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A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7 mm (0.1457 in.) will experience only elastic deformation up to a tensile load of 2000 N (449.6 lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.41 mm (0.01614 in.).
A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7 mm (0.1457 in.) will experience only elastic deformation up to a tensile load of 2000 N (449.6 lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.41 mm (0.01614 in.).
Themaximum lengthof the specimen before deformation is; 235.6 mmWe are given;Elastic modulus; E = 107 GPa = 107 × 10⁹ PaTensile Load; P = 2000 Noriginal diameter; d = 3.7 mm = 0.0037 mAllowable elongation; ΔL = 0.41 mm = 0.00041 mFormula forstressof the cylinder is;σ = P/AWhere;P =Tensile Load= 2000 NA =Cross-sectional area= πd²/4 = π(0.0037)²/4 = 1.0752 × 10⁻⁵ m²Thus;σ = 2000/(1.0752 × 10⁻⁵)σ = 186 × 10⁶ Paσ = 186 MPaFormula for thestrainis gotten from;Elastic Modulus = Stress / StrainE = σ/∈Thus;∈ = σ/E∈ = 186 × 10⁶/(107 × 10⁹)∈ = 1.74 × 10⁻³To find theoriginal length, we will use the formula;∈ = ΔL/LWhere L is original length.Thus;L = ΔL/∈L = 0.00041/(1.74 × 10⁻³)L = 0.2356 mL = 235.6 mmRead more aboutElastic Modulusat;brainly.com/question/14468674...