1、HRB400柱(梁)式门力I箍筋级别 : HPB300输出柱(梁)式门框的详细计算过程 : 是、计算依据:门框侧墙L一I I门框上临空墙人防门洞口 门框alB a2 F混凝土结构设计规范 GB50010 -2010人民防空地下室设计规范 GB50038 -2005建筑结构静力计算实用手册浙江大学编写,中国 建筑工业出版社出版三、计算结果:普通门框墙梁式门框墙荷载导算过程:单个门扇宽度:a = B + 2 Ma= 1200 + 2 100 = 1400 mm门扇高度:b = H + 2 Ma = 2000 + 2 100 = 2200 mm门扇宽度度与高度之比: a / b = 0.636,查人
2、防规范表475-2:上下门框反力系数: a = 0.3664左右门框反力系数: b = 0.4327门扇传给上下门框的等效静荷载标准值 :qa =工xqe1 xa / 1000门扇传给左右门框的等效静荷载标准值=0.3664 200.0:qb = % xqe1 xa / 1000=0.4327 200.0仅00 / 1000 = 102.58 kN/m仅00 / 1000 = 121.16 kN/m左侧门框墙的计算过程:左侧门框墙可以按悬臂墙来计算。L1 = L - da / 3 = 400 - 100 / 3 = 367 mmL2 = L - da = 400 - 100 = 300 mmM
3、 = qb XL1 / 1000 + qe1 XL22 / (2 1册=121.16 367 / 1000 + 200 300% (2 1(06) = 53.4 kN.mV = qb + qe1 XL2 = 121.16 + 200 30(X= 181.2 kN左侧门框墙钢筋计算面积:左侧门框墙配筋左侧门框墙配筋面积混凝土轴心受压强度设计值:混凝土材料综合调整系数:453 mm2C8100503 mm2ft = 1.57 N/mm2 d = 1.50调整后的混凝土轴心受压强度设计值 :ftd = ft Xld X0.8 = 1.57截面有效高度:H0 = H - c - 10 = 300 -
4、15 - 10 = 275 mm:h = 11.50 0s = 1.88 N/mm2截面抗剪承载力:R = 0.7 梳ftd xhV H / 2,门框上档墙采用梁式门框。L1 = b1 - da = 2050 - 100 = 1950 mmL2 = L - b1 = 2050 - 2050 = 0 mm梁式门框跨度:梁式门框线荷载Lb = 2100 mmq = qa + qe1 凡1 / 1000 + qe2 MinL2,Lb / (21000)=102.58 + 200 1950 / 1000 + 200 Min (0,2100 / (2 1000)=492.58 kN/mV = q XLb
5、 / (2 1000) = 492.5821*02 / (12 1峡)=181.0 kN.m2100 / (2 1000) = 517.2 kN以下是柱(梁)式门框正截面受弯的详细计算过程: 混凝土轴心受压强度设计值: fc = 16.7 N/mm2混凝土轴心受拉强度设计值: ft= 1.57 N/mm2混凝土轴心受压强度标准值: fck = 23.4 N/mm2钢筋抗拉强度设计值: fy = 360 N/mm2钢筋受压强度设计值: fy = 360 N/mm2钢筋弹性模量: Es = 200000 N/mm2最大配筋率:max = 2.4 %混凝土材料强度综合调整系数: rd1 = 1.5钢
6、筋材料强度综合调整系数: rd2 = 1.2fcd = fc Md1 = 16.70 1为0 = 25.05 N/mm2调整后的混凝土轴心受拉强度设计值 :ftd = ft Md1 = 1.57 1为0 = 2.36 N/mm2调整后的混凝土轴心受压强度标准值 :fckd = fck xrd1 = 23.40 1.50 = 35.10 N/mm2调整后的钢筋抗拉强度设计值 :fyd = fy2 = 360 位0 = 432 N/mm2调整后的钢筋受压强度设计值 :fyd = fy吊2 = 360仅20 = 432 N/mm2系数: :-1 = 1混凝土立方体轴心受压强度标准值: fcuk =
7、35 N/mm2 4二0.8非均匀受压时的混凝土极限压应变 :Ecu = 0.0033 - (fcuk - 50) 10 0.0033 时,取 Ecu = 0.0033。相对界限受压区高度:彳=口/ 1 + fy / (Es XEcu)=0.8 / 1 + 432.0 / (200000 0.0033) = 0.4835H0 = H - c2 - 20 = 300 - 15 - 20 = 265 mmAomax = S R1 - 0.5 力发 0.4835 (1X 0.5 0.4835) = 0.3666Mb = Aomax Xj Mc XB XH02 = 0.36 66 1W 25.05 2
8、050 2652 = 1322124000.00 N.mM Mb ,按单筋矩形截面梁进行设计。受压区高度:x = H0 - H02 - 2 MK / (必 XFc XB)1 / 2=265 - 265 2 - 2 便1023800.00 / (1.0 25.05X 2050) 1 / 2 = 14 mm受压区高度与有效高度之比 :之=x / H0 = 14 / 265 = 0.052受拉钢筋配筋面积:As =冈xfc XB改/ fy=1.0 25.05 2050 14 / 432.0 = 1623 mm2人防构件受压区构造配筋面积 :As = As / 3 = 1623 / 3 = 541 m
