1、所以可只在 D 轴线上取一个开间的外纵墙作为计算单元, 其受荷面积为: 3.3 3=9.9 。2 确定计算面积:每层墙的控制截面位于墙的顶部梁(或板)的底面和墙低的底面处。因为墙的顶部梁(或板)的底面处,梁(或板)传来的支撑压力产生的弯矩最大,且为梁(或板)端支承处,其偏心承压和局部变压均为不利。而墙底的底面处承受的轴向压力最大。所以此处对截面: 1-1 6-6 的承受力分别进行计算。3 荷载计算:取一个计算单元,作用于纵墙的荷载标准值如下:层面横荷载:4.799.9+3 3.3=56.82 kN女儿墙自重:5.243.3 0.6=10.38 kN二、三四楼面活荷载:3.4 3+33.13=4
2、3.05 kN屋面活荷载:2.0 9.9=19.8 kN二、三四层楼面活荷载:3 9.9=29.7 kN二、三四层墙体和窗自重:5.24 ( 3.3 3.3-2.11.5 )+0.25 2.1 1.5=41.35 kN一层墙体和窗自重:( 3.75 3.3-2.1 1.5 )+0.25 1.5=49.13 kN4 控制截面的内力计算: 第四层: 第四层截面 1-1 处:由屋面荷载产生的轴向力涉及值应考虑两种内力组合,由可变荷载效应控制的组合, G =1.2, Q =1.4 则N1(1) =1.2( 56.82+10.38)+1.4 19.8=108.36 kNN11(1) =1.256.82+
3、1.419.8=95.91 kN由永久荷载效应控制的组合: G =1.2, Q =1.4( 2 )N1 =1.35 0.7 19.8=110.13 kNN11 =1.350.719.8=96.12 kN因为本教学楼采用 MU10,M5 砂浆砌筑,查表 24 得,砌体的抗压强度设计值 f=1.5MPa。屋(楼)面均设有刚性垫块, 0 f 0, 1 =5.4,此时刚性垫块上表面处梁端有效支承长度a0,b =5.4 hc =5.4 500 =99mmf 1.5M1(1) = N11(1) (y-0.4 a0,b )=95.91(0.12-0.40.099)=7.72 kN /mM1(2) = N11
4、(2) (y-0.4 a0,b )=96.120.099)=7.73 kN /m(1)M 1(1)N1(1) = 7.72 108.36=0.072me1=e1(2)= M 1(2)(2) = 7.73 110.13=0.071mN1 第四层截面 2-2 处轴向力为上述荷载 Nl 和本层墙自重之和N2(1) =108.36+1.241.35=158 kNN2(2) =110.13+1.35 41.35=165.96 kN 第三层 第三层截面 3-3 处:4轴向力为上述荷载 N2 和本层楼盖荷载 N3l 之和( 1 )N3l =1.243.05+1.4 29.7=93.24 kNN3(1) =1
5、58+93.24=251.24 kN0(1) = 15810-30.366MPa, 0(1) f = 0.366 1.5=0.2440.241.8查表 3-5,(11)=5.74,则:500a0, b =5.74=105mm1.5M 3(1)= N3l(1)( )(y-0.4 a01,b=93.240.105)=7.28 kN /me(1)= M 3(1)(1) =7.28 =0.029mN 3251.24N3l(2) =1.3543.05+0.7 1.4 29.7=87.23 kNN3(2) =165.96+87.23=253.19 kN0(2)= 165.96=0.385MPa(2)f =
6、 0.385 1.5 =0.266 查表 3-5=5.8=5.8=106ma0,bM 3(2)= N3l(2) (y-0.4 a0,b )=87.230.106)=6.77 kN /me =(2) =6.77 253.19=0.027mN3 第三层截面 4-4 处轴向力为上述荷载 N3 与本层墙自重之和,5N4(1) =251.24+1.241.35=300.86 kNN4(2) =253.19+1.3541.35=309.02 kN 第二层: 第一层截面 5-5 处轴向力为上述荷载 N4 和本层楼盖荷载之和N5l(1) =93.24 kNN5(1) =300.86+93.24=394.1 k
7、N0(1) = 300.86 10-31.8)=0.697MPa(0.24f = 0.697=0.465 查表 1(1) =6.30a0,b =6.30=115mmM 5(1) = N5(1)l (y- 0.4a01,b0.115)=6.90 kN /mM5(1)N 5(1) =6.90e5394.1=0.018mN5l(2) =87.23 kNN5(2) =309.02+87.23=396.25 kN0(2) =309.2=0.716MPaf = 0.716=0.478 查表 3-5 1=6.35a0,b =6.35=116mmM 5(2)= N5l(2) (y- a0 ,b0.116)=6
8、.42 kN /m6=0.016m(2) = 6.42396.25N 5 第二层截面 6-6 处轴向力为上述荷载 N5 与本层本层墙自重之和N6(1) =394.1+1.2 41.35=443.72kNN6(2) =396.25+1.3541.35=452.07 kN 第一层: 第一层截面 7-7 处N 7 l ( 1) =93.24 第一层截面 6-6 处窗间墙受压承载力验算:第一组内力: N6(1) =453.06 kN, e6(1) =0第二组内力: N6(2) =462.58 kN, e6(2) =0e h =0, =15.63 查表 3-1, =0.73fA =0.731.5? 1.
