For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
his material elongate when a true stress of 414 MPa (60050 psi) is applied if the original length is 520 mm (20.47 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.
1 answer:
Answer:
24.348 mm
Explanation:
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K = d / €^n
Note: d refers to the Greek letter epsilon.
K = 345 / 0.02⁰.²² = 816 mPa
The true strain based on the stress of 414 mPa =
€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576
The actual relationship between true strain and length is given by
€ = ln(Li/Lo)
To isolate Li by rearranging,
Li = Lo.e^€
Li = 520e⁰.⁰⁴⁵⁷⁶
Li = 544.348 mm
The elongation can be calculated from
Change in L = Li - Lo = 544.348 - 520 change in L = 24.348 mm.
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Answer:
The rate at which the root beer level is decreasing is 0.08603 cm/s.
Explanation:
The formula for the volume of the cone is:
Where V denotes the cone's volume
r indicates the radius
h signifies the height
The ratio of radius to height remains consistent throughout the cone.
Thus, we have r = d / 2 = 10 / 2 cm = 5 cm
h is 13 cm
Consequently, r / h = 5 / 13
r = {5 / 13} h
Additionally, we differentiate the volume expression in relation to time:
Given that = -4 cm³/sec (the negative sign indicates outflow)
h equals 10 cm
Hence,
The rate at which the root beer level is decreasing is 0.08603 cm/s.
Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².