Beam Deflection Formulae
Upcoming SlideShare
Loading in...5
×
 

Like this? Share it with your network

Share

Beam Deflection Formulae

on

  • 4,039 views

 

Statistics

Views

Total Views
4,039
Views on SlideShare
4,039
Embed Views
0

Actions

Likes
0
Downloads
34
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Beam Deflection Formulae Presentation Transcript

  • 1. BEAM DEFLECTION FORMULAEBEAM TYPE SLOPE AT FREE END DEFLECTION AT ANY SECTION IN TERMS OF x MAXIMUM DEFLECTION 1. Cantilever Beam – Concentrated load P at the free end Pl 2 Px 2 Pl 3 θ= y= ( 3l − x ) δ max = 2 EI 6 EI 3EI 2. Cantilever Beam – Concentrated load P at any point Px 2 y= ( 3a − x ) for 0 < x < a Pa 2 6 EI Pa 2 θ= δ max = ( 3l − a ) 2 EI Pa 2 6 EI y= ( 3x − a ) for a < x < l 6 EI 3. Cantilever Beam – Uniformly distributed load ω (N/m) ωl 3 ωx 2 ωl 4 θ= 6 EI y= 24 EI ( x 2 + 6l 2 − 4lx ) δ max = 8 EI 4. Cantilever Beam – Uniformly varying load: Maximum intensity ωo (N/m) ωol 3 ωo x 2 ωo l 4 θ= 24 EI y= 120lEI (10l 3 − 10l 2 x + 5lx2 − x3 ) δ max = 30 EI 5. Cantilever Beam – Couple moment M at the free end Ml Mx 2 Ml 2 θ= y= δ max = EI 2 EI 2 EI
  • 2. BEAM DEFLECTION FORMULASBEAM TYPE SLOPE AT ENDS DEFLECTION AT ANY SECTION IN TERMS OF x MAXIMUM AND CENTER DEFLECTION 6. Beam Simply Supported at Ends – Concentrated load P at the center Pl 2 Px ⎛ 3l 2 ⎞ l Pl 3 θ1 = θ2 = y= ⎜ − x 2 ⎟ for 0 < x < δ max = 16 EI 12 EI ⎝ 4 ⎠ 2 48 EI 7. Beam Simply Supported at Ends – Concentrated load P at any point Pb(l 2 − b 2 ) y= Pbx 2 ( l − x2 − b2 ) for 0 < x < a Pb ( l 2 − b 2 ) 32 θ1 = 6lEI 6lEI δ max = 9 3 lEI at x = (l 2 − b2 ) 3 Pb ⎡ l 3⎤ ⎢ b ( x − a ) + (l − b ) x − x ⎥ 3 Pab(2l − b) y= 2 2 θ2 = 6lEI 6lEI ⎣ ⎦ δ= Pb 48 EI ( 3l 2 − 4b2 ) at the center, if a > b for a < x < l 8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m) ωl 3 ωx 3 5ωl 4 θ1 = θ2 = 24 EI y= 24 EI ( l − 2lx2 + x3 ) δmax = 384 EI 9. Beam Simply Supported at Ends – Couple moment M at the right end Ml 2 l Ml δmax = at x = θ1 = 9 3 EI 3 6 EI Mlx ⎛ x 2 ⎞ y= ⎜1 − ⎟ Ml 6 EI ⎝ l 2 ⎠ Ml 2 θ2 = δ= at the center 3EI 16 EI 10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity ωo (N/m) 7ωol 3 ωo l 4 θ1 = δ max = 0.00652 at x = 0.519 l ωo x 360 EI ω l3 y= 360lEI ( 7l 4 − 10l 2 x 2 + 3x4 ) ωol 4 EI θ2 = o δ = 0.00651 at the center 45 EI EI