Young's Modulus of Elasticity Calculator: Steel • Aluminum • Fiberglass

$$\sigma_{\text{true}} = \sigma_{\text{engineering}} (1 + \varepsilon_{\text{engineering}})$$

$$\varepsilon_{\text{true}} = \ln(1 + \varepsilon_{\text{engineering}})$$

$$I_x = \frac{b h^3}{12}$$

$$S_x = \frac{I_x}{c} = \frac{b h^2}{6}$$

where c = h/2 (distance from neutral axis to extreme fiber)

$$I_x = 2 \left[ \frac{b_f t_f^3}{12} + (b_f \cdot t_f) \cdot y^2 \right] + \frac{t_w h_{web}^3}{12}$$

where:

• $y = \frac{d}{2} - \frac{t_f}{2}$ (distance from centroid to flange centroid)
• $h_{web} = d - 2t_f$ (web height)

$$E = \frac{\sigma}{\varepsilon}$$

where:

• E = Young's Modulus (units: Pa, MPa, GPa, psi, ksi)
• σ (sigma) = Stress (force per unit area)
• ε (epsilon) = Strain (dimensionless, change in length / original length)

$$\sigma = \frac{F}{A}$$

where:

• F = Applied axial force (N, kN, lbf)
• A = Cross-sectional area (m², mm², in²)

$$\varepsilon = \frac{\Delta L}{L_0}$$

where:

• ΔL = Change in length (elongation or compression)
• L₀ = Original length (gauge length)

$$E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L / L_0} = \frac{F \cdot L_0}{A \cdot \Delta L}$$

This is the most practical form for experimental tensile testing.

$$k = \frac{EA}{L_0}$$

where:

• k = Axial stiffness (N/m) - represents the "spring constant"
• EA = Axial rigidity (product of elastic modulus and area)

Relationship to force-deflection:

$$F = k \cdot \Delta L$$

This is Hooke's Law for axial members, useful for vibration and dynamic analysis.

$$\delta_{max} = \frac{P L^3}{48 E I}$$

where:

• δ_max = Maximum deflection at midspan
• P = Point load applied at center
• L = Span length (distance between supports)
• I = Moment of inertia (second moment of area)

$$\delta_{max} = \frac{P L^3}{3 E I}$$

Note: Cantilever deflection is 16× larger than simply supported for same P, L, E, I

General Definition:

$$I_x = \int y^2 \, dA$$

Or for discrete areas:

$$I_x = \sum (I_{x,i} + A_i y_i^2)$$

This is the parallel axis theorem, where y_i is distance from component centroid to overall neutral axis.

$$S_x = \frac{I_x}{c}$$

where:

• S_x = Section modulus
• c = Distance from neutral axis to extreme fiber (usually h/2 for symmetric sections)

Usage: Section modulus is used in bending stress calculations:

$$\sigma_{bending} = \frac{M}{S_x}$$

Step 1: Identify known values

• σ = 150 MPa
• ε = 0.00075 (ratio)

Step 2: Apply formula

$$E = \frac{\sigma}{\varepsilon} = \frac{150 \text{ MPa}}{0.00075} = 200,000 \text{ MPa} = 200 \text{ GPa}$$

Step 3: Verify reasonableness

✓ Result of 200 GPa matches typical structural steel E value
✓ Strain of 0.075% is well within elastic region

Step 1: Calculate cross-sectional area

$$A = \frac{\pi d^2}{4} = \frac{\pi (10 \text{ mm})^2}{4} = 78.54 \text{ mm}^2$$

Step 2: Convert to consistent units (SI base units)

• F = 5,000 N
• A = 78.54 mm² = 78.54 × 10⁻⁶ m² = 7.854 × 10⁻⁵ m²
• L₀ = 500 mm = 0.5 m
• ΔL = 0.51 mm = 0.00051 m

Step 3: Apply combined formula

$$E = \frac{F \cdot L_0}{A \cdot \Delta L} = \frac{5000 \text{ N} \times 0.5 \text{ m}}{7.854 \times 10^{-5} \text{ m}^2 \times 0.00051 \text{ m}}$$

$$E = \frac{2500}{4.005 \times 10^{-8}} = 6.24 \times 10^{10} \text{ Pa} = 62.4 \text{ GPa}$$

Step 4: Verify reasonableness

⚠️ Expected aluminum E ≈ 69 GPa, but we got 62.4 GPa (10% low)
Possible causes: measurement error, not pure aluminum 6061, temperature effects, or specimen damage
Result is within typical experimental uncertainty range

Step 1: Convert to consistent units (SI)

• E = 200 GPa = 200 × 10⁹ Pa
• I = 18.5 in⁴ × (0.0254 m/in)⁴ = 7.701 × 10⁻⁶ m⁴
• L = 12 ft = 12 × 0.3048 m = 3.658 m
• P = 5,000 lbf × 4.448 N/lbf = 22,240 N

Step 2: Apply formula for simply supported beam, midpoint load

$$\delta_{max} = \frac{PL^3}{48EI} = \frac{22240 \times (3.658)^3}{48 \times (200 \times 10^9) \times (7.701 \times 10^{-6})}$$

$$\delta_{max} = \frac{1,087,732}{73,929.6 \times 10^3} = 0.01471 \text{ m} = 14.71 \text{ mm}$$

Step 3: Check serviceability

Deflection limit (typical): L/360 = 3658mm / 360 = 10.16 mm
⚠️ Actual deflection (14.71 mm) exceeds L/360 limit
→ Consider larger section or reduce load

Step 1: Calculate moment of inertia

$$I_x = \frac{bh^3}{12} = \frac{50 \times (200)^3}{12} = \frac{50 \times 8,000,000}{12} = 33,333,333 \text{ mm}^4$$

$$I_x = 33.33 \times 10^6 \text{ mm}^4 = 3.333 \times 10^{-5} \text{ m}^4$$

Step 2: Calculate section modulus

$$S_x = \frac{I_x}{c} = \frac{I_x}{h/2} = \frac{bh^2}{6} = \frac{50 \times (200)^2}{6} = 333,333 \text{ mm}^3$$

$$S_x = 3.333 \times 10^5 \text{ mm}^3 = 3.333 \times 10^{-4} \text{ m}^3$$