福大结构力学课件第5章影响线

第5章影响线5-1静力法作单跨梁的影响线5-2机动法作影响线5-3间接荷载作用下的影响线5-4桁架影响线5-5影响线应用5-6简支梁绝对最大弯矩5-1

静力法做单跨梁的影响线FyA11FyBFQC1ab1b/la/lMC1

支反力：以向上为正，正的画在上面。yAF

=

(l

-

x

)/

lABx5-1-1简支梁yx

FP=1abClFyB

=

x

/

l2

剪力：绕隔离体顺时针转动为正，正的画在上面。yAFQC

=

FFQC

=

-FyBx

˛

[0,

ax

˛

[a,

l3

弯矩：以下侧受拉为正，正的画在上面。MC

=

FyAaMC

=

FyBbx

˛

[0,

ax

˛

a,

labl5-1-2悬臂梁FyA1FQCMC2

剪力CM

=

a

-

xABxyx

FP=1abCl1x

˛

[0,

ax

˛

[a,

l1

支反力FyA

=

1MA

=

-xMAlFQC

=

0FQC

=

FyA3

弯矩MC

=

0x

˛

[0,

ax

˛

[a,

ll-axFyA

1FQCC截面弯矩AByx

FP=1abClFyBl1

ll

+

l1l1FQC

=

FyAyBFQC

=-Fx

˛

[0,

ax

˛

a,

l

+

l1

b/la/l1

1ll5-1-3伸臂梁支反力FyA

=

(l

-

x

)/

lFyB

=

x

/

l简支部分截面内力C截面剪力MCMC

=

FyAax

˛

a,

l

+

l1

MC

=

FyBbx

˛

[0,

al1a

laDcl1ablxD截面弯矩AByx

FP=1abClc3

伸臂部分截面内力D截面剪力Dl1dFQD1MD

=

l

+

c

-

xFQC

=

1

x

˛

l

+

c,

l

+

l1

FQC

=

0x

˛

[0,

l

+

cDM

=

0x

˛

[0,

l

+

cx

˛

l

+

c,

l

+

l1

dMD影响线基本规律1剪力影响线在截面两侧平行,数值差值为1；2弯矩影响线在截面两侧的夹角为1；伸臂梁影响线1截面在简支梁部分时，其影响线是将简支梁影响线延长到伸臂部分；2截面在伸臂部分时，其影响线与伸臂梁相同；5-1-4其它类型FyA

1FyB1xByAx

FP=1l/2Cl/2x

˛

[0,

l

/

2x

˛

[l

/

2,

lx

˛

[0,

l

/2x

˛

[l

/

2,

l

l2MB1FQC支反力FyA

=

(l

-

x

)/

lFyA

=

0FyB

=

1

-

FyAMB

=

-FyBl

/2MB

=

x

-

l截面剪力FQC

=

-FyBFQC

=

0x

˛

[0,

l

/

2x

˛

[l

/

2,

lQ

ByBF

L=

-FQ

BF

L112a

1

2a121212aFyA1

支反力yAF

=

-1/

2aFyBMBFQCFQ

C=

FyAaxMP=1ByAxaCaFyB

=

1/

2a2

截面剪力MC

=

FyBaMC

=

FyAax

˛

[0,

ax

˛

[a,

3a2a4aMAxyBx

FP=11aaaaaaa23AC1

支反力AM

=

x

-

4aaQ1F

=

-1x

˛

[0,

a2

截面内力M1

=

x

-

aM2

=

2a

-

x3M

=

x

-

3aFN2

=

-1FQ3

=

-1a4aM1M22a3aM33ax

˛

[0,

a右侧受拉为正※横轴是荷载移动的范围M1

=

2

-

2ax

˛

[a,

2ax

˛

[2a,

4a]2

截面内力M1

=

a

-

xxM1※横轴是荷载移动的范围xCyx

FP=11aa2aaaBDAyDF2a1

支反力FyD

=

0=

x

-

2ax

˛

[0,

2ax

˛

[2a,

4a]FyD1FQ1

=

1右侧受拉为正x

˛

[a,

2aFQ1

=

1

-

FyDx

˛

[2a,

4a]a1FQ15-2机动法作影响线d2d1d1

+d2

=1FP=1CABabl1

P

PyAF

1-

F

d

=

0dPPyAF

=

d1FyAAFP=1

BFQCFQCPdd1

+

FQ

C

d2

-

FP

dP

=

0FQ

CFQC=

dP

/

(d1

+d2

=

dPa

lb

lFP=1CAFyABCAFP=1

BablFQCFP=1CABabl1aa1

+a

2

=1AB2adPMC

=

dP

/

(a1

+a2ab

lMC

a1

+

MC

a

2

-

FP

dP

=

0=

dPMCABMCFP=1撤去相应的约束。

使体系沿约束的正向发生单位位移，则荷载作用点的位移图即为该量值的影响线。机动法做影响线的理论基础刚体虚功原理机动法做影响线的步骤MC1例CBFP=1abAlFyA1FyA11MAFQCFQC

