结构设计原理复习题 及答案. 下载本文

内容发布更新时间 : 2024/11/9 5:13:26星期一 下面是文章的全部内容请认真阅读。

560?106?11.5?300?x(740?0.5x)?11.5?(600?300)?100?(740?50) x?139.2mm <?bh0?0.56?740?414.4mm

fsd13、解:1)h0?700?60?640mm

2)fcdbh(h0?'f'f4)As?fcdbx?fcdh'f(b'f?b)?11.5?300?139.2?11.5?100?300?2947mm2

28080)?552kn.m<?0Md?580kn.m 2h'f2)?11.5?1000?80(640? 属于第二类T形截面

h'f''x)?fcd(bf?b)hf(h0?)3)?0Md?fcdbx(h0?22

580?106?11.5?250?x(640?0.5x)?11.5?(1000?250)?80?(640?40)

x?97.7mm <?bh0?0.56?640?358.4mm

11.5?250?97.7?11.5?80?750?3467mm2

fsd28014、解:1)h0?800?73.4?726.6mm

4)As?fcdbx?fcdh'f(b'f?b)?2)fcdb'fh'f?11.5?600?100?690000N.mm?0.69KN.m fsdAs?280?2945?0.8246KN.m

fcdb'fh'f?0.69KN.m<fsdAs?0.8246KN.m 故为第二类T形截面

280?2945?11.5(600?300)?100?139mm

fcdb11.5?300x<?bh0?0.56?726.6?406.9mm

'''4)Mu?fcdbx(h0?x)?fcd(bf?b)hf(h0?0.5hf)

2?11.5?300?139(726.6?139)?11.5(600?300)?100?(726.6?0.5?100)

2?548.5KN.m

15、解:1)h0?600?44.2?555.8mm

3)x?fsdAs?fcd(b'f?b)h'f?2)fcdbfhf?11.5?400?80?368000N.mm?0.368KN.m fsdAs?280?1473?0.41244KN.m

''KN.m 故为第二类T形截面 fcdb'fh'f?0.368KN.m<fsdAs?0.41244280?1473?11.5(400?250)?80?95.5mm

fcdb11.5?250x<?bh0?0.56?555.8?311.2mm Mu?fcdbx(h0?x)?fcd(b'f?b)h'f(h0?0.5h'f)24)?11.5?250?95.5(555.8?95.5)?11.5(400?250)?80?(555.8?0.5?80)

2?210.7KN.m5)Mu?210.7KN.m>?0Md?200KN.m 安全 16、解:1)h0?800?42.6?757.4mm

3)x?fsdAs?fcd(b'f?b)h'f?2)fcdbfhf?11.5?600?100?690000N.mm?0.69KN.m fsdAs?280?1520?0.4256KN.m

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''fcdb'fh'f?0.69KN.m>fsdAs?0.4256KN.m 故为第一类T形截面

3)x?fsdAs280?1520??61.7mm<?bh0?0.56?757.4?424.1mm '11.5?600fcdbf'4)Mu?fcdbfx(h0?x)?11.5?600?61.7(757.4?61.7)?309.3KN.m

225)Mu?309.3KN.m<?0Md?450KN.m 不安全

Md240?10617、解:1) e0???600mm 3Nd400?10'2)as?as?50mm h0?450mm

?e0?1.034?600?620mm

N400?103 3) ????0.242 < ?b?0.56

fcdbh09.2?400?450 按大偏心受压构件设计

' 4) x??h0?0.242?450?109mm > 2as?100mm

h?as?620?250?50?820mm 22Nes?fcdbh0?(1?0.5?)' 5) As?As??151mm32 'fsd(h0?as) es??e0?'6)As?As?1513mm2>0.002bh?0.002?400?500?400mm

218、解:1)

l02500??4.2 <5 故不考虑附加偏心增大系数的影响,取??1.0 h600'2) as?as?45mm h0?600?45?555mm

Md330?106e0???717mm 3Nd460?10 ?e0?1.0?717?717mm

N460?103 3) ????0.18 < ?b?0.56

fcdbh011.5?400?555 按大偏心受压构件设计

' 4) x??h0?0.18?555?99.9mm > 2as?90mm

h?as?717?300?45?972mm 22Nes?fcdbh0?(1?0.5?)' 5) As?As? 'fsd(h0?as) es??e0?460?103?972?11.5?400?5552?0.18(1?0.5?0.18) ?280(555?45)?1505mm2

