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Chapter 4
Analysis of 2D trusses

4.1 Introduction

This chapter deals with the static analysis of two dimensional trusses, which are
basically bars oriented in two dimensional cartesian systems. A transformation of
coordinate basis is necessary to translate the local element matrices (stiﬀness ma-
trix, force vector) into the structural (global) coordinate system. Trusses support
compressive and tensile forces only, as in bars. All forces are applied at the nodes.
After the presentation of the element formulation, some examples are solved by
MATLAB codes.

4.2 2D trusses

In ﬁgure 4.1 we consider a typical 2D truss in global x − y plane. The local system
of coordinates x − y  deﬁnes the local displacements u1 , u2 . The element possesses
two degrees of freedom in the local setting,

u = [u1 u2 ]
T
(4.1)

while in the global coordinate system, the element is deﬁned by four degrees of
freedom
uT = [u1 u2 u3 u4 ] (4.2)
The relation between both local and global displacements is given by

u1 = u1 cos(θ) + u2 sin(θ) (4.3)

u2 = u3 cos(θ) + u4 sin(θ) (4.4)

A.J.M. Ferreira, MATLAB Codes for Finite Element Analysis: 51
Solids and Structures, Solid Mechanics and Its Applications 157,

c Springer Science+Business Media B.V. 2009
52 4 Analysis of 2D trusses

u4

x u2
u3

u2

θ
u1
u1

y

x

Fig. 4.1 2D truss element: local and global degrees of freedom

where θ is the angle between local axis x and global axis x, or in matrix form as

u = Lu (4.5)

being matrix L deﬁned as  
l m0 0
L= (4.6)
0 0 l m
The l, m elements of matrix L can be deﬁned by the nodal coordinates as
x2 − x1 y2 − y1
l= ; m= (4.7)
Le Le
being Le the length of the element,

Le = (x2 − x1 )2 + (y2 − y1 )2 (4.8)

4.3 Stiﬀness matrix

In the local coordinate system, the stiﬀness matrix of the 2D truss element is given
by the bar stiﬀness, as before:
 
 EA 1 −1
K = (4.9)
Le −1 1
4.5 First 2D truss problem 53

In the local coordinate system, the strain energy of this element is given by
1 T  
Ue = u Ku (4.10)
2
Replacing u = Lu in (4.10) we obtain
1 T T 
Ue = u [L K L]u (4.11)
2
It is now possible to express the global stiﬀness matrix as

K = LT K L (4.12)

or ⎡ ⎤
l2 lm −l2 −lm

EA ⎢ lm m2 −lm −m2 ⎥
K= ⎥ (4.13)
Le ⎣ −l2 −lm l2 lm ⎦
−lm −m2 lm m2

4.4 Stresses at the element

In the local coordinate system, the stresses are deﬁned as σ = E. Taking into
account the deﬁnition of strain in the bar, we obtain
 
u − u1 E u1 E
σ=E 2 = [−1 1] = [−1 1]u (4.14)
Le Le u2 Le

By transformation of local to global coordinates, we obtain stresses as function of
the displacements as
E E
σ= [−1 1]Lu = [−l −m l m]u (4.15)
Le Le

4.5 First 2D truss problem

In a ﬁrst 2D truss problem, illustrated in ﬁgure 4.2, we consider a downward point
force (10,000) applied at node 1. The modulus of elasticity is E = 30e6 and all
elements are supposed to have constant cross-section area A = 2. The supports
are located in nodes 2 and 4. The numbering of degrees of freedom is illustrated
in ﬁgure 4.3.

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