11. Frame analysis - Rice University · 2020-01-13 · Frame analysis 11.1 Planar frames: If the bar member, which carries only loads parallel to its axis, and a beam, which carries

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

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11. Frame analysis ......................................................................................................................................... 1

11.1 Planar frames .................................................................................................................................... 1

11.2 Frame member reactions .................................................................................................................. 7

11.3 Enhanced frame post-processing* .................................................................................................... 9

11.4 Space frames ................................................................................................................................... 12

11.5 Numerically integrated frame members ........................................................................................ 12

11.6 Summary ........................................................................................................................................ 14

11.7 Exercises .......................................................................................................................................... 17

11. Frame analysis

11.1 Planar frames: If the bar member, which carries only loads parallel to its axis, and a

beam, which carries only loads transverse to its axis, are combined the new element is the so-

called beam-column or frame member which is illustrated in Fig. 11.1-1. The one dimensional

version of that has three generalized nodal degrees of freedom ( 𝑛𝑔 = 3): an axial displacement

vector, a transverse displacement vector, and a rotation vector (perpendicular to the other two),

in that order. Analogous to a planar truss, which is the extension of a simple two force axial

member to a planar assembly that resists loads by axial loads only, a frame is an assembly of

beam-column members that adds the ability to also transmit moments (couples) through its

joints. In other words, a planar frame is a combination of individual beam-column elements that

resist loadings by a combination of bending, axial member forces, and transverse (shear) forces.

Therefore, a planar frame is a more efficient load transferring structure than a planar truss

structure.

Figure 11.1-1 A plane frame member with axial (1) and transverse (2) displacements,

and an in-plane rotation (3) at each node.


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Each beam component of the member has a cross-section with two principle axes associated

with the second-moments of inertia, 𝐼𝑧𝑧 and 𝐼𝑦𝑦. For the coordinate system shown above the

principle inertia is the former: 𝐼 ≡ 𝐼𝑧𝑧 = ∫ 𝑦2

𝐴𝑑𝐴. Its defining axis is the z-axis coming out of

the paper and lies in the horizontal plane. Conversely, the minor inertia y-axis lies in the vertical

plane of the paper. Any changes in the two inertia axis directions becomes important if the beam

is part of a space frame member located in general three-dimensional space, and they are

discussed later.

Recall that the stiffness of the linear axial two-node bar element depends on the ‘axial

stiffness’, 𝐸𝐴 𝐿⁄ , of the member and it has the equilibrium equation of

𝐸𝐴

𝐿[ 1 −1−1 1

] {𝑢1

𝑢2} = {

𝐹1

𝐹2} +

𝐿

6[2 11 2

] {𝑤𝑥1

𝑤𝑥2} + 𝐸𝐴𝛼∆𝑇𝑥 {

−1 1

} (11.1-1)

[𝑺𝒂]{𝜹𝒂} = {𝑭 𝒂} + {𝑭𝒘

𝒂 } + {𝑭𝑻𝒂}

where 𝐸 is the modulus of elasticity, A is the cross-sectional area, L is the length, 𝐹𝑥 denotes the

axial force at node k, 𝑤𝑥𝑘 is the axial load per unit length at node k, 𝛼 is the coefficient of

thermal expansion, and ∆𝑇𝑥 is the constant increase in temperature along the axis of the bar.

These axial effects are associated with the first and third degrees of freedom of a frame element

thus a vector subscript, 𝐵𝑎𝑟 = [1 4], to insert its terms into the combined matrices.

