8. C-13
Example C-3
Determine the design
compressive strength
(c Pn) for W 12 x 65
column shown below,
(Fy = 50 ksi)?
From properties:
Ag =19.1 in2
rx = 5.28 in
ry = 3.02 in
ksi
40.225
50
x
0.8045
F
0.658
F
3.2)
(E
Equ.
ksi)
22
(
F
0.44
F
ksi
96.2
(54.55)
x29000
π
r
Kl
E
π
F
31.79
3.02
12
x
8
x
1
r
L
K
54.55
5.28
12
x
24
x
1
r
L
K
y
96.2
50
cr
y
e
2
2
2
2
e
y
y
y
x
x
x
=
=
=
=
=
=
=
=
=
=
=
Solution:
c Pn = 0.9 x Fcr Ag = 0.9 x 40.225 x 19.1 = 691.5 kips
A) By direct LRFD
(controls)
9. C-14
B) From Table (4.22) LRFD
Evaluate = = 54.55
Enter table 4.22 (page 4 – 318 LRFD)
cFc = 36.235 ksi (by interpolation)
Pn = Fc x Ag = 692.0 kips
C) From (Table 4.1 LRFD)
max
r
Kl
ft
13.7
1.75
1x24
r
r
L
K
(KL)
y
x
x
x
y =
=
=
Enter table (4.1 ) page 4.17 LFRD with (KL)y = 13.7
Pn = 691.3 kips (by interpolation).
10. C-15
Example C-4
Solution:
Find the maximum load capacity
(Pn) of the W 14 x 53 (A-36)
column shown in figure ?
P
P
25 ft.
15 ft.
10 ft.
A
C
A
B
C
x x
x
x
x-axis Lx = 25 ft, kx = 0.8, rx = 5.89 in.
y-axis
Section (AB) Ly = 15 ft, ky = 0.8, ry = 1.92 in.
Section (BC) Ly = 10 ft., ky = 1.0, ry = 1.92 in.
75
1.92
12
15
0.8
r
Kl
41
5.98
12
25
0.8
r
Kl
max
y
x
=
=
=
=
Enter table (4-22) , Fc = 24.1 ksi
Column capacity Pn = Fcr Ag = 24.1 x 15.6 = 376 kips
(controls)
13. C-15
Uso de tablas de la parte 4 del Manual del AISC
Find the maximum load capacity
(Pn) of the W 14 x 53 (A-992)
column shown in figure ?
P
P
25 ft.
15 ft.
10 ft.
A
C
A
B
C
x x
x
x
14.
15. Una columna de 22 pies de largo
en un edificio existente forma
parte de un marco arriostrado
en los dos planos xx y yy de la
sección transversal. Esta
empotrada en la base y articulas
en la parte superior, alrededor
de ambos ejes . Además, la
columna esta soportada
lateralmente en uno de sus
lados de forma perpendicular a
su eje débil a una altura de 12
píes de la base. La columna es
una sección W8x35 de acero
A572 Grado 65. Determine la
resistencia de la columna.
Efectué todas las verificaciones
16.
17.
18.
19. C-23
The effective length factor (K) was introduced in page (C-7) for six ideal conditions,
these are not encountered in practical field conditions. LRFD commentary provides
both real conditions and standard ideal conditions (C-C2.2) (page 16.1-239 to 242)
Braced Frames:
No lateral movement is allowed
(0.5 < K < 1.0) (sideway prevented)
Unbraced Frames:
Lateral movement possible
(1.0 < K < 20.0) (sideway allowed)
a) Diagonal
bracing
b) Shear Walls
(masonry,
reinforcement concrete
or steel plate)
21. C-24
GA = 𝜏𝑎
∑Ic/Lc
∑Ig/Lg
GB = 𝜏𝑎
∑Ic/Lc
∑Ig/Lg
* For fixed footing G = 1.0
* For pinned support G = 10.0
where
A is top of column
where
B is bottom of column
22.
23. C-25
In the rigid frame shown below, Determine Kx for columns
(AB) & (BC). Knowing that all columns webs are in the plane.
Column (AB):
Joint (A):
0.94
169.2
1586
1830/18
1350/20
1070/12
833/12
/L
I
/L
I
G
g
g
c
c
A
=
=
+
+
=
=
Example C – 8 :-
Solution:
A
B
C
W24 x 55
W24 x 55
W24 x 68
W24 x 68
W12
x
120
W12
x
120
W12
x
96
12'
15'
12'
18'
20'
20'
24. C-26
For joint B,:-
0.95
169.2
160.5
169.2
1070/15
1070/12
/L
ΣI
/L
ΣI
G
g
g
c
c
=
=
+
=
=
From the alignment chart for sideways uninhibited, with GA = 0.94 and GB = 0.95,
Kx = 1.3 for column AB.
Column (BC):
For joint B, as before,
G = 0.95
For joint C, at a pin connection the situation is analogous to that of a very
stiff column attached to infinitely flexible girders – that is, girders of zero
stiffness. The ratio of column stiffness to girder stiffness would therefore
be infinite for a perfectly frictionless hinge. This end condition is only be
approximated in practice, so the discussion accompanying the alignment
chart recommends that G be taken as 10.0.
From the alignment chart with GA = 0.95 and GB = 10.0, Kx = 1.85 for column BC.
25. The given frame is unbraced, and bending is about the x axis of each member.
The axial dead load supported by column AB is 155 kips, and the axial live load is
460 kips. Fy = 50 ksi. Determine Kx for member AB. Use the stiffness reduction
factor if aplicable.
Combination 1: 1.4D
Combination 2: 1.2D + 1.6L + 0.5(Lr or S or R)
Combination 3: 1.2D + 1.6(Lr or S or R) + (L or 0.5W)
Combination 4: 1.2D + 1.0W + L + 0.5(Lr or S or R)
Combination 5: 1.2D + 1.0E + L + 0.2S
Combination 6: 0.9D + 1.0W
Combination 7: 0.9D + 1.0E
