9. truss analysis - rice university · a truss is just multiple axial bars joined together, but...

Lecture Notes: Finite Element Analysis, J.E. Akin, Rice University, Copyright. 2017-20. All rights reserved.

1

9. Truss analysis

9. TRUSS ANALYSIS ............................................................................................................................................. 1

9.1 PLANAR TRUSS .................................................................................................................................................................... 1

9.2 SPACE TRUSS ................................................................................................................................................................... 11

9.3 SUMMARY ....................................................................................................................................................................... 12

9.4 EXERCISES........................................................................................................................................................................ 15

9.1 Planar truss: The differential equation for the equilibrium of an elastic bar (above)

showed that it has only axial forces and axial displacements due to applied forces and/or

temperature changes. A truss is a structural system composed of straight “two force members”

defined as having only two equal and opposite end forces acting along the axis of the member

and thereby causing only axial displacement of the member. These members are joined (in

theory) only at their ends to span the space between concurrent support points. Clearly, the “two

force members” are the axial bars governed by the element matrices developed from the

differential equation of equilibrium. A truss is just multiple axial bars joined together, but they

are not all aligned in the same direction and usually more than two bars are connected at a joint.

The design of a truss usually requires a series of analyses. Historically, the first is usually just

based on the assumed geometry (truss mesh), a constant assumed area for each member, and

neglects the weight of the truss. Today, the FEA easily allows the inclusion of the member

weights in the first analysis. Multiple vertical and horizontal load cases are considered to

determine the maximum tension force and maximum compression force in each member. Then,

each member is assigned a new area (and weight) consistent with material stress limits in tension

and compression, and each load case is re-run. In the final iteration the compression members are

checked for buckling and additional braces may be attached at the center of long compression

members to reduce their effective buckling length. An application script to carry out automated

planar truss studies is supplied with the name Planar_Truss.m.

Modern trusses continue to have the centroidal axes of the truss members intersect at a single

point. However, the members are often joined to plates that lie in the plane of the truss. Such

joints can actually transmit bending moments (from distributed weight or wind loads), so the

final step in a design is re-check the truss by modeling it as a planar frame which includes its

bending moments of inertia.

The question here is how can the bar matrices developed in the prior chapter be used to

analyze a truss structure? The key to the process is to recall that the displacement of the bar is a

vector quantity. Previously, when two or more bars were joined together all of the displacement

vectors at the joints were collinear. Consider a planar truss where all of the bars lie in the x-y

plane. Some of the bars must be inclined relative to the x-axis, as in Fig. 9.1-1. That means that

the bar displacement vectors which previously were all directed along the x-axis now have

components in both the x- and y-directions. In other words, the bar matrices need to be revised to

account for being inclined in the x-y plane.


2

Figure 9.1-1 Direction cosines for an inclined truss member

The position vector defining the length, 𝐿, between the two ends of the truss member now

includes horizontal and vertical components, ∆𝑥 and ∆𝑦, that combine to form a right triangle

that defines that total length: 𝐿2 = ∆𝑥2 + ∆𝑦2and the direction cosines with respect to the x- and

y-axes: cos 𝜃𝑥 = ∆𝑥/𝐿 and cos 𝜃𝑦 = ∆𝑦/𝐿. Of course, those length components depend on the

differences in the x- and y-coordinates of the two end nodes. Let the axial displacement of the

bar now be denoted as 𝑢𝑏. It is collinear with the length of the bar. Therefore, the displacement

vector at either end of the member also now has two components, 𝑢𝑥 and 𝑢𝑦, when viewed in

the x-y plane containing the truss. From similar triangles, the two system (global) displacement

components are 𝑢𝑥 = 𝑢𝑏 cos 𝜃𝑥 = 𝑢𝑏 ∆𝑥/𝐿 and 𝑢𝑦 = 𝑢𝑏 cos 𝜃𝑦 = 𝑢𝑏 ∆𝑦/𝐿.

Before extending the bar stiffness matrix to a truss stiffness matrix, consider what the above

observations require as changes to a computer model of the structure. Now, both the x- and y-

coordinates of each node will be required. That means that the physical space dimension control

integer has increased to two, 𝑛𝑠 = 2. Each truss node now has two generalized displacements,

𝑛𝑔 = 2, and the number of unknowns has basically doubled at both the element and system

levels. There are still only two nodes on the element, 𝑛𝑛 = 2, but the number of independent

degrees of freedom on the element has increased to four, 𝑛𝑖 = 𝑛𝑔 × 𝑛𝑛 = 2 × 2 = 4. Likewise,

if the truss system has a total of 𝑛𝑚 nodes then the total number of displacement equations for

the truss, 𝑛𝑑 = 𝑛𝑔 × 𝑛𝑚, has doubled. Also, vector subscripts for gathering and scattering

element data double because their number is directly proportional to 𝑛𝑔. Furthermore, when

specifying an essential boundary condition (known displacement) at a node there has to be a way

to distinguish between horizontal or vertical components.

