three-member buttress frame - university of british columbiacivil-terje.sites.olt.ubc.ca › files...

Three-member Buttress FrameThis frame, shown below, is analyzed here using the stiffness method. It is interesting to compare the procedure of establishing the stiffness matrix with transformation matrices, versus establishing it by hand, setting degrees of freedom (DOFs) equal to one, one at a time.

A

q

B

C D

P

L L

L

L

Input values in kN and mThe members are European steel beams with HEB360 cross-section carrying load about its strong axis, resulting in these material and cross-section parameter values:

values = Ε → 200 × 106, A → 18 100 × 10-6, Ι → 431 900 000 × 10-12,

L → 3.0, P → 10, q → 3;

Classical stiffness methodIn the classical stiffness method we establish the stiffness matrix by hand, by setting DOFs equal to one, one at a time. This allows us to easily neglect axial deformations and reduce the number of DOFs. However, as seen in this example, getting it right for inclined members can be hard. Notice in the figure below that the vertical DOF that could have appeared at C is linearly dependent on the horizontal DOF at B. For that reason there are only three DOFs when axial deformations are neglected:

Professor Terje Haukaas The University of British Columbia, Vancouver terje.civil.ubc.ca

Examples Updated January 29, 2018

1

2

3

A

B

C D

Stiffness matrix

The first column of the stiffness matrix is obtained by setting the first DOF equal to unity while fixing the others, still observing zero axial deformations:

1.0

1.0

12

A

B

C D

The expression below for the stiffness coefficient K11 has four contributions: 1) shear force in AB; 2) shear force in BC due to its total displacement 2 1/2 decomposed at B into the horizontal direction by the factor 1 / 2 ; 3) the same shear force in BC decomposed into the vertical direction at C, which also is taken by the first DOF; 4) shear force in CD, which also contributes to the first DOF:



K11 =12 Ε Ι

L3+

12 Ε Ι

2 L23

21

2

1

2+

12 Ε Ι

2 L23

21

2

1

2+12 Ε Ι

L3;

K21 = -6 Ε Ι

L2+

6 Ε Ι

2 L22

21

2;

K31 = -6 Ε Ι

L2+

6 Ε Ι

2 L22

21

2;

Second column:

1.0

K12 = -6 Ε Ι

L2+

6 Ε Ι

2 L22

1

2+

6 Ε Ι

2 L22

1

2;

K22 =4 Ε Ι

L+

4 Ε Ι

2 L2;

K32 =2 Ε Ι

2 L2;

Third column:



1.0

K13 = -6 Ε Ι

L2+

6 Ε Ι

2 L22

1

2+

6 Ε Ι

2 L22

1

2;

K23 =2 Ε Ι

2 L2;

K33 =4 Ε Ι

L+

4 Ε Ι

2 L2;

Collect those values into a matrix:

Kclassical = {{K11, K12, K13}, {K21, K22, K23}, {K31, K32, K33}};Simplify[Kclassical, L > 0] // MatrixForm // N

32.4853 Ε Ι

L3- 1.75736 Ε Ι

L2- 1.75736 Ε Ι

L2

- 1.75736 Ε Ι

L26.82843 Ε Ι

L1.41421 Ε Ι

L

- 1.75736 Ε Ι

L21.41421 Ε Ι

L6.82843 Ε Ι

L

which yields:

Load vector

Fclassical = P - q L,q L2

12, 0;



Introduce values for material and geometry parameters

KclassicalNum = Kclassical /. values // N;KclassicalNum // MatrixForm

103 929. -16 866.7 -16 866.7-16 866.7 196 613. 40 719.9-16 866.7 40 719.9 196 613.

which yields:

FclassicalNum = Fclassical /. values;FclassicalNum // MatrixForm

1.2.250

which yields:

Solve the equilibrium equations

uClassical = (Inverse[KclassicalNum].FclassicalNum) // N

0.0000114241, 0.0000127685, -1.66442 × 10-6which yields:

Use the slope-deflection equation to get end moments

The general form of this equation is:

MNF =2 Ε Ι

L2 θN + θF - 3

Δ

L + FEM

That gives the following end moments:

MAB =2 Ε Ι

L2 × 0 + uClassical[[2]] - 3

uClassical[[1]]

L/. values

0.0774232which yields:

MBA =2 Ε Ι

L2 uClassical[[2]] + 0 - 3

uClassical[[1]]

L/. values

0.81272which yields:

Notice that the length of BC is 2 L2 and that its total displacement perpendicular to its original

chord position is twice u1 decomposed by the factor 1 / 2 :



MBC =

2 Ε Ι

2 L22 uClassical[[2]] + uClassical[[3]] + 3

2 12

uClassical[[1]]

2 L2-

q L2

12/. values

-0.81272which yields:

MCB =

2 Ε Ι

2 L22 uClassical[[3]] + uClassical[[2]] + 3

2 12

uClassical[[1]]

2 L2-

q L2

12/. values


MCD =2 Ε Ι

L2 uClassical[[3]] + 0 - 3

uClassical[[1]]

L/. values


MDC =2 Ε Ι

L2 × 0 + uClassical[[3]] - 3

uClassical[[1]]

L/. values


Those end moments give the following bending moment diagram:



MAB

MBA

MBC

MCB MCD MDC

qL2

8 qL2

8

Computational stiffness methodNow the selection of DOFs is easy: we automatically place three DOFs at each node, and we establish the stiffness matrix by assembling contributions from the three individual elements:



ub Fb kb BASIC

ub=Tblul Fl=TblFb

ul Fl kl=TblkbTbl

ul=Tlgug Fg=TlgFl

ug Fg kg=TlgklTlg

ug=Tgaua Fa=TgaFg

ua Fa

ua=Tafuf Ff=TafFa

uf Ff Kf=TafKaTaf

LOCAL

GLOBAL

ALL

FINAL

T

T

T

T

T

T

T

2 1

3

4 6

5 3 1

2

3 1

2

4 6 5

1

2

3

4

5

6

7

8 9

10

11

12

1

2

3

4 5

6

Ka = TgaT kgTga

Num. el.∑

Basic stiffness matrix

This is the element stiffness matrix that we “take for given,” namely that we consider known from either looking at the slope-deflection equation or solving the differential equation for a beam element:

Kbasic = Ε A

L, 0, 0, 0,

4 Ε Ι

L,2 Ε Ι

L, 0,

2 Ε Ι

L,4 Ε Ι

L;

MatrixForm[Kbasic]

A ΕL

0 0

0 4 Ε ΙL

2 Ε ΙL

0 2 Ε ΙL

4 Ε ΙL

which yields:



Transformation from basic to local

Transformation matrix:

Tbl = {-1, 0, 0, 1, 0, 0}, 0, -1

L, 1, 0,

1

L, 0, 0, -

1

L, 0, 0,

1

L, 1;

MatrixForm[Tbl]

-1 0 0 1 0 0

0 - 1L

1 0 1L

0

0 - 1L

0 0 1L

1

which yields:

Pre and post multiplying the stiffness matrix from above to obtain the new matrix:

Klocal = Transpose[Tbl].Kbasic.Tbl;MatrixForm[Klocal]

A ΕL

0 0 - A ΕL

0 0

0 12 Ε Ι

L3- 6 Ε Ι

L20 - 12 Ε Ι

L3- 6 Ε Ι

L2

0 - 6 Ε Ι

L24 Ε ΙL

0 6 Ε Ι

L22 Ε ΙL

- A ΕL

0 0 A ΕL

0 0

0 - 12 Ε Ι

L36 Ε Ι

L20 12 Ε Ι

L36 Ε Ι

L2

0 - 6 Ε Ι

L22 Ε ΙL

0 6 Ε Ι

L24 Ε ΙL

which yields:

Transformation from local to global

Generic transformation matrix:



Tlg = {{Cos[θ], Sin[θ], 0, 0, 0, 0},{-Sin[θ], Cos[θ], 0, 0, 0, 0},{0, 0, 1, 0, 0, 0},{0, 0, 0, Cos[θ], Sin[θ], 0},{0, 0, 0, -Sin[θ], Cos[θ], 0},{0, 0, 0, 0, 0, 1}};

MatrixForm[Tlg]

Cos[θ] Sin[θ] 0 0 0 0-Sin[θ] Cos[θ] 0 0 0 0

0 0 1 0 0 00 0 0 Cos[θ] Sin[θ] 00 0 0 -Sin[θ] Cos[θ] 00 0 0 0 0 1

which yields:

Transformation of member AB (Element 1)

Tlg1 = Tlg /. θ →90

180π;

K1global = Transpose[Tlg1].Klocal.Tlg1;MatrixForm[K1global]

12 Ε Ι

L30 6 Ε Ι

L2- 12 Ε Ι

L30 6 Ε Ι

L2

0 A ΕL

0 0 - A ΕL

0

6 Ε Ι

L20 4 Ε Ι

L- 6 Ε Ι

L20 2 Ε Ι

L

- 12 Ε Ι

L30 - 6 Ε Ι

L212 Ε Ι

L30 - 6 Ε Ι

L2

0 - A ΕL

0 0 A ΕL

0

6 Ε Ι

L20 2 Ε Ι

L- 6 Ε Ι

L20 4 Ε Ι

L

which yields:

Transformation of member BC (Element 2)



Tlg2 = Tlg /. θ →45

180π;

K2global = Transpose[Tlg2].Klocal.Tlg2;MatrixForm[K2global]

A Ε2 L

+ 6 Ε Ι

L3A Ε2 L

- 6 Ε Ι

L33 2 Ε Ι

L2- A Ε2 L

- 6 Ε Ι

L3- A Ε2 L

+ 6 Ε Ι

L33 2 Ε Ι

L2

A Ε2 L

- 6 Ε Ι

L3A Ε2 L

+ 6 Ε Ι

L3- 3 2 Ε Ι

L2- A Ε2 L

+ 6 Ε Ι

L3- A Ε2 L

- 6 Ε Ι

L3- 3 2 Ε Ι

L2

3 2 Ε Ι

L2- 3 2 Ε Ι

L24 Ε ΙL

- 3 2 Ε Ι

L23 2 Ε Ι

L22 Ε ΙL

- A Ε2 L

- 6 Ε Ι

L3- A Ε2 L

+ 6 Ε Ι

L3- 3 2 Ε Ι

L2A Ε2 L

+ 6 Ε Ι

L3A Ε2 L

- 6 Ε Ι

L3- 3 2 Ε Ι

L2

- A Ε2 L

+ 6 Ε Ι

L3- A Ε2 L

- 6 Ε Ι

L33 2 Ε Ι

L2A Ε2 L

- 6 Ε Ι

L3A Ε2 L

+ 6 Ε Ι

L33 2 Ε Ι

L2

3 2 Ε Ι

L2- 3 2 Ε Ι

L22 Ε ΙL

- 3 2 Ε Ι

L23 2 Ε Ι

L24 Ε ΙL

which yields:

Transformation of member CD (Element 3)

Tlg3 = Tlg /. θ → 0;K3global = Transpose[Tlg3].Klocal.Tlg3;MatrixForm[K3global]

A ΕL

0 0 - A ΕL

0 0

0 12 Ε Ι

L3- 6 Ε Ι

L20 - 12 Ε Ι

L3- 6 Ε Ι

L2

0 - 6 Ε Ι

L24 Ε ΙL

0 6 Ε Ι

L22 Ε ΙL

- A ΕL

0 0 A ΕL

0 0

0 - 12 Ε Ι

L36 Ε Ι

L20 12 Ε Ι

L36 Ε Ι

L2

0 - 6 Ε Ι

L22 Ε ΙL

0 6 Ε Ι

L24 Ε ΙL

which yields:

Contribution to Kall from Element 1

Transformation matrix that “glues” the element into the right position in the All configuration:

Tga1 = {{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0},{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}};



K1all = Transpose[Tga1].K1global.Tga1;MatrixForm[K1all]

12 Ε Ι

L30 6 Ε Ι

L2- 12 Ε Ι

L30 6 Ε Ι

L20 0 0 0 0 0

0 A ΕL

0 0 - A ΕL

0 0 0 0 0 0 0

6 Ε Ι

L20 4 Ε Ι

L- 6 Ε Ι

L20 2 Ε Ι

L0 0 0 0 0 0

- 12 Ε Ι

L30 - 6 Ε Ι

L212 Ε Ι

L30 - 6 Ε Ι

L20 0 0 0 0 0

0 - A ΕL

0 0 A ΕL

0 0 0 0 0 0 0

6 Ε Ι

L20 2 Ε Ι

L- 6 Ε Ι

L20 4 Ε Ι

L0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0

which yields:



Tga2 = {{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0}};



K2all = Transpose[Tga2].K2global.Tga2 /. L → 2 L2 ;Simplify[K2all, L > 0] // MatrixForm

0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0

0 0 0Ε A L2+6 Ι

2 2 L3

Ε A L2-6 Ι

2 2 L33 Ε Ι

2 L2-

Ε A L2+6 Ι

2 2 L3-

Ε A L2-6 Ι

2 2 L33 Ε

2 L2

0 0 0Ε A L2-6 Ι

2 2 L3

Ε A L2+6 Ι

2 2 L3- 3 Ε Ι

2 L2-

Ε A L2-6 Ι

2 2 L3-

Ε A L2+6 Ι

2 2 L3- 3 Ε

2

0 0 0 3 Ε Ι

2 L2- 3 Ε Ι

2 L22 2 Ε Ι

L- 3 Ε Ι

2 L23 Ε Ι

2 L22 ΕL

0 0 0 -Ε A L2+6 Ι

2 2 L3-

Ε A L2-6 Ι

2 2 L3- 3 Ε Ι

2 L2

Ε A L2+6 Ι

2 2 L3

Ε A L2-6 Ι

2 2 L3- 3 Ε

2

0 0 0 -Ε A L2-6 Ι

2 2 L3-

Ε A L2+6 Ι

2 2 L33 Ε Ι

2 L2

Ε A L2-6 Ι

2 2 L3

Ε A L2+6 Ι

2 2 L33 Ε

2 L2

0 0 0 3 Ε Ι

2 L2- 3 Ε Ι

2 L22 Ε ΙL

- 3 Ε Ι

2 L23 Ε Ι

2 L22 2

L

0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0

which yields:



Tga3 = {{0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0},{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0},{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}};



K3all = Transpose[Tga3].K3global.Tga3;MatrixForm[K3all]

0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 A ΕL

0 0 - A ΕL

0 0

0 0 0 0 0 0 0 12 Ε Ι

L3- 6 Ε Ι

L20 - 12 Ε Ι

L3- 6 Ε Ι

L2

0 0 0 0 0 0 0 - 6 Ε Ι

L24 Ε ΙL

0 6 Ε Ι

L22 Ε ΙL

0 0 0 0 0 0 - A ΕL

0 0 A ΕL

0 0

0 0 0 0 0 0 0 - 12 Ε Ι

L36 Ε Ι

L20 12 Ε Ι

L36 Ε Ι

L2

0 0 0 0 0 0 0 - 6 Ε Ι

L22 Ε ΙL

0 6 Ε Ι

L24 Ε ΙL

which yields:

Total stiffness matrix in the All configuration

Here we sum over all elements, and the result is a bit too messy to display:

Kall = K1all + K2all + K3all;

Transformation from All to Final

This is the transformation matrix that introduces the boundary conditions:

Taf = {{0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0},{1, 0, 0, 0, 0, 0},{0, 1, 0, 0, 0, 0},{0, 0, 1, 0, 0, 0},{0, 0, 0, 1, 0, 0},{0, 0, 0, 0, 1, 0},{0, 0, 0, 0, 0, 1},{0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0},{0, 0, 0, 0, 0, 0}};



Kfinal = Transpose[Taf].Kall.Taf;MatrixForm[Kfinal] // PowerExpand // Simplify

Ε 2 A L2+6 8+ 2 Ι

4 L3Ε A L2-6 Ι

2 2 L3

3 -4+ 2 Ε Ι

2 L2-

Ε A L2+6 Ι

2 2 L3

Ε A L2-6 Ι

2 2 L3

Ε 4+ 2 A L2+6 2 Ι

4 L3- 3 Ε Ι

2 L2-

Ε A L2-6 Ι

2 2 L3

3 -4+ 2 Ε Ι

2 L2- 3 Ε Ι

2 L2

2 2+ 2 Ε Ι

L- 3 Ε Ι

2 L2

-Ε A L2+6 Ι

2 2 L3-

Ε A L2-6 Ι

2 2 L3- 3 Ε Ι

2 L2

Ε 4+ 2 A L2+6 2

4 L3

-Ε A L2-6 Ι

2 2 L3-

Ε A L2+6 Ι

2 2 L33 Ε Ι

2 L2

Ε A L2-6 Ι

2 2 L3

3 Ε Ι

2 L2- 3 Ε Ι

2 L22 Ε ΙL

- 3 Ε Ι

2 L2

which yields:

For reference, the DOFs for that stiffness matrix is:

1

3

2

4

6

5

Comparison with stiffness matrix from hand calculation

To enable the comparison we need to reduce Kfinal in two ways. First, we must remove the two displacement-DOFs 2 and 4 because they are locked when axial deformations are neglected:



T1 = {{1, 0, 0, 0},{0, 0, 0, 0},{0, 1, 0, 0},{0, 0, 0, 0},{0, 0, 1, 0},{0, 0, 0, 1}};

K1 = Transpose[T1].Kfinal.T1;MatrixForm[K1] // PowerExpand // Simplify

Ε 2 A L2+6 8+ 2 Ι

4 L33 -4+ 2 Ε Ι

2 L2-

Ε A L2-6 Ι

2 2 L33 Ε Ι

2 L2

3 -4+ 2 Ε Ι

2 L22 2+ 2 Ε Ι

L3 Ε Ι

2 L22 Ε ΙL

-Ε A L2-6 Ι

2 2 L33 Ε Ι

2 L2

Ε 2 A L2+6 8+ 2 Ι

4 L33 -4+ 2 Ε Ι

2 L2

3 Ε Ι

2 L22 Ε ΙL

3 -4+ 2 Ε Ι

2 L22 2+ 2 Ε Ι

L

which yields:

Next, we must enforce a linear coupling of the remaining two displacement-DOFs, namely number 1 and 5 in the figure above:

T2 = {{1, 0, 0},{0, 1, 0},{1, 0, 0},{0, 0, 1}};

K2 = Transpose[T2].K1.T2;Simplify[K2, L > 0] // MatrixForm // N

32.4853 Ε Ι

L3- 1.75736 Ε Ι

L2- 1.75736 Ε Ι

L2

- 1.75736 Ε Ι

L26.82843 Ε Ι

L1.41421 Ε Ι

L

- 1.75736 Ε Ι

L21.41421 Ε Ι

L6.82843 Ε Ι

L

which yields:

That matrix is compared visually with the matrix we obtained earlier by hand calculation:

Simplify[Kclassical, L > 0] // MatrixForm // N

32.4853 Ε Ι

L3- 1.75736 Ε Ι

L2- 1.75736 Ε Ι

L2

- 1.75736 Ε Ι

L26.82843 Ε Ι

L1.41421 Ε Ι

L

- 1.75736 Ε Ι

L21.41421 Ε Ι

L6.82843 Ε Ι

L

which yields:

And just to make sure we did not miss any difference, we let Mathematica calculate the difference; they are the same:



And just to make sure we did not miss any difference, we let Mathematica calculate the difference; they are the same:

K2 - Kclassical

{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}which yields:

Load vector in the Final configuration

Ffinal = P, -q L

2,q L2

12, 0, -q L,

q L2

12-q L2

12;

Introduce values

FfinalNum = Ffinal /. values;FfinalNum // MatrixForm

10-4.52.250

-9.0

which yields:

KfinalNum = Kfinal /. values // N;KfinalNum // MatrixForm

471 799. 419 834. -37 226.7 -433 408. -419 834. 20 360.419 834. 1.64007 × 106 -20 360. -419 834. -433 408. -20 360.-37 226.7 -20 360. 196 613. -20 360. 20 360. 40 719.9-433 408. -419 834. -20 360. 1.64007 × 106 419 834. -20 360.-419 834. -433 408. 20 360. 419 834. 471 799. -37 226.720 360. -20 360. 40 719.9 -20 360. -37 226.7 196 613.

which yields:

Solve the equilibrium equations

uFinal = (Inverse[KfinalNum].FfinalNum) // N

0.0000319065, -0.0000113314, 0.000020219,

8.23709 × 10-6, -0.0000100621, -9.7171 × 10-6which yields:

Compare with solution above: 0.0000114241, 0.0000127685, -1.66442×10-6



Forces in Element 1 (AB) by matrix multiplication

Using this approach, the values that come out are 1) axial force, 2) bending moment at the bottom, 3) bending moment at the top:

Fbasic1 = Kbasic.Tbl.Tlg1.Tga1.Taf.uFinal /. values

{-13.6733, -0.673046, 0.491297}which yields:

Forces in Element 2 (BC) by matrix multiplication


{-19.1127, 2.78462, 1.0607}which yields:

Forces in Element 3 (CD) by matrix multiplication


{-9.93942, -0.539708, 0.0198676}which yields:

Adding “particular solution”

Note that the fixed end moment from distributed element loads are included in the slope-deflection equation used earlier. However, they are not automatically included in the formula F=Ku use a few lines above. This means that not only the “zero to zero parabola” must be added to those end moments, but the full fixed-fixed bending moment diagram for the member. If we used the slope-deflection equations then the fixed-end moments would be included in the slope-deflection equation, but it is not included in F=Ku.

The fixed-end moments are:

q L2

12/. values

2.25which yields:

The maximum amplitude of the parabola is:



q L2

8/. values

3.375which yields:

Effect of neglecting axial forceThe solutions presented above for the hand calculation approach and the transformation matrix approach are somewhat different. The reason is that we neglected axial deformations in the first approach, and included them in the second. To compare, we resolve the system of equations in the second approach, leaving the cross-sectional area as a free variable:

Knew = Kfinal /. A → area /. values // N;uNew = (Inverse[Knew].FfinalNum) // N;F1new = Kbasic.Tbl.Tlg1.Tga1.Taf.uNew /. A → area /. values;

Here is a comparison of the horizontal displacement at B; it is observed that the solutions converge when the cross-sectional area grows bigger, i.e., when the axial deformations become smaller:

0.2 0.3 0.4 0.5Cross-section area

0.000015

0.00002

0.000025

0.00003

Displacement

Without axial deformation

With axial deformation

The same observation is made with regards to the bending moment at A:



0.2 0.3 0.4 0.5Cross-section area

-0.6

-0.4

-0.2

Bending moment at A

Without axial deformation

With axial deformation



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