Deflection of a Simply Supported BeamFull description
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DE VARIIS LATINIS GRAECISQVE COLLOQVENDI FORMVLIS ENCHIRIDION IN ANGLICAM LINGVAM CONVERSIS ΠΕΡΙ ΠΟΙΚΙΛΩΝ ΡΩΜΑΙΚΩΝ ΤΕ ΚΑΙ ΕΛΛΗΝΙΚΩΝ ΤΗΣ ΟΜΙΛΙΑΣ ΡΗΤΡΩΝ ΕΝΧΕΙΡΙΔΙΟΝ ΕΙΣ ΤΗΝ ΑΝΓΛΙΚΗΝ ΓΛΩΤΤΑΝ ΜΕΘΕ...
DE VARIIS LATINIS GRAECISQVE COLLOQVENDI FORMVLIS ENCHIRIDION IN ANGLICAM LINGVAM CONVERSIS ΠΕΡΙ ΠΟΙΚΙΛΩΝ ΡΩΜΑΙΚΩΝ ΤΕ ΚΑΙ ΕΛΛΗΝΙΚΩΝ ΤΗΣ ΟΜΙΛΙΑΣ ΡΗΤΡΩΝ ΕΝΧΕΙΡΙΔΙΟΝ ΕΙΣ ΤΗΝ ΑΝΓΛΙΚΗΝ ΓΛΩΤΤΑΝ ΜΕΘΕ...Full description
Fluid Mech 2 Formulae SheetFull description
MMAN2600 Formulae
Descripción: problemas resueltos
tablas de slope deflection rigidezDescripción completa
deflection
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
Pl 2
θ=
y=
2 EI
Px 2
6 EI
( 3 l− x)
Pl 3
δmax =
3 EI
2. Cantilever Beam – Concentrated load P at any point
y=
Pa 2
θ=
2 EI
3. Cantilever Beam – Uniformly distributed load
y= ω
6 EI Pa 2
6 EI
y=
6 EI
4. Cantilever Beam – Uniformly varying load: Maximum intensity
θ=
( 3 a− x)
for 0 < x < a
δmax =
( 3 x − a)
for a < x < l
Pa 2
6 EI
( 3l − a )
(N/m)
ωl 3
θ=
Px 2
ωol 3 24 EI
y=
ω x 2 24 EI ωo
ωo x 2
(
)
x2 + 6 2l − 4 lx
δmax =
ωl 4 8 EI
(N/m)
(10 l − 10 l 120lEI 3
2
x+ 5 lx2 − x3
)
δmax =
ωol 4 30 EI
5. Cantilever Beam – Couple moment M at the free end
θ=
Ml EI
y =
Mx 2
2 EI
δmax =
Ml 2
2 EI
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 = θ2 =
Pl 2
16 EI
=y
Px ⎛ 3l 2
⎜
12 EI ⎝ 4
−
2
⎞
x⎟ for 0 <
⎠
l
Pl 3
δmax =
2
48 EI
7. Beam Simply Supported at Ends – Concentrated load P at any point
θ1 = θ2 =
Pb(l 2 − b 2 )
6lEI Pab(2l − b) 6lEI
y=
Pbx
2
− x2 − b2 ) for 0 < x < a
⎤ y= ( x − a) + ( l2 − b2 ) x− x3 ⎥ ⎢ 6lEI ⎣ b ⎦ for a < x < l
8. Beam Simply Supported at Ends – Uniformly distributed load
θ1 = θ2 =
(l
6lEI Pb ⎡ l
ωl 3
ω
y=
24 EI
δ max =
(
32
at x=
9 3 lEI
3
δ=
)
Pb l 2 − b 2
Pb
48 EI
(3l
2
(l − b) 2
( 24 EI
l − 2 lx + x )
3
2
δmax =
3
5ωl 4 384 EI
9. Beam Simply Supported at Ends – Couple moment M at the right end
θ1 = θ2 =
6 EI Ml
y =
Ml⎛x ⎜1 − 6 E⎝I
θ2 =
x⎞
2
2 ⎟ l⎠
δ=
3 EI
10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity
