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2024 고체(1) 과제 02 solution

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From the stress-strain diagram Fig. a, the modulus of elasticity for the steel alloy is
𝐸
1
=
290𝑀𝑝𝑎−0
0.001−0
;
E = 290(103 )𝑀𝑝𝑎
When the specimen is unloaded, its normal strain recovered along line AB, Fig. a,
Which has a gradient of E. Thus
500
500𝑀𝑝𝑎
Elastic Recovery =
=
= 0.001724 𝑚𝑚/𝑚𝑚
𝐸
290(103 )𝑀𝑝𝑎
Thus, the permanent set is
∈𝑝 = 0.08 − 0.001724 = 0.07828 𝑚𝑚/𝑚𝑚
Then, the increase in gauge length is
∆L = ∈𝑝 𝐿 = 0.07828(50) = 3.914𝑚𝑚
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