¡¶ÍÁÁ¦Ñ§Óë»ù´¡¹¤³Ì¡·¸´Ï°×ÊÁϺʹð°¸--¼ÆËãÌâÒªµã

·¢²¼Ê±¼ä : ÐÇÆÚÁù ÎÄÕ¡¶ÍÁÁ¦Ñ§Óë»ù´¡¹¤³Ì¡·¸´Ï°×ÊÁϺʹð°¸--¼ÆËãÌâÒªµã¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁ

.

44£®µØ»ùÍÁ³Êˮƽ³É²ã·Ö²¼£¬×ÔÈ»µØÃæÏ·ֱðΪ·ÛÖÊÕ³ÍÁ¡¢Ï¸É°ºÍÖÐÉ°£¬µØÏÂˮλÓÚµÚ

Ò»²ãÍÁµ×Ã棬¸÷²ãÍÁµÄÖضȼûͼ¡£ÊÔ¼ÆËãͼÖÐ1µã¡¢2µãºÍ3µã´¦µÄÊúÏòÓÐЧ×ÔÖØÓ¦Á¦¡££¨8·Ö£©

Ìâ 44 ͼ ½â£º

?cz1'?0

?cz2'?18.8?1.5?28.2kPa

?2'?19.6?10?9.6kN/m3

?cz3'?18.8?1.5?9.6?2.5?52.2kPa

45. ÒÑ֪ijÍÁÑù??18?£¬c=15kPa£¬³ÐÊܵĴóСÖ÷Ó¦Á¦·Ö±ðΪ?1=400kPa£¬?3=180kPa£¬ÊÔ

ÅжϸÃÍÁÑùÊÇ·ñ´ïµ½¼«ÏÞƽºâ״̬£¿£¨8·Ö£© ½â£º

ÒÑÖªÍÁµÄ¿¹¼ôÇ¿¶ÈÖ¸±ê¦Õ¡¢c¼°×îСÖ÷Ó¦Á¦¦Ò3 ¼Ù¶¨ÍÁÌå´¦ÓÚ¼«ÏÞƽºâ״̬£¬¸ù¾Ý¼«ÏÞƽºâÌõ¼þ£º

?1f??3tan2(45???/2)?2ctan(45???/2)

2???? ?180tan(45?18/2)?2?15tan(45?18/2)

?382.3 kPa

ÒòΪ?1?400kPa??1f£¬ËùÒÔÍÁÌå´¦ÔÚÆÆ»µ×´Ì¬¡£

46. ijÖùÏ·½Ðε×Ãæ»ù´¡£¬µ×Ãæ±ß³¤2.8 m£¬»ù´¡ÂñÉî1.8 m£¬Öù×÷ÓÃÔÚ»ù´¡¶¥ÃæµÄÖáÐÄ

.

.

ºÉÔرê×¼×éºÏÖµFk=1250KN,µÚÒ»²ãÍÁΪճÍÁ£¬ºñ¶È3.8 m£¬ÌìÈ»ÖضÈ?=18 kN/m3£»µØ»ù³ÐÔØÁ¦ÌØÕ÷Öµfak=160 kPa,Éî¶ÈÐÞÕýϵÊý¦Çd=1.6£»Õ³ÍÁ²ãÏÂΪÓÙÄà²ã£¬ÌìÈ»ÖضÈ

?=17kN/m3£»µØ»ù³ÐÔØÁ¦ÌØÕ÷Öµf=80 kPa£¬Éî¶ÈÐÞÕýϵÊý¦Ç=1.0£¬È¡µØ»ùѹÁ¦À©É¢

akd½Ç?=22¡ã¡£ÊÔ·Ö±ðÑéËãµØ»ù³ÖÁ¦²ã¼°ÏÂÎÔ²ãÊÇ·ñÂú×ã³ÐÔØÁ¦ÒªÇó¡££¨14·Ö£©

Ìâ 46 ͼ

½â£º

£¨1£© ÑéËã³ÖÁ¦²ã³ÐÔØÁ¦

fa?fak??d?m(d?0.5)?160?1.6?18?(1.8?0.5)?197.4kPa

Pk?Fk?Gk1250?20?2.8?2.8?1.8??195.4kPa?faA2.8?2.8

¹Ê³ÖÁ¦²ã³ÐÔØÁ¦Âú×ãÒªÇó

£¨2£© ÑéËãÈíÈõÏÂÎÔ²ã³ÐÔØÁ¦

ÈíÈõÏÂÎԲ㶥Ã渽¼ÓÓ¦Á¦£º

Pz?bl(Pk??md)

(b?2ztan?)(l?2ztan?)2.8?2.8?(195.4?18?1.8)?65.5kPa,

(2.8?2?2?tan22?)2 ?ÈíÈõÏÂÎԲ㶥Ãæ×ÔÖØÓ¦Á¦£º

pcz??m(d?z)?18?3.8?68.4kPa

ÈíÈõÏÂÎÔ²ã³ÐÔØÁ¦ÌØÕ÷Öµ£º

fa?fak??d?m(d?0.5)?80?1.0?18?(3.8?0.5)?139.4kPa·Ö£©

.

.

