Lecture 12: Grid Elements and Grid Analysis.mech420/Lecture12.pdf · Lecture 12: Grid Elements and...

MECH 420: Finite Element Applications

Lecture 12: Grid Elements and Grid Analysis.

The grid element equations in terms of elemental coords:

3 2 3 2

11

1

2 21

23 2 3 2

2

2

2 2

12 0 6 12 0 6

ˆˆ 0 0 0 0ˆˆ

6 0 4 6 0 2ˆˆ

12 0 6 12 0 6ˆˆ 0 0 0 0

6 0 2 6 0 4

yy

x

z

y

x

z

EI EI EI EIL L L L

GJ GJ df L Lm EI EI EI EIm L L L L

EI EI EI EIfL L L Lm

GJ GJmL L

EI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎧ ⎫⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪

−⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎬ ⎢ ⎥⎪ ⎪ − − −⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥−⎪ ⎪⎩ ⎭ ⎢ ⎥⎢ ⎥⎢ − ⎥⎣ ⎦

1

1

2

2

2

ˆˆ

ˆˆ

x

z

y

x

z

d

φφ

φφ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

x

y

z

It is obvious that we need to apply global coordinates in solving grid problems.

2

3



The three-dimensional rotation matrix, .

The grid element, by convention, lies in the xz plane of the global frame.

This is defined in Logan’s steps 1 through 4 when we created ourgrid element.

R

ˆˆ ˆ( ) ( ) ( )ˆ ˆ ˆˆˆ ˆˆ ˆ ˆ( ) ( ) ( )ˆˆ ˆ ˆˆ ˆ( ) ( ) ( )

x x x

y y y

z z z

i i j i k ix x x c c c xy R y i j j j k j y c c c yz z z c c c zi k j k k k

θ ψ φθ ψ φθ ψ φ

⎡ ⎤⋅ ⋅ ⋅⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= = ⋅ ⋅ ⋅ =⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⋅ ⋅ ⋅⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭⎢ ⎥⎣ ⎦

ˆ, , angles between and the , , and global directions respectivelyˆ, , angles between and the , , and global directions respectively

ˆ, , angles between

x y z

x y z

x y z

i i j k

j i j k

θ θ θ

ψ ψ ψ

φ φ φ

≡

≡

≡ and the , , and global directions respectivelyk i j k



For grid analysis, orientation changes are only about the y direction.

Angle of rotation from the global to the elemental

2π θ−

zz

ˆ, y y

Plan View

ˆˆ ˆ( ) ( ) ( ) ˆˆˆ ˆ ˆ( ) ( ) ( )ˆ ˆˆ ˆ( ) ( ) ( )

i i j i k ix xy i j j j k j yz zi k j k k k

⎡ ⎤⋅ ⋅ ⋅⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪= ⋅ ⋅ ⋅⎢ ⎥⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⋅ ⋅ ⋅⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦

ˆ ˆ( ) ( ) 0ˆ ˆ( ) ( ) 0ˆ( ) 1

ˆ( ) cos( ) cos( )cos( ) sin( )sin( ) sin( )2 2 2

ˆ( ) cos( ) sin( )2

ˆ( ) cos( )ˆ( ) cos( )

i j j i

k j j k

j j

i k

k i

i i

k k

π π πθ θ θ θ

π θ θ

θ

θ

⋅ = ⋅ =

⋅ = ⋅ =

⋅ =

⋅ = − − = − − = −

⋅ = − = +

⋅ =

⋅ =

Sense of θ defined by RHR (y-axis).



For rotations θx, θy, or θz about the X, Y, or Z axes, respectively, of the global reference frame, the elemental coordinates are related to the global coordinates by:

ˆ1 0 0ˆ0ˆ0

x xy c s yz s c z

θ θθ θ

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭

ˆ0ˆ0 1 0ˆ0

x c s xy yz s c z

θ θ

θ θ

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

ˆ0ˆ0ˆ0 0 1

x c s xy s c yz z

θ θθ θ

−⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭

Rotation about only the x axis.

Rotation about only the y axis.

Rotation about only the z axis.



