Truss Analysis: Stiffness Matrix & Displacements

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________________________________________________________________________
PROBLEM (*13.17) A two-bar plane truss has its support at joint 1 that settles vertically by an
amount of u = 15 mm downward when loaded by a horizontal concentrated load P (Fig. P13.17).
Determine:
(a) The global stiffness matrix of each element.
(b) The system stiffness matrix.
(c) The nodal displacements.
(d) The support reactions.
Given: E = 210 GPa, A = 5 x 10-4 m2 , P = 10 kN
*SOLUTION
We have EG PaA==×
210 5 10 42
m
v
2
Table P13.17 Data for the truss of Fig.P13.17
100109042
64.048.036.08.06.013.5351
)( 22
o
o
scscscmLengthElement
θ
( a ) Apply Eq.(13.14):
uv u
11 2
=
64.048.064.048.0
48.036.048.036.0
64.048.064.048.0
48.036.048.036.0
5
][ 1AE
k
uvuv
1133
=
1010
0000
1010
0000
4
][ 2AE
k
( b ) Global Stiffness Matrix
uv uvu
11 223
v
3
=
625.2000625.20
000000
00344.1008.1344.1008.1
00008.1756.0008.1756.0
625.20344.1008.1969.3008.1
00008.1756.0008.1756.0
10][ 7
K
Continued on next slide
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( c )
17
1
0.015
0.756 1.008
10
10,000 1.008 3.969
x
F
v
⎧⎫⎡ ⎤
=
⎨⎬ ⎨
⎢⎥
⎩⎭⎣ ⎦
⎩⎭
7
or
37 1
10 10 (1.008 10 )( 0.015) 3.696 10 v×= × − + × 10.0044vm
=
or
77
11
(0.756 10 )( 0.015) 1.008 10
x
Fv+× 169.4
x
FkN
=
( d ) Support reactions:
2
27
3
3
0.756 1.008 69
1.008 1.344 0.015 92.1
10 0 0 0.0044 0
0 2.625 115.5
x
y
x
y
F
FkN
F
F
−−⎧⎫ ⎡⎤
⎪⎪ ⎢⎥
−− −
⎧⎫
⎪⎪ ⎪ ⎪
⎢⎥
==
⎨⎬ ⎨ ⎬
⎢⎥
⎩⎭
⎪⎪ ⎪ ⎪
⎢⎥
⎪⎪ ⎪ ⎪
−−
⎣⎦
⎩⎭
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
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