砌体结构课程设计实例文档格式.docx
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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:
4.79
×
9.9+3×
3.3=56.82kN
女儿墙自重:
5.24
3.3×
0.6=10.38kN
二、三四楼面活荷载:
3.4
3+3×
3.13=43.05kN
屋面活荷载:
2.0
9.9=19.8kN
二、三四层楼面活荷载:
3
9.9=29.7kN
二、三四层墙体和窗自重:
5.24×
(3.3×
3.3-2.1
1.5)+0.25×
2.1×
1.5=41.35kN
一层墙体和窗自重:
(3.75×
3.3-2.1×
1.5)+0.25×
1.5=49.13kN
4>
控制截面的内力计算:
第四层:
①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=108.36kN
N11
(1)=1.2×
56.82+1.4×
19.8=95.91kN
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=110.13kN
N11=1.35×
0.7×
19.8=96.12kN
因为本教学楼采用MU10,M5砂浆砌筑,查表2—4得,砌体的抗压强度
设计值f=1.5MPa。
屋(楼)面均设有刚性垫块,0f0,1=5.4,此时刚性垫块上表面处梁
端有效支承长度
a0,b=5.4hc=5.4×
500=99mm
f1.5
M1
(1)=N11
(1)(y-0.4a0,b)=95.91×
(0.12-0.4×
0.099)=7.72kN/m
M1
(2)=N11
(2)(y-0.4a0,b)=96.12×
0.099)=7.73kN/m
(1)
M1
(1)
N1
(1)=7.72108.36=0.072m
e1
=
e1
(2)
=M1
(2)
(2)=7.73110.13=0.071m
N1
②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+1.2×
41.35=158kN
N2
(2)=110.13+1.35×
41.35=165.96kN
第三层
①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=1.2×
43.05+1.4×
29.7=93.24kN
N3
(1)=158+93.24=251.24kN
0
(1)=158
10-3
0.366MPa,0
(1)f=0.3661.5=0.244
0.24
1.8
查表3-5,
(11)=5.74,则:
500
a0,b=5.74×
=105mm
1.5
M3
(1)
=N3l
(1)
()
)
(y-0.4a01,b
=93.24×
0.105)=7.28kN/m
e
(1)
=M3
(1)
(1)=
7.28=0.029m
N3
251.24
N3l
(2)=1.35×
43.05+0.7×
1.4×
29.7=87.23kN
N3
(2)=165.96+87.23=253.19kN
0
(2)
=165.96
=0.385MPa
(2)
f=0.3851.5=0.266查表3-5
=5.8
=5.8×
=106m
a0,b
M3
(2)
=N3l
(2)(y-0.4a0,b)
=87.23×
0.106)
=6.77kN/m
e=
(2)=6.77253.19=0.027m
N3
②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,
5
N4
(1)=251.24+1.2×
41.35=300.86kN
N4
(2)=253.19+1.35×
41.35=309.02kN
第二层:
①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)=93.24kN
N5
(1)=300.86+93.24=394.1kN
0
(1)=300.8610-3
1.8)
=0.697MPa
(0.24
f=0.697
=0.465查表1
(1)=6.30
a0,b=6.30×
=115mm
M5
(1)=N5
(1)l×
(y-0.4a01,b
0.115)
=6.90kN/m
M
5
(1)
N5
(1)=
6.90
e5
394.1
=0.018m
N5l
(2)=87.23kN
N5
(2)=309.02+87.23=396.25kN
0
(2)=
309.2
=0.716MPa
f=0.716
=0.478查表3-51
=6.35
a0,b=6.35×
=116mm
M5
(2)
=N5l
(2)×
(y-a0,b
0.116)
=6.42kN/m
6
=0.016m
(2)=6.42
396.25
N5
②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=443.72kN
N6
(2)=396.25+1.35×
41.35=452.07kN
第一层:
③第一层截面7-7处
N7l
(1)=93.24②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)=453.06kN,e6
(1)=0
第二组内力:
N6
(2)=462.58kN,e6
(2)=0
eh=0,=15.63查表3-1,=0.73
fA=0.73×
1.5×
?
