武汉理工大学毕业设计
图3.1梯形荷载、三角形荷载分布等效图
屋面梁上线荷载:(恒载=梁自重+板传荷载)
q边1=(1-2?2+?3)×6.34×2.1×2+3.93=23.27+3.93=27.20KN/M
楼面梁上线荷载:(恒载=梁自重+板传荷载+填充墙重)
q边2=(1-2?2+?3)×4.08×2.1×2+3.93+15.5=14.97+15.5=30.47KN/M
中框梁(BC)承担的由屋面板、楼面板传来的荷载为三角形,为计算方便,按支座弯矩等效原则,将其化为矩形分布,则:
屋面梁上线荷载:(恒载=梁自重+板传荷载)
q中1=5/8×6.34×1.2×2+2.61=9.51+2.61=12.12KN/M
楼面梁上线荷载:(恒载=梁自重+板传荷载)
q中2=5/8×4.08×1.2×2+2.61=6.12+2.61=8.73KN/M
3.3 框架柱节点集中荷载计算
3.3.1 A、D轴柱纵向集中荷载的计算 顶层柱恒载=女儿墙自重+梁自重+板传荷载
16
武汉理工大学毕业设计
=6.35×4.2+3.93×(4.2-0.5)+5/8×6.34×2.1×4.2
=26.67+14.54+34.95=76.16KN
标准层柱恒载=墙自重+梁自重+板传荷载
=6.32×(4.2-0.5)+3.93×(4.2-0.5)+5/8×4.08×2.1×4.2 =23.38+14.54+22.49=60.41KN 基础顶面荷载=底层外纵墙自重+基础梁自重
=16.02×(4.2-0.5)+3.13×(4.2-0.5)=70.86KN
3.3.2 B、C轴柱纵向集中荷载的计算 顶层柱恒载=梁自重+板传荷载
=2.61×(4.2-0.5)+ 5/8×6.34×2.1×4.2+[1-2(1.2/7.8)2+(1.2/7.8)3]
×6.34×1.2×4.2
=9.66+34.95+30.55=75.16KN 标准层柱恒载=墙自重+梁自重+板传荷载
=15.5×(4.2-0.5)+2.61×(4.2-0.5)+ 5/8×4.08×2.1×
4.2+[1-2(1.2/7.8)2+(1.2/7.8)3]×4.08×1.2×4.2
=57.35+9.66+22.49+19.66=109.16KN 基础顶面荷载=底层内墙自重+基础梁自重
=27.75×(4.2-0.5)+3.31×(4.2-0.5)=114.92KN 恒载作用下框架受荷简图如下(图 3.2):
17
武汉理工大学毕业设计
恒载作用下结构计算简图
(均布荷载单位:KN/m;集中荷载单位:K N)
3.4 恒载作用下框架的固端弯矩
采用力矩分配法计算框架弯矩
屋面层梁的固端弯矩:
M11AB?MCD??12q边1l2=?12?27.20×7.82=-137.9KN?m
M12BA?MDC?12q边1l=137.9KN?m
M??11BC12q中1l2??12×12.12×2.42=-5.82KN?m
18
3.2
图武汉理工大学毕业设计
MCB?
1q中1l2=5.82KN?m 12标准层梁的固端弯矩:
11MAB?MCD??q边2l2=??30.47×7.82=-154.48KN?m
12121MBA?MDC?q边2l2=154.48KN?m
1211MBC??q中2l2??×8.73×2.42=-4.19KN?m
12121MCB?q中2l2=4.19KN?m
12
3.5 节点分配系数计算
为各杆件的相对线刚度与交汇于此结点的所以杆件的线刚度之和的比值,公式如下:
?ik?iik?iik (4-1)
i为修正分层法的误差,一般层柱的线性刚度降低0.9,相应的传递系数取1/3,底层柱的线性刚度不降低,相应的传递系数取1/2.
A轴各节点: A6点:
??
A6B6=1/(1+1.05×0.9)=0.51
=1.05×0.9/(1+1.05×0.9)=0.49
A6A5A5、A4、A3、A2点:
???A5B5=1/(1+1.05×0.9+1.05×0.9)=0.35 =1.05×0.9/(1+1.05×0.9+1.05×0.9)=0.33 =1.05×0.9/(1+1.05×0.9+1.05×0.9)=0.33
A5A4A5A6A1点:
19
武汉理工大学毕业设计
???
A1B1=1/(1+1.05×0.9+0.63)=0.39
=1.05×0.9/(1+1.05×0.9+0.63)=0.37 =0.63/(1+1.05×0.9+0.63)=0.24
A1A2A1A0B轴各节点: B6点:
???????B6A6=1/(1+1.58+1.05×0.9)=0.28 =1.58/(1+1.58+1.05×0.9)=0.49 =1.05×0.9/(1+1.58+1.05×0.9)=0.27
B6C6B6B5B5、B4、B3、B2点:
B5A5=1/(1+1.58+1.05×0.9+1.05×0.9)=0.22 =1.05×0.9/(1+1.58+1.05×0.9+1.05×0.9)=0.21 =1.05×0.9/(1+1.58+1.05×0.9+1.05×0.9)=0.21 =1.58/(1+1.58+1.05×0.9+1.05×0.9))=0.35
B5B4B5B6B5C5B1点:
????B1A1=1/(1+1.58+0.63+1.05×0.9)=0.24 =1.58/(1+1.58+0.63+1.05×0.9)=0.38 =0.63/(1+1.58+0.63+1.05×0.9)=0.15 =1.05×0.9/(1+1.58+0.63+1.05×0.9)=0.23
B1C1B1B0B1B23.6 恒载作用下框架的内力计算
3.6.1 恒荷载作用下内力分析采用力矩二次分配法,计算图见图3.3。
20