EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Chapter 2, Problem 2.64P
To determine
The load acting on specimen at fracture.
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A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find
(i)The stress at elastic limit.(ii) Young’s modulus.(iii) Percentage elongation. (iv)Percentage reduction in area.
(v)Ultimate tensile stress.
A mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained.
- Yield Strength as 88 MPa
- Maximum Strength as 212 MPa
- Fracture Strength as 152 MPa
- Percentage of Elongation as 63 %
- Percentage of Reduction in area as 39%
Determine the Fracture load
The Final length in mm =
The Final area in mm2 =
Sketch the typical engineering stress-strain diagram behaviour to fracture point, F and the tensile strength TS, together with the geometry of the deformed specimen at various point along the curve.
Chapter 2 Solutions
EBK MANUFACTURING PROCESSES FOR ENGINEE
Ch. 2 - Prob. 2.1QCh. 2 - Prob. 2.2QCh. 2 - Prob. 2.3QCh. 2 - Prob. 2.4QCh. 2 - Prob. 2.5QCh. 2 - Prob. 2.6QCh. 2 - Prob. 2.7QCh. 2 - Prob. 2.8QCh. 2 - Prob. 2.9QCh. 2 - Prob. 2.10Q
Ch. 2 - Prob. 2.11QCh. 2 - Prob. 2.12QCh. 2 - Prob. 2.13QCh. 2 - Prob. 2.14QCh. 2 - Prob. 2.15QCh. 2 - Prob. 2.16QCh. 2 - Prob. 2.17QCh. 2 - Prob. 2.18QCh. 2 - Prob. 2.19QCh. 2 - Prob. 2.20QCh. 2 - Prob. 2.21QCh. 2 - Prob. 2.22QCh. 2 - Prob. 2.23QCh. 2 - Prob. 2.24QCh. 2 - Prob. 2.25QCh. 2 - Prob. 2.26QCh. 2 - Prob. 2.27QCh. 2 - Prob. 2.28QCh. 2 - Prob. 2.29QCh. 2 - Prob. 2.30QCh. 2 - Prob. 2.31QCh. 2 - Prob. 2.32QCh. 2 - Prob. 2.33QCh. 2 - Prob. 2.34QCh. 2 - Prob. 2.35QCh. 2 - Prob. 2.36QCh. 2 - Prob. 2.37QCh. 2 - Prob. 2.38QCh. 2 - Prob. 2.39QCh. 2 - Prob. 2.40QCh. 2 - Prob. 2.41QCh. 2 - Prob. 2.42QCh. 2 - Prob. 2.43QCh. 2 - Prob. 2.44QCh. 2 - Prob. 2.45QCh. 2 - Prob. 2.46QCh. 2 - Prob. 2.47QCh. 2 - Prob. 2.48QCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.61PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71PCh. 2 - Prob. 2.72PCh. 2 - Prob. 2.73PCh. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. 2.76PCh. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. 2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. 2.82PCh. 2 - Prob. 2.83PCh. 2 - Prob. 2.84PCh. 2 - Prob. 2.85PCh. 2 - Prob. 2.86PCh. 2 - Prob. 2.87PCh. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - Prob. 2.90PCh. 2 - Prob. 2.91PCh. 2 - Prob. 2.92PCh. 2 - Prob. 2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - Prob. 2.99PCh. 2 - Prob. 2.100PCh. 2 - Prob. 2.101P
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- A mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained. - Yield Strength as 88 MPa - Maximum Strength as 212 MPa - Fracture Strength as 152 MPa - Percentage of Elongation as 63 % - Percentage of Reduction in area as 39% The yield load in N= The Ultimate load in N = The Fracture load in N =arrow_forwardThe data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13mm and a gage length of 50mm. At fracture, the elongation between the gage marks was 3.0mm and the minimum diameter was 10.7mm. Plot the conventional stress-strain curve for the steel and determine the propotional limit, modulus of elasticity (i.e the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 50mm, and percent reduction area. TENSILE-TEST DATA Load(kN) Elongation(mm) 5 0.005 10 0.015 30 0.048 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracturearrow_forwardAn engine part is being tested with a load of 65 000 lb. The allowable tensile stress is 10000 psi ,modulus of elasticity 40×106 psi . If the original length of specimen is 40 inches with elongation not exceeding 0.0155 inches, what diameter of the specimen is rejected.arrow_forward
- In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm The true stress at fracture is less than engineering stress at fracture True Falsearrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 65 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 36.1 mm - Final Diameter after fracture = 4.25 mm & - Ultimate stress = 401 MPa SOLUTION: Initial Diameter (in mm) = Ultimate Load (in N) =arrow_forwardA mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained. - Yield Strength as 88 MPa - Maximum Strength as 212 MPa - Fracture Strength as 152 MPa - Percentage of Elongation as 63 % - Percentage of Reduction in area as 39% Determine the Final length, Final area, Yield load, Ultimate loadarrow_forward
- A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 60 % - Percentage of Reduction in Area = 36 % - Final length after fracture = 35.2 mm - Final Diameter after fracture = 3.54 mm & - Ultimate stress = 439 MPa iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =arrow_forwardA metallic rod with an initial diameter of 10 mm and an initial length of 50 mm is subjected to the tensile test. After the fracture, the final length was measured as 51.8 mm, and the final diameter was measured as 9.5 mm.(a) Calculate modulus of elasticity, ultimate tensile strength, elongation at fracture in %,reduction of area in %, true stress at maximum load, true strain at maximum load, strain hardening exponent, strength coefficient.?arrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 66 % - Percentage of Reduction in Area = 38 % - Final length after fracture = 34.6 mm - Final Diameter after fracture = 4.43 mm & - Ultimate stress = 364 i) Initial Length (in mm) = iv) Initial Diameter (in mm) =arrow_forward
- The Highest load sustained druing an uniaxial tensile testing experiment is 7,500lb. If the original cross section has a diamter of 0.25in, what is the ultimate tensile strength? (please also make a drawing)arrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 65 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 36.1 mm - Final Diameter after fracture = 4.25 mm & - Ultimate stress = 401 MPa SOLUTION: Initial Length (in mm) = Final Area (in mm2) = Initial Area (in mm2) =arrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: i) Initial Length (in mm) = ii) Final Area (in mm2) = iii) Initial Area (in mm2) =arrow_forward
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