PROBLEM 12.70
Knowing that the allowable normal stress for the steel used is 24 ksi,
consent of McGraw–Hill Education.
PROBLEM 12.71
Two rolled–steel channels are to be welded back to back and used to
support the loading shown. Knowing that the allowable normal stress
for the steel used is 30 ksi, determine the most economical channels that
can be used.
SOLUTION
0: 12 9(20) (6)(2.25)(3) 0
MA
Σ= −+ + =
consent of McGraw–Hill Education.
PROBLEM 12.72
Two metric rolled–steel channels are to be welded along their edges
and used to support the loading shown. Knowing that the allowable
normal stress for the steel used is 150 MPa, determine the most
economical channels that can be used.
SOLUTION
PROBLEM 12.73
Two L4 × 3 rolled-steel angles are bolted together and used to support
the loading shown. Knowing that the allowable normal stress for the
steel used is 24 ksi, determine the minimum angle thickness that can
consent of McGraw–Hill Education.
PROBLEM 12.73 (Continued)
Angle section
3
(in )S
1
L4 3 2
××
1.87
3
L4 3 8
××
1.44
1
L4 3 4
××
0.988 Smallest allowable thickness:
3in.
8
t
=
consent of McGraw–Hill Education.
PROBLEM 12.74
A steel pipe of 100–mm diameter is to support the loading shown.
Knowing that the stock of pipes available has thicknesses varying
from 6 mm to 24 mm in 3–mm increments, and that the allowable
normal stress for the steel used is 150 MPa, determine the minimum
wall thickness t that can be used.
SOLUTION
max
0: (1)(1.5) (1.5)(1.5) (2)(1.5) 0 6.75 kN m
6.75 kN m
AA A
A
MM M
MM
∑=−− − − = =− ⋅
= = ⋅
( )
363 3 3
max
min 6
all
3 64
min
min min 2 min
2
44
min 2 1max
4 4 4 6 64
1max 2 min
1max
min 2 1max
6.75 10 N m 45 10 m 45 10 mm
150 10 Pa
(50)(45 10 ) 2.25 10 mm
4
44
(50) (2.25 10 ) 3.3852 10 mm
42.894 mm
50 4
M
S
I
S I cS
c
I cc
c cI
c
t cc
σ
π
ππ
−
×⋅
== =×=×
×
= = = ×=×
= −
=− = − ×= ×
=
=−=−2.894 7.106 mm
=
9 mmt=
consent of McGraw–Hill Education.
PROBLEM 12.75
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 170 MPa,
SOLUTION Continued
Shape
33
(10 mm )S
W690 125×
3490
W610 101×
2520 ← Lightest wide flange section:
W610 101×
W530 150×
3720
W460 113×
2390
consent of McGraw–Hill Education.
PROBLEM 12.76
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 24 ksi, select
consent of McGraw–Hill Education.
PROBLEM 12.70
Knowing that the allowable normal stress for the steel used is 24 ksi,
consent of McGraw–Hill Education.
PROBLEM 12.71
Two rolled–steel channels are to be welded back to back and used to
support the loading shown. Knowing that the allowable normal stress
for the steel used is 30 ksi, determine the most economical channels that
can be used.
SOLUTION
0: 12 9(20) (6)(2.25)(3) 0
MA
Σ= −+ + =
consent of McGraw–Hill Education.
PROBLEM 12.72
Two metric rolled–steel channels are to be welded along their edges
and used to support the loading shown. Knowing that the allowable
normal stress for the steel used is 150 MPa, determine the most
economical channels that can be used.
SOLUTION
PROBLEM 12.73
Two L4 × 3 rolled-steel angles are bolted together and used to support
the loading shown. Knowing that the allowable normal stress for the
steel used is 24 ksi, determine the minimum angle thickness that can
consent of McGraw–Hill Education.
PROBLEM 12.73 (Continued)
Angle section
3
(in )S
1
L4 3 2
××
1.87
3
L4 3 8
××
1.44
1
L4 3 4
××
0.988 Smallest allowable thickness:
3in.
8
t
=
consent of McGraw–Hill Education.
PROBLEM 12.74
A steel pipe of 100–mm diameter is to support the loading shown.
Knowing that the stock of pipes available has thicknesses varying
from 6 mm to 24 mm in 3–mm increments, and that the allowable
normal stress for the steel used is 150 MPa, determine the minimum
wall thickness t that can be used.
SOLUTION
max
0: (1)(1.5) (1.5)(1.5) (2)(1.5) 0 6.75 kN m
6.75 kN m
AA A
A
MM M
MM
∑=−− − − = =− ⋅
= = ⋅
( )
363 3 3
max
min 6
all
3 64
min
min min 2 min
2
44
min 2 1max
4 4 4 6 64
1max 2 min
1max
min 2 1max
6.75 10 N m 45 10 m 45 10 mm
150 10 Pa
(50)(45 10 ) 2.25 10 mm
4
44
(50) (2.25 10 ) 3.3852 10 mm
42.894 mm
50 4
M
S
I
S I cS
c
I cc
c cI
c
t cc
σ
π
ππ
−
×⋅
== =×=×
×
= = = ×=×
= −
=− = − ×= ×
=
=−=−2.894 7.106 mm
=
9 mmt=
consent of McGraw–Hill Education.
PROBLEM 12.75
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 170 MPa,
SOLUTION Continued
Shape
33
(10 mm )S
W690 125×
3490
W610 101×
2520 ← Lightest wide flange section:
W610 101×
W530 150×
3720
W460 113×
2390
consent of McGraw–Hill Education.
PROBLEM 12.76
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 24 ksi, select
consent of McGraw–Hill Education.