Principle of Virtual Work for Trusses

Principle of Virtual Work for Trusses

A truss is like a series of interconnected springs in 2-D since the force in each hb bar iis along l th the axis i off th the b bar , causing i each hb bar tto extend t d or compress, just like a spring. The spring constant, k, for a bar is given by

EA k= L

E is the Young's modulus for the bar ( a material property) given in GPa (GigaPascals) in the SI system, where 1 Pa = 1 Newton/m2. For example, example for steel steel, E = 210 GPa GPa=210x10 210x109 Pa A is the cross-sectional area of the bar (in m2) L is the length of the bar (in m2) In the English system E is measured in psi =lb/in2, A is measured in in2 , and L is measured in inches inches. For steel steel, E = 30x106 psi. psi

One difference between a truss and our previous 1-D spring examples is that the displacements at the ends of each bar are all in different directions

We need to place all of these displacements in the same "global" coordinate system along the x- and y-axes

Utj

t s Usjj

Uyj

jth node Uxj Uti

Usi θ

Uyi θ

x

ith node x

Uxi θ is the angle that the bar makes with the positive x-axis x axis

t

s

Utj Usj Fsj

Uti

Usi θ

Fsi In the s-t coordinates we have the stiffness matrix for each element (bar):

⎧ Fsi ⎫ ⎡ k ⎪F ⎪ ⎢ ⎪ ti ⎪ ⎢ 0 ⎨ ⎬= ⎪ Fsj ⎪ ⎢ −k ⎪⎩ Ftj ⎪⎭ ⎣⎢ 0

0 −k 0

0

0 0

k 0

0 ⎤ ⎧U si ⎫ ⎪ ⎪ 0 ⎥⎥ ⎪U ti ⎪ ⎨ ⎬ 0 ⎥ ⎪U sj ⎪ ⎥ 0 ⎦ ⎪⎩U tj ⎪⎭

Fsi = k (U si − U sj ) which is the same as

Fti = 0

Fsj = k (U sj − U si ) Ftj = 0

⎧ Fsi ⎫ ⎡ k ⎪F ⎪ ⎢ ⎪ ti ⎪ ⎢ 0 ⎨ ⎬= ⎪ Fsj ⎪ ⎢ − k ⎪⎩ Ftj ⎪⎭ ⎢⎣ 0

0 −k 0

0

0 0

k 0

0 ⎤ ⎧U si ⎫ ⎪ ⎪ 0 ⎥⎥ ⎪U ti ⎪ ⎨ ⎬ 0 ⎥ ⎪U sj ⎪ ⎥ 0 ⎦ ⎪⎩U tj ⎪⎭

again this relationship simply comes from the virtual work principle

δ W = Fsiδ U si + Ftiδ U ti + Fsjδ U sj + Ftjδ U tj 2⎤ ⎡1 = δ E = δ ⎢ k (U sj − U si ) ⎥ ⎣2 ⎦

= k (U sj − U si )(δ U sj − δ U si )

Fsi = k (U si − U sj ) Fti = 0

Fsj = k (U sj − U si )

Ftj = 0

We need to transform the stiffness matrix below, which is given in the s-t coordinates to one given in the global x-y coordinates coordinates,

⎡k ⎢0 % ⎤=⎢ ⎡⎣K e⎦ ⎢ −k ⎢ ⎣0

0 −k 0 0 0 0

k 0

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎦

The easiest way to do this is to note that the strain energy of each element (bar ) can also be written as

{ }

1 % E= U 2

T

{ }

% ⎤ U % = 1 {U U U ⎡⎣K e⎦ si ti sj 2

2 1 = k (U sj − U si ) 2

⎡k ⎢0 U tj } ⎢ ⎢ −k ⎢ ⎣0

0 −k 0

0

0

k

0

0

0 ⎤ ⎧U si ⎫ ⎪ ⎪ 0 ⎥⎥ ⎪U ti ⎪ ⎨ ⎬ ⎥ 0 ⎪U sj ⎪ ⎥ 0 ⎦ ⎪⎩U tj ⎭⎪

Uyi Uti

Usi θ

Uxi

ith node To transform the displacements at the ith node we have

⎧U si ⎫ ⎡ c s ⎤ ⎧U xi ⎫ ⎨ ⎬=⎢ ⎥ ⎨U ⎬ U s c − ⎦ ⎩ yi ⎭ ⎩ ti ⎭ ⎣ and at the jth node

⎧U sj ⎫ ⎡ c s ⎤ ⎧U xj ⎫ ⎨ ⎬=⎢ ⎥ ⎨U ⎬ U − s c ⎦ ⎩ yj ⎭ ⎩ tj ⎭ ⎣

c = cos θ s = sin θ

We can write these relations together as

⎧U si ⎫ ⎡ c s 0 0 ⎤ ⎧U xi ⎫ ⎪U ⎪ ⎢ ⎥ ⎪U ⎪ − 0 0 s c ⎪ ti ⎪ ⎢ ⎪ yi ⎪ ⎥ = ⎨ ⎬ ⎨ ⎬ U ⎢ ⎥ 0 0 c s ⎪U xjj ⎪ ⎪ sjj ⎪ ⎪⎩U tj ⎪⎭ ⎢⎣ 0 0 − s c ⎥⎦ ⎪⎩U yj ⎪⎭ or as

{U% } = [Q]{U} Pl i thi Placing this relationship l ti hi iinto t th the strain t i energy expression i we fifind d

{ }

{ }

1 % T % % = 1 {U}T [Q ]T [ K ][Q ]{U} E = U ⎡⎣K e ⎤⎦ U e 2 2 1 T = {U} [ K e ]{U} 2

[K e ] = [Q] ⎡⎣K% e ⎤⎦ [Q] T

which shows that

[ K e ] = [Q ]

T

% ⎤ [Q ] ⎡⎣K e⎦

Expanding this out we find

⎡ c2 cs −c 2 −cs ⎤ ⎢ 2 2⎥ − − cs s cs s ⎥ [K e ] = k ⎢⎢ 2 2 −c −cs c cs ⎥ ⎢ 2 2 ⎥ cs s ⎦ ⎣ −cs − s element (bar) stiffness matrix in the global x-y coordinates We now need to assemble these stiffness matrices into a global stiffness matrix [Kg] for the truss, apply the boundary conditions, and solve for the displacements {U} . Then we can find the external forces on all the nodes (pins) in the x-y coordinates (including the reactions) from

⎡⎣K g ⎤⎦ {U} = {Fn }

the contribution to the global Kg matrix for an element containing the ith and jth nodes U1 K g ( 2i − 1, 2i − 1) K g ( 2i − 1, 2i ) K g ( 2i, 2i − 1) K g ( 2i, 2i )

K g ( 2i − 1, 2 j − 1) K g ( 2i − 1, 2 j ) K g ( 2i, 2 j − 1) K g ( 2i, 2 j )

U 2i −1

U xi

U 2i

U yi

K g ( 2 j − 1, 2i − 1) K g ( 2 j − 1, 2i ) K g ( 2 j , 2i − 1) K g ( 2 j , 2i )

K g ( 2 j − 1, 2 j − 1) K g ( 2 j − 1, 2 j ) K g ( 2 j , 2 j − 1) K g ( 2 j, 2 j )

U 2 j −1

U xj

U2 j

U yj

UN

K g ( 2i − 1, 2i − 1) = k11 K g ( 2i − 1, 2i ) = k12

etc etc.

components of the Ke stiffness matrix for an element

Finally, we can find the force in each bar from

⎧U xi ⎫ ⎪U ⎪ ⎪ yi ⎪ Fsj = k {−c − s c s} ⎨ ⎬ ⎪U xj ⎪ ⎩⎪U yj ⎭⎪ since

Fsj = k (U sj − U si )

U sj

Fsj U si

Fsi

+ if tensile - if compressive

and we have simply used the transformation of displacements to express this force in terms of the x and y components of the displacements