结构力学1课件18结力07十八

27

F

a31

P3112

FP

aM(kN.m)FPABFPCDa2EI3aEI3aM(kN.m)FP

2735PF

a24

F

a35

PABFPCDaEI3aEI3a第六章力法§6-4

荷载作用下超静定结构计算例1:用力法计算图示连续梁例2:用力法计算图示排架。qLLEI

EIqEILL2EIM(kN.m)2

5

16ql16

3

ql

2M(kN.m)41

ql

2214ql例3:用力法计算图示刚架。lqlEIEIM

(ql

2

)

1

32

7

64EIq31

EIqEI3EIM

(ql

2

)

1

16

3

32M

(ql

2

)

1

16

3

321.与静定结构相比,超静定结构使截面最大弯矩降低,整个结构内力分布趋于均匀;小结:荷载作用时,超静定结构的内力分布与相对刚度有关;刚度大的部分,承担了较多的弯矩；调整各杆刚度比可使内力重分布。例1:用力法计算图示连续梁3127

FP

a3112

FP

aM(kN.m)FPABFPCDa2EI3aEI3aM(kN.m)FP27

F

a35

P3524

FP

aABFPCDaEI3aEI3aFPM

P3FP

a例2:用力法计算图示排架。qLLEI

EIqEILL2EIM(kN.m)2

5

16ql16

3

ql

2M(kN.m)41

ql

2214qlM

Pql

2

/

2例3:用力法计算图示刚架。lqlEIEIM

(ql

2

)

1

32

7

64EIq31

EIqEI3EIM

(ql

2

)

1

16

3

32M

(ql

2

)

1

16

3

32MP21

ql

2EIEIq

1

14

1

M

(ql

2

)28

5

56例4:用力法计算图示桁架,各杆EA

等于常数。基本方程:基本基结本构未基知本量体系方法一：D

1

=

d11

X

1

+

D

1

P

=

0FPFPa

aa

aaFPFPx111d

=2

)

FP

aEA111d1PDX

=

-2

)

a

EAF

N

1

F

N

1L

=

(3

+

2EA1PEAD=

F

N

1FNP

L

=

(4

+

2=

-FP

=

-1.17FP4

+

2

23

+

2

21

11

1 1

PD

=

d

X

+

D

=

0X1=1F

N

1

(kN

)0.5-0.5-10.5FPFNP

(kN

)0-FPFP

-FPFP例4:用力法计算图示桁架,各杆EA

等于常数。基本方程:FPFPa

aa

aa1D

=

d

X

+

D

=

0X1=1F

N

1

(kN

)-0.511

1 1

P-10.50.5FPNPF

(kN

)-FPFP

-FPFP例4:用力法计算图示桁架,各杆EA

等于常数。基本方程:FPFPa

aa

aaX1

=

-1.17FPFN

=

FN1

+

FNP

=

F

N1

X1

+

FNPFPFN

(kN

)(FP

)FP

0.17-0.59-0.830.41方法二：D

1

=

d11

X

1

+

D

1

P

=

011EAd=

F

N

1

F

N

1LF

aPEA=

-N

1NP1PF

F

LEAD

=111d1PDX

=

-=FP

=

0.26FP

1

1+

2

2P„

0.17F?EA=

(1

+

2

2

)

a

F

PFPx1x1x1x1F

P基本体系2FPNPF

1(kN

)

2

22-

22

F

N

1

(kN

)-

12P2

F-2

P2

F-2P

D=0

吗?1

F变形条件仍为:方法二：=

D

lC

D力法方程的实质为：“C、D两结点的相对位DCD移 等于所拆除杆的拉（压）变形

DlCD

”1X1XCD

1EADl

=

-

2a

xx1x12222-2

F

N

1

(kN

)-

1F

PFPFNP

(kN

)2

P2

F-2

P2

F-21

FP基本体系2d11

X

1

+

D

1

PFPx1x1ABCF

PD1PEAD=

-

FP

aEA11d

=

(1+

2

2

)

a11

EA1+

2aD1PX

=

-d1=FP

=

0.17FP3

+

2

2方法三：11EAd

=PF

aEA=-EAF

N

1

F

N

1L

=

(3

+

2

2

)

aN

1NP1PF

F

LEAD

=111d1PDX

=

-1=FP

=

0.17FP3

+

2

2D1

=

?D

1

=

d11

X

1

+

D

1

P

=

022-2-

12

F-2P2

F2

P

-21

FPF

P基本体系3x1FPx1x1

x122F

N

1

(kN

)F

PFNP

(kN

)FP0方程(a),(b)的形式及物理意义不同,系数也不相同,但内容等效。比较：1(a)EA=

-

2a

Xd11

X1

+

D1PEA11d

=

(1+

2

2

)

a1PEAD=

-

FP

a=

0

(b)D1

=

d11

X1+

D1PEA11d

=

(3

+

2

2

)

a1PEAD=

-

FP

aX1X1基本体系2FPx1x1ABCF

PDFP基本体系3Fx1Px1例5.

力法解图示桁架，EA=常数。D1

=

0=

0d11

X1

+

D1P112)

a

EAd

=

F

N1

F

N1l

=

4(1+1P2)

FP

aEAEAD

=F

N1FNPl

=

2(1+EAX1

=

-FP

/2FN

=

F

N1

X1

+

FNP-

210011-

2-1/2-1/2

aFP1/21/2a-

2

/

22

/

2位移条件X1FP基本体系·FPX1

=111

-

21F

N1FP0

01FN

(·FP

)例5.

力法解图示桁架,EA=常数。PP00

0-

2

P0PNPN11X1

=11

11

-

2-

21-P/2-P/2aP/2aP

-

2

/

2P/22

/

2X1P基本体系Paa各杆内力会

发生什么变化？例6:用力法计算图示组合结构。D1

=

d11

X1

+

D1P

=

0FPa

a

a基本方结程构:基本未知量基本体系aaFPEIEAx1x1F

N

1、M

1

1aa-

21

a

-

1d11

=

(1+

2)P2FFPFNP、M

PFP-FPPF

a/2PF

a/24

8aaEA

3EI+6EI-

P

4F

5F

a22

)

P

EAD1P

=

-(1+=

0基本方程:

D1

=

d11

X1

+

D1PaEA

3EI4

8a+d11

=

(1

+

2

)1111PPd

18DX

=

-=

7

F

aEA3EIa2=（4

1

+

2）设3a=EI7F

a26EI=

-

P

EA

6EI4F

5F

a21P2)

P

-

P

D

=

-(1+aaaaaFPEIEAx1F

N

1、M

1

1aa-

21

a

-

1P2FFPFNP、M

PFP-FPFPa/2FPa/2FN(kN)FPFN

=

FN1

+

FNP

=

F

N1

X1

+

FNPM

=

M1

+

M

P

=

M

1

X1

+

M

P1811

FP1

F

a9

P

7

F

a18

PM(kN.m)1111PPd

18DX

=

-=

7

F

ax1F

N

1、M

1FP

1aa-

21

a

-

1P2FFPFNP、M

PFP-FPFPa/2FPa/2例6:用力法计算图示组合结构。EAa2=3EI（4

1

+

2）设基本方程:D1

=

d11

X1

+

D1P

=

0x1F

N

1、M

1-1aM

PFPPF

aEA

3EI

EI4a

8a3

3a3d11

=

(1

+

2

)

+

=11F

a36EID1P

=

-

P

1111PPN

1d

18DX

=

-=

11

F

=

FaaaaaFPEIEA分析：a2=3EI（4

1

+

2）设EA

3EId11

=

(1

+

2

)

+11F

a3D1P

=

-

P

116EIX1

=

18

FP3a3d11

=

EI18P跨中弯矩

=

7

F

a3EI=EA（4

1

+

2）

4a2设11124PX=

F4a3d11

=13=

0EAEA（4

1

+

2）设11116PX=

F3EIEI8a3d11

=24516EA设（4

1

+

2）

¥11d

¥PFPFPaM

P1x

=

?跨跨中中弯弯矩矩==?FP

a1x

=

?跨跨中中弯弯矩矩==?FP

a11xX

==0

?跨跨中中弯弯矩矩==F?ax1

F

N

1、M

14a

8a3-1例6:用力法计算图示组合结构。D1

=

d11

X1

+

D1P

=

0a分析：EA=3EI（4

1

+

2）

4a2设124PX

=

11

F=

0EA（4

1

+

2）设116PX

=

11

F

¥EA设（4

1

+

2）1X

=

0PFPFPFP24

1

FP

aF

aP16

3

FP

a16

5

FP

aPF

aaaaaaFPEIEAd11

X1

+

D1P

=

-X1

/

k例7.

求作图示梁的弯矩图。1111kX

=

-(d

+

)D1PEI11=

-X

/

kDl

3d11

=

6EI5Pl

3=

-24EID1PM

=

M1

X1

+

MP1D

=

?例7.

求作图示梁的弯矩图。1111kX

=

-(d

+

)D1PEIl

3d11

=

6EI5Pl

3=

-24EID1P321M

=

M1

X1

+

MP,

X

=

25

ql

(

›

)l

3k

=

10

EI当4X

=

5

ql

(

›

)1k

fi

0当1X

=

064BM

=

7

ql

2当k

fi

¥8BM

=

-

1

ql

22BM

=

1

ql

2？例7.

求作图示梁的弯矩图。EI思考：如果选取图示基本结构，写出力法方程，并求系数。=

0d11

X1

+

D1

PM1