钢结构第三章习题答案
3-2 解: (1)
l w 1=
N 1
2×0.7h f1f f w
N 1=0.65×540=351 (KN)
h f max =1. 2×10=12 (mm)
h f min =1. 51=1. 5=5. 6(mm) h f max =1. 2×10=12 (mm) 取h f =10mm
l w 1
351×103
==156. 7mm 2×0.7×10×160
l 1=l w 1+2h f =156. 7+2×10=176.7 =177 (mm)
8h f <177<60h f
N 2=0. 35×540=189 (KN)
h f min =1. 51=1. 5=5. 6(mm ) h f max =10-(1~2) =9或8(mm ) 取h f =8mm
l w 1
189×103
==105. 47mm l 2=l w 2+2h f =105. 47+2×8=121. 47 取125 mm 2×0.7×8×160
(2)
N 3=2×0.7×h f b βf f f w =2×0.7×6×100×1.22×160=163.968 (KN)
N 1=∂1N -
N 3163.968
=0. 65×540-=269. 016(KN ) 22N 3163.968=0. 35×540-=107. 016(KN ) 22
N 2=∂2N -
l w 1
N 1269. 016⨯103
===200. 16(mm) w
2×0.7×6⨯1602×0.7h f1f f
l
1=l w 1+h f =200. 16+6=206.16=210 (mm)
l w 1
N 2107. 016⨯103
===79. 625(mm) w
2×0.7×6⨯1602×0.7h f2f f
l 2=l w 2+h f =79. 625+6=85. 625=90(mm)
(1)
h f min ==1.5≈6.7mm h f max =1.2t 1=1.2⨯14=16.8mm 取h f =10mm
540≈449.3kN 540≈299.5kN d +d
M =N x |d 2-12|=0kN
m
2
l w =d 1+d 2-2h f =170+170-2⨯10=320mm N x =
N x 449.3⨯103
σf ==≈100.3MPa
2h e l w 2⨯0.7⨯10⨯320299.5⨯103
τf ==≈66.9MPa
2h e l w 2⨯0.7⨯10⨯320N y
N =N y =N =
≈106MPa
h f =10mm , N x =449.3kN , N y =299.5kN , l w =320mm , f f w =160MPa M =N x
|d 2-I z =2⨯
d 1+d 2150-190
|=449.3⨯|190-|=8.986kN m 22
σf max
1
⨯0.7⨯10⨯3203≈3823⨯104mm 412N x M l w 449.3⨯1038.986⨯106320=+=+⨯≈137.9M Pa 42h e l w I z 22⨯0.7⨯10⨯3203823⨯102N y
299.5⨯103
τf ==≈66.9MPa
2h e l w 2⨯0.7⨯10⨯320
=≈131.3MPa
M =Fe =98⨯0. 12=11. 76(KN ⋅m )
h f 2max =12-(1~2) =11或10(mm ) 取h f 2=10mm h f 3max =12-(1~2) =11或10(mm ) 取h
f 3=10mm
形心距腹板底端的距离X
2⨯18. 6⨯0. 7⨯(18. 6/2) +2⨯0. 7⨯5. 9⨯(18. 6+0. 35) +0. 7⨯14⨯20. 85x ==13. 674(cm )
2⨯18. 6⨯0. 7+0. 7⨯5. 9⨯2+0. 7⨯14
⎡0.7⨯18.63⎤
I w =2⨯⎢+(13.674-9.3) 2⨯18.6⨯0.7⎥+2⨯0.7⨯5.9⨯5.2762+0.7⨯7.1762=1983.505(cm4)
12⎣⎦
11. 76⨯106w
σ1=⨯75. 26=44. 62(MPa )
1983. 505⨯10
σmax
11. 76⨯106=⨯136. 74=81. 072(MPa )
98⨯103
τ1==37. 63(MPa )
2⨯186⨯0. 7⨯10
(
σf 1281. 0722
) +τ2=() +37. 632=76. 367(MPa )
则f f 2, f f 3均满足要求。
弯矩主要由翼缘承担,剪力F 主要由腹板承担,焊缝f f 1基本不承担力,因此按构造配置。
h f 1max =12-(1~2) =11或10(mm ) h f min =1. 51=1. 5=5. 2(mm ) 取h f 1=10mm
