sm13 6

________________________________________________________________________
PROBLEM (*13.6) A stepped aluminum bar 1-5 is fixed at the left end and given a displacement
Δ
at the right end as depicted in Fig. P13.6. Determine:
(a) The system stiffness matrix.
(b) The displacements of nodes 2, 3, and 4.
(c) The nodal forces and reaction at support.
*SOLUTION
We have
AE L the same for all elements. Thus, Eq. (13.1):
1234
11
[] [] [] [] 11
AE
kkkk L
−
⎡
⎤
====
⎢
⎥
−
⎣
⎦
(a) By superposition, we have
11000
12100
[] 01210
00121
00011
AE
KL
−
⎡⎤
⎢⎥
−−
⎢⎥
⎢⎥
=−−
⎢⎥
−
−
⎢⎥
⎢⎥
−
⎣⎦
(b)
11
22
33
44
55
11000
12100
01210
00121
00011
x
x
x
x
x
Fu
Fu
AE
Fu
L
Fu
Fu
−
⎧⎫ ⎧
⎡⎤
⎪⎪ ⎪
⎢⎥
−−
⎪⎪ ⎪
⎢⎥
⎪⎪ ⎪
⎢
=−−
⎨⎬ ⎨
⎢⎥
⎪⎪ ⎪
−−
⎢⎥
⎪⎪ ⎪
⎢⎥
−
⎪⎪ ⎪
⎣⎦
⎩⎭ ⎩
⎫
⎪
⎪
⎪
⎥
⎬
⎪
⎪
⎪
⎭
(1)
Boundary conditions: 0u1
=
and u5Δ. We have 234 0
xxx
FFF
=
==.
=
Hence,
2
3
4
0
012100
20 01210
0 00121
u
AE u
Lu
⎧
⎫
⎪
⎪
−−
⎧⎫ ⎡ ⎤
⎪
⎪
⎪⎪ ⎪⎪
⎢⎥
=−−
⎨⎬ ⎨⎬
⎢⎥
⎪⎪ ⎪⎪
⎢⎥
−−
⎩⎭ ⎣ ⎦
⎪
⎪
Δ
⎪
⎪
⎩⎭
Continued on next slide
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This equation may be rewritten, after transposing the product of the appropriate
stiffness coefficients by the known displacement (
Δ
) to the left side. In so doing,
2
3
4
0210
0121
012
u
AE u
L
AE u
L
⎧⎫
⎪⎪ −
⎡⎤
⎪⎪ ⎧⎫
⎪
⎪
⎢⎥
=− −
⎨⎬ ⎨
⎢⎥
⎪⎪ ⎪
⎢⎥
Δ−
⎣⎦
⎪⎪
⎩⎭
⎬
⎪
⎩⎭
Solving
234
3
42 4
u
ΔΔ Δ
===
uu
(c) Equations (1) give then
1
2
3
4
5
11000 0 4
12100 4 0
01210 2 0
0012134 0
00011 4
x
x
x
x
x
F
FAE AE
FLL
F
F
−−
⎧⎫ ⎡⎤⎧⎫⎧
⎪⎪ ⎢⎥⎪⎪⎪
−− Δ
⎪⎪ ⎢⎥⎪⎪⎪
Δ
⎫
⎪
⎪
⎪ ⎪⎪ ⎪
⎢⎥
==
−− Δ
⎨ ⎬⎨ ⎬
⎢⎥
⎪⎪ ⎪ ⎪⎪ ⎪
−−Δ
⎢⎥
⎪⎪ ⎪ ⎪⎪ ⎪
⎢⎥
−ΔΔ
⎪⎪ ⎪ ⎪⎪ ⎪
⎣⎦⎩⎭⎩⎭
⎩⎭
⎪⎪
⎨⎬
The reaction is
11
4
RAE
L
=Δ←
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.