第1章工程概 .-4-----------------------章框架荷載計算--- .-37-----章框架內力計算----章荷載效應組合---章框架側移驗算---章延性框架設計---------- .-78-------------第1 設計要師總建筑面積6000~7000??2，5層。需設有綜合（約200??2）1個、多功??2師設計條；底層凈高：4.2??；其它層凈高：3.7??；內粉刷：20mm厚石灰砂漿，噴內墻涂料201:2.5樓面構造：10mm厚水泥砂漿面層20mm1:2.5水泥砂漿找平層粉底（噴涂料

葉列平,趙作周,樊健生.《混凝土結構(下冊)》.,滕智明,朱金銓.《混凝土結構及砌體結構(上冊)》,中國建筑工業羅福午,方鄂華等.《混凝土結構及砌體結構(下冊)》,中國建筑工業社,《建筑結構荷載規范（GB50009-《混凝土結構設計規范（GB50010-《建筑抗震設計規范（GB50011-《建筑結構制圖標準》（GB/T50105-第2 建筑平面設82.8??，寬14.6??，長寬比為5.67:1，考慮到結構層數較低，可略微超過《建筑混5:1筑面積為6044.4??2。縱向設18個開間，每個開間為4.6??。首層設大門與側門兩個主要出。考慮到辦、衛生間、綜合、小會議室、中會議室、多功能廳、咖啡廳、討論區和傳達室等。房間布置如下表所示。建筑平面圖見圖2.1.1到圖2.1.3。圖2.1.12.1.22.1.31一層1一層1一層1一層1三層3二層(1)三層(1)四層3二層(1)三層(1)四層一層(6)二層(7)三層(7)四層(7)五層一層(7)二層(8)三層(8)四層(8)五層一層(2)二層(4)三層(4)四層(4)五層5一層(1)二層(1)三層(1)四層(1)五層建筑立面設立面采用灰色瓷磚裝飾，窗戶均采用1.8??×1.8??推拉窗，有利于采光。結構方案設828??14.6??結合建筑功能要求進行柱網布置，按軸線布置四排柱子，柱間距為分別為：6??,2.6??,6??。由于建筑東端為綜合和多功能廳，為滿足使用空間要求，不列柱子，其中軸線17、19處不設置房屋內部柱（空間使用要求。本結構的柱網布2.3.1第3 材料參????=0.550，????,??????=0.399????/????/????/????/3.0×3.1.2????/??′/????/2.0×2.1×????????=0.45????/????=0.451.43/300=????????=??????????/????=0.5514.3/300=雙向板布置方0.8，在計算與折減板相鄰的板時，將已知配筋支座處雙向板屋蓋設 JS-063.2.13.2.14600????×60004600????×60004600????×26004600????×2600的??1、??2板進行設計。根據《混凝土結構設計規范》(GB50010-2010)9.1.2，雙向板的跨厚比不應大于40，對于??1、??2，雙向板短跨計算跨度4600????，則板不應小于115????，取雙向板板厚?=120????。樓板保護層厚度按照規范要求選取，短邊為20????，長邊為30????。截面有效高度：?0??=120?20=100????,0??=120?30=90每米樓板最小配筋面積：????????=????????×?×1??=252依據《建筑結構荷載規范》(GB50009-2012)5、7、8A確定屋面荷載，屋面板板厚120????，建筑做法如前所述。13.2.20.02×200.10×6.50.12×140.12×2520mm粉底（噴涂料0.02×172—????=市基本雪壓??0=建筑屋面積雪分布系數????=????=0屋面風荷載體型系數????=3????=1.2????+1.4????=1.2×6.37+1.4×0.5=8.34????=1.35????+1.4×0.7×????=1.35×6.37+1.4×0.7×0.5=9.09取設計值較大的最不利荷載組合進行設計，??=9.09????=????+????=6.37+0.5=6.87取該荷載組合進行正常使用極限狀態的撓度、裂縫驗算，????=6.87綜上，屋面荷載的標準值為6.87????/??2，設計值為9.09????/??2B1????=4600?200=4400????=6000?200=5800??=????/????=1.318??=0.576，??=1.8??=B1區格為兩鄰邊簡支，兩鄰邊固支板，短跨支座配筋和長跨支座配筋均與下面計算的B4板????=???????0??????=300×0.95×100×276.7=7.89?????′????=???????0??????=300×0.95×100×276.7=8.76?????′????=(??????2×(3???1)?12???????′?????)?(2??+2??+′9.09×4.42×(3×1.318?1)?12?1.318×7.89?=2×1.318+2×0.576+1.8×=5.00?????(???????= (????2/????=175.4<8@180,????=??????=112.3<8@180,B2????=4600?200=4400????=6000?200=5800??=????/????=1.318??=0.576，??=1.8??=B1區格為一短邊簡支，三邊固定板，????/????<1.50.8。長跨支座配筋與下文板內????/4后截斷。0.8???? 3??????? ×2??+2????+2??+0.8×9.09× 3×1.318? 2×1.318+2×1.318×1.8+2×0.576+0.576×=3.19?????(???????= (????2/????=111.93<8@180,????=??????=64.47<8@180,????=????=??????= 8@180,B3????=2600?200=2400????=4600?200=4400??=????/????=1.83??=0.274，??=1.8??=1.5<????/????<2.0板內????/4后截斷。????=???????0??????=7.1?????′??????×(3???2????=0.9× ?????)?(2??+2????+′9.09×2.42×(3×1.83?=0.9× ?7.1)?(2×1.83+2×1.83×1.8+2×=1.04?????(???????= (????2/????=38.52<8@180,????=??????=13.6<8@180,????=????=??????= 69.26<8@180,B4????=2600?200=2400????=4600?200=4400??=????/????=1.83??=0.274，??=1.8??=板內????/4后截斷。??????×(3???2????=0.9× ?????)?(2??+2????+′9.09×2.42×(3×1.83?=0.9× ?7.1)?(2×1.83+2×1.83×1.8+2×=1.04?????(???????= (????2/????=38.52<8@180,????=????=??????= 69.26<8@180,????=??????=13.60<8@180,????′=????′′=??????=24.32<8@180,根據規范，沿支座配置8@200,251.5????2構造鋼筋，伸出支座????/4； 雙向板樓蓋設 JS-073.2.23.2.44600????×26004600????×26004600????×26004600????×60004600????×2600樓板厚度與屋面板保持一致，?=120????截面有效高度：?0??=120?20=100????,0??=120?30=90每米樓板最小配筋面積：????????=????????×?×1??=252依據《建筑結構荷載規范》(GB50009-2012)5A確定樓面荷載，樓面板板厚120????，建筑做法如前所述。1