FQCMAl1MCbMC1例CBAFyA1FyAMA1MA1aa

aFP=1a

aCBFP=1AaaaaQ

AF

LQ

AF

LQ

AF

LQ

AF

RQ

AF

RQ

AF

R11

1/23/2111/2例CBFP=1AFQCMCaaaaCBFP=1AaaaaMCFQ

CFQ

C

11/21/21/21/21a/2a/2a/2111/2Q

BQ

BF

L

F

LQ

BF

L1/2例CBFP=1AMCMA1a

aa

a1MCaaMA13aaFQBFQC

FQCCBFP=1Aaaaaa/2a/21/21/2a/21例DFP=1BAaaaaaaaCEFFyAMDQ

DF1/21/21/21/23/213/2aa例DFP=1BAaaaaaaaCEFFyBMBQ

BF

L11/21/2例DFP=1BAaaaaaaaCEFMEFyCFQ

Ea/2a121/21/2111/22aaa2a2a例DFP=1BAaaaaaaaMABMEF

CQ

BF

R1111/2a/2aa/2a2aa2a12例DFP=1BAaaaaa

a

C

EaFyAMDQ

BF

RMBa2a3aa3a2a3a例DFP=1BAaaaaaaCMEFaEMDFM

LGHMHMG5-3间接荷载作用下的影响线要求：一、理解相关概念。二、熟练掌握画法。三、理解公式的物理意义。AEDBC

FFP1FP2AD

EBC

FMF=yCFP=1AEDBC

FMF=yDFFP=1MFyCyDMF

=

FP1yC

+

FP

2

yDAEDBC

FFP=1ddddAEDFP=1横梁B主梁纵梁dC

FdddxFP=1yEFCDMy=

d

-

x

+

xyddd-xdxdyDyCyE先假定没有纵横梁，将FP=1当做直接荷载，做出相应的影响线；从各结点引出竖线与直接荷载作用下的影响线相交，将所得的交点在每一纵梁范围内用直线相连。间接荷载影响线绘制的步骤：a

