Simple Stresses and Strains - Cheat Sheet
Your last-minute reference for the foundational chapter of Strength of Materials. All formulas, definitions, and exam-critical facts condensed for 30-minute final review before the Civil Engineer Licensure Examination.
Normal (Axial) Stress
Formulas
σ = P / A
Meaning
σ = normal stress (Pa or MPa); P = axial force (N); A = cross-sectional area (mm²)
Watch out
Always use consistent units (N and mm² give MPa). Tensile stress is positive (+); compressive is negative (−). This is AVERAGE stress; St. Venant's principle says local stress concentrations die out ~one member-width from load point.
When to use
Force acts perpendicular to and through the centroid of the section.
Important facts
- Tensile stress (pull) is positive; compressive stress (push) is negative.
- Stress is a point property; it can vary across a section (except for centroidal axial load).
- Maximum stress often governs design, not average stress.
- For non-uniform sections, substitute the relevant area for that section into σ = P/A.
Key definitions
Normal Stress
Example
A steel rod pulled in tension: 60 kN ÷ 500 mm² = 120 MPa tensile stress.
Stress perpendicular to the resisting area, caused by axial tension or compression.
St. Venant's Principle
Example
Stress at a hole in a plate rises significantly, but 300 mm away the stress returns to P/A.
Local stress concentrations near load application points and holes vanish within ~one member-width; average stress formula applies at remote sections.
Centroidal Load
Example
Load applied at the geometric center of a circular cross-section.
Force passes through the centroid of the cross-section, producing uniform (average) stress without bending.
Diagrams to know
- Free-body diagram of a bar in tension with stress block.
Shear Stress
Formulas
τ = V / A (single shear)
Meaning
τ = shear stress (Pa or MPa); V = shear force (N); A = area resisting shear (mm²)
Watch out
Single shear: use the full area A. Do NOT divide by 2. One cut plane = one area.
When to use
Force is parallel (tangent) to the area; the connector is cut on ONE plane only.
τ = P / (2A) (double shear)
Meaning
τ = shear stress (MPa); P = applied force (N); A = cross-sectional area of connector (mm²)
Watch out
Must clearly identify double shear before applying the factor of 2 in the denominator. Missing this doubles your stress computation incorrectly.
When to use
The connector is cut on TWO planes (e.g., bolt in a lap joint with two load paths).
Important facts
- Shear stress acts PARALLEL to the area, NOT perpendicular.
- In riveted and bolted connections, shear is the primary failure mode.
- Double shear reduces the shear stress by half compared to single shear for the same force.
- For keys and pins: τ_avg = P/(A_key) or τ_avg = P/(2A) if two shear planes.
Key definitions
Shear Stress
Example
A bolt connecting two plates in a lap joint carries shear stress where the plates overlap.
Stress parallel (tangential) to the resisting area, caused by forces that try to cut through a member.
Single Shear
Example
A 20 mm bolt between two plates: A = π(20)²/4 = 314 mm²; τ = P/314.
Connector (bolt, rivet, pin) is cut on ONE plane; shear area = cross-sectional area of connector.
Double Shear
Example
A 20 mm bolt in a three-plate assembly (outer plates and a central gusset): τ = P/(2 × 314) = P/628.
Connector is cut on TWO parallel planes (two load paths); effective shear area is doubled.
Diagrams to know
- Single-shear lap joint diagram (one cut plane).
- Double-shear lap joint diagram (two cut planes).
Bearing Stress
Formulas
σ_b = P / A_b = P / (d × t)
Meaning
σ_b = bearing stress (MPa); P = force (N); d = diameter of bolt/pin (mm); t = thickness of plate (mm); A_b = projected bearing area (mm²)
Watch out
Bearing area is the PROJECTED area d×t (rectangle), NOT the circular cross-section π d²/4. This is the contact pressure on the hole wall.
When to use
Bolt or pin presses against the side of a hole in a plate; governs hole crushing and tear-out.
Important facts
- Bearing is a contact/crushing stress — the bolt pushes sideways on the hole wall.
- Bearing area is ALWAYS d × t (diameter × thickness), regardless of bolt shape or material.
- Allowable bearing stress (e.g., per AISC 360) is typically higher than allowable shear stress.
- Hole tear-out (edge failure) occurs when the edge distance is too small; bearing check alone is not sufficient.
- In composite bolted joints, check ALL three: shear in bolt, bearing in plate, and edge tear-out.
Key definitions
Bearing Stress
Example
20 mm bolt on 12 mm thick plate: σ_b = P/(20 × 12) = P/240 mm²; if P = 60 kN, σ_b = 250 MPa.
Contact pressure between a bolt (or pin) and the surface of a hole; governs hole deformation and tear-out.
Projected Bearing Area
Example
Bolt d = 25 mm; plate t = 10 mm → A_b = 250 mm² (not the bolt's 491 mm² cross-section).
The rectangular shadow (d × t) cast by a bolt on a plate surface; represents the contact zone between bolt and hole wall.
Diagrams to know
- Bolt-hole assembly showing bearing area (d × t rectangle on hole wall).
Strain and Hooke's Law
Formulas
ε = δ / L
Meaning
ε = normal strain (dimensionless); δ = axial deformation (elongation or shortening) (mm); L = original length (mm)
Watch out
Strain is dimensionless (a ratio). δ and L must have the same units. Positive ε = elongation; negative ε = contraction.
When to use
Quantify the fractional change in length due to axial loading.
σ = E × ε (Hooke's Law)
Meaning
σ = normal stress (MPa); E = modulus of elasticity (Young's modulus) (GPa or MPa); ε = normal strain (dimensionless)
Watch out
E varies widely (steel ~200 GPa, aluminum ~70 GPa, concrete ~20–30 GPa). Outside the proportional limit, Hooke's law does NOT apply. Use consistent units: if E is in GPa, convert stress to GPa or E to MPa.
When to use
Within the linear (proportional) limit of the stress–strain curve; relates stress and strain elastically.
τ = G × γ (Shear Hooke's Law)
Meaning
τ = shear stress (MPa); G = shear modulus (modulus of rigidity) (GPa); γ = shear strain (radians, dimensionless)
Watch out
γ is in RADIANS (not degrees). For small angles, γ ≈ tan(γ). G ≈ E/2.6 for typical steel (G ≈ 80 GPa).
When to use
Linear elastic shear; relates shear stress and shear angle change.
Common values
Value
200 GPa = 200,000 MPa
Symbol
E_s
Quantity
Modulus of Elasticity (Structural Steel)
Value
70 GPa
Symbol
E_al
Quantity
Modulus of Elasticity (Aluminum)
Value
20–30 GPa (varies with strength)
Symbol
E_c
Quantity
Modulus of Elasticity (Concrete)
Value
80 GPa
Symbol
G
Quantity
Shear Modulus (Structural Steel)
Value
0.27–0.30
Symbol
ν
Quantity
Poisson's Ratio (Steel)
Value
250 MPa
Symbol
F_y
Quantity
Yield Strength (A36 Steel)
Value
400–450 MPa
Symbol
F_u
Quantity
Ultimate Strength (A36 Steel)
Important facts
- Hooke's Law (σ = E·ε) applies ONLY within the proportional limit; do NOT use it beyond that point.
- The linear region (Hooke's law region) is where elastic reversible deformation occurs.
- Yield point marks the start of plastic (permanent) deformation in ductile materials.
- Ultimate strength is NOT the same as yield strength; ultimate is higher and marks the maximum load.
- Shear modulus G is related to E by G = E / [2(1 + ν)], where ν is Poisson's ratio.
- For steel, G ≈ 80 GPa (E = 200 GPa, ν ≈ 0.27).
Key definitions
Strain
Example
A 1000 mm rod elongates 2 mm → ε = 2/1000 = 0.002 = 0.2%.
Fractional deformation per unit length; dimensionless ratio of displacement to original length.
Modulus of Elasticity (Young's Modulus)
Example
Structural steel: E = 200 GPa; aluminum alloy: E ≈ 70 GPa; concrete: E ≈ 20–30 GPa.
Material stiffness constant; slope of the linear (elastic) region of the stress–strain curve.
Proportional Limit
Example
For mild steel, ≈ 200–250 MPa; beyond this, the curve deviates from linearity.
Highest stress at which stress and strain are directly proportional (σ ∝ ε); end of the linear region on the stress–strain diagram.
Elastic Limit
Example
For mild steel, very close to the proportional limit; approximately 200–240 MPa.
Highest stress with no permanent set (plastic deformation) upon unloading; material returns fully to original shape.
Yield Point (Yield Stress)
Example
Mild steel (Grade 40 or A36): F_y ≈ 250 MPa; Grade 60 (A572): F_y ≈ 345 MPa.
Stress at which a ductile material yields; strain increases significantly with little or no increase in stress (the 'plateau' on the curve).
Ultimate Strength
Example
Mild steel: F_u ≈ 400–450 MPa (higher than yield).
Maximum stress the material can sustain; peak of the stress–strain curve.
Rupture Strength (Fracture Stress)
Example
For ductile mild steel, lower than ultimate strength due to necking; for brittle materials, fracture and ultimate coincide.
Stress at which the material fractures and fails completely.
Shear Strain
Example
In torsion or direct shear: γ = change in angle (rad). For small deformations, γ ≈ displacement/height.
Change in angle (in radians) of an originally right angle within the material; measured for shear loading.