The rotation matrix applied in grid analysis is…

Consider the vector quantities that are being transformed from global to elemental coordinates.At node 1:

ˆ0ˆ0 1 0ˆ0

x c s xy yz s c z

θ θ

θ θ

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

11

1 1

1 1

ˆ0ˆ0 1 0ˆ0

xx

y y

z z

ff c sf ff s c f

θ θ

θ θ

⎧ ⎫⎧ ⎫ ⎡ ⎤ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎪ ⎪⎩ ⎭

In grid analysis the vertical loads are aligned with the global y axis. There is no need to transform the transverse forces or the corresponding displacements.

0.0

0.0



Consider the moments applied at the grid element node points.At node 1:

Applying this transformation at both nodes for the element load and displacement vectors…

1 1

1 1

1 1

ˆ0ˆ0 1 0ˆ0

x x

y y

z z

m c s mm mm s c m

θ θ

θ θ

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

0.0

1 1

1 1

ˆˆ

x x

z z

m mc sm ms c

θ θθ θ

⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦⎩ ⎭ ⎩ ⎭



3 2 3 2

1

1

2 21

23 2 3 2

2

2

2 2

12 0 6 12 0 6

0 0 0 0

6 0 4 6 0 2

12 0 6 12 0 6

0 0 0 0

6 0 2 6 0 4

y

x

zG

y

x

z

EI EI EI EIL L L L

GJ GJf L L

m EI EI EI EIm L L L LTf EI EI EI EI

L L L LmGJ GJmL L

EI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎧ ⎫⎧ ⎫ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ −⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎨ ⎬⎬ ⎢ ⎥⎪ ⎪ ⎪⎪ − − −⎢ ⎥⎪ ⎪ ⎪⎪ ⎢⎪ ⎪ ⎪⎪ ⎢⎩ ⎭ −⎩ ⎭ ⎢⎢⎢ −⎣ ⎦

1

1

1

2

2

2

1 0 0 0 0 00 0 0 00 0 0 0

; 0 0 0 1 0 00 0 0 00 0 0 0

y

x

zG G

y

x

z

dc ss c

T Td

c ss c

φ θ θφ θ θ

θ θφθ θφ

⎧ ⎫⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥−⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎨ ⎬⎬ ⎢ ⎥⎪ ⎪ ⎪⎪ ⎢ ⎥⎪ ⎪ ⎪⎪ −⎢ ⎥⎥ ⎪ ⎪ ⎪⎪ ⎢ ⎥⎥ ⎣ ⎦⎩ ⎭⎩ ⎭⎥⎥⎥

k

df

ˆ TG Gf T k T d⎡ ⎤= ⎣ ⎦

k



A note on “transforming rotations.”Sequences of finite rotation angles are not vectors.

2xπφ =

x

y

z

2zπφ = −

x

y

z

and then z xφ φ

x

y

z

and then x zφ φ

a vector prior to a sequence of active rotations.the same vector after the rotations.

rr≡′ ≡

r ′

r

r ′



Vectors are a superposition of scalar quantities that are each associated with a direction. Rotation sequences are NOT commutative and thus can’t be vectors.

Rotations must be expressed using rotation matrices. Examples are the 3x3 matrices we have looked at in this course.

x z z xi k k iφ φ φ φ+ ≠ + Rotation vectors DO NOT exist.



But if the two angles were both very small…

and then :0 1 0 00 0

0 0 1 0Note:

1 0 0 00 00 0 0 1

z x

z z

z z x x

x x

z z

x x z z

x x

x c s xy s c c s yz s c z

x c s xy c s s c yz s c z

φ φφ φφ φ φ φ

φ φ

φ φφ φ φ φφ φ

′ −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥′ = −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪′ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭

′ −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥′ ≠ −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪′ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭

Can’t change the order of matrix multiplication.



0 1 0 0 1 0 0 00 0 0

a

00 0

nd

1 0 0 0 0 1

:

1 0

z z z z

z z x x x x z z

x

z

x x

z

z

x

x x

z

cd sd x cd sd xsd cd cd sd y cd sd sd cd y

sd cd z sd cd z

d dxyz

dd

φ φ φ φφ

φ φ φ φ

φ φ φ φ φ φ φφ φ φ

φφ

φ

→ →

′⎧ ⎫ − −⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− = −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭

⎪ ⎪′ =⎨ ⎬⎪ ⎪′⎩ ⎭

= −0 0

1 00 1 0 0

0

z

x z x

x x

x

z

x x d xd y y d d y

d z z d z

x d xy yz d z

φφ φ φ

φ φ

φ

φ

⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= + −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + ×⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭

The rotations would form a vector entity. But this is only strictly true in the case of infinitesimal rotations


Lecture 12: Frame Analysis Example.

