A tensile bar has an initial 2.000-in. gauge length and a circular cross-sectionalarea of 0.2 in.?. The loads and corresponding gauge lengths recorded in a tensiletest are show below. After fracture the bar diameter was 0.370 in.Gauge length(in.)Load(Ib)2.00006,00012,00015,00018,00021,00024,00025,00022,0002.00172.00342.00422.00882.0432.302.55B.00а)b)c)Plot the data as engineering stress versus engineering strain.Compute the modulus of elasticity.Determine the yield strength at a strain offset of 0.002.

Given:- gauge length=2.000 in cross sectional area =0.2in2 bar diameter =0.370 in strain…

Answered: A tensile bar has an initial 2.000-in.…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A 13 mm diameter tensile specimen has 50 mm gauge length. If the load corresponding to the 0.2% offset is 6800 kg, the yield stress will be nearly 1. 31 kg/mm² 2. 43 kg/mm² 3. 51 kg/mm² 4. 63 kg/mm² X
During a tensile test on a specimen the following results were obtainedload(KN) 15 30 40 50 55 60 65 70 75 80 82 80 70extension(mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 5.08 7.62 12.7 16 19.05 22.9diameter of gauge length=19mmdiameter at fracture=16.49mmgauge length=100mmgauge lengthat fracture=121mmplot the complete load extension graph and the straight line portion to an enlarged scale .hence determinethe modulus of elasticitythe percentage elongationthe percentage reduction in areathe nominal stress at fracturethe actual stress at fracturethe tensile strength
A cylindrical tensile specimen has an undeformed diameter of 8.6 mm. It is subjected to a tensile load of 100 kN. The elastic modulus (E) is 10 GPa. Poisson's ratio is 0.25. Calculate the change in the diameter (in mm).
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4. The following creep data were obtained for a metal alloy tested under a constant load and at a constant temperature (the initial length of the unloaded sample was 2.0 inches): Sample Length (in) Time (hr) 2.004 2.01 100 2.02 200 2.03 400 2.045 1000 2.075 2000 2.135 4000 2.193 6000 2.23 7000 2.30 8000 (fracture) a.) Plot the strain versus time (in hours) curve for this material. b.) Determine the steady-state creep rate.
A brass specimen of the circular cross-section is fractured at 151 kN force and the final length of the specimen at fracture is 48 mm. The fracture strength of the specimen is found to be 72 kN/mm?. The percentage of elongation of the specimen is 44 %. Determine the following (i) Diameter of the specimen, ii) Initial length of the specimen, iii) Stress under an elastic load of 15 kN, iv) Young's Modulus if the elongation is 1.5 mm at 15 kN (v) Final diameter if the percentage of reduction in area is 21 %. ( Initial Cross-sectional Area (in mm?) The Diameter of the Specimen (in mm) Initial Length of the Specimen (in mm) Stress under the elastic load (in N/mm?) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)
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DURING THE STRESS-STRAIN TEST, THE UNIT DEFORMATION AT A STRESSOF 30 MPa WAS OBSERVED TO BE 110 x 10-6 mm/mm AND AT A STRESS0F 130 MPa IT WAS 520 X 10-6 mm/mm. IF THE PROPORTIONAL LIMITWAS 250 MPa, WHAT IS THE MODULUS OF ELASTICITY? WHAT IS THESTRAIN CORRESPONDING TO A STRESS OF 150 MPa?
A tensile test is performed on a brass specimen 10 mm indiameter using a gage length of 50 mm. When the tensile load Preaches a value of 20 kN, the distance between the gagemarks has increased by 0.122 mm. a) What is the modulus of elasticity E of the brass? b) If the diameter decreases by .00830 mm,what is Poisson's ratio? 10mm PO 50mm. P
ngineering / Civil Engineering / Q&A Library/1 a. a. During a tensile test on a speci... 1 a. During a tensile test on a specimen the following results were obtaineo Load (kN) 15 30 40 50 55 60 Extension 1.03 2.00 2.73 3.42 3.80 4.10 (mm) Diameter at fracture -16.5mm Diameter of gauge length =19mm Gauge length = 150mm Load at fracture - R0KN Gauge length at fracture = 170mm Ultimate load-120kN Plot the load/extension graph to an appropriate scale. Hence determine the i. Modulus of elasticity ii. Percentage elongation iii. Percentage reduction in area iv. The nominal stress at fracture
PROBLEMS 203. During a stress-strain test, the unit deformation at a stress of 5000 psi Wap ODserved to be 0.000167 in./in. and at a stress of 20,000 psi it was 0.000667 in./in. If the proportional limit is 30,000 psi, what is the modulus of elasticity? What is the stress corresponding to a strain of 0.0002 in./in.? Would these re- sults be valid if the proportional limit were 18,000 psi? Explain. E = 30 X 106 psi; S 6000 psi %3D Ans. 204. A uniform bar of length L, cross-sectional area A, and weight W is sus- WL If pended vertically from one end. Show that its total elongation is & = 2AE wL? its weight per unit volume is w, show also that & 2E A steel rod having a cross-sectional area of sq in. and a length of 600 ft is suspended vertically. It supports a load of 5000 lb at the lower end. If steel 30 X 106 psi, find the total elongation in the rod. Ans. 8 = 2.64 in. 205 weighs 490 lb/cu ft and E Hint: Use the result of Prob. 204. steel wire 30 ft long, hanging vertically, supports a…

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A tensile bar has an initial 2.000-in. gauge length and a circular cross-sectional area of 0.2 in.?. The loads and corresponding gauge lengths recorded in a tensile test are show below. After fracture the bar diameter was 0.370 in.