Diagrams to know
- Stress–strain diagram for mild steel (showing proportional limit, yield, ultimate, rupture).
- Linear region of stress–strain curve highlighting Hooke's law slope.
Axial Deformation
Formulas
δ = (P × L) / (A × E)
Meaning
δ = total axial deformation (elongation or shortening) (mm); P = axial force (N); L = length of member (mm); A = cross-sectional area (mm²); E = modulus of elasticity (MPa)
Watch out
CRITICAL FORMULA — most-tested in the chapter. Units must be consistent: N and mm² → stress in MPa, and E in MPa. Product AE is axial rigidity (stiffness). If P, A, or E changes along length, sum contributions δ = Σ(P_i × L_i)/(A_i × E_i).
When to use
Calculate elongation or shortening of a member under constant axial load, constant cross-section, and uniform material.
δ = ∫₀ᴸ [P(x) / (A(x)·E)] dx
Meaning
δ = total deformation with continuous variation in P(x) or A(x) along length.
Watch out
Most common case: bar hanging under its own weight. Set up limits and integrate; final answer is always positive (elongation).
When to use
Member carries distributed load (own weight) or has continuously varying cross-section (tapered bar).
δ = Σ (P_i × L_i) / (A_i × E_i)
Meaning
δ = total deformation summed over i segments; each segment has constant P_i, L_i, A_i, E_i.
Watch out
Each segment must be constant within itself. If the load changes mid-section, split that section further. Keep careful track of signs: elongation is positive, shortening is negative.
When to use
Multi-segment bar: different loads, areas, materials, or E values in different sections. Break into segments and sum.
Important facts
- δ is directly proportional to load P and length L.
- δ is inversely proportional to area A and modulus E.
- For the same force, a longer bar deforms more; a larger area deforms less.
- Doubling the modulus (e.g., using a stiffer material) halves deformation — this is why steel (E=200) is preferred over aluminum (E=70).
- In composite members (steel + concrete) sharing a load, use δ_s = δ_c (compatibility) to find load distribution.
Key definitions
Axial Deformation (Axial Strain / Deformation)
Example
A 500 mm steel rod (A = 400 mm², E = 200 GPa) carries 50 kN → δ = (50,000 × 500)/(400 × 200,000) = 0.3125 mm elongation.
The total change in length (δ) of a member due to applied axial force, expressed as δ = PL/(AE).
Axial Rigidity
Example
Steel: A = 500 mm², E = 200 GPa → AE = 100,000 GPa·mm². Aluminum: A = 800 mm², E = 70 GPa → AE = 56,000 GPa·mm². Steel is stiffer.
The product A·E, representing the stiffness of a member to axial deformation; higher rigidity means less deformation for the same load.
Diagrams to know
- Multi-segment bar (three sections with different P, A, E) showing how to compute total δ as sum of three terms.
Poisson's Ratio and Related Elastic Constants
Formulas
ν = −ε_lateral / ε_axial
Meaning
ν = Poisson's ratio (dimensionless); ε_lateral = lateral (transverse) strain; ε_axial = axial (longitudinal) strain
Watch out
Negative sign: when axial strain is positive (elongation), lateral strain is negative (contraction). In formulas, use ν as positive. For incompressible materials (rubber), ν = 0.5.
When to use
Quantify the ratio of lateral contraction to axial elongation; typical values: steel ≈ 0.27–0.30, concrete ≈ 0.15–0.20.
G = E / [2(1 + ν)]
Meaning
G = shear modulus (GPa); E = modulus of elasticity (GPa); ν = Poisson's ratio
Watch out
If ν and E are given, compute G. For steel (E=200, ν=0.3): G = 200/[2(1.3)] ≈ 77 GPa (close to 80). Keep units consistent.
When to use
Relate shear modulus to Young's modulus and Poisson's ratio; essential for torsion and combined-stress chapters.
K = E / [3(1 − 2ν)]
Meaning
K = bulk modulus (GPa); relates hydrostatic pressure to volumetric strain; also called incompressibility modulus.
Watch out
As ν → 0.5, K → ∞ (incompressible). K is always positive; 1 − 2ν > 0 requires ν < 0.5.
When to use
Problems involving hydrostatic pressure or volumetric change; less common in basic Strength of Materials but fundamental in mechanics of materials.
Common values
Value
0.27–0.30
Symbol
ν_s
Quantity
Poisson's Ratio (Steel)
Value
0.33
Symbol
ν_al
Quantity
Poisson's Ratio (Aluminum)
Value
0.15–0.20
Symbol
ν_c
Quantity
Poisson's Ratio (Concrete)
Value
~160 GPa
Symbol
K_s
Quantity
Bulk Modulus (Steel)
Important facts
- Poisson's ratio is always between 0 and 0.5 for stable materials (except exotic composites).
