CELE Strength of Materials — Simple Stresses and StrainsRevision Notes

Formulas

Exam Tips

• Board exams frequently give diameter — remember A = π d² / 4; compute A first and keep it in mm².
• When a problem says 'allowable stress,' immediately write σ = P/A ≤ σ_allow and solve for the unknown.
• Verify units in every step: N ÷ mm² = MPa. If an answer looks like 122 000 MPa, you forgot to convert kN to N.
• For tapered bars or stepped bars, identify each segment separately and sum axial forces from a free-body diagram.
• Remember F.S. is always ≥ 1; if you get F.S. < 1, the member has failed.

Key Points

• Normal stress σ acts perpendicular to the cross-sectional area — the force 'stabs through' the section.
• The formula σ = P/A gives the AVERAGE stress; it is valid only at sections remote from load points (St. Venant's principle — local disturbances die out within ~1 member-width).
• Sign convention: TENSILE stress is POSITIVE (+); COMPRESSIVE stress is NEGATIVE (−).
• The load P must pass through the centroid of the area — otherwise bending is superimposed.
• Units check: P in N, A in mm² → σ in MPa. Never mix kN with mm² without converting.
• St. Venant's principle is the justification for using σ = P/A everywhere except near holes, notches, and load points.
• Allowable (working) stress = σ_yield / F.S. or σ_ultimate / F.S., depending on the design standard.

Definitions

Section Title

Normal (Axial) Stress

Common Mistakes

• Mixing units — applying P in kN and A in mm² without converting: always use P in N and A in mm² to get σ in MPa.
• Forgetting that σ = P/A is an AVERAGE — actual peak stresses near holes or notches can be 2–3× higher (stress concentration).
• Using the gross area instead of the NET area when a member has bolt holes (net area = gross area − hole area).
• Treating compressive stress as positive — by convention compression is negative.
• Applying σ = P/A when the load is eccentric (off-centroid) — eccentricity induces bending, requiring combined-stress analysis.

Formulas

Exam Tips

• The phrase 'in double shear' is your signal to use τ = P/(2A) — watch for it in board exam stems.
• For bearing, the projected area is always d × t — draw a sketch of the bolt-in-hole to confirm which t applies.
• A quick check: shear stress uses the bolt's CIRCULAR area (πd²/4); bearing stress uses the RECTANGULAR projected area (d×t).
• When finding the 'safe load' on a connection, compute allowable P for each failure mode separately, then take the MINIMUM.
• NSCP 2015 Table 502.3.2 lists allowable shear stresses for A325 and A490 bolts — memorize the common values for exam readiness.

Key Points

• Shear stress τ acts PARALLEL to (tangent to) the resisting area — the force tries to 'slice' through the material.
• SINGLE shear: the connector is cut on ONE plane → τ = P / A (one bolt area).
• DOUBLE shear: the connector is cut on TWO planes → τ = P / (2A) — the load is split equally between two shear planes.
• Bearing stress σ_b is a CONTACT PRESSURE on the projected (rectangular) area between the bolt shank and the plate hole: A_b = d × t.
• Bearing area uses the PROJECTED area — NOT the bolt's circular cross-section — to represent the actual contact pressure distribution.
• In a bolted or riveted connection, three failure modes must be checked: tensile rupture of the plate, shear failure of the bolt, and bearing/crushing of the plate.
• NSCP 2015 Section 502 and AISC 360 Chapter J both prescribe allowable stresses for bolts in shear and bearing.

Definitions

Section Title

Shear Stress and Bearing Stress

Common Mistakes

• Using single-shear formula (τ = P/A) for a double-shear connection — this doubles the computed stress and is unconservative.
• Using the bolt's circular area (πd²/4) instead of the projected area (d×t) for bearing stress calculation.
• Forgetting to check ALL three failure modes (plate tensile rupture, bolt shear, and plate bearing) in a connection problem.
• Using the total plate thickness when the bolt bears on only one plate in a multi-plate sandwich.
• Confusing shear stress in the bolt with shear stress in the plate — they act on different areas.