For each truss element the connecting nodes are gathered and used to get the nodal x-

coordinates in order to compute the element length. The connections are used again with the

control integers to generate the system degree of freedom numbers for the element. That is done

with the get_element_index.m script given in Fig. 7.2-1. That script is always used for each

element being assembled. Because 𝑛𝑔 = 2 the system equation number for degree of freedom j

at system node number N now is 𝑟𝑜𝑤 (𝑗) = 𝑛𝑔(𝑁 − 1) + 𝑗, 1 ≤ 𝑗 ≤ 𝑛𝑔. For a planar truss the

degree of freedom j = 1 corresponds to a horizontal displacement and j = 2 corresponds to the

vertical displacement.

The truss equilibrium analysis requires solving for all of the horizontal and vertical

displacements. Since the node displacement components, 𝑢𝑥 and 𝑢𝑦, are now the primary

unknowns they need to be related to the bar displacement, 𝑢𝑏, by using the orientation (direction

cosines) of the bar element. From geometry, at any node the local bar axial displacement is

related to the global x- and y-components by: 𝑢𝑏 = 𝑢𝑥 cos 𝜃𝑥 + 𝑢𝑦 cos 𝜃𝑦. This identity can be

written in matrix form

{𝑢𝑏}𝐿 = [cos 𝜃𝑥 cos 𝜃𝑦] {𝑢𝑥𝑢𝑦}𝐺 or 𝒖𝑳 = 𝒕(𝜃) 𝒖𝑮. (9.1-1)


3

Since the direction cosines are actually calculated from the length components of the truss it is

practical to re-write the direction cosines and the member length components and the node

displacement transformation as cos 𝜃𝑥 ≡ 𝐶𝑥 = ∆𝑥 𝐿⁄ , cos 𝜃𝑦 ≡ 𝐶𝑦 = sin 𝜃𝑥 = ∆𝑦 𝐿⁄

{𝑢𝑏}𝐿 = [𝐶𝑥 𝐶𝑦] {𝑢𝑥𝑢𝑦}𝐺=1

𝐿[∆𝑥 ∆𝑦] {

𝑢𝑥𝑢𝑦}𝐺or 𝒖𝑳 = 𝒕(𝜃) 𝒖𝑮

In Ex. 8.3-1 the structural stiffness matrix of the two-noded bar was shown to be

𝑺𝑒 =𝐸𝑒𝐴𝑒

𝐿𝑒[ 1 −1−1 1

]

That stiffness was obtained from the element’s scalar contribution to the integral equilibrium

form

𝐼𝑆 = 𝒖𝑳𝑻𝑺𝑒𝒖𝑳 = {

𝑢𝟏𝑢𝟐}𝑏

𝑻 𝐸𝑒𝐴𝑒

𝐿𝑒[ 1 −1−1 1

] {𝑢𝟏𝑢𝟐}𝑏

In order to convert that original bar relation to a truss member stiffness relation the node

transformation above must be applied to both of the end nodes of the bar:

{𝑢𝟏𝑢𝟐}𝑏

= [cos 𝜃𝑥0

cos 𝜃𝑦0

0cos 𝜃𝑥

0cos 𝜃𝑦

] {

𝑢𝑥1𝑢𝑦1𝑢𝑥2𝑢𝑦2

} (9.1-2)

𝒖𝑳𝑒 = [𝑻(𝜃)] 𝒖𝑮

𝑒 = [𝒕(𝜃) 𝟎

𝟎 𝒕(𝜃)] 𝒖𝑮

𝑒

2 × 1 = (2 × 4)(4 × 1)

In other words, the four truss components of displacement are reduced to the two bar end

displacements by multiplying by a rectangular geometry transformation matrix.

