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PROBLEM 16.12
A column of effective length L can be made by gluing together
identical planks in either of the arrangements shown. Determine the
ratio of the critical load using the arrangement a to the critical load
using the arrangement b.
SOLUTION
PROBLEM 16.13
A compression member of 7-m effective length is made by welding
together two
L152 102 12.7
××
angles as shown. Using
200 GPa,E=
determine the allowable centric load for the
member if a factor of safety of 2.2 is required.
SOLUTION
Angle L152 × 102 × 12.7:
2
64 64
3060 mm
7.20 10 mm 2.64 10 mm
50.3 mm 25.3 mm
xy
A
II
yx
=
=×=×
= =
Two angles:
6 64
(2)(7.20 10 ) 14.40 10 mm
x
I= ×= ×
6 2 64
2[(2.64 10 ) (3060)(25.3) ] 9.197 10 mm
y
I= ×+ = ×
6 4 64
min 9.197 10 mm 9.197 10 m
y
II −
==×=×
22 9 6 3
cr 22
(200 10 )(9.197 10 ) 370.5 10 N 370.5 kN
(7.0)
e
EI
PL
ππ
−
××
== =×=
cr
all
370.5
. . 2.2
P
PFS
= =
all 168.4 kNP=
consent of McGraw-Hill Education.
PROBLEM 16.14
A single compression member of 27-ft effective length is obtained by
connecting two C8 × 11.5 steel channels with lacing bars as shown.
Knowing that the factor of safety is 1.85, determine the allowable
centric load for the member. Use
6
29 10 psiE= ×
and
4.0 in.d=
SOLUTION
2
PROBLEM 16.15
A column of 22-ft effective length is to be made by welding two
9 0.5-in.×
plates to a W8 × 35 as shown. Determine the allowable
centric load if a factor of safety 2.3 is required. Use
6
29 10 psi.×
SOLUTION
4
4
PROBLEM 16.16
A column of 3-m effective length is to be made by welding together two C130 × 13
rolled-steel channels. Using
200 GPa,E=
determine for each arrangement shown the
allowable centric load if a factor of safety of 2.4 is required.
SOLUTION
For channel C130 × 13:
2
1700 mm
A=
48.0 mm
f
b=
64
3.70 10 mm
x
I= ×
64
0.260 10 mm
y
I= ×
12.1 mmx=
Arrangement (a):
6 64
(2)(3.70 10 ) 7.40 10 mm
x
I= ×=×
6 2 62
2[0.260 10 (1700)(12.1) ] 1.0178 10 mm
y
I= ×+ = ×
6 64
min 1.0178 10 mm 1.0178 10 m
y
II −
==×=×
2 29 6 3
min
cr 22
(200 10 )(1.0178 10 ) 223 10 N 223 kN
(3.0)
e
EI
PL
ππ
−
××
= = =×=
cr
all
223
. . 2.4
P
PFS
= =
all
93.0 kNP=
Arrangement (b):
6 4 64
(2)(3.70 10 ) mm 7.40 10 mm
x
I=×=×
6 2 64
2[0.260 10 (1700)(48 12.1) ] 4.902 10 mm
y
I= ×+ − = ×
6 4 64
min
4.902 10 mm 4.902 10 m
y
II
−
==×=×
229 6 3
min
cr 22
(200 10 )(4.902 10 ) 1075 10 N 1075 kN
(3.0)
e
EI
PL
ππ
−
××
= = =×=
cr
all
1075
. . 2.4
P
PFS
= =
all
448 kNP=
consent of McGraw-Hill Education.
PROBLEM 16.17
Knowing that
5.2 kN,P=
determine the factor of safety for the structure
shown. Use
200E=
GPa and consider only buckling in the plane of the
structure.
SOLUTION
PROBLEM 16.18
Members AB and CD are 30-mm-diameter steel rods, and members BC
and AD are 22-mm-diameter steel rods. When the turnbuckle is
tightened, the diagonal member AC is put in tension. Knowing that a
factor of safety with respect to buckling of 2.75 is required, determine
the largest allowable tension in AC. Use
200E=
GPa and consider
only buckling in the plane of the structure.
