103地方特考 土壤力學與基礎工程 第二題 - 土木
By Lauren
at 2014-12-16T11:45
at 2014-12-16T11:45
Table of Contents
103地方特考 土壤力學與基礎工程 第二題
感覺題目第三小題有一些瑕疵
我只把要問的部分出來
題目前兩小題問的是滲透性問題
下面將第三小題有關的參數貼出來跟大家討論
Given: 土壤飽和單位重 γsat = 20kN/m^3
孔隙比 e = 0.7
土粒比重 Gs = 2.7
Find : 該土壤含水比 w
方法一:
假設 Vsolid = 1m^3
則 1. 由 Gs = 2.7 得 Wsolid = 2.7 * 9.81kN/m^3 *1m^3
= 26.487kN
2. 由 e = 0.7 得 Vvoid = 0.7m^3
因為是飽和土樣故 Vvoid = Vwater
所以 Wwater = 0.7m^3 * 9.81kN/m^3 = 6.867 kN
Wwater 6.867kN
土壤含水比 w = ──── = ──── = 0.259
Wsolid 26.487kN
方法二:
假設 Vsolid = 1m^3
則 1. 由 Gs = 2.7 得 Wsolid = 2.7 * 9.81kN/m^3 *1m^3 = 26.487kN
2. 由 e = 0.7 得 Vvoid = 0.7m^3
3. Wwater = Wsolid * w = 26.487w kN
Wsolid + Wwater 26.487 (1+w) kN kN
γsat = ───────── = ───────── = 20 ─
Vsolid + Vvoid 1+0.7 m^3 m^3
20 * 1.7
or 1+w = ───── = 1.284
26.487
or w = 0.284
兩種方法求得的土壤含水比不相同
我的看法是出題者的失誤
26.487kN + 6.867kN kN
如果修改 γsat = ───────── = 19.62 ─
1 m^3 + 0.7 m^3 m^3
這題就完美了
考古題中93年高考第一題也有類似的失誤
版上各位先進們看法如何呢?
--
感覺題目第三小題有一些瑕疵
我只把要問的部分出來
題目前兩小題問的是滲透性問題
下面將第三小題有關的參數貼出來跟大家討論
Given: 土壤飽和單位重 γsat = 20kN/m^3
孔隙比 e = 0.7
土粒比重 Gs = 2.7
Find : 該土壤含水比 w
方法一:
假設 Vsolid = 1m^3
則 1. 由 Gs = 2.7 得 Wsolid = 2.7 * 9.81kN/m^3 *1m^3
= 26.487kN
2. 由 e = 0.7 得 Vvoid = 0.7m^3
因為是飽和土樣故 Vvoid = Vwater
所以 Wwater = 0.7m^3 * 9.81kN/m^3 = 6.867 kN
Wwater 6.867kN
土壤含水比 w = ──── = ──── = 0.259
Wsolid 26.487kN
方法二:
假設 Vsolid = 1m^3
則 1. 由 Gs = 2.7 得 Wsolid = 2.7 * 9.81kN/m^3 *1m^3 = 26.487kN
2. 由 e = 0.7 得 Vvoid = 0.7m^3
3. Wwater = Wsolid * w = 26.487w kN
Wsolid + Wwater 26.487 (1+w) kN kN
γsat = ───────── = ───────── = 20 ─
Vsolid + Vvoid 1+0.7 m^3 m^3
20 * 1.7
or 1+w = ───── = 1.284
26.487
or w = 0.284
兩種方法求得的土壤含水比不相同
我的看法是出題者的失誤
26.487kN + 6.867kN kN
如果修改 γsat = ───────── = 19.62 ─
1 m^3 + 0.7 m^3 m^3
這題就完美了
考古題中93年高考第一題也有類似的失誤
版上各位先進們看法如何呢?
--
Tags:
土木
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