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钢结构设计基本原理课后答案-- 肖亚明合肥工业大学出版社出版 (肖亚明编)第三章1.解:Q235钢、=160N/mm2、N=600kN采用侧面角焊缝f 最小焊脚尺寸:f f 角钢肢背处最大焊脚尺寸:f f角钢肢尖处最大焊脚尺寸:h乞t-(1〜2)=10-(1〜2)=9~8mmf角钢肢尖和肢背都取
h=8mmf查表3-2得:©=0.65、K=0.352N =0.65600=390kN,N^KN=0.35600=210kN1 2所需焊缝计算长度:N1
390103
二217.63mmw1 20.7hfwffN2
20.78160210103 -117.19mmw2 20.7hfff焊缝的实际长度为:
w 20.78160w1 h=I2h二217.63 28二233.63mm,取240mmw1 1 2h=117.1928=133.19mm,取140mm。2=lw2 ffh=6mmf正面角焊缝承担的内力为:N=0.7h'I:f=0.7621001.22160=163.97kN3 f侧面角焊缝承担的内力为:N=QN—2/2=0.65"00—163.97/2=308.01kN1N=K2 2
N-N3
/2=0.35600-163.97/2=128.02kN所需焊缝计算长度:N 308.01! i
229.17mmw1 20.7hfwff1 N2w22O.7hfwff
20.76160128.0210320.76160
=95.25mm焊缝的实际长度为:11=l1w1
h二229.仃6=235.17mm,取240mm。f12工lw2hf2.解:Q235钢、 fT=160N/mm2(1) 取d=d=170mm1 2
6=101.25mm,取110mm。角焊缝受到轴力和剪力的共同作用,且均作用在角焊缝形心处15角焊缝受到轴力为:N=600 499.23kN.1.51角焊缝受到剪力为:V=600
.1.5
332.82kNl 170-2h=340-2hw=2 f
1_2 .2499.2310320.7hl1.4h(340-2h「‘f 2fw f f
二1.22160二f由上式求得: h5.55mmf332.82103
-f
=160N/mm220.7hl1.4h(340-2h) ffw f f由上式求得:最小焊脚尺寸
h-4.49mmfh-1.5t=1.5 20=6.71mmf maxf故取焊脚尺寸为h=7mmf3499.2310
=156.26N/mm2乞-f
fw=195.2N/mm2f20.7hlfw
1.47(340-27)332.82103
104.18兰=160N/mm2f20.7h1fw
1.47(340-27).匚/Jf
2二.156.26/1.22
104.182
=165.10N/mm2 f「=160N/mm2ffh=8mm可满足要求。f(2)改取d=150mm,d=190mm1 2角焊缝受到轴力、剪力和弯矩的共同作用15角焊缝受到轴力为:N=600———=499.23kN1角焊缝受到剪力为:V=600——' =332.82kNJ1.5+12角焊缝受到弯矩为:M=Ne=499.2320=9984.6kNmm=9.9846106Nmme h=0.78=5.6mm,匚=150190-2h=340-28= e 35.6324 6 4二31.7410mm12W2I231.7410Www
6=195.96103mm3最大正应力为:
324一324N M499.23 1
9.984610620.7hl W 20.78324 195.96103fw w二188.53N/mm2
::1.22160=195.2N/mm2V20.7hlV20.7hl332.8210320.78324-91.72-fw=160N/mm2f.二/J2 二.188.53/1.22f 2
91.722
=179.70N/mm2
fw=160N/mm2f3.解:Q235钢、 fT=160N/mm2角焊缝①
不满足要求。最小焊脚尺寸:h_1.5t -1.5一12=5.2mmf ma:最大焊脚尺寸:h-1.2tf最大焊脚尺寸:h-1.2tfmin=1.212=14.4mmf角焊缝①受到轴力和弯矩的共同作用角焊缝受到轴力为:N=F=100kN角焊缝受到弯矩为:M=Ne=10020=2kNmh=0.76=4.2mm,l=200-2h=200-26=188mme w fI=2竺^8^w
=4.6516mm412Ww最大正应力为:
_2lw一188
2
65106188
49.4810mm3 3M20.7hl+—fw
100103
2106+---------103.74N/mm2角焊缝②③
— 20.76188 49.48103::1.22160=195.2N/mm2满足要求。f角焊缝的有效截面如图所示取hf1JVZ
=8mm12[/////
Id——x5.6角焊缝有效截面形心位置:y=y2
=[1345.6(192122.8) 255.45.6(192-2.8)21925.6192/2]/(1345.6255.45.621925.6)=136.03mm剪力由腹板焊缝承担,腹板面积为:2A=21925.6=2150.4mmwx轴的惯性矩为:I-1345.6(192-136.03122.8)2 255.45.6(192-136.03-2.8)2w21925.6(136.03-192/2)225.61923/12=15.56106mm4I15.56I15.5606=21.15104mm3W192-136.03125.6Ww2 h2
615.5610=27.81104mm3192-136.03腹板底边缘处的截面模量:弯矩:M=100120=12kNmw弯矩:M=100120=12kNmwy15.56106=11.44104mm3136.03由弯矩得最大应力为:MWMW1210611.441042=104.90N/mm3腹板的剪应力为:二F/Aw
3 2=10010/2150.4=46.50N/mm翼缘和腹板连接处的折算应力;込〕2込〕2+46.502<1.22丿=97.75N/mm2::=160N/mm24.解:Q235钢、查附表1-2得ftW=185N/mm2, w 2f=125N/mm,vfw=215N/mm2c焊缝有效截面形心位置:y=y2二[12612(188 6)18812188/2]/(12612 18812)=134.13mmx轴的惯性矩为:■j■j22口12l=12612(18812-134.13-6)2 18812(134.13-188/2)2 121883/12x=15.70106mm4翼缘焊缝边缘的截面模量:l15.70106xVW|:
-23.83104mm3y 18812-134.131翼缘和腹板连接处的截面模量:
15.70106188—
=29.14104mm3腹板底边缘处的截面模量:弯矩:M-120F
W二」3y32
15.70106134.13
=11.70104mm3MW1
120F23.8310
4
乞ft
=185N/mm2得:F乞18523.83104/120=367.38kN由腹板下边缘处焊缝压应力M 120F11.70104
w<fc
=215N/mm2得:F乞21511.70104/120=209.62kNw.二F/A二F/(18812)空f,=125N/mm2w得:F<12518812282.00kN腹板下端点正应力、剪应力均较大,由腹板下端点的折算应力120F\J1.70X10120F\J1.70X104丿F18812乞1.1ftw=1.1185=203.5N/mm2综上得该连接所能承受的最大荷载为
F158.84kN5.解:Q345B钢、f5.解:Q345B钢、fwf=200N/mm2,h=8mmf角焊缝有效截面的形心位置:X二為=(21975.6197/2)/(21975.65005.6)=43.41mmx截面的惯性矩:I=21975.62502wxI=21975.6(197/2-43.41)2wy5.650043.412=19.11106mm4
y25.61973/12J=Iwx
3 3 4I=(196.2319.11)10=215.3410mmwy扭矩:T=(300 205-43.41)F=461.59F由扭矩产生的最大应力TTrTTry461.59F2506=535.8810“F_fw=200N/mm2f得:F乞200/535.8810^-373.22kN461.59F
吟也=329.2310十215.34106< f=1.22200=244N/mm2f f"得:F乞244/329.2310“=741.12kNF产生的应力F_
=199.7410--
fw 2A_21975.6500w
f f 二244N/得:F乞244/199.7410$=1221.59kN由最大应力点的折算应力(T)2
329.23x10(F+199.74x10》F"1.22
(535.8810』F)2=689.3210
6冬fT=200N/mm2得:F岂200/689.3210』=290.14kN综上得该连接所能承受的最大荷载为 F=290.14kN。6.解:Q235B钢、fv
=140N/mm2,fc
=305N/mm‘单个螺栓的抗剪承载力设计值:二d2f:=2 4tf;=2214305=93.94kN508045 458050二N min(N^N)=93.94kN二;in :一侧所需的螺栓数目为:一侧所需的螺栓数目为:n=N/N =700/93.94=7.45(个),取n=8min螺栓的布置如图所示:净截面强度验算:螺栓群最外端净截面强度:二二N/An二二N/An=70010/[(360-423.5)14]=187.97N/mm<f-215N/mm|_净截面强度满足要求。i_i_7.解:Q235钢、f=140N/mm,vf=305N/mm,fc t=170N/mm单个螺栓的抗剪承载力设计值:
2兀X222=14
140=53.22kNNo tfbc
=2218305=120.78kN=min(N,N)=53.22kNVb :单个螺栓的抗拉承载力设计值:■:de2二N? b二t
AfJ=303170=51.51kNe(1) 牛腿下支托承受剪力时F0320N= 2
0.133F乞N
=51.51kN1 y 2; ;
160
2 t320)得:FE386.32kN故支托承受剪力时该连接所能承受得最大荷载设计值为 F386.32kN牛腿下支托不承受剪力时由受力最大的螺栓承受的拉力N MyL^
F[603*
0.133F空N
=51.51kNtm'y;2
22(802
1602-2402
2 t3202)得: <386.32kN由NV 二F/n二F/10=0.1F空N:=120.78kN得:F<1207.80kN由剪力和弯矩共同作用下2「咤二吐]2「咤二吐]2J I---■=0.00319^1+-综上得支托承受剪力时该连接所能承受得最大荷载设计值为F=313.51kN8.解:Q345B钢螺栓采用M22C级,fvb=140N/mm2,螺栓的布置如图所示单个螺栓的抗剪承载力设计值:
fb=385N/mm2c22二22二b-f=1 140=53.22kNv4 4N:二d'tfc:=2210385=84.70kNN min(Nb,N)=53.22kNmin= :螺栓群承受剪力及扭矩的共同作用扭矩:T=Fe=120(0.30.050.105/2)=48.3kNm受力最大的螺栓承受的应力为:N1Tx
Tyi-2-2
48.^10^20022
二42.45kN2~'X'yi
1052.524(1002
2002)N48.3x10N°
x52.5
=11.14kN1Ty
1052.52
4(1002
2002)N1Vy
F120n10
12kN.N21TX
N N1Ty 1Vy
二.42.45
2 11.1412满足要求,故采用M22
=48.35kN:::Nm^-53.22kN8.8级,M20摩擦型连接预拉力P=125kN,接触表面用喷丸处理,丄-0.45单个螺栓的抗剪承载力设计值:N/=0.9njP=0.920.45125=101.3kN一侧所需的螺栓数目为:n二與700=6.9n=7(2)承压型连接fv=250N/mm2,
fb=470N/mm2c
N101.3V螺栓的布置如图所示单个螺栓的抗剪承载力设计值:dd22Jrx2024N:二d'tf;=2014470=131.6kNNmin=min(N:,Nb)=131.6kN一侧所需的螺栓数目为:10.10.9
700Nb 131.6i
-5.3个,取n=6摩擦型连接预拉力P=190kN,接触表面用喷丸处理,」•-0.50①角钢肢与牛腿连接n=580mm50mm单个螺栓的抗剪承载力设计值:N"=0.9nUP=0.9x2x0.50xl90=171.00kNf此处连接承受剪力和扭矩的共同作用,则受力最大的螺栓承受的应力为:
T350(180一55)=43.75kNN =N/n=350/5=70kNvyTr 43.75106y
=109.38kNTx xH 222(8021602)2{(NTX
+(NVy
=J(109.38$+(70$=129.86kNv满足要求。
=171.00kNN7② 角钢肢与柱翼缘连接n=10,对称布置在牛腿两侧,栓距为80mm50mm此处连接承受剪力和弯矩的共同作用,弯矩受力最大的螺栓承受的拉力为:
M=350180=63kNmMy 631061N —t 2 2
78.75kN::0.8^0.8190=152kNm'y2
4(802
1602)单个螺栓的抗剪承载力设计值:fNb=0.9n」(P—1.25NJ=0.910.50(190—1.2578.75)=41.20kNfN=N/n=350/10=35kN::N?=41.20kNv承压型连接
满足要求。f;二310N/mm2①角钢肢与牛腿连接
fb=590N/mm2,fc t
=500N/mm2n=480mm50mm单个螺栓的抗剪承载力设计值:2=2310=235.56kN=2310=235.56kNN:=d'tfbc
=2220590=259.6kNNmin二min(N:,N:)=235.56kNT=350(180-55)=43.75kNm则受力最大的螺栓承受的应力为:二N/n=350/4=87.5kNTr 43.75106N yTX 2(402
1201202)
=164.06kN..N2Tx
N2= 164.06Vy
87.52
=185.94kN::: =235.56kN满足要求。②角钢肢与柱翼缘连接n=8单个螺栓的抗剪承载力设计值:
80mm50mmbfv4
22二224
310=117.78kNN;=d'tfj=2220590=259.6kNNm^-min(N:N:)=117.78kN此处连接承受剪力和弯矩的共同作用,弯矩受力最大M=350180=63kNm的螺栓承受的拉力为:63106Nt 4(402
1201202)
=82.03kNN=N/n=350/8=43.75kN::N/1.2=216.33kNv :‘82.03‘82.03151.50‘43.75丫 I二0.661第四章三.计算题1.解:Q235截面I净截面面积:nA=210汇(45n
+402
+45—2汇21.5)=3094.06mm2N/代—f,得:N—Anf=3094.06215=665.22kNu
10045-21.5)=3370mm
An=2
10(45nEf 由应力•:;=N/A 得:N^Af=nEf 综上得此拉杆所能承受的最大拉力为:查型钢表得:i =3.05cm,i -x ymin
N=665.22kN4-2I'丨-250由长细比,=l/iU,得:l^<iJ.1-3.05250=762.5cm0所以,此拉杆容许达到的最大计算长度为:l。=762.5cm。2.解:Q235钢,N=2400kN,l=6mAB柱为两端铰接轴心受压柱截面几何特性:A=2320164008=13440mm2840012
32 6 4232016208=485.6910mm31632012
6=87.38106
4mm4i二「 I/A=.485.69106/13440=190.1mmx xi=.Iy
/A=.87.38106/13440=80.63mmyoy61080.63oy61080.633=74.40::I’I一150长细比:=ox=331.56::丨-1501'xi190.1整体稳定:由长细比查表得稳定系数为:
申=0.930(b申=0.723b类)2400103A
x =186.10/x
2::f-215N/mm2x 0.96013440324001022f二215N/mmy轴整体稳定不满足要求;由以上计算结果知,此柱在y支撑来满足稳定承载力的要求。3.解:Q235钢,I=10m(1)截面a截面几何特性:A=2x320x20+320x10=16000mm231032012
232020170
2 6 4=397.2310mm2012
3=109.23106mm4i=I/A=397.23106
/16000=157.56mmx x二i .I/^.109.23106/16000=82.62mm二y y33长细比:oy10103 =121.03::I’I-15082.62X 由长细比查表得稳定系数为:护=0.788(boy10103 =121.03::I’I-15082.62X 得:N乞 Af=0.37516000205=1230kNmina(2)截面b截面几何特性:
N1230kNA=24001640038400 240016
16000mm22 6 4=596.4410mm=2 16400312
=170.67106mm4ifjlx/A=596.44106/1600^193.07mmx二il/A .170.67二y= y
/16000=103.28mm长细比:
l10 oxi-193.07oy1010oy10103=96.8:::i'丨=150103.28
=51.8::丨」150由长细比查表得稳定系数为:牛由整体稳定 f牛
化=0.848(b类),申y=0.476(c类)得:N乞 Af=0.47616000215=1637.44kNmin故截面b的柱所能承受的最大轴心压力设计值为: N=1637.44kN4.解:Q235钢,N=1000kN,l=3m,l=6m0x 0y取'=80, =80x 'yx 查表得:半=0.688(b类),=0.688(bx 所需截面面积:—
31000―10
6760.41mm20.688乂215i=l 二3000/80=37.5mmx ox/'xi=ly oy
/■=6000/80=75mmy由各截面回转半径的近似值查附表 9-1得:i=0.28h,i^0.24bx贝U h=37.5/0.28=133.93mm,b=75/0.24312.5mm选取截面220012512截面验算:查型钢表得:A=237.9=75.8cm2,i=3.57cm,i^9.54cmox310ox3103二84.0::I'I-15035.7oy610oy6103=62.9::I'1-15095.4查表得:®x=0.661(b类),半y=0.792(b类)00 10旦」°
3=199.58N/mm
f215N/mm:A0.66175802: x满足要求,故选用220012512。5.解:Q235钢,N=1000kN,l=6m,l=3m0x 0y热轧工字钢查表得:\=0.783 (a类),®y=0.688(b类)所需截面面积:A=
f0.688215
NJ000
"760.41mm2i =6000/80=75mmx=lox/'xi -l/'-3000/80=37.5mmy oy y由各截面回转半径的近似值查附表9-1得:,i=0.39hi=0.20b,x y贝U h=75/0.39=192.31mm,b=37.5/0.20=187.5mm选取截面45a(h=450mm,b=150mm)截面验算:b/h=0.333v0.8查型钢表得:A=102cm2,i=17.7cm,i^2.89cmxloxi177=oxi177=33.9::I'I-150=103.8::I'I-150oy护oy护310328.9(a类),(b查表得:查表得:x=0.955类)®y=0.530N1000103A0.53010200184.98N/mm22::f=215N/mm22x满足要求,故选用45a。3取■X=80'y=80查表得:®x=0.688(b类),®y=0.578(c类)所需截面面积:所需截面面积:N1000103AA28047mm0.578x215i /Z.y=3000/80=37.5mmy=loy由各截面回转半径的近似值查附表9-1得:i=0.43h,i=0.24bx y贝U h=75/0.43=174.4mm,b=37.5/0.24=156.25mm200mm12mm250mm10mm截面几何特性:A=220012 25010=7300mm22200121312=95.39106mm4310250122 =16106mm412i二.l/A二95.39106/7300=114.31mmx xi二|/A=.16106/7300=46.82mmy y长细比:长细比:"i+.31=52.5「丄150x由长细比查表得稳定系数为:
玖=0.844(b类),甲 =0.681(c类)N_1000103Ax0.6817300
=201.15N/mm2
y::f=215N/mm2y局部稳定:b_
满足要求95 ).135/fy
=100.164.1=16.41t 127.92::(100.1hp_:=250=25<(25
0.5
max
).135/fy
=250.564.1=57.05oy3103=oy3103=64.1::I’I-15046.82w满足要求刚度、整体稳定及局部稳定均满足要求,故选用此截面++:■hrl1■1il113■1■■1■11■■1*・11・H111■■■1IIFli2设计柱脚,基础混凝土的强度等级为 C20O/a22nO仁=9.6N/mm2,锚栓采用d=20mm,则其孔面积约为5000mm1) 底板所需面积
000139.6
435000=109167mm2底板宽度:底板长度:
B=200210270二360mmL二A/B109167/360=303.24mm,取360mm采用BL=360360,靴梁、肋板的布置如图所示基础对底板的压应力为:二q N/A二n
=1000103
3/(3602
2 2-5000)=8.02N/mm2三边支承板:区格①bJa=43/200=0.215:::0.3,可按悬臂板计算M=1/2qb;=1/28.0243=7414.5Nmm区格②1b /a^70/2500.28::0.3,可按悬臂板计算M^1/2qb;^^1/28.0270^19649Nmm四边支承板(区格③):b/a=250/95=2.63,查表4-8得。=0.1121 1M^■qa2
=0.1128.02952
=8131Nmmmax底板厚度t_.6M /f=619649/205=23.98mm,取t=24mm。2)靴梁计算max靴梁与柱身的连接按承受柱的压力计算,为四条侧面角焊缝连接,取fh=10mm,则焊缝的计算长度为fN40.7hf取靴梁高即焊缝长度260mm
100010340.710160
=223mmf靴梁与底板的连接焊缝传递柱压力,取h=10mm,所需焊缝总计算长度为fN1.220.7hf焊缝的实际计算总长度超过此值。
1000103 =732mm1.220.710160靴梁作为支承于柱边的双悬臂简支梁,悬伸部分长度 l=43mm,取其厚t10mm,底板传给靴梁的荷载q=Bq/2=3608.02/2=1443.6N/mm靴梁支座处最大剪力
Vmax
Fl=1443.643=62074.8N靴梁支座处最大弯矩Mmax
=1/2q1
=1/21443.6432
=1334608.2NmmV.^1.5h
max=1.5
62074.8
=35.81N/mm2::f^125N/mm2ht260X0max
61334608.22
2 2二11.84N/ 二215N/靴梁强度满足要求。3)肋板计算
10260将肋板视为支承于靴梁的悬臂梁,悬伸长度为l=70mmt=10mm偏安全假设为q二Lq/2=3608.02/2=1443.6N/mm2肋板与底板的连接为正面角焊缝,取h=12mm,焊缝强度为qqlzw0.71.22hlfw140.87N/mm2::fw=160N/mm2f0.71.2212肋板与靴梁的连接为侧面角焊缝,所受肋板的支座反力为R=1443.670=101052Nf取h=7mm,所需焊缝长度为f取肋板高度130mm肋板支座处最大剪力
R20.7hff f
101052 =64.45mm20.77160V=ql=1443.670T01052Nmax 2肋板支座处最大弯矩M=1/2ql2max 2
=1/21443.6702
=3536820NmmV 101052T.5 5益八
=116-60N/m^
v=125N/mm2肋板强度满足要求
63536820max10130
-125.57N/mm22~
=215N/mm2I11-—I—121010
_70U③①④-U③①④-G②201070前面柱脚板件区格弯矩相差较大,重新设置肋板如上图基础对底板的压应力为:3 2 2q=N/An=100010/(360-5000)=8.02N/mm三边支承板:区格①bi/ai=43/200=0.215:::0.3,可按悬臂板计算M=1/2qbi i
22=1/28.0243=7414.5Nmm区格②
,b/a=70/120=0.58 P=0.0688,i iM二■qa=0.06888.02i20=7945.57Nmm3 i四边支承板(区格③):b/a=250/95=2.63,查表4-8得:=0.ii2i i22M4=:qa0.ii28.02958i3iNmm两边支承板(区格④):a^:!ii02702T30.38mm,bi
=ii070/130.38=59.06mmb/a^59.06/i30.38二-0.049iM二pa;=0.0498.022
=6680.24Nmm底板厚度t_.6M/f=68i3i/205=i5.42mm,取t=20mm。max6.解:Q235钢,N=2400kN,l=6my(1)选取分肢截面尺寸(按实轴稳定条件确定)取’=60y查表得:\=0.807(b类)所需截面面积:截面验算:
N24001032A 13832mm2%f0.807715i =l /■二6000/60100mmy oy yy查型钢表得:A=268.09=136.18cm2,i=13.63cm,i^2.70cm,z^2.37cm,yJ=496.7cm4长细比:Ioyyl由长细比查表得稳定系数为:
44.0:::〔•I-150—0.882(b类)3N24001030.88213618满足要求,故选用2[36b。
199.82N/mm:::
2f=215N/mm\A(2)确定分肢间距(按虚轴稳定条件确定)取缀条.565,查得A=5.42cm2TAJ W7盟叔二 i -二 x=lox/‘x查附表9-1得:ix
:0.44bbix
/0.44=150.0/0.44=340.91mm,取b=340mm。两槽钢翼缘间净距为340-29^144mm100mm,满足构造要求。验算虚轴稳定:Ix=2
L吟号-ZI212
=2 496.7104 6809(170-237)2iI3014110x=148.77mmiI3014110x=148.77mmxA一,613618=l/ix ox x
