Beam Deflection Formulae

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

θ1 =

δmax =

 Ml

7ωo l

ωo

 Ml 2

9 3 EI 

 Ml

45 EI 

y=

ωo x 360lEI 

at  x =

l

3

2

16 EI 

at the center 

(N/m)

3

360 EI  ωol 3

( 7 l4 − 10 l2 x2 + 3 x4 )

δmax = 0.00652 δ = 0.00651

3

− 4b 2 ) at the center, if  a > b

(N/m)

ω x

2

ωol 4  EI 

ωol 4  EI 

at  x = 0.519 l

at the center