§5.2 Rigid Plane Frame Examples.Pg.# 192-210 contains Examples 5.1 through 5.4.Here we look at P.5.3 pg.#241.

Find the nodal displacements at node 2 and the reaction forces at node 1. Draw the V and M diagrams for element #1. Select a channel section that ensures the bending stress is 66% of the yield stress for A36 steel.



Before embarking on a long solution process – target the necessary calculations.Only concerned with node 1 reactions and node 2 deflections.Problem has a few stages:

Solve for the deflections at node 2.Recover the element nodal loads on element 1 (including reactions at node 1).Using the element nodal loads draw the V and M diagrams (Vconstant and M linear).Based on the peak M value and allowable bending stress, size a cross section.

Involves some criterion on the allowable stress levels and some tabulated cross section geometries.



Could carry the channel geometry through as a variable or use guess and check…

x

y

z

Neutral axis of channel

4

1.920 in6.000 in13.1 inz

bhI

===

C6×8.2 American Standard Channel


Element #1:

The 1st three equations will be discarded until the recovery stage and only the bottom 3x3 submatrix will be used in the assembly.


(1)1 3 2 3 2(1)

1(1) 2 21

(1)2

(1)2(1)

3 2 3 22

2 2

0 0 0 0

ˆ 0 12 6 0 12 6ˆ

0 6 4 0 6 2ˆˆ

0 0 0 0ˆ

0 12 6 0 12 6ˆ

0 6 2 0 6 4

x

y

z

x

y

z

AE AEL L

EI EI EI EIfL L L L

f EI EI EI EIm L L L L

AE AEfL Lf

EI EI EI EIm L L L L

EI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢⎢⎧ ⎫ −⎢⎪ ⎪⎢⎪ ⎪

−⎢⎪ ⎪⎪ ⎪ ⎢=⎨ ⎬ ⎢⎪ ⎪ −⎢⎪ ⎪ ⎢⎪ ⎪ ⎢ − − −⎪ ⎪⎩ ⎭ ⎢⎢⎢ −⎣ ⎦

1

1

1

2

2

2

ˆˆ

ˆˆˆ

ˆ

x

y

z

x

y

z

dd

dd

φ

φ

⎥⎧ ⎫⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎥ ⎨ ⎬⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎩ ⎭⎥⎥

(1)

(1)

1.00.0

CS

=

=0.0

0.0

0.0



Element #2:

The full 6x6 element #2 equations need to be filled out.

2 2 2 22 2 2 2

(2)2 2 2 22

2 2 2(2)

2(2)2

(2)3

(2)3(2)3

12 12 6 12 12 6

ˆ 12 6 12 12 6ˆ

6 6ˆ 4ˆ

ˆ

ˆ

x

y

z

x

y

z

I I I I I IAC S A CS S AC S A CS SL LL L L L

I I I I If AS C C A CS AS C CL LL L Lf

I Im E I SL LLf

fm

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − − − + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎧ ⎫ ⎛ ⎞ ⎛ ⎞+ − − − +⎪ ⎪ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪

⎪ ⎪⎪ ⎪ −=⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

2

2

2

32 2

2 23

2 2 32

ˆ

ˆ

ˆ2ˆ

12 12 6ˆ

ˆ12 6

SYM 4

x

y

z

x

y

z

d

d

C IdI I IAC S A CS S dLL L

I IAS C CLLI

φ

φ

⎡ ⎤⎢ ⎥⎢ ⎥

⎧ ⎫⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ⎨ ⎬⎢ ⎥ ⎪ ⎪⎢ ⎥⎛ ⎞ ⎪ ⎪+ −⎜ ⎟⎢ ⎥ ⎪ ⎪⎝ ⎠⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎩ ⎭+ −⎢ ⎥

⎢ ⎥⎣ ⎦

(2)

(2)

0.7070.707

CS

=

=



Element #3:

The last 3 equations are being discarded in this particular problem (homogeneous conditions at Node ‘3’). Only the top 3x3 submatrix needs be computed.

