梁式樓梯計(jì)算書(shū)項(xiàng)目名稱_____________構(gòu)件編號(hào)_____________日期_____________設(shè)計(jì)_____________校對(duì)_____________審核_____________計(jì)算時(shí)間2010年5月29日（星期六）8:19一、設(shè)計(jì)示意圖二、基本資料 1．設(shè)計(jì)規(guī)范《建筑結(jié)構(gòu)荷載規(guī)范》（GB50009—2001）（2006年版）《混凝土結(jié)構(gòu)設(shè)計(jì)規(guī)范》（GB50010—2002） 2．幾何參數(shù)樓梯類型:梁式A型樓梯(/)約束條件:兩端簡(jiǎn)支斜梯段水平投影長(zhǎng)度:L1=5100mm梯段凈跨:Ln=L1=5100mm樓梯高度:H=3000mm 樓梯寬度:W=1400mm踏步高度:h=166mm 踏步寬度:b=300mm樓梯級(jí)數(shù):n=18（級(jí)）梯段板厚:C=130mm上部平臺(tái)梯梁寬度:b1=300mm下部平臺(tái)梯梁寬度:b2=250mm樓梯梁寬度:b3=200mm樓梯梁高度:h3=350mm 3．荷載參數(shù)樓梯混凝土容重:b=26.00kN/m3樓梯欄桿自重:qf=0.50kN/m樓梯均布活荷載標(biāo)準(zhǔn)值:q=3.50kN/m2可變荷載組合值系數(shù):c=0.70可變荷載準(zhǔn)永久值系數(shù):q=0.50 4．材料參數(shù)混凝土強(qiáng)度等級(jí):C30混凝土抗壓強(qiáng)度設(shè)計(jì)值:fc=14.30N/mm2混凝土抗拉強(qiáng)度標(biāo)準(zhǔn)值:ftk=2.01N/mm2混凝土抗拉強(qiáng)度設(shè)計(jì)值:ft=1.43N/mm2混凝土彈性模量:Ec=3.00×104N/mm2主筋強(qiáng)度等級(jí):HRB335(20MnSi) fy=300.00N/mm2主筋彈性模量:Es=200000N/mm2箍筋強(qiáng)度等級(jí):HPB235(Q235) fyv=210.00N/mm2箍筋彈性模量:Es=210000N/mm2受拉縱筋合力點(diǎn)到梯梁底邊的距離:as=35mm三、荷載計(jì)算過(guò)程 1．樓梯幾何參數(shù)梯段板與水平方向夾角余弦值:cos=EQ\F(b,\R(b2+h2))=\F(300,\R(3002+1662))=0.87梯梁的計(jì)算跨度:L0=Min{Ln+(b1+b2)/2,1.05Ln}=Min{5100+(300+ 250)/2,1.05×5100}=Min{5375,5355}=5355mm梯段板的凈寬度:B=W-2b3=1400-2×200=1000mm梯段板的計(jì)算跨度:B0=Min{B+b3,1.05B}={1000+200,1.05×1000} =Min{1200,1050}=1050mm梯段板折算到水平投影后的計(jì)算厚度:T=(h+2×C/cos)/2=(166+2×130/0.87)/2=232mm 2．荷載設(shè)計(jì)值 2.1均布恒載標(biāo)準(zhǔn)值 2.1.1L1段梯板自重gk1'=b×T/1000=26.00×232/1000=6.02kN/m2gk1=gk1'×L1/Ln=6.02×5100/5100=6.02kN/m2 2.1.2L1段梯板面層自重gk3'=c1×C2×(H+L1)/L1/1000=26.00×0×(3000+5100)/5100/1000 =0.00kN/m2gk3=gk3'×L1/Ln=0.00×5100/5100=0.00kN/m2梯段斜板恒荷載標(biāo)準(zhǔn)值:gkx=gk1'+gk3'=6.02+0.00=6.02kN/m2梯板折算永久荷載標(biāo)準(zhǔn)值:gkb=gk1+gk3=6.02+0.00=6.02kN/m2考慮到面層重量等對(duì)梯梁的作用,梯梁自重放大系數(shù)b取1.1梯梁自重的標(biāo)準(zhǔn)值:gkL=b×b×b3×h3×[Ln+L1×(1/cos-1)]/Ln/1000000 =1.1×26.00×200×350×[5100+5100×(1/0.87-1)]/5100/1000000 =2.29kN/m梯梁上永久荷載標(biāo)準(zhǔn)值:gk=gkb×B/2000+gkL+qf=6.02×1000/2000+2.29+0.50=5.80kN/m 2.2均布荷載設(shè)計(jì)值由活荷載控制的梯段斜板荷載設(shè)計(jì)值:pxL=1.2gkx+1.4q=1.2×6.02+1.4×3.50=12.13kN/m2由恒荷載控制的梯段斜板荷載設(shè)計(jì)值:pxD=1.35gkx+1.4cq=1.35×6.02+1.4×0.70×3.50=11.56kN/m2最不利的梯段斜板荷載設(shè)計(jì)值:px=Max{pxL,pxD}=Max{12.13,11.56}=12.13kN/m2由活荷載控制的梯梁荷載設(shè)計(jì)值:pL=1.2gk+1.4qW/2/1000 =1.2×5.80+1.4×3.50×1400/2/1000=10.39kN/m由恒荷載控制的梯梁荷載設(shè)計(jì)值:pD=1.35gk+1.4cqW/2/1000 =1.35×5.80+1.4×0.70×3.50×1400/2/1000=10.23kN/m最不利的梯梁荷載設(shè)計(jì)值:p=Max{pL,pD}=Max{10.39,10.23}=10.39kN/m四、正截面承載能力計(jì)算 1．梯段斜板計(jì)算梯段斜板按簡(jiǎn)支計(jì)算,其最大彎矩在跨中:Mmax=EQ\F(1,8)pxB\o(0,2)=\F(1,8)×12.13×10502×10-6=1.67kN·m/mh0=T-25=232-25=207mm 1)相對(duì)界限受壓區(qū)高度bcu=0.0033-(fcu,k-50)×10-5=0.0033-(30-50)×10-5=0.0035>0.0033取cu=0.0033按規(guī)范公式(7.1.4-1)b=EQ\F(1,1+\F(fy,Escu))=EQ\F(0.80,1+\F(210,2.10×105×0.00330))=0.61 2)受壓區(qū)高度x按規(guī)范公式(7.2.1-1),EQA\o(s,')=0,EQA\o(p,')=0MEQ=1fcbx\b(h0-\F(x,2))EQx=h0-\r(,h\o(0,2)-\F(2M,1fcb))EQ=207-\r(,2072-\F(2×1.67×106,1.00×14.30×1000))=0.57mm <bh0=0.61×207=127.09mm,按計(jì)算不需要配置受壓鋼筋 3)受拉鋼筋截面積As按規(guī)范公式(7.2.1-2)1fcbx=fyAs得As=EQ\F(1fcbx,fy)=\F(1.00×14.30×1000×0.57,210.00)=38mm2梯段斜板實(shí)際配置鋼筋為A12@150每米板寬內(nèi)配筋面積As=754mm2 4)驗(yàn)算配筋率EQ=\F(As,bh)=\F(754,1000×232)×100%=0.32%<max=2.50%不超筋EQ\b\rc\}(\a\al\co1\vs3\hs3(min=0.2%,min=0.45×1.43/210.00=0.306%))min=0.306%<滿足最小配筋率要求 2．梯梁配筋計(jì)算h0=h3-as=350-35=315mm跨中最大彎矩,Mmax=EQ\F(1,8)pL\o(0,2)=\F(1,8)×10.39×53552×10-6=37.24kN·m 1)相對(duì)界限受壓區(qū)高度bcu=0.0033-(fcu,k-50)×10-5=0.0033-(30-50)×10-5=0.0035>0.0033取cu=0.0033按規(guī)范公式(7.1.4-1)b=EQ\F(1,1+\F(fy,Escu))=EQ\F(0.80,1+\F(300,2.00×105×0.00330))=0.55 2)受壓區(qū)區(qū)高度x按規(guī)范公式(77.2.11-1),EQA\o(s,')=00,EQA\o(p,')=00MEQ=1fcbx\b(h00-\\F(x,2)))EQx=h0--\r((,h\o(0,2)-\F(22M,1fcb))EQ=315-\r(,3152-\\F(2×377.24×1006,1.000×14.330×200)))=44.447mm <bh0==0.555×315=1773.255mm,,按計(jì)算算不需要配配置受壓鋼鋼筋 3)受拉鋼鋼筋截面積積As按規(guī)范公式(77.2.11-2)1fcbx=fyyAs得As=EQ\F(1fcbbx,fy)=\F(11.00×144.30×2000×44.447,3000.000)=424mm2梯梁中間截面實(shí)實(shí)際配置受受拉鋼筋為為3B16,實(shí)際配筋筋面積為6603mmm2 4)驗(yàn)算配配筋率EQ=\F(AAs,bh)==\F((603,,200×3550)××1000%=0.866%<max==2.550%不超筋筋EQ\b\rc\}(\a\al\co1\vs3\hs3(min=00.2%,,min==0.445×1.433/3300.000=0.2114%)))minn=00.2144%<滿足最小配筋率率要求五、斜截面承載載能力計(jì)算算 1．箍筋V=0.5ppL0cos==0.55×110.399×55355/10000××0.887=24.33kN 1)復(fù)核截截面條件按規(guī)范公式(77.5.11) 0.25cfccbh0=00.25×1..00××14..30××2000×3315==2255.2××1033NV=24.33kN<2225.2kN,截面尺寸寸滿足要求求 2)驗(yàn)算構(gòu)構(gòu)造配筋條條件按規(guī)范公式(77.5.77-1) 0.7ftbbh0=00.7××1.443×200×3115=63.11×1103N>>V=224.3kkN按構(gòu)造配置箍筋筋實(shí)際配置雙肢箍箍筋A(yù)8@2000EQ\b\bc(\f(Asv,s))=\\f(2×500.27,,200))=00.50227)六、跨中撓度驗(yàn)驗(yàn)算 1．荷載效應(yīng)的標(biāo)準(zhǔn)準(zhǔn)組合值pk=gk++0.55qW/10000=55.80+0..5×3.500×11400/10000==8.225kNN/mMk=EQ\F(1,8)pkL\\o(0,2)=\F(11,8)×8..25××535552×100-6=29.557kNN·m 2．荷載效應(yīng)的準(zhǔn)永永久組合值值pq=gk++0.55qqW/10000=55.80+0..5×0.500×33.50×14400//10000=7.022kN//mMq=EQ\F(1,8)pqL\\o(0,2)=\F(11,8)×7..02××535552×100-6=25.118kNN·m 3．撓度驗(yàn)算 1)裂縫間縱縱向受拉鋼鋼筋應(yīng)變不不均勻系數(shù)數(shù):由規(guī)范公式（88.1.33-3）,縱向受拉拉鋼筋的應(yīng)應(yīng)力:sk=EQ\F(Mkk,0.887h0As)=EQ\F(29566838.33,0.87×315×603.19)=1178.886N//mm2對(duì)矩形截面受彎彎構(gòu)件,有效受拉拉混凝土截截面面積::Ate=0..5bh=0.5×2000×350=355000mm2按規(guī)范公式（88.1.22-4）計(jì)算縱縱向鋼筋配配筋率:te=EQ\F(Ass,Ate)==\F((603,,350000)==17..23××10--3混凝土抗拉強(qiáng)度度標(biāo)準(zhǔn)值::ftk=2.011N/mmm2按規(guī)范公式（88.1.22-2）=1.1-0..65EQ\F(ftk,tesk)==1.11-00.65×\FF(2.001×10000,17..23400×1178.886)==0.668 2)鋼筋彈性性模量和混混凝土彈性性模量的比比值:EE=EQ\F(Es,EEc)=\F(22000000,300000))=66.67 3)受壓翼緣緣面積與腹腹板有效面面積的比值值:f'對(duì)于矩形截面,,f'=0.000 4)縱向受拉拉鋼筋配筋筋率:=EQ\F(As,bbh0)=\F(6603,2200××3155)=0.00095744 5)受彎構(gòu)件件的短期剛剛度:Bs由規(guī)范公式（88.2.33-1）,Bs=EQ\F(EsAsh\o(0,2),1..15++0.22+\\F(6EE,1++3.55f')) =EQ\F(200000×603×3152,1.115×0.688+00.2++\F((6×6.677×00.0099574,,1+3.5×0..00)))×110-9=87798.006kNN·m2 6)考慮荷載載長(zhǎng)期效應(yīng)應(yīng)組合對(duì)撓撓度增大的的影響系數(shù)數(shù):根據(jù)混凝土結(jié)構(gòu)構(gòu)設(shè)計(jì)規(guī)范范8.2..5條規(guī)定定:'=0,取=22.00 7)受彎構(gòu)件件的長(zhǎng)期剛剛度:B根據(jù)規(guī)范公式（8.2.2）,可得B=EQ\F(Mk,Mq(-1)+Mk)Bs =EQ\F(29.57,25.18×(2.00-1)+29.57)×8798.06=47751.889kNN·m2 8)跨中撓度度:ff=EQ\F(5,384)\F(pkL\\o(0,4),Bcosscoss)=EQ\F(5,384)×\F(8.25×53554,47551.899×00.8722)×10-99=244.28mm 9)容許撓度度:EQLx=\F(LL0,coss)=\F(55355,,0.877)=61200mm因計(jì)算跨度Lxx小于70000mm,,所以容容許撓度[[L0]=EQ\F(Lx,2000)=\F(66120,,200))=30..60mmm跨中最大撓度小小于容許撓撓度,滿足要求求。七、裂縫寬度驗(yàn)驗(yàn)算 1．荷載效應(yīng)的標(biāo)準(zhǔn)準(zhǔn)組合值pk=gk++0.55qW/10000=55.80+0..5×3.500×11400/10000==8.225kNN/mMk=EQ\F(1,8)pkL\\o(0,2)=\F(11,8)×8..25××535552×100-6=29.557kNN·m 2．裂縫寬度驗(yàn)算 1)裂縫間縱縱向受拉鋼鋼筋應(yīng)變不不均勻系數(shù)數(shù):由規(guī)范公式（88.1.33-3）,縱向受拉拉鋼筋的應(yīng)應(yīng)力:sk=EQ\F(Mkk,0.887h0As)=EQ\F(29566838.33,0.87×315×603.19)=1178.886N//mm2對(duì)矩形截面受彎彎構(gòu)件,有效受拉拉混凝土截截面面積::Ate=0..5bh=0.5×2000×350=355000mm2按規(guī)范公式（88