=6000/148.77=40.3mm40.32ox=150
+27
13618
=44.3£k】查得:
“0.881(b类)24001030.8811361827.0故分肢的刚度、强度和整体稳定无需进行验算,均满足要求斜缀条稳定、,、,Af.f13618215柱剪力:yV34.44kN“N185.235VV/28534.44/2斜缀条内力:11si-sinvsin45…24.35kN'1=101丿'i-[(340-223.7)/sin45°]/11.0=37.62:::丨1-150min查表得:「=0.908(b类)1-5得:=0.60.0015'扎=0.60.001537.62=0.6563吐_24.3510A一0.908542
=49.48N/mm2
::f=0.656215=141.04N/mm2满足要求。f缀条无孔洞削弱,不必验算强度。缀条的连接角焊缝采用两面侧焊,按要求取hf=5mm
=0.85则焊缝的计算长度:kN
0.724.35103w 0.750.850.724.35103w 0.750.851600.324.35103w0.750.85160w1 0.7hff fkN21 =15.34mmw2 0.7hff f肢尖与肢背焊缝长度均取50mm。柱中间设置横隔,与斜缀条节点配合设置。7.解:Q235钢,N=1500kN,I=7.5m(1)选取分肢截面尺寸(按实轴稳定条件确定)取=60取y查表得:\=0.807(b类)所需截面面积: =
1500
3~~10
8645mm2®f0.807汉215yi9-1得:9-1得:i:y0.38h,故h:i/0.38=125/0.38=328.95mmy
二7500/60=125mm选取截面2[32a截面验算:查型钢表得:查型钢表得:A=248.5=97cm,i=12.44cm,2.yh=2.51cm,z=2.24cm°,41=304.741l7.5103l7.5103ioy一124.4y=60.3:::I■丨=150由长细比查表得稳定系数为: 3=0.806(b类)3150010 2 20.8069700满足要求,故选用2[32a。
=191.86N/mm:::二215N/mm确定分肢间距(按虚轴稳定条件确定)ox='y
=60.3max0.5 =0.560.3=30.2,取「=30max= ;-2x
二60.32
-302
=52.3i=l/x ox x
=7500/52.3=143.40mm查附表9-1得:ix
0.44b,故b:i/0.44=143.40/0.44=325.91mm,取xb=320mm。两槽钢翼缘间净距为320-288=144mm100mm,满足构造要求。验算虚轴稳定:l二丫詬=3025.1=753mm,取缀板净距l=760mmo1 o1=2沪。 21=2304.7104
4850(160-22.4)2=189.75=189.75=139.86mm'x=lox x
=7500/139.86=53.6mmox「x2=ox
53.62
302
=61.4」1-150查得:「X=0.800(b类)
N1500103=0.8009700
=193.30N/mm2 2=215N/mm满足要求。=215N/mm打=30<:0.7^^乂=0.561.4=30.7及40故分肢的刚度、强度和整体稳定无需进行验算,均满足要求。缀板设计2纵向高度h
c(320-222.4)=183.47mm,3 b、 c 320—2氏224厚度tb
40
6.88mm,取htb b
=200mm8mm。相邻缀板净距l=760mm,相邻缀板中心矩l^lh^760-200=960mm。o1 o1缀板线刚度之和与分肢线刚度比值为:「/c_
2 (8 203/12)/(320-2 22 4七.21 6I" 304.7104/960缀板刚度满足要求。柱剪力: V舊235 y9700 215=24.54kNi85il缀板弯矩: Mlb1
Vli V
24.54炙02
=5889.6kNmm缀板剪力:
24.549602(320-222.4)
=42.80kN6M 65889.6b1
=110.43N/mm2
:f=215N/mm:bth 8200b1.5V_1.542.8 b1 103th 8200b b
=40.12N/mm
:fv=
22125N/mm满足缀板的强度要求。缀板焊缝计算。采用三面围焊,计算时可偏安全地仅考虑端部纵向焊缝,取焊脚尺寸h=7mm,l^200mm。则fA=0.7hw f
l^0.77200=980mm20.772002二32666.67mmW”aM 5889.6103bi =180.29N/mm2mm2W mm2
:: f-f
=1.22160=195.2N/428010lV .1=A w
43.67N/mm2
fw=160N/mm2f在弯矩和剪力共同作用下焊缝的折算应力为"80.29
2〕+43.672
=154.10N/mm2cf「=160N/mmf焊缝强度满足要求。四.计算题1.解:Q235钢,l=5.5m荷载标准值:qk
第五早=10.524=34.5kN/m何载设计值:何载设计值:q=24=46.2kN/m11弯矩设计值:Mxql246.25.52=174.69kNm88证梁的整体稳定由抗弯强度要求的截面模量为:M 174.69106W xnx 1.05215
=773.82103mm3x 查型钢表选用I36a,截面几何特性:W=878cm3,I=15796cm4,质量q= 60.0kg/x 强度验算:Mx
174.69
0.61.255002
/8=192.44N/mm2
::f=215N/mm21.05878103满足要求。挠度验算:u5 q』3 5乂(34.5+0.6)汉55003 1 b】1l-384EIx 3842.0610515796104 一428I-250满足要求。故选用此截面。假定梁的受压翼缘无可靠的侧向支承按整体稳定确定梁截面145〜163之间,均布荷载作用在梁上翼缘,自由长度il=5.5m,由附表3-2查政体稳定系数\=0.660 0.6,所以i\=1.07-喑=1.07严=0.643\0.660x所需毛截面抵抗矩:W Mxn IX
174.6910
61263.63103mm3f0.643215x查型钢表选用145a, 截面几何特性:W-1433cm3,I32241cm4,质x量q= 80.4kg/mM174.6910M174.69106X0.8041.255002/8=118.52N/mm2二215N/mm满足要求整体稳定验算:174.691060.8041.255002/8=193.54N/mm=215N/mm=193.54N/mm=215N/mmbxW0.643X433汉103满足要求。故选用此截面。(3)假设梁的跨度中点处受压翼缘设置一可靠的侧向支承,一定 此梁的整体稳定保证。按整体稳定确定梁截面I22〜I40L,=5.5/2=2.75m3-2查整体稳定系数\=2.1 0.6,所以“07寤5詈=0.936b所需毛截面抵抗矩:所需毛截面抵抗矩:10%飞厂0.936^15“68®亦查型钢表选用I36c,截面几何特性: W=964cm3,I17351cm4,质量xq=71.3g/mM 174.69106 0.7131.255002/81.05x964x103=175.78N/mm:::f=215Nx 21.05x964x103=175.78N/mm:::f=215N满足要求整体稳定验算:6 2174.6910 0.7131.25500/8=197.19N/mm2::f=215N/mm2Wbx 0.936x964x103满足要求故选用此截面2.解:Q235钢,I=15m,!)-1/400截面形心位置
420201230320 201012008620 …x 670mm420汉20+320汉20+1200汉88120032212 81200(670-620)42020(1240-670-10)2 6 432020(670-10)=6598.0810mm抗弯强度6” l 6598.08x
=11.576106mm3W —x11240-670 570I 6598.08106W x
=9.848106mm3x2 6708
6703.2152 4005=2090kNmM 2090106x_W 1.059.848
-202.12N/mm2::f=205N/mm2xnx2
满足要求。抗剪强度中和轴处截面的面积矩:2 3 3S=42020(1240-670-10) (1240-670-20)8/2=591410mmxo上翼缘与腹板交界处的面积矩:S=42020(1240-670-10)=470410mmx1下翼缘与腹板交界处的面积矩:S=32020(670-10)=4224103mm3x2支座处剪力最大,所受最大剪应力为:VS (4003.27.5)103xo-6
5914103
247.50N/mm:仁=120N/mmIt 6598.08106
ci"'_x
满足要求。z 取支撑长度a=0.2m,贝Ul=a5h=200520=z '■F1.0400105cc
2 2166.67N/mm::f=205N/mmlt 3008zw满足要求,无需设置支撑加劲肋。折算应力在集中力作用处弯矩、剪力、局部压应力均较大集中力作用处:Mx=4005 3.27.55-3.252.5=2080kNmV=4003.27.5-3.25=408kNMxMx20801011.576106-179.68N/mm2Wnx1VS 4081034704103t6598.081068=36.36N/mmt6598.081068=36.36N/mmI 0.282=1.07
=0.806: =1.07-
一0282b 1.07b二所需毛截面抵抗矩: M二x
73.810
425.88103mm3f0.806215二=166.67N/mm3二c腹板与上翼缘交界处的折算应力为:£2三3;=;;179.682
166.672
-166.67179.68 36.362(5)挠度
h77.31N/mm222
::1.1f=1.1205=225.5N/mm2满足要求。_,M』_
(2090/1.2)10615000
_1.!]_1l10EI
102.061056598.08106x
520I4003.解:Q235Bffl,l=6m1Mmax
满足要求。(61.2
301.4)6=73.8kNm5考虑到葫芦轮子对轨道梁下翼缘的磨损,所需3 3 3W=425.8810/0.9=473.210mmxx查型钢表选用128a,截面几何特性:W=508cm3,I^7115cm4,质x量q=43.5kg/m强度验算:Mx
73.8106
6000.43512
/8=
/rW215N/mmWxnx
1.05508103整体稳定验算:600/8600/873.81060.43512。2=低蒯/rW 215N/mm0.806508103满足要求。故选用此截面。4.解:Q235B钢,l=9m截面几何特性:A=2200128800=11200mm2xW
满足要求。bx8800二I二x 12
2200124062
=1132.55106mm4Iy=2
12200126
=16106mm46 3W=1132.5510/412=2.74910mmix
=Jly
/A二16106/11200=37.80mm1 12M 0.91.292x8 4
1101.29=307.94kNm梁在跨中无侧向支承,集中荷载作用于梁上翼缘y=ly/iy=9000/37.80=238.10h=824mm,h=12mm900012=0.655:::2.0bh 200824i查附表3-1得:—0.730.18=0.730.180.655=0.8479-嚟14;jcc,”4320x11200x824L『238.102、=0.8479汇---2I
---------=<'1+--------238.102x2.749汉106<4.4x824丿M 307.9410x 厂405.86N/mm2f=215N/mm:bWx 0.2762.749106整体稳定不满足要求。Q345B钢厂‘筈阳■/ 4.4h 235345=0.8479心笔
11200竿
2『238.10x12、235238.10°2.749X06 ------------Ix——=0.188 <4.4^824丿3456xM 307.94106x
2595.84N/mmf=310N/mm
整体稳定不满足要求。查附表3-1得:—2.23-0.28=2.23-0.280.655=2.0466432011200824 ‘238.10汉二 12y2.0466=0.6660.6
238.10
2610 I4.4汉824=1.07-0.282/0.666=0.646b307.9410622W 0.6462.749106 =173.40N/mm:f=215N/mmbx整体稳定满足要求跨度中点增设上翼缘侧向支承/i=4500/37.80=119.05y=|y yh=824mm,鮎=12mm3-1得:'=1.754320Ahb4320Ahbyt124.4h2=1.752
432011200824 519.05x12、119.052-1.9250.6
2.749106
\、4.4x824;: =1.07-0.282/1.925=0.924b307.94106bW< 0.9242.749
=121.23N/mm2
::f=215N/mm2整体稳定满足要求5.解:Q235钢,I=6m,')-l/200075疋J1+252檩条沿屋面坡向间距为
025
2.5
—^=0.808m选用[10,自重0.10kN/m,加拉条后取0.15kN/m截面几何特性:3 4 3,W=39.7cm l,x
=198cm,Wymin
=7.8cm,ix
=3.94cm,i
y=1
.42cm檩条线荷载标准值为:q=0.75k檩条线荷载设计值为:
5二1.131N/mmq=0.751.20.808 1.40.75=1.432N/mmq=1.4322.5/1x
2.52
=1.330N/mm弯矩设计值为:
q=1.4321/、1y
=0.532N/mmM -1/81.33062x
=5.985kNmM=1/80.532y
23=0.598kNmMM5.985106 0.598106x y3 3Wxnx
W1.0539.710 1.27.810yny=207.46N/mm2::f=215N/mm2挠度验算:垂直于屋面方向的挠度
满足要求。v5qj3 5 1.131X2.5/J1+2.52
x60003
1H1I384EIx
384
2.06105198104不满足要求。
138I200故选用此截面强度满足要求,而挠度不能满足要求。6.解:Q235钢选择次梁截面将次梁设计为简支梁,受均布荷载作用次梁的荷载线标准值为:q=(318)3=63kN/mk次梁的荷载线设计值为:q=(1.231.418)3=86.4kN/md次梁的最大弯矩标准值:
M1/863xk=
=283.5kNm次梁的最大弯矩设计值:
Mx=1/886.462
=388.8kNm次梁的整体稳定能得到保证,故按强度计算所需净截面抵抗矩为M 388.8106W xnx 1.05310
-1194.47103mm3查型钢表选用HN446199812,梁自重为66.7kg/m,W=1300cm3,x4l=29000cmxM=1/886.46x
221/81.20.6676=392.4kNm强度验算:
392.4106xW 1.051300x
=287.47N/mm2::满足要求
=310N/mm22-1得'=l/250三_5q』3
_5三(63+0.667)
_1v1l384EIx
3842.0610529000104334I200满足要求。主梁跨间次梁传递给主梁的集中力为:F二(630.667)6= 382.0kNkF=(86.4+1.2x0.667)x6=523.2kN则主梁的支座反力为: R=382.03=1146.0kNkR=523.23=1569.6kN主梁的跨中最大弯矩为:M=1146.09-191.09-382.06-382.03=5157.0kNmxkM=1569.69-261.69-523.26-523.23=7063.2kNmx主梁的整体稳定能得到保证,故按强度计算所需净截面抵抗矩为10 6w①不考虑腹板屈曲后强度腹板高度hw
W叫=7063.2=22803nx
f1.05295x主梁无净空要求,据刚度条件得梁的最小高度:2-1得!丁/400,故hmin— 6
29518000 §
=1585.07mm1.34106!J1.34106梁的经济高度为:h=2Ws xw选取梁的腹板高度为腹板厚度tw
-2(22803103)0.41800mm。
=1753.27mmt=h/3.5二1800/3.5=12.12mm,取t=10mmw w w翼缘尺寸
lw二
228031031800
--101800=9668.33mm6取翼缘宽度为 460mm,厚度为22mm。截面几何特性:A=180010 246022=38240mm2lx=10180012Wx
32 10 4246022911=2.16610mmio2.16610 =23.49106mm4922梁自重为考虑到加劲肋等乘以增大系数 1.2。g=1.27838240二3.58kN/mkg=1.21.27838240=4.30kN/m则主梁的支座反力为:R=382.031/23.5818=1178.22kN则主梁的支座反力为:R=382.031/23.5818=1178.22kN,R=523.2汉3+1/2>^4.30>d8=1608.3kN主梁的跨中最大弯矩为:M=1146.09-191.09-382.06-382.031/83.58182xk-5301.99kNm2M=1569.69-261.69-523.26-523.231/84.30182x=7237.35kNm强度验算:MxW
723735.
=293.43N/mm626
2f二295N/mmxnx
1.0523.49106满足要求。刚度验算:
V 3ht 1608.310w 180010
=89.35kN/mm2::-170kN/mm2满足要求。.Mlkx
5301.99106
1800010EIx
102.06105
2.16610
468I-400故选用此截面。②考虑腹板屈曲后强度
满足要求。截面尺寸取腹板:h=1800mm,t=8mm;翼缘:b=480mm,t=24mmw w截面几何特性:A=18008 248024=37440mm2818003
2480249122
=2.3051010mm4Wx'3050
24.9510mm46924梁自重为78kN/m3,考虑到加劲肋等乘以增大系数g^1.27837440=3.5kN/mk
1.2。g=1.21.27837440=4.2kN/md则主梁的支座反力为: 瓦=382.031/23.518=1177.5kNR=523.2x3+1/2x4.2x18=1607.4kNM=1146.09-191.09-382.06-382.03-1/83.5182xk=5331.15kNmM=1569.69-261.69-523.26-523.231/84.2182x-7233.3kNm强度验算:Mx
67233.310
2=276.11N/mm::f=295N/6htww刚度验算:
1607.410318008
满足要求。=111.62kN/mm::fv满足要求。
=170kN/mm25331.15106
18000
1」丨1腹板局部稳定计算:
102.061052.30510满足要求。
495I一400在支座处和每个次梁处设置支撑加劲肋,另在距梁端部较近范围内设置横向加劲肋,其构造如图所示。严严1r{F{lF123~r~1050105010009002000E 3000 y *t *匕*对板段①/+a=900mm,a1
/h::1,0
s4lj4+5.34(1800/900)6
1.321.2,支Ht的作用。梁端头采用如图所示构造。V=htf/wwv
1212=18008170/1.32.'s
=1754.38kNR=1607.4kN对板段②左侧截面剪力
满足要求。V=1607.4-523.23/2-4.24=805.8kN2相应弯矩2M =(1607.4-261.2)4-523.21-1/24.24=4828kNm2M-480241824295=6198.68kNmf,C=2000mm /h,°
h/t,1 ° w,s
二2.2744(1800/2000)2V=htfwwv
1212/h=1800汉8
=915.33kN>V.
