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Large axial forces - Poznań

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POZNAŃ UNIVERSITY OF TECHNOLOGY<br />

INSTITUTE OF STRUCTURAL ENGINEERING<br />

DIVISION OF STRUCTURAL MECHANICS<br />

EXERCISE NO. 3<br />

INITIAL STABILITY AND STATIC ANALYSIS WITH LARGE AXIAL<br />

FORCES FOR PLANE FRAMES – MATRIX METHOD<br />

Name<br />

Academic year<br />

________________________________<br />

________________________________<br />

For the frame No. ___ :<br />

1. Find the distribution of <strong>axial</strong> <strong>forces</strong> in the frame subjected to the reference loading (λ = 1) using any of the first-order<br />

methods of static analysis.<br />

2. Calculate the critical loading factor λ cr using the discretisation 1 beam – 1 element. Draw the buckling mode.<br />

3. Calculate nodal displacements and internal <strong>forces</strong> taking into account the second-order effects for the loading corresponding<br />

to ____ % of the calculated critical loading. Carry out 1 iteration.<br />

Geometric stiffness matrices for beam elements<br />

1. Clamped-clamped beam 2. Hinged-clamped beam<br />

~<br />

K<br />

g<br />

N<br />

=<br />

30l<br />

⎡0<br />

0 0 0 0 0 ⎤<br />

⎢<br />

⎥<br />

⎢<br />

0 36 3l<br />

0 − 36 3l<br />

⎥<br />

⎢<br />

2<br />

2<br />

0 3l<br />

4l<br />

0 − 3l<br />

− l ⎥<br />

⎢<br />

⎥<br />

⎢0<br />

0 0 0 0 0 ⎥<br />

⎢<br />

⎥<br />

⎢<br />

0 − 36 − 3l<br />

0 36 − 3l<br />

⎥<br />

⎢<br />

2<br />

2⎥<br />

⎣0<br />

3l<br />

− l 0 − 3l<br />

4l<br />

⎦<br />

~<br />

K<br />

g<br />

N<br />

=<br />

30l<br />

⎡0<br />

0<br />

⎢<br />

⎢<br />

0 36<br />

⎢0<br />

0<br />

⎢<br />

⎢0<br />

0<br />

⎢0<br />

− 36<br />

⎢<br />

⎢⎣<br />

0 6l<br />

0<br />

0<br />

0<br />

0<br />

0<br />

0<br />

0 0 0 ⎤<br />

0 − 36 6l<br />

⎥<br />

⎥<br />

0 0 0 ⎥<br />

⎥<br />

0 0 0 ⎥<br />

0 36 − 6l<br />

⎥<br />

⎥<br />

2<br />

0 − 6l<br />

6l<br />

⎥⎦<br />

3. Clamped-hinged beam<br />

~<br />

K<br />

g<br />

⎡0<br />

0 0 0 0<br />

⎢<br />

⎢<br />

0 36 6l<br />

0 − 36<br />

⎢<br />

2<br />

N 0 6l<br />

6l<br />

0 − 6l<br />

= ⎢<br />

30l<br />

⎢0<br />

0 0 0 0<br />

⎢<br />

⎢<br />

0 − 36 − 6l<br />

0 36<br />

⎢⎣<br />

0 0 0 0 0<br />

0 ⎤<br />

0<br />

⎥<br />

⎥<br />

0⎥<br />

⎥<br />

0 ⎥<br />

0<br />

⎥<br />

⎥<br />

0 ⎥⎦


1 2 3<br />

2 2<br />

2<br />

2<br />

1 1<br />

1<br />

1<br />

k = ___ EA1<br />

1<br />

1<br />

2<br />

4 5 6<br />

2 2<br />

2<br />

1<br />

2<br />

1<br />

1<br />

1<br />

2<br />

2<br />

1<br />

1<br />

7 8 9<br />

2 2<br />

2<br />

2<br />

2<br />

1<br />

k = ___ EA1<br />

1<br />

1<br />

2<br />

1<br />

1<br />

10 11 12<br />

2 2<br />

2 2<br />

2<br />

1<br />

1 1<br />

1<br />

1<br />

1<br />

1<br />

13 14<br />

2<br />

1<br />

1<br />

2<br />

1<br />

2<br />

2<br />

1<br />

Element cross-sections:<br />

1 − ____________________<br />

2 − ____________________

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