- Most engineering materials: 0.25 ≤ ν ≤ 0.35 (e.g., steel ≈ 0.28, aluminum ≈ 0.33, concrete ≈ 0.15–0.20).
- The three elastic constants (E, G, ν) are not independent; knowing two determines the third.
- G is always less than E; for steel, G ≈ E/2.5.
- Incompressibility (ν = 0.5) implies K → ∞; rubber and some elastomers approach this limit.
Key definitions
Poisson's Ratio
Example
Steel ν = 0.28: a 1 m long rod stretched 10 mm contracts laterally by 10 × 0.28 = 2.8 mm (proportional to the change in cross-section).
The negative ratio of lateral (transverse) strain to axial strain; a material constant that quantifies the lateral contraction accompanying axial elongation.
Shear Modulus (Modulus of Rigidity)
Example
Structural steel: G ≈ 80 GPa; much lower than E because the material shears more easily than it stretches.
Material stiffness in shear; slope of the linear region of the shear stress–shear strain curve (τ = G·γ).
Bulk Modulus (Incompressibility Modulus)
Example
Water: K ≈ 2.2 GPa (highly compressible); steel: K ≈ 160 GPa (nearly incompressible).
Resistance to volumetric compression under hydrostatic pressure; ratio of pressure to fractional volume change.
Diagrams to know
- Cube element under axial load showing axial elongation and lateral contraction (Poisson effect).
Reactions or equations
G = E / [2(1 + ν)]
Note
Relates all three elastic constants; fundamental identity in mechanics of materials.
Conditions
Linear elastic, isotropic material under small deformations.
Generalized Hooke's Law (Biaxial / Triaxial Stress)
Formulas
ε_x = (1/E)[σ_x − ν(σ_y + σ_z)]
Meaning
ε_x = strain in x-direction; σ_x, σ_y, σ_z = normal stresses in three orthogonal directions; ν = Poisson's ratio.
Watch out
Each strain is affected by ALL three stresses: direct stress contribution (σ_x/E) MINUS Poisson contraction from the other two [−ν(σ_y + σ_z)/E]. Similar equations for ε_y and ε_z (cyclic permutation).
When to use
Multi-axial stress state; compute the strain in one direction accounting for the Poisson effect from perpendicular stresses.
ε_y = (1/E)[σ_y − ν(σ_z + σ_x)]
Meaning
Strain in y-direction due to triaxial stress; same form as ε_x with cyclic index permutation.
Watch out
The formula is cyclic: x→y→z→x. Do NOT confuse the order of the Poisson terms.
When to use
y-component of strain in a triaxial state.
ε_z = (1/E)[σ_z − ν(σ_x + σ_y)]
Meaning
Strain in z-direction due to triaxial stress.
Watch out
Complete the cyclic pattern; all three stresses contribute to each strain.
When to use
z-component of strain in a triaxial state.
ε_v = ε_x + ε_y + ε_z = [(1 − 2ν)/E](σ_x + σ_y + σ_z)
Meaning
ε_v = volumetric strain (dilatation); sum of all three normal strains.
Watch out
Volumetric strain depends on the SUM of all three stresses. As ν → 0.5, the coefficient (1−2ν) → 0, meaning the material becomes incompressible. If all three stresses are equal (hydrostatic pressure p), then ε_v = 3(1−2ν)p/E.
When to use
Find the total fractional volume change under multi-axial loading.
Important facts
- The generalized Hooke's law is linear: each strain is a linear combination of all three stresses.
- For uniaxial stress (σ_x ≠ 0, σ_y = σ_z = 0): ε_x = σ_x/E, ε_y = ε_z = −ν·σ_x/E (the familiar result).
- Volumetric strain ε_v is linked to the bulk modulus K by: ε_v = (σ_x + σ_y + σ_z) / (3K).
- For incompressible materials (ν = 0.5), ε_v = 0 even under non-zero stresses → volume is constant.
- The generalized law is the foundation for failure theories (von Mises, Tresca) in later chapters.
Key definitions
Triaxial Stress State
Example
A corner of a concrete dam experiences water pressure on two faces and internal stress from weight; σ_x ≠ σ_y ≠ σ_z.
A point carrying three independent normal stresses σ_x, σ_y, σ_z in three orthogonal directions; each stress causes direct strain in its direction and Poisson contraction in the other two.
Biaxial Stress State
Example
Thin-walled pressure vessel wall: hoop stress σ_θ and axial stress σ_a; plane stress in the wall surface.
A point under stress in two directions (e.g., σ_x and σ_y); the third stress is zero (σ_z = 0), as in plane stress.
Volumetric Strain (Dilatation)
Example
Under hydrostatic compression (all three stresses equal), ε_v < 0 (volume decreases).
