TRUSSES
CHAPTER 4
§ Trusses are multi-dimension structures
composed of one-dimensional elements.
We derived a 1-D element in Chapter 3,
let’s see if it can be used to solve truss
problems.
2-D Stiffness Matrix for a Rod
ym
1
2
i
j
K21
Δ1=1
unit
K11
3
4
xm
K41
K31
§ The (xm,ym) coordinates are local
coordinates for each truss element. xm
originates at node i and proceeds through
node j. Local zm is always normal to the
plane of the truss. Local ym is established
using right hand rule.
Direct Stiffness Method
§ The stiffness term KMN has the
meaning: force at the M degree-offreedom due to unit displacement at
the N degree-of-freedom. So first set
D.O.F. #1 = 1 unit, and restrain all
others. From strength of materials
we know: ΔL F
AE
=
and F =
ΔL
L AE
L
Thus:
ym
0
Δ1=1
unit
EA/L
0
EA/L
EA
K11 =
L
EA
K13 = −
L
K12 = 0
K14 = 0
xm
§ Now set D.O.F. #2 = 1 unit,
restrain all others (the rod has
no bending stiffness).
K 21 = 0 K 22 = 0
K 23 = 0 K 24 = 0
§ Now set D.O.F. #3 = 1 unit,
restrain all others.
K 31 = − EA K 32 = 0
L
EA
K 33 =
K 34 = 0
L
§ Finally, set D.O.F. #4 = 1 unit,
restrain all others.
K 41 = 0 K 42 = 0
K 43 = 0 K 44 = 0
Here u and v are local deformations defined
with respect to the local xm, ym coordinates.
§ So
ui vi
EA
[K ] = ⎡ 1
L ⎢ 0
⎢− 1
⎢⎣ 0
uj
vj
0 − 1 0⎤
0 0 0⎥
0 1 0⎥
0 0 0⎥⎦
§ We have developed [K] in a
local x-y coordinate system.
We must transform [K] to
global coordinates such that
the direct stiffness assembly
method can be used to form
[K] for the entire structure in
one coordinate system.
x
3
4
y
4
Y
3
⇒ 2
1
1 2
Local x-y
Global X-Y
Project vector X1
onto the x & y axes:
x1 = X1 cos θ
y1 = -X1 sin θ
y
Y
x1
y1
X
Project vectorY1
onto x & y axes:
y1 = Y1 cos θ
x1 = Y1 sin θ
y
x
X1 θ
X
Y
Y1
y1
x1
x
θ
X
§ In matrix form, we have:
⎧x1 ⎫ = ⎡ cos θ sin θ ⎤ ⎧X1 ⎫
⎨ y ⎬ ⎢− sin θ cos θ⎥ ⎨ Y ⎬
⎩ 1 ⎭ ⎣
⎦ ⎩ 1 ⎭
[t ]
where the transformation
matrix [t] can transform a
vector VXY with components
X1 and Y1 in global
coordinates
(XY) to a vector
v
Vxy in a local coordinate
system (xy) with components
x1 and y1.
§ Define
Displacement of a node
⎧δ x ⎫
Δ
⎨δ ⎬ = {δ}= in local coordinates
⎩ y ⎭
Displacement of a node
⎧Δ X ⎫ = {Δ}Δ
=
⎨Δ ⎬
in global coordinates
⎩ Y ⎭
⎧f x ⎫
Δ
Forces
at
a
node
in
=
⎨f ⎬ = {f } local coordinates
⎩ y ⎭
⎧Fx ⎫
Forces
at
a
node
in
Δ
⎨F ⎬ = {F}= global coordinates
⎩ y ⎭
Y
y
x
δy
δx
ΔY
ΔX
X
§ The work performed by given
forces must be the same
regardless of the coordinate
system chosen:
1 T
1
T
{f } {δ} = {F} {Δ}
2
2
§ Note: {δ} = [t ]{Δ}
is the Displacement Transform
and the Force Transform is:
{f } = [t ]{F}
§ Also we developed K in local
coordinates:
{f } = [K ]{δ}
§ We seek: {F} = [K]{Δ} in global
coordinates.