3.2.50.01×200.02×200.12×2520mm粉底（噴涂料0.02×172 荷載值????/??2 衛生間、2.5 31衛生間、：????=1.2????+1.4????=8.23????/??2走廊、門廳、樓梯：????=1.2????1.4????=9.63????/??2多功能廳、機房：????=1.2????+1.4????=8.93????/??2會議室：????=1.2????+1.4????=7.53????/??22教室、衛生間、：????=1.35????+1.4×0.7×????=走廊、門廳、樓梯：????=1.35????+1.40.7×????=8.75多功能廳、機房：????=1.35????+1.40.7????=8.26會議室：????=1.35????+1.40.7×????=7.28取設計值較大的最不利荷載組合進行設計，對教室、衛生間、，????2，對會議室，??=7.53衛生間、：????=????+????=3.94+2.50=6.44????/??2走廊、門廳、樓梯：????=????+????=3.943.507.44????/??2會議室：????=????+????=3.94+2.50=7.28????/??2多功能廳、機房：????=????+????=3.943.0=6.94B1????=2600?200=2200????=4600?200=4400??=????/????=1.833??=0.298，??=1.8??=B1板位于走廊區域，??=9.63????/??21.5<????/????<20.9板內????/4后截斷。0.9???? 3??????? ×2??+2????+2??+0.9×9.63× 3×1.833? 2×1.833+2×1.833×1.8+2×0.298+2×0.298×=1.37?????(???????= (????2/????=48.07<8@180,????=??????=15.90<8@180,????=????=??????= 8@180,????′=????′′=??????=8@180,B2????=2600?200=2400????=4600?200=4400??=????/????=1.833??=0.298，??=1.8??=9.63????/??2，1.5<????/????<20.9板內????/4后截斷。????=0,????=???????0??????=300×0.95×90×276.7=7.1????? (3???1)?????=0.9× 2??+2??+9.63× ×3×1.833?1?=0.9 2×1.833+2×0.298+2×1.833×=1.00?????(???????= (????2/????=35.09<8@180,????=??????=11.60<8@180,????=????=??????= 8@180,B3????=2600?200=2400????，????=4600?200=4400??=????/????=1.833??=0.298，??=1.8??=B3板位于走廊區域，??=9.63????/??2，1.5<????/????<20.9。板內????/4后截斷。????=′????=????=???????0??????=300×0.95×90×276.7=7.1????? (3???1)?????′???=0.9× 2??+2??+9.63× ×3×1.833?1?7.1?7.1=0.9 2×1.833+2×0.298+2×1.833×=0.55?????(???????= (????2/????=19.30<8@180,????=??????=6.38<8@180,????=??????=8@180,B4????=4600?200=4400????,????=6000?200=5800??=????/????=1.318??=0.576，??=1.8??=B4板位于衛生間、、會議室、多功能廳、機房等區域，取荷載最大的機房、多功能廳區格計算，??=8.93????/??2????/????<1.50.8板內????/4后截斷。????=???????0??????=300×0.95×90×276.7=7.1?????′ (3???1)?????=0.8× 2??+2????+8.93× ×3×1.318?1?=0.8 2×1.318+2×1.318×1.8+2×=3.32?????(???????= (????2/????=204.2<8@180,????=??????=74.45<8@180,????=????=??????= 8@180,????=??????=′8@180,B5????=4600?200=4400????=6000?200=5800??=????/????=1.318??=0.576，??=1.8??=B5板位于衛生間、、會議室、多功能廳、機房等區域，取荷載最大的機房、多功能廳區格計算，??=8.93????/??2。????/????<1.50.8板內????/4后截斷。????=???????0??????=300×0.95×100×276.7=7.89?????