 × 10⁻⁶ /°C. For aluminum: E = 70 GPa, α ≈ 23 × 10⁻⁶ /°C. For concrete: E_c = 4700√f'c MPa (NSCP 2015/ACI 318).
SELF-WEIGHT BAR ELONGATION: Use δ = WL/(2AE) = γL²/(2E). The factor of ½ is tested; candidates who forget it double the correct answer.
SIGN CONVENTION CONSISTENCY: Establish one sign convention and stick with it throughout a problem. Recommended: rightward or downward = positive for forces; elongation = positive for deformation. At the end, a positive δ means elongation, negative means shortening.
REVIEW RA 544 (CIVIL ENGINEERING LAW): The board may include questions on who can legally sign and seal structural computations, definition of civil engineering practice, scope of the profession, and CE liability. RA 544 Section 10 defines civil engineering as including the design, construction, and maintenance of structures.
STRESS–STRAIN CURVE LABELS: Know all five landmark points (proportional limit, elastic limit, yield point, ultimate strength, fracture strength) and be able to describe the difference between ductile (steel) and brittle (concrete, cast iron) materials.
PAST BOARD EXAM PATTERN: The PRC CE board typically includes 2–4 problems on simple stresses and strains per examination. Expect one axial deformation problem, one connection problem (shear + bearing), one thermal or indeterminate problem, and occasionally a Poisson/elastic constants problem. Speed and accuracy in unit conversion is often the deciding factor between correct and incorrect answers.

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In summary

Simple Stresses and Strains is simultaneously the most fundamental and the most heavily tested chapter in the entire Strength of Materials subject. Every formula here — σ = P/A, τ = V/A, σ_b = P/(dt), δ = PL/(AE), σ_T = EαΔT — reappears in every subsequent chapter of the PRC CE Licensure Examination syllabus, from torsion and bending to column design and RC/steel structural design. The four competencies that separate passers from non-passers in this chapter are: (1) UNIT DISCIPLINE — always N and mm so that stress is in MPa; (2) CONNECTION COMPLETENESS — running all three checks (shear, bearing, net tension) without prompting; (3) THERMAL LOGIC — recognizing that free expansion never causes stress, and only restraint does; and (4) INDETERMINATE THREE-STEP EXECUTION — always writing equilibrium + compatibility + force-deformation, never attempting to solve with equilibrium alone. As a licensed civil engineer practicing in the Philippines under RA 544, you will encounter these stresses in every structural computation you sign and seal — from designing the hanger rods of a gymnasium roof (axial tension, δ = PL/AE) to specifying the expansion joints in a bridge deck (thermal deformation) to checking the bolt pattern of a steel girder connection (shear + bearing + net section, per NSCP 2015 and AISC 360). The board examination tests these skills at exactly the level of daily professional practice. Drill the six practice problems in this chapter until you can solve each one in under eight minutes. Master the visual decision flowcharts for thermal and indeterminate problems. Memorize the material constants for steel, aluminum, and concrete. Then move confidently to the next chapter — torsion — knowing that every concept there is simply an extension of τ = V/A and Hooke's Law in a circular geometry.

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