工程力學(xué) 第二版 (王斌耀 顧惠琳 著) 機(jī)械工業(yè)出版社 習(xí)題解答

理論力學(xué)習(xí)題解答靜力學(xué)

3練習(xí)冊(cè)CDHEGHP1P2P1P2ABCDFGFHFCFDFDFAFBFHMG

=0

FH×2

–

P1×1

–

P2×5

=0FH

=

50kNMC

=0

FD×6

–

FH×1

=0FD

=

8.33kNMB

=0

FD×9

–

FA×3–

P1×3

–

P2×7

=0FA

=

–

48.34kNFy

=0

FA

+

FB

+FD

–F1

–F2

=0FB

=100kNP11

習(xí)題：2l

=3

m

有主次之分的結(jié)構(gòu)P12

習(xí)題：3ABCDEPPCDEPPFCyFCxFCyFCxFAyFAxMAFEMC

=0

1√5P×2

–

P×2

–P×2+FE×=02√5√2FE=483NFx

=0

=553NFCx

=

P

–FE1√52√51√5FCy

=P

–P+FE1√2=894NFy

=0

Fx

=0

=447NFAx

=

P

–FCxFy

=0

=106NFAy

=

P

–FCyMA

=0

MA

=

P×

–P×1

–

FCy×125=606N·mPyPx有主次之分的結(jié)構(gòu)ABCDEGq321FAxFAyF3GF3F1F2jjMA

=0

–

7b

q×b

–

F3×h

=027F3

=–

26.1kNFx

=0

FAx=F3Fy

=0

FAy=7b

qFx

=0

F2

–

F3

=053F2

=–

41.8kNFy

=0

F1

+

F2

=054F1

=32.7kNP13

習(xí)題：1有主次之分與無主次之分的結(jié)構(gòu)二力桿=–

26.1kN=28kNMFjFBOABFBFABFABMOAjjFOyFOxFx

=0

FAB

–

F

=0√52FAB=200N√5MO

=0

M=(FAB

+

FAB)×l√51√52=60N·m也可使用力偶系平衡特性P13

習(xí)題：2l

=10

cm

FABFAB機(jī)構(gòu)P14

習(xí)題：3CFBFCyFCxFBjj2jFBCAFBFAyFAxMAMC

=0

FB

=

FFx

=0

FAx

=

F√32Fy

=0

FAx

=–

F12MA

=0

MA

=–

Fl也可使用三力匯交定理有主次之分的結(jié)構(gòu)P14

習(xí)題：4PBCDOFByFBxFDqMB

=0

FDcosq×6L–P×6L

=0FD

=1500NFx

=0

FBx

+FDcosq

=0FBx

=–

1200NFy

=0

FBy

+FDsinq

–P

=0FBy

=300Nsinq

=35cosq

=45有主次之分與無主次之分的結(jié)構(gòu)PABCDOFAyFAxFCMA

=0

FC×6L

–

P×6L

=0FC=1200NFCE[E]FEBFEDF[C]FCEFCB[B]FEBFBDFBA[D]FDAFBDFDOFEDFFCE

=FCB

=0Fx

=0

F+

FEBsin45o

=0FEB=–

3kN√2Fy

=0

FED

+

FEBsin45o

=0FED=

3kNFx

=0

FBD

+

FEBsin45o

=0FBD=

3kNFy

=0

FBA

–

FEBsin45o

=0FBA=–

3kNFx

=0

FBD

+

FDAsin45o

+

F=0FDA=–

6kN√2FDO=

9kNFy

=0

FED

–

FDAsin45o

–

FDO

=0P15

習(xí)題：1零桿判定節(jié)點(diǎn)法順序拉力假設(shè)CEBDAOFFlll(圖)P15

習(xí)題：22ABCFCDE1FDHGDE1FDHGF2F4F3FEFAyF2F1H∥

∥

∥

∥

∥

∥

∥

∥

MA

=0

FE

=kN34FEMD

=0

F2=-2kNFx=0

F1=-2kNFAx零桿判定截面法截面P17

習(xí)題：2FSFNqF45oBAPllMA

=0

Fsinq×2lcos45o

=P×lcos45o

+Fcosq×2lsin45oF

=2(sinq

–

cosq)

PFy

=0

FN+

Fsinq

=PFN

=P2(sinq

–

cosq)

sinq

–

2cosq

Fx

=0

Fcosq

=FS臨界平衡令=fSFN即tanjm

×

P2(sinq

–

cosq)

sinq

–

2cosq

P=2(sinq

–

cosq)

cosq

cosq

=2(sinq

–

cosq)

tanjm

∴

tanq

=tanjm

1

+

2tanjm或受力圖qjm幾何法qBQqAFF1F2FSFNF3F4FSFNFy

=0

FNcosq

+

FSsinq

=Q而

|

FS|

≤

fSFN≤

FN≤cosq

+

fSsinq

Qcosq

–

fSsinq

QFx

=0

FNsinq

–

FScosq

=F且

|

FS|

≤

fSFNQ

≤

F

≤

Qcosq

–

fSsinq

sinq

+

fScosq

cosq

+

fSsinq

sinq

–

fScosq

(1)(2)(3)Fmin

=Qcosq

+

fSsinq

sinq

–

fScosq

Fmax

=Qcosq

–

fSsinq

sinq

+

fScosq

令Fmin

≤0q

≤

arctanfS設(shè):