Chapter 4
4.1 The degrees of freedom at each node of a truss element correspond to its axial displacement.
4.2 An element with axial force, shear force, and bending moment would be considered a beam
4.3
2
dN
dx
4
dN
dx
x
N1
2
dN
dx
N3
4
dN
dx
0
1
1
0
0.2L
0.896
0.32
0.104
0.4L
0.648
–0.12
0.352
0.6L
0.352
–0.32
0.648
0.8L
0.104
–0.28
0.896
1.0L
0
0
1.00
where by Equation (4.1.7)
3 2 3
1(2 3 )x x L L
138
4.4
(0)v
=
1
v
= a4
(0)dv
dx
=
1
= a3
22
2
2
2
1
–
aL
1
v
a
2
2 1 1
1()
22
L
v L v
= a1
3
1
2L
21
2
LL
2
3
L
21
22
11
a2L2 =
2 2 2 1 1 1 1
22v v L L v L v
12
3 (2 )
Let
2
L
= l
Element 1–2 Element 2–3
3
EI
l
22
22
66
12 12
6 4 6 2
66
12 12
6 2 6 4
ll
l l l l
ll
l l l l
3
EI
l
22
22
66
12 12
6 4 6 2
66
12 12
6 2 6 4
ll
l l l l
ll
l l l l
1
1
2
2
3
3
?
0
0
?
?
y
y
y
F
M
FP
M
F
M
=
1
22 1
2
32 2 2 2
3
22
3
0
6 6 0 0
12 12
?
6 4 6 2 0 0
?
6 0 6
12 24 12
?
6 2 0 8 6 2
0
0 0 6 6
12 12
0
0 0 6 2 6 4
v
ll
l l l l
v
EI ll
ll l l l l
v
ll
l l l l
0
0
22
1
2
3
22
2
4 6 2
2 0 8
l l l
EI lv
lll
Rearrange
0
0
P
=
22 1
22 2
3
2
4 2 6
2 8 0
60
24
l l l
EI ll
lv
l
{d
} = – [k

–1 [k

] {d
}
{d
} =
1
2
=
3
22
22
7
4 2 6
96
2 8 0
Pl
l l l
EI
ll
=
3
2
17
0.2857 0.0714 6
96
0.0714 0.1429 0
Pl
l
EI
l
1
2
2
2
22
–
–
32
8
32 128
PL
Pl
EI
EI
Pl PL
EI EI
Substituting back in the global matrix equation we have
1
1
2
2
3
3
y
y
y
F
M
F
M
F
M
=
2
3
2
–
22 8
–7
96
32 2 2
32
22
0
6 6 0 0
12 12
6 4 6 2 0 0
6 0 6
12 24 12
6 2 0 8 6 2
0 0 6 6
12 12
0
0 0 6 2 6 4
0
Pl
EI
Pl
EI
Pl
EI
ll
l l l l
EI ll
ll l l l l
ll
l l l l
F1y =
3 3 3 3
1
33
6 7 6 10 5
8 8 32 32 16
y
EI Pl Pl Pl EI Pl P
F
EI EI EI EI
ll
20M
4.6
22
3
2
66
12 12
6
42
6
12
Symmetry 4
LL
EI L
LL
LL
L
Boundary conditions
v2 =
2 = 0
0
P
=
2
1
1
321
6
12
2
64
v
EI PL
L
EI
LLL
3
13
PL
vEI
F1y =
32
312 6
32
EI PL PL
L
EI EI
L
F1y = – P
Similarly M1 = 0
F2y = P
M2 = – PL
4.7
322
2
86
2
6
12
Symmetry 4
LLL
L
L
L
E = 30 106, I = 200 in.4, L = 20 ft = 240 in.
{F} = [K] {d}
1
1
2
2
3
3
10
0
?
0
?
?
y
y
y
F
M
F
M
F
M
= [K]
1
1
2
2
3
3
v
v
v
where v2 = v3 =
3 = 0
500
0
0





=
1
6
22
1
32
2
2
66
12
30 10 (200) 642
(240) 68
2
L L v
LLL
LL
L










(1)
Solving for the displacements we have
1 = 0.0036 rad,
2 = 0.0012 rad, v1 = – 0.672 in.
Substituting in the equation {F} = [K] {d} we have
1
1
2
2
3
3
y
y
y
F
M
F
M
F
M
=
22
6
22
2
1.344 in.
6 6 0 0
12 12
6 0 0 0.0072
42
30 10 (200) 0 6 0
24 12
(240) 8 6 0.0024
2
60
12
Symmetry 0
4
LL
L
LL
L
LL
L
L
L








F1y = – 500, M1 = 0,
F2y = 1250 lb, M2 = 0, F3y = – 750 lb, M3 = 60 kip-in.
Element 1–2
1
1
2
2
y
y
f
m
f
m
=
22
3
2
6 6 1.344
12 12
6 0.0072
42
60
12
0.0024
4
LL
EI L
LL
LL
L







