sm13 13

________________________________________________________________________
PROBLEM (13.13) A planar truss containing 7 members of axial rigidity AE is supported at joints 1
and 5 as shown in Fig. P13.13. Determine:
(a) The stiffness matrix for each element.
(b) The system stiffness matrix.
(c) The force-displacement equations.
SOLUTION
Table 13.13 Data for the truss of Fig. P13.13.
22 2
1, 4 60 1 2 3 4 1 4 3 4 3 4
2,5 60 123414343
3,6,7 0 1 0 1 0 0
o
o
o
Element Length c s c s c
L
L
L
θ
−−− 4−
(a) For elements 1 and 4:
1,4
13 13
3333
[] 413 13
3333
EA
kL
⎡⎤
−−
⎢⎥
−
−
⎢⎥
=⎢⎥
−−
⎢⎥
⎢⎥
−−
⎣⎦
For elements 2 and 5:
2,5
13 13
3333
[]
413 13
3333
EA
kL
⎡⎤
−−
⎢⎥
−−
⎢⎥
=⎢⎥
−−
⎢⎥
⎢⎥
−−
⎣⎦
For elements 3, 6 and 7:
3,6,7
40 40
00 00
[]
40 40
4
00 00
EA
kL
−
⎡⎤
⎢⎥
⎢⎥
=⎢⎥
−
⎢⎥
⎣⎦
Continued on next slide
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(b, c) We assemble a set of seven of stiffness equation to obtain [K].
Because ,
125
0vuv===
{
}
{
}
[]FK
δ
= become then
1 1
3 2
2 2
3
4 4
4 4
5 5
14 1 3 0 0 0 0
1114 3 3 3 4 0 0
33333 3 0 0 0
03 333330
4040 3114331
00 0 333333
00 00 1 314
x
x
y
x
y
x
Fu
Fu
Fv
EA
Wv
L
Fu
Fv
Fu
⎡⎤
+− −
⎧⎫ ⎧
⎢⎥
⎪⎪ ⎪
−++ − −
⎢⎥
⎪⎪ ⎪
⎢⎥
−−+−
⎪⎪ ⎪
⎢⎥
⎪⎪ ⎪
⎢⎥
=−+− −
⎨⎬ ⎨
⎢⎥
⎪⎪ ⎪
⎢⎥
−−++−−
⎪⎪ ⎪
⎢⎥
⎪⎪ ⎪
⎢⎥
−− +
⎪⎪ ⎪
⎢⎥
⎩⎭ ⎩
−+
⎢⎥
⎣⎦
⎫
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎭
4
The total 7x7 stiffness matrix can be reduced to a 7x4 matrix by using the
following antisymmetrical conditions:
, and
152
,uuuu=− =− 24 vv
=
.
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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