(3)3 3 2 3 2(3)

3(3) 2 23

(3)4

(3)4(3)

3 2 3 24

2 2

0 0 0 0

ˆ 0 12 6 0 12 6ˆ

0 6 4 0 6 2ˆˆ

0 0 0 0ˆ

0 12 6 0 12 6ˆ

0 6 2 0 6 4

x

y

z

x

y

z

AE AEL L

EI EI EI EIfL L L L

f EI EI EI EIm L L L L

AE AEfL Lf

EI EI EI EIm L L L L

EI EI EI EIL L L L

⎡ ⎤−⎢ ⎥⎢⎢⎧ ⎫ −⎢⎪ ⎪⎢⎪ ⎪

−⎢⎪ ⎪⎪ ⎪ ⎢=⎨ ⎬ ⎢⎪ ⎪ −⎢⎪ ⎪ ⎢⎪ ⎪ ⎢ − − −⎪ ⎪⎩ ⎭ ⎢⎢⎢ −⎣ ⎦

3

3

3

4

4

4

ˆˆ

ˆˆˆ

ˆ

x

y

z

x

y

z

dd

dd

φ

φ

⎥⎧ ⎫⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎥ ⎨ ⎬⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎩ ⎭⎥⎥

(3)

(3)

1.00.0

CS

=

=

0.0

0.0

0.0



The layout of the assembled system is…

(1) (2)2 22(1) (2)

2 22(1) (2)2 22

(2) (3)3 3 3

(2) (3)33 3(2) (3)23 3

ˆ ˆ0ˆ ˆ2000ˆ ˆ0ˆ ˆ0

2000 ˆ ˆ0 ˆ ˆ

x xx

y yy

z zz

x x x

yy y

zz z

f fFf fFm mm

F f fF f fm m m

⎧ ⎫+ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ +− ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ +⎪ ⎪ ⎪ ⎪ ⎣ ⎦= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎡+⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢⎪ ⎪ ⎪ ⎪ ⎪ ⎪− +⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎩ ⎭ ⎪ ⎪⎩ ⎭ ⎣+⎩ ⎭

2

2

2

3

3

3

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

x

y

z

x

y

z

d

d

d

d

φ

φ

⎧ ⎫⎡ ⎤⎡ ⎤ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎨ ⎬⎢ ⎥⎤⎢ ⎥ ⎪ ⎪⎢ ⎥⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥⎢ ⎥ ⎪ ⎪⎦⎣ ⎦⎣ ⎦ ⎩ ⎭

Element #3

Element #1 Element #2



Calculating the physical and geometric system values:

Element #1:

Element #2:

Element #3:

Elements #1 and #3 should have identical element equations.

x

2ˆ 24 ksi3

ˆ is the distance from N.A. to surface; is the 2nd moment of area of the section.

x Y

zz

Mc c II

σ σ

σ

< =

= →

2 3 52 3

12 6 lbf0.03032 in ; 1.092 in ; 4.028 10 in

z zI I EL L L

= = = ×

2 3 52 3

12 6 lbf0.008528 in ; 0.5789 in ; 2.136 10 in

z zI I EL L L

= = = ×

2 3 52 3

12 6 lbf0.03032 in ; 1.092 in ; 4.028 10 in

z zI I EL L L

= = = ×



The assembly process leaves us with (imposing the BC’s simultaneously):

1

1

51

2

2

2

0 12.24 2.554 0.8745 2.573 2.554 0.87452000 2.695 3.523 2.554 2.573 0.87450 323.0 0.8745 0.8745 55.97

100 12.24 2.55 0.8745

2000 2.69 3.5230 323.0

x

y

z

x

y

z

FFMFFM

⎧ ⎫ − − − −⎧ ⎫ ⎡⎪ ⎪ ⎪ ⎪ ⎢− − − −⎪ ⎪ ⎪ ⎪ ⎢⎪ ⎪ −⎪ ⎪ ⎢= =⎨ ⎬ ⎨ ⎬ ⎢⎪ ⎪ ⎪ ⎪ ⎢⎪ ⎪ ⎪ ⎪− ⎢⎪ ⎪ ⎪ ⎪ ⎢⎩ ⎭ ⎣⎩ ⎭ i

2

2

2

3

3

3

x

y

z

x

y

z

dd

dd

φ

φ

⎧ ⎫⎤⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎨ ⎬⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎦ ⎩ ⎭

92

2

29

3

3

3

3.01 10 in0.402 in

0.00666 rad3.30 10 in

0.402 in0.00666 rad

x

y

z

x

y

z

dd

dd

φ

φ

−

−

⎧ ⎫ ⎧− × ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪ −⎪ ⎪=⎨ ⎬ ⎨ ⎬×⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪

⎩ ⎭⎩ ⎭

Note the magnitude of the rotations.