Formulas

Exam Tips

• Memorize E_steel = 200 GPa, E_aluminum = 70 GPa, and G_steel ≈ 77 GPa — these appear in almost every SoM problem.
• For ductile steel, proportional limit ≈ elastic limit ≈ yield point in most exam problems — use yield stress unless told otherwise.
• The 0.2% offset method: draw a line parallel to the initial slope starting at ε = 0.002; where it intersects the curve is the yield strength.
• The stress–strain curve for brittle materials (concrete, cast iron) has NO distinct yield point and fractures at a relatively low strain.
• When a board exam problem gives two of {E, G, ν}, always solve for the third using G = E/[2(1+ν)] before proceeding.

Key Points

• Normal strain ε is DIMENSIONLESS — it is the ratio of deformation to original length (ε = δ/L).
• Shear strain γ is also dimensionless — it is the change in the right angle (in radians) of an element.
• The stress–strain diagram for mild (low-carbon) steel has five landmark points: proportional limit, elastic limit, upper/lower yield point, ultimate strength, and fracture/rupture.
• Hooke's Law (σ = Eε and τ = Gγ) applies ONLY in the linear-elastic range below the proportional limit.
• E (Young's modulus) for structural steel ≈ 200 GPa; for aluminum ≈ 70 GPa; for concrete ≈ 17–30 GPa.
• G (shear modulus) for steel ≈ 77 GPa; related to E and ν by G = E / [2(1+ν)].
• Poisson's ratio ν = −ε_lateral / ε_axial; for steel ν ≈ 0.27–0.30; for concrete ν ≈ 0.15–0.20.
• For incompressible materials (rubber), ν → 0.5 and the bulk modulus K → ∞ (no volume change).

Definitions

Section Title

Strain, Hooke's Law, and the Stress–Strain Diagram

Common Mistakes

• Applying Hooke's law (σ = Eε) beyond the proportional limit — the formula is invalid in the plastic range.
• Confusing proportional limit with yield strength — they are close but NOT the same; yield strength may involve permanent set.
• Using E in GPa without converting to MPa when computing strain: if σ is in MPa, E must also be in MPa (200 000 MPa, not 200 GPa).
• Forgetting that Poisson's ratio makes lateral strains OPPOSITE in sign to axial strain (contraction under tension).
• Computing G independently instead of using G = E/[2(1+ν)] — the three constants are NOT independent.

Formulas

Exam Tips

• Always draw a free-body diagram (FBD) to find P at each segment — use method of sections and ΣF = 0 at each cut.
• For a bar with multiple external loads at intermediate points, the internal force changes at each load point — you must compute it segment by segment.
• The product PL/AE must be dimensionally consistent: N × mm / (mm² × MPa) = N × mm / (mm² × N/mm²) = mm. ✓
• If asked for the displacement of a specific point (not just the total elongation), sum δ from the FIXED support to that point.
• Board exam shortcut: if all segment materials are the same (E is constant), factor it out: δ = (1/E) × Σ(P_i L_i / A_i).

Key Points

• The axial deformation formula δ = PL/AE is derived by combining σ = P/A, ε = δ/L, and σ = Eε.
• AE is called the AXIAL RIGIDITY of the member — the higher AE, the less it deforms for a given load.
• For STEPPED or SEGMENTED bars (changing P, A, or E), compute δ for each segment separately and ALGEBRAICALLY SUM: δ_total = Σ(P_i × L_i)/(A_i × E_i).
• Sign convention for summing: tensile elongation is POSITIVE (+), compressive shortening is NEGATIVE (−).
• For a continuously varying load (e.g., self-weight of a hanging bar) or continuously varying area (tapered bar), INTEGRATE: δ = ∫₀ᴸ P(x)/[A(x)E] dx.
• For a bar hanging under its own weight: δ = γL²/(2E) where γ is the unit weight of the material.
• The formula applies to any axially loaded member in the elastic range — rods, cables, columns, truss members.

Definitions

Section Title

Axial Deformation (δ = PL/AE)

Common Mistakes

• Forgetting to convert L to mm when A is in mm² and E is in MPa — inconsistent units are the #1 computational error.
• Using the same P for all segments without re-drawing the free-body diagram — internal forces change at every point of load application.
• Adding all elongations as positive — compressive segments SHORTEN and must be given a negative sign before summing.
• Applying δ = PL/AE to a non-prismatic bar without integrating — the formula assumes constant P, A, and E.
• Confusing axial deformation (δ) with strain (ε) — δ is in mm, ε is dimensionless.