§ Substituting into our work equation
yields:
1
1
T
T
T
{δ} [K] {δ} = {Δ} [K]{Δ}
2
2
§ Originally we defined our [t] to
transform the displacement or forces
at a node, since our force and
displacement vectors for our
element involve 2 nodes we must
expand our transformation matrix:
0
0 ⎤
⎡ cos θ sin θ
⎢ − sin θ cos θ
0
0 ⎥
⎥
[T ] = ⎢
0
cos θ sin θ ⎥
⎢ 0
⎢⎣ 0
0
− sinθ cos θ ⎥⎦
§ So:
T
T
T
{Δ} [T] [K][T]{Δ} = {Δ} [K]{Δ}
§ And thus: [K] = [T]T [K][T]
will transform our stiffness
matrices in local coordinates to
the global coordinate system.
§ This product can be determined
symbolically as:
ui
vi
uj
vj
⎡ C2θ
CθSθ
−C2θ −CθSθ ⎤
⎢
⎥
S2θ
−CθSθ −S2θ ⎥
EA ⎢ CθSθ
⎡ K ⎤ =
⎥
⎣ ⎦ L ⎢
2
2
⎢ −C θ −CθSθ
C θ
CθSθ ⎥
⎢
⎥
2
2
CθSθ
S θ ⎥⎦
⎢⎣ −CθSθ −S θ
§ The stiffnesses are now defined in
terms of global DOF ui , vi , uj
and vj and can be included into
the global [K] for the structure.
Example
EA = 100 lb
3
1 lb
For all elements
φ = 60o
3
1
φ
φ
2
φ
2
1
100"
§ For all 3 members:
⎡ 1
[K]xy = ⎢
0
local
⎢
⎢−1
⎢⎣ 0
0 −1 0 ⎤
0 0 0 ⎥
⎥
0 1 0 ⎥
0 0 0 ⎥⎦
§ For rod 1
Y
y
1
1
2
θ = 0°
[K]1 =
x
u1 v1 u2
⎡ 1 0 −1
⎢ 0 0 0
⎢
1
⎢−1 0
⎢⎣ 0 0 0
X
v2
0 ⎤
0 ⎥
⎥
0 ⎥
0 ⎥⎦
§ For rod 2
x
3
Y
2
θ =120°
2
y
u2
[K] 2
=
X
v2
u3
v3
.433⎤
⎡ .25 −.433 −.25
⎢−.433 .75
.433 −.75 ⎥
⎢
⎥
.433 .25 −.433⎥
⎢−.25
⎢⎣ .433 −.75 −.433 .75 ⎥⎦
§ For rod 3
x
3
Y
3
y
1
[K] 3 =
θ = 60°
X
u1
v1
u3
v3
.433 −.25 −.433⎤
⎡ .25
⎢ .433 .75 −.433 −.75 ⎥
⎢
⎥
.433⎥
⎢−.25 −.433 .25
⎢⎣−.433 −.75
.433 .75 ⎥⎦
§ What if we chose
node 3 as the
origin rather than
node 1 on rod 3?
3
Y
y
3
θ = 240°
X
1
x
u3
v3
u1
v1
.433
−.25 −.433 ⎤
⎡ .25
⎢ .433
⎥
.75
−
.433
−
.75
⎥
⎡⎣K ⎤⎦ = ⎢
3 ⎢ −.25 −.433
.25
.433 ⎥
⎢
⎥
.75 ⎦
⎣ −.433 −.75 .433
§ Note the indices on the degrees of
freedom have changed. Also note
Ku3,v1 from the previous slide is
equal to Ku3,v1 on this slide, thus
after assembly the structure [K] will
be the same.