′????=???????0??????=300×0.95×90×276.7=7.1?????′ (3???1)???????′???=0.8× 2??+8.93× ×3×1.318?1?1.318×7.89?7.1=0.8 2×1.318+2×=5.29?????(???????= (????2/????=8@180,????=??????=118.80<8@180,根據規范，沿支座配置8-200,251.5????2構造鋼筋，伸出支座????/4； 單向板布置方柱或墻，取單位板寬??=1000????作為計算單元進行設計計算。板剛度遠小于次梁8時，框架梁作根據《混凝土結構設計規范》(GB50010-2010)9.1.2，單向板的跨厚比不應大于30，最小板厚為60????，偏保守的取板的厚度為80????。框架橫梁截面：??×?=200????×500橫向次梁截面：??×?=200????×300????縱向次梁截面：??×?=200????×500板荷載計3.3.1屋面板/單向板屋蓋設B15.2，B2板長寬比為2.26<3B2B1板，可近似按座處配置足夠構造筋滿足受力要求，詳圖請參見JS-11。3.3.1邊跨：????1=1150?100?100=950????中間跨：????2=1150100100=950按等跨連續板計算，取????=950??????=????????(?????2B?=80????，取截面有效高度?0=60????，??=1000????；最小????????????????1??160按構造配筋時，鋼筋直徑不小于8????200????。根據下列計算屋面板配筋： ?? ,ξ=1?√1?2??,?? ????? ??/?????ξ????/BBC長向支座處，為承擔實際存在的負彎矩，配8@200的負彎矩構造筋，距支座????/4處截斷。嵌固在承重墻內的板，由于墻體實際約束作用，存在一定負彎矩，配置8-200的構造鋼筋，兩邊嵌固在墻內的板角處，在????/458????/4在垂直于受力鋼筋方向布置分布鋼筋6@250單向板樓蓋設B22.26<3B2板位于走廊區域，面積遠小B1板，可近似按單向板計算，盡管對長跨方向偏于激進，但可通過配置一定的分布鋼筋并在長向支座處配置足夠構造筋滿足受力要求，詳圖請參見JS-11。3.3.3B17跨連續梁、2852跨連續梁，B2????1950????????2950????。按等跨連續板計算，取????=950????。荷載統一取走廊荷載設計值，??=8.43??=????????(?????2B?=80????，取截面有效高度?0=60????，??=1000????；最小????????????????1??160 ?? ,ξ=1?√1?2??,?? ????? ??/?????ξ????/BBC長向支座處，配8-200????/4嵌固在承重墻內的板，配置8-200的構造負彎矩鋼筋，伸出墻邊的長度為????/4兩邊嵌固在墻內的板角處，在????/4范圍內雙向布置負彎矩構造筋5 8，距支座????/4處截斷。在垂直于受力鋼筋方向布置分布鋼筋6-250。分別與其他次梁或框架梁組叉梁系，次梁位置參見單向板屋面布置方案及單向板樓面布置方案。其中???1在樓蓋和屋蓋中尺寸相同，有雙向板部分荷載計算可?????1梁支承于框架橫梁及墻上，計算跨度取凈跨??????1：????1=6000100100=5800????，截面200????300????2=2600100100=2400????，截面200????300=4600100100=4400????，截面200????500 200× =200×3003/14400=15.15>A????1=5.8 ????2=2.40????1=5.8D?????2梁支承于主梁及墻上，計算跨度取凈跨??????2：????1=6000100100=5800????，截面200????300????2=2600100100=2400????，截面200????300A????2=2.40B????1=5.8C從屬板面積小于25??2，不必考慮荷載折減，折算荷載與板略有不同。觀察可知，2.94×1.225×0.2×0.2220mm17×0.02×(0.22×2+0.2)3.5×1.21.35×4.85+1.4×0.7×4.21.2×4.85+1.4×4.29.49+1.2×1.32??均布荷載設計值??11.70????/??作用下，作次梁??3??/??????/000??均布荷載設計值??11.70????/??作用下，作次梁??4??=?0.9537.61=?35.73????????????/001次梁截面寬度：??=200翼緣寬度：??′??=min(??0?3??+??????+12?′??)=200+12×80=1160??35????，????????????????????，??????1,??2?0=300?35=265????，????????=285.6????2，????=53.05?′? =????′?′(????)=296.01????? 面計算。根據下列計算正截面配筋：????

2??=??/?????ξ????/ ξ<ξ??=0.550 ??=??/?????ξ????/ ξ<ξ??=0.550 方案比3.4.1長寬板6板板6板63.4.23.4.3向板方案具有更好的經濟性與適用性。故最終采向板屋蓋/樓蓋方案。第4 框架結構布置與計算簡6榀橫向框架進行設計。考慮到結構東側有綜合實驗圖4.1.1梁柱尺寸估梁截面尺 1=

)

=478~860????，?? ????=6.0

4KL2：??=2.6

=(

~1)??=144~260 ????≥41KL3：??=4.6

=(

~1)??=256~460 ????≥4 KL4：??=14.6??，?= )??=811~1460 ????≥4 KL4：??=13.8??，?= )??=766~1380 ????≥4柱截面尺恒載標準值為12??????，活載標準值為2.5??????，估算底層中柱設計軸力：恒載標準值：0.5×(6+8.6)×4.6×12×5=2014.8????/??2活載標準值：0.5(68.64.62.55=419.75軸力設計值：????=1.352014.81.40.7419.75=3131.34????≥

1.2× =350359.720.75×??600600梁柱特征參????按混凝土全截面計算，除底層外，柱的線剛度需要乘以折減0.9，令????=0.9????????應考慮樓板作為受壓翼緣4.2.1截面????/???=(???200×2.08×4.17×2.08×200×2.08×4.17×4.81×,600×1.08×2.16×6.61×????1,????2(2~5層600×1.08×1.94×7.88×4.3荷載計4.3.1(5層0.02×20=0.10×6.5=0.12×14=0.12×25=0.02×17=0.01×20=20mm1:2.50.02×20=0.12×25=0.02×17=4.3.2梁25×0.20×0.38=17×0.02×0.76=25×0.20×0.38=17×0.02×0.76=25×0.20×0.38=17×0.02×0.76=25×0.30×0.58=17×0.02×1.16=25×0.30×0.58=17×0.02×1.16=10×0.18×3.7=17×2×0.02×3.7=2~510×0.18×3.7=17×2×0.02×3.7=10×0.30×3.7=12mm1:320×0.012×3.7=20×0.008×3.7=(17×0.02+0.5)×3.7=2~510×0.30×3.7=12mm1:320×0.012×3.7=20×0.008×3.7=(0.34+0.5)×3.7=柱1~525×0.6×0.6=0.34×4×0.6=圖 1?2??2+????3????3不考慮房屋布置帶來的對南北兩側KL1梁的荷載差異。由于多功能廳和機房布由于KL4和KL5不進行手算，因此不在本節計算其荷載。4.3.42.16+0.744××4.6=0.744×0.5×=0.744×0.4×=2.16+0.625××2.6=0.625×0.5×=0.625×0.4×=2.16+0.625×0.625×0.5×0.625×0.4××2.3+0.891×+0.891×0.5×+0.891×0.4××1.3===2.16+0.744××4.6+9.21=0.744×2.5×=?1-42.16+0.625××2.6=0.625×3.5×=?2.16+0.625×0.625×2.5××2.3+0.891×+0.891×2.5×?×1.3+9.21==4.3.811.97×4.6=0.72×4.6=0.58×4.6=19.34×4.6=1.30×4.6=1.04×4.6=1~423.33×4.6=3.59×4.6=?22.25×4.6=6.49×4.6=? 作相對較小，可忽略不計。因此，計算中僅考慮橫向水平作用。4.3.956.37×14.6×82.8=2.16×14.6×17+4.74×14.6×2+4.74×13.8×2+2.1682.8×2+2.16×69×2=9.82×3.7×72=15.15×2×14.6+15.15×2×82.8=9.21×6×23+9.21×4.6×28=0.5×0.4×14.6×82.8=2~43.94×(14.6×82.8?2×4.6×6)=2.16×14.6×17+4.74×14.6×2+4.74×13.8×2+×82.8×2+2.16×69×2=9.82×3.7×72=15.15×2×14.6+15.15×2×82.8=9.21×6×24+9.21×4.6×34=0.5×(3.0×12+3.5×2.6)×82.8=13.94×(14.6×82.8?2×4.6×6)=2.16×14.6×17+4.74×14.6×2+4.74×13.8×2+×82.8×2+2.16×69×2=9.82×4.9×72=17.81×2×14.6+17.81×2×82.8=10.42×6×23+10.42×4.6×26=0.5×(3.0×12+3.5×2.6)×82.8=0.85×????(???1.58×5.70×11.32×8.91×1.58×1.49×11.32×1.28× 5.16× 5.16× 5.16× 5.16× 5.57× ??1=1.7??0√????=1.7×0.7×√432.11=0.78抗震設防烈度為8度，多遇，水平影響系數最大值????????=0.16。計算相????<??1<5????，處于雙曲線段，結構阻尼比??0.05???2=1+0.08+