9、m2人防构件受压区构造配筋应满足受拉钢筋最小配筋率: As = Asmin = 1538 mm2计入受压钢筋的受压区高度 :x = (fy xAs - fy) / (口1 Mc XB)=(432.0 1623 - 432.0 1538) / (1.0 25.05 2050)=1 mm计入受压钢筋的受压区高度与有效高度之比 :=x / H0 = 1 / 265 = 0.003加强梁计算配筋面积: 1623.1 mm2加强梁采用对称配筋: 21C12(2375.1 mm2)以下是柱(梁)式门框斜截面受剪的详细计算过程:混凝土轴心受压强度设计值: ft = 1.57 N/mm2箍筋抗拉强度设计值:
10、fyv = 270 N/mm2 d1 = 1.5钢筋材料综合调整系数: d2 = 1.35fcd = fc Md1刈.8 = 16.70 1.50 0.8 = 20.04 N/mm2ftd = ft Md1 X0.8 = 1.57 1.50 0.8 = 1.88 N/mm2调整后的箍筋抗拉强度设计值 :fyvd = fyv Md2 = 270保35 = 365 N/mm2混凝土保护层厚度: c = 15 mm截面有效图度:H0 = H - c - 20 = 300 - 15 - 20 = 265 mm混凝土强度影响系数: ic = 1对矩形截面,截面的腹板高度: Hw = H0 = 265 m
11、m hb = Hw / B = 0.1292683 , hb qVx = (0.25 帜 Xfc XB XH0) / 1000=(0.25 1.0 20.04 2050 265) / 1000 = 2721.68 kNV Vx,满足截面受剪限制条件。L/H0 B / 2,左侧门框墙米用柱式门框。L1 = c1 - da = 1500 - 100 = 1400 mmL2 = L - c1 = 400 - 1500 = -1100 mm柱式门11 高度:Lh = h1 + H + h2 = 500 + 2000 + 2050 = 4550 mm柱式门框线荷载 :q = qb + qe1 XL1 /
12、 1000 + qe2 4MinL2,Lh / (2 1000)=135.38 + 200 1400 / 1000 + 200 Min - 1100,4550 / (2 1000)4Min L2,Lb / (2 1000)=122.40 + 200 1950 / 1000 + 200 Min 0,1500 / (2 1000)=512 kN/m梁式门框传给加强柱的集中力 ;F = q Lb/ (2 1000)=512 1500 / (2 1000) = 384.3 kN集中力距柱式门框底距离 ;y1 = h1 + H + b1 / 2 = 500 + 2000 + 2050 / 2 = 352
13、5 mm集中力距柱式门框顶距离 ;y2 = h2 - b1 / 2 = 2050 - 2050 / 2 = 1025 mmM = q XLh2 / (12 1峡)+ F 内1 汉2 / (Lh2 X103)=305.38 4550 2 / (12 1 06) + 384.3 352 5 2 X1025 / (45 5 0 2 X103) = 763.3 kN.mV = q XLh / (2 1000) + F 必2 X(1 + 2 yW Lh) / Lh2=305.38 4550 / (2 1000) + 384.3 35252 1 + 2 1025 / 4550) / 4550 2=1029
14、.3 kN ; = fck 3 = 23.40 1.50 = 35.10 N/mm2fyd = fy Md2 = 360 位0 = 432 N/mm2为2 = 360仅20 = 432 N/mm2 * = 1 )=0.8q = 口/ 1 + fy / (Es XEcu)H0 = H - c2 - 20 = 1500 - 15 - 20 = 1465 mmAomax = 4 R1 - 0.5 4)发 0.4835 (1X 0.5 0.4835) = 0.3666Mb = Aomax X5 f 汨 XH。2=0.3666 1M 25.05 便00 14652 = 29566050000.00 N.mx = H0 - H02 - 2 MK / (1 XFc XB)1 / 2=1465 - 1465 2 - 2 763270300.00 / (1.0 25.05X 100) 1 / 2 = 14 mm= x / H0 = 14 / 1465 = 0.010As = 5 xfc XB改/ fy=1.0 25.05 1500 14 / 432.0 = 1212 mm2As Asmin , As = Asmin = 5625 mm2 = As / 3 = 5625 / 3 = 1875 mm2 = Asmin = 5625 mm2) / (如
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