# Calculation of Beam Deflection

August 21, 2017 | Author: api-3848892 | Category: Bending, Solid Mechanics, Acceleration, Materials, Structural Engineering

#### Description

〔Technical Data〕

〔Technical Data〕

Calculation of Deflection

Calculation of Beam Deflection Load bending moment diagram

1

Deflection （δ）

3

δmax.＝

W  R 3EI

( ) 2 (σt)max.＝(σr)max.＝± 3P 3m＋12 R 8mt

Also, the central deflection δmax. can be expressed as follows： δmax.＝

W  r

M

The circumferential stress σt and radial σr can be expressed as follows at the center.

t

M max.＝W  R

2

Maximum stress, Maximum deflection

1.Disk receiving uniform load and having supported the perimeter

r

δmax.

W

Bending moment （M）

δmax. （Deflection） R

M max.＝

δmax.＝

w  R4 8EI

E：Young's modulus，

2.Disk receiving uniform load and having been fixed to the perimeter

1 w  r 2 2

M

w  R2 2

W

δmax. （Deflection）

R

δ   r

M max.＝ M

4

1 W  R 4

δmax.＝

W  R 48 EI

1 r 4 W

W

3

3.Disk receiving uniform load on the concentric circle having supported to the perimeter

P （Uniform load） 2r 0 R

p R

5

1 r 8 W  1 r 8 W

δmax.＝

2r

W  R3 192 EI

4.Disk receiving uniform load on the concentric circle having been fixed to the perimeter P （Uniform load）

w

2r 0

r

M

M max.＝ 1 8

1 w  R2 8

δmax.＝

5 wR 4 384EI R

wr 2

5.Rectangular plate receiving uniform load and having supported the perimeter

w

1 M max.＝ w  R2 12

r 1 12 1 24

δmax.＝

wR 4 384 EI

wr 2 wr 2

X 2b

M

6 δ

p

t

δ

t

M

1 W  R 8

0

2R

M max.＝

7

O

Y

r

Young's modulus σ E＝ ε （Vertical elastic modulus）

1 w  R2 8

δmax.＝

w  R4 184.6EI

When the central deflection r0 is smaller than R, δmax. is expressed as follows： 3(m−1)(3m＋1)PR δmax.＝ 2t3 4πEm

2

Wherein, P:Total ただしP…同心円上の総荷重でP＝π   load on the concentric circle. r 02p 1 R：Radius of plate， t：Plate thickness, E：Young's modulus， ：Poisson's ratio m The peripheral stress can be expressed as follows： 3P r 0 2 σr＝± 3P 1− r 0 2 σt ＝± 1− 2 2πt 2  2πmt 2 2R 2R2 2 ( ) and the central one is as follows： σt ＝σr＝±3 m＋1 2P log R ＋ r 02 2πmt r 0 4R The central deflection δmax. can be expressed as follows：

δmax.≒

3(m−1)(7m＋3)PR2 16πEm 2 t 3

Wherein, P:Total ただしP…同心円上の総荷重でP＝π   load on the concentric circle. r 02p 1 R…Radius of plate, t…Plate thickness, E…Young's modulus …Poisson's ratio m The X-axis direction stress Pb2 （σx） max.＝α 1 2 at center O can be expressed as follows： t The deflection Pb4 at the center O can be expressed as follows：δmax.＝β1 Et 3 a/b

1.0

1.5

2.0

3.0

4.0

α1

1.150

1.950

2.440

2.850

2.960

3.000

β1

0.709

1.350

1.770

2.140

2.240

2.280

（σx)max.＝α2

1 wr 2 8 9 w 2 128 r

ε＝Deflection I＝Cross-sectional secondary moment σ＝Rectangular stress

1 ：Poisson's ratio m The central stress can be expressed as follows： 3(m＋1)P m R m−1 r 02 （σt)max.＝(σr)max.＝± ＋log − 2πmt 2 m＋1 r 0 m＋1 4R2

6.Rectangular plate receiving uniform load and The stress to X-axis direction at the center A of the longer side can be expressed as follows： having been fixed to the perimeter Pb2 Pb4 δmax.＝β2 2 Et

X A 2b

M

M max.＝

The peripheral stress can be expressed as follows： 3PR2  3PR2 σt＝± （σr)max.＝± 4mt 2 4t 2 and the central one is as follows： 3(m＋1)PR （σt)max.＝σr＝± 8mt 2 The central deflectionδmax. can be expressed as follows： 2−l)PR 4 3(m  δmax.＝ 16Em 2 t3 Wherein, P：Load，R：Radius of plate，t：Plate thickness,

E：Young's modulus

2a w

δ

1113

δmax.

δ   r

1 ：Poisson's ratio m

E：Young's modulus，

t

3

3(m−1)(5m＋1) PR 4 16Em2 t 3

O A 2a

Y

t2

a/b

1.0

1.5

2.0

α2

1.231

1.817

1.990

2.000

β2

0.221

0.384

0.443

0.454

E：Young's modulus

1114