SOLUTION
22
(3.5) (2.25) 4.1608 m
AC
L=+=
Joint C:
2.25
0: 0
4.1608
1.84926
Σ= − =
=
x BC AC
AC BC
FF T
TF
3.5
0: 0
4.1608
Σ= − =
y CD AC
FF T
1.1888
AC CD
TF=
Members BC and AD:
443 4 94
22 11.499 10 mm 11.499 10 m
4 2 42
BC
BC
d
I
ππ
−
= = =×=×
229 9 3
,cr 22
2.25 m
(200 10 )(11.499 10 ) 4.4836 10 N
(2.25)
BC
BC
BC BC
L
EI
FL
ππ
−
=
××
= = = ×
,cr 33
,all ,all
1.6304 10 N 3.02 10 N
..
BC
BC AC
F
FT
FS
==×=×
Members AB and CD:
443 4 94
30 39.761 10 mm 39.761 10 m
4 2 42
CD
CD d
I
ππ
−
= = =×=×
2
229 9 3
,cr 2
3.5 m
(200 10 )(39.761 10 ) 6.4069 10 N
(3.5)
CD
CD
CD
CD
L
EI
FL
ππ
−
=
××
= = = ×
, cr 33
,all ,all
2.3298 10 N 2.77 10 N
..
CD
CD AC
F
FT
FS
==×=×
Smaller value for
,allAC
T
governs.
,all 2.77 kN
AC
T=
consent of McGraw-Hill Education.
PROBLEM 16.19
A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A
and by sets of rollers at B and C that prevent rotation of the strut in the plane of the
figure. Knowing that
3 ft,
AB
L=
4 ft,
BC
L=
and
1 ft,
CD
L=
determine the allowable
load P using a factor of safety with respect to buckling of 3.2. Consider only buckling
in the plane of the figure and use
6
10.4 10 psi.E= ×
SOLUTION
33 4
2
cr 2
2
cr min
all 2
max
11
(1)(1) 0.083333 in
12 12
()
.. ( . .)( )
e
e
I bh
EI
PL
PEI
PFS FS L
π
π
= = =
=
= =
Portion AB:
0.7 (0.7)(3) 2.1 ft
e AB
LL= = =
Portion BC:
0.5 (0.5)(4) 2.0 ft
e BC
LL= = =
Portion CD:
2 (2)(1) 2.0 ft
ee
LL= = =
max
( ) 2.1 ft 25.2 in.
e
L= =
26 3
all 2
(10.4 10 )(0.083333) 4.21 10 lb
(3.2)(25.2 )
P
π
×
= = ×
all 4.21 kipsP=
PROBLEM 16.20
A 1-in.-square aluminum strut is maintained in the position shown by a pin
support at A and by sets of rollers at B and C that prevent rotation of the strut in
the plane of the figure. Knowing that
3
AB
L=
ft, determine (a) the largest
values of
BC
L
and
CD
L
that can be used if the allowable load P is to be as large
as possible, (b) the magnitude of the corresponding allowable load. Consider
only buckling in the plane of the figure and use
6
10.4 10
E= ×
psi.
SOLUTION
33 4
11
(1)(1) 0.083333 in
12 12
I bh
= = =
(a) Equivalent lengths:
AB:
0.7 2.1ft 25.2 in.
e AB
LL= = =
BC:
0.5
e BC
LL=
2.1
0.5
BC
L=
4.20 ft=
BC
L
CD:
2
e CD
LL=
2.1
2
CD
L=
1.050 ft
=
CD
L
(b)
22 6
cr
all 22
3
(10.4 10 )(0.083333)
.. ( . .) (3.2)(25.2)
4.21 10 lb
e
PEI
PFS FS L
ππ
×
= = =
= ×
all
4.21kipsP=
consent of McGraw-Hill Education.