0.05?=1.0，??=0.9 =0.3+???? 0.35??1= )??2????????=( ×1.0×0.16= 結構總水平作用：??????=??1??????=0.078×71522.61=5563.39頂部附加作用系數：????=0.08??1+0.01=頂部附加水平作用：?????=??????????=0.072×5563.39=402.79????????(×103?????54321風荷????=（1）基本風壓：建筑位于地區，基本風壓??0=0.45風振系數：由于該建筑外形簡單，高度小于30??，取????=型系數為-0.5，因此取????=0.8+0.5=1.34.3.13標高寬度54321第5 框架內力計算說圖 豎向恒載下內力計圖 5.2.2(????剪力圖 軸力圖5.2.3為了驗證手算結果，使用SAP2000建模計算結構彎矩圖如下5.2.4(????·便，僅選取KL1進行計算如下（為表述簡便，表中值為彎矩絕對值：5.2.1????????????5????1????14????1????13????1??????1??2????1??????1??1????1??????1??對比手算與電算結果可見，KL1梁彎矩值平均相對誤差制大10%以內，可KL2KL2梁的豎向活載下內力計 圖 圖 5.3.2(????剪力圖 軸力圖5.3.35.4作用下內力計55.16×45.16×35.16×25.16×15.57×510-1010104101010101010310101010101021010-101001-0-0 5.4.1模型計算簡圖層數置y(?????(????·(????·(????·5邊---中4邊---中3邊---中2邊---中1邊---中 圖5.4.2D值法計算彎矩圖(????? 圖5.4.3Sap2000計算彎矩圖(?????圖5.4.4剪力圖 圖5.4.5軸力圖便，僅選取KL1進行計算如下（為表述簡便，表中值為彎矩絕對值：5.2.1????????????5????1????14????1????13????1??????1??2????1??????1??1????1??????1??對比手算與電算結果可見，KL1梁彎矩值平均相對誤差制大20%，手算結第6 計算原理及計算簡6.1.1??= ???? = ????，?? ???????=????0，????=????0???

，????=0.8(?????

)，????=0.8(????? ??0 ??0 調幅后檢查是否滿足要求：(????+????)+????≥??， ????≥2 框架內力標準6.2.1(?????(????? 3=126.2.2(?????(?????(?????(?????注：重力荷載3=10.5框架內力組=1.2恒載1.41.35恒載1.40.7活載北側控制=1.2重力荷載+1.3北側荷載南側控制=1.2重力荷載+1.3南側荷6.3.1????(?????????(?????????(?????無可變荷載永久荷載有-6.3.2????(?????????(?????????無可變荷載無永久荷載有-第7 梁柱彎曲變形引起的側慮荷載起控制作用。忽略桿件剪切及軸向變形，利用D值法計算梁柱彎曲變形7.1.155.16×45.16×35.16×25.16×15.57×柱軸向變形引起的側向側抗震變形驗54321第8 抗震等框架梁設寸為200????500????，跨度為2.6??；取保護層厚度??=??′=35?????0=整系數取0.75，受剪時取0.85。混凝土選用C30HRB400HRB3358.2.114.31.43支座處：????????=??????{0.300.65????/????0.3%????,??????=??????????300????2跨中處：??????????????{0.250.55????/????0.25????,????????????????250????2最大配筋率：????????=2.5%,????,??????=???????????=2500????2正截面抗彎配跨中彎矩由無作用時的永久荷載控制：????=85.69???????=200????,?0=465????????

0???????0

85.69×= =0.14<????,??????=14.3×200×ξ=1?√1?2????=1?√1?2×0.14=??=ξ???????0=0.15×14.3×200×465=553.33 3C18，沿全長拉通配置。????=763????2553.33????2ξ=???????? 360× =0.21<??=0.518 14.3×200× 2C18，????′=508.68????2左支座由作用組合控制：????????=138.29???????，????????=?260.86???????；梁端截面采筋截面，下部鋼筋由正彎矩控制，考慮??????=0.75。??

0.75×138.29×=

=670.01

360×(465?3C18，沿全長拉通配置,????=763????2>670.01????2。負彎矩作用下，受壓鋼筋??′=763????2，配置受拉鋼筋：????

???????????????′??(?0??????

=0.75×260.86×106?360×763×430=14.3×200× ξ=1?√1?2????=1?√1?2×0.13=0.13<????=

+

=0.13×14.3×200×465+763=1259.534C22，????=1520????2??=0.13<0.35??′??=763=0.502>0.3 右支座由作用組合控制：????????=135.06???????，????????=?264.09???????；取下部鋼筋沿梁長直通，與跨中相同為：3C18，????=763????2。負彎矩作用下，受壓鋼筋??′=763????2????

