PROBLEM 16.21
Column ABC has a uniform rectangular cross section and is braced in the xz
plane at its midpoint C. (a) Determine the ratio b/d for which the factor of
safety is the same with respect to buckling in the xz and yz planes. (b) Using
the ratio found in part a, design the cross section of the column so that the
factor of safety will be 3.0 when
4.4P=
kN,
1
L=
m, and
200E=
GPa.
PROBLEM 16.22
Column ABC has a uniform rectangular cross section with
12b=
mm
and
22d=
mm. The column is braced in the xz plane at its midpoint C
and carries a centric load P of magnitude 3.8 kN. Knowing that a factor
of safety of 3.2 is required, determine the largest allowable length L.
Use
200E=
GPa.
SOLUTION
33
cr
2
cr 2cr
( . .) (3.2)(3.8 10 ) 12.16 10 N
e
e
P FS P
EI EI
PL
P
L
ππ
= = ×= ×
= =
Buckling in xz–plane.
cr
eEI
LL P
π
= =
3 3 34
94
11
(22)(12) 3.168 10 mm
12 12
3.168 10 m
I db
−
= = = ×
= ×
99
3
(200 10 )(3.168 10 ) 0.717 m
12.16 10
L
π
−
××
= =
×
Buckling in yz-plane.
cr
222
e
eLEI
L LL P
π
= = =
3 3 34
94
99
3
11
(12)(22) 10.648 10 mm
12 12
10.648 10 m
(200 10 )(10.648 10 ) 0.657 m
212.16 10
I bd
L
−
−
= = = ×
= ×
××
= =
×
π
The smaller length governs.
0.657 mL=
657 mmL=
consent of McGraw–Hill Education.
PROBLEM 16.23
A
W8 21×
rolled–steel shape is used with the support and cable
arrangement shown. Cables BC and BD are taut and prevent
motion of point B in the xz plane. Knowing that
24L=
ft,
determine the allowable centric load P if a factor of safety of 2.2 is
required. Use
6
29 10
E= ×
psi.
SOLUTION
PROBLEM 16.24
Each of the five struts shown consists of a solid
steel rod. (a) Knowing that the strut of Fig. (1) is
of a 0.8–in. diameter, determine the factor of
safety with respect to buckling for the loading
shown. (b) Determine the diameter of each of
the other struts for which the factor of safety is
the same as the factor of safety obtained in part a.
Use
6
29 10 psiE= ×
.
PROBLEM 16.24 (Continued)
PROBLEM 16.25
A steel pipe having the cross section shown is used as a column. Using the
allowable stress design, determine the allowable centric load if the effective
length of the column is (a) 18 ft, (b) 26 ft. Use
36 ksi
Y
s
=
and
6
29 10 psi.E= ×
PROBLEM 16.26
A rectangular structural tube having the cross section shown is used as a
column of 5–m effective length. Knowing that
250 MPa=
Y
σ
and
E = 200 GPa, use allowable stress design to determine the largest centric
load that can be applied to the steel column.
PROBLEM 16.27
Using allowable stress design determine the allowable centric load for a column of 6–m effective length that is
made from the following rolled–steel shape: (a)
W200 35.9×
, (b)
W200 86.×
Use
250 MPa
Y
σ
=
and
200 GPa.E=
.
PROBLEM 16.28
A
W8 31×
rolled–steel shape is used for a column of 21–ft effective length. Using allowable stress design,
determine the allowable centric load if the yield strength of the grade of steel used is (a)
36 ksi,
Y
s
=
(b)
50 ksi.
Y
s
=
Use
6
29 10 psi.E= ×
PROBLEM 16.29
A column having a 3.5–m effective length is made of sawn lumber with a
114 140-mm
×
cross section.
Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is
7.6 MPa
C
σ
=
and the adjusted modulus is
2.8 GPa,E=
determine the maximum allowable centric load for
the column.
PROBLEM 16.22
Column ABC has a uniform rectangular cross section with
12b=
mm
and
22d=
mm. The column is braced in the xz plane at its midpoint C
and carries a centric load P of magnitude 3.8 kN. Knowing that a factor
of safety of 3.2 is required, determine the largest allowable length L.
Use
200E=
GPa.
SOLUTION
33
cr
2
cr 2cr
( . .) (3.2)(3.8 10 ) 12.16 10 N
e
e
P FS P
EI EI
PL
P
L
ππ
= = ×= ×
= =
Buckling in xz–plane.
cr
eEI
LL P
π
= =
3 3 34
94
11
(22)(12) 3.168 10 mm
12 12
3.168 10 m
I db
−
= = = ×
= ×
99
3
(200 10 )(3.168 10 ) 0.717 m
12.16 10
L
π
−
××
= =
×
Buckling in yz-plane.
cr
222
e
eLEI
L LL P
π
= = =
3 3 34
94
99
3
11
(12)(22) 10.648 10 mm
12 12
10.648 10 m
(200 10 )(10.648 10 ) 0.657 m
212.16 10
I bd
L
−
−
= = = ×
= ×
××
= =
×
π
The smaller length governs.
0.657 mL=
657 mmL=
consent of McGraw–Hill Education.
PROBLEM 16.23
A
W8 21×
rolled–steel shape is used with the support and cable
arrangement shown. Cables BC and BD are taut and prevent
motion of point B in the xz plane. Knowing that
24L=
ft,
determine the allowable centric load P if a factor of safety of 2.2 is
required. Use
6
29 10
E= ×
psi.
SOLUTION
PROBLEM 16.24
Each of the five struts shown consists of a solid
steel rod. (a) Knowing that the strut of Fig. (1) is
of a 0.8–in. diameter, determine the factor of
safety with respect to buckling for the loading
shown. (b) Determine the diameter of each of
the other struts for which the factor of safety is
the same as the factor of safety obtained in part a.
Use
6
29 10 psiE= ×
.
PROBLEM 16.24 (Continued)
PROBLEM 16.25
A steel pipe having the cross section shown is used as a column. Using the
allowable stress design, determine the allowable centric load if the effective
length of the column is (a) 18 ft, (b) 26 ft. Use
36 ksi
Y
s
=
and
6
29 10 psi.E= ×
PROBLEM 16.26
A rectangular structural tube having the cross section shown is used as a
column of 5–m effective length. Knowing that
250 MPa=
Y
σ
and
E = 200 GPa, use allowable stress design to determine the largest centric
load that can be applied to the steel column.
PROBLEM 16.27
Using allowable stress design determine the allowable centric load for a column of 6–m effective length that is
made from the following rolled–steel shape: (a)
W200 35.9×
, (b)
W200 86.×
Use
250 MPa
Y
σ
=
and
200 GPa.E=
.
PROBLEM 16.28
A
W8 31×
rolled–steel shape is used for a column of 21–ft effective length. Using allowable stress design,
determine the allowable centric load if the yield strength of the grade of steel used is (a)
36 ksi,
Y
s
=
(b)
50 ksi.
Y
s
=
Use
6
29 10 psi.E= ×
PROBLEM 16.29
A column having a 3.5–m effective length is made of sawn lumber with a
114 140-mm
×
cross section.
Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is
7.6 MPa
C
σ
=
and the adjusted modulus is
2.8 GPa,E=
determine the maximum allowable centric load for
the column.