Formulas

Exam Tips

• Thermal problems follow a clear pattern: (1) compute free δ_T, (2) compare to any gap, (3) if restrained, use σ_T = EαΔT or the gap equation.
• The keyword 'no gap' or 'snugly between rigid walls' immediately signals a FULLY RESTRAINED problem → σ_T = EαΔT.
• In a board exam, if they ask 'what temperature rise will just close the gap?', set δ_T = g → ΔT = g/(αL), no stress at that exact ΔT.
• For railway rails: NSCP 2015 and practical design require expansion joints spaced to limit thermal stress — understand the engineering rationale.
• Thermal and mechanical loads combine in indeterminate structures: the compatibility equation has both a δ_T term and a PL/AE term on opposite sides.

Key Points

• A temperature change ΔT causes thermal deformation: δ_T = αLΔT, where α is the coefficient of thermal expansion.
• FREE thermal expansion or contraction produces DEFORMATION but ZERO STRESS — no stress without restraint.
• If the member is FULLY RESTRAINED (rigid walls at both ends, no gaps), thermal stress develops: σ_T = EαΔT.
• Thermal stress is COMPRESSIVE when temperature RISES (member wants to expand but cannot) and TENSILE when temperature FALLS.
• For PARTIAL restraint (a gap g between the bar and the wall): if δ_T ≤ g → no stress; if δ_T > g → stress builds only after the gap is closed: αLΔT − σL/E = g, solve for σ.
• α for structural steel ≈ 11.7 × 10⁻⁶ / °C (sometimes given as 12 × 10⁻⁶ / °C in exam problems).
• α for concrete ≈ 10–12 × 10⁻⁶ / °C; for aluminum ≈ 23 × 10⁻⁶ / °C.
• Thermal problems in statically indeterminate structures combine thermal deformation with mechanical deformation in the compatibility equation.

Definitions

Section Title

Thermal Stress and Deformation

Common Mistakes

• Assuming thermal stress occurs simply because temperature changes — stress requires PHYSICAL RESTRAINT preventing free deformation.
• Using ΔT in Kelvin instead of °C — temperature DIFFERENCES are numerically identical in K and °C, so this is not actually an error, but students sometimes confuse absolute temperature with temperature change.
• Forgetting the sign of thermal stress: temperature RISE with full restraint → COMPRESSIVE stress (member pushes against the walls).
• In gap problems, forgetting to check if δ_T > g before computing stress — if the gap is larger than the free thermal expansion, no stress develops.
• Using the wrong α value — board exams may give α = 12 × 10⁻⁶/°C for steel; use the value given in the problem.

Formulas

Exam Tips

• Write the three steps explicitly on your scratch paper: (1) ΣF = 0, (2) compatibility δ₁ = δ₂ or Σδ = 0, (3) δ = PL/AE substitution — then solve simultaneously.
• For composite columns, the shortcut ratio P₁/P₂ = A₁E₁/A₂E₂ (same length, same ΔL) saves time on multiple-choice board exam problems.
• The phrase 'rigid bar supported by two wires' immediately signals indeterminate — use moment equilibrium at one support plus compatibility (geometric relationship between wire elongations from bar rotation).
• Thermal + mechanical indeterminate: the sum of free thermal elongation and mechanical elongation (from the reaction) must equal the constraint — set δ_T − PL/AE = 0 for a fully restrained bar that heats up.
• Double-check by verifying equilibrium with your computed forces — a quick sanity check catches arithmetic errors before moving on.