The fractional change in volume: ε_v = ΔV/V₀ = ε_x + ε_y + ε_z; governs whether the material expands or contracts.
Poisson Contraction / Poisson Effect
Example
A rod stretched axially (σ_x > 0, σ_y = σ_z = 0) shrinks laterally by amount ν·ε_x.
The lateral strain (contraction or expansion) induced by an axial strain via the negative Poisson ratio term.
Diagrams to know
- Cube element under triaxial stress (σ_x, σ_y, σ_z) showing direct strains and lateral contractions.
Thermal Stress and Deformation
Formulas
δ_T = α × L × ΔT
Meaning
δ_T = free thermal deformation (elongation if ΔT > 0) (mm); α = coefficient of thermal expansion (/°C); L = original length (mm); ΔT = temperature change (°C)
Watch out
FREE expansion means ZERO stress; thermal deformation ≠ thermal stress. If δ_T is not restrained, the member simply gets longer (or shorter) with no stress. Only RESTRAINT creates thermal stress.
When to use
Calculate the change in length when a member is FREE to expand or contract (no restraint).
σ_T = E × α × ΔT
Meaning
σ_T = thermal stress in a fully restrained bar (MPa); E = modulus of elasticity (MPa); α = coefficient of thermal expansion (/°C); ΔT = temperature change (°C)
Watch out
CRITICAL: Restraint is required for thermal stress. If the bar is free, σ_T = 0. For full restraint: the bar wants to expand by δ_T = α·L·ΔT but cannot, so a compressive stress builds up to prevent the expansion. Sign: ΔT > 0 (heating) → compressive stress (negative) in a fully restrained bar.
When to use
Find the stress when a member is FIXED at both ends (or ends are immovable) and the temperature changes.
α·L·ΔT − (σ·L)/E = gap (partial restraint)
Meaning
For partial restraint (e.g., a clearance g that must be closed): thermal expansion minus elastic compression equals the allowed movement.
Watch out
Set up the compatibility equation: free thermal deformation minus elastic contraction from thermal stress equals the closure required (g). Solve for σ.
When to use
Bar has some freedom to move (initial gap g) before restraint takes effect.
Common values
Value
11.7 × 10⁻⁶ /°C
Symbol
α_s
Quantity
Coefficient of Thermal Expansion (Structural Steel)
Value
23.6 × 10⁻⁶ /°C
Symbol
α_al
Quantity
Coefficient of Thermal Expansion (Aluminum)
Value
10–14 × 10⁻⁶ /°C
Symbol
α_c
Quantity
Coefficient of Thermal Expansion (Concrete)
Important facts
- FREE thermal expansion produces ZERO stress. Only restraint (fixed ends, wall confinement) creates thermal stress.
- Heating (ΔT > 0) causes expansion; if restrained, this produces COMPRESSIVE stress.
- Cooling (ΔT < 0) causes contraction; if restrained, this produces TENSILE stress.
- Thermal stress is independent of the rate of heating; it depends only on ΔT and the material's E and α.
- Composite materials with different α values (e.g., concrete reinforced with steel) develop internal stresses due to temperature change.
- In railroad design, gaps (rails cooled, then heated) prevent excessive thermal stress (NSCP 2015, design code references).
Key definitions
Thermal Deformation (Free Thermal Expansion / Contraction)
Example
Steel rail 12 m long, α = 11.7 × 10⁻⁶/°C, heated from 20°C to 50°C (ΔT = 30°C): δ_T = 11.7e−6 × 12,000 × 30 = 4.21 mm free expansion (no stress).
The change in length due to temperature change when the member is unconstrained; given by δ_T = α·L·ΔT.
Thermal Stress
Example
Same rail FIXED at both ends: σ_T = 200,000 × 11.7 × 10⁻⁶ × 30 = 70.2 MPa compressive stress (the rail is forced to stay the same length).
Stress developed in a member due to temperature change when the member is restrained from expanding or contracting freely.
Coefficient of Thermal Expansion
Example
Structural steel: α ≈ 11.7 × 10⁻⁶/°C; aluminum: α ≈ 23.6 × 10⁻⁶/°C (nearly double); concrete: α ≈ 10–14 × 10⁻⁶/°C.
Material constant (α) representing fractional change in length per unit temperature change (/°C or /K).
Diagrams to know
- Bar fixed at both ends: free thermal expansion blocked → compressive stress (shows force equilibrium).
- Bar with partial restraint: gap g closes before support takes effect.
Reactions or equations
Total deformation = δ_mechanical + δ_thermal
Note
Superpose the effects: δ_total = (P·L)/(A·E) ± α·L·ΔT, with sign of thermal term depending on heating (positive, expansion) or cooling (negative, contraction).
