EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Chapter 2, Problem 2.58P
To determine
The expression for the toughness of material.
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A brass specimen of the circular cross-section is fractured at 151 kN force and the final length of the specimen at fracture is 49 mm. The fracture strength of the specimen is found to be 74 kN/mm2. The percentage of elongation of the specimen is 42 %.
Determine the following
(i) Diameter of the specimen
ii) Initial length of the specimen
iii) Stress under an elastic load of 16 kN
iv) Young's Modulus if the elongation is 1.6 mm at 16 kN
(v) Final diameter if the percentage of reduction in area is 20 %
solve:
Stress under the elastic load (in N/mm2) =
Young's Modulus of the Specimen (in N/mm2) =
Final Area of the Specimen at Fracture (in mm) =
A brass specimen of the circular cross-section is fractured at 151 kN force and the final length of the specimen at fracture is 49 mm. The fracture strength of the specimen is found to be 74 kN/mm2. The percentage of elongation of the specimen is 42 %.
Determine the following
(i) Diameter of the specimen
ii) Initial length of the specimen
iii) Stress under an elastic load of 16 kN
iv) Young's Modulus if the elongation is 1.6 mm at 16 kN
(v) Final diameter if the percentage of reduction in area is 20 %
solve:
Final Diameter of the Specimen after Fracture (in mm) =
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 =
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 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 brass specimen of the circular cross-section is fractured at 159 kN force and the final length of the specimen at fracture is 51 mm. The fracture strength of the specimen is found to be 85 kN/mm2. The percentage of elongation of the specimen is 42 %. Determine the following 4-Stress under an elastic load of 14 kN 5-Young's Modulus if the elongation is 1.5 mm at 14 kN 6-Final diameter if the percentage of reduction in area is 27 %.arrow_forwardA tensile specimen havign a diameter of 6 mm and a gauge length of 50 mm was tested to fracture and the stress-strain diagram is show. Poisson's ratio of material, v = 0.3 Determine the load P that the sample is subjected to when a stress of 300 MPa is applied. What is the instantaneous diameter of the sample given a longitudinal strain of 0.002?arrow_forward
- A 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 MPa Determine the Initial length, Initial diameter and Maximum load. i) Initial Length (in mm) = ii) Final Area (in mm2) = iii) Initial Area (in mm2) = iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =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 Diameter (in mm) = Ultimate Load (in N) =arrow_forwardA cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of m (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm (0.260 in.), and the fractured gauge length is 72.14 mm (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.arrow_forward
- A cylindrical steel pressure vessel ___171__ mm in diameter with a wall thickness of 5 mm is subjected to an internal pressure of 3 MN/m². Calculate a- the tangential and longitudinal stresses in the steel. b- To what value may the internal pressure be increased if the stress in the steel is limited to 80 MN/ m²? c-If the internal pressure were increased until the vessel burst, sketch the type of fracture that would occurarrow_forwardA cylindrical steel pressure vessel D-575 mm in diameter with a wall thickness of 5 mm is subjected to an internal pressure of 3 MN/m². Calculatea- the tangential and longitudinal stresses in the steel.b- To what value may the internal pressure be increased if the stress in the steel is limited to 80 MN/ m²?c-If the internal pressure were increased until the vessel burst, sketch the type of fracture that would occurarrow_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: iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =arrow_forward
- A 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_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 MPa Determine the Initial length, Initial diameter and Maximum load. ii) Final Area (in mm2) = iii) Initial Area (in mm2) =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 1) Final Area (in mm2) = 2) Initial Area (in mm2) = 3) Ultimate Load (in N) =arrow_forward
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