结构力学答案同济

Ⅰ`Ⅰ`Ⅱ(ⅡⅢ W=5×3-4×2– ⅠⅠⅡⅢ(ⅡⅠⅡ

ⅢⅢW=4×3-3×2- ⅠⅡⅢⅠⅠⅡ （ⅡⅢ）ⅢⅢⅠⅡ

ⅢⅡⅢ ⅡⅠⅢ

ⅡⅡⅠ

ABCDEF

24 4AABCD ４ ４４Q４

ABCDEF ABCDEF AMAB E H AA

M 4 42 AD 6BCDABCDA BCBCAD72 724 EAB 4244 N 0 484484

Q15NMADAD qlAEBCAEBCFD

q(lx)x1qx2ql MBC中MBq(lx)

1ql1 x x18

Q209(4.53)RF6RFQME0.52092459405,REMCF453135,MCD0.5209MBA0.520911M11M1QME4.254242MK3.51.50.252对A点求矩：RB724252.5RB对C点求矩：2420.52HB4HBVA3.5(),HA

5.75

244.25MM1

80380,

8063HC对F点求矩:VC2023304)2

803443M

Q433841Q433

对A点求矩416142VB8VB对C点求矩44142H

6

43

8(),V

MC0VB2Fp(),ME02HBMB03FP2a2aHH2FP2aVFHHFP(),VF2FPHD4FP(),VD88888844444444448+-+---++8888可知HB4KN(),VB4KNHI4KN(),VI4KN(),MA42810N qa2qa2 a HC 1.5qa(aA

qa1.5aHAa0HAH0, qa2, 2

1

A

由Fx0知

FAC FBC520FBCF0FF33F BC F3F27.5KN(),FBD3KN,F14KNM0Fx4 MC0FN4

MB0,2FN2aFN1

F0,

4a 2Fa N N2a联立方程解得：FaF2N13N FNN 又易求得杆4=13 FFF

P，FN2 方法一：利用对称性和称性E21DAB E21DAB FFFF 2222212 22 2可求得F17 8 FA2

14

综上，F2F,F5

P 8 1C DD212N N 由F点平衡知,FF又F0,FN12N N F1F,F5FF5 8P N (b)

2EA

由平衡条件知:F2F3,F4 又3F3FF4F23 23

5 F5F，即 F 24 N 再对BF3a4aF3 F5F,

5

6 65 再分析C节点,不难得到FN2 8用同样的方法分析22 可求得F50.5FP,F6FP,F7 FP,F80. A最后用节点法分析E节点,得

0414 3222CDB3141123一. 1234二.再求出 然后可求出FN1 三.由MB0,可求得FC0.75FP四.分析截面右半部分4

0,可求得x1FP

F由节点法,对C分析可求得FN2 F4A D2A2CFBE5FD2A2CFBE5由对称性 由F0,F F FP0F5 由对称性有FF5 再由节点法分析C,D两节点容易求出 1F,F1F 4 2C取截面左侧分析由F0,F 1F0F13 3 1 再由节点法分析A,B节点马上可以求得F1=FP,F13F0,FF1F13F 0FF 2 取截面右侧,由MC0F22dF4dFPdF43FP,F2再由节点法分析D,E节点马上可以求得FDE=2FP,F3

取图示体，对A点取1

F FPD 0,FaF

2aF

1a0F 5 F=-2F, 1F,F 2 1qA

q F

1

0,1qa2 a2qa2a 再分析节点

22qa, 2qa, 1

1

1DEDEFABC0

1qaq2q

,NBF

1qa由对称性

1

由MA1F求得FHG

2FP22(FP1FP2 2(FP1FP2 2(FP1FP2 2P 1 MCFQEC2a(FP12FP2)a a1Fa

C M0,3qa2aF a0C 用节点法分析G节点，易得F=2qa,

考虑DB

由F0 3qa, 3qa2qa5 34344ACBal坐标原点设在A处，由静力平衡可知MAxFQA当FP在C点以左时，MC0FQC0x当FP在C点以右时，MCxaaxFQC1x MC的影响 B以A

假设F

x/ F(la),(xa)x(la/l),(0x a

FRAa,(x

alx,(lxlF

xcos,(0x x x

)cos,(ax (1a)1l

由M0知 341(7x)0 355 NCD NCD12 32(5x),(0x

NCD

E 2,(5xNCDMx3,(0xM 0,(3x 3 1,(0x3)x,(0x

473 ,(3x x,(3x35 4 4FRF QCDDEACB x1,

x4

,

1814

9489498ACB FL、FR、 FL1,(0xa),FR0,(0x

0,(ax0,(0x

1,(ax4a,(5axx4a,(5ax

1x,(0x

x,(0xAEBAEB

x,(0x

x,(0x M

x,(ax2a),

x,(2ax

3 0,(2ax 5x,(4ax 1

1

M2FF11FF1

1

AIFBCGDH1312EFFFEIFGJHACDBFFFF 3 F F

MA015dFRB7d1(5dFx,

x,(0x2d 22d,(2dx5d

当0xa

0

1xa

FBFBE

aFBEFBE ME=(2a-x),FNEFBFBEG

当3ax4a时，由MG0M=x-4a,Fx3ax3, 4 a

1

x当0x8(C点以左)时，取1-1

0

[(10xx)(1x)10]/2 当12x20(D点以右)(1x)

0

x 当0x8时，取1-1截面左侧分析2由F01x sin451知 2x2y由F0

4 01(8dx) 8d 1 1 当0x3d时，取11截面右侧分析F0F 0 当4dx8d时，取11截面左侧分析F0 5x y当0x4d时，取22

M0 4d 2d0 M0 3dF2d 当5dx8d时,取22M0 4dF2d0 x M0 5dF2d 5 5455858轮压为FP3=FP4=250kN，轮距及车挡限位的最小车距如图所示。1 F置于B点 FL 285 250( PCR 55.42KN P 300300a

FR

285250(23127P7.92KN PCR 300 此时R=2851285250(23127) FL 285 PCR 79.2KNP 60a

FR

250250P37.525KN PCR 60 此时R=250+23128525019 综上所述，Rmax14.5(xx4.5)(0x 10 x (0x7.5) (13.5 A=21852x102 (7.5x12)0.15x22.7x7.425(7.5x12)[118[

13.5x]4.5(12x

0.9x14.175(12x5 5当7.5x

0.3x2.70x9。此时A=2.79-810.15-当0x7.5

1.87.54.05 Amax4.725S=qA=4.72556264.6KN,此时x91123AKB求FN1FN2FN3影响线时,只需求得当FP11作用于AB中点时杆3的轴力。求MK的影响线，需求得当FP1作用于AB中点与K点时MK的值。首先，用静力法求得当FP1作用于AB中点时FN1FN2

FL=

F=F FF=F 17 17(12F F25F5(12F 17

N1N2

2FNCD M0F=2F,即12FQC2FF 于是 2,F251.49F171.37F 3 FF1MF42(以下侧受拉为正 1123AKB F17 =- F 2 F M0F8112F16 1M(a)F4 1在K点右侧

1

FL(a) 1 1

AAKAKB

KC

B

0

8

MB0FYA1 1M(b) 121.5FR(b) FL(b) M11.52.5FR11

FL13

F的影响

M=1.4420+102.421.44101.231.44 FR200.6102.420.6101.230.4

106

106MF3.66KNmFRF10 综上所述M64.8658.8KN FR1810 MK的影响 M(2qa1a22qaa)2qa2 22

12

K 根据对称性，FR

22 C100(62)50(10MB0FyA 12

3MK

(62)504355.6KN3 113MC100350FR A50375 20375 50375

MB

FyAFR(50.375)/10Mmax370(50.375)2001.51411.25KNFR A50375 20375 50375

MC

MAFyBFR(50.375)/10Mmax370(50.375)1001.510033002.52001.75100(1.751.00.25)1400KN5-1试回答：用单位荷载法计算结构位移时有何前提条件？单位荷载法是否可用于超静定结构的位移计 a aC aFNCDFNCE0,RARB FNBEFNAD

2FPFNBCFNACFPFDE 121212121F 2(2F) FNFNPl 2 21(FP)2a6.83F EAA＝30cm2，E＝20.6×106N/cm2，FP＝98.1kNC点竖向位移ΔyC。4P45F545k对A节点FNAD=-5FPFNAE对E节点F5 F5 4 41k1k5对A节点FNAD=-

FNAEFNFNPl1(12F2515F6(-5)(5F)25 11.46cm

- - - --4-4-444884444-44-414 4

2

1

4 4FNFNPl

(12 2 ABABlq(x)q2q1x M(x)1qx2q2q1 2 1 1l q11

M(x)Mp(x)dx

EI(

qx3

1x4 =1(q2l411q1l4EI BCBCAllllqll7l745ql4

MP

M q 1

l l qll ql l ll l l) qlEI EIBBAOM()1(Rsin)212R(1cos)M()212

2 0 EI1[2(Rsin 0=(8-3)-1.42(逆时针)

R(1cosAqAqROBqdsM() qRdRsin()qR2(1cos0M()R 1

(1cos)RsinRd 0 EIM()M()dsEI0

（b）（c）（d）（f）11 M(x)5xM(x)231

(0x(3x 01 0ycEIM(x)M(x)dxEI 6MPM

3

EEA1A6

6

(231)1261236

(132162(3)1 +2

612

2kNB2kNBCAD 612323 6 xc (21822182230423018423042366436630)62EI+6(2366)1262623918

k kk181218122MPMds1FF1(11231)10(212 k EI 1(210161)4(2126)1141613)11 (EI EI =

27(顺时针)位移：（a）h＝l/10，求ΔxB；（b）h＝0.5m，求ΔCD（C、D点距离变化。CDCDABllLL111LL11Mtt1t26030

10t FNdst 0 h

1=301l

(l22l2 =30l(102l2)/

00A00A000BN54 54 54 FNdstMdst5 0 h +t55t(1)12t(14324 M M b hb h 11h1h00 FRC[(1a11h1h00 C′1C′10 FR FRC[1C3C]3C1C 2 2 2 2 3 C 3

4a1

4a 2a 4a24a12a 9次超静定(g)IIII只需通I

Cl3

23上图l

M11X11p 1 2 2 ll l

14l

2

3 26EI

2

EI 32 l 7Fl3 1p6EI223lFpl3lFp3 14l

X1

7Fl 0X1 MM1X1M1F 1F1F QQ1X11 Q l l Ml3 MpPF1FpPF211X112X21p21X122X22pMM1X1M2X2MQQ1X1Q2X2

1F3BCBCDA 1616 M11X11pMM1X1MCDCDABq

1111qq2q2 q2q2

qa

qa M11X11pMM1X1M11 M M用图乘法求出11,12,221p211X112X21p21X122X22pEDACEDACB1 36 M

1112

66

233233266233233

62332332661

31618032221 EI22

3 2EI

3 XX

2700

X1X X MCA1803255390KNmMCB18032553120KNMCD6530KNII II AB110kN·m N M M 13233 2332992392 EI 6E 210329109103102 11X11p0X1MDA31.29106.13KNmMDC31.293.87KNmMIIGABC1 M 2332992392 6E 2693625.2 6E

6E

266

266 1134533 2345294053405 12

6456 EI2p

46E

5EI

EI X111.6XEI X XX X1

X

MAD405917.39248.49KNMBF68.69917.39104.37KNmMFE317.3952.17KNmMCG68.6952.14KN

1a1a212 12

6-6

ABABlEA=Cl11 l

2l2l2l

2

F1p

2Fpl2lFpllkk

2lX 0X211 Fl2F2l3F 3FpMaaa a EA= EA=∞ EA=∞ABC 22解 22原结构 22 29 9

2 M19992243 3 1199F92243

3 X 0X111 94

94

4

4M 整体结构MqBADClBADClq

1

ql8 1l12 EI1 11lql21

ql

ql1p 1 EI ql11X11p0X1MM1X1Mql

ql qlqlqlqlql CCFBAEDlql q1

q2q2l 1 M11

EI 11l21l

EI 1l11l113lEI 11lql2l3ql1 EI

1lql

2l

EIl

1 ql

X12EIX28EI X112 1 X3lX

X

qlql

ql9

ql

ql9

ql

ql

qlCICIDIABIIFEql9

qlMG

p取1/2结构 p

p

①

FFp

2

Fp Fp F22 2Fp2F Fp

Fp2Fp2Fp FFp

Faaa2 Fp2 F

FpF2Fp2

Fp2Fp2

Fp2F2

aFaa aEI=常 22 22212 FaM2 1

2a2a 2

EI a22aFaaFa5F 6EI

11X11pM

3EIX112EIa3EIX1X1485F48 5F48 7F 24h4FP h 2 2

D 6-10AA-BCD- -l--l-lA 6-11图示平面链杆系各杆l及EA均相同，杆AB的制作长度短了 C题6-12 题6-13 q6-15CABRRRCABRRR6-16同济大学结构力学第7章位移法习题答 个角位 3个角位移，1个线位 4个角位移，3个线位 3个角位移，1个线位 2个线位 3个角位移，2个线位 kk A A iCiBlll3iZ

1ql36位移型方

Mp

1ql3

1ql23qlZ1M1181ql6ql25qlM DC3EIZ 位移型方r5EI,R 15EIZ35 Z M7M图(KN EA= EA=∞ EA=∞ABC Z1 1121

r4EI,R4

1 Z M9 9 4 2 4F 2EA/2EA/54EA/5r

Z1

4 a4 aF35a15FpMp位移型方r2EA/a,R6 1 5EAZ6F5 5Z MFpMD D ABll Z1 EA 2 r21

Z2rr l14 00R001 R2pr21Z1r22Z2R2prEA

2,rr l 4 r

2 l 4 llZ1

lZ2 l

画M212121222212 212ACEACEF 6m 23

23

3

1323

1163R1pR1p位移型方r21Z1r22Z2R2pr2EI,rr1

116Z115.47,Z23311871Mr21M1iiir22M2RMp位移型方r21Z1r22Z2R2p

4R1p30KN,R2pZ301,Z40 11

MCAEFDBCAEFDBii M1 2M2位移型方

Mpr21Z1r22Z2R2pr11i,rr

4R1p0,R2pll

Z6.316, 12.63M

Eq Eq l2lllllZ1l lllllZ2lM21181ql RM位移型方r21Z1r22Z2R2pr13EI,rr

l

R1ql2,R1 2Z66ql3,Z211 3600 360000008ql0278ql0231ql00176ql80kN80kN·m10kN·mABCD 4m44Z1412

M1M

r21Z1r22Z2R2pr5EI,rr1

78

R1p45KNm,R2pZ138.18,Z2M M Mq

4444

4Z1Z11 2230r112EI,rr4 41 1 9

149RR1R1p20KNR2p0 4EIZ20

EIZ29Z83.381,

71.47 M

DBCDBCA 解：CDCy=1Cx4,CCD知 rEI,rrEI, 4

r22

4EI

9EI, r3EI3EI27 R1p10KNm,R2p0,R3pMD027EI3EI3EI9EI9EIr33

128 8EIZEIZ3EIZ10

Z117.9/ 27EIZ

58.5/4Z110Z 3EIZ27Z0.055EIZ6.25

Z3285.6/ 1160 CB a2a2aaa46EI46EIM1M

0

a9EI

2a18EIa

9218EIPP14M0PaRa

2

11Z 192921 5292292929292

Mq 4EI k= M1Mp2l2lZ11l1ql

1ql8

1qlr10EI8EI

M10EI9EI /211

2l2 R1ql1ql2/l71

85ql55ql4118 7-9qFCGqqFCGqDBEAaa a3i3iZr1ql21ql82iiZ1ql8ql1ql2iql M 22r117i,r12r21i,r228iR5qa2,R13qa21p 2p Z153 159ql

