N
N
1
2
3
4
?
450 N
0
?
x
x
x
x
F
F
F
F
1 1 0 0
1 4 3 0
0 3 4 1
0 0 1 1
1
2
3
4
0
?
?
0
u
u
u
u
0 = – 3 u2 + 4 u3 u2 =
4
3
u3 u2 = 1.33 u3
450 N = 40000 (1.33 u3) – 30000 u3
N
1
2
?
0
x
x
F
F
11000
1 2 1 0 0
1
2
0
?
u
u
1
?
x
F
1 1 0
1
0
u
2
x
f
11
0.025
(1)
2
100 N
x
f
Element (2)
22
xx
ff
1 –1
0.025
(2)
2
200 N
x
f
0.025
2.15
[k(1)] =
500 500
500 500
; [k(2)] =
500 500
500 500
; [k(3)] =
1000 1000
1000 1000
1
2
?
?
x
x
F
F
500 0 500 0
0 500 500 0
1
2
0
0
u
u
f
f
Element (2)
2
x
f
500 500
0
2
x
f
1.0 kN
15 0 0
p
x
2.0
1000
p
x
= 2000 x – 1000 = 0 x = 0.5 in. yields minimum
p as table verifies.
(b)
1
p =
1
2
(2000) x2 – 3924 x = 1000 x2 – 3924 x
p
x
= 2000 x – 3924 = 0
1
2
(d)
p =
1
2
(400) x2 – 981 x
p
x
1
2
2.19
1
2
p =
1
2
(500) x2 – 1000 x
p
x
x = 2.0 in.
2.20
F = k
2 (x =
)
dU = F dx
x
p
x
22
3
1e
1
2
1
2
1
2
– f1x(1) u1 – f2x(1) u2 – f2x(2) u2
– f3x(2) u3 – f2x(3) u2 – f4x(3) u4
1
p
u
= – k1 u2 + k1 u1 – f1x(1) = 0 (1)
2
p
u
= k1 u2 – k1 u1 – k2 u3 + k2 u2 – k3 u4
+ k3 u2 – f2x(1) – f2x(2) – f2x(3) = 0 (2)
3
p
u
= k2 u3 – k2 u2 – f3x(2) = 0 (3)
4
p
u
= k3 u4 – k3 u2 – f4x(3) = 0 (4)
In matrix form (1) through (4) become
11
1 1 2 3 2 3
22
33
00
00
00
kk
k k k k k k
kk
kk
1
2
3
4
u
u
u
u
=
(1)
1
(1) (2) (3)
2 2 2
(2)
3
(3)
4
x
x x x
x
x
f
f f f
f
f
(5)
1000 1000 0 0
1000 2000 500 –500
0 500 500 0
0 500 0 500
1
2
3
4
0
0
0
u
u
u
u
=
1
3
4
8000
x
x
x
F
F
F
(6)
For reactions and element forces, see solution to Problem 2.10
2.22 Solve Problem 2.15 by P.E. approach
p =
3
1e
p (e) =
1
2
k1 (u3 – u1)2 +
1
2
k2 (u3 – u2)2
1
