Stiffness Matrix Method

Coursework 4

Local Axes For the Elements:

Degrees Of Freedom:

There are 8 unknowns as shown in the figure above. 1

Calculation Of Displacements Section Properties : Element ac cb ab cd

1st Node a c a c

2nd Node c b b d

θ 53 307 0 0

sinθ 0.8 -0.8 0 0

cosθ 0.6 0.6 1 1

I (m4) 0.00068 0.00137 0.00068

E kN/m2 30000000 30000000 210000000 30000000

Local Stiffness Matrices for the Elements: Local Stiffness Matrix - Element ac uxa

uya

θa

uxc

uyc

θc

270000 0

0 243

0 1215

-270000 0

0 -243

0 1215

0 -270000 0 0

1215 0 -243 1215

8100 0 -1215 4050

0 270000 0 0

-1215 0 243 -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 -342000 0 0

2466 0 -493 2466

16440 0 -2466 8220

0 342000 0 0

-2466 0 493 -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

2

A (m2) 0.09 0.114 0.0004 0.09

l(m) 10 10 12 6

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.

3

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

129480 172890

-970 730

-97360 -129480

-129480 -172890

-970 730

-970 -97360 -129480 -970

730 -129480 -172890 730

8100 970 -730 4050

970 97360 129480 970

-730 129480 172890 -730

4050 970 -730 8100

Global Stiffness Matrix - Element cb uxc

uyc

θc

uxb

uyb

θb

123440 -163920

-163920 219060

1970 1480

-123440 163920

163920 -219060

1970 1480

1970 -123440 163920 1970

1480 163920 -219060 1480

16440 -1970 -1480 8220

-1970 123440 -163920 -1970

-1480 -163920 219060 -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 131140 -1970 -123440 163920 -1970 0 0 0

θB -1970 16440 1970 1480 8220 0 0 0

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

YC 163920 1480 -34440 393075 4125 0 -1125 3375

θC -1970 8220 2940 4125 38040 0 -3375 6750

XD 0 0 -450000 0 0 450000 0 0

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

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

θB -1970 16440 1970 1480 8220 0 0 0

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

YC 163920 1480 -34440 393075 4125 0 -1125 3375

θC -1970 8220 2940 4125 38040 0 -3375 6750

XD 0 0 -450000 0 0 450000 0 0

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

θD 0 0 0 3375 6750 0 -3375 13500

Displacements

Force

XB θB XC YC θC XD YD θD

0 0 0 -150 -150 0 -150 150

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

5

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

fxa fya ma fxc fyc mc

Local Forces ac 118.8 -51.2536 -166.9653 -118.8 51.2536 -345.5703

kN kN kNm kN kN kNm

fxc fyb mc fxb fyb mb

Local Forces cb 253.08 -55.5738 -555.2939 -253.08 55.5738 -0.4439

kN kN kNm kN kN kNm

xa xb

fxc fyc mc fxd fyd md

Local Forces ab -110.1 110.1

meter meter

Local Forces cd 0 149.7375 749,25 0 -149,7375 149,1750

kN kN kNm kN kN kNm

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

fxc fyc mc fxd fyd md

Local Forces cd 0 300 900 0 0 0

7

kN kN kNm kN kN kNm

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.)

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