【文章內(nèi)容簡介】 載力計算的基本規(guī)定 Relative Compression Depth eyec uxnbh00hxycucunb ??? eee 0hx bb ?? ycucubeebe???Relative pression depth is only related to the material behaviors. 0hx nbb?ycucueebe??scuyEfeb??1第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 三、相對界限受壓區(qū)高度 eyec uxnbh00hxycucunb ??? eee 0hx bb ?? ycucubeebe???相對界限受壓區(qū)高度僅與材料性能有關(guān)，而與截面尺寸無關(guān) 0hx nbb?ycucueebe??scuyEfeb??1第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 表 5 3 相對界限受壓區(qū)高度 ?b和 as , m a x混凝土強度等級 ≤ C5 0 C6 0 C7 0 C8 0?b0 .5 5 0 0 .5 3 1 0 .5 1 2 0 .4 9 3HRB3 3 5鋼筋as ,m a x0 .3 9 9 0 .3 9 0 0 .3 8 1 0 .3 7 2?b0 .5 1 8 0 .4 9 9 0 .4 8 1 0 .4 6 2HRB4 0 0鋼筋as ,m a x0 .3 8 4 0 .3 7 5 0 .3 6 5 0 .3 5 6對配置無明顯屈服點鋼筋的截面，其界限相對受壓區(qū)高度 ?b=？ 第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 表 5 3 相對界限受壓區(qū)高度 ?b和 as , m a x混凝土強度等級 ≤ C5 0 C6 0 C7 0 C8 0?b0 .5 5 0 0 .5 3 1 0 .5 1 2 0 .4 9 3HRB3 3 5鋼筋as ,m a x0 .3 9 9 0 .3 9 0 0 .3 8 1 0 .3 7 2?b0 .5 1 8 0 .4 9 9 0 .4 8 1 0 .4 6 2HRB4 0 0鋼筋as ,m a x0 .3 8 4 0 .3 7 5 0 .3 6 5 0 .3 5 6When the reinforcement in the section has no obvious yield point, what about balanced relative pression depth? ?b=？ Relative Compression Depth Table 第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 達到界限破壞時的受彎承載力為適筋梁 Mu的上限 )(20m a x, bbcu bhfM ??a ??)(m a x, bbs ??a ??ycbb ffa??? ??m a xm a x?? ?★ 適筋梁的判別條件 這幾個判別條件是等價的 本質(zhì)是 b?? ?20m a x, bhf cs aa ??20m a x,m a x, bhfMM csu aa ???b?? ?m a x,20/ scs bhfM aaa ??第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 The upper limit of the flexural capacity Mu of the proper reinforced beams )(20m a x, bbcu bhfM ??a ??)(m a x, bbs ??a ??ycbb ffa??? ??m a xm a x?? ?★ Criterion of proper reinforced beams These criterions are equivalent The nature is b?? ?20m a x, bhf cs aa ??20m a x,m a x, bhfMM csu aa ???b?? ?m a x,20/ scs bhfM aaa ??第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 Minimum Reinforcement Ratio Mcr=Mu ftsc = Ececetuech/4h/32247342bhfhhhbfM tktkcr ??????? ?? )( 0 xhAfM syku ??)(20 ?? ?? bhf ykC = a fcbxTs= ssAsM a fcx = b xnyktksffbhA m i n ???approximately ? = h= 第五章 受彎構(gòu)件 正截面受彎承載力計算的基本規(guī)定 四、最小配筋率 Minimum Reinforcement Ratio Mcr=Mu ftsc = Ececetuech/4h/32247342bhfhhhbfM tktkcr ??????? ?? )( 0 xhAfM syku ??)(20 ?? ?? bhf ykC = a fcbxTs= ssAsM a fcx = b xnyktksffbhA m i n ???近似取 ? = h= 正截面受彎承載力計算的基本規(guī)定 yktksffbhA m i n ???ytsffbhA i n ???第五章 受彎構(gòu)件 ftk /fyk=◆ 同時不應(yīng)小 于 % ◆ 對于現(xiàn)澆板和基礎(chǔ)底板沿每個方向受拉鋼筋的最小配筋率不應(yīng)小于 %。 正截面受彎承載力計算的基本規(guī)定 yktksffbhA m i n ???ytsffbhA i n ???第五章 受彎構(gòu)件 ftk /fyk=◆ In addition,it is no less than % ◆ For the castinsite slabs and foundation slabs, the minimum tensile reinforcement of either direction should not less than %. 第五章 受彎構(gòu)件 正截面受彎承載力計算 Flexural Analysis and Design of Beams Singly Reinforced Section ◆ Basic Formulae )2()2( 00xhAfxhbxfMMAfbxfsycusyc??????aa0020200)( )(hAfhAfbhfbhfMMAfhbfssysycscusyc??aa??a?a???????????C = a fcbxTs= ssAsM a fcx = b xn第五章 受彎構(gòu)件 正截面受彎承載力計算 受彎構(gòu)件正截面承載力計算 Flexural Analysis and Design of Beams 一、單筋矩形截面 Singly Reinforced Section ◆ 基本公式 Basic Formulae )2()2( 00xhAfxhbxfMMAfbxfsycusyc??????aa0020200)( )(hAfhAfbhfbhfMMAfhbfssysycscusyc??aa??a?a???????????C = a fcbxTs= ssAsM a fcx = b xn第五章 受彎構(gòu)件 正截面受彎承載力計算 ◆ Application Conditions m a x,20m a x,m a x,m a x00 sscsuycbsbbbhfMMffbhAhxaaaaa???????????????或或To avoid brittle fracture due to over reinforcement. To avoid brittle fracture due to under reinforcement. bhA s m in??第五章 受彎構(gòu)件 正截面受彎承載力計算 ◆ 適用條件 (Application Conditions ) m a x,20m a x,m a x,m a x00 sscsuycbsbbbhfMMffbhAhxaaaaa???????????????或或防止超筋脆性破壞 To avoid brittle fracture due to over reinforcement. 防止少筋脆性破壞 To avoid brittle fracture due to under reinforcement. bhA s m in??第五章 受彎構(gòu)件 正截面受彎承載力計算 ★ Checking Calculation Given that: section size b， h(h0), reinforcement As and the material strength fy, fc To determine： flexural capacity MuM Indeterminate： pression depth x and flexural capacity Mu Basic formulae： )2()2( 00xhAfxhbxfMAfbxfsycusyc?????aax≥?bh0時， Mu=？ 20m a x,m a x, bhfM csu aa ??As?minbh， ？ It is usually happened when the constructing quality is not good enough and the strength of concrete does not reach the predicted value. 第五章 受彎構(gòu)件 正截面受彎承載力計算 ★ 截面 復(fù)核 (Checking Calculation) 已知： 截面尺寸 b， h(h0)、截面配筋 As，以及材料強度 fy、 fc 求： 截面的受彎承載力 MuM 未知數(shù)： 受壓區(qū)高度 x和受彎承載力 Mu 基本公式： )2()2( 00xhAfxhbxfMAfbxfsycusyc?????aax≥?bh0時， Mu=？ 20m a x,m a x, bhfM csu aa ??As?minbh， ？ 這種情況在施工質(zhì)量出現(xiàn)問題，混凝土沒有達到設(shè)計強度時會產(chǎn)生。 第五章 受彎構(gòu)件 正截面受彎承載力計算 ★ Design of Beam Section Given that: bending moment M To determine： section size b， h(h0), reinforcement As and the material strength fy、 fc Indeterminate： pression depth x, b, h(h0), As, fy and fc Basic formulae： A total of two No unique solution