The scalar integral contribution to equilibrium must be the same in any coordinate system. So

it is the same in the global system:

𝐼𝑆 = 𝒖𝑳𝑻 𝑺𝑒 𝒖𝑳 = 𝒖𝑮

𝑻 𝑺𝑮𝒆 𝒖𝑮 (9.1-3)

1 × 1 = (1 × 2)(2 × 2)(2 × 1) = (1 × 4)(4 × 4)(4 × 1) Substituting the bar transformation:

𝐼𝑆 = ([𝑻(𝜃)] 𝒖𝑮𝑒)𝑻 𝑺𝑒 ([𝑻(𝜃)] 𝒖𝑮

𝑒) = 𝒖𝑮𝑻 𝑺𝑮

𝒆 𝒖𝑮

𝐼𝑆 = 𝒖𝑮𝑻 [[𝑻(𝜃)]𝑻 𝑺𝑒 [𝑻(𝜃)]] 𝒖𝑮 = 𝒖𝑮

𝑻 𝑺𝑮𝒆 𝒖𝑮

gives the global truss stiffness as

𝑺𝑮𝒆 = [𝑻(𝜃)]𝑻 𝑺𝑒 [𝑻(𝜃)] (9.1-4)

4 × 4 = (4 × 2)(2 × 2)(2 × 4)


4

It is probably easier to evaluate the matrix 𝑺𝑒, since we already have it, and 𝑻(𝜃), and have

the computer do the multiplications, but the truss stiffness can be written in detail in terms of the

length components as:

𝑺𝑮𝒆 =

𝐸𝐴

𝐿

[ 𝐶𝑥𝐶𝑥 𝐶𝑥𝐶𝑦𝐶𝑥𝐶𝑦 𝐶𝑦𝐶𝑦

−𝐶𝑥𝐶𝑥 −𝐶𝑥𝐶𝑦−𝐶𝑥𝐶𝑦 −𝐶𝑦𝐶𝑦


𝐶𝑥𝐶𝑥 𝐶𝑥𝐶𝑦 𝐶𝑥𝐶𝑦 𝐶𝑦𝐶𝑦 ]

=𝐸𝐴

𝐿3[ ∆𝑥∆𝑥 ∆𝑥∆𝑦∆𝑥∆𝑦 ∆𝑦∆𝑦

−∆𝑥∆𝑥 −∆𝑥∆𝑦−∆𝑥∆𝑦 −∆𝑦∆𝑦

−∆𝑥∆𝑥 −∆𝑥∆𝑦−∆𝑥∆𝑦 −∆𝑦∆𝑦

∆𝑥∆𝑥 ∆𝑥∆𝑦 ∆𝑥∆𝑦 ∆𝑦∆𝑦

] (9.1-5)

As a spot check, degenerate the truss into a single horizontal bar. Then, ∆𝑦 = 0 and ∆𝑥 = 𝐿

and the truss stiffness becomes

𝑺𝑮𝒆 =

𝐸 𝐴

𝐿3[ 𝐿

2 0 0 0

−𝐿2 0 0 0

−𝐿2 0 0 0

𝐿2 0 0 0

] =𝐸 𝐴

𝐿 [ 1 0 0 0

−1 0 0 0

−1 0 0 0

1 0 0 0

]

which shows that a horizontal bar member has no vertical stiffness. Enforcing the EBCs that the

two vertical displacements are zero reduces the effective horizontal truss member stiffness back

to that of the bar

𝑺𝑒 =𝐸 𝐴

𝐿 [ 1 −1−1 1

]

Bars and truss members can be subjected to thermal loads. For the bar it was shown in section

8.4 that the thermal load is

𝒄𝜶𝒆 = 𝐸 𝐴𝛼 ∆𝑇 {

−1 1} (9.1-6)

This bar load is also transformed to a truss member load by using the scalar work term in the

governing integral form: 𝐼𝑐 = 𝒖𝑳𝑻 𝒄𝑒 = 𝒖𝑮

𝑻 𝒄𝑮𝒆 . Substituting the rectangular member orientation

transformation matrix gives:

𝒄𝑮𝒆 = [𝑻(𝜃)]𝑻 𝒄𝑒 (9.1-7)

4 × 1 = (4 × 2)(2 × 1)

which converts any thermal load in a truss member to

𝒄𝜶𝒆 = 𝐸𝐴𝛼 ∆𝑇

{

−𝐶𝑥−𝐶𝑦 𝐶𝑥 𝐶𝑦 }

=𝐸𝐴𝛼 ∆𝑇

𝐿{

−∆𝑥−∆𝑦

∆𝑥∆𝑦

}

When a truss has all members receive the same temperature change and is free to expand it just

changes shape and no net member forces develop. However, if the truss is not free to expand

then a temperature change in even one member can produce addition axial forces in some or all

of the other members.


5

To form a kinematically stable truss the truss elements and/or the support connections must

form a structure made up of triangular regions. A region with only four truss elements combined

into the shape of a quadrilateral will be kinematically unstable and would collapse under its own

weight. Mentioning the weight, for trusses it is assumed that the weight of a truss element is split

equally at its two ends as vertical forces:

𝒄𝜸𝑮 =

𝛾 𝐴 𝐿

2{

0−1 0−1

} =𝑤 𝐿

2{

0−1 0−1

} (9.1-8)

where 𝛾 denote the specific weight of the material, and 𝑤 = 𝛾𝐴 is the weight per unit length.

Now that each node has two degrees of freedom the packed integer code that flags the

essential boundary condition at each node can have the following packed values: 00 a free joint,

10 only u is given, 01 only v is given, 11 both u and v are given (a pin joint), etc. Usually the

given values are zero, but any small deflection value can be imposed as required by the

application. If any unit value in the flag is replaced with a 2 then that means that component is

coupled to another degree of freedom with a type 2 multi-point-constraint (MPC)

The application software for planar truss analysis, with limited graphical support, is supplied

in the application library as Planar_Truss.m. As another illustration of its application it was

applied to a seven bar truss shown in Fig. 9.1-6. All of the members had the same properties. The

steel truss was pinned at the lower left, supported by a horizontal roller at the lower right. The

top point carried a vertical load, the middle left point a horizontal load, and the weights were

omitted. These data are superimposed on the node displacement plot on the left of that figure.

The right portion of the figure shows the computed axial forces in each bar member, with

negative being compression. To review the inputs to the general finite element library, the input

files are summarized in Fig. 9.1-7.

Example 9.1-1 Given: The two bar steel truss in Fig. 9.1-2 supports a downward vertical sign

load of 𝑃 = 20 𝑘𝑁 at their junction (node 3). The two member connections to the vertical wall

are pinned against displacement. Both elements have the same cross-sectional area, 𝐴 =0.001 𝑚2. Neglect the member weights. Find the deflection of the load, and the system

reactions. Check the reaction forces using statics. The truss data are:

Node 1 2 3 Element Connections ∆𝑥 (𝑚) ∆𝑦 (𝑚) 𝐿 (𝑚) 𝐶𝑥 𝐶𝑦

X(m) 0 0 3 1 2, 3 3 0 3 1 0

Y(m) 4 0 0 2 1 ,3 3 -4 5 3/5 -4/5

DOF 1,2 3,4 5,6

Solution: First, note that the two wall connections for the third side of a stable triangle. Steel has

an elastic modulus of 𝐸 = 200𝑒9 𝑁 𝑚2⁄ . The vertical sign load is at node 3 in the negative

vertical direction. It corresponds to system DOF number of 2*(3-1) +2 = 6. The null external

horizontal point load there is at DOF number is 2*(3-1) +1 = 5. So the system external point load

and reaction force vector is: 𝒄𝑷𝑻 = [𝑅𝑥1 𝑅𝑦1 𝑅𝑥2 𝑅𝑦2 0 −𝑃]

The numerical values of the data can be substituted to obtain the numerical solution, or an

analytic solution is available from a symbolic script by letting the inclined member have a length

𝐿 so that the length of the horizontal member is 3𝐿 5⁄ . Those results are rather messy and not


6

very practical. As an educational compromise, substitute the numerical direction cosine values

and use the inclined as a reference length. The horizontal truss member stiffness and loads are

𝑺ℎ =𝐸 𝐴

(3𝐿 5⁄ )[

1 0 0 0

−1 0 0 0

−1 0 0 0

1 0 0 0

] =125 𝐸 𝐴

75 𝐿[

1 0 0 0

−1 0 0 0

−1 0 0 0

1 0 0 0

],

𝒄𝜸𝒉 =

𝑤(3𝐿 5⁄ )

2{

0−1 0−1

}, 𝒄𝜶𝒉 = 𝐸𝐴𝛼 ∆𝑇 {

−1 0 1 0

}

The inclined stiffness is

𝑺𝑖 =𝐸 𝐴

25 𝐿[

9 −12−12 16

−9 12 12 −16

−9 12 12 −16

9 −12−12 16

] =𝐸 𝐴

75 𝐿[

27 −36−36 48

−27 36 36 −48

−27 36 36 −48

27 −36−36 48

]

Since the reactions are required the full 6 × 6 equilibrium system will be assembled even though

only the partition at node 3 is needed to find the displacements:

𝐸 𝐴

75 𝐿

[ 27 −36−36 48

0 00 0

−27 36 36 −48

0 00 0

125 0 0 0

−125 0 0 0

−27 36 36 −48

−125 0 0 0

152 −36 −36 48]

