(Suppose you need to design a tension test machine capable of testing specimens that have nominal ultimate stresses as high as σu = 100 ksi . How much force must the machine be capable of generating? Assume the testing specimen has the ASTM shape shown. Answer for this is 19.6 kip) (If the maximum nominal strain is ϵf = 0.7 just before the test specimen fractures and the test machine operates by moving only one grip, how far must that grip be designed to travel? The total length of the deforming part of the specimen is 3 in. Answer for this is 2.10 in) Do not know if this info is needed but this was the other 2 parts

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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(Suppose you need to design a tension test machine capable of testing specimens that have nominal ultimate stresses as high as σu = 100 ksi . How much force must the machine be capable of generating? Assume the testing specimen has the ASTM shape shown. Answer for this is 19.6 kip)

(If the maximum nominal strain is ϵf = 0.7 just before the test specimen fractures and the test machine operates by moving only one grip, how far must that grip be designed to travel? The total length of the deforming part of the specimen is 3 in. Answer for this is 2.10 in)

Do not know if this info is needed but this was the other 2 parts

 

Force (kip)
14
12
10
8
4
2
04
2.00
stress =
strain =
2.20
ksi
2.40
Length (in)
stress =
strain =
2.60
2.80
ksi
stress =
strain =
ksi
Transcribed Image Text:Force (kip) 14 12 10 8 4 2 04 2.00 stress = strain = 2.20 ksi 2.40 Length (in) stress = strain = 2.60 2.80 ksi stress = strain = ksi
●
Question 1.png
72°F
Sunny
|
D
W
:
d₂ = 0.5 in.
L₁= 2 in.
H
o
Q Search
+
100%
00
▬▬
0
X
W
Question 12.png
H
@
IA
⠀
T
Q
You have built the majority of the tension testing machine, but much of the instrumentation is still being assembled. To test the machine, you perform a test on a steel specimen with known properties. The machine provides you with
the given load data, and you manually record the lengths between the marks on the specimen at each point using an extensometer to obtain the table of data shown below.
L (in)
2.0012
2.0025
2.0350
2.0798
2.1313
2.1937
2.2804
3.51
7.09
7.87
9.84
10.80
2.4774
12.37
8.83
11.77
P (kip)
Use these results to calculate the stress and strain at the yield point, the ultimate strength point, and the fracture point.
57%
2.5791
11.80
<
2.6657
10.81
0
4) O
2.7281
9.81
10:53 AM
10/6/2023
2.7497
9.42
Transcribed Image Text:● Question 1.png 72°F Sunny | D W : d₂ = 0.5 in. L₁= 2 in. H o Q Search + 100% 00 ▬▬ 0 X W Question 12.png H @ IA ⠀ T Q You have built the majority of the tension testing machine, but much of the instrumentation is still being assembled. To test the machine, you perform a test on a steel specimen with known properties. The machine provides you with the given load data, and you manually record the lengths between the marks on the specimen at each point using an extensometer to obtain the table of data shown below. L (in) 2.0012 2.0025 2.0350 2.0798 2.1313 2.1937 2.2804 3.51 7.09 7.87 9.84 10.80 2.4774 12.37 8.83 11.77 P (kip) Use these results to calculate the stress and strain at the yield point, the ultimate strength point, and the fracture point. 57% 2.5791 11.80 < 2.6657 10.81 0 4) O 2.7281 9.81 10:53 AM 10/6/2023 2.7497 9.42
Expert Solution
Step 1: Problem 1

Given data:

Ultimate stress, σu = 100 ksi

diameter of the specimen, d = 0.5 in (from the given image)


Solution:

The stress is given by,

straight sigma subscript straight u equals fraction numerator Force space left parenthesis straight P right parenthesis over denominator Area space left parenthesis straight A right parenthesis end fraction
straight P equals straight sigma subscript straight u cross times open parentheses straight pi over 4 cross times straight d squared close parentheses
substituting space all space the space known space values comma
straight P equals 100 cross times open parentheses straight pi over 4 cross times 0.5 squared close parentheses
bold P bold equals bold 19 bold. bold 635 bold space bold kip

The force must the machine be capable of generating can be calculated as 19.635 kip.


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