104ql5543ql17767 ABCABCDE 2

3 rEIR300KN 14EIZ300 Z1300

CCAB 245

rEI4EI21 R1p20KN21EIZ20 Z EAll N125lZ1r18PN45M 11 r12EIx212EI132EI,R4 5Z1

1 3NN8222M 50kN Z 2 rr3294923498 rZ24323

r43

3r

Mpr8EI,rr4EI,r20 Z25

33666 33CECEFDBA Z111

MpB

l l Δl

6i6i

l

r16i,rr6i,rr6i,r16i,r

l287EI 193EIl373EI 467EI 试用位移法求作下列结构由于温度变化产生的M图。已知杆件截面高度ABCABC题7-13l

ll

20

l

3453453

Mt（2）r5EIR95EIR11 5EIZ95EI Z DACFDACFBEllll题7-14l同济大学结构力学第8章矩阵位移法习题答BCBCD ①1①12②3③4x局部坐标系（ij①1l10②2l10③3l104112233441F 1

Fy2

Fy3

M44

2 l 2 6l

4l2- k①

k①k① l3 - 12-6l 22 2l2-

② - ②

2EI 2l2- l2k②k② l36- -3l 33 l

③ - ③

2EI 2l2- l2k③k③ l36- -3l 44 l

2

-

k①k①k②k

2EI

-6l -3l- -

l k k22 23②

3

k

- -3l k③

l

- l2 k

l

6-

-3l2l2 ν1θ1θ3θ4000k2EI-

l l3121245

l l

ll x局部坐标系（ij①1l01②2l01③3l011145T0005故22345k①2EIl3

2

4l ②EI ②EI 4l22ll3k③EI4l

l2

l3l24l2-k2EI

-0

l l3 l

l2 l 2l2 BCBCD y1①③⑤y1①③⑤⑥3②2l④l l局部坐标系（ij①1l10②3l10③1l0-④2l0-⑤222⑥1222211223344F 0-