Conditions
When both a mechanical load and temperature change act on a member.
Statically Indeterminate Axial Members
Formulas
Equilibrium: ΣF = 0 (sum of internal forces equals applied load)
Meaning
Free-body diagram force balance; gives one equation but multiple unknowns → need additional compatibility.
Watch out
Equilibrium alone is insufficient to solve an indeterminate problem. You will have MORE unknowns than equations from statics alone; indeterminate degree = (number of unknowns − number of equilibrium equations).
When to use
Always the first step; write the sum of forces (internal member forces must balance external applied load).
Compatibility: δ_1 = δ_2 = ... = δ_n (or sum = 0 for fixed-end bars)
Meaning
Deformation constraint; members in series shorten equally; bar fixed at both ends has zero net deformation.
Watch out
Identify the constraint: if two materials share a bar, they must shorten the same amount (δ_1 = δ_2). If a bar is fixed at both ends, the total deformation is zero (δ_total = 0). This is the KEY equation in indeterminate problems.
When to use
Second condition; relate deformations of different segments (or materials) based on physical constraints.
Force–deformation: δ_i = (P_i · L_i) / (A_i · E_i) (for each segment or material)
Meaning
Substitute the deformation formula into the compatibility equation.
Watch out
Keep track of which segments have which properties. In a composite column (steel core + concrete shell), each material has its own A and E.
When to use
Third step; express deformations in terms of unknown forces (P_1, P_2, etc.) and known geometry/material properties.
Important facts
- Method: (1) Equilibrium (ΣF = 0), (2) Compatibility (δ constraints), (3) Force–deformation (δ = PL/AE), then solve simultaneously.
- In composite members, the stiffer material (higher E) attracts proportionally more load.
- Load distribution in composite members: P_i ∝ A_i · E_i (axial rigidity). Stiffer material takes more of the load.
- For a bar fixed at both ends and heated: thermal stress = E·α·ΔT (fully restrained). If the ends can move slightly, the stress is reduced.
- Two-material column under central load: use equilibrium P = P_1 + P_2 and compatibility δ_1 = δ_2 to find P_1 and P_2; stress in each is σ_i = P_i / A_i.
Key definitions
Statically Indeterminate Member
Example
Bar fixed at both ends and pulled at the center; three unknown reactions but only one equilibrium equation (ΣF = 0).
Structure or member with more unknowns than equilibrium equations; requires compatibility (deformation) conditions to solve.
Degree of Indeterminacy
Example
Bar fixed at both ends: 2 end reactions, 1 applied load → 2 unknowns from statics → D = 2 − 1 = 1 (once indeterminate).
Number of excess unknowns: D = (number of unknowns) − (number of independent equilibrium equations). Also called 'redundancy.'
Compatibility Condition / Kinematic Condition
Example
Composite column: steel and concrete must shorten together → δ_steel = δ_concrete → (P_s·L)/(A_s·E_s) = (P_c·L)/(A_c·E_c).
Equation relating deformations based on geometric constraints; essential for solving indeterminate problems.
Composite Member / Composite Section
Example
Concrete column reinforced with steel rods: steel carries more load per unit area because E_steel >> E_concrete.
Bar made of two or more different materials bonded together (e.g., steel rebar in concrete); forces distribute based on stiffness (A·E).
Diagrams to know
- Bar fixed at both ends with central load: FBD showing three unknowns (R_1, R_2, P).
- Composite column (steel + concrete) under axial load: separate free bodies showing load distribution.
- Two-material composite: stress blocks (σ_1 = P_1/A_1, σ_2 = P_2/A_2) with equal deformations.
Reactions or equations
P_1 + P_2 = P_applied (equilibrium)
Note
Forces in the two materials sum to the total applied load.
Conditions
Two materials (or segments) in series carrying the same applied load.
δ_1 = δ_2 (compatibility)
Note
They must shorten (or lengthen) by the same amount: (P_1·L)/(A_1·E_1) = (P_2·L)/(A_2·E_2).
Conditions
Two materials bonded together, deforming jointly.
P_1 / (A_1·E_1) = P_2 / (A_2·E_2) (combined for equal deformation)
Note
Immediately tells load ratio: P_1/P_2 = (A_1·E_1)/(A_2·E_2) → load shares with axial rigidity.
Conditions
Simplification of compatibility when lengths are equal.
Allowable (Working) Stress and Safety
Formulas
σ_allow = σ_y / F.S. (or σ_u / F.S.)
Meaning
σ_allow = allowable (working) stress (MPa); σ_y = yield stress (MPa); σ_u = ultimate stress (MPa); F.S. = factor of safety (dimensionless)
Watch out
Some codes use yield-based (safety against plastic deformation); others use ultimate-based (safety against fracture). For ductile steel in tension, typical F.S. = 1.5–2.0 on yield; for brittle materials or compression, may be 2.0–3.0 on ultimate. Always check the applicable code (NSCP 2015, AISC 360).