The two-node beam equilibrium equation depends of the ‘bending stiffness’, 𝐸𝐼 𝐿3⁄ , and has

the form

𝐸𝐼

𝐿3 [

12 6𝐿 6𝐿 4𝐿2

−12 6𝐿−6𝐿 2𝐿2

−12 −6𝐿 6𝐿 2𝐿2

12 −6𝐿−6𝐿 4𝐿2

] {

𝑣1

𝜃1𝑣2

𝜃2

} +𝑁

30 𝐿[

36 3𝐿 3𝐿 4𝐿2

−36 3𝐿−3𝐿 −𝐿2

−36 −3𝐿 3𝐿 −𝐿2

36 −3𝐿−3𝐿 4𝐿2

] {

𝑣1

𝜃1𝑣2

𝜃2

}

= {

𝑉1

𝑀1

𝑉2

𝑀2

} +𝐿

60[

21 9 3𝐿 2𝐿 9 21

−2𝐿 −3𝐿

] {𝑤𝑦1

𝑤𝑦2} +

𝐸𝐼𝛼∆𝑇𝑦

𝑡{ 0 1

0−1

}, (11.1-2)

[𝑺𝒃]{𝜹𝒃} = {𝑭 𝒃} + {𝑭𝒘

𝒃 } + {𝑭𝑻𝒃}

but with the axial force, 𝑁, identically zero unless a prior iteration has provided the axial force in

the bar element. The other items are 𝐼 the moment inertia of the area evaluated at its centroid, 𝑉𝑘

is the external transverse shear force applied at node k, 𝑀𝑘 is the external couple (moment)

applied at node k, 𝑤𝑦𝑘 is the transverse line load at node k, 𝑡 is the depth of the beam, and ∆𝑇𝑦 is

the increase in temperature from the top of the beam through its depth. For now, the axial force

𝑁 is taken as zero, until it becomes necessary to study buckling (instability) of frames. The

degrees of freedom of the bar are identified by the vector subscript 𝐵𝑒𝑛𝑑 = [2 3 5 6]. In other words, the combined element degrees of freedom are


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𝜹𝒆(𝑩𝒂𝒓) = 𝜹𝒆 (14) = {

𝑢1

𝑢2} , 𝜹𝒆(𝑩𝒆𝒏𝒅) = 𝜹𝒆 (

2356

) = {

𝑣1

𝜃1𝑣2

𝜃2

}

𝜹𝒆𝑇 = [𝑢1 𝑣1 𝜃1 𝑢2 𝑣2 𝜃2]. (11.1-3)

Letting 𝑲 be the combined frame 6 × 6 stiffness matrix, it is made up of, and programmed as the

scattering of the 2 × 2 axial stiffness matrix, 𝑲(𝑩𝒂𝒓,𝑩𝒂𝒓) = 𝑺𝒂(: , ∶), and the scattering of the

4 × 4 axial stiffness matrix, 𝑲(𝑩𝒆𝒏𝒅,𝑩𝒆𝒏𝒅) = 𝑺𝒃(: , ∶). In the provided application script

Plane_frame_MPC.m the axial and bending stiffnesses are written in the above forms, and then

combined using vector subscripts. That makes it easier to understand where the entries in matrix

𝑲 came from. The completed stiffness in the local coordinate axes (with element length 𝐿) is

𝑲𝑳𝒆 =

𝐸

𝐿

[

𝐴 0 00 12𝛽 𝐿⁄ 6𝛽0 6𝛽 4𝐼

−𝐴 0 0 0 −12𝛽 𝐿⁄ 6𝛽 0 −6𝛽 2𝐼

−𝐴 0 0 0 −12𝛽 𝐿⁄ −6𝛽 0 6𝛽 2𝐼

𝐴 0 0 0 12𝛽 𝐿⁄ −6𝛽 0 −6𝛽 4𝐼 ]

, 𝛽 ≡𝐼

𝐿. (11.1-4)

Likewise, the combined external point loads are scattered the same way:

𝑭𝑻 = [𝐹1 𝑉1 𝑀1 𝐹2 𝑉2 𝑀2]

as are the initial thermal loads

𝑭𝑻 = 𝐸𝛼[−𝐴∆𝑇𝑥 0 𝐼∆𝑇𝑦 𝑡⁄ 𝐴∆𝑇𝑥 0 − 𝐼∆𝑇𝑦 𝑡⁄ ]

The element resultant forces, and moments, are written as a rectangular matrix times the

input nodal line loads as {𝑭𝒘𝒃 } ≡ [𝑹𝒃]{𝒘𝒃} and it probably makes sense to calculate the resultant

axial and bending (generalized) forces and scatter them. But, since the line load components will

come in as a vector in the property data the element rectangular matrix could also be assembled

and simply post-multiplied by that load vector. The assembled 6 × 4 rectangular load transfer

matrix and the line load inputs for the frame element are

{𝑭𝒘𝒆 } ≡ [𝑹𝒆]{𝒘𝒆}, 𝑹𝒆 =

𝐿

60

[

20 0 100 21 00 3𝐿 0

092𝐿

10 0 20 0 9 0 0 −2𝐿 0

021

−3𝐿]

, 𝒘𝒆 = {

𝑤𝑥1

𝑤𝑦1

𝑤𝑥2𝑤𝑦2

}. (11.1-5)

Usually, several of the frame members in a structure do not lie parallel to the global axis. For

any inclined frame member the relationship between the above local coordinate equations and

their form for assembly into the system matrix equations of equilibrium depends on its


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orientation in the plane, as sketched in Fig. 9.11-2. As with a truss element, one must consider

how the nodal degrees of freedom transform from the element axes to the global axes.

Figure 11.1-2 An inclined frame member

The rotation of the frame member at node k, 𝜃𝑘, is unchanged by rotation of the coordinate

system about a parallel axis, so a unity term is added to the truss transformation rule. The frame

transformation matrices for the generalized displacements at a node (similar to 9.6-2) are:

{𝜹}𝐿 = {𝑢𝑣𝜃}

𝐿

= [

𝑐𝑜𝑠 (𝜃𝑥) 𝑐𝑜𝑠(𝜃𝑦) 0

−𝑐𝑜𝑠(𝜃𝑦) 𝑐𝑜𝑠 (𝜃𝑥) 0

0 0 1

] {𝑢𝑣𝜃}

𝐺

or 𝜹𝑳 = 𝒕(𝜃) 𝜹𝑮

Applying the same transformation to both ends gives the element transformation law

𝜹𝑳𝑒 = [𝑻(𝜃)]𝜹𝑮

𝑒 = [𝒕(𝜃) 𝟎

𝟎 𝒕(𝜃)] 𝜹𝑮

𝑒,

6 × 1 = (6 × 6)(6 × 1)

[𝑻(𝜃)] =

[ 𝑐𝑜𝑠 (𝜃𝑥) 𝑐𝑜𝑠(𝜃𝑦) 0


0 0 1

𝟎

𝟎



0 0 1]

Finally, the transformed global coordinate form of the combined frame member stiffness matrix

and force vectors is

𝑲𝑮𝒆 = [𝑻(𝜃)]𝑻 𝑲𝑳

𝒆[𝑻(𝜃)], 𝑭𝑮𝒆 = [𝑻(𝜃)]𝑻𝑭𝑳

𝒆 (11.1-6)

The above element generation, combination, and assembly are illustrated in the script segment in

Fig. 11.1-3, which is taken from the application script Planar_Frame.m.

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

essential boundary conditions at each node can have the following packed values: 000 a free


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joint, 100 only u is given, 010 only v is given, 001only the slope given, 110 both u and v are

given (a pinned joint), 111 all are given (usually a fixed joint), etc.

Usually the given boundary condition flag 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 multipoint

constraint (MPC). For example, some frames contain hinges that release the moment there.

Hinges are modeled by two nodes having the same coordinates and having the vertical

displacements constrained to be equal and also having the horizontal displacements constrained

to be equal. The rotations at the two nodes become independent (they no longer transmit a

moment). Then there are two type-2 MPCs at those nodes.


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Figure 11.1-3 Combining axial and bending arrays to form a frame member stiffness


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11.2 Frame member reactions: To design buildings utilizing frame members it is

important to be able to determine the element reactions (member end forces and moments) and

or to graph its moment and shear diagrams. Once the entire structure has been solved to give all

of the nodal displacements each element is post-processed and its subset of displacements is

gathered back to the element. Assuming that the element is in equilibrium with those

displacements, then (11.11-1) and/or (11.11-2) are re-arranged to solve for the generalized forces

at its end points. In other words, the saved element stiffness matrix, in global coordinates, is

multiplied by the element displacements to give a vector of generalized forces, say {𝑭𝜹𝒆} ≡

[𝑺𝒆]{𝜹𝒆}, so that the equilibrium of the generalized element forces becomes {𝑭𝜹𝒆} = {𝑭

𝒆} +{𝑭𝒘

𝒆 } + {𝑭𝑻𝒆} which is simply rearranged to find the member end reactions as

{𝑭 𝒆} = {𝑭𝜹

𝒆} − {𝑭𝒘𝒆 } − {𝑭𝑻

𝒆}, (11.2-1)

which are also in global coordinates.

To find the member axial force and the two end point transverse shear forces it is necessary

to transform the global element reactions back to the local member coordinates. The

transformation matrices have the property of being ‘orthogonal matrices’ which means that their

inverse is simply their transpose: [𝑻(𝜃)]−1 = 𝑻(𝜃)𝑇, so the array used to rotate from an inclined

member axis to the global member axis gets re-used. The above element reactions recovery in

first the global system and then in the member local coordinate system are illustrated in the

script segment in Fig. 11.2-1, which is taken from the application script Planar_Frame.m.

Consider the statically determinate plane frame in Fig. 11.2-2. It was analyzed numerically

using frame elements consisting of two-node cubic bending element combined with two-node

linear axial elements. The analysis used values of EI = 1,000, and A = 1. The Planar_Frame.m.

script utilized the data in Fig. 11.2-3 and numerically gives the system displacements as:

Node, 3 displacements per node (m, radian, m)

1 0.0000e+00 0.0000e+00 3.7190e-01

2 -1.3674e+00 -2.1500e-01 1.1903e-02

3 -1.5274e+00 -1.7667e-01 3.8569e-02

4 -1.5036e+00 -8.8333e-02 -1.4306e-03

5 -1.5065e+00 0.0000e+00 -1.4306e-03

which match the analytically exact values. That script recovers the system reaction forces as

Recovered Reactions at Displacement BCs

Node, DOF (1 & 2=force, 3=couple), Value

1 1 30.0000

1 2 35.8333

5 2 44.1667

which are also exact (check them with Newton’s equations for static equilibrium).

xxx


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Figure 11.2-1 Recovering the frame member global and local reaction forces and moment


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Figure 11.2-2 A determinate plane frame structure

Figure 11.2-3 Data files describing the plane frame of Fig. 11.2-2 w signs

11.3 Enhanced frame post-processing*: Today, it is practical to execute the Wilson

static condensation in Section 7.8, for any element, using symbolic software to complete the

algebra. Applying that static condensation symbolically to the quintic elastic stiffness matrix

surprisingly gives the same elastic stiffness matrix as cubic beam element. Also, the symbolic

condensation of the quintic element trapezoidal line load (with w2 = (w1 + w3)/2) matrix gives

the same matrix as the cubic beam element resultant trapezoidal line load. Likewise, static

condensation of the quintic element thermal load matrix gives the matrix for the cubic beam


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element thermal load resultant. In other words, the classic cubic beam element matrices are the

statically condensed version of the three-noded quintic beam element matrices.

