Force (N) time (s) Elongation (t0.05 mm/s) Engineering stress(Mpa) Engineering Strain 0 0 0 0 0 1500 5 0.25 53.05978069 0.00625 5000 10 0.5 176.8659356 0.0125 9250 20 1 327.2019809 0.025 11250 25 1.25 397.9483551 0.03125 11750 30 1.5 415.6349487 0.0375 11500 32.5 1.625 406.7916519 0.040625 12000 35 1.75 424.4782455 0.04375 14250 50 2.5 504.0679165 0.0625 15250 60 3 539.4411036 0.075 16000 70 3.5 565.970994 0.0875 16750 85 4.25 592.5008843 0.10625 17250 100 5 610.1874779 0.125 17500 132.5 6.625 619.0307747 0.165625 17250 170 8.5 610.1874779 0.2125 17000 180 9 601.3441811 0.225 16500 190 9.5 583.6575875 0.2375 16250 195 9.75 574.8142908 0.24375 16000 200 10 565.970994 0.25 15000 210 10.5 530.5978069 0.2625 13250 220 11 468.6947294 0.275 12697.1 222.5 11.125 449.1368942 0.278125 Initial Length (mm) Initial Diameter(mm) Area = pir^2 40 6 28.27 Maximum load capacity: 50,000...

80.2K

Verified Solution

Question

Mechanical Engineering

Force (N)
time (s)
Elongation (t0.05 mm/s)
Engineering stress(Mpa)
Engineering Strain
0
0
0
0
0
1500
5
0.25
53.05978069
0.00625
5000
10
0.5
176.8659356
0.0125
9250
20
1
327.2019809
0.025
11250
25
1.25
397.9483551
0.03125
11750
30
1.5
415.6349487
0.0375
11500
32.5
1.625
406.7916519
0.040625
12000
35
1.75
424.4782455
0.04375
14250
50
2.5
504.0679165
0.0625
15250
60
3
539.4411036
0.075
16000
70
3.5
565.970994
0.0875
16750
85
4.25
592.5008843
0.10625
17250
100
5
610.1874779
0.125
17500
132.5
6.625
619.0307747
0.165625
17250
170
8.5
610.1874779
0.2125
17000
180
9
601.3441811
0.225
16500
190
9.5
583.6575875
0.2375
16250
195
9.75
574.8142908
0.24375
16000
200
10
565.970994
0.25
15000
210
10.5
530.5978069
0.2625
13250
220
11
468.6947294
0.275
12697.1
222.5
11.125
449.1368942
0.278125
Initial Length (mm)
Initial Diameter(mm)
Area = pir^2
40
6
28.27
Maximum load capacity: 50,000 N Test rate: 3 mm/min - 0.05 mm/sSteel A Steel B Initial length, Lo 40 mm 40 mm Initial diameter, do6 mm 6 mm Final length, Lf 48 mm 52 mm Final diameter, df 4.2 mm3.1 mmÂ Â
From the data given in excel file, calculate:
(b) Tensile strength, modulus of elasticity, modulus ofresilience, % elongation and % reduction in area.
(c) Plot both engineering stress strain curve and true stressstrain curve.
(d) Compare the tensile behaviour of steel A and B in oneparagraph

Answer & Explanation Solved by verified expert

4.3 Ratings (662 Votes)

A ENGINEERING STRESSSTRAIN CURVEB TRUE STRAIN ln 1 ENGINEERING STRAIN TRUE STRESS ENGINEERING STRESS 1 ENGINEERING STRAIN C TRUE STRESSSTRAIN CURVEDTENSILE STRENGTH MAX LOAD INITIAL AREA 17500 2827 61903 N See Answer

Get Answers to Unlimited Questions

Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!

Membership Benefits:

Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!

Become a Member

Force (N)	time (s)	Elongation (t*0.05 mm/s)	Engineering stress(Mpa)	Engineering Strain
0	0	0	0	0
1500	5	0.25	53.05978069	0.00625
5000	10	0.5	176.8659356	0.0125
9250	20	1	327.2019809	0.025
11250	25	1.25	397.9483551	0.03125
11750	30	1.5	415.6349487	0.0375
11500	32.5	1.625	406.7916519	0.040625
12000	35	1.75	424.4782455	0.04375
14250	50	2.5	504.0679165	0.0625
15250	60	3	539.4411036	0.075
16000	70	3.5	565.970994	0.0875
16750	85	4.25	592.5008843	0.10625
17250	100	5	610.1874779	0.125
17500	132.5	6.625	619.0307747	0.165625
17250	170	8.5	610.1874779	0.2125
17000	180	9	601.3441811	0.225
16500	190	9.5	583.6575875	0.2375
16250	195	9.75	574.8142908	0.24375
16000	200	10	565.970994	0.25
15000	210	10.5	530.5978069	0.2625
13250	220	11	468.6947294	0.275
12697.1	222.5	11.125	449.1368942	0.278125
Initial Length (mm)	Initial Diameter(mm)	Area = pi*r^2
40	6	28.27