1.8×
0.24×
103=473.04kN>
462.58kN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740mm×
240mm×
240mm的预制刚性垫块
01
(1)
=0.097MPa,N5
(1)l=93.24kN,a0,b=115mm
N0
Ab=0.697×
0.1776×
106=123.79kN
+N5l=123.79+93.24=217.79kN
E=N5l×
(y-0.4a0,
(1)b)/(N0+N5l)
0.115)/217.03=0.033m
e
=0.033/0.24=0.1375,=H0
h
查表3-1得,=0.815
7
A0=(0.74+2×
0.24)×
0.24=0.2928
A0
Ab=1.629
=1+0.35
1.649
1=1.282<
1=0.8
=1.02b
1fAb=0.815×
1.026×
1.5×
103=222.76kN>
217.03kN所以满
足要求
对于第二组内力:
0
(2)=0.716MPa
N5
(2)l=87.23kN
a0,b=116mm
(2)
由于a0,b基本接近且N2l较小,所以才有此垫块亦能满足局压承载力的要
求。
N7=443.72+93.24=536.96kN
0
(1)=443072*103/(0.24*1.8)=1.027MPa
f=1.027/1.5=0.685查表
=7.283
a0,b=7.283×
=132.9mm
M7
=N7l
e7=
/N7
=0.012m
N7l=87.23kN
(2)
N=309.02+87.23=396.25kN
0
(2)=452.07*103
0.24*1.8
=1.046MPa
f=1.046/1.5=0.698查表3-51
(2)=7.35
8
=7.35×
=134.19mm
M7=N
(y-a0,b
0.13419)
=5.79kN/m
e7==
/N
=0.011m
④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
N8=536.96+1.2×
49.13=595.96kN
N8=539.30+1.35×
49.13=605.626kN
5>
第四层窗间墙承载力验算
①第四层截面1-1处窗间墙受压承载力验算:
N1
(1)=108.36kN,e1
(1)=0.072m
N1
(2)=110.13kN,e1
(2)=0.071m
对于第一组内力:
eh=0.0720.24=0.3
且e=0.0720.6y=0.6×
0.12=0.0721
=H0
=3.3
=13.75
H
查表3-1,
=0.275
fA=0.275×
1.8×
0.24×
103=178.2kN>
108.36kN满足要求
eh=0.0710.24=0.296
E=0.071<
0.6×
0.12=0.072
查表3-1,=0.278
fA=0.278×
103=180.14kN>
110.13kN满足要
9
求
②第四层截面2-2处窗间墙受压承载力验算:
N2
(1)=158kN,e
(1)2=0
(
2)
=0
N2
=165.96kN,e2
·
eh=0,=13.75,查表3-1,=0.78
fA=0.78×
103=505.4kN>
165.96kN满足要求。
③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=0.24×
0.74=0.1776m2
0=10.38=0.024MPa,N1l=96.12kN,a0,b=99mm;
1.80.24
N0=0Ab=0.024N/mm2×
0.1776×
106=4.26kN
N0+N1l=4.26+96.12=100.4kN
e=N1l(y-0.4a0,b)/(N0+N1l)
=96.12×
(0.12-0.4×
0.099)/100.4
=0.072m
=0.072/0.24=0.3,=H
3查表3-1得=0.48
0.24=0.2928㎡
Ab=1.649
=1.026
1fAb=0.48×
10.26×
103
=131.20kN>
N0+N1l=100.4kN,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×
240×
300mm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:
①第三层截面3-3处窗间墙受压承载力验算:
N3
(1)=251.24kN,e3
(1)=0.029m
N3
(2)=253.19kN,e3
(2)=0.027m
eh=0.029/0.24=0.12
且e=0.0290.6y=0.6×
0.12=0.072m
=Hh=3.3/0.24=13.75
查表3-1,=0.525
fA=0.525×
103=340.2kN>
251.24kN满足要求
eh=0.027/0.24=0.11且e=0.027<
0.6,=13.75
查表3-1,=0.53
fA=0.53×
103=343.44kN>
253.19kN满足要
②第三层截面4-4处窗间墙受压承载力验算:
N4
(1)=300.86kN,e4
(1)=0
N4
(2)=309.02kN,e4
(2)=0
=0,
=13.75,查表3-1,=0.78
103=505.4kN>
309.02kN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
=3.24kN
(1)
A
=65kN
0b=0.3660.1776
+N3l=65+93.24=158.24kN
11
e=N3l(y-0.4a0,b)/(N0+N3l)
0.105)/158.24
=0.046m
=0.046/0.24=0.19,=H
查表3-1得:
=0.69
1fAb=0.69×
=188.60kN>
N0+N3l=158.24kN满足要求
0
(2)=0.385MPa
N3l
(2)=87.23kN
a0,
(2)b=106mm
由于第二组内力与第一组内力相近,且N3
(2)l=87.23kN更小,这对局部变
压更有利,所以才有740×
240×
300mm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:
①第二层截面5-5处窗间墙受压承载力验算:
N5
(1)=394.1kN,e5
(1)=0.018m
N5
(2)=396.25kN,e5
(2)=0.016m
对于第一组内力:
=0.018/0.24=0.075且e0.6y=0.072m
=3.75/0.24=15.63
查表3-1,=0.62
12
fA=0.62×
103=402kN>
394.1kN满足要求
对于第二组内力:
e
=0.016/0.24=0.067且e
0.6y=0.072m,=15.63
查表3-1,=0.63
fA=0.63×
1.24×
103=408.24kN>
396.25kN满足要
②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)=452.07kN,e6
(2)=0
eh=0,=13.75查表3-1,=0.78
452.07kN满足要求