=805.8kN满足要求。对板段③62 12二V htf二u wwv
=18008170/2.55.'s
=796.09kN右侧截面剪力
V=1607.4-523.25/2-4.29=261.6kN3a^3000mm,a/h1 01
h/t° w
=2.551.2415.344(1800/3000)2,V:::0.5V3 u相应弯矩M二M7233.3kNm3180/8.345=5153235卜=(1—0.2/飞)/ 180/8.345=5153235b3 31022.30510Meue
Wf=1.050.93624.95x
6295=7236.97kNmM^7233.3kNm满足要求。其他板段可不验算。计算主梁翼缘连接焊缝翼缘与腹板间产生的沿梁单位长度的水平剪力为1VS1v1 二-1tw3=48024912=10506.24103mm3角焊缝有效截面上承受的剪应力为+ V VS1-f
1 .,wff20.7hf
1.4hIf x所需焊脚尺寸为hf
60741031050624103=2.62mmJ一斗 .=2.62mmJ01.4Ix
f; 1.42.305
200f 由1.524=7.35mm込h込1.289.6mm,取h=8mmf 考虑腹板屈曲后强度,设计主梁加劲肋(略)设计主次梁连接,次梁连接与主梁侧面(略)第六章三.计算题1.解:Q235钢,l°x=l°y=1=2.6m,N=78kN,M-2.4kNm选用截面2 90566查型钢表得截面的几何特性为:2A=28.56=17.12cm,i2x
=2.88cm i,y,
=2.24cm,zx
=2.95cm2 2 4二l iA=2.8817.12=142cm二x xl二i:A=2.242y
17.12=85.90cm89I 竺=48.14卅zx 2.95 x刚度:
■=l/ix ox
=2600/28.8=90.28I-150::」丨-/i=2600/22.4=116.07 ::」丨-■y=loy y满足要求。强度:xM 7810xA W 1712
62.410
=93.04N/3
=215N/mm2xnx
mm2::f满足要求。2.解:Q235钢,l =15m,I =5m,N=900kNox oy截面几何特性:A=23201264010=14080mm23I 10640 26x 12 232012326=1034.6510mmIx2^f^65.541mm4l.I/A=-1034.65106/14080二271.08lxy=.Iy/A=.65.5410/14080=68.22M =1/410015=375kNmx刚度:
=l/i=15000/271.08=55.33 150'x ox x=l/i=5000/68.22=73.29::〔■1-150y oy yM 900103 375x
332
=178.52N/mm2
::f二215N/mm2iwnx
14080
1.051034.65106满足要求由长细比查表得: 玖=0.832(b类),®y=0.730(b类) 弯矩作用平面内稳定: =1.0FXN 二空22061014080=8500.72kNEX1.1':N
1.155.332---+A W(1—0.8N/N900103
1.0375106332_0.832140801.051034.65106(1-0.8900/8500.72)=202.03N/mm2::f=215N/mm弯矩作用平面外稳定:中间段飞".0
满足要求。=1.07-b
22/44000=1.07-73.29/44000=y900103 1.03751063320.730140800.9481034.65106局部稳定:
=214.49N/mm2:f=215N/mm2满足要求。155=12.92::t13对于腹板,柱长中间截面处:
3 375
320=_N.Ms 900103=
1034.65106-max
AW 14080 mmn nx 2=63.92115.98=179.90N/N M 900103x
375106
320-min
A Wn
14080 1034.65=63.92—115.98-52.06N/mm2、二 max一二min 四心鸣 1.29<1.6-max 179.90hot
640=64:161.29
0.555.3225=73.3满足要求故此柱的承载力满足要求。°y3.解:Q235钢,I*=4m,l=2m,N=20kN°y选用截面116查型钢表得截面的几何特性为:,,2 3,,,IA=26.1cm i,x
=6.57cm iy=
1.89cmWx=
141cm=l/i=4000/65.7=60.88:::〔■I-150ox x=l/i=2000/18.9=105.821-150oy yy由长细比查表得:®x=0.879(a类),申=0.518(b类)y
M=28kNmx弯矩作用平面内稳定:弯矩作用平面内稳定::mx=1.0N 二二
206
103
261°=
.56kNN丸EX 1.1N丸2
1 1M1.160.882 Pxx节 W<(1-0.8N/N)X EX20103
1.028106 -0.87926101.05141103(1-0.820/1301.56)=200.20N/mm2::f=215N/mm2弯矩作用平面外稳定:Lx=1.0
满足要求=1.07-b
2 2/44000=1.07-105.82/44000=0.816yN SxMx_20103 1.028106A Wy b
0.51826100.816141103=258.15N/mm2f=215N/mm2不满足要求。F=25kNM=1/4254=25kNmxN MFxx飞 W<(1-0.8N/N)X EX20103 1.02510630.87926101.05141103(1—0.820/1301.56)=179.68N/mm2:::f=215N/mm2弯矩作用平面外稳定:■-=0.65tx
满足要求。N M-txxtxA W
20103 0.65251060.51826100.816141103y bx=156.03N/mm2::f=215N/mm2满足要求。4.解:Q235钢,l=3m,I®=3m,N=80kN,M=40kNmox选用截面120a查型钢表得截面的几何特性为:2 3A=35.5cm,ix
=8.16cm,iy
=2.11cm,Wx
=237cm=l/ix=3000/81.6=36.76:::丨,IT50x ox■y=l°y
/i=3000/21.1=142.18::由长细比查表得:碎由长细比查表得:碎x=0.948(a类),砕y=0.336(b
I-150h弯矩作用平面内稳定:
=065Fx二
2EA二2
206103
3550
=4855.70kNEx 1.1-x"1.136.762N. 1Mmx xA W(1-0.8N/N)x X X EX80103 0.6540106= r 0.94835501.05237103(1-0.880/4855.70)=129.65N/mm2:f=215N/mm2弯矩作用平面外稳定:Lx=0.65
满足要求。b=1.07-■y/44000=1.07T42.18/44000=0.610N M 80103 0.6540106-txxtxx
AW-y b
0.33635500.610237103=246.91N/mm2 f=215N/mm2不满足要求。°x 5.解:Q235钢,l=29.3m,|=18.2m,N=2800kN,°x 2I63a查型钢表得截面几
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