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第 8 章 位 移 法 PowerPoint PPT Presentation


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第 8 章 位 移 法. 目的要求 1. 熟练掌握位移法基本未知量的确定和基本结构的建立、位移法的 典型方程及其物理意义、位移法方程中的系数和自由项的物理意 义及其计算、弯矩图的绘制。 2. 熟记常用的形常数和载常数。 3. 熟练掌握由弯矩图绘制剪力图和轴力图的方法。 4. 掌握利用对称性简化计算。 5. 重点掌握荷载作用下超静定结构的内力计算,了解其它因素下的 计算。 6. 位移法方程有两种建立方法,写典型方程法和写平衡方程法。要 求熟练掌握一种,另一种了解即可。 7. 知道位移法既能解超静定结构也能解静定结构。.

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第 8 章 位 移 法

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8

8

1.

2.

3.

4.

5.

6.

7.


8

8-1

8-1(a)BBABCBfBABBC8-1(b)(c)ABfBBCfBF .


8

8-1


8 1 a b 8 1 d b

()

(a)

()

(b)

8-1(a)B8-1(d)B:

(c)


8 1 b c

8-1(b)(c) :

8-1(e)

8-1(a)

(1)

(2)

(3)


8

8-2

1

8-2(a)()AB()8-2(b)


8

MFFSF

8-2

2

1


8

8-3


8

8-3(a)8-3(b) X3

(a)

(b)


8

(b) FX1=MAB X2=MBA ()(b)

(8-1)

(8-1)AB

AB

(8-2)


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(8-2)

F

2

8-4(a)ABAAB8-4(b)

(8-3)


8

8-4


8

(8-4)

(8-3)8-4(a)(8-3)(8-1)MBA=0()

(a)


8

(a)(8-1)

(a)

(b)

(b)

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(8-3)


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8-4(a)jBjA DAB ()8-18-3 1 8-5(a)CDjC jD


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CDCDC=D= jCjD3n jnl, 8-5(d)EE,


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8-5(b)(e)8-5(a)(d)8-6(a)(b)CD


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8-6

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8-1()


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8-7(a)EGZ1Z2Z38-7(b)8-7(c)FBFEFF, 8-7(d)


8

8-7


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8-4

8-8(a)CZ1CDZ28-8(b)Z1Z2()() R1R2

(a)

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8

R21R22R2P8-8(c) (d)(e)(a)

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R21+R22+R2P =0 (b)

(b)R()

()r11r12r21r22(b)

(c)


8

8-8


8

(8-5)


8

8-9


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(8-5)rii()rij() rij =rjiRiP

Z1 Z2Zn


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8-8(a)8-1

8-9(a)(b)(c)8-9(d)

(e)(f)(g)(h)(i) , ,

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8

r21r22R2P8-1Fx=0

(c)


8

8-10(a)(b)(c)M


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FN

M

Fs


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(1) nj nl

(2)

(3)

(4) MP

(5) Z1Z2Zn

(6) MFSFN


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8-18-11(a)E

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(2) 8-11(b)

(3)

(4) 8-11(c)(d)(e)8-11(f)(g)(h)


8

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()


8

8-11


8

(6) MFSFN8-11(i)(j)(k)MM


8

8-28-12(a)MBB=0.5cm,EI=3105kNm2

8-12


8

M8-1

(1)

(2) 8-12(a)

(3)

r11Z1+R1=0

(4)

8-12(c)(d)r11R1


8

BB=0.005m,8-1

MC=0

r11=38i, R1=-0.009i

(5) Z1

38iZ1-0.009i=0

Z1=2.3710-4()


8

(6) M8-12(e)

8-5

8-13a1Z112Z21M1=0Fx=0

M1=M13+M12=0 (a)

Fx=FS13+FS24=0 (b)

8-18-3


8

8-1

(a)(b)


8

8-13


8

8-6

1

8-14(a)8-14(b) CD=0 8-14(c) CD02 8-14(d)(e)8-14(b)(c)


8

8-14


8

2

8-15(a)8-14(b) =0 8-15(c)

8-15(d)(e)13


8

8-15


8

8-38-16(a)M

1/28-16(b)

(1)

(2) 8-16(c)i=EI/12

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(3)

r11Z1+R1P=0

(4) MP8-16(d)(e)r11R1P

r11=9i R1P=-15


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(5) Z1

9iZ1-15=0 Z1=5/3i()

(6) M AEMEDM8-16(a)8-16(f)


8

8-16


8

8-48-17(a)MEI=

1/48-17(b)

(1)

(2) 8-17(c)

(3)

r11Z1+R1P=0

(4)

MP8-17(d)(e)r11R1P

r11=2i R1P=Fa/8

(5) Z1


8

2iZ1+Fa/8=0 Z1=-Fa/16i()

6 8-17(a)M8-17(f)


8

8-17


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