Step 7: Recovery.Start with the elemental nodal loads.Pull out element #1 and consider its equilibrium:

(1)11

(1)11

(1)1 15(1)

2 2

(1)2 2(1)2 2

ˆˆ2.4 0 0

ˆˆ 0 0.0303 1.0917ˆˆ 0 1.0917 26.2

4.025 10ˆ ˆˆ ˆ

ˆˆ

xx

yy

z z

x x

y y

z z

dfdf

mf df dm

φ

φ

⎧ ⎫⎧ ⎫ −⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪ −⎪ ⎪ ⎪ ⎪⎢ ⎥= ×⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

0.0

0.0

0.0

(1)1

(1)1

3(1)1

ˆ 0.0 lbfˆ 2000 lbf

106.9 10 lbf inˆ

x

y

z

ffm

⎧ ⎫ ⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪× ⋅⎩ ⎭⎪ ⎪⎩ ⎭



The remaining elemental node 2 loads can be obtained by an FBD (the element must be in static equilibrium).

x

y

(1)1ˆ 106,900 lbf inzm = ⋅

(1)1 2000 lbfyf = (1)

2 2000 lbfyf = −

( )(1) (1) (1)2 1

ˆ 0 :ˆ ˆ2000 37060 lbf in

z

z z

mm L m

=

∴ = − = ⋅∑

(1)2ˆ 37,060 lbf inzm = ⋅

Note: no portion of the external 2000 lbf load is applied to element #2.

1

1 2



For element #2 it only remains to calculate the elemental nodal moments.

Using the element #2 equations in terms of global components…(2) (2)2 2(2) (2)3 3

(2)2(2)

2(2)

52(2)

3(2)

3(2)3

ˆ

ˆ

ˆ 0.4094 0.4094 52.4 0.4094 0.4094 26.22.136 10

ˆ 0.4094 0.4094 26.2 0.4094 0.4094 52.4

z z

z z

x

y

z

x

y

z

m m

m m

ffmffm

=

=

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢− −⎪ ⎪ = ×⎨ ⎬ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢

− −⎪ ⎪ ⎣⎩ ⎭

… … … … … …… … … … … …

… … … … … …… … … … … …

2

2

2

3

3

3

x

y

z

x

y

z

dd

dd

φ

φ

⎧ ⎫⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎨ ⎬⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎦ ⎩ ⎭

(2)2(2)3

ˆ 37,060 lbf inˆ 37,060 lbf in

z

z

m

m

= − ⋅

= ⋅Was there any need to

even evaluate these values? Could you have

immediately deduced them from the previous

slide?



The FBD for element #2 becomes…

x

y

(2)2ˆ 37,060 lbf inzm = − ⋅

(2)2 0.0 lbfyf = (2)

3 0.0 lbfyf =

(2)3ˆ 37,060 lbf inzm = ⋅



Recalling that:

The FBD of element #3 can be drawn right away:

(2) (3)3 3 3y y yF f f= +

x

y

(3)2ˆ 37,060 lbf inzm = − ⋅

(3)3 2000 lbfyf = − (3)

3 2000 lbfyf = +

(3)3ˆ 106,900 lbf inzm = − ⋅

( )(3) (1) (1)3 1

ˆ 0 :ˆ ˆ2000 106,900 lbf in

z

z z

mm L m

=

∴ = − − = − ⋅∑



The peak bending moment in the system (neglecting any concentrated bend effects at the interconnections) is 106,900 lbfin.

x

y

z

Neutral axis of channel

3106,900 1 6.00 24.5 10 psi13.1 2

MAXMAXx

z

M cI

σ = = ⋅ = ×

4

1.920 in6.000 in13.1 inz

bhI

===

Made a good choice.

Note: we have only considered bending stresses – we assumed the transverse loads were applied through the shear centre, ‘SC’.

Lecture 12: Grid Elements and Grid Analysis.mech420/Lecture12.pdf · Lecture 12: Grid Elements and...

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