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Coursework 4

Local Axes For the Elements:

Degrees Of Freedom:

There are 8 unknowns as shown in the figure above.

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Calculation Of Displacements

Section Properties :

Element 1st

Node 2nd

Node θ sinθ cosθ I (m4) E kN/m2 A (m2) l(m)

ac a c 53 0.8 0.6 0.00068 30000000 0.09 10

cb c b 307 -0.8 0.6 0.00137 30000000 0.114 10

ab a b 0 0 1 - 210000000 0.0004 12

cd c d 0 0 1 0.00068 30000000 0.09 6

Local Stiffness Matrices for the Elements:

Local Stiffness Matrix - Element ac

uxa uya θa uxc uyc θc

270000 0 0 -270000 0 0

0 243 1215 0 -243 1215

0 1215 8100 0 -1215 4050

-270000 0 0 270000 0 0

0 -243 -1215 0 243 -1215

0 1215 4050 0 -1215 8100

Local Stiffness Matrix - Element cb

uxc uyc θc uxb uyb θb

342000 0 0 -342000 0 0

0 493 2466 0 -493 2466

0 2466 16440 0 -2466 8220

-342000 0 0 342000 0 0

0 -493 -2466 0 493 -2466

0 2466 8220 0 -2466 16440

Local Stiffness Matrix - Element cd

uxc uyc θc uxd uyd θd

450000 0 0 -450000 0 0

0 1125 3375 0 -1125 3375

0 3375 13500 0 -3375 6750

-450000 0 0 450000 0 0

0 -1125 -3375 0 1125 -3375

0 3375 6750 0 -3375 13500

ab

uxa uxb

7700 -7700

-7700 7700

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Transformation Matrices for the Elements :

Transformatin Matrix - Element ac

uxa uya θa uxc uyc θc

0.6 -0.8 0 0 0 0

0.8 0.6 0 0 0 0

0 0 1 0 0 0

0 0 0 0.6 -0.8 0

0 0 0 0.8 0.6 0

0 0 0 0 0 1

Transformatin Matrix - Element cb

uxc uyc θc uxb uyb θb

0.6 0.8 0 0 0 0

-0.8 0.6 0 0 0 0

0 0 1 0 0 0

0 0 0 0.6 0.8 0

0 0 0 -0.8 0.6 0

0 0 0 0 0 1

Transformation Matrices for elements ab and cd are not needed because they are on the

same axes with the global.

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Global Stiffness Matrices for the Elements :

Transformation Matrix × Local Stiffness Matrix × Transpose Of Transformation Matrix

(Matrix Calculations are made using MATLAB)

Global Stiffness Matrix - Element ac

uxa uya θa uxc uyc θc

97360 129480 -970 -97360 -129480 -970

129480 172890 730 -129480 -172890 730

-970 730 8100 970 -730 4050

-97360 -129480 970 97360 129480 970

-129480 -172890 -730 129480 172890 -730

-970 730 4050 970 -730 8100

Global Stiffness Matrix - Element cb

uxc uyc θc uxb uyb θb

123440 -163920 1970 -123440 163920 1970

-163920 219060 1480 163920 -219060 1480

1970 1480 16440 -1970 -1480 8220

-123440 163920 -1970 123440 -163920 -1970

163920 -219060 -1480 -163920 219060 -1480

1970 1480 8220 -1970 -1480 16440

Local Stiffness Matrix - Element cd

uxc uyc θc uxd uyd θd

450000 0 0 -450000 0 0

0 1125 3375 0 -1125 3375

0 3375 13500 0 -3375 6750

-450000 0 0 450000 0 0

0 -1125 -3375 0 1125 -3375

0 3375 6750 0 -3375 13500

ab

uxa uxb

7700 -7700

-7700 7700

The columns and rows with gray background are not needed when creating the Global Stiffness

Matrix because those displacements are zero.

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Global Stiffness Matrix for the Structure :

Global Structure Matrix

XB θB XC YC θC XD YD θD

131140 -1970 -123440 163920 -1970 0 0 0

-1970 16440 1970 1480 8220 0 0 0

-123440 1970 670800 -34440 2940 -450000 0 0

163920 1480 -34440 393075 4125 0 -1125 3375

-1970 8220 2940 4125 38040 0 -3375 6750

0 0 -450000 0 0 450000 0 0

0 0 0 -1125 -3375 0 1125 -3375

0 0 0 3375 6750 0 -3375 13500

Calculation Of The Displacements by Using The Known Forces On The Nodes :

The uniformly distributed load can be act as two forces and two moments on nodes c and d to

do the same effect on the total structure.

