Embed
Email

Analysis

Document Sample
Analysis
Shared by: HC111122235629
Categories
Tags
Stats
views:
1
posted:
11/22/2011
language:
Malay
pages:
20
Beam_curved.xls

To determine deflection of a curved beam

By Alex Slocum, 12/29/03, last modified 09/21/04 by Xue'en Yang

Enters numbers in BOLD, Results in RED

Schematic



Beam dimensions and properties Values

If round cross-section (enter "yes" or "no") yes Mo

Outside diameter, Do (mm) 30 F x

Inside diamter, Di (mm) 25

If rectangular cross-section no

Outside height, ho (mm) 20 Fy

Outside width, wo (mm) 10

Inside height, hi (mm) 18

Inside width, wi (mm) 8 Circular cross-section

Radius of curvature, Rr (mm) 100

Angle, phi (degrees) 90

Angle, phi (radius) 1.571

Modulus of elasticity, E (N/mm^2) 200000

Poisson ratio, v 0.29

Shear modulus, G (Pa) 77519 Rectangular cross-section

Moment of inertia, I (mm^4) 20586

Distance farthest fiber to neutral axis, cc (mm) 15

Cross section area, A 216

EA 4.32E+07

GA 1.67E+07

EI 4.12E+09

Loading

Point load, Fx (N) 10

Point load, Fy (N) 76

Moment, Mo (N*mm) 22

X direction displacement (microns) Axial Shear Bending Total

Due to Fx 0.018 0.047 0.865 0.930

Due to Fy 0.000 0.000 9.230 9.230

Due to Mo 0.000 0.000 0.031 0.031

Total 0.018 0.047 10.125 10.190

Y direction displacement (microns) Axial Shear Bending Total

Due to Fx 0.000 0.000 1.214 1.214

Due to Fy 0.138 0.357 14.498 14.992

Due to Mo 0.000 0.000 0.053 0.053

Total 0.138 0.357 15.766 16.260

Slope (milliradians) Bending

Due to Fx 0.014

Due to Fy 0.185

Due to Mo 0.001

Total 0.199



Equations

P  F x cos   F y sin  V  F x sin   F y cos  M  F xR 1  co

P P P V V V M

 cos    sin  0  sin   cos  0  R 1  co

F x F y M o F x F y M o F x



U 

F x cos   F y sin  cos  

F xsin   F y cos  sin  

F x R 1

  

2 2 2

x  Rd  Rd 

F x EA GA

0 0 0

axial shear

P  F x cos   F y sin  V  F x sin   F y cos  M  F xR 1  co

P P P V V V M

 cos    sin  0  sin   cos  0  R 1  co

F x F y M o F x F y M o F x



U 

F x cos   F y sin  cos  

F xsin   F y cos  sin  

F x R 1

  

2 2 2

x  Rd  Rd 

F x EA GA

0 0 0

axial shear









 x  R  F x  2  sin 2   F y  1  cos     R   2  sin 2   1  cos 2    1 

2



     F x   F y  

EA   4   2   axial GA   4   2  EI 

shear





U 

 F x cos  sin   F y sin  

F x cos  sin   F y cos  





  

2 2 2

F xR

y  Rd  Rd 

F y EA GA

0 0 0

axial shear









 y  R   F x  1  cos    F y  2  sin 2    R   1  cos 2   2  sin 2 

2

 1

     F x   F y 

EA   2   4  GA   2   4  EI

axial shear





U F xR 1  cos    F yR sin   M o



 

M o





0 EI

Rd

bending







  1  2F   sin    2F 1  cos    M R 

R x

EI

R y o 

bending

Mo Rsin

R(1-cos)

F x



Fy  R



cular cross-section 









angular cross-section









Check: Bending (from

Roark) for phi = p/2

Bending ratio

0.865 1

9.230 1

0.031 1

10.125 1



1.214 1

14.498 1

0.053 1

15.766 1



0.014 1

0.185 1

0.001 1

0.199 1







M  F xR 1  cos    F yR sin   M o ds  Rd

M M M

0  R 1  cos    R sin  1

F x F y M o



F x R 1  cos    F y R sin  1  cos    M oR 1  cos  

 2







2 2



Rd  Rd

0 EI

shear bending

M  F xR 1  cos    F yR sin   M o ds  Rd

M M M

0  R 1  cos    R sin  1

F x F y M o



F x R 1  cos    F y R sin  1  cos    M oR 1  cos  

 2







2 2



Rd  Rd

0 EI

shear bending







 1  cos 2    1  3F  6  8sin   sin 2   3F  1  2 cos   cos   

2 

 R x   M o R   sin   

2

  R y

 2  EI   4   2  

shear bending





 

F x R sin  1  cos    F y R sin   M oR sin 



2 2 2 2



Rd  Rd

0 EI

shear bending







2  sin 2   1  3F  1  2 cos   cos    3F  2  sin 2  

2 

 R x   M o R 1  cos   

2

 R y

4  EI   2   4  

shear bending


Related docs
Other docs by HC111122235629
AFE IIt Project Status
Views: 0  |  Downloads: 0
graham2
Views: 9  |  Downloads: 0
Indonesia2001
Views: 2  |  Downloads: 0
Calculator Notes
Views: 0  |  Downloads: 0
THE GEOSPHERE AND GEOCHEMISTRY
Views: 6  |  Downloads: 0
PENGOLAHAN PUSTAKA
Views: 30  |  Downloads: 0
Slide 1
Views: 0  |  Downloads: 0
2624 FRONT Males and Females
Views: 1  |  Downloads: 0
By registering with docstoc.com you agree to our
privacy policy

You are almost ready to download!

You are almost ready to download!