f c w =215MPa , f t w =185MPa , f v w =125MPa , V =F , M =e F =0.12F 126⨯12⨯(176+12+6) +176⨯12⨯y =
126⨯12+176⨯12
176
≈132mm
176
I z =126⨯12⨯(176+12+6-132) 2+176⨯12⨯(-132) 2≈990.1⨯104mm 4
2
S z =126⨯12⨯(176+12+6-132) =93744mm 3
S z 1=126⨯12⨯(176+12+6-132) +12⨯(176-132) ⨯
176-132
=105360mm 3
2
τmax
VS z 1F ⨯103⨯105360w ==≈0.887F ≤f =125MPa ⇒F ≤140.9kN v 4
I z t 990.1⨯10⨯12
M 0.12F ⨯106
σ1=h 1=⨯(176+12+12-132) ≈0.824F ≤f t w =185MPa ⇒F ≤224.5kN 4
I z 990.1⨯10M 0.12F ⨯106
σ3=
h 3=⨯132≈1.6F ≤f c w =215MPa ⇒F ≤134.4kN 4
I z 990.1⨯10腹板上最靠近翼缘处的焊缝
M 0.12F ⨯106
σ=h =⨯(176-132) ≈0.533F 4
I z 990.1⨯10VS z F ⨯103⨯93744τ==≈0.789F 4
I z t 990.1⨯10⨯12
=≈1.467F
解:h 1=400mm , h 2=305mm
h e 1=0.43cm , h e 2=0.56cm
l 2
30.5⨯0.56⨯30.5=x 0==10.22cm
2l 2h e 2+l 1h e 12⨯30.5⨯0.56+40⨯0.42
2l 2h e 2
I x =
1
⨯0.42⨯403+2⨯0.56⨯30.5⨯202=15904cm 412
2
1⎛30.5⎫I y =⨯2⨯0.56⨯30.53+2⨯0.56⨯30.5⨯ -10.22⎪+0.42⨯40⨯10.222=5267.12cm 4
12⎭⎝2 I p =I x +I y =21171. 1c 42m
e 2=h 2-x 0=30.5-10.22=20.28cm r x =20.28cm , r y =20cm
T =F (e 1+e 2)=F (39.5+20.28)=59.78F =0.5978F (KN ⋅M ) 0.5978F ⨯106⨯200τ===0.5647F 4
I P 21171.12⨯10
T A
T ry
T rx 0.5978F ⨯106⨯202.8σ===0.5726F 4
I P 21171.12⨯10
T A
F F ⨯103
σ===0.19623F
∑h w h e 400⨯0.42+2⨯0.56⨯30.5V A
T
σf =σA +σV A =
0.76883F
=≤f f w =160 则F ≤189.08KN
解:
V =374kN , M =1122kN mm , f w c =215MPa , f w t =185MPa , f w v =125MPa , f =215MPa I 1z =
12⨯8⨯(1000-8⨯2) 3+2⨯252⨯14⨯(500+14
2
) 2≈24.5⨯108mm 4S 14
z =252⨯14⨯(500+2) =1788696mm 3
S 14492
z 1=252⨯14⨯(500+2) +8⨯492⨯2
=2756952mm 3
τ=VS z 1374⨯103⨯2756952w
max I =24.5⨯108
⨯12≈52.6MPa
1=3=I h 1=8
⨯(500+14) ≈235.4MPa
z 24.5⨯10σ1=235.4MPa >f =215MPa (不合格)
σ3=235.4MPa >f =215MPa (不合格
) 腹板上最靠近翼缘处的焊缝
M 1122⨯106
σ=I h =8⨯492≈225.3MPa
z 24.5⨯10τ=VS z 374⨯I =103⨯178869624.5⨯108⨯8
≈34.1MPa
z t =≈232.9MPa >1.1f w t =1.1⨯185=203.5MPa (不合格) ∴焊缝强度不符合要求
解:V =100KN , M =1000KN ⋅M , 加引出板
I x =
1
⨯1.4⨯1003+2⨯40⨯2⨯512=532826.667cm 4 12
S x 1=40⨯2⨯51=4080cm 3
100⎛⎫
100⨯103⨯ 4080+⨯14⨯100⎪⨯103
V 2w =s ==99.31Mpa
I t 532826.667⨯10⨯14
M h 100⨯1061020=⋅=⨯=95.72Mpa
I 2532826.667⨯102
τmax
σmax
较点正应力与较大剪应力验算: 500
