Title | Beam Formulas |
---|---|
Author | Doodoo Hado |
Course | Engineering Computations |
Institution | University of Technology Sydney |
Pages | 5 |
File Size | 596.3 KB |
File Type | |
Total Downloads | 72 |
Total Views | 168 |
Formulas for Assignment...
BEAM DEFLECTION FORMULAE BEAM TYPE SLOPE AT FREE END DEFLECTION AT ANY SECTION IN TERMS OF x 1. Cantilever Beam – Concentrated load P at the free end 2
2
Pl 2EI
θ=
MAXIMUM DEFLECTION
y=
Px (3l − x ) 6 EI
3
δmax =
Pl 3EI
2. Cantilever Beam – Concentrated load P at any point
Px 2 (3 a − x ) for 0 < x < a 6 EI 2 Pa y= ( 3x − a ) for a < x < l 6EI
y=
2
Pa 2EI
θ=
2
δ max =
Pa ( 3l − a ) 6 EI
3. Cantilever Beam – Uniformly distributed load ω (N/m)
ωl3 6EI
θ=
ωx 2 ( x2 + 6l 2 − 4lx ) 24 EI
δmax =
ωl 4 8EI
ωo x 2 (10l3 −10l2 x + 5lx2 − x3 ) 120lEI
δ max =
ωo l 4 30EI
Mx 2 2 EI
δmax =
Ml 2 2EI
y=
4. Cantilever Beam – Uniformly varying load: Maximum intensity ωo (N/m)
θ=
ωol 3 24EI
y=
5. Cantilever Beam – Couple moment M at the free end
θ=
Ml EI
y=
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
θ1 = θ 2 =
Pl 2 16 EI
y=
Px ⎛ 3l 2 l ⎞ − x 2 ⎟ for 0 < x < ⎜ 12 EI ⎝ 4 2 ⎠
δ max =
Pl 3 48EI
7. Beam Simply Supported at Ends – Concentrated load P at any point
Pb (l 2 − b 2 ) θ1 = 6lEI Pab(2l − b) θ2 = 6lEI
Pbx 2 ( l − x2 − b2 ) for 0 < x < a 6lEI 3 Pb ⎡ l y= ( x − a ) + (l 2 − b 2 ) x − x 3 ⎤⎥ 6lEI ⎢⎣ b ⎦ for a < x < l y=
δmax = δ=
Pb ( l 2 − b2 )
32
9 3 lEI
at x =
(l
2
− b2) 3
Pb (3l 2 − 4b 2 ) at the center, if a > b 48 EI
8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m)
θ1 = θ 2 =
ωl 3 24EI
y=
ωx 3 ( l − 2lx 2 + x 3 ) 24 EI
δmax =
5ωl 4 384 EI
9. Beam Simply Supported at Ends – Couple moment M at the right end
Ml θ1 = 6EI Ml θ2 = 3 EI
y=
2 Mlx ⎛ x ⎞ ⎜ 1− 2 ⎟ 6EI ⎝ l ⎠
δmax = δ=
Ml 2 l at x = 3 9 3 EI
Ml 2 at the center 16EI
10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity ωo (N/m)
7ωol 3 360 EI ω l3 θ2 = o 45E I
θ1 =
y=
ωo x (7l 4 − 10l 2x 2 + 3x 4 ) 360lEI
δmax = 0.00652
ωo l 4 at x = 0.519 l EI
ω l4 δ = 0.00651 o at the center EI
http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf...