【正文】 ΛpFs166。ΛpFs166。Λp= 1 . Fs= 1 . 2Fs0Fs166。Λp= 1 . Fs= 1 . 2 166。Λp= 1 . Fs= 1 . 2 166。0. 0 m15. 0 m①( a)( c)P = 0. Fs= 0 . 998P = 300 kN / m Fs= 1. 287(Fs= 1 . 357 , Co n v e n t i o n a l )Fs= 1. 5 P = 485 kN / m算例 1 算例 2 T a b l e 1. V a l ue s o f f a c t o r s o f s a fe t y a n d l o a d f a c t o r s? B= 0 . 0 ? B= 0 . 2 5 ? B= 0 . 5 ? B= 0 . 7 5 ? B= 1 . 0No . 166。 Γ= 18 kN /m3c = 15. kP a? ?= 28 ?L a y e r ②166。 sr Frr1?? ?? ?? ( 2 7 a )? ?cuF su?? ??1 。 ??應力修正： 簡化式： ? ?? ????????baxpcPdxwks ??? ( 1 3 a ) ? ?? ??????baypPdxws ??? ( 1 3 b )? ? ? ? ? ? ? ? ? ?? ?? ? ? ????????????????????????cpypcpxpbaccccxxPyyPdxgsywkxxwssys?????? ( 1 3 c )? ? ??? ? ??????????????????baspbabasxpxsdxBFdxcuFPFdxFs ??????? 111 ( 1 5 a )? ?? ?? ? ???????????????????babaspbasypysdxBsFdxcusFPFdxFs ??????? 1111 ( 1 5 b )? ? ?? ?? ????????????????baspbasppcbasdxBrFdxcurFMMdxFrr ??????? ???? 111 ( 1 5 c )極限平衡方程 安全系數(shù)解答 pp???? ?? 0 ????? 1???? Bcu pc令 0?根據(jù)無錨作用下邊坡穩(wěn)定計算結果確定，建議用Spencer法 ??? ? ????????????????????????????bacsxpxbabassdxFPFdxFsdxFs ??????????1 112211 ( 18a ) ??? ? ???????????????????????????bacsypybabassdxsFPFdxFsdxFs ??????????1 11112211 ( 18b )? ? ?? ?????????babappcbabacbasMMdxrdxrdxrdxrdxrF?????????????????????22112211 ( 18c )33233223223223 ????????????????????????????????? pqqpqqtFs所需錨固力解答 sx Fs1?? ???? 。ΛpP1166。Γkc166。Λp p0166。Β1166。Θ 180。Σppr166。 本方法 ：利用無限楔的彈性應力解答計算錨固力引起滑面正應力，再與無錨固力作用下滑面正應力疊加，經過調整修正后使邊坡滿足極限平衡條件，直接得到已知錨固力的安全系數(shù)，或給定的安全系數(shù)所需的錨固力。 K ara f i at h錨固邊坡穩(wěn)定性計算分析方法 傳統(tǒng)方法 ：將錨固力作為邊坡面外力直接計入所處條塊，再用傳統(tǒng)條分法直接計算安全系數(shù)。 K ara f i at hT k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 yt?vTEE?v(R )PwT(R )E(R )?vPwT?u?(R )u??vPwTEu? k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0?vPwEu? yt(R ) yt yt ytMet h o d 1 1 :Co rreri aMet h o d 1 2 :Ri g o ro u s J an b uMet h o d 1 0 :Sarm a ( III )Met h o d 9 :Sarm a ( II )?vTEE?v(R )PwT(R )E(R )?v(R )PwT(R )E(R )?u?(R )u??(R )?v(R )PwT(R )E(R )u??(R ) k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0?v(R )PwT(R )E(R )u??(R ) yt(R ) yt(R ) yt(R ) yt(R )Met h o d 3 :Si m p l i fi ed J an b uMet h o d 4 :Co rp s o f E n g i n ee rsMet h o d 2 :Si m p l i fi ed Bi s h o pMet h o d 1 :O rd i n aryT k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 yt?vTEE?v(R)PwT(R)E(R)?vPwT?u?(R)u??vPwTEu? k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0?vPwEu? yt(R) yt yt ytMe t h o d 1 1 :Co rreri aMe t h o d 1 2 :Ri g o ro u s J an b uMe t h o d 1 0 :Sarm a ( III )Me t h o d 9 :Sarm a ( II )TE?v?vTE(R ) k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0 2 . 0 1 . 0 0 1 . 0 2 . 0 k Pa 8 0 0 6 0 0 4 0 0 2 0 0 0 2 0 0k N / m 8 0 0 0 6 0 0 0 4 0 0 0 2 0 0 002 0 0 0