Key Points

• A problem is statically INDETERMINATE when the number of unknown forces EXCEEDS the number of independent equilibrium equations.
• Degree of indeterminacy = number of unknowns − number of equilibrium equations.
• The three-equation system for solving indeterminate axial problems: (1) Equilibrium (ΣF = 0), (2) Compatibility (geometric deformation condition), (3) Force–deformation (δ = PL/AE for each member).
• Typical indeterminate configurations: bar fixed at BOTH ends, composite columns (steel + concrete sharing load), rigid bar supported by multiple wires.
• KEY INSIGHT: In a parallel composite member (two materials sharing a common load in PARALLEL), BOTH members undergo EQUAL deformation (δ₁ = δ₂). They share the load in proportion to their axial rigidities (A₁E₁ and A₂E₂).
• The STIFFER member (higher AE) attracts proportionally MORE load — this is the fundamental result of indeterminate analysis.
• For a bar fixed at both ends with a load P at an intermediate point: the reactions R_A and R_B are found by ΣF = 0 plus the compatibility condition that the total bar deformation is zero (R_A × a = R_B × b, rearranged).
• Thermal effects in indeterminate members: the free thermal deformation is counteracted by an induced mechanical force — set up compatibility with both δ_T and PL/AE terms.

Definitions

Section Title

Statically Indeterminate Axial Members

Common Mistakes

• Using EQUAL STRESS (σ₁ = σ₂) instead of EQUAL DEFORMATION (δ₁ = δ₂) as the compatibility condition for parallel composite members — this is wrong unless A₁E₁ = A₂E₂.
• Forgetting to apply ΣF = 0 AND the compatibility equation simultaneously — either alone is insufficient.
• Getting the compatibility condition backwards for a doubly-fixed bar: the total deformation should be ZERO (bar returns to original position), not equal to PL/AE.
• Assuming the stiffer material carries less load — it is the OPPOSITE: higher AE → higher P for the same δ.
• In thermal indeterminate problems, forgetting to include BOTH the thermal deformation AND the mechanical deformation (reaction force) in the compatibility equation.

Formulas

Exam Tips

• In PRC board exams, biaxial Hooke's law appears most often in pressure vessel problems — learn to recognize 'hoop stress + longitudinal stress' as a biaxial state.
• For biaxial problems, always write BOTH strain equations; you may need either or both depending on what the question asks.
• The volumetric strain formula is a quick check: if ν = 1/3, then (1 − 2ν) = 1/3 and the relationship simplifies nicely.
• When given three strains and asked to find stresses (inverse problem), solve the three simultaneous equations for σ_x, σ_y, σ_z using Cramer's rule or substitution.
• Remember: for uniaxial tension, ε_y = ε_z = −νε_x — the lateral strains are EQUAL (for isotropic material) and opposite in sign.

Key Points

• When stresses act in more than one direction simultaneously, each strain has a DIRECT component (from its own stress) and a POISSON component (from the perpendicular stresses).
• For BIAXIAL stress (σ_z = 0, as in thin plates and pressure vessel walls): ε_x = (σ_x − νσ_y)/E and ε_y = (σ_y − νσ_x)/E.
• For TRIAXIAL stress (all three normal stresses present): each of the three strain equations has all three stresses.
• Volumetric strain (dilatation) ε_v = ε_x + ε_y + ε_z = [(1−2ν)/E](σ_x + σ_y + σ_z).
• For ν = 0.5 (rubber, incompressible): ε_v = 0 → no volume change, confirming incompressibility.
• The hydrostatic stress case (σ_x = σ_y = σ_z = p): ε_v = 3p(1−2ν)/E = p/K where K = E/[3(1−2ν)] is the bulk modulus.
• Generalized Hooke's law is the foundation for pressure vessel analysis (cylindrical, spherical), biaxial bending, and principal stress calculations.

Definitions

Section Title

Generalized Hooke's Law (Biaxial and Triaxial Stress)

Common Mistakes

• Forgetting to SUBTRACT the Poisson terms — many students write ε_x = σ_x/E + ν(σ_y/E) instead of ε_x = [σ_x − ν(σ_y + σ_z)]/E.
• Using plane stress equations (σ_z = 0) for problems that clearly have three non-zero stresses — always verify which stresses are zero.
• Confusing plane stress (σ_z = 0 but ε_z ≠ 0) with plane strain (ε_z = 0 but σ_z ≠ 0) — different assumptions lead to different equations.
• Computing ε_v as ε_x alone — volumetric strain requires summing all THREE normal strains.
• Applying biaxial formulas in problems that are actually uniaxial — check the loading condition before choosing which Hooke's law to use.