0

k①EA

k②k①EA

l-

0

l-

0 1

03

2

04

0

0k③EA k④k③EA l 0

l 0 1

1

1---

1--- 2

2

k⑤

EA 2同理k⑥k⑤

EA 2 1

1-

-

2

2

1

1

2

214

2

- 2 4 4 244 4

4 0

kEA

4 -- --

-1ll- -

4- -2

0 - - -

4

4 2444 444 - -

-

42- k11223344

0-

01 112 3 3 4 4

EA 4

2 -Fl 2 p 4 4

4 4 442

4 4

0.5578

-

2 p 4 l 4 44 44

2

3

4 -

1---

2 EA

- 2.13542.1354F

2

2 2l 1 EA0.78882

2

0.7888

1 2

TF5 p 22 2

10.7888

-

T22TF0

p0p2EA 0

k②k①

EA140k① l 2

4

l 4 -1

4 -1k④EA 0- k⑤EA 2

k⑥EA 2 l

2l- 1 - 10 2 20 4 4

EA 4

2 -Fl 2 p 4 4

4 度系数为k。y②3③y②3③①1 局部坐标系（ij①1201②223212T222T3F200-30 ll ll k①

0 0 4EI

l0000003EA l 3 12EI 4

l

l k②

l l 3EI

3 12EI

3EI

l

4

l l

l3 33 l lk③

4EI l l

l 3 12EI

4 l3 l

3l 23l2 230KN

F 3EI 3 12EI 3EIF

3

4ll3

-4ll3 l 4ll3

33

4EIl l l l3EA15EI 3 3EI 3

EA

4 2

3

3EI

5EA9

33EI2

3EI

334 2 2 2 333 3 2 3

30KNm4EIc 3 -

4

15kND④

3 G 71 4 7 局部坐标系（ij①25②3610③660-④1301⑤4601622356F1510000 2EIk①

4EI l 2EIk②

4EI l 6EI k③

2 ll 4EI 6EIk④l

2 ll 4EI 6EIk⑤l

2 ll 4EI 5

2

1

1EI 2

62

2 0 306

0

12 1解得：31.81

-6

- -

6

82.06

82.06

0

0 6

- 2

1EI 120.52 20.52F36

3 7 6-

EI

-

16016

0

66

4

3qCACAyxyx 2①Al局部坐标系（ij①1l②2l10T22T

F0

-12 EA32 4 4l5l35

l

-24

l 5 5l2k①

EA12EI

EA4

3

6EI3EI

-18 l3 5 5 l 5 6EI

l

l

k②

6EIl l

l2l l -2425l kkEI25l

5l2

5l

0k2ql 2 2 12

2 FC x②3 局部坐标系（ij①1410②230③5410 F5KNk①12EI l3 -6EIlk③ll

2 - 4EIl -6EIkl l2 - 4EIl l22 l22 4EI 0- 4 l

-6423

-

0 1 Q5 l EI EI

- 0

M5 5

l2 -64EI-1Q- - -63 4

M4

l -ll

0 Q1 EI - 2l 0

M F

l2 - - -63 - Q2

0

M28KN 8KN 4KN

8KN

4KN 30kND E + 11

2 N N 3

320F3,322

112EI 1 k①2 2EI 4EI 1 l 22k②34EI3EI4l

2011 3 162

203133EI2-

- 3 130

20

8

1

11

-3

2 -80 2

2EI

2

11 -10

0 Fy1 - M 20

- 1

- 1

01 1 2 - 3

4011 - 2 F②3EI

3211 20 -

-40

31443

11 11M2.称构

Q 33①3y24 1x N N 3

320

M 3 M1 22,F 2 3 4 25

y34y32211 3384 3841k①EI 1 33222k②EI2k③EI

23324433388 1 3

4 20

3 42 3 3

3

0 84 1

99161

25 143.54EI23.21EI37.08

，4111.50-

- 3

20

8 1

4 -3.25① ①

2

32EI

22 -10

0 -Fy1 - - M 20 1

3 13 2Fy2

-

30 - M

2

- 2 2

32 F

-

20 -

20.58M 9M3

33 -2 2

- 4 3 - 3Fy3

84 18.25 3

EI

1 2

34.733 30 Fy4

4M - 80 38.274

- 13 3 M Q M Q义，求出该特殊单元在图示坐标系中的刚度矩阵元素k33和k31。xa xa RR1 i

M2i0

4

727i72

771

92EIC ABC AB y ② 223T F6KN,8KNm,4KN

2

EA

2

2 k①2EA 2 2EA 2 EI

3

EI EI k②

2 23 136

③ ③ 2EA

8 2 2 1 2EA 2 24 2 2 4 2

1

28KN

4KNm3 32EA 2