PROBLEM 16.12
A column of effective length L can be made by gluing together
identical planks in either of the arrangements shown. Determine the
ratio of the critical load using the arrangement a to the critical load
using the arrangement b.
SOLUTION
PROBLEM 16.13
A compression member of 7-m effective length is made by welding
together two
L152 102 12.7
××
angles as shown. Using
200 GPa,E=
determine the allowable centric load for the
member if a factor of safety of 2.2 is required.
SOLUTION
Angle L152 × 102 × 12.7:
2
64 64
3060 mm
7.20 10 mm 2.64 10 mm
50.3 mm 25.3 mm
xy
A
II
yx
=
=×=×
= =
Two angles:
6 64
(2)(7.20 10 ) 14.40 10 mm
x
I= ×= ×
6 2 64
2[(2.64 10 ) (3060)(25.3) ] 9.197 10 mm
y
I= ×+ = ×
6 4 64
min 9.197 10 mm 9.197 10 m
y
II −
==×=×
22 9 6 3
cr 22
(200 10 )(9.197 10 ) 370.5 10 N 370.5 kN
(7.0)
e
EI
PL
ππ
−
××
== =×=
cr
all
370.5
. . 2.2
P
PFS
= =
all 168.4 kNP=
consent of McGraw-Hill Education.
PROBLEM 16.14
A single compression member of 27-ft effective length is obtained by
connecting two C8 × 11.5 steel channels with lacing bars as shown.
Knowing that the factor of safety is 1.85, determine the allowable
centric load for the member. Use
6
29 10 psiE= ×
and
4.0 in.d=
SOLUTION
2
PROBLEM 16.15
A column of 22-ft effective length is to be made by welding two
9 0.5-in.×
plates to a W8 × 35 as shown. Determine the allowable
centric load if a factor of safety 2.3 is required. Use
6
29 10 psi.×
SOLUTION
4
4
PROBLEM 16.16
A column of 3-m effective length is to be made by welding together two C130 × 13
rolled-steel channels. Using
200 GPa,E=
determine for each arrangement shown the
allowable centric load if a factor of safety of 2.4 is required.
SOLUTION
For channel C130 × 13:
2
1700 mm
A=
48.0 mm
f
b=
64
3.70 10 mm
x
I= ×
64
0.260 10 mm
y
I= ×
12.1 mmx=
Arrangement (a):
6 64
(2)(3.70 10 ) 7.40 10 mm
x
I= ×=×
6 2 62
2[0.260 10 (1700)(12.1) ] 1.0178 10 mm
y
I= ×+ = ×
6 64
min 1.0178 10 mm 1.0178 10 m
y
II −
==×=×
2 29 6 3
min
cr 22
(200 10 )(1.0178 10 ) 223 10 N 223 kN
(3.0)
e
EI
PL
ππ
−
××
= = =×=
cr
all
223
. . 2.4
P
PFS
= =
all
93.0 kNP=
Arrangement (b):
6 4 64
(2)(3.70 10 ) mm 7.40 10 mm
x
I=×=×
6 2 64
2[0.260 10 (1700)(48 12.1) ] 4.902 10 mm
y
I= ×+ − = ×
6 4 64
min
4.902 10 mm 4.902 10 m
y
II
−
==×=×
229 6 3
min
cr 22
(200 10 )(4.902 10 ) 1075 10 N 1075 kN
(3.0)
e
EI
PL
ππ
−
××
= = =×=
cr
all
1075
. . 2.4
P
PFS
= =
all
448 kNP=
consent of McGraw-Hill Education.
PROBLEM 16.17
Knowing that
5.2 kN,P=
determine the factor of safety for the structure
shown. Use
200E=
GPa and consider only buckling in the plane of the
structure.
SOLUTION
PROBLEM 16.18
Members AB and CD are 30-mm-diameter steel rods, and members BC
and AD are 22-mm-diameter steel rods. When the turnbuckle is
tightened, the diagonal member AC is put in tension. Knowing that a
factor of safety with respect to buckling of 2.75 is required, determine
the largest allowable tension in AC. Use
200E=
GPa and consider
only buckling in the plane of the structure.
