Question

Part A The material for the 54-mm-long specimen has the stress- strain diagram shown. (Figure 1) If $P = 150 \text{ kN}$ is applied and then released, determine the permanent elongation of the specimen. Express your answer to three significant figures and include the appropriate units. $\delta_p = 0.129 \text{ mm}$ Incorrect; Try Again; 27 attempts remaining

          Part A
The material for the 54-mm-long specimen has the stress-
strain diagram shown.
(Figure 1)
If $P = 150 \text{ kN}$ is applied and then released, determine the permanent elongation
of the specimen.
Express your answer to three significant figures and include the
appropriate units.
$\delta_p = 0.129 \text{ mm}$
Incorrect; Try Again; 27 attempts remaining
        
Show more…
Part A
The material for the 54-mm-long specimen has the stress-
strain diagram shown.
(Figure 1)
If P = 150  kN is applied and then released, determine the permanent elongation
of the specimen.
Express your answer to three significant figures and include the
appropriate units.
= 0.129  mm
Incorrect; Try Again; 27 attempts remaining

Added by Eric J.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The material for the 54-mm-long specimen has the stress-strain diagram shown. (Figure 1) If P = 150 kN is applied and then released, determine the permanent elongation of the specimen. Express your answer to three significant figures and include the appropriate units. δp = 0.129 mm Previous Answers Incorrect; Try Again; 27 attempts remaining Part A The material for the 54-mm-long specimen has the stress-strain diagram shown. (Figure 1) If P = 150 kN is applied and then released, determine the permanent elongation of the specimen. Express your answer to three significant figures and include the appropriate units. δp = 0.129 mm Submit Previous Answers Request Answer Incorrect; Try Again; 27 attempts remaining Provide Feedback Figure Next> 1 of 1 (MPa) 20 mm 500 450 (mm/mm) 0.00225 0.03
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Transcript

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00:02 Hello everybody, we know that applied force divided by area is equal to sigma that is stress here a is the area now strain from the graph we see is 0 .0025 now stress for that given strain is equal to 450 into 10 to the power 6 that much pascal now we know young's modulus y is equal to stress by strain rather wrong longitudinal stress by longitudinal strain and that will be equal to 18 into 10 to the power 10 that much pascal now area a is equal to pi d square by 4 here d is 0 .02 that much meter so it will be equal to 3 .14 into 10 to the power minus 4 meter square this is the area so stress for the given applied force that is 100 kilo newton so force is 100 into 10 to the power 3 newton divided by the area that is 3 .14 into 10 to the power minus 4 and that is 318 megapascal now within elastic limit we see that the stress is there so from the graph if we find x that is del l by l that is the given strain so we can find out the elongation elongation is equal to x into the initial length l and that is given as 0 .046 meter into x so this is the value of elongation if you find x from the graph we can easily find out the value of the elongation.
02:10 So that's it...
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