When to use
Design check: computed stress must not exceed allowable stress. Choose F.S. based on code (NSCP 2015, AISC 360).
Common values
Value
1.5–2.0
Symbol
F.S.
Quantity
Typical Factor of Safety (Ductile Steel, ASD)
Value
2.0–3.0
Symbol
F.S.
Quantity
Typical Factor of Safety (Brittle Material / Compression)
Value
0.6 × F_y ≈ 150 MPa
Symbol
σ_allow
Quantity
Allowable Tensile Stress (A36 Steel, per AISC ASD)
Important facts
- Allowable stress design (ASD) is the traditional method; compare computed stress to σ_allow and require computed σ ≤ σ_allow.
- Modern codes (AISC 360, NSCP 2015) often use Load and Resistance Factor Design (LRFD), which applies factors to loads and resistances separately; same principle, different implementation.
- For ductile materials (mild steel), failure is often taken as yield (inelastic deformation); F.S. on yield is typical.
- For brittle materials (cast iron, concrete), failure is fracture; F.S. on ultimate is used.
- The choice of F.S. depends on confidence in loads, material properties, and analysis method. Typical ranges: 1.5–2.0 (ductile, well-known loads) to 3.0–4.0 (brittle, uncertain loads).
Key definitions
Allowable Stress (Working Stress)
Example
Structural steel A36: σ_y = 250 MPa, F.S. = 1.67 → σ_allow = 250/1.67 ≈ 150 MPa (per AISC).
The maximum average stress permitted in a member under service loads, computed as yield (or ultimate) stress divided by a safety factor.
Factor of Safety (Safety Factor)
Example
F.S. = 2 means the allowable stress is half the failure stress; a 50% overload before failure (in a simple analysis).
Dimensionless ratio (>1) of the failure stress to the allowable stress; accounts for uncertainties in loads, materials, and analysis.
Must remember
- δ = PL/(AE) is the WORKHORSE formula. Master it: elongation is proportional to load and length, inversely proportional to area and stiffness. Use it in nearly every problem.
- Distinguish single shear (τ = P/A) from double shear (τ = P/2A). A single careless error here costs points on bolted/riveted connection problems.
- Bearing stress is σ_b = P/(d·t), NOT the bolt's circular cross-section area. The projected area d×t (diameter × thickness) is the contact rectangle on the hole wall.
- Free thermal expansion δ_T = α·L·ΔT produces ZERO stress. Only RESTRAINT creates thermal stress σ_T = E·α·ΔT. This is a classic confusion point.
- Hooke's Law (σ = E·ε) applies ONLY within the proportional (elastic) region. Beyond yield, the material is plastic and the curve is non-linear. Do NOT use Hooke's law for post-yield behavior.
- In indeterminate problems, use three steps: (1) Equilibrium ΣF = 0, (2) Compatibility (equal deformations or zero net deformation), (3) Force–deformation δ = PL/(AE). Solve simultaneously.
- In composite members (steel + concrete, or two different materials), load shares with axial rigidity AE. Stiffer material takes proportionally more load: P ∝ A·E for equal strain.
- Poisson's ratio ν is the NEGATIVE ratio of lateral strain to axial strain. For most metals, 0.25 ≤ ν ≤ 0.35. Higher ν (approaching 0.5) means more lateral contraction for the same axial strain.
- Generalized Hooke's law in 3D: each strain depends on ALL three stresses via ε_x = (1/E)[σ_x − ν(σ_y + σ_z)], etc. This is the foundation for combined-stress problems.
- Always check units. N and mm² give stress in MPa; E in MPa or GPa (convert consistently). A common error: mixing mm and m, or forgetting to convert GPa to MPa.
Last minute tips
- In bolted/riveted connections, ALWAYS check three items: (1) shear in the bolt/rivet, (2) bearing in the plate, and (3) net-area tension (if there's a hole row). Missing one is a common exam mistake.
- For thermal problems, ask yourself: 'Is the member free to move?' If YES → zero stress, only deformation. If NO (restrained) → stress develops. Use equilibrium + compatibility if partially restrained (gap problem).
- When you see 'composite column' or 'two materials in series,' immediately set up: equilibrium P = P_1 + P_2, compatibility δ_1 = δ_2, then substitute δ = PL/(AE) for each. Solve the system—this is a standard indeterminate pattern.
- Remember the relationship G = E/[2(1+ν)]: if you're given two elastic constants, you can find the third. For steel, E ≈ 200 GPa and ν ≈ 0.28 → G ≈ 80 GPa. Memorize these typical values.