Tong’s Theorem [5] shows that the cubic element will give analytically exact nodal solutions

when there is no axial load and no elastic foundation. Thus, the symbolic condensations mean

that the recovery of the quintic internal node degrees of freedom will be exact when post-

processing a classic frame member. The Wilson condensation matrices for a quintic beam are

W =L

1024 EI[5L2 0

0 28] , Tk =

1

8 L[−4L −L2 −4L L2

12 2L −12 2L],

Tf =L3

3840 EI{5L(w2 + w1)

2(w2 − w1)} (11.3-1)

To illustrate the enhancement procedure for the above plane frame the enhanced post-processing

of the horizontal element is detailed:

1. Gather the global degrees of freedom from nodes 2 and 3 of the cubic frame:

δGT = [−1.3674 0.21500 0.01190 −1.5274 −0.17667 0.03857]

2. Transform them back to the local member axis: δL = Re(θ)δG

T

𝑹𝒆(𝜃) = [𝒕(𝜃) 𝟎𝟎 𝒕(𝜃)

] , 𝒕(𝜃) = [ cos (𝜃) sin (𝜃) 0−sin (𝜃) cos (𝜃) 0

0 0 1

] , 𝜃 = 0

δLT = [−1.3674 0.21500 0.01190 −1.5274 −0.17667 0.03857]

3. Extract the end bending degrees of freedom (2, 3, 5,6)

δbT = [0.21500 0.01190 −0.17667 0.03857]

4. Gather the element’s input local trapezoidal line load values

{w1

w2} = {

−5−15

}

5. Compute what the condensed out quintic element forces would have been

Fa =L

420{112(w1 + w2)−8L(w1 − w2)

} = {−42.6667−12.1905

}

6. Substitute the member length into the stiffness transformation matrix

Tk =1

8 L[−4L −L2 −4L L2

12 2L −12 2L] =

1

64[−32 −64 −32 64

12 16 −12 16]

7. Evaluate the square work matrix

W =L

1024 EI[5L2 0

0 28] = [

2.5000 00 0.21875

] × 10−3

8. Compute the transformed force vector


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Tf =L3

3840 EI{5L(w2 + w1)

2(w2 − w1)} = {

−0.1066667−0.0026667

}

9. Recover the center node displacements of a quintic beam

δa = TF − Tkδb = {−0.32916766667−0.00809779167

}

10. Loop over points along the quintic beam to graph any or all of the quintic equations

for its deflection, slope, moment, or shear for enhanced recover graphs.

Repeat that enhancement process of all members in the frame. This enhancement process has,

within typical round-off errors, extracted the exact solution for the frame from the approximate

solution of the cubic frame members which are shown in Fig. 11.3-1. Enhanced axial

displacements and forces were also obtained, but not plotted here due to their simplicity.

Figure 11.3-1 Enhanced post-processing from cubic/quintic frame members


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11.4 Space frames: A space frame is made up of frame members that are connected together

in three-dimensional space. That means that each node now has three displacement components

and three rotational components. A space frame member is required to resist three force

components and three moment components at each node, see Fig. 11.3-1. The bar element carries

the axial force, say 𝐹𝑥, a torsional shaft element carries the moment about the axis, say 𝑀𝑥, the

previous beam element carries the first transverse force, say 𝐹𝑦, and the moment, say 𝑀𝑧,

perpendicular to 𝐹𝑥 and 𝐹𝑦, and a second beam element, with the same axis, is set in the plane

perpendicular to the first beam and it transmits the transverse force, say 𝐹𝑧, and the moment, say

𝑀𝑦, perpendicular to the axis and 𝑀𝑧. The second beam has a different moment of inertia since it

also lies at the centroid of the cross-sectional area, but is in a plane perpendicular to that of the

first beam section.

The analysis of a space frame is very similar to that of the planar frame. If the axes of the two

principle inertia axis continue to lie in the horizontal plane and vertical plane, as they did for

planar frames, then the nodal displacements and rotations transform into three-dimensional space

just like the space truss displacements did with Eq. 8.7-X.

If a space frame member is rotated about its axis causing the principle inertia axes to no

longer be in the horizontal and vertical planes then more detailed user data required in order to

correctly orientate the axis and inertia planes of the frame member in three-dimensional space.

