ABSTRACT
The derivation of the stiffness matrix for a bar element is applicable to the solution of pin-connected trusses. The bar element is assumed to have a constant cross-section area A, uniform modulus of elasticity E, and initial length L. The bar is subjected to tensile forces along the local axis that are applied at its ends. There are two coordinate systems: a local one (xˆ, yˆ) and a global one (x, y). The nodal degrees of freedom are the four local displacements: dˆ1x, dˆ1y, dˆ2x, and dˆ 2y. The strain-displacement relationship is obtained from Hooke’s law
x xEσ = ε (2.1)
where
=
ˆd ˆdx u
x ε
(2.2)
xA Tσ = (2.3)
uˆ is the axial displacement in the xˆ-direction T is the tensile force
Note that the bar element cannot sustain shear forces. Substituting σx and εx into Hooke’s law:
ˆd d 0
ˆ ˆd d
u AE
x x ⎛ ⎞
=⎜ ⎟⎝ ⎠ (2.4)
Assuming a linear displacement along the local x-axis of the bar, the displacement function can be written as
1 2 ˆ ˆu a xa= + (2.5)
Now, the displacement function is expressed as a function of the nodal displacement dˆ1x and dˆ2x, which can be achieved by evaluating û at the nodes, and solving for a1 and a2, as follows
( ) = =1 1ˆˆ 0 xu d a (2.6)
( ) = = +2 2 1ˆ ˆˆ x xL d a L du (2.7) and solving for a2
ˆ ˆ x xd da
L
−
=
(2.8)
The displacement function becomes
ˆ ˆ ˆ ˆ ˆx x
xu d d
d L
x ⎛ ⎞
−
= +⎜ ⎟⎝ ⎠
(2.9)
In matrix form, the above equation can be written as
ˆ ˆ
d u N N
d
⎧ ⎫⎪ ⎪ = ⎨ ⎬⎪ ⎪⎩ ⎭
(2.10)
with the shape functions:
ˆ N
x
L = −
(2.11)
ˆ N
L
x =
(2.12)
The strain-displacement is
−
ˆˆ ˆd ˆd
d
L
u d
x ε
The stiffness matrix is derived as follows
xT A= σ (2.13)
x xd dT AE L
⎛ ⎞ −
= ⎜ ⎟⎝ ⎠
Also, the nodal force at node number 1 should have a negative sign, as follows:
xf T= − (2.14)
ˆ ˆ( )ˆ
AE
L f
d d =
−
(2.15)
On the other hand, the nodal force at node number 2 should have a positive sign, as follows:
xf T= (2.16)
2 1 2 ˆ ˆ( ) ˆ
AE
L d d f =
−
(2.17)
Expressing the nodal forces in the x-direction in a matrix form
1 1 ˆ 1 1 ˆ 1 1
ˆ
f AE
Lf
d
d
⎧ ⎫ ⎧ ⎫ −⎡ ⎤⎪ ⎪ ⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎢ ⎥ −⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
(2.18)
and similarly for the nodal forces in the y-direction
ˆ ˆ1 1 ˆ ˆ1 1
Af d
f
E
L d
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩
−
⎭− (2.19)
Since ˆ ˆ ˆf k d= , the stiffness matrix for a bar element in a local coordinate can be written as
1 1
1 1 ˆ AE
L k
−⎡ ⎤ = ⎢ ⎥
−⎣ ⎦ (2.20)
The local coordinate system is always chosen to represent an individual element, while the global coordinate system is chosen for the whole structure. In order to relate the global displacement components to a local one, the transformation matrix is used. Figure 2.1 shows the general displacement vector with local and global coordinate systems.