{

𝑢𝑥1𝑢𝑦1𝑢𝑥2𝑢𝑦2𝑢𝑥3𝑢𝑦3}

=

{

𝑅𝑥1𝑅𝑦1𝑅𝑥2𝑅𝑦20−𝑃}

+𝑤𝐿

2

{

0−10

−3/50

−1 − 3/5}

+ 𝐸𝐴𝛼 ∆𝑇

{

−3/5 4/5−1 0

1 + 3/5−4/5 }

Note that the rows and columns associated with 𝑢𝑦2 are all zero. If there is no EBC applied to

that degree of freedom the system stiffness will be singular. In that case, the system would be

kinematically unstable since the horizontal bar could rotate about node 3. The pinned boundary

condition prevents that rotation and renders the system non-singular. Since no gravity (𝛾 = 0) or

temperature change (∆𝑇 = 0) was given the free displacements satisfy

𝐸 𝐴

75 𝐿[152 −36−36 48

] {𝑢𝑥3𝑢𝑦3

} = {0−𝑃

}

Multiplying by the inverse of the stiffness gives

{𝑢𝑥3𝑢𝑦3

} =−𝑃𝐿

80 𝐸𝐴{36152

}


7

a node deflection down and to the left. Multiplying all of the displacements times the fixed

partition of the stiffness gives the reactions as:

{

𝑅𝑥1𝑅𝑦1𝑅𝑥2𝑅𝑦2}

=𝑃

4{

−3 4 3 0

}

The results are sketched in Fig. 9.1-2. By inspection, the sum of the horizontal forces is zero, as

is the sum of the vertical forces, and the sum of the moments about any of the three nodes.

Figure 9.1-2 Two bar planar truss with a point load

Figure 9.1-3 Two bar planar truss with self-weight

Example 9.1-2 Given: For the truss given in Ex. 9.1-2 subject to a point load determine the local

axial displacements of each truss member and the axial force in each bar. Solution: Eq (9.1-2)

gives the two displacements of a bar in terms of the four displacements of the truss. For the

inclined member (element 1) those displacements are


= [cos 𝜃𝑥0

cos 𝜃𝑦0

0cos 𝜃𝑥

0cos 𝜃𝑦

] {


} =𝟏

𝟓[30

−4 0

03

0−4]

𝑃𝐿

80 𝐸𝐴{

00−36−152

}

{𝑢𝟏𝑢𝟐}1

=𝑃𝐿

4 𝐸𝐴{05}


8

The mechanical strain in that element is 𝜀 = 𝑩𝒆𝒖𝒆 = 1

𝐿[−1 1]

𝑃𝐿

4 𝐸𝐴{05} =

5

4

𝑃

𝐸𝐴 and the force in

the element is 𝐹 = 𝐸𝐴 𝜀 = 5𝑃 4⁄ tension. Of course, since it is the only member at the pin

support the pin reaction must be providing the same force. The reaction magnitude at node 1 is

𝑅1 = √𝑅𝑥2 + 𝑅𝑦

2 = √(−3𝑃 4⁄ )2 + 𝑃2 = 5𝑃 4⁄ . Likewise, for the horizontal member (element 2)

the displacements are

{𝑢𝟏𝑢𝟐}2

= [10

00

01

00]𝑃𝐿

80 𝐸𝐴{

00−36−152

} =𝑃𝐿

80 𝐸𝐴{0−36

}

Its strain is 𝜀 =1

(3𝐿 5⁄ )[−1 1]

𝑃𝐿

80 𝐸𝐴{0−36

} =−3

4

𝑃

𝐸𝐴 and the force is 𝐹 = −3 𝑃 4⁄ , compression.

By inspection that agrees with the reaction shown in Fig. 9.1-3.

Example 9.1-3 Given: The truss in Ex. 9.1-2 is subjected only to a temperature increase of ∆𝑇.

Find the displacement of the free node and the reaction forces. Solution: Starting with the system

matrices only the thermal load vector has changed:

𝒄𝜶𝑻 = 𝐸𝐴𝛼 ∆𝑇 [−3 4 − 5 0 (3 + 5) (−4 + 0)]/5

Retaining the free partition, the equilibrium equations are

𝐸 𝐴

75 𝐿[152 −36−36 48

] {𝑢𝑥3𝑢𝑦3

} =𝐸𝐴𝛼 ∆𝑇

5{8−4}

Thus, the free displacements are

{𝑢𝑥3𝑢𝑦3

} =𝐿 𝛼 ∆𝑇

5{ 3−4}

The support reaction portions of the system matrices are

𝐸𝐴

75 𝐿[

−27 36−125 0

36−48

00

]𝐿 𝛼 ∆𝑇

5{ 3−4} =

𝐸𝐴𝛼 ∆𝑇

5{

−3 4−5 0

} =

{

𝑅𝑥1𝑅𝑦1𝑅𝑥2𝑅𝑦2}

+𝐸𝐴𝛼 ∆𝑇

5{

−3 4−5 0

} = 𝟎

Therefore the external reactions are identically zero. Also the forces in each truss member are

zero because, for this connectivity set, each member is free to expand. Node 3 of the inclined

member moves on an arc, centered at support node 1, of radius (1 + 𝛼 ∆𝑇 )𝐿 and node 3 of the

horizontal member moves on an arc, centered at support node 2, of radius (1 + 𝛼 ∆𝑇 ) 3𝐿 5⁄ .