9、80.24 103 =473.04 kN 462.58 kN 满足要求梁端支撑处(截面 5-5 )砌体局部受压承载力验算:梁端设置尺寸为 740mm240mm 240mm的预制刚性垫块01(1)=0.097MPa , N5(1)l =93.24 kN, a0 ,b =115mmN0Ab =0.6970.1776106 =123.79 kN+ N5 l =123.79+93.24=217.79 kNE= N5 l ( y-0.4 a0,(1)b ) /( N0 + N5 l )0.115)/217.03=0.033me=0.033/0.24=0.1375 , = H 0h查表 3-1 得, =0
10、.8157A0 =(0.74+20.24) 0.24=0.2928A0Ab =1.629=1+0.351.6491 =1.282217.03 kN 所以满足要求对于第二组内力: 0(2) =0.716MPaN5(2)l =87.23 kNa0, b =116mm( 2)由于 a0, b 基本接近且 N 2l 较小,所以才有此垫块亦能满足局压承载力的要求。N 7 =443.72+93.24=536.96 kN0(1) =443072*10 3 /(0.24*1.8)=1.027MPaf =1.027/1.5=0.685 查表=7.283a0,b =7.283=132.9mmM 7= N 7le7
11、 =/ N 7=0.012mN 7l =87.23 kN(2 )N =309.02+87.23=396.25 kN0(2) = 452.07* 10 30.24 * 1.8=1.046MPaf =1.046/1.5=0.698 查表 3-5 1(2) =7.358=7.35=134.19mmM7 =N( y- a0, b0.13419)=5.79 kN /me7 =/ N=0.011m 第一层截面 8-8处轴向力为上述荷载 N 与本层本层墙自重之和N 8 =536.96+1.249.13=595.96kNN 8 =539.30+1.3549.13=605.626 kN5第四层窗间墙承载力验算
12、第四层截面 1-1 处窗间墙受压承载力验算: N1(1) =108.36 kN, e1(1) =0.072m N1(2) =110.13 kN, e1(2) =0.071m对于第一组内力: e h =0.072 0.24 =0.3且 e=0.072 0.6y=0.60.12=0.0721= H 0= 3.3=13.75H查表 3-1,=0.275fA=0.2751.8 0.24 103 =178.2 kN 108.36 kN满足要求 e h =0.071 0.24 =0.296E=0.071110.13 kN 满足要9求 第四层截面 2-2 处窗间墙受压承载力验算: N2(1) =158 kN
13、, e(1)2 =0(2 )=0N2=165.96 kN, e2 e h =0, =13.75, 查表 3-1, =0.78fA =0.78 103 =505.4 kN 165.96 kN满足要求。 梁端支撑处(截面 1-1)砌体局部受压承载力验算: 300mm 的预制刚性垫块。Ab = ab bb =0.240.74=0.1776m2 0 = 10.38 =0.024MPa , N1l =96.12 kN, a0,b =99mm;1.8 0.24N0 = 0 Ab =0.024 N / mm2 0.1776 106 =4.26 kNN0 + N1l =4.26+96.12=100.4 kNe
14、= N1l (y-0.4 a0, b )/( N0 + N1l )=96.12( 0.12-0.40.099)/100.4=0.072m=0.072/0.24=0.3, = H3 查表 3-1 得 =0.48 0.24=0.2928 Ab =1.649=1.0261 f Ab =0.4810.26 103=131.20 kN N0 + N1l =100.4 kN , 满足要求。对于第一组内力,由于 a0, b 相等,梁端反力略小些,对结构更有利。因此采用 740 240300mm 的刚性垫块能满足局压承载力的要求。10(6)第三层窗间墙承载力验算: 第三层截面 3-3 处窗间墙受压承载力验算:
15、 N3(1) =251.24 kN, e3(1) =0.029m N3(2) =253.19 kN, e3(2) =0.027m e h =0.029/0.24=0.12且 e=0.029 0.6y=0.60.12=0.072m= H h =3.3/0.24=13.75查表 3-1, =0.525fA =0.525 103 =340.2 kN 251.24 kN 满足要求 e h=0.027/0.24=0.11且 e=0.027253.19 kN 满足要第三层截面 4-4 处窗间墙受压承载力验算: N4(1) =300.86 kN, e4(1) =0 N4(2) =309.02 kN, e4(
16、2) =0=0,=13.75, 查表 3-1, =0.78103 =505.4 kN 309.02 kN 满足要求梁端支承处(截面 3-3 )砌体局部受压承载力验算:梁端尺寸设置尺寸为 740mm 300mm的预制刚性垫块。01 =0.366MPa , N3(1)l= 3.24 kN( 1)A=65 kN0 b =0.366 0.1776+ N3l =65+93.24=158.24 kN11e=N3l (y-0.4 a0,b )/( N0 + N3l )0.105)/158.24=0.046m=0.046/0.24=0.19, = H查表 3-1 得: =0.691 f Ab =0.69=18
17、8.60 kN N0 + N3l =158.24 kN 满足要求 0(2) =0.385MPaN3l(2) =87.23 kNa0,(2)b =106mm由于第二组内力与第一组内力相近,且 N3(2)l =87.23 kN 更小,这对局部变压更有利,所以才有 740240300mm 的预制刚性垫块能满足局部受压承载力的要求。(7)第二层窗间墙的承载力验算:第二层截面 5-5 处窗间墙受压承载力验算: N5(1) =394.1 kN, e5(1) =0.018m N5(2) =396.25 kN, e5(2) =0.016m对于第一组内力 :0.018/0.24=0.075且 e 0.6y=0.072m=3.75/0.24=15.63查表 3-1, =0.6212fA =0.62103 =402 kN 394.1 kN 满足要求对于第二组内力 : e=0.016/0.24=0.067且 e0.6y=0.072m, =15.63查表 3-1, =0.63fA =0.63 1.2410 3 =408.24 kN 396.25 kN 满足要第二层截面 6-6 处窗间墙受压承载力验算: N6(1) =443.72kN, e6(1) =0 N6(2) =452.07 kN, e6(2) =0e h =0, =13.75 查表 3-1, =0.78452.07 kN 满足要求
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