Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 10, Problem 10.16SP
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A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress
and strain data obtained during the test are shown. Determine the true fracture stress if the final diameter of the specimen at the
location of the fracture was 0.350 in.
Stress (ksi)
120
Upper scale
Lower scale
H
0.0 0.020 0.040 0.060 0.080 0.100 0.120
0.0 0.002 0.004 0.006 0.008 0.010 0.012
Strain (in./in.)
90
60
30
0-
300 ksi
O 126 ksi
EA
O 237 ksi
O 202 ksi
O 261 ksi
A 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.
A 200mm long rod with a diameter of 50mm is loaded with a 4kN weight. The rod final length 201.314mm and final diameter 42mm the load at fracture is 2.5kN and the maximum load is 3.5KN.
1. Calculate the stress
2. Calculate the strain
3. Calculate limit of proportionality
Chapter 10 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 10 - A 916 - in. - diameter steel rod is tested in...Ch. 10 - A concrete cylinder 150 mm in diameter was tested...Ch. 10 - Prob. 10.3PCh. 10 - The data from the tension test of a steel specimen...Ch. 10 - An 18-in.-long titanium alloy rod is subjected to...Ch. 10 - ASTM A36 steel rods are used to support a balcony....Ch. 10 - A 450-mm-long AISI 1020 steel rod is subjected to...Ch. 10 - A tension member in a roof truss is composed of...Ch. 10 - A short, solid, compression member of circular...Ch. 10 - A main cable in a large bridge is designed for a...
Ch. 10 - Test results of a steel specimen indicated an...Ch. 10 - A concrete canoe in storage is supported by two...Ch. 10 - A load is applied to a rigid bar that is...Ch. 10 - Prob. 10.14CPCh. 10 - Write a program that will allow a user to input...Ch. 10 - A 12 - in. - diaiíct.cr structural nickel steel...Ch. 10 - Compute the modulus of elasticity of a copper...Ch. 10 - A concrete cylinder 6 in. in diameter was tested...Ch. 10 - An aluminum bar 2 in. by 12 - in. in cross section...Ch. 10 - During a tensile test of a steel specimen, the...Ch. 10 - A 12.5-mm-diameter steel rod was subjected to a...Ch. 10 - Prob. 10.22SPCh. 10 - A standard steel specimen having a diameter of...Ch. 10 - 10.24 A tension member in a structure is composed...Ch. 10 - A pair of wire cutters is designed to operate...Ch. 10 - Calculate the end bearing length required for a...Ch. 10 - Design a 3-m-long rod subjected to a tensile load...Ch. 10 - The collar bearing shown is subjected to a...Ch. 10 - A 10-ft-long steel member is subjected to a...Ch. 10 - Two steel bars A and B support a load P, as shown....Ch. 10 - Prob. 10.31SP
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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.arrow_forwarda 200mm long rod with a diameter of 50mm is loaded with a 4kN weight.the rod finale length 201.314mm and finale diameter 42mm.the load at the fracture is 2.5kN and the maximum load is 3.5kN. Calculate to third decimal and determine the folliwing 1.stress 2.strain 3.limit of proportionality 4.percentage of reduction in area 5.percentage elogationarrow_forwardThe table below shows the deformation data that resulted from applying a pure tensile load to a brass alloy rod with an initial length of 30 mm and a diameter of 10 mm. This rod was subjected to necking and beyond necking deformation. The data at fracture is shown in the last row. Applied Load (N) Length (mm) Diameter (mm) 75,100 33.8 68, 250 34.9 50,200 35.7 6.7 Solve for the TRUE STRESS/ES and TRUE STRAINS. 8.3 7.8arrow_forward
- A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the proportional limit. Stress (ksi) 40 30- 20 10- 0- 0.0 0.0 Upper scale 0.020 0.002 22 ksi O 15 ksi E 19 ksi O 10 ksi 13 ksi Lower scale- 0.040 0.060 0.080 0.100 0.004 0.006 0.008 0.010 Strain (in./in.) 0.120 0.012arrow_forwardIn 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 initial diameter of specimen is * O 25 022 20 O26arrow_forwardIn 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 % reduction in area is * O 15.819% 18.819 O 17.819 O 20.819arrow_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 % Elongation of specimen is O 18.79 17.79 O 20.79 O 16.79 Karrow_forward5. A cylindrical sample has a gauge length of 50.80 mm and original diameter of 12.8 mm. The sample is pulled with a high load that led to fracture. At the point of fracture, the diameter is 6.60 mm, and the fractured gauge length is 72.14 mm. Calculate the: (a) Ductility in term of percent elongation. (b) Ductility in term of percent reduction in area.arrow_forwardA tensile test specimen has a starting gage length 50 mm and a cross-sectional area 200 mm2. During the test, the specimen yields under a load of 30,000 N (this is the 0.2% offset) at a gage length of 52 mm.The maximum load of 63,000 N is reached at a gage length of 57 mm just before necking begins. Final fracture occurs at a gage length of 63.5 mm. Determine (a) yield strength, (b) modulus of elasticity, (c) tensile strength, (d) engineering strain at maximum load, and (e) percent elongation.arrow_forward
- How can I solve for the %0.2 offset strain and the calculated stress? Original diameter : 6mm Original length : 30mm Fracture diameter : 4.54mmarrow_forwardA tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the ultimate strength. Stress (ksi) 60 50- 40- 30- 20 10- 0 0 0 Upper scale O 66 ksi 28 ksi O 72 ksi O 53 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)arrow_forwardASAParrow_forward
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