Page 328 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 328
Sec. 10.4 Element Stiffness and Element Mass in Global Coordinates 315
We now impose the condition of zero displacement for joints 1 and 3, which
wipes out columns 1, 2, 5, and 6, leaving the equation
-16 - 1 2 '
^ ly -12 - 9
' b ' ‘ ( EA 16 + 31.25 12
- p [251 12 9
- 31.25 0
0 0
The middle two rows are
47.25 12
12 9
which can be inverted to
«2 1 ('25/1 1 9 ■12
v j i £ 4 /2 8 1 .2 5 12 47.25
Thus, the horizontal and vertical deflections of joint 2 are
PI
“2 ( 281.25£4
PI
'^2 = ( 28L 25E 4 EA
With these values, the reaction forces at pins 1 and 3 are
1 EA 1 PI
( 257) 'EA EA = 1.333P
K = l.OOOT
K = -1.333P
0
Of course, these reaction forces are easily found by taking moments about the
fixed pins; however, this example illustrates the general procedure to be followed
for a more complex structure.
