sm13 14

________________________________________________________________________
PROBLEM (13.14) A vertical concentrated load P = 5 kips is applied at joint 2 of the two-bar plane
truss supported as shown in Fig. P13.14. Each member has axial rigidity AE = 6(106 ) lb. Determine:
(a) The global stiffness matrix of each bar.
(b) The system stiffness matrix.
(c) The nodal displacements.
(d) The support reactions.
(e) The axial forces in each bar.
SOLUTION
We have AE=6(106) lb.
Table P13.14 Data for the truss of Fig.P13.14
00101042
64.048.036.08.06.013.5351
)( 22
o
o
scscscftLengthElement θ
( a ) Use Eq.(13.14):
uv u
11 2
v
2
1
0.36 0.48 0.36 0.48
0.48 0.64 0.48 0.64
[] 0.36 0.48 0.36 0.48
5(12)
0.48 0.64 0.48 0.64
AE
k
−−
⎡⎤
⎢⎥
−−
⎢⎥
=⎢⎥
−−
⎢⎥
−−
⎣⎦
uv uv
22 3 3
2
10 10
00 00
[] 10 10
4(12)
00 00
AE
k
−
⎡⎤
⎢⎥
⎢⎥
=⎢⎥
−
⎢⎥
⎣⎦
(b) Global Stiffness Matrix
uvuvu
11223
v
3
0.072 0.096 0.072 0.096 0 0
0.096 0.128 0.096 0.128 0 0
0.072 0.096 0.322 0.096 0.25 0
[] 0.096 0.128 0.096 0.128 0 0
12
0 0 0.250 0 0.25 0
000000
AE
K
−−
⎡⎤
⎢⎥
−−
⎢⎥
⎢⎥
−− −
=⎢⎥
−−
⎢⎥
⎢⎥
−
⎢⎥
⎣⎦
Continued on next slide
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( c ) Boundary Conditions: .0
3311
=
=
=
=
vuvu
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎭
⎬
⎫
⎩
⎨
⎧
2
2
2
2
128.0096.0
096.0322.0
12 v
u
AE
F
F
y
x
2
2
4 3 0 15,000 0.03
12 12 .
3 10.063 5,000 50,375 0.10
uin
vAE AE
−
⎧⎫ ⎡⎤⎧⎫⎧⎫⎧
==
⎨⎬ ⎨ ⎬⎨ ⎬ ⎨ ⎬
⎢⎥
−−− −
⎣⎦⎩⎭⎩⎭⎩
⎩⎭
⎫
=
⎭
⎫
⎪
⎭
( d )
1
1
3
3
0.072 0.096 3.75
0.096 0.128 15,000 5
0.250 0 50,315 3.75
00 0
x
y
x
y
F
Fkips
F
F
−−⎧⎫
⎡⎤⎧
⎪⎪
⎢⎥⎪
−−
⎧⎫
⎪⎪ ⎪ ⎪
⎢⎥
==
⎨⎬ ⎨ ⎬⎨ ⎬
⎢⎥
−−−
⎩⎭
⎪⎪ ⎪ ⎪
⎢⎥
⎪⎪ ⎪ ⎪
⎣⎦⎩
⎩⎭
( e ) Use Eq.(13.15)
[]
12
0.03
0.6 0.8 6.2 ( )
0.10
5(12)
AE
Fk
⎧⎫
==−
⎨⎬
−
⎩⎭ ipsC
[]
23
0.03
1 0 3.75 ( )
0.10
4(12)
AE
Fk
−
⎧⎫
==−
⎨⎬
⎩⎭ ipsC
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.