???????????????′??(?0??????

=0.75×264.09×106?360×763×430=14.3×200× ξ=1?√1?2????=1?√1?2×0.13=0.14<????=

+

=0.13×14.3×200×465+763=1275.574C22，????=1520????2??=0.14<0.35??′??=763=0.502>0.3 中跨支座由作用組合控制：????????=146.80???????，????????=?189.42??????；梁端截面采筋截面，下部鋼筋由正彎矩控制，考慮??????=0.75??

0.75×146.80×=

=711.24

360×(465?3C18，沿全長拉通配置,????=763????2>711.24????2負彎矩作用下，受壓鋼筋??′=763????2????

???????????????′??(?0??????

=0.75×189.42×106?360×763×430=14.3×200× ξ=1?√1?2????=1?√1?2×0.04=0.04<????=

+

=0.13×14.3×200×465+763=908.974C20，????=1256????2??=0.04<0.35??′??=763=0.602>0.3 跨中彎矩由無作用時的永久荷載控制：????=?20.84???????=200????,?0=465????????

0???????0

20.84×= =0.03<????,??????=14.3×200×ξ=1?√1?2????=1?√1?2×0.03=??=ξ???????0=0.03×14.3×200×465=126.66 2C18直通，????=508.682C16直通，????8.2.3 注：彎矩單位為??????。各梁支座處配筋均滿足??<0.35，??′/??>0.3 斜截面抗剪配計，其余各段梁配筋以列表形式給出。對抗剪驗算，取??????=0.85。????+

138.29+

????=

??+??????=1.2× +30.35×1.2

=188.20 (0.2???? ) ×0.2×1.0×14.3×200×465=312.92????>

?? ????=188.20????，對一般框架梁，??????=

0.85×188.2×103?0.42×1.43×200×1.25×300× =箍筋加密區長度：1.51.5500750????，800????500????。????????=??????(8×280.25×500,100)=100????，??=100??????=1.005>2B8，??????100.5????2，????????????/????0.50%????????

0.85×176.1×103?0.42×1.43×200×1.25×300× =??=200????，??????>0.54??=1082B10，??????157????2，??????=??????/????0.39????????8.2.4????????1

(0.2?????????????0)=312.92????>框架梁配筋構????/3處截斷，兩根通長配置，作為跨中截面構造鋼筋，形成鋼骨架。對500????，仍應設置梁腹縱筋2C14。框架柱設??=??′=35?????0=60035=565????C30混凝土，縱筋選用HRB400級鋼筋，箍筋選用HRB335級鋼筋，基本材料參數見8.2.1總配筋率：????????=5%，????????=0.85%每側配筋率：????????=0.2%,????,??????=???????????=720最小配箍率：??????,??????=??????{????????/??????，0.6%軸壓比驗跨比不大于2時，軸壓比限值為0.7。驗算各框架柱軸壓比??如下表所示：8.3.1??????=是是是是是是是是是是KZ1-1一層邊柱：????=155.59??????，????=368.64??????，??=?1658.61????，?? 97.68????。二層邊柱：????=140.98??????，??1243.50????。∑????=1.5∑????=1.5×260.86=391.3>155.59+140.98=???? ×391.3=

???? ×391.3= ????1.5368.64552.96。根據《混凝土結構設計規范》6.2.20，框架結構現澆樓????=1.0??=4900600????600????????=0.2887?=173.21

=28.82>34?12

??)= =0.5??????=0.5×14.3×360000=2.07>1，取??=1 1243.5× ?? =350.60 =20 =??+??=373.60 1

????

1

4900 ??????=1+1300?? (?

=????=0.7+

??=0.7+0.3

=柱端截面的彎矩設計值：??′??= ??????=1.083×0.840????=0.909????< ???? 柱截面按對稱配筋，??′??=??????=????????=0.80×1243.5×

14.3×600× =0.205<????=??=??+????a=373.60+600?35=638.6 ??′??=???? ??′ ? 0.8×1243.5×103×638.6?14.3