a2aFyA11/2ME例ABa

aCa

a

a

aDE1/2a/2a/21F

RQ

B3/4a/2MGFQGFyB1例AGCDFFP=1B

Ea

a

a

a

a

a

a

a

a

aa/2

a/41/21/21/41/23/2FQD1例AGCDFFP=1B

Ea

a

a

a

a

a

a

a

a

a1/2F

LQ

E11/21F

RQ

E111/21/21/2a

a

a

aa

a

a

a1a1FyB例ABCFP=1DaFyAMA右侧受拉为正MBaa

a

a

aa

a

a

a例ABCFP=1DaM

LDaaFN15-4桁架的影响线FN1FyAFyB1122dPddddAC

F

=1DEBFN31解F

N1

=

-

2

F

yBx

˛

[A

,

D

]F

N1

=

-

2

F

yAx

˛

[D

,

B

]2FN2N

2F

=

2

FyAx

˛

[D

,

B

]dAEBFN1FN3FyAFyB11PC

F

=1DN

2F

=

-

2

FyBx

˛

[A

,

C

]1FN32γα13yBhX

=

-

4d

Fx

˛

[A,

4]13yAhX

=

-

2d

Fx

˛

[4,

B]FN13

=

X13

/

cos

b例ABh1

β24366d5

7910P8

F

=1解4dh

cos

bhcos

bFN132d4d3hcos

bFyAFyB11γα例ABh24366d5

7910P8

F

=11

β解FN14FyAFyB1114

yB=

-

6d

+

a

FY2d

+

aax

˛

[A,2]Y

14

=

F2d

+

a

yAx

˛

[4,

B] 6d

+

a

(2d

+

a)sinaN14

14F

=Y

/

sinaa(2d

+

a

)sinaah=

sin

ba

+

2dC( 6d

+

a

6 2d

+

a)sina2a3(2d

+

a

)sina1singγα例ABh243566d78910FP=11

β解FyAFyB11aCY45

=

FyBx

˛

[A,

4]N45singY=

45F45yAY

=

-Fx

˛

[6,

B]singFN45113sing12singγα例ABh24366d5

7910P8

F

=11

β解FN34

=

-FN13

sin

b4dh

cos

bhcos

bFN132d4d3hcos

b3h4d

tg

b4dtg

bh2dtg

bhN34F132解a.荷载在下弦x

˛

[A,1x

˛

[2,

ByBFN15

=

Fb.荷载在上弦注：强调荷载在上、下弦移动时影响不完全相同。1FyA

1FyB例FP=1AB6548FP=17FN15

=

-FyA11N15F

Dx

˛

[4FN15

=

0x

˛

[5,

8FN15

=

-FyA1N15F

U3/45-5影响线的利用—利用影响线求某一量值1

集中荷载作用MC

=

FP1

y1

+

FP

2

y2FQC=

-FP1

y1¢+

FP

2

y2¢MCy1y2FQCy1y2FP1FP2CABDLQ

DF=-FP1

y1¢+FP

2

y上F

R

=

-F y

¢-

F

yQ

D

P1

1

P

2

下FQD下y上y1y

¢xy(

x)QCDEq(

x) d

x y(

x)F

=2.均布荷载作用qD

C

E2A

(

x)A1

(

x)y(x)

d

xED=

q=

qA=

q(-A1

+

A2

)BDAq=10kN/mFP=20kNC1.2

1.2

1.2

1.2

1.2QCF

=

q(-A1

+

A2

)

+

FP

y0.2

+0.40.6

+0.2=10·(-·1.2

+·2.4)22+20·0.4=14kN0.40.20.60.40.2例解二判断最不利荷载的位置1

可动均布荷载作用：可以任意断续地布置M

max=

qA1M

min

=

-q(

A2

+

A3)CA12A3AMCABqqq2

移动集中荷载：一组互相平行而且间距保持不变的荷载Z

(

x

+

Dx)

=

FR1

(y1

+

Dy1

+

FR2

(y2

+

Dy2Dy1

=

Dx

tana1

Dy2

=

Dx

tana

2+

FR3

(y3

+

Dy3Dy3

=

Dx

tana

3R13tana1

+

FR

2

tana2

+

FR

3

tanaDxfi

0Z(

x

+

Dx)

-

Z

(x

Z

¢=

lim

=

FDxFR1yFR2FR3y2y1

3xZ

(

x)

=

FP

i

yi

=

FR

i

yR

i

=

FR1

yR1

+

FR

2

yR

2

+

+FR

3

yR

3a1a

23ay荷载组稍向左移时Z

¢(x

-0>

0稍向右移时

Z

¢(x

+

0

<

0若则荷载组在当前位置时Z

(x=

ZmaxZ

¢=

FR1

tana1

+

FR

2

tana2

+

FR

3

tana3考察FR1y1FR2FR3y2y3x只有荷载组移动时,有1个荷载越过了影响线顶点,才有可能导致导数的变化.a1a

23ay则，荷载组稍向左移时(LR

i

iF

tana

>

0Z

¢x

-

0

=FR1y1FR2FR3y2y3xa12aa

3假设：这个荷载为FPcrFPcryR1F

LR

3=

F

，F

L

=

F

，F

L

=

FR1

R

2

R

2

R

3荷载组稍向右移时Pcr=

F

，F

R

=

F

-

FR1

R

2

R

2R1F

RF

RR3

=

FR3

+

FPcr(Z

¢x

+

0

=<

0RRiF

tanaimaxZ

=

Z

(x

=P

i

iF

y荷载组移动到FPcr位于影响线顶点时,8m1.00.754m6m5×1.5myqFP1

FP2

FP3

FP4

FP5x已知：FP1=FP2=FP3=FP4=FP5=90kN，q=37.8kN。确定：荷载最不利位置和Z的最大值。30mtana1

=

1

8tana

2

=

-0.25

4tana

3

=

-0.75

6荷载组稍向左移时R1R2R3F

LF

LF

L=

4

·90

kN

=

360

kN=

90

kN+

37.8

kN/

m·1m

=

127.8

kN=

37.8

kN/

m·6

m

=

226.8

kN1.00.75yFP1FP4=FPcrqxFP2

FP3

FP50.9060.815×1.5m3.5m1m6m假设FP4=FPcrR1R2R3F

RF

RF

R=

3

·90

kN

=

270

kN=

2

·90

kN+

37.8

kN/

m·1m

=

217.8

kN=

37.8

kN/

m·6

m

=

226.8

kN=

8.7

kN荷载组稍向右移时(LR

iF

tana1460.75

=

360

kN·

+127.8

kN

8iZ

¢x

-

0

=0.25

·

- +

226.8

kN·

-

(RR

itana146217.8

kN

0.75

=

270

kN·

+8=

-8.2

kNiZ

¢x

+

0

=F0.25

·

- +

226.8

kN·

-

此位置即为荷载的临界位置Z

=

90

kN·

3.556.5

+1

+

90

kN·0.906

8

+

8

+

822+37.8

kN

/m·

0.81

+

0.75

·1m+

0.75

·6

m

=

455

kN

10.75yFP5=FPcrqFP1

FP2

FP3

FP4x5×1.5m2m荷载组稍向左移时R1R2R3F

LF

LF

L=

5

·90

kN

=

450

kN=

37.8

kN/

m·

2.5

m=

94.5

kN=

37.8

kN/

m·6

m

=

226.8

kN2.5m6m假设FP5=FPcr0.906R1R2R3F

RF

RF

R=

4

·90

kN

=

360

kN=

90

kN+

37.8

kN/

m·2.5

m

=

184.5

kN=

37.8

kN/

m·6

m

=

226.8

kN=

22

kN荷载组稍向右移时(LR

iF

tana1460.75

=

450

kN·

+

94.5

kN

8iZ

¢x

-

0

=0.25

·

- +

226.8

kN·

-

(RR

itana146184.5

kN

0.75

=

360

kN·

+8=

5.1kNiZ

¢x

+

0

=F0.25

·

- +

226.8

kN·

-

Z

=

90

kN·

2

+

3.5

+

+8

8

85 6.5

+1

+

822+37.8

kN

/m·

0.906

+

0.75

·

2.5

m+

0.75

·6

m

=

444.5

kN

yFP1

FP2FP3=FPcrqxFP4FP55×1.5m5m荷载组稍向左移时R1R2R3F

LF

LF

L=

3

·90

kN

=

270

kN=

2

·90

kN

=

180

kN=

37.8

kN/

m·5.5

m

=

207.9

kN5.5m假设FP3=FPcr10.906

0.813

0.688R1R2R3F

RF

RF

R=

2

·90

kN

=

180

kN=

3

·90

kN

=

270

kN=

37.8

kN/

m·5.5

m

=

207.9

kN(LR

iF

tana1460.75

=

270

kN·

+180

kN

8iZ

¢x

-

0

=

=

-3.5

kN荷载组稍向右移时0.25

·

- +

207.9

kN·

-(RR

itana1460.75

=

180

kN·

+

270

kN

8=

-40.5

kNiZ

¢x

+

0

=F0.25

·

- +

207.9

kN·

-

8Z

=

90

kN·

5

+

6.5

+1

+

0.906

+

0.813

8

2+37.8

kN

/m·

0.688

·5.5

m=

445.6

kN25m15m130kN

100kN100kNBAC6.88m9.38m7.50m6m0.38mxy=

0.625=

-0.3759.38tana1

=

159.38tana

2

=

-

25R1R2F

LF

L=

50

kN+130

kN

=

180

kN=

50

kN+100

kN+

50

kN

=

200

kN70kN15m15m50kN4m50kN4m50kN5m

4m50kN4m(LCR

i

iF

tanaM

¢x

-

0

==

200

kN·0.625

-

200

kN·0.375=

50

kN=

50

kN=

130

kN+

50

kN+100

kN+

50

kN

=

330

kNR1R2F

RF

R(=

50

kN·0.625

-

330

kN·0.375RCR

i

iM

¢x

+

0

=F

tana=

-92.5

kNMC

=

FP

i

yi=

50

kN·6.88

m+130

kN·9.38

m+

50

kN·7.5

m+100

kN·6

m+

50

kN·0.38

m=

2557.4

kN

m25m15mAC6.25m9.38m7.88m2.25m0.75mxyR1R2F

LF

L=

100

kN+

50

kN+130

kN

=

280

kN=

70

kN+100

kN+

50

kN

=

220

kN130kN100kN50kNB100kN50kN4m50kN4m

5m70kN4m3.75m15m100kN4m15m(LCR

i

iF

tanaM

¢x

-

0

==

280

kN·0.625

-

220

kN·0.375=

92.5

kN=

100

kN+

50

kN

=

150

kN=

130

kN+

70

kN+100

kN+

50

kN

=

350

kNR1R2F

RF

R(RR

i

iF

tana=

150

kN·0.625

-

350

kN·0.375CM

¢x

+

0

==

-37.5

kNMC

=

FP

i

yi=

100

kN·3.75

m+

50

kN·6.25

m+130

kN·9.38

m+70

kN·7.88

m+100

kN·2.25

m+

50

kN·0.75

m=

2720

kN

mMC

max

=

2720

kN

myB

max1=

478.5

·0.125

+

478.5

+324.5

·0.758

=

784.3

kNF已知：FP1=FP2=478.5kN，FP3=FP4=324.5kN试求：B支座的最大反力。AB

C6m6m5.25m4.8mFP1FP2

FP31.45mFP4xy10.7580.1251266解

tana

=

1

，tana

=

-

1R1R2F

LF

L=

2

·487.5

kN

=

975

kN=

324.5

kN(LyBR

i

iF

¢x

-

0

=F

tana=

108.42

kN=

487.5

kN

=

487.5

kN=

487.5

kN+

324.5

kN

=

812

kNR1R2F

RF

R(RRix

+0

==

-54.08kNyBiF

tanaF¢假设FP2=FPcrFyB

max

2

=

478.5

·0.758+

324.5

·1

+324.5

·0.2

=

752.1kN1.45m5.25m4.8mFP1FP2

FP3FP4xy10.7580.125R1R2F

RF

R=

487.5

kN

=

487.5

kN=

2

·324.5

kN

=

649

kN(RRix

+0

==

-26.92kNyBiF

tanaF¢假设FP3=FPcrR1F

LR2F

L=

487.5

kN+

324.5

kN=

812

kN=

324.5

kN(x

-

0

==

81.25

kNLR

iF

tanayBF

¢i0.2FyB

max

=

max（FyB

max1，FyB

max

2）=

784.3

kN已知：FP1=FP2=FP3=FP4=FP求：FQCmax，FQCmin解（1）FQCmax3m6mCABQC

maxP=

F

2

+F+P

1

1

=

4

F

3

2

6

3FP1FP43.5mFP2

FP31.5m3.5mxy2/31/31/21/6（2）FQCminFP3

FP43.5mxy2/31/35/18FP13FQ

C

min

=-5-6简支梁的绝对最大弯矩定义：在移动荷载作用下，简支梁各截面最大弯矩值中的最大值。间接荷载作用直接荷载作用将各截面最大弯矩值按前述方法求出然后进行比较。根据绝对最大弯矩的定义，可知截面1FPcrFPcrM1maxM2maxFPcrMnmax截面2……

……

……截面nMmax=max[M1max，M2max，…，

Mnmax]设FPi作用在截面i

时产生MmaxFPi的特点当移动到其他位置时，在其作用处产生的弯矩总是小于其移动到i截面时，在i截面产生的弯矩Mi。若已知FPi，则可通过求极值的方法确定i截面进而求得其对应的Mi，则Mi=Mmax。FR—梁上所有荷载的合力；a—FR与FPi之间的距离；yAlF

=

FR

(l

-

x

-

a)=

FR

(l

-

x

-

a)

x

-

MMi

=

FyA

x

-

MlM——FPi

左侧所有荷载对

FPi作用点的力矩和。对一组荷载M是常数。2

2l

ax

=

-FP1

FP2FPiFRFPn-1

FPnxa

l-a-x

l

2

l

22a

a2d

Mid

x

l=

FR

(l

-

2

x

-

a)

=

0FR距右端的距离2

2即FPi与FR对称位于梁中点。l

-

a

-x

=

l

-

aA由于①重复计算比较麻烦；②绝对最大弯矩通常发生在梁中点附近。故，设想，使梁中点发生最大弯矩的荷载就是使梁产生绝对最大弯矩的荷载。（一般情况下，与实际情况一致）所以，实际步骤如下：判断使梁中点发生最大弯矩的临界荷载FPcr；移动荷载组，使FPcr与梁上全部荷载的合力FR对称于梁的中点，再算出此时