4 2

1

2EA8

4 1

2 2 1 3 8KN 1 2

4KNm 4 8

34

3 3 1 6

2

2 2

32

4KNm 43 3 同济大学结构力学第9静定结构的实用计算方法与概AB EI=6，则iAB1,iBC

41 0.474131.531.5 0.534131.5ABC--0---0--0B13i

1

mAB245.960167.05EI

33 0.3633333141

41 0.1633333141 0.1233333141ABC000-00-0B13i

1

mAB37.2013.6EI BC EI=8,iABiBC1BABC

Pablb32441248KN

9l216258KNMBC ABC0-

-

60kN 2m2m 6m 2m2mEI=6，则

41 0.54141

4 314 BC7MAB80KNmMBA80KNMBCMCB40 MCD 802140KNABC-00----------------

A 11,

4BA1

1 44,

1 1

1 24MAB

64KNM

2442128KN3245250KN245250KN

2452200KN32452100KNABCD---------------最后弯 - - -CDEABCDEAB 4

414 4

4141 DE 3 4141 2428KNM

8KN3ACDE00000-------------

k

3B3

K41EIm24M

12,

9,

16,

4, DEB00-900----5------5--5AAB

4EI41/22/ 3另

2, 1CAACBB1

CBA0421-0-0-00-0-00449-449-4BABB 2

2671-00290-00030450-4

-1-0-0-4转动弹簧，刚度系数为kθ=4i。

B

BB 3

3i3i1l4i(其中iEI MCB

3

16i

MBCM 4i6i11l 444 2i6i11lSBAM

444283

MAB ABC0000733M4AiBiC kkθA 根据杆BC端，可得M4iBCkBCBA 根据杆BA端，可得k

4ikθ 将②式代入①式得：M4iθBC 4ik4i4iBC4iθBC 4i 4i μ1μ 4i CEDBCEDBllqlllSBD3i,SBA0,SBCBD0.6,BA0,BCBC0---000210-0021--00FGBFGBCDA

EI/414 112 4

1AB88

14SBFSAE

EI/212 121 2

121

1

8BC85kNAAB1-1

8 -1

0110-9799BCBFB2/118/11F1 2028-00 0020-10 2048 QF5ql62.5KNQF3ql37.5KNQF22.5KN 8 MFql2125KN kkkk 3EI9EI9EI3EI r1r 8 Fr7070 Fr70210 7010125212.5KN821010262.5KN8M图KNr1r 8 108210QF Fr8.961.12KN,M

108 3.2KNPFr8.963.36KN,MF

8212.8KNM图KN H

rrrr FQbFQcFQdFQe 15KNb

d并 c

12EI1则kbkckdkeka124EI kdekdke

kbckb

2 1 ra1rbcdeFQa45/315KN FFFF1F 15KN

2 M图KNAA GBe bc并联与deak3EI 12EI 12EI r39, 120,rr12024a159 159 15939r120151,r12015d159395 15939Qa4.97KN,QbQc4.64KN,Qd1.16KN,QeM图KNMEME LBDHAK FEI1=∞Rrrr1 R305585KN

a并ac 并

ekabc233kde22

1 4 83 18

85945KN

85

FF140

2408FF340 M图KN

ql MC

ll2l2

Pl

MDCDEMDEMEDMDEMDBM,

MBA aaa aBB端弯矩为Fpa4B端实际弯矩应介于两者之间。MBD

2Ml ll1F

3F1F1F82 可以看出M 1 ABFpl,Fpl 且 1F 2 M 平衡。所以AB杆与ED杆的弯矩与杆平行。1 31 A

a77

371 1 31 1 1 1 141 11 1 14

1433 74737 5 h hhlh121212 112 A M1F 4DDFBGAHMMABMBAMqMqqlh2hhh + D2D2EM2BCA点无转角，对于剪力静定杆而言，无转角则无弯矩，所以DB杆中无弯矩。MDBD杆、BCMMMD本结构无线位移，D、B两结点又无转角，DB杆、BADB杆、BA81ql8构如图b所示。xxABl

Px22ll 11+ FyBFLFRMM l2l2 ①++-+FyB②②11-③③④M1⑤1 ymmmmm 器（阻尼系数为c），梁上受三角形分布动力荷载作用，试用不同的方法建立体系的运动方程。AkB Cc 取A点体，A结点力矩为：

1mlal2l1mal3 1ql2l1ql2t 3t3AMIMpMs01

kalcal2

qq2 2 整理得：ma 11ql3tBC133AA1l为：ql2lklllc mxxdxl3t maka3caqt 弹簧的刚度系数为k，B处阻尼器的阻尼系数为c，试建立体系自由振动时的运动方程。 mkD mc∞kaaaaa22 2ka32取 体，MF0R2a2amx2dx3 R2ma234取 体：MAk3amx2dxca24ka23Ra0R15ma325ka2k