Figure 11.4-1 Degrees of freedom of a space frame member

11.5 Numerically integrated frame members: The results of important studies should

be validated by some other technique like analytical solutions, experiments, or a different type of

finite element analysis. There are long irregularly shaped solids carrying bending loads that can

be modelled as frame members where the area, moment of inertia tensor, and/or modulus of

elasticity varies along the length of the solid. They can be approximated as numerically

integrated space frames.

One such structure is human long bones. The data for long bones are extracted from

computer tomography (CT) images taken at multiple scan planes. The extracted data give the


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inner and outer perimeters of the bone and thereby give the centroid, cross-sectional area, and the

moment of inertias about orthogonal axes in each plane slice. Every pixel in the CT slice gives

the density of the bone at that point. There is an equation that converts bone density to bone

modulus of elasticity. The elastic modulus value, 𝐸, at every point in the slice is used to obtain

the average modulus. A space frame element is established between the centroids of two adjacent

slices. Then the slice area, inertias, and average modulus at the two (or three) member nodes are

interpolated to define the variations along the length of the member.

Such space frame models are more efficient than using solid models and can give similar

accuracy. The element stiffness must be evaluated by numerical integration. Then the standard

constant property bending stiffness matrix generalizes from (10.4-2) to

𝑺𝑏𝑒 = ∫ 𝑩𝑏

𝑒(𝑥)𝑇

𝐿𝑒 𝐸 (𝑥) 𝐼 (𝑥) 𝑩𝑏𝑒(𝑥) 𝑑𝑥

with 𝐸(𝑥) = 𝑯(𝑟) 𝑬𝑒, 𝐼(𝑥) = 𝑯(𝑟) 𝑰𝑒. The axial stiffness matrix generalizes to

𝑺𝑎𝑒 = ∫ 𝑩𝑎

𝑒(𝑥)𝑇

𝐿𝑒

𝐸 (𝑥) 𝐴 (𝑥) 𝑩𝑎𝑒(𝑥) 𝑑𝑥

with 𝐴(𝑥) = 𝑯(𝑟) 𝑨𝑒. Those terms are integrated numerically:

𝑺𝑏𝑒 = ∑ 𝑩𝑏

𝑒(𝑟𝑞)𝑇𝑛𝑞

𝑞=1 {𝑯(𝑟𝑞) 𝑬𝑒} {𝑯(𝑟𝑞) 𝑰

𝑒} 𝑩𝑏𝑒(𝑟𝑞) |𝐽| 𝑤𝑞

𝑺𝑎𝑒 = ∑ 𝑩𝑎

𝑒(𝑟𝑞)𝑇𝑛𝑞

𝑞=1 {𝑯(𝑟𝑞) 𝑬𝑒} {𝑯(𝑟𝑞) 𝑨

𝑒} 𝑩𝑎𝑒(𝑟𝑞) |𝐽| 𝑤𝑞. (11.5-1)

A numerically integrated space frame model was used as a validation for a three-dimensional

solid element study of an artificial hip replacement. The space frame model ran in seconds, but

the solid model ran for an hour, mainly because the modulus of elasticity was different at every

point and required a higher than normal number of integration points in each solid element.

Figure 11.5-1 shows a comparison of the maximum stress results from the space frame (dashed)

and the stresses from the solid model xxxx

Xxx

Figure 11.5 -1 Maximum stresses from a bone space frame model and a solid model


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11.6 Summary:

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

Classic beam (𝑛𝑔 = 2 , 𝑛𝑝 = 1, 𝑛𝑟 = 1, 𝑛𝑠 = 1) quantities:

𝐸𝐼 𝑣′′′′(𝑥) = 𝑓(𝑥) load per unit length , Differential equation

𝐸𝐼 𝑣′′′(𝑥) = 𝑉(𝑥) transverse shear force, Natural boundary condition

𝐸𝐼𝑣′′(𝑥) = 𝑀(𝑥) bending moment, Natural boundary condition

𝑣′(𝑥) = 𝜃(𝑥) slope, Essential boundary condition

𝑣(𝑥) = deflection, Essential boundary condition

Beam element degrees of freedom: 𝜹𝒆𝑻 = [𝑣1 𝜃1 𝑣2 𝜃2 ⋯ 𝜃𝑛𝑛]

Beam with a transverse force per unit length, 𝑓(𝑥):

𝑑2

𝑑𝑥2 [𝐸𝐼(𝑥)𝑑2𝑣

𝑑𝑥2] − 𝑓(𝑥) = 0

Beam on an elastic foundation (BOEF) with a stiffness of 𝑘(𝑥).

𝑑2


𝑑𝑥2] + 𝑘(𝑥)[𝑣 − 𝑣∞] − 𝑓(𝑥) = 0

Beam with a constant axial load, N

𝑑2


𝑑𝑥2] − 𝑁𝑑2𝑣

𝑑𝑥2 − 𝑓(𝑥) = 0

Beam with a variable axial force, 𝑁(𝑥)

𝑑2


𝑑𝑥2] −𝑑

𝑑𝑥[𝑁(𝑥)

𝑑𝑣

𝑑𝑥] − 𝑓(𝑥) = 0

Beam with an axial force and an elastic foundation:


15

𝑑2


𝑑𝑥2] −𝑑

𝑑𝑥[𝑁(𝑥)

𝑑𝑣

𝑑𝑥] + 𝑘(𝑥)[𝑣 − 𝑣∞] − 𝑓(𝑥) = 0

Bending stiffness:

𝑺𝑬𝑒 = ∫

𝑑2𝑯(𝑥)

𝑑𝑥2

𝑇

𝐿𝑒 𝐸𝐼𝑒 𝑑2𝑯(𝑥)

𝑑𝑥2 𝑑𝑥 ≡ ∫ 𝑩𝑒(𝑥)𝑇

𝐿𝑒 𝐸𝐼𝑒𝑩𝑒(𝑥)𝑑𝑥

Foundation stiffness and load:

𝑺𝒌𝑒 = ∫ 𝑯(𝑥)𝑇

𝐿𝑒 𝑘𝑒𝑯(𝑥) 𝑑𝑥, 𝒄∞𝑒 = ∫ 𝑯(𝑥)𝑇

𝐿𝑒 𝑣∞𝑒 𝑑𝑥

Axial load stiffness(es): 𝑺𝑵𝑒 = ∫

𝑑𝑯(𝑥)

𝑑𝑥

𝑇

𝐿𝑒 𝑁𝑒 𝑑𝑯(𝑥)

𝑑𝑥𝑑𝑥, 𝑺𝒘

𝑒 = −∫ 𝑯(𝒙)𝑇

𝐿𝑒 𝑑𝑁

𝑑𝑥 𝑑𝑯(𝑥)

𝑑𝑥𝑑𝑥

Line load resultant: 𝒄𝒇𝑒 = ∫ 𝑯(𝑥)𝑇

𝐿𝑒 𝑓𝑒(𝑥)𝑑𝑥 = ∫ 𝑯(𝑥)𝑇

𝐿𝑒 𝒉(𝑥)𝒇𝒆𝑑𝑥 ≡ 𝑹𝒆𝒇𝒆

Transverse thermal moment:

𝒄𝜶𝒆 = ∫ 𝑩𝒆𝑻

𝐿𝑒 𝐸𝐼𝑒(𝑥)𝛼𝑒∆𝑇(𝑥)

𝑡(𝑥)

𝑒

𝑑𝑥 =𝛼𝑒∆𝑇𝑒𝐸𝐼𝑒

𝑡𝒆 ∫ 𝑩𝒆𝑻

𝐿𝑒 𝑑𝑥

Element mass matrix: 𝒎𝑒 = ∫ 𝑯(𝑥)𝑇𝜌𝑒

𝐿𝑒 𝐴𝑒𝑯(𝑥) 𝑑𝑥

Force and moment point sources: 𝒄𝑷𝒆𝑻 = [𝑉1 𝑀1 𝑉2 𝑀2 ⋯ 𝑀𝑛𝑛]

Two-node beam stiffness:

𝐸𝐼

𝐿3 [

12 6𝐿 6𝐿 4𝐿2

−12 6𝐿−6𝐿 2𝐿2

−12 −6𝐿 6𝐿 2𝐿2

12 −6𝐿−6𝐿 4𝐿2

] {

𝑣1

𝜃1𝑣2

𝜃2

}

Two-node beam line load and thermal load:

𝐿

60[

21 9 3𝐿 2𝐿 9 21

−2𝐿 −3𝐿

] {𝑤𝑦1

𝑤𝑦2} +

𝐸𝐼𝛼∆𝑇𝑦

𝑡{ 0 1

0−1

}

Two node planar frame local stiffness:

𝑲𝑳𝒆 =

𝐸

𝐿

[

𝐴 0 00 12𝛽 𝐿⁄ 6𝛽0 6𝛽 4𝐼

−𝐴 0 0 0 −12𝛽 𝐿⁄ 6𝛽 0 −6𝛽 2𝐼

−𝐴 0 0 0 −12𝛽 𝐿⁄ −6𝛽 0 6𝛽 2𝐼

𝐴 0 0 0 12𝛽 𝐿⁄ −6𝛽 0 −6𝛽 4𝐼 ]

, 𝛽 ≡𝐼

𝐿

Two node planar frame local line load and thermal load resultants:


16

𝐿

60

[

20 0 100 21 00 3𝐿 0

092𝐿

10 0 20 0 9 0 0 −2𝐿 0

021

−3𝐿]

{

𝑤𝑥1

𝑤𝑦1

𝑤𝑥2𝑤𝑦2

}

𝑭𝑻 𝑻 = 𝐸𝛼[−𝐴 ∆𝑇𝑥 0 𝐼 ∆𝑇𝑦 𝑡⁄ 𝐴 ∆𝑇𝑥 0 − 𝐼 ∆𝑇𝑦 𝑡⁄ ]

Planar frame transformation from local (L) to global (G) coordinates:

𝑲𝑮𝒆 = [𝑻(𝜃)]𝑻 𝑲𝑳

𝒆[𝑻(𝜃)], 𝑭𝑮𝒆 = [𝑻(𝜃)]𝑻𝑭𝑳

𝒆

Two node planar frame transformation matrix:

[𝑻(𝜃)] =

[ 𝑐𝑜𝑠 (𝜃𝑥) 𝑐𝑜𝑠(𝜃𝑦) 0


0 0 1

𝟎

𝟎



0 0 1]


17

11.7 Exercises

11.8 Subject Index axial force, 2 axial stiffness, 2 bar member, 1 beam on an elastic foundation, 14 bending stiffness, 2, 15 boundary condition flag, 5 cassic beam, 14 data files, 9 degrees of freedom, 14 element axes, 4 element length, 3 element reactions, 7 external couple, 2 fixed joint, 5 foundation stiffness, 15 free joint, 5 global axes, 4 inclined member, 4

load transfer matrix, 3 mass matrix, 15 member end forces, 7 modulus of elasticity, 2 multipoint constraint, 5 Newton’s equations, 7 planar frame, 1 point load, 3 principle inertia axes, 12 resultant forces, 3 space frame, 2, 12 static equilibrium, 7 summary, 14 temperature, 2 thermal load, 3 thermal moment, 15 vector subscript, 2

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11. Frame analysis - Rice University · 2020-01-13 · Frame analysis 11.1 Planar frames: If the bar member, which carries only loads parallel to its axis, and a beam, which carries