單擊此處編輯母版標(biāo)題樣式,,單擊此處編輯母版文本樣式,,第二級(jí),,第三級(jí),,第四級(jí),,第五級(jí),,*,,*,,第二章,,、判斷題,,1.瞬變體系的計(jì)算自由度一定等零,,2有多余約束的體系一定是幾何不變體系,,3.三個(gè)剛片用不在同一條直線上的三個(gè)虛鉸兩兩相連,,,則組成的體系是無多余約束的幾何不變體系。(,,,,第二章,,選擇填空,,體系的計(jì)算自由度W是保證體系為幾何不變的,,A條件,,A必要B充分C非必要D.必要和充分,,2.圖示結(jié)構(gòu)為了受力需要一共設(shè)置了五個(gè)支座鏈桿,對(duì),,于保持其幾何不變來說有個(gè)多余約束,其中第個(gè)鏈,,桿是必要約束,不能由其他約束來代替。,,,,第二章,,3圖a屬幾何A體系。,,A不變,無多余約束B不變,有多余約束,,C可變,無多余約束D可變,有多余約束,,圖b屬幾何B體系,,A不變,無多余約束B不變,有多余約束,,C可變,無多余約束D可變,有多余約束,,,,第三章,,1靜定結(jié)構(gòu)的內(nèi)力與桿件的截面尺寸和材料性質(zhì)有關(guān)。x,,2.圖示剛架AD中截面C的彎矩等于(,,A.q2(左拉)B.q(右拉),,C.qa2(左拉)D.qa2(右拉),,,,第三章,,靜定結(jié)構(gòu)在支座移動(dòng)時(shí),會(huì)產(chǎn)生:C,,A.內(nèi)力;,,B.應(yīng)力,,C.剛體位移;,,D.變形。,,2.比較圖a、圖b際示兩種情況:其內(nèi)力,,B,,支座水平位移,,A,,A.相同,不等B.不相同,不等,,C.相同,相等D.不相同,相等,,,,第三章,,3圖a所示結(jié)構(gòu)彎矩圖形狀正確的是,,(A),,,,第五章,,例5-1試用結(jié)點(diǎn)法求三角形桁架各桿軸力。,,10 kN,,OkN,,10kN,,5kN,,5KN,,2 mmx 4=8 m,,20 kN,,20 KNT,,解:(1)求支座反力,,F=0,,F=20 KN,,FR=20kN (T),,(2)依次截取結(jié)點(diǎn)A,G,E,C,畫出受力圖,,,由平衡條件求其未知軸力,,,,第五章,,10 kN,,10 kN,,10KN,,5kN,,5kN,,SkN,,FNAE,,m,,A,,20 KN,,2m×4=8m,,20 kN,,20 kN,,取A點(diǎn)為隔離體,由,,∑X=0 Fu cos a+FG=0,,∑Y=020kN-5kN+F,,NAE SIn a=0,,有,,FRg=-15kNx√5=-155kN(壓),,所以,,NG=- fnaecosa=(-155)×=30kN(拉),,,,第五章,,10 KN,,10kN,,I0 KN,,5kN,,5 kN,,m,,2m×4=8m,,20 kN,,20 KN,,G ENGD,,取G點(diǎn)為隔離體,,∑X=0,,NGD,,30kN(拉),,∑Y=0,,NGE,,0,,,,第五章,,10kN,,10 kN,,5kN,,5kN,,m,,2m4=8m,,20 KN,,20kN1,,取E點(diǎn)為隔離體,由,,Ex=0 FNEC COS a+ FNED COS &-FNEA COS a=0,,NEC2NO=-15√5kN,,F,,2Y=0 FNEC Sin a-FNED Sin a-FNEA Sin a-10kN=0,,NED,,聯(lián)立解出Fc=-105kN(壓),,F=-5√5kN(壓),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,