1y
1
m
= 0
2y
f
= 500 lb
2
m
= – 120,000 lb-in.
2
2
3
f
f
4.8
1
1
2
2
3
3
y
y
y
F
M
F
M
F
M
=
22
32
22
22
6 6 0 0
12 12
6 6 0 0
42
6 0 6
12 24 12
6 0 8 6
22
0 0 6 6
12 12
0 0 6 6
24
LL
LL
LL
EI LL
LL L L
LL
LL
LL
LL
1
1
2
2
3
3
v
v
v
(1)
2
0
500
0
y
F







=
322
33
22
3
0.5 in.
06
24 12
0 8 6 2
66
12 12
66
24
L
EI LL
L
v
LLL
LL
LL
(2)
3
EI
L
=
64
3
(30 10 psi)(200 in. )
(240 in.)
= 434
lb
in.
0
500
0





= 217
22
2
3
22
3
– 0.5 in.
0 8 6 2
66
12 12
66
24
LL
L
LL
v
LL
LL
(3)
Solving (3)
v3 = – 1.922 in.
2 = – 0.004325 rad
3 = – 0.006725 rad
Back substituting into (1)
F1y = – 99 lb
M1 = – 96,250 lb in.
F2y = 599 lb
Element 1
1
1
y
f
m
1
22
1
0
66
12 12
0
66
42
v
LL
EI LL
LL





147
4.9
Figure P4-9
1
1
5000
0
5000
0
y
F
M
3
EI
L
22
2
2 2 2 2
3
22
3
0 0 0
24 12 24 12
8 0 0 0
12 12 4
36 6 6
24 12 12
66
12 4 12 2
0 0 6 6
12 12
0 0 6 6
24
LL
L
L L L
v
LL
L
LL
L L L L
v
LL
LL
LL
Solving the last four equations of (A)
v2 = – 0.315 in.
148
3 = – 0.0135 rad
4.10
22
3
2
6
12
Symmetry 4
LL
L
22
66
12 12
6
42
LL
EI L
LL
66
12 12
6
42
LL
EI L
LL
3
2
6
12
Symmetry 4
LL
L
Boundary conditions
2
20000 N m
M


322
(3) 08
L
– 0.006428 = 24 v2 v2 = – 2.68 10–4 m
1
1
2
2
3
3
y
y
y
F
M
F
M
F
M
=
22
4
94
25
22
22
0
6 6 0 0
12 12
0
6 6 0 0
42
2.68 10
(210 10 )(4 10 ) 6 0 6
12 24 12
36 0 8 6 8.93 10
22
0 0 6 6
12 12 0
0 0 6 6
24
0
LL
LL
LL
LL
L L L
LL
LL
LL
LL









 














F1y = 3.1 106 (– 12 (−2.68 10–4) + 6(3) (8.93 10–5)) F1y = 15000 N
M1 = 3.1 106 (– 6(3) (−2.68 10–4) + 2(3)2 (8.93 10–5)) M1 = 20000 N m
Similarly
F2y = – 20000 N
M2 = 20000 N m
F3y = 5000 N
Element 1–2
1
1
y
f
m
94 22
0
66
12 12
0
(210 10 )(4 10 ) 66
42
LL
LL
LL




 
4.11
Using symmetry
12
66
12
27 9 27 9
6
42
3 9 3
6
12
27 9
4
3
Symmetry
12
12 6 6
88
44
63
4
2 4 2
6
12
84
4
2
Applying the boundary conditions
v1 = v2 =
3 = 0 we have
1
0
M


42
33 1
–3
10
22
0
1
1
2
2
y
y
f
m
f
m
= (70 109) (1 10–4)
4
4 2 2
9 3 9 3
2
2 4 2 3
3333
4 2 2
4
9 3 9 3 3
2
2 2 4
3333
0
1.904 10
0
3.809 10
















1y
f
= – 8890 N
1
m
= 0
2
2
3
0 = – 6v2 + 4(240)
2 v2 = 160
2
– 4000 = 419.56 [(12 + (2.38) 160
2 – 6(240)
2]
2 = – 0.01106 rad
v2 = 160(– 0.001106) v2 = – 1.772 in.
Beam element
1
1
y
f
m



622
6 6 0
12 12
(29 10 )(200) 6 6 0
42
LL
LL
LL

6
6
1 1 2 2
3
3 3 3
2 2 2 2 1
3
322
vv
ll
v
EI ll
2
3
22
3
12 12
24 24
12 12
48
v
ll
ll
ll
2
2
2
3
3
2
4
2
12
wl
wl
wl
wl
wl
wl
=
1
22 1
2
32 2 2 2
3
22
3
0
1.5 1.5 1.5 1.5 0 0
0
1.5 2 1.5 0 0
?
1.5 15 25.5 10.5 12
24
?
1.5 10.5 10 12 4
?
0 0 12 12
24 24
0
0 0 12 4 12 8
v
ll
l l l l
v
EI l l l
ll l l l l l
v
ll
l l l l
Adding third row equation to fifth row equation we have
33
22
l wl wl
EI
= 25.5v2 + 10.5l
2 – 24v3 – 24v2 – 12l
2 + 24v3

4
2wl
EI
= 1.5v2 – 1.5l
2
5
4
3
2
EI
v2 =
4
4
3
wl
EI
+ l
2 (A)
Multiplying fourth row equation by –2 and third row equation by l and adding we have
32
3
l wl wl