The forces that should be acted would be V1 = V2 = -150 kN, Mc = -150 kNm and Md = 150 kNm

Global Structure Matrix Displacements Force

XB θB XC YC θC XD YD θD

131140 -1970 -123440 163920 -1970 0 0 0 XB 0

-1970 16440 1970 1480 8220 0 0 0 θB 0

-123440 1970 670800 -34440 2940 -450000 0 0 XC 0

163920 1480 -34440 393075 4125 0 -1125 3375 YC -150

-1970 8220 2940 4125 38040 0 -3375 6750 θC -150

0 0 -450000 0 0 450000 0 0 XD 0

0 0 0 -1125 -3375 0 1125 -3375 YD -150

0 0 0 3375 6750 0 -3375 13500 θD 150

Solving the Global Matrix × Displacements = Force by Matlab gives us the Displacements:

Displacements

XB θB XC YC θC XD YD θD

0.0143 0.0234 0.0074 -0.0061 -0.0441 0.0074 -0.6705 -0.1333

meter rad meter meter rad meter meter rad

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Local Displacement Vectors:

= Transpose of Transformation Matrix × Global Displacement Vector Of Element

(Matrix multiplications made my MATLAB)

Global Displacement Vector For Element ac

XA 0 meter

YA 0 meter

θA 0 rad

XC 0.0074 meter

YC -0.0061 meter

θC -0.0441 rad

Local Displacement Vector For Element ac

xa 0 meter

ya 0 meter

θa 0 rad

xc -0.0004 meter

yc -0.0096 meter

θc -0.0441 rad

Global Displacement Vector For Element cb

XC 0.0074 meter

YC -0.0061 meter

θC -0.0441 rad

XB 0.0143 meter

YB 0 meter

θB 0.0234 rad

Local Displacement Vector For Element cb

xc 0.0093 meter

yc 0.0023 meter

θc -0.0441 rad

Xb 0.0086 meter

yb 0.0114 meter

θb 0.0234 rad

Global Displacement Vectors are same with the Local Displacement Vectors for the elements

cd and ab as they are on the same axes.

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Calculating the Local Nodal Forces:

= Local Stiffness Matrix × Local Displacement Vector

Local Forces ac

fxa 118.8 kN

fya -51.2536 kN

ma -166.9653 kNm

fxc -118.8 kN

fyc 51.2536 kN

mc -345.5703 kNm

Local Forces cb

fxc 253.08 kN

fyb -55.5738 kN

mc -555.2939 kNm

fxb -253.08 kN

fyb 55.5738 kN

mb -0.4439 kNm

Local Forces ab

xa -110.1 meter

xb 110.1 meter

Local Forces cd

fxc 0 kN

fyc 149.7375 kN

mc 749,25 kNm

fxd 0 kN

fyd -149,7375 kN

md 149,1750 kNm

We need to subtract the forces we have loaded on the nodes to get the correct nodal forces

for cd:

Local Forces cd

fxc 0 kN

fyc 300 kN

mc 900 kNm

fxd 0 kN

fyd 0 kN

md 0 kNm

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Nodal Forces : (in kN and kN m)

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Axial Force :

Member ab : Dark Gray : +110.01 kN

(in tension)

Member ac : Light Gray : -118.80 kN

(in compression)

Member cb : Light Gray : -253.08 kN

(in compression)

No Axial Force on member cd.

Shear Force :

No shear force on element ab.

Member ac : Light Gray : - 51.26 kN

Member cb : Dark Gray : -55.57 kN

Member cd : Dark Gray : 300.00 kN to a zero value

at the d

Moment Diagram :

Diagram on the tension side of the element.

No Moment on member ab.

Member ac: The moment on node a is in clockwise

direction. At node c it’s rotating the node c in

clockwise direction. There is no external load on the

frame element causing a constant shear force thru

the element and that causing the moment diagram

to be linear.

Member bc: Moment on node b is zero since there is

a roller support there. Moment on node c is bending

the node in clockwise direction.

Member cd : Moment at node c is rotating at counter-clockwise direction. So the total

moment on node c is zero. (The other two moments on this node were in clockwise

direction.) Node d is a free end so the moment there is zero. There is uniformly distributed

load on the frame element causing a non constant – but linear shear force which causes the

moment diagram to be parabolic. The moment is never on the other side of the frame

element since there is no contra flexure.

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Reaction Forces:

Reaction Forces at node b (Roller Support)

Vb = 253.08 × 0.8 + 55.57 × 0.6 = 235.80 kN

Hb = 0 ( Roller Support )

Mb = 0 ( Roller Support )

Reaction Forces at node a

Va = 50 × 6 – 235.80 = 64.20 kN

Ha = 0

Ma = 166.96 kN m

Comments On The Coursework : The main difference between the slope deflection method

and the stiffness matrix method is that in the stiffness matrix method, the axial deflections

are also considered.

With both methods, the displacements are found first, and then the internal forces are

derived from the displacements, using the material and section properties, such as the

Young Modulus, Moment Of Inertia and the area of the section. (Area of the section only for

the stiffness matrix.)