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  • 2010-01-04
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t T
发表于 2010-06-21 14:36 |只看楼主
                                                                                            ›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
2下列表格中的数据为做CBR试验十所测得的结果，请将所需数据填在表格空白处，并计算CBR2.5和CBR5，P判断该试验是否重做。›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(已知：量力环校正系数C=200N/0.01mm,贯入杆直径d=50mm›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
荷载测力计百分表读数R›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
单位压力P(KPa)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
百分表读数(0.01mm›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
贯入量Lmm›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
0.9›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(92)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
60.5›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(0.61)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
1.8›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(183)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
106.5›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(1.07)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
2.9›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
（›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
151›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
4.0›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
194›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
4.8›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
240.5›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
5.1›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
286›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
解：根据贯入杆的单位压力计算公式›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
P=C*R/A›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
则P=200X0.9/1963=0.092 ›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
MPa›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
0.092X1000=9（KPa）›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
P=200X1.8/1963=0.183 MPa›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
0.183X1000=183 (MPa)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(其中 A=1/4Πd2›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
=1/4X3.14X502 =1963›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
请问，贯入量是如何计算的？›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
1 某工地地基原状土直剪试验（快剪）结果如下表，试求该地基的强度指标›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
σ (KPa)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
100›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
200›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
300›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
τ( KPa)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
100›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
108›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
115›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
解:›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
（1）直剪试验在直角坐标σ~τ关系图是线性的，表达式是τ =σtgΦ+C›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
(2)作图如下：›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
（3）tgΦ=15%200=0.075›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
Φ=4.3o›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
C=100-100tgΦ›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
C=92.5(KPa)›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
请问我这样做是正确的吗？那么›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
（3）中的tgΦ=15%200中的15和200是怎样确定的？›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
C=100-100tgΦ中的两个100是怎样确定的？›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
›u¯ÓÓ^4ä$(bbs.3c3t.com8óâ?w\4&
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