SOLUTION
22
(3.5) (2.25) 4.1608 m
AC
L=+=
Joint C:
2.25
0: 0
4.1608
1.84926
Σ= − =
=
x BC AC
AC BC
FF T
TF
3.5
0: 0
4.1608
Σ= − =
y CD AC
FF T
1.1888
AC CD
TF=
Members BC and AD:
443 4 94
22 11.499 10 mm 11.499 10 m
4 2 42
BC
BC
d
I
ππ
−
= = =×=×
229 9 3
,cr 22
2.25 m
(200 10 )(11.499 10 ) 4.4836 10 N
(2.25)
BC
BC
BC BC
L
EI
FL
ππ
−
=
××
= = = ×
,cr 33
,all ,all
1.6304 10 N 3.02 10 N
..
BC
BC AC
F
FT
FS
==×=×
Members AB and CD:
443 4 94
30 39.761 10 mm 39.761 10 m
4 2 42
CD
CD d
I
ππ
−
= = =×=×
2
229 9 3
,cr 2
3.5 m
(200 10 )(39.761 10 ) 6.4069 10 N
(3.5)
CD
CD
CD
CD
L
EI
FL
ππ
−
=
××
= = = ×
, cr 33
,all ,all
2.3298 10 N 2.77 10 N
..
CD
CD AC
F
FT
FS
==×=×
Smaller value for
,allAC
T
governs.
,all 2.77 kN
AC
T=
consent of McGraw-Hill Education.
PROBLEM 16.19
A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A
and by sets of rollers at B and C that prevent rotation of the strut in the plane of the
figure. Knowing that
3 ft,
AB
L=
4 ft,
BC
L=
and
1 ft,
CD
L=
determine the allowable
load P using a factor of safety with respect to buckling of 3.2. Consider only buckling
in the plane of the figure and use
6
10.4 10 psi.E= ×
SOLUTION
33 4
2
cr 2
2
cr min
all 2
max
11
(1)(1) 0.083333 in
12 12
()
.. ( . .)( )
e
e
I bh
EI
PL
PEI
PFS FS L
π
π
= = =
=
= =
Portion AB:
0.7 (0.7)(3) 2.1 ft
e AB
LL= = =
Portion BC:
0.5 (0.5)(4) 2.0 ft
e BC
LL= = =
Portion CD:
2 (2)(1) 2.0 ft
ee
LL= = =
max
( ) 2.1 ft 25.2 in.
e
L= =
26 3
all 2
(10.4 10 )(0.083333) 4.21 10 lb
(3.2)(25.2 )
P
π
×
= = ×
all 4.21 kipsP=
PROBLEM 16.20
A 1-in.-square aluminum strut is maintained in the position shown by a pin
support at A and by sets of rollers at B and C that prevent rotation of the strut in
the plane of the figure. Knowing that
3
AB
L=
ft, determine (a) the largest
values of
BC
L
and
CD
L
that can be used if the allowable load P is to be as large
as possible, (b) the magnitude of the corresponding allowable load. Consider
only buckling in the plane of the figure and use
6
10.4 10
E= ×
psi.
SOLUTION
33 4
11
(1)(1) 0.083333 in
12 12
I bh
= = =
(a) Equivalent lengths:
AB:
0.7 2.1ft 25.2 in.
e AB
LL= = =
BC:
0.5
e BC
LL=
2.1
0.5
BC
L=
4.20 ft=
BC
L
CD:
2
e CD
LL=
2.1
2
CD
L=
1.050 ft
=
CD
L
(b)
22 6
cr
all 22
3
(10.4 10 )(0.083333)
.. ( . .) (3.2)(25.2)
4.21 10 lb
e
PEI
PFS FS L
ππ
×
= = =
= ×
all
4.21kipsP=
consent of McGraw-Hill Education.
