PROBLEM 11.102
Portions of a
11
–
22
in.×
square bar have been bent
to form the two machine components shown.
Knowing that the allowable stress is 15 ksi,
determine the maximum load that can be applied
to each component.
SOLUTION
PROBLEM 11.103
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
SOLUTION Continued
P Pec
This image cannot currently be displayed.
consent of McGraw–Hill Education.
PROBLEM 11.104
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress
σ
m will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
SOLUTION
2 2 2 2 22
32
1 11
12 2 6
Dbd d Db
I
I bd c d bd
c
=+=−
= = =
(a)
σ
m is the minimum when
I
c
is maximum.
22 2 3
22
1 11
()
6 66
13 1
0
66 3
IbD b Db b
c
dI Db b D
db c
= −= −
= −= =
22
12
33