BEAM DEFLECTION FORMULAE BEAM 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 2 2 Pl EI θ= ( ) 2 3 6 Px y l x EI = − 3 max 3 Pl EI δ = 2. Cantilever Beam – Concentrated load P at any point 2 2 Pa EI θ= ( ) 2 3 for 0 6 Px y a x x a EI = − < < ( ) 2 3 for 6 Pa y x a a x l EI = − < < ( ) 2 max 3 6 Pa l a EI δ = − 3. Cantilever Beam – Uniformly distributed load ω (N/m) 3 6 l EI ω θ= ( ) 2 2 2 6 4 24 x y x l lx EI ω = + − 4 max 8 l EI ω δ = 4. Cantilever Beam – Uniformly varying load: Maximum intensity ω o (N/m) 3 o 24 l EI ω θ= ( ) 2 3 2 2 3 o 10 10 5 120 x y l lx lx x lEI ω = − + − 4 o max 30 l EI ω δ = 5. Cantilever Beam – Couple moment M at the free end Ml EI θ= 2 2 Mx y EI = 2 max 2 Ml EI δ =
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BEAM DEFLECTION FORMULAE
BEAM TYPE SLOPE AT FREE END DEFLECTION AT ANY SECTION IN TERMS OF x MAXIMUM DEFLECTION1. Cantilever Beam – Concentrated load P at the free end
2
2PlEI
θ = ( )2
36Pxy l xEI
= − 3
max 3PlEI
δ =
2. Cantilever Beam – Concentrated load P at any point
2
2Pa
EIθ =
( )2
3 for 06Pxy a x x aEI
= − < <
( )2
3 for6Pay x a a x l
EI= − < <
( )2
max 36Pa l a
EIδ = −
3. Cantilever Beam – Uniformly distributed load ω (N/m)
3
6lEIω
θ = ( )2
2 26 424
xy x l lxEI
ω= + −
4
max 8lEIω
δ =
4. Cantilever Beam – Uniformly varying load: Maximum intensity ωo (N/m)
3o
24lEI
ωθ = ( )
23 2 2 3o 10 10 5
120xy l l x lx xlEI
ω= − + −
4o
max 30lEI
ωδ =
5. Cantilever Beam – Couple moment M at the free end
MlEI
θ = 2
2Mxy
EI=
2
max 2Ml
EIδ =
BEAM DEFLECTION FORMULAS
BEAM 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
2
1 2 16Pl
EIθ = θ =
223 for 0
12 4 2Px l ly x xEI
⎛ ⎞= − < <⎜ ⎟
⎝ ⎠
3
max 48Pl
EIδ =
7. Beam Simply Supported at Ends – Concentrated load P at any point
2 2
1( )6
Pb l blEI−
θ =
2(2 )
6Pab l b
lEI−
θ =
( )2 2 2 for 06Pbxy l x b x alEI
= − − < <
( ) ( )3 2 2 3
6for
Pb ly x a l b x xlEI b
a x l
⎡ ⎤= − + − −⎢ ⎥⎣ ⎦< <
( )3 22 2
max 9 3
Pb l b
lEI
−δ = at ( )2 2 3x l b= −
( )2 2 at the center, if 3 448
Pb l bEI
δ = − a b>
8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m)
3
1 2 24lEI
ωθ = θ = ( )3 2 32
24xy l lx xEI
ω= − +
4
max5
384lEI
ωδ =
9. Beam Simply Supported at Ends – Couple moment M at the right end
1 6MlEI
θ =
2 3MlEI
θ =
2
216Mlx xy
EI l⎛ ⎞
= −⎜ ⎟⎝ ⎠
2
max 9 3Ml
EIδ = at
3lx =
2
16Ml
EIδ = at the center
10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity ωo (N/m)
3o
17360
lEI
ωθ =
3o
2 45lEI
ωθ =
( )4 2 2 4o 7 10 3360
xy l l x xlEI
ω= − +
4o
max 0.00652 lEIω
δ = at 0.519x l=
4o0.00651 l
EIω
δ = at the center
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