Structural Analysis Methods I: Beam Elements - Mechanics of Structures: Variational and Computational Methods - page 236

Hint: For this system of four differential equations, use the first step of length

k 1 = f 1 ( x 0 , w

0 ,

θ

0 ,V 0 ,M 0

)

f 1 x 0 + 2 ,

k 1
2

m 1
2

n 1
2

r 1
2

k 2 = w 0 + , θ 0 + ,V 0 + ,M 0 +

f 1 x 0 + 2 ,

k 2
2

m 2
2

n 2
2

r 2
2

k 3 = w 0 + , θ 0 + ,V 0 + ,M 0 +

k 4 = f 1 ( x 0 + , w + k 3 , θ + m 3 ,V 0 + n 3 ,M 0 + r 3 )

0 0

m 1 = f 2 ( x 0 , w 0 , θ 0 ,V 0 ,M 0 )

. etc.

with

1
EI M,

f 1 =− θ , f 2 = f 3 =− p z ( x ) = p 0 , f 4 = V

Initial conditions:

w( 0 ) = w 0 , θ( 0 ) = θ 0 , V ( 0 ) = V 0 , M ( 0 ) = M 0

EI M 0

k 1 = ( − θ ) m 1 =

0

EI M 0

= EI M 0 p 0

1
2

1
2

k 2 = − θ − m 2 +

0

M 0

V 0

M 0 V 0

p 0

1
2

EI 1

2

EI 1

2

1
2

k 3 = − θ − + m 3 = + +

0

M 0 V 0

p 0 = EI M 0 + V 0 p 0

EI 1

2

1
2

1
2

k 4 = − θ − + + m 4 +

0

n 1 = p 0 r 1 = V 0

r 2 = V 0 + p 0

1

n 2 = p 0

2

V 0

p 0

1
2

n 3 = p 0 r 3 = +

n 4 = p 0 r 4 = ( V 0 + p 0 )

Answer:

1
6 (

w = w + k 1 + 2 k 2 + 2 k 3 + k 4 )

0

3
6 EI V 0

2
2 EI M 0

4
24 EI p 0

= w − θ − − −

0 0

1
6 (

θ = θ + m 1 + 2 m 2 + 2 m 3 + m 4 )

0

2
2 EI V 0 +

3
6 EI p 0

EI M 0 +

= θ 0 +

1
6 (

V =

+ + + + )

V 0 n 1 2 n 2 2 n 3 n 4

= V 0 + p 0

1
6 (

M

=

M 0 + r 1 + 2 r 2 + 2 r 3 + r 4 )

1
2

2 p 0

= V 0 + M 0 +