§ Now use direct stiffness
assembly to form [K] for the
entire truss in global
coordinates!
u1
v1
u2
v2
u3
v3
−1
0
−.25 −.433⎤
⎡ 1.25 .433
⎢ .433
.75
0
0
−.433 −.75 ⎥
⎢
⎥
0
1.25 −.433 −.25 .433 ⎥
⎢ −1
[K] = ⎢
0
0
−.433 .75
.433 −.75 ⎥
⎢
⎥
.5
0 ⎥
⎢ −.25 −.433 −.25 .433
⎢⎣−.433 −.75 .433 −.75
0
1.5 ⎥⎦
{Δ} = ⎧ u1 ⎫
⎪ v1 ⎪
⎪⎪u ⎪⎪
2
⎨ v ⎬
⎪ 2 ⎪
u
⎪ 3 ⎪
⎪⎩ v3 ⎪⎭
⎧0 ⎫
⎪0 ⎪
⎪ ⎪
⎪0 ⎪
{F} = ⎨ ⎬
⎪0 ⎪
⎪1 ⎪
⎪ ⎪
⎩0 ⎭
§ After constraining u1, v1 and
v2 we can solve:
⎡ 1.25 − .25 .433⎤ ⎧⎪u 2 ⎫⎪ ⎧⎪0⎫⎪
0 ⎥ ⎨ u 3 ⎬ = ⎨1⎬
⎢− .25 .5
1.5 ⎥⎦ ⎪⎩ v3 ⎪⎭ ⎪⎩0⎪⎭
⎢⎣ .433 0
⎧⎪u 2 ⎫⎪ ⎧⎪ 0.5 ⎫⎪
⎨ u 3 ⎬ = ⎨ 2.25 ⎬ in
⎪⎩ v3 ⎪⎭ ⎪⎩− .1443⎪⎭
Member Loads and Stresses
§ We now know {Δ} for our
truss. How do we compute
loads and stresses?
Procedure
§ Extract the deformations for
the nodes i and j from
{Δ} for a given element. Now
compute local deformations
using [T ] : {δ} = [T][Δ]
⎧u xyi ⎫ ⎡ ⎤ ⎧u XYi ⎫
⎪⎪ v ⎪⎪ ⎢ ⎥ ⎪⎪ v
⎪
xyi
XYi ⎪
⎨ u ⎬ = ⎢T ⎥ ⎨ u XYj ⎬
⎪ xyj ⎪ ⎢ ⎥ ⎪
⎪
⎪⎩ v xyj ⎪⎭ ⎣ ⎦ ⎪⎩ v XYj ⎪⎭
§ For member 1
⎧ u1 ⎫ = ⎡1 0 0
⎪⎪ v ⎪⎪
⎢0 1 0
1
⎢
⎨ ⎬
u
⎢0 0 1
⎪ 2 ⎪
⎪⎩ v2 ⎪⎭ xy ⎢⎣0 0 0
0⎤ ⎧ 0 ⎫
⎧ 0 ⎫
⎪⎪ 0 ⎪⎪
0⎥ ⎪⎪ 0 ⎪⎪
⎥ ⎨ ⎬
= ⎨ ⎬
0⎥ ⎪.5⎪
⎪.5⎪
1⎥⎦ ⎪⎩ 0 ⎪⎭ XY ⎪⎩ 0 ⎪⎭xy
§ We know: [K]xy{δ}={f}:
⎡ 1
⎢ 0
⎢− 1
⎢⎣ 0
0 − 1 0⎤ ⎧ 0 ⎫ ⎧− .5⎫
0 0 0⎥ ⎪ 0 ⎪ = ⎪ 0⎪ = {f }
local
0 1 0⎥ ⎨.5⎬ ⎨ .5⎬
0 0 0⎥⎦ ⎪⎩ 0 ⎪⎭ ⎪⎩ 0⎪⎭
1
.5
1
2
.5
§ So element 1 is subjected to .5
lbs of tension and a tensile stress
.
5
of σ = PSI.
A
§ Follow identical procedure for
elements 2 and 3 . From a
design perspective, we now
have:
1. Deformations
2. Member loads (we should
check buckling)
3. Stresses!
Temperature Effects and Mismatch
§ In local coordinates, a truss element
is a 1-dimensional element. As
discussed earlier:
⎧−1⎫
⎪⎪ 0 ⎪⎪
{θ}xy = E e Ae α e ΔT ⎨ ⎬
⎪ 1⎪
⎪⎩ 0 ⎪⎭
are the loads imposed by
temperature at nodes in the local
element coordinate system. Using
our 2-D force transformation: ⎧−Cθ⎫
⎪ −Sθ⎪
⎪
⎪
T
{θ}XY = [T ] {θ}xy = Ee Ae αe ΔT ⎨ ⎬
⎪ Cθ⎪
⎪⎩ Sθ⎪⎭
We now have forces in global
coordinates that can be assembled
with the systeme force
vector:
F = ∑ f + T e + θe + P
(e
)
§ We realize imperfection always
exists to some degree in the
manufacturing process. What if
some of the truss members were
not of the design lengths? This
can be modeled by letting:
Δ
L
ε0 =
Le
where ΔL is the error in length.
This can now replace the thermal
strain αΔT as:
⎧ −1⎫
⎪
⎪
0
ΔL ⎪ ⎪
{θ}xy = E e Ae
⎨ ⎬
L e ⎪ 1⎪
⎪⎩ 0 ⎪⎭
§ This is the simple truss problem we
worked earlier.
EA = 100 lb
3
1 lb
For all elements
φ = 60o
3
1
φ
φ
2
φ
2
1
100"
§ What would be the effect of heating
one element or by having a
manufacturing error of 0.01” in one
of the 100” members?
§ The answer is very little. This truss is
internally determinate. Heating one
member or introducing some
manufacturing will change the total
deformations but the internal loads
will be due only to the 1 lb force.
§ Thus the truss must be internally
indeterminate before an
expansion or contraction of a
element will affect the internal
loads in the truss.
Symmetry
§ Many components we deal with
have some degree of symmetry.
§ If the components are symmetric
in terms of structure, loads, and
constraints it may be possible to
reduce the size of the problem
we must solve.
§ If there is one axis of symmetry
you may be able to model only
half of a problem, if there are
two axes of symmetry you may
half to model only a quarter of
the problem,etc.
§ You must watch to ensure that
constraints are applied properly
and that any members that are
shared between the two
symmetric pieces of structure are
treated properly.
§ Problem 4.10 is good example:
30,000 lb
20 ft
20 ft
(10)
(10)
(10)
15 ft
(12.5)
(12.5)
(10)
(10)
(12.5)
(12.5)
15 ft
Axis of
Symmetry
§ Symmetric Structures
– share the symmetric load
– share the stiffness of shared
members
– nodes on the axis of symmetry are
allowed to displace only in the
direction of the axis of symmetry.
15,000 lb
20 ft
4
5
(10)
3
3
15 ft
(12.5)
(5)
4
(10)
1
15 ft
(12.5)
2
2
1
§ Dealing with Constraints which are
not aligned with the global
coordinate system. 30,000 lb
This is not
a
symmetric
structure
due to the
constraints!
1
d
30o
u1 = d cos30°
v1 = d sin30°
u1
v1
u1
v1
=d =
or
−
=0
cos30°
sin30°
cos30° sin30°
This is a multi-point constraint
equation. Enforce it using Penalty or
the Lagrange method.
Space Trusses
§ 3D trusses are a simple extension of
2D planar trusses.
§ Each element has stiffness only on
the local x axis defined by the 2
nodes which are coupled by the
element.
§ In 3D global coordinates, the
element will contribute to the
structures stiffness in the X, Y and Z
dimensions.
⎡ l2
lm
ln
−l2 −lm − ln ⎤
⎢
⎥
2
2
⎢ lm m
mn −lm −m
−mn ⎥
⎢
⎥
2
2
mn
n
− ln −mn −n ⎥
Ee A e ⎢ ln
K global =
⎢
⎥
2
Le ⎢ −l2 −lm − ln
l
lm
ln ⎥
⎢
⎥
2
2
−mn lm m
mn ⎥
⎢ −lm −m
⎢
2
2 ⎥
−
ln
−
mn
−
n
ln
mn
n
⎣
⎦
§ You are given code with your text to
solve 3D truss problems.
§ Cessna 421C twin Continental
Lycoming 375 hp piston engines
§ Cessna 425 twin Pratt and
Whitney 450 hp turbines
§ Example: A space truss for a
turbine for a turboprop aircraft.
§ Analysis
– Flight loads
• Up gust, down gust, etc.
• Thermal loads?
• Seizure?
– Landing loads
– Manufacturing effects