型钢截面计算例题

【】。钢HPB235Q235。习题4-1图解选择I22aA bt,t wh220mm,A 4213mm2,

A'

402mm2s ss s sf f'N/mm2,f N/mm2,

N/mm2y y s ca'(66022050)mm330mm,E 2.06105N/mm2s ss计算压区高度：当中和轴在翼缘上通过时，有fAfA fthf

fA fthx y s s sf sws y s s sf swsfb fbc c1353 x180.6mm0.8a's

0.8330mm264mm属于第二种情况，即中和轴不通过，此时fAfA f'A' f

x y s s ss y s

s ss

mm253.4mm0.8a'fsf

fb fbc c

196mmx1 s0.003Ess

1 1050.8(hafsf

) 0.8(660

mm326mmx1 s0.003Ess

1 2150.0032.06105以，不考虑上翼缘xf(A th)fAf'

x s sf ws y s y sfbc

mm180.6mm

th xM fA(haa')fh(A ws)f'A'(a'a')f(a' )u y s r s ss sf 2 y s s r c s 2[210)214.3m2m。C30，筋HPB235级筋，Q235。习题4-2图解选择I40aA bt,t

400mm,w sA mm2,f N/mm2,E 5N/mm2,s A402mm2,f f'N/mm2,

N/mm2,s s y y ca a'mm,a a'mms s r r压区高度：当中和轴在上冀缘通过时，有fAfA fthf'

fA fthx y s s sf sws y s s sf swsfb fbc c10.5 x383mm

a'50mms属于第一种情况，即中和轴通过腹板，时f(AA')fthf(A'A' a't)x y s s sw c s sf swf(b1.25t)2.5ftc w sw20510.550014.3(4025010.5) mm126.6mm14.3(2501.2510.5)2.520510.5M f

(ha)f

(ha)2s

f

(xa')ft

(xa')2s u s sf s sw 2 s sf s sw 2fA (ha)f'A'(xa)fb2y sf r y s r c 2[205126.650)20510.550126.6)2220550)20510.550)22126.614.3250126.62]kNm377.9kNm2

N0m×0(由于空间高限制HPB235I25a普通热轧工字Q235等级C30试验斜剪切能力。习题4-3图解 查0fc

14.3N/mm2ft

1.43N/mm2。I25a工字

250mm,

8mm,

13mm所以s w th (250213)mm

mm,

N/2w s上与裁之和为最大剪力

gq15)/m/m1 1V (gq)l 28.11

105.4kN2 2h (46030)mm430mm0则0.7fbhfth (0.71.432504302158Nt 0 sw w331070N

0.4fbh 0.414.3250430NN

614.9c 0V 105.4kN0.4fc 00m×0配a46。试验斜抗剪力并配置箍。习题4-4图C30

14.3N/mm2,f

1.43N/mm2。c tI36a工字

7630mm2,h360mm,

10mm,s s w15.8mm,f 215N/mm2t sh 215.8)328.4mmwh (46030)mm430mm0gq16)/m/m最大剪力为1 1V (gq)l 9kN=127.0kN2 20.4fbh

0.414.3250430N

614.9c 0V 127.0kN0.4f

614.9kNc 0所以符合要求则0.7fbh0.58ftht 0 swwN

V

127.0kNφ8@200。解【例5＝00N=6，C30．HPB235型Q235。验算其正截面强度。解习题4-5图A,f N/mm2,

210N/mm2,s s c yE 103N/mm2,I45a,A bt18mm2mm2,t

s wA mm2,f N/mm2,E 5N/mm2,s a a'

50mm,a

a'

35mm,s s r 计算受压区高度：受压区界限高度为

0.8(h

) 0.8(550xb 1

s fs 1

215

mm297mm先

0.003Ess

0.003206103x N

fthswfb2.5ftc sw135010321555011.5mm227.7mmx

297mm14.34002.521511.5 b1.2a'

x

x确属大偏心受压。则s bl1 1 (0)2Kl

1

1 (

1.1260e h 1 20h

550K K1 2

1.0d0.417x0.417227.7mm95mm而h 1 1 2N ffA(haa')f[Aht( h2 d20 2

c y s r r s sf s w2 s 31N 14.322 21 2N

[2700

4502 9522 3MNe0

19481.1260.37kNm812kNm500kNm0M0=6，为55Ia解 A,f N/mm2,

N/mm2,s s c yE 103N/mm2,I45a,A bt18mm2mm2,t

s wA mm2,f N/mm2,E 5N/mm2,s a a'50mm,a a'35mm,s s r re M0 N

3001063200103

mm94mm0.8(hax s

) 0.8(550

mm297mmsb 1 f 1s

2150.003Ess

0.003206103先按大偏心值Nftx sw

103

mm383mmxfb2.5ftc s

11.5 b故再按小偏心受压算。型钢未屈服区高度d0.417x型钢受拉翼缘应力 0.003Esf

0.8(hasx

1]103[

1) 0.003Es s

0.8(harx

x x1]103[

1)x x0K 39401 h 0l 15K 1.0则0h 2l1 1 (0)2KKl01400e h 1 20h1 1

94550

)21承载能力d 1Nff'A'f[A' a' )t]A [A (h

c y s y sf s 2 w s s sf sf 2 r)x)2630(11.25x x 2经整理得8.92108N7.91032.96106 （a）xhN 1.25ff'A'a')fA'a')h0 2 c y s r s sf sA(ha

1.25

A (h

1.25s s r sf sf s1 1 1f5a')2t fd2t (ha5)2t2 s s w 6 s w 3 sf s wN 1.2514.321.25x1618(x

1.25 50)211.52

005)26 3 x

6.551011N(1.25x158)24476.371063.89109xxNMNe0

367kNm300kNm3N10N0NH068555C3筋量。C30混凝土得fc=ft=N/mm2h0=500-50mm=450mm=00N=0N>0符 H 2h0

2

2.取2.5kN

0.070.5

fbht

0.72.50.5

1.43

9kNQ235钢 f swHPB235钢 f 取N=0N=0N>0A S

0.7

fbht

1.3 t1.5

f hsw

0.056N

f hyv0

1.3 38000090000

2.51.582153000.05677200／21045mm0.837mm若采φ8双肢箍

A ＝250.3mm2＝100.6SV配钢箍φ8@100

S 100 mm119 mm0.8375-84-8

C40f 19.5Nmm2fc

0Nm25fa

315N mm2

Ha005

W 2.896106mm3Aax

1.78104mm2,

1104

M。HZ450a-440x300x11.5x21.020解：由

1.0,1b b

0.811

4-8图 0.8 0.517f f1 y a 1

36031520.003Es

20.0032.0610515021(28021h ha0

mm) 280211 2 f x ysy

( h

)1

12721017b 1 0f af

(x

)

(0.8)1113890a h 1 0.5170.8b 1 0

11.25

11.25x，10 1

20 1N [2.5h

[2.5(0.3950.917)]11.5762.3315aw 1 2 w0 aN

f

f'

f'

AA

1c ys a s s a 0.579xh0

b0.579762.3441.4mm。