ÆäÖÐ?m?18kN/m3

Pcz?Pz?68.4?65.5?133.9kPa?faz?139.4kPa

ÈíÈõÏÂÎÔ²ã³ÐÔØÁ¦ÒàÂú×ãÒªÇó¡£

44£®Óû·µ¶ÇÐÈ¡ÍÁÑù£¬ÆäÌå»ýΪ100cm3£¬ÖÊÁ¿Îª185g£¬ÊÔÑùºæ¸ÉºóÖÊÁ¿Îª145g¡£ÈôÒÑÖª¸ÃÍÁµÄÍÁÁ£±ÈÖØΪ2.73£¬ÊÔÇó¸ÃÍÁµÄÌìÈ»Ãܶȡ¢º¬Ë®Á¿¼°¿×϶±È¡£(Ë®µÄÃܶÈÈ¡1.0g£¯cm3)

£¨6·Ö£©

½â£º

ÌìÈ»ÃÜ¶È ??m?185?1.85g/cm3

V100º¬Ë®Á¿ ??m?ms185?145??100%?27.6% ms145G(1??)2.73(1?0.276)?1??1?0.88 ¿×϶±È e?s?1.8545. .ijµ²ÍÁǽ¸ßΪ5.0m£¬Ç½ºóÌîÍÁ·ÖΪÁ½²ã£¬µÚÒ»²ãΪÖÐÉ°£¬ÆäÌìÈ»ÖضÈ?=19.0kN£¯m3£¬ÄÚĦ²Á½Ç?=14¡ã£»µÚ¶þ²ãΪճÐÔÍÁ£¬ÆäÌìÈ»ÖضÈ?=18.5kN£¯m3£¬ÄÚĦ²Á½Ç?=20¡ã£¬Õ³¾ÛÁ¦c=8kPa£¬ÊÔÓÃÀÊ¿ÏÀíÂÛÇóÖ÷¶¯ÍÁѹÁ¦EaµÄ´óС¡££¨10·Ö£©

½â£º

Ìâ45 ͼ

.

.

14?20?2?)?0.610 Ka2?tan(45?)?0.490 Ka1?tan(45?222?Aµã£º

paA?0

BµãÉϽçÃæ £º paBÉÏ?19?2?0.610?23.2KPa

BµãϽçÃæ £º paBÏÂ?19?2?0.490?2?8?0.7?7.4KPa

Cµã £º paC?(19?2?18.5?3)?0.49?2?8?0.7?34.6KPa Ö÷¶¯ÍÁѹÁ¦ Ea?23.2?2/2?(7.4?34.6)?3/2?86.2KN

46. ij¡°ÀÃβ¥¡±¹¤³Ì£¬¼¸ÄêÇ°½öÊ©¹¤ÖÁÒ»²ã±ãÍ£¹¤£¬ÏÖ×¢Èë×ʽð»Ö¸´½¨É裬µ«Ðè¸Ä±ä¸ÃÂ¥µÄʹÓù¦ÄÜ¡£ÏÖ³¡¿±²ìµÃÖª£¬Ä³Öù²ÉÓøֽî»ìÄýÍÁ¶ÀÁ¢»ù´¡£¬»ùµ×³ß´çΪl¡Áb=4m¡Á3m£¬ÂñÉî×´¿ö¼°µØ»ù³ÖÁ¦²ã¡¢ÏÂÎÔ²ãÇé¿ö¼ûͼ¡£¼ÙÉèÌìÈ»µØÃæÓë¡À0.00µØÃæÆëƽ£¬ÊÔ¸ù¾Ý³ÖÁ¦³ÐÔØÁ¦ÒªÇóºËËã¸Ã»ù´¡ÄܳÐÊܶà´óµÄÊúÏòÁ¦F£¬²¢¾Ý´ËÑéËãÏÂÎÔ²ã³ÐÔØÁ¦ÊÇ·ñÂú×ãÒªÇó¡££¨14·Ö)

½â£º

£¨1£©È·¶¨³ÖÁ¦²ã³ÐÔØÁ¦£ºb=3 m£¬ÎÞÐë¿í¶ÈÐÞÕý

¸ù¾Ýp?fa£¬»ù´¡ÄܳÐÊÜ×î´óµÄÊúÏòÁ¦F

Ìâ46 ͼ

fa?fak??d?m(d?0.5)?160?1.6?18?(1.8?0.5)?197.4kPa F?(fa??Gd)bl?(197.4?20?1.8)?3?4?1936.8kN

£¨2£©ÑéËãÈíÈõÏÂÎÔ²ã³ÐÔØÁ¦

µ±F?1936.8kNʱ£¬×÷ÓÃÔÚ»ùµ×ÉÏѹÁ¦ÎªµØ»ù³ÐÔØÁ¦£¬ ÔòÈíÈõÏÂÎԲ㶥Ã渽¼ÓÓ¦Á¦£º

bl(p??md)

(b?2ztan?)(l?2ztan?)3?4?(197.4?18?1.8)?76.4kPa, ?(3?2?2tan22?)(4?2?2tan22?)pz?ÈíÈõÏÂÎԲ㶥Ãæ×ÔÖØÓ¦Á¦£º

pcz??m(d?z)?18?3.8?68.4kPa

ÈíÈõÏÂÎÔ²ã³ÐÔØÁ¦ÌØÕ÷Öµ£º

fa?fak??d?m(d?0.5)?80?1.0?18?(3.8?0.5)?139.4kPa.

ÁªÏµºÏͬ·¶ÎÄ¿Í·þ£ºxxxxx#qq.com(#Ì滻Ϊ@)