電大《混凝土結(jié)構(gòu)設(shè)計原理》形成性考核冊答案

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁眉頁腳。本資料屬于網(wǎng)絡整理，如有侵權(quán)，請聯(lián)系刪除，謝謝！《混凝土結(jié)構(gòu)設(shè)計原理》形成性考核冊作業(yè)1、2、3、4參考答案11至3ADAD5C）D）ABCDBABCD8AABCD1/15×）2）．×）．×））8（×）10（2/15R24至5AICICx2a'C3/15D1、√）×）））√）×）?bb4/15（bh250mm600mmMmf300N/mm2，f14.3N/mm2，f1.43N/mm2，1.0，0.5501bycth60035565mm02M22161061.014.3xhh1)1)119.6mmfbh20021c00.550310.75mmhb01.014.3250119.6fbx1A1425mm2cf300sy1.43f0.450.453000.2145%0.2%min0.2145%tfyA0.95%0.2145%smm2)3（As5/15bh250mm600mm①；②①查教材附錄知：對于一類環(huán)境，可取梁的保護層厚度c，級鋼筋ffy360N/2f11.9N/mm2，Amm2。'ycsfAy3601520184mmxsfb111.92501chha20sxMfbx(h)2u1c01841.011.9250184(564)22.5810Nmm82.58108Nmm。9.6N/mm2。②fcfA3601520mm19.6250xyfbs1cxMfbx(h)2u1c022821.09.6250228(564)2.4610Nmm82.46108Nmm。bh250mm550mm4f300N/y，f210N/yv11.9N/mmf1.27N/mmt1.0，，22，f2，2cchhha55035515mmwos6/15h5152502.064wb0.25fbh0.251.011.9250515NkNV250kNccomaxV0.7fbh0.71.27250515114458N114.5kNV250kNctomax(2501031.25210515AVV1.0023svcs1.25fhyv0（Ammn2，1250.3As1100.37S250取s10050.321.27210Abs2501000.24f0.240.145%sv0.402%tfsvsv,minyvqbhl5mn8q。f210N/yv，f14.3N/mm，f1.43N/mm，1.0，222ctchhha55035515mmwosqnAV0.7fbh1.25fho1stoyv250.30.71.432005151.2521010310367999.3171102.3N5152001V171102.3N2nq171102.32/568440.9N68.4437/156至9CBDACA7（A8/151級、（2√））√）bb×）×））√）√）√）9/15hxb02x2asAsAsAsA及AssNNAAss410/15答：odNfN/2300/Nmmy2fclH5600mm0，(1692000l/b5600/40014，查附表1N0sfAc1692mm2A0.92f300，得y1692A1.06%0.6%,3%s而A4004002bhN，11/15300N/mm2f。yN234000As780mm2f300y4（A804mm2s4至C和DBCA（B12/15（）））)）9Vcs)））13/15614/151C25f11.5MPa，f280MPa，0.56，1.00cdsdba40hh4055040510mm0。sx2Mxhh200fbd0cd21.016010611.52005105102162.2mmh0.56510