12MpM1图（B点处作用一附加约束MMt 24M Mk3EI，Rml3 1 m72EI24Mt ll2l2m M1ll2P2

M2y11M112Fp(t

22 11

3

lll2 惯性力矩为my y myl my3EIy

16Fpt a2M11 a2 2 图乘得：f11EI22a22aa3a

23 m2 34由此根据弯矩平衡可求得P k9 m

于是两者并联的柔度系数为并6EI

llll

13l3

1

2f224a5

3

2EA13a lml l 3 5

31131133 lM1 M2 M131233132316213EI32l2

l l3 2

3

19364l 1

lnyk

ln1.188解：AF 已知0.20.02.0.75F1Fsint，A9 2 49 2F1 16 16F9 9

916

40.022并绘制最大动力弯矩图。设 ABABl

，

2 2y

sint sint即幅值为

1

Mmax

mA BmA Bll2l2l1l1M1

M2

如所示弯矩图，图乘后，f1124EI,f223EI,f12f2148EI fFfFsintfmyfFsint 11 11 y24EIy5Fsin 其中224EIP*5

5

1

=

114= ytC36EIsin

fFfPsintfm fP 21 t

m2

12ytB2

P*

1 ytC=36EIsin * ytBf21P2 2Pf22sin 2 = P 48EI

Pl325 = 1sint3EI32

1 72 =Pl31322 3EI

1 121=

3EI128

llm1lmM1

M2

281 121簧的刚度系数为k。m3l2ll2l22222ml39已知图示体系为静定结构，具有一个自由度。设为B点处顺时针方向转角为坐标。 l3lm33l k 3

2q0 ml2k2l292mk9

1

9l8 k1k3

中柱

kmkm

126106N126106N63m3 T

t10.1

1

8105v18104 1

vt1510m/1mv21 1810551032111062

yst

0.007711066 1 2Q h h第一阶段（0tt1 t t m0P0ZFP0tZsintZm0 t 1 1sin st 1 2

t ttsin2Tt

t T tys

t T

M1

T 0.05yy

nn

ln

mk（2）ktnmkk

2n2t2tn

23.

1421.223103N/2EI12kEI3.79106Nm2设某单自由度体系在简谐荷载FP(t)=Fsint作用有阻尼强迫振动，试问简谐荷载频率分AF

y

22

1 2

222

2

2 2

22f0 41 2421，显然要求 1 1 10得 1 2

2h

1 令h122 22显然要求h1最小 则h12 0ll 21l2l21l2l2M1 l 2l12ll2

M2 （1）EIf1122l2

f11 21ll2llll

f12f21

1 4EIm2A10A2 1 0A

2A2令 3D0

10-310110,212

11111 llllllPllPl2l21211ll2l1l2l1211

EI

3 2112

2l2lEI 32

21

2l

l EI2 32 12 1 m2A16EImA2 1

2A2 令 2D

ml3ml3解得：ml3ml3A212.41412.773，A222.4142 k= 11/11/1/1M1

M2

l/2（1）f113EI，f2212EI，f12f21 1 3EIm2A1 mA2 1 mA1 2A2 令 4D5

13-2175225115.227,22

28

21.773A121A2211

aa 1111212121211121M1 1a31 11

M2 6 2

/ /k2

48 1 2/ 2/k2/ 48 1a31 11 6 2

/ /k2

48m f11f

12fm21

22 mmam1ma3 244a40a203111.045a,2

1 2

m1AmA1112 12 a 1fmAfm A21 1 2 1

11 0.2301 0.2292 A EI

12 ma48EI1 3.62511A 111 EI

12 ma48EI

a1 a1 aaaa1l2l1l2ll2l2l21M1 M21l1l2M3 （1）f112EI，f222EI，f12f21f33 1 2EIm2A16EImA2