()
2
e
M
= 0.25 wl 2,
()
3
e
y
F
= – 0.5 wl,
()
3
e
M
= 1.85 wl2
The equation
158
4.15
Applying the boundary conditions
v1 =
1 = v3 =
3 = 0
wl
2
0
24
v
EI
EI
3
y
2
4
()
1
()
1
()
2
e
y
e
e
y
F
M
F
4
22
24
0
6 6 0 0
12 12
0
4 6 2 0 0
06
24 12
wl
EI
ll
l l l
EI l
4 12
3
12
F2y = – wl – (– wl) = 0, M2 = 0 – 0 = 0
wl wl wL
2 2 2
wl wl wL
24 24
EI EI
2384
4.16
After applying the boundary conditions
v1 = v3 =
2 = 0 we have
2
2
–
48
2
48
wL
wL
wL
=
2
1
2
3
2
3
30
33
24
03
L
L
EI L L v
LLL
3
1
4
2
3
3
24
5
384
24
wL
EI
wL
vEI
wL
EI
161
1
1
3
3
y
y
F
M
F
M
22
22
–
44
11
––
48 48
–
4433
48 48
,0
2
,0
2
wL
wL
y
wL wL
wL wL
y
wL wL
wL
FM
wL
FM
4.17
Total [K] for the whole beam
4
EI
L
22
22
33
22
2 3 2
33
22
22
33
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
LL
LL
L L L
L L L
LL
LL
LL
LL
LL
{F} = [K] {d}
After applying boundary conditions
2
212
wL
wL
M
32
3
2
22
86
2
LL
L
EI L L v
4.18
Wdistributed =
0( ) ( )
Lw x v x dx
and
Wdiscrete =
1 1 2 2 1 1 2 2yy
m m f v f v
0( ) ( )
Lw x v x dx
=
0
Lwx
L
[a1x3 + a2x2 + a3x + a4]
3
1 2 1 2
32
0
21
Lwx v v x
LLL
2
1 2 1 2 1 1
2
31
( ) (2 )v v x x v dx
L
L
=
4
1 2 1 2
32
0
21
( ) ( )
Lwv v x
LLL
3
1 2 1 2
2
31
( ) (2 )v v x
L
L
2
11
v x v x dx
1 2 1 2 1 2 1 2
22
2 1 3 1
3
ww
L L L
LL
4 3 2
11
00
0
4 3 2
L L L
x w x w x
v
LL
4
1 2 1 2
32
21
5
wL vv
LL
32
1 2 1 2 1
21
31
( ) (2 )
4 3 2
wL wL wL
v v v
L
L
2
23
wL wL wL
4.19
Work equivalent load system
1
22 1
2
0
6 6 0 0
12 12
0
4 6 2 0 0
12 12 6 6 6
12
v
ll
l l l
ll v
EI l
1
1
14
220
(3)
y
wl
y
F
M
F
1
1
2
y
y
F
M
F
2
4
3
20
22 30
77
240 10
0
6 6 0 0
12 12
0
4 6 2 0 0
06
24 12
wl
wl
wl wl
EI
ll
l l l
EI l
4.20
{F0} = [K]{d}
2
3
01 20
01 30
wL
y
wL
F
M
1
22
1
0
66
12 12
0
66
42
v
LL
EI LL
LL
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
After imposing the boundary conditions and using work equivalence
v1 =
1 = v2 = 0, we have in {F} = [K]{d}
2
12
wL
wL
22
2
8 6 2 (1)
L L L
EI L L v
2
EI
wL
EI
2y
3
4
2
Element 2–3
2
2
2
y
f
m
32
2
22
812
0
66
12 12
66
42
wL
wL wL
EI
LL
EI LL
LL
{F} = [K]{d}
2
2
20
6
20
20
wL
wL
wL
=
3
2
1
24
2
3
2
3
60
4(1)
66(2)
0 6 (3)
4
KL
EI
L
L
EI LLv
LLL
since
1 = –
3 we can ignore Equation (3)
Multiplying (1) by 3 and (2) by L and adding we have
2
–3
20
wL
=
3
EI
L
[12L2
1 – 18Lv2]
2
–6
20
wL
=
3
EI
L
4
2
12
12 24 KL
Lv
EI
30.00235 rad
The reactions can be found by the global matrix {F} = [K] {d} – {F0}
1
1
2
y
y
F
M
F
3
22
24
6 6 0 0
12 12 0
6 6 0 0
42 0.00235
0.0081
KL
LL
LL
LL
EI LL
2
7
20
20
6
20
wL
wL
wL
1
2
2
y
f
m
2
33
20
2
30
6 0.0081
12
Symmetry 0
4
wL
wL
LL
L
1 1 2 2
25.9kN, 0, 4.05 kN, 35.6 kN m
yy
f m f m
Element 2–3
2
2
3
3
y
y
f
m
f
m
=
2
2
3
20
22 30
37
20
2
20
0.0081
66
12 12
60
42
60
12
0.00235
4
wL
wL
wL
wL
LL
EI L
LL
LL
L
2 3 3 3
4.05, 35.6 kN m, 25.9 kN, 0
yy
f m f m
Force in spring
FS =
10kN 0.0081 m
m
= – 8.1 kN
4.23
Global stiffness matrix of the beam
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
After imposing the boundary conditions v1 =
1 = v3 = 0 in {F} = [K] {d}
wL
3
22
0 6 (1)
24
v
L
EI LL
1
1
2
2
3
3
y
y
y
F
M
F
M
F
M
22
32
22
22
6 6 0 0 0
12 12
6 6 0 0 0
42
6 0 6 1.2569
12 24 12
6 0 8 6 0.003491
22
0 0 6 6 0
12 12
0 0 6 6 0.01396
24
LL
LL
LL
EI LL
LL L L
LL
LL
LL
LL
–
2
2
2
12
2
12
7500 lb
225,000 lb in.
15,000 lb
0
7500 lb
225,000 lb in.
wL
wL
wL
wL
wL
F2y = 0, M2 = 0
Element 1–2
2
EI
1
1
y
f
m
EI
1
22 1
18750 lb
6 6 0 7500
12 12
1350 k in.
6 6 0 225000
42
y
f
LL
m
LL
LL
[K] =
22
322
2
6 6 0 0
12 12
6 0 0
42
06
24 12
86
2
6
12
Symmetry 4
LL
LL
EI L
LLL
L
L
L
After applying the boundary conditions
v1 =
1 = v2 = 0 in {F0} = [K] {d}
–13333.33 ft lb
222
(1)
86
2
LL
L
v
– 0.441379 = 8L2
2 – 6L (– 2.73103 10–1) + 2L2 (– 3.22758)
–2
21.29655 10 rad
()
1
e
y
2
2
6EI
L
64
2
6(29 10 )(150 in. )
(120)in.
M
()
e
2EI
L
64
2(29 10 )(150 in. )
120 12
2
()
e
12 6EI EI
()
2
e
M
=
22
2 2 3
3 3 3
8 6 2L EI LEI L EI
v
L L L
= – 13333.33 ft lb
()
3
e
y
2 3 3
3 3 3
6 12 6LEI LEI LEI
L L L
()
3
e
M
=
22
2 3 3
3 3 3
2 6 4L EI LEI L EI
v
L L L
= 26,666.67 ft lb
Global forces {F} = {F(e)} – {F0}