Chapter 3
3.1
(a)
(1)
k
=
11
1
11
11
AE
L
(2) 22
2
11
11
AE
(3)
k
=
33
3
11
ˆ
11
AE
L
1 1 1 1
11
1 1 1 1 2 2 2 2
1 1 2 2
–
––
00
0
A E A E
LL
A E A E A E A E
L L L L
u2 =
1
3
PL
AE
AE
1
PL
2
PL
3
AE
6
3(1)(10 10 )
u2 = 6.66 10–4 in.
2
3
PL
AE
u3 = 13.34 10–4 in.
L
L
L
3
AE
3P
F4x =
2
3
(2000) F4x = – 1333.4 lbs
Going back to the local system and substituting
2
x
(1)
1x
L
10
(1)
1x
= – 666.6 psi (C)
(1)
2x
=
2
Eu
L
=
6
10 10
10
(6.66 10–4)
(1)
2x
= 666.6 psi (T)
3
3
x
f
xA
–4
3
11 13.34 10
AL u
E
L
6
10 10
10
(2)
2x
E
L
6
10 10
10
(2)
2x
= – 666.6 psi (C)
(2)
3x
E
L
6
10 10
10
(2)
3x
= 333.33 psi (T)
2
x
f
4(1)
2
11 1.19 10 10000 N
x
f
Element 2–3
2
3
x
x
f
f
= 84 106
(2)
4
2
(2)
4
3
10000 N
1 1 1.19 10
11 10000 N
2.38 10
x
x
f
f
4
4
0
2.38 10
110000 N
x
3.3
[k1–2] = 3 106
11
11
11
2
x
f
3
f
11
11
11
[k1–2] = 4 106
11
11
11
11
11
11
22
11
22
Global [K] = 5 106
–1
3
22
11
22
1 1 0
1
0
[k2–3] = [k2–4] = 106
0.667 0.667
0.667 0.667
f
4
x
x
f
f
1
2
x
x
f
f
= 1.5 106
(1)
1
5(1)
2
0130 lb
11
11
8.70 10 130 lb
x
x
f
f
2
x
f
(2)
5
2
130 lb
1 1 8.70 10
x
f
33
1
2
x
x
f
f
(1)
1
3(1)
2
040 kN
11
11 0.50 10 40 kN
x
x
f
f
2
3
x
x
f
f
= 140 102
(2)
3
2
(2)
3
3
40 kN
1 1 0.50 10
11 40 kN
3.356 10
x
x
f
f
3.9
11
kN
2
3
?
x
x
F
2
3
0 1 1 0.025 m
u
4
5 kN
4.2 10
= 2 u2 – 1(0.025)
u2 = 0.01244 m
0.01244
(1)
2
x
f
(1)
2
1 1 0.01244 522.5 kN
x
f
(2)
2
(2)
3
x
x
f
f
= 4.2 104
(2)
2
(2)
3
527.5 kN
1 1 0.01244
1 1 0.025 527.5 kN
x
x
f
f
3.10
11
11
11
1
2
3
4
?
16kN
0
?
x
x
x
x
F
F
F
F
= 103
1
2
3
4
0
7 7 0 0
?
7 14 7 0
?
0 7 9 2
0
0 0 2 2
u
u
u
u
Element (1)
1
2
x
x
f
f
1
32
0
1 1 13.10 kN
1 1 13.10
1.870 10
x
x
f
f
Element (2)
2
3
x
x
f
f
3
2
33
1 1 1.870 10 2.90 kN
1 1 2.90
1.454 10
x
x
f
f
Element (3)
3
4
x
x
f
f
= 2 103
33
4
1 1 2.90
1.454 10 kN
1 1 2.90
0
x
x
f
f
013.10 kN
[k1–2] = [k2–3] = [k2–4] = [k2–5] = 2.1 107
11
11
u =
0
PL
AE
ln (L + x)
( 1000) (20)
Two elements
A
4
L
= A0
4
1
L
L
= A0
1
14
=
5
4
A0
14
3
4
3
4
3
7
2
11
4L
[k(2)] =
0
2
11
7
11
4L
AE
3
x
P
F
55 1
44
05 5 7 7 2
4 4 4 4
27730
44
0
0
L
u
AE u
u
0
2
L
AE
(
5
4
u1
5
–4
u2) = – P (1)
0
AE
5
5
4
u1 =
0
– 24
35
PL
AE
5 24
PL
u1 = –
6
(1000) (20) 24
35
2(10 10 )
31
uu
1 3 2
2( 2 )u u u
C =
1
2
, S =
1
2
EA
22
22
–C CS C CS
S CS S
1 1 1 1
1 1 1 1
in.
(b)
C = 0, S = 1
6
0000
15 10 1 00
11
15 0000
00
11
[K] =
6
0000
10 00
11
40000
00
11
lb
in.
(c)
3
2
1
2
[K] =
3 3 3 3
4 4 4 4
33
11
64
4 4 4 4
3 3 3 3
4 4 4 4
33
11
4 4 4 4
(210 10 )(4 10 )
3
3 3 3 3
33
11
3 3 3 3
33
11
kN
m
(d)
41
0.883 0.321 0.883 0.321
0.321 0.883 0.321 0.883
0.883 0.321 0.883 0.321
0.321 0.883 0.321 0.883
N
m
3.16 (a)
S = 0.707 u2 = 0.25 in. v2 = 0.75 in.
u
1
1
u
= 0.3536 in.
2
2
(b)
3
2
1
2
u
1
1
2
3
2
1
2
1
u
= 0.433 in.
2
1
3
3
1
2
3.17
u1 = 0.0 u2 = 5.0 mm E = 210 GPa
v1 = 2.5 mm v2 = 3.0 mm A = 10 10–4 m2
L = 3 m
(a) We know that {d} = [T] {d}
00
00
00
00
CS
SC
CS
SC
C = cos 120° = – 0.5, S = sin 120° = 0.866
1
2
u
v
u
v
=
0.0
0.5 0.866 0 0
0.0025
0.866 0.5 0 0
0.005
0 0 0.5 0.866
0.003
0 0 0.866 0.5
1
2
u
v
u
v
=
0.002165
0.00125
0.000098
0.00583
m =
2.165
1.25
0.098
5.830
mm
1
2
u
v
u
v
=
0.866 0.5 0 0 0.0
0.5 0.866 0 0 0.0025
0 0 0.866 0.5 0.005
0 0 0.5 0.866 0.003
1
2
u
v
u
v
=
1.25
2.165
3.03
5.098
mm
3.18
(a)
=
E
L
[– C – S C S]
1
1
2
2
, 45
u
v
u
v
2
2
(b) C =
3
2
, S =
1
2
, E = 210 GPa, L = 3 m, θ = 30°
33
11
0.25
0
6
210 10
1
1
2
2
2
2
0
10
x
y
x
x
y
y
f
f
f
f
f
f
1
1
2
2
u
v
u
v