Chapter 5
5.1
Element (1)
22
3.61 4.79 4 3.61 4.79 4
4.79 6.41 3 4.79 6.41 3
4 3 2000 4 3 1000
–4
0
0.6 0.8 0 0 0 0
0
0.8 0.6 0 0 0 0
0
0 0 1 0 0 0
0.0139
0 0 0 0.6 0.8 0
0
0 0 0 0.8 0.6 0
0 0 0 0 0 1 0.2775 10
3
3
3
2
2
2
4150 lb
2.3 lb
1387.5 lb in.
– 4150 lb
– 2.3 lb
0
x
y
x
y
f
f
m
f
f
m
Equilibrium check
5.2
2
12 I
L
= 0.104,
6I
L
= 12.5,
E
L
= 125,000
0.104 0 12.5
10 0
Symmetry 2000
Element (2)
C(2) = 1, S(2) = 0
10 0 0 10 0 0
0.104 12.5 0 12.5
2000 0 12.5 1000
E = 30 × 106 psi
A = 10 in.2
I = 500 in.4
u1 = v1 =
1 = 0, u4 = v4 =
4 = 0
Global equations {F} = [K] {d} are
2
2
2
10.104 0 12.5 10 0 0
10.104 12.5 0 12.5
4000 0 –12.5 1000
u
v
4000
0
0
(3)
{}f
=
(3) (3) (3)
[ ] [ ] { }k T d
=
3
3
3
4
4
4
1710 lb
1995 lb
205, 000 lb in.
1710 lb
1995 lb
274,000 lb in.
x
y
x
y
f
f
m
f
f
m
Free body diagram of frame
(using local force results)
Check equation
ft
ft
5.3
2
I
Assume channel section, C6 8.2
Element (1)
C = 1 S = 0
12 I
Element (3)
C = cos 0° = 1, S = sin 0° = 0
(3) (4)
2.4 0 0 2.4 0 0
0.0303 1.092 0 0.0303 1.092
52.4 0 1.092 26.2
(1)
1
(1)
1
(1)
1
(1)
2
(1)
2
(1)
2
x
y
x
y
f
f
m
f
f
m
= 4.028 105
2.4 0 0 2.4 0 0
0.0303 1.092 0 0.0303 1.092
52.4 0 1.092 26.2
2.4 0 0
0.0303 1.092
Symmetry 52.4
–9
–3
0
0
0
3.008 ×10
0.402
6.663×10
=
(1)
1
(1)
1
(1)
1
(1)
2
(1)
2
(1)
2
0
2000 lb
106,900 lb in.
0
2000 lb
37060 lb in.
x
y
x
y
f
f
m
f
f
m
Element (2)
1.204 1.196 0.409 1.204 1.196 0.409
1.204 0.409 1.196 1.204 0.409
52.4 0.409 0.409 26.2
–9
–3
3.008 10
0.402
6.66 10
=
(2)
2
(2)
2
(2)
2
(2)
3
(2)
3
(2)
3
x
y
x
y
f
f
m
f
f
m
=
0
0
37060 lb in.
0
0
37060 lb in.
Element (3)
4.028 105
2.4 0 0 2.4 0 0
0.0303 1.092 0 0.0303 1.092
52.4 0 1.092 26.2
2.4 0 0
0.0303 1.092
Symmetry 52.4
–9
–3
3.3 10
0.402
6.66 10
0
0
0
=
(3)
33
(3)
33
(3)
33
(3)
44
(3)
44
(3)
44
xx
yy
xx
yy
ff
ff
mm
ff
ff
mm
=
0
2000 lb
– 37060 lb in.
0
2000 lb
–106900 lb in.
1x
F1y =
(1)
1y
f
= 2000 lb
M1 =
(1)
1
m
= 106900 lb in.
F4x =
(3)
4x
f
= 0
F4y =
(3)
4y
f
= 2000 lb
M4 =
(3)
4
m
= – 106900 lb in.
106900 (3 in.)
Mc
Element (1)
41
~
xx
L
20
44.7
40
44.7
2
12 I
L
= 0.0334,
6I
L
= 8.949,
E
L
= 55.93
Global equations
0
1500 kip in.
582 0 896
Symmetry 517900
4
4
4
u
Solving
u4 = 0.445 10–2 in., v4 = – 0.123 10–1 in.
4 = – 0.290 10–2 rad
Element forces
0
Element (1) [T] {d}
447 0 0 477 0 0
0 1.868 500.5 0 1.868 500.5
0 500.5 179000 0 500.5 89490
447 0 0 447 0 0
0 1.868 500.5 0 1.868 500.5
0 500.5 89490 0 500.5 179000
–2
–2
00
00
00
0.90193 10 0
0
0.94808 10
0
0.2895 10
1
1
1
4
4
4
x
y
x
y
f
f
m
f
f
m
=
4.04 kip
1.43 kip
254 kip in.
4.04 kip
1.43 kip
513 kip in.
Element (2)
Similarly
2
2
2
4
4
4
x
y
x
y
f
f
m
f
f
m
=
5.82 kip
–1.45 kip
– 260 kip in.
– 5.82 kip
1.45 kip
– 519 kip in.
Element (3)
{}f
4
4
4
3
3
3
x
y
x
y
f
f
m
f
f
m
=
–1.78 kip 0
1.17 kip 10
– 468 kip in. –1500
1.78 kip 0
–1.17 kip 10
– 236 kip in. 1500
=
1.78 kip
8.83 kip
1032 kip in.
1.78 kip
11.17 kip
1736 kip in.
Free-body diagrams of each element
216
Equilibrium at node 4
M4 = – 513 kip in. – 519 kip in. + 1032 kip in. 0
Fy = sin 26.57° (1.45 – 1.43) – 8.83 + sin 63.43° (4.04 + 5.82) 0
Reactions
Support node 1
217
Fy = 4.04 sin 63.43° – 1.43 sin 26.57° = 2.96 kip
Fx = 4.04 cos 63.43° + 1.43 cos 26.57° = 31 kip
M = 254 kip in.
Reactions support node 2
Fx = 1.45 cos 26.57° – 5.82 cos 63.43° = – 1.31 kip or
Reactions support node 3
Already in global x–y directions
Fy = 11.17 kip, Fx = 1.78 kip , M = 1736 kip in.
5.5
Element 1–2 (1)
2
12 I
L
E
L
6I
L
After imposing the boundary conditions u1 = v1 =
1 and u3 = v3 =
3 = we have
214.2 319.8 160
2
12 I
L
Assembled global equations.
0
1200
1047.5 319.8 160
122134.4
2
2
2
u
Solving
u2 = 0.0562 in.
Element forces
Element (1)
0
0
0
0.1179 in.
0.1462 in.
0.00965 rad
[k ] [T] {d} –{f 0} =
693.38 0 0 693.38 0 0
0.8875 192 0 0.8875 192
55380 0 192 27690
221
Replacement (Equivalent) force system
Element 1–2 (1)
2
12 I
L
E
L
3
kip
in.
6I
L
Since u1 = v1 =
1 = 0
2 2 2
442.82 441.06 220.97
E
L
uv
0 3.01 434.03
0 434.03 83333.28
Element 4 –2 (3)
12 I
6I
E