'6)As?As?1505mm2>0.002bh?0.002?400?600?480mm2

l2000?5 故可不考虑附加偏心增大系数的影响,即取??1.0 19、解:1)0?h400'2) as?as?45mm h0?400?45?355mm

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Md185?106e0???569mm

Nd325?103 ?e0?1.0?569?569mm

N325?103 3) ????0.265 < ?b?0.56

fcdbh011.5?300?355 按大偏心受压构件设计

' 4) x??h0?0.265?355?94.1mm > 2as?90mm

h?as?569?200?45?724mm 22Nes?fcdbh0?(1?0.5?)' 5) As?As? 'fsd(h0?as) es??e0?325?103?724?11.5?300?3552?0.265(1?0.5?0.265) ?280(355?45)?1559mm2

'6)As?As?1559mm2>0.002bh?0.002?300?400?240mm2

l2800?4.7<5 故不考虑附加偏心增大系数的影响,取??1.0 20、解:1)0?h600'2) as?as?45mm h0?600?45?555mm

Md240?106e0???600mm 3Nd400?10 ?e0?1.0?350?600mm

N400?103 3) ????0.209 < ?b?0.56

fcdbh011.5?300?555 按大偏心受压构件设计

' 4) x??h0?0.209?555?116mm > 2as?90mm

h?as?600?300?45?855mm 22Nes?fcdbh0?(1?0.5?)' 5) As?As?

fsd(h0?as') es??e0?400?103?855?11.5?300?5552?0.209(1?0.5?0.209) ?280(555?45)?1002mm2

'6)As?As?1002mm2>0.002bh?0.002?300?600?360mm2

21、解:1)

l02000??5 故可不考虑附加偏心增大系数的影响,即取??1.0 h400'2) as?as?45mm h0?400?45?355mm

Md185?106e0???569mm

Nd325?103 ?e0?1.0?569?569mm>0.3h0?0.3?355?106.5mm

可按大偏心受压进行设计 es??e0?

h?as?569?200?45?724mm 228

3)取???b?0.56

2Ndes??b(1?0.5?b)fcdbh0 A? ''fsd(h0?as)'s325?103?724?0.56(1?0.5?0.56)11.5?300?3552 ?280(355?45)?691.2mm2 > 0.002bh?0.002?300?400?240mm2

'fcdb?bh0?fsdAs'??0Nd4)As?

fsd11.5?300?0.56?355?280?691.2?1?325?103 ?

280mm2 > 0.002bh?0.002?300?400?240mm2 ?1980l3000?5 故不考虑附加偏心增大系数的影响,取??1.0 22、解:1)0?h600'2) as?as?45mm h0?600?45?555mm

Md420?106e0???700mm

Nd600?103 ?e0?1.0?700?700mm

N600?103 3) ????0.235 < ?b?0.56

fcdbh011.5?400?555 按大偏心受压构件设计

' 4) x??h0?0.235?555?130mm > 2as?90mm

h?as?700?300?45?955mm 22Nes?fcdbh0?(1?0.5?)' 5) As?As? 'fsd(h0?as) es??e0?600?103?955?11.5?400?5552?0.235(1?0.5?0.235) ?280(555?45)?1955mm2

'6)As?As?1955mm2>0.002bh?0.002?400?600?480mm2

23、解:1)

l04000??8 >5 应考虑附加偏心增大系数的影响,即取??1.034 h500Md300?1062) e0???750mm 3Nd400?10'as?as?50mm h0?450mm

?e0?1.034?750?775.5mm >0.3h0?0.3?450?135mm

可按大偏心受压进行设计

h?as?775.5?250?50?975.5mm 2 3)取???b?0.56

es??e0? 29

2Ndes??b(1?0.5?b)fcdbh0 A? 'fsd(h0?as')'s400?103?975.5?0.56(1?0.5?0.56)9.2?400?4502 ?280(450?50)?801mm2 > 0.002bh?0.002?400?500?400mm2

'fcdb?bh0?fsdAs'??0Nd4)As?

fsd9.2?400?0.56?450?280?801?1?400?103 ?

28042 > 0.002bh?0.002?400?500?400mm2 ?268mml2000?5 故可不考虑附加偏心增大系数的影响,即取??1.0 24、解:1)0?h400'2) as?as?45mm h0?400?45?355mm

Md159?106e0???530mm

Nd300?103 ?e0?1.0?530?530mm>0.3h0?0.3?355?106.5mm

可按大偏心受压进行设计 es??e0?h?as?530?200?45?685mm 23)取???b?0.56

's2Ndes??b(1?0.5?b)fcdbh0 A? 'fsd(h0?as')300?103?685?0.56(1?0.5?0.56)11.5?300?3552 ?280(355?45)?348mm2 > 0.002bh?0.002?300?400?240mm2

'fcdb?bh0?fsdAs'??0Nd4)As?

fsd11.5?300?0.56?355?280?348?1?300?103 ?

28062 > 0.002bh?0.002?300?400?240mm2 ?172mm

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