σ1=σmax ⋅=92.04mpa
520
V sx 1100⨯103⨯4080⨯103
τ1===55.17mpa 4
I t 532826.667⨯10⨯
14
=132.67mpa
3-9
解:f t b =170MPa , f v b =140MPa , f c b =305MPa , f =215MPa (1)
N =n v
44
N c b =d ∑t f c b =24⨯min{10+10, 20}⨯305=146.4kN N
b min b v
πd 2
f =2⨯
b v
π⨯242
⨯140≈126.7kN
=min{126.7,146.4}=126.7kN
N 1350b
==112.5kN
N 1b =
(2)
1-1:A n 1=t ∑l =20⨯(400-3⨯25.5) =6470mm 2
' 2
A n 1=t ∑l =20⨯(40⨯2+4⨯5⨯25.5) ≈8101mm
N 1350⨯103 σn 1==≈208.7MPa
A n 16470N 1350⨯103
σ=' =≈166.6MPa
A n 18101
'
n 1
2-2:A n 2=t ∑l =20⨯(400-5⨯25.5) =5450mm 2
N -
σn 2=(3)
A n =2t ∑l =2⨯10⨯(400-4⨯25.5) =5960mm 2
n 19N 1350⨯103⨯=≈185.8MPa
N 1350⨯103
σn 1==≈226.5MPa >f =215MPa (不合格)
A n 15960
(4) f t b =40MPa , f v b =250MPa , f c b =470MPa , f =215MPa , 设孔径为d +1.5=21.5mm
N =n v
f =2⨯⨯250≈157.1kN 44
N c b =d ∑t f c b =20⨯min{10+10, 20}⨯470=188kN
b N min =min{157.1,188}=157.1kN
b v
πd 2
b v
π⨯202
N 1350b
==112.5kN
1-1:A n 1=t ∑l =20⨯(400-3⨯21.5) =6710mm 2N 1b =
' 2 A n =t l =20⨯(40⨯2+4⨯5⨯21.5) ≈8501mm ∑1
N 1350⨯103 σn 1==≈201.2MPa
A n 16710N 1350⨯103
σ=' =≈158.8MPa
A n 18501
'
n 1
2-2:A n 2=t ∑l =20⨯(400-5⨯21.5) =5850mm 2
N -
σn 2=
n 19N 1350⨯103⨯=≈173.1MPa
拼接板:A n =2t ∑l =2⨯10⨯(400-4⨯21.5) =6280mm 2
N 1350⨯103
σn 1===215MPa =f =215MPa (合格)
A n 16280
3-10
解:
(1)一个螺栓抗剪承载力设计值
N =n v b v πd 23.14⨯222f =2⨯⨯140⨯10-3=106.38KN 44b
v
一个螺栓承压承载力设计值
b N c =d ∑tf c v =22⨯14⨯305⨯10-3=93.94KN
连接所需螺栓数
n =N 540==5.7个,取6个 b N min 93.94
A n =A -ndot =17.2⨯2-2⨯23.5⨯1=29.7cm 2
N 540⨯103
22 δ===181.8N /mm
边距50mm ,中距80mm
(2)采用承托
令N 过螺栓群中心
N x =F sin θ=540N y =F cos θ=540=449.307KN =299.54KN
承托承受全部 N y =299. 5 K 4N
查表8-1得A c =303mm 2
N t =A c f t =330⨯170⨯10=51.51KN
n =N x 449.307==8.7个 取10个 N t b 51.51b -3
边距50mm ,中距80mm
设承托板与柱翼缘连接焊缝为两面侧焊
h f =10mm 高200mm
1.25⨯299.54⨯103
τ=1.25==148.58
若不采用承托,则螺栓也要承担剪力
3.14⨯222
N =f =⨯140⨯10-3=53.19KN 44
N c b =d ∑t f c b =22⨯20⨯305⨯10-3=134.2KN b v b v πd 2
一个螺栓的拉力
N N t =x =449.307=44.93KN
一个螺栓的剪力
N v =N y
n
=299.5410=299.54KN
偏心距与3-3⑵相同
e =190-170=20mm
M =N x ⋅e =449.307⨯0.02=8.986KN ⋅m
设承托板,刚剪力由承托承担
b N b
t =A e f t =51.51KN
取12个 N x 449.307n =b ==8.7N t 51.51
