# experiment

**Topics:**Beam, Deflection, Cantilever

**Pages:**8 (353 words)

**Published:**November 24, 2013

BEAM TYPE

SLOPE AT FREE END DEFLECTION AT ANY SECTION IN TERMS OF x

1. Cantilever Beam – Concentrated load P at the free end

Pl 2

2 EI

θ=

y=

MAXIMUM DEFLECTION

Px 2

( 3l − x )

6 EI

δ max =

Pl 3

3EI

2. Cantilever Beam – Concentrated load P at any point

Px 2

( 3a − x ) for 0 < x < a

6 EI

Pa 2

y=

( 3x − a ) for a < x < l

6 EI

y=

Pa 2

θ=

2 EI

δ max

Pa 2

=

( 3l − a )

6 EI

3. Cantilever Beam – Uniformly distributed load ω (N/m)

ωl 3

6 EI

θ=

δ max =

ωl 4

8 EI

ωo x 2

(10l 3 − 10l 2 x + 5lx2 − x3 )

120lEI

δ max =

ωo l 4

30 EI

Mx 2

2 EI

δ max =

Ml 2

2 EI

y=

ωx 2

( x 2 + 6l 2 − 4lx )

24 EI

4. Cantilever Beam – Uniformly varying load: Maximum intensity ωo (N/m)

θ=

ωol 3

24 EI

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

Pl 2

θ1 = θ2 =

16 EI

⎞

Px ⎛ 3l 2

l

− x 2 ⎟ for 0 < x <

y=

⎜

12 EI ⎝ 4

2

⎠

δ max =

Pl 3

48 EI

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

Pb ⎡ l

3

2

2

3⎤

y=

⎢ b ( x − a ) + (l − b ) x − x ⎥

6lEI ⎣

⎦

for a < x < l

y=

δ max =

δ=

Pb ( l 2 − b 2 )

32

9 3 lEI

at x =

(l

2

− b2 ) 3

Pb

( 3l 2 − 4b2 ) at the center, if a > b

48 EI

8. Beam Simply Supported at Ends – Uniformly distributed load ω (N/m)

θ1 = θ2 =

ωl 3

24 EI

y=

ωx 3

( l − 2lx2 + x3 )

24 EI

δmax =

5ωl 4

384 EI

9. Beam Simply Supported at Ends – Couple moment M at the right end

Ml

θ1 =

6 EI

Ml

θ2 =

3EI

y=

Mlx ⎛ x 2 ⎞

⎜1 − ⎟

6 EI ⎝ l 2 ⎠

δmax =

δ=

Ml 2

l

at x =

9 3 EI

3

Ml 2

at the center

16 EI

10. Beam Simply Supported at Ends – Uniformly varying load: Maximum intensity ωo (N/m)

7ωol 3

360 EI

ω l3

θ2 = o

45 EI

θ1 =

y=

ωo x

( 7l 4 − 10l 2 x 2 + 3x4 )

360lEI

δ max = 0.00652

δ = 0.00651

ωo l 4

at x = 0.519 l

EI

ωol 4

at the center

EI

http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf

file:///G|/BACKUP/Courses_and_seminars/0MAE4770S12/url%20for%20beam%20formulas.txt[1/23/2012 12:15:35 PM]

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