五、 均一可靠度導向之部分係數率定
6.2 建議
1. 本文因未收集足夠的地下水與地震力參數資料，故在分析上做了一 些假設，若是有能以較符合實際的參數去分析，其結果可能更較接 近實際情形，也可將地震力視為一隨機變數，加上地震力之部分係 數考量。
2. 率定與分析上重要的基礎為資料庫的完整。資料庫的建立應包含基 本參數的收集和設計案例的結果。受資料之限制，本文目前只能採 用虛擬的資料庫來施作。參數之變異性資料除了可供後人做邊坡設 計之研究，亦可供其他需考量參數變異性之研究使用。但現今的實 驗結果多是以平均值來考量，而能提供有關參數的分佈型態、變異 性資料有限。故此處也建議後續之的資料需包含完整的試驗結果，
以提供後續研究統計的考量。
3. 本文對平面滑動未達上浮破壞機制的率定上，即使分析結果對不同 弱面傾角分別考量，也無法使結果有極好的回歸關係。所以對部分 係數與影響因子的歸納上，後續仍有研究的空間存在，探討對部分 係數是否有更好的分析方式、更適合的因子考量。若是以影響之因 子加以分析，此因子宜兼顧設計上的使用，才能使率定之部分係數 能配合實際的設計。
4. 受於時間限制，本文只進行平面破壞中低變異性度參數的研究，而 楔形破壞的機制與平面滑動相同，故理當可仿照該本文之流程對楔 形破壞率定一組部分係數。但楔形破壞又因需考量多組弱面之位態 之影響，故結果之分析上恐也會複雜許多。
5. 以率定的觀點來看，對應目標可靠度βT來率定部分係數之結果，應 取設計結果可靠度為βT之案例來率定。但對相同可靠度之案例，以 不同設計方式率定之部分係數也不全然會相同。故後續的研究上，
還可探討不同設計方式、設計限制對率定部分係數之影響。
6. 部分係數的制定形式，除了本文談到的兩種方式（平均值與函數 式），也可採用如圖 5.10 中查圖或查表的方式要求。使用者可計算 設計中影響因子的值（圖 5.10 中為參數佔總阻抗、載重之比例），
並對照圖中不同目標安全性決定需採用之部分係數，並分析該設計 是否能達到此安全性。
7. 一般傳統的 FSD 設計需考量參數在常時、高水位及地震作用下的 三種狀態，並檢核對應之安全係數是否足夠，而部分係數設計之參 數則簡化成只需考量名目值做設計。此處建議將原本參數的多狀態 考量（常時、高水位等），轉成對部分係數較完整的考量（如設計 方式二的函數式考量），以改善設計結果對全等級之均一性。
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附 錄
D-4 B10 21.5 裕大花園別墅 D 區附件三：
0 6 55 20 1 0 0 N
0 30 55 30 6 5 2 N
50T@1m, 2rows
80T@1m 60 50 40 9 7 1.5 Y
0 4 55 30 1 0 0 N
0 25 55 40 3 8 1.5 Y
0 2 55 40 1 0 0 N
0 25 55 10 5 5 2 N
50T@1m, 2rows
80T@1m 55 50 40 8 7 1.5 Y
0 60 55 10 12 5 2 N
0 60 33 20 12 5 2 Y
50T@1m 60 40 30 10 6 2 Y
50T@1m, 2rows
80T@1m 60 50 40 10 6 1.5 Y
表 B-4 虛擬設計結果（c、tanφ 平均值＝30kPa、0.65）
0 18 55 30 2 9 2 N
50T@1m, 3rows
80T@1m 50 50 40 7 7 2 Y
0 55 55 10 11 5 2 N
0 55 55 20 11 5 2 N
50T@1m, 4rows 55 45 30 8 7 4 Y
50T@1m, 4rows
80T@1m 55 50 40 8 7 2 Y
0 60 55 10 12 5 2 N
0 60 55 20 12 5 2 N
50T@1m, 6rows 60 45 30 12 5 3 Y
50T@1m, 4rows
80T@1m, 2rows 60 50 40 7 9 2.5 Y
0 14 55 40 2 7 2 N
50T@1m, 2rows
80T@1m 45 50 40 5 9 2.5 Y
0 50 55 10 10 5 2 N
0 50 33 20 10 5 2.5 Y
50T@1m
80T@1m 50 40 30 6 8 4 Y
80T@1m, 3rows 50 50 40 6 8 2.5 Y
0 55 55 10 11 5 2 N
0 55 33 20 14 4 2.5 Y
50T@1m, 3rows
80T@1m 55 40 30 11 5 2 Y
50T@1m, 4rows
80T@1m 55 50 40 8 7 2 Y
0 60 55 10 12 5 2 N
0 60 30 20 12 5 3 Y
50T@1m, 6rows 60 40 30 10 6 2.5 Y
50T@1m, 6rows
80T@1m 60 50 40 7 9 2.5 Y
0 10 55 40 2 5 2 N
0 45 55 10 9 5 2 N
0 8 45 40 1 0 0 Y
50T@1m, 2rows
80T@1m 35 40 30 5 7 3.5 Y
50T@1m, 2rows
80T@1m 35 50 40 6 6 2 Y
0 40 55 10 8 5 2 N
50T@1m, 2rows 40 55 20 8 5 2 Y
80T@1m, 3rows 40 40 30 6 7 3.5 Y
50T@1m, 2rows
80T@1m, 2rows 40 50 40 6 7 2 Y
50T@1m, 2rows
80T@1m, 4rows 50 40 30 10 5 2.5 Y
50T@1m, 2rows
80T@1m, 4rows 50 50 40 10 5 1.5 Y
0 55 55 10 11 5 2 N
50T@1m, 4rows
80T@1m 55 50 20 11 5 2 Y
0 4 55 20 1 0 0 N
80T@1m 25 40 30 4 4 2 Y
50T@1m, 2rows
80T@1m 35 40 30 9 6 3 Y
50T@1m, 2rows
80T@1m 35 50 40 5 6 2 Y
0 40 55 10 8 5 2 N
50T@1m, 2rows
80T@1m 40 30 20 10 4 3 Y
80T@1m, 3rows 40 40 30 5 6 3 Y
50T@1m, 4rows
80T@1m 40 50 40 5 5 1.5 Y
50T@1m, 4rows
80T@1m, 3rows 50 40 30 10 5 2.5 Y
表 B-9 虛擬設計結果（c、tanφ 平均值＝10kPa、0.45）
80T@2m 18 40 30 3 6 3 Y
50T@1m, 2rows
80T@1m 35 50 40 4 6 2 Y
50T@1m, 6rows
80T@1m 45 40 30 9 5 2.5 Y
50T@1m, 2rows
80T@1m, 4rows 45 50 40 5 5 1.5 Y
0 50 55 10 10 5 2 N
80T@1m, 6rows 50 30 20 25 5 4 Y
50T@1m, 4rows
80T@1m, 4rows 50 40 30 10 7 3.5 Y
50T@1m, 11rows 50 55 40 7 6 3 Y
0 55 55 10 11 5 2 N
50T@1m, 12rows 55 30 20 14 5 4 Y
50T@1m, 12rows 55 40 30 11 5 2.5 Y
50T@1m, 12rows 55 50 40 7 6 2 Y
0 60 55 10 12 5 2 N
50T@1m, 15rows 60 30 20 15 5 4 Y
80T@1m, 8rows 60 40 30 7 8 4 Y
80T@1m, 8rows 60 50 40 7 6 2 Y
作者簡歷
姓名 ：郭文鳴
英文姓名 ：Wen-Ming Kuo
出生日期 ：民國 72 年 1 月 26 日
學歷 ：基隆市私立二信中學畢業（民國 84 年～87 年）
台北市立建國高級中學畢業（民國 87 年～90 年）
國立交通大學土木工程學系畢業（民國 90 年～94 年）
國立交通大學土木工程學系碩士班大地工程組畢業（民國 94 年
～96 年）
電子信箱：[email protected]