边距50mm ,中距80mm
设承托板与柱翼缘连接角焊缝为两侧焊缝
h f =10mm 高200mm
1.25⨯299.54⨯103
τ=1.25==148.58N /mm 2
2y i
ny 1
N max =4(102+1202+2002)12⨯200=93mm >e N x M 449.3075.986⨯103=+y =+⨯200=45.465KN
3.14⨯222
N =f =⨯140⨯10-3=53.19KN 44
N c b =d ∑t f c b =22⨯20⨯305⨯10-3=134.2KN b v b v πd 2
一个螺栓最大拉力 N t =45. 46KN 5
N v =
一个螺栓剪力为
N y n
=299.54=24.96KN
(3)P =150KN
u =0. 35
N v b =0. 9nfup =0. ⨯9⨯20. ⨯35=150KN 94. 5n =N 540==5.7b N v 94.5
取n =6 边距50mm ,中距80mm
A n =A -n 0d t m =17. 2⨯2-2⨯2. 2⨯1=2c 30
N 540⨯103
22 δ===180N /mm
(4)设螺栓群心与焊缝中心重合
① 设置承托板
P =155KN
u =0. 35
N t b =0. 8P =0⨯. 81=55KN 124
N x 499. 307n ≥b ==3. 6N t 124
取n=6 边距70mm 中距100mm
承托板焊缝设为h f =10mm l 1=200mm
1.25V 1.25⨯299.54⨯103
τf ===148.58N /mm 2
h e l w 0.7⨯10⨯2200-2⨯10② 不设承托,螺栓承受剪力
取 n =14 边距50mm 中距80mm
499.307N t ==37.44KN 14
299.54N v ==21.4KN
b N v =0.9nfu (p -1.25N t )=34.083KN
N v N t +b
当作用偏心拉力时 e =20mm
① 设承托板,承托板焊缝h f =10mm l 1=200mm
取n=6个 ,边距70mm 中距 100mm ,分两排,每排3个
N max N x Fey 1449.307449.307⨯0.02⨯103=+=+⨯8022n y i 64⨯80
=103KN
N t b =0.8P =0.8⨯155=124KN
N max
② 不设承托板
取n=10个 边距50mm 中距80mm ,分两排,每排5个
N x My 1449.307449.307⨯0.02⨯103
N 1=+=+⨯160222n y i 104⨯80+160=56.16KN
449.3078.986⨯103
N 5=-⨯160=33.698KN 22104⨯80+160N 2=50.5445KN
N 3=44.93KN
N 4=39.3135KN
∑N
∑N ti b
v =449.3KN =0.9⨯1⨯0.35(10⨯155-1.25⨯449.3)
=311.34KN >N y =299.54KN
3-11
解:
(1) P =150kN , μ=0.35
与牛腿连接的螺栓:
V =F =220kN , M =eF =0.2⨯220=44kN m N t max M 44⨯103=y =⨯150=66kN
50)]150∑N v b =0.9n f μ(nP -1.25∑N ti ) =0.9⨯1⨯0.35⨯[8⨯150-1.25⨯(2⨯66+2⨯66⨯
=308.7kN >V =220kN (合格
)
与连接板连接的螺栓:
V =F =220kN , T =eF =(0.2-0.055) ⨯220=31.9kN m N T
x max Ty 31.9⨯103⨯150===95.7kN 2222x i +y i 0+2⨯(50+150)
Tx =0kN 22x i +y i N T y max =
b N y =V 220==55kN n 4
≈110.4kN
(2) f t b =400MPa , f v b =250MPa , f c b =470MPa , A e =
245mm 2 与牛腿连接的螺栓:
N =n v f =1⨯⨯250≈78.5kN 44
N c b =d ∑t f c b =20⨯min{20,20}⨯470=188kN N =b
t b v πd 2b v π⨯202πd e 24f t b =245⨯400=98kN
N c b 188N 220N v ===27.5kN
≈0.76
) 与连接板连接的螺栓:
N =n v 44
N c b =d ∑t f c b =20⨯min{14, 20+20}⨯470=20⨯14⨯470=131.6kN b N min =min{157.1,131.6}=131.6kN
b ≈110.4kN