- On the exam, when computing deformation δ = PL/(AE), pause and double-check: Is P constant over the entire length? If load changes mid-bar, break it into segments and sum. If bar hangs under its own weight, integrate. This step alone prevents half the errors.
Comparison tables
Single Shear vs. Double Shear
| Feature | Single Shear | Double Shear |
|---|---|---|
| Number of Cut Planes | One | Two |
| Shear Area (A_shear) | Cross-sectional area of connector (A) | Twice the cross-sectional area (2A) |
| Shear Stress Formula | τ = P / A | τ = P / (2A) |
| Stress for Same Force | Higher (all force on one plane) | Lower (force split over two planes) |
| Common Application | Rivets between two plates (lap joint) | Bolt between three plates (central gusset or hanger) |
| Maximum Load Capacity | Lower (limited by A) | Higher (load shared, same material) |
Normal Stress vs. Shear Stress vs. Bearing Stress
| Stress Type | Direction Relative to Area | Formula | Common Failure Mode |
|---|---|---|---|
| Normal (Axial) | Perpendicular (⊥) to resisting area | σ = P / A | Tensile rupture or compressive crushing |
| Shear | Parallel (∥) to resisting area | τ = V / A (single) or V / (2A) (double) | Shear failure / cut-through (rivets, bolts, keys) |
| Bearing | Perpendicular, contact pressure on hole wall | σ_b = P / (d × t) | Hole crushing / tear-out |
Free Thermal Expansion vs. Restrained Thermal Stress
| Condition | Deformation (δ_T) | Thermal Stress (σ_T) | Outcome |
|---|---|---|---|
| Free (Unrestrained) | δ_T = α·L·ΔT (present) | σ_T = 0 (no stress) | Member expands/contracts freely; no internal stress |
| Fully Restrained (Fixed Ends) | δ_T = 0 (expansion blocked) | σ_T = E·α·ΔT (develops) | Thermal expansion prevented; internal stress (compression if heated, tension if cooled) |
| Partially Restrained (Gap g) | δ_T reduced by gap closure | σ_T develops only after gap closes | Member moves until gap is consumed, then stress builds |
Elastic vs. Plastic Deformation (Stress–Strain Diagram Regions)
| Region | Stress Level | Strain Behavior | Permanent Set? | Hooke's Law Valid? |
|---|---|---|---|---|
| Linear (Elastic) Region | Below proportional limit (~σ_y for ductile steel) | Directly proportional to stress (ε ∝ σ) | No (recovers fully on unload) | Yes (σ = E·ε applies) |
| Yield Plateau (Ductile Only) | At yield point (σ_y); small stress increase → large strain | Strain increases rapidly with little stress change | Yes (permanent deformation) | No (non-linear) |
| Strain-Hardening (Plastic) Region | Beyond yield, up to ultimate (σ_u) | Large inelastic strain; curve rises | Yes (significant plastic deformation) | No (non-linear and non-proportional) |
| Fracture / Rupture | At rupture stress (lower than ultimate for ductile materials due to necking) | Material fails; no more stress increase | Yes (material breaks apart) | N/A (material fractured) |
Material Elastic Constants (Typical Values for Common Materials)
| Material | E (GPa) | G (GPa) | ν (Poisson's Ratio) | α (×10⁻⁶ /°C) |
|---|---|---|---|---|
| Structural Steel (A36, A572) | 200 | 80 | 0.27–0.30 | 11.7 |
| Aluminum Alloy | 70 | 26 | 0.33 | 23.6 |
| Concrete | 20–30 | 8–12 | 0.15–0.20 | 10–14 |
| Cast Iron | 100–150 | 40–60 | 0.25 | 10–12 |
| Brass | 100–110 | 37–41 | 0.34 | 18–19 |
Diagrams
Simple Stresses and Strains — Concept Map
Hierarchical concept map showing how normal stress, shear stress, and bearing stress relate to strain, deformation, and material properties. Emphasizes the core formula δ = PL/(AE) and its application in indeterminate problems.
Solving an Indeterminate Axial Member Problem
Step-by-step flowchart for solving indeterminate axial problems. Emphasizes the three-step method: equilibrium, compatibility, force–deformation, and simultaneous solution. Includes decision points for common constraint types.
Stress vs. Strain: Which Formula to Use?
Decision tree guiding students to the correct formula based on what is asked: normal stress (σ = P/A), shear stress (single or double), bearing stress, or deformation. Color-coded by category (stress vs. strain).
Stress–Strain Diagram: Material State Transitions
State diagram showing the progression of a ductile material (e.g., mild steel) through the stress–strain curve: elastic region, yield point, strain-hardening (plastic flow), ultimate strength peak, necking, and fracture. Each state annotates key material behavior.