Where those two arcs intersect is where node 3 displaces to, without causing any member forces.

The free expansion leading to no member forces found here would not occur if any other

restrained members were present in the truss. For example, if a third bar was added to the truss

between node 3 and a new node 4 at the mid-point between nodes 1 and 2, then an increase in

temperature would cause non-zero reactions well as non-zero member forces. Such thermal

induced reaction forces are shown in Fig. 9.1-3, and they clearly satisfy Newton’s equations of

equilibrium by inspection.


9

Figure 9.1-3 Three- and two-bar (right) reactions for uniform temperature increase

Example 9.1-4 Given: The truss given previously in Fig. 9.1-4 has an inclined roller support that

requires a multipoint constraint (MPC). Omitting the weight and the thermal loads, numerically

determine the displacements, the reactions, and the member axial loads. Check the reactions with

Newton’s Laws. Solution: The property data in Fig. 7.7-2 were edited to set the specific weight

and temperature change to zero. The data sets were submitted to the planar truss script

Planar_Truss.m which is included in the separate Application Library. The computed system

displacements and reactions are:

Computed Solution: System Reactions Axial Force

Node, 2 displacements per node Node, DOF, Reaction Value Elem, Value

1 0.0 0.0 1 1 -36000 1 0.0

2 0.229713 0.0 1 2 -48000 2 -1e5

3 0.0861728 0.114796 3 1 -64000 3 6e4

3 2 48000

Checking, the sum of the x-forces are ∑𝐹𝑥 = 100,000 − 36,000 − 64,000 = 0, as required.

The sum of the y-forces are ∑𝐹𝑦 = −48,000 + 48,000 = 0, as required. Note that the reactions

at node 3 from the MPC give a vector resultant of 80,000 N perpendicular to the inclined support

surface. Checking the moments about node one (with CCW positive) gives

∑𝑀𝑧 = −100,000 𝑁 × 4 𝑚 + 48,000 𝑁 × 3 𝑚 + 64,000 𝑁 × 4 𝑚 = 0

The external load and reactions are shown in Fig. 9.1-4. Note that the ratio of the displacement

components at node 3 is 0.7507 which is a 0.1% error in the MPC equation.


10

Figure 9.1-4 Loads and reactions in a truss with a MPC

Figure 9.1-5 Truss member axial force values for Fig. 9.1-4


11

Figure 9.1-6 Displacement vectors (left) and member axial forces in a planar truss

Figure 9.1-7 Data files for the seven bar planar truss model

9.2 Space Truss: Unlike beam and frame members, for any truss member only the value of its

cross-sectional area matters. The orientation of the area about the local member axis does not

matter. That means that the extension from planar truss analysis to space truss analysis is


12

relatively simple. The z-coordinates of the nodes are required to define their locations and their

differences on a member define a third direction cosine required for an expanded transformation

matrix to relate the local bar axial displacements to the six displacement components of the space

truss member. Therefore (9.1-2) expands to be:


= [cos 𝜃𝑥 cos 𝜃𝑦 cos 𝜃𝑧 0 0 0

0 0 0 cos 𝜃𝑥 cos 𝜃𝑦 cos 𝜃𝑧]

{

𝑢𝑥1𝑢𝑦1𝑢𝑧1𝑢𝑥2𝑢𝑦2𝑢𝑧2}

(9.2-1)

The local bar stiffness matrix expands in the same way as (9.1-4) and (9.1-7):

𝑺𝑮𝒆 = [𝑻(𝜃)]𝑻 𝑺𝑒 [𝑻(𝜃)]

6 × 6 = (6 × 2)(2 × 2)(2 × 6)

𝒄𝑮𝒆 = [𝑻(𝜃)]𝑻 𝒄𝑒

6 × 1 = (6 × 2)(2 × 1)

Of course, point forces at any node can be applied in any or all of the three directions. Likewise,

any or all of the three displacement components at a node can have prescribed values. That also

means that the packed boundary condition flag assigned to each node now contains three digits:

the left-most flags the x-displacement; the center one flags the y-displacement, etc. A packed

flag value of 000 means that the joint is completely free, a value of 100 means only the x-

displacement is prescribed. A value of 111 means all displacements at the node are prescribed.

Clearly, any mixture of prescribed displacements can be flagged by other combinations of the

three digits.

An application script Space_Truss.m was obtained from the planar truss script

Planar_Truss.m by adding less than 1% of new calculations to utilize the z-coordinates to form

the member transformation matrix an to extend helpful plots from 2-D to 3-D.

9.3 Summary 𝑛_𝑏 ⟷ 𝑛𝑏 ≡ Number of boundary segments

𝑛_𝑑 ⟷ 𝑛𝑑 ≡ Number of system unknowns = 𝑛_𝑔 × 𝑛_𝑚

𝑛_𝑒 ⟷ 𝑛𝑒 ≡ Number of elements

𝑛_𝑔 ⟷ 𝑛𝑔 ≡ Number of generalized DOF per node

𝑛_𝑖 ⟷ 𝑛𝑖 ≡ Number of unknowns per element = 𝑛_𝑔 × 𝑛_𝑛

𝑛_𝑚 ⟷ 𝑛𝑚 ≡ Number of mesh nodes

𝑛_𝑛 ⟷ 𝑛𝑛 ≡ Number of nodes per element

𝑛_𝑝 ⟷ 𝑛𝑝 ≡ Dimension of parametric space

𝑛_𝑞 ⟷ 𝑛𝑞 ≡ Number of total quadrature points

𝑛_𝑟 ⟷ 𝑛𝑟 ≡ Number of rows in the 𝑩𝒆 matrix (and material matrix)

𝑛_𝑠 ⟷ 𝑛𝑠 ≡ Dimension of physical space

b = boundary segment number e = element number

⊂ =subset ∪ = union of sets

Boundary, element, and system unknowns: 𝜹𝒃 ⊂𝑏 𝜹𝒆 ⊂𝑒 𝜹


13

Boolean extraction arrays: 𝜹𝒃 ≡ 𝜷𝒃 𝜹, 𝜹𝒆 ≡ 𝜷𝒆 𝜹

Geometry: Γ𝑏 ⊂ Ω𝑒 ≡ Boundary segment Ω𝑒 ≡ Element domain

Ω = ∪𝒆 Ω𝑒 ≡ Solution domain Γ = ∪𝒃 Γ

𝑏 ≡ Domain boundary

Interpolation: 𝑢(𝒙) = 𝑯(𝑟) 𝜹𝒆 = 𝑢(𝒙)𝑻 = 𝜹𝒆𝑻𝑯(𝑟)𝑻

Local gradient: 𝜕𝑢(𝑟) 𝜕𝑟⁄ = (𝜕𝑯(𝑟) 𝜕𝑟⁄ ) 𝜹𝒆

Physical gradient: 𝜕𝑢(𝑥) 𝜕𝑥 =⁄ (𝜕𝑢(𝑟) 𝜕𝑟⁄ )(𝜕𝑟 𝜕𝑥⁄ ) = (𝜕𝑢(𝑟) 𝜕𝑟⁄ ) (𝜕𝑥 𝜕𝑟⁄ ) −1

𝜕𝑢(𝑥) 𝜕𝑥 =⁄ [(𝜕𝑮(𝑟) 𝜕𝑟⁄ ) 𝒙𝒆]−1 (𝜕𝑯(𝑟) 𝜕𝑟⁄ ) 𝜹𝒆 ≡ 𝑩(𝑟) 𝜹𝒆

Local Line Element Stiffness Matrix:

𝑺𝒌𝒆 = ∫

𝑑𝑯(𝑟)

𝑑𝑥

𝑇

𝐸(𝑥) 𝐴(𝑥)𝑑𝑯(𝑟)

𝑑𝑥

𝐿𝑒 𝑑𝑥

A symmetry plane has a zero normal displacement.

Thermal loads for linear quadratic bar:

𝒄𝜶𝒆 = 𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒 {

−1 1}

Equilibrium equations for a linear bar element:

𝐸𝑒𝐴𝑒

𝐿𝑒[ 1 −1−1 1

] {𝑢1𝑢2}

= {𝑃1𝑃2} +

1

6𝐿𝑒[2 11 2

] {𝑞1𝑞2}

+ 𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒 {−1 1}

Planar truss element: 𝑛𝑔 = 2, 𝑛𝑛 = 2, 𝑛𝑖 = 4, stiffness, thermal and gravity loads

cos 𝜃𝑥 ≡ 𝐶𝑥 = ∆𝑥 𝐿⁄ , cos 𝜃𝑦 ≡ 𝐶𝑦 = sin 𝜃𝑥 = ∆𝑦 𝐿⁄ , cos 𝜃𝑧 ≡ 𝐶𝑧 = ∆𝑧 𝐿⁄

Transforming from local bar to global truss matrices

𝑺𝑮𝒆 = [𝑻(𝜃)]𝑻 𝑺𝑒 [𝑻(𝜃)], 𝒄𝑮

𝒆 = [𝑻(𝜃)]𝑻 𝒄𝑒, [𝑻(𝜃)] = [𝒕(𝜃) 𝟎𝟎 𝒕(𝜃)

]

Planar truss bar without previous axial load:


𝐿𝑒[ 1 −1−1 1

], 𝒄𝑒 = 𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒 {−1 1}, 𝒕(𝜃) = [cos 𝜃𝑥 (cos 𝜃𝑦 = −sin 𝜃𝑥)]


= [cos 𝜃𝑥0

cos 𝜃𝑦0

0cos 𝜃𝑥

0cos 𝜃𝑦

] {


}

Global planar stiffness matrix, thermal and gravity load vectors:


14

𝑺𝑮𝒆 = [𝑻(𝜃)]𝑻

𝐸𝑒𝐴𝑒

𝐿𝑒[ 1 −1−1 1

] [𝑻(𝜃)] =𝐸𝑒𝐴𝑒

𝐿𝑒

[ 𝐶𝑥𝐶𝑥 𝐶𝑥𝐶𝑦𝐶𝑥𝐶𝑦 𝐶𝑦𝐶𝑦



𝐶𝑥𝐶𝑥 𝐶𝑥𝐶𝑦 𝐶𝑥𝐶𝑦 𝐶𝑦𝐶𝑦 ]

,

(4 × 2)(2 × 2)(2 × 4) = (4 × 4)

𝒄𝜶𝒆 = [𝑻(𝜃)]𝑻𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒 {

−1 1} = 𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒

{

−𝐶𝑥−𝐶𝑦 𝐶𝑥 𝐶𝑦}

, 𝒄𝜸

𝒆 =𝛾 𝐴 𝐿

2{

0−1 0−1

}

(4 × 2)(2 × 1) = (4 × 1)

Space truss bar without previous axial load:


𝐿𝑒[ 1 −1−1 1

], 𝒄𝑒 = 𝐸𝑒𝐴𝑒𝛼𝑒∆𝑇𝑒 {−1 1}, 𝒕(𝜃) = [cos 𝜃𝑥 cos 𝜃𝑦 cos 𝜃𝑧]


= [cos 𝜃𝑥 cos 𝜃𝑦 cos 𝜃𝑧 0 0 0

0 0 0 cos 𝜃𝑥 cos 𝜃𝑦 cos 𝜃𝑧]

{

𝑢𝑥1𝑢𝑦1𝑢𝑧1𝑢𝑥2𝑢𝑦2𝑢𝑧2}

Planar truss bar with previous tension load N; axial stiffness and initial (geometric) stiffness:


𝐿𝑒[ 1 0 0 0

−1 0 0 0

−1 0 0 0

1 0 0 0

] , 𝑺𝒊𝒆 =

𝑁𝑒

𝐿𝑒[ 0 0 0 1

0 00 −1

0 0 0 −1

0 00 1

]

𝒕(𝜃) = [𝐶𝑥 𝐶𝑦 0 0−𝐶𝑦 𝐶𝑥 0 0

]


15

9.4 Exercises

9.5 List of Examples Ex 9.1-1 Two bar truss displacements and reactions, for point force, L2 Ex 9.1-2 Two bar truss, element axial displacements and reactions, L2 Ex 9.1-3 Two bar truss displacements and reactions, due to temperature change, L2 Ex 9.1-4 Three bar truss with inclined roller and point load, L2

9.6 Index application library, 9 bar, 2 buckling, 1 conduction, 13 direction cosines, 2 displacement, 1 displacement transformation, 3 displacement vector, 1 displacements, 7 equilibrium, 2, 6 essential boundary condition, 2 Exercises, 15 get_element_index.m, 2 gradient, 13 gravity load, 13 inclined roller, 9 interpolation, 13 kinematically unstable, 6 L3_C0, 13 L4_C0, 13 member weight, 1 MPC, 5, 9 msh_bc_xyz.txt, 11 msh_ebc.txt, 11 msh_load_pt.txt, 11 msh_properties.txt, 11 msh_typ_nodes.txt, 11

Newton’s Laws, 9 packed flag, 12 packed integer code, 5 pin support, 5, 6, 8 planar truss, 1, 9, 13 Planar_Truss.m, 1, 9 point load, 5 reaction, 5 roller support, 5, 9 seven bar truss, 5 space truss, 11 Space_Truss.m, 12 specific weight, 5 steel, 5 stiffness matrix, 2, 3 summary, 12 temperature, 4 thermal load, 4, 8, 13 transformation matrix, 4 truss element, 13 truss element stiffness, 3 truss member, 2 truss member force, 10 two bar truss, 5 two force member, 1 union, 12

9. truss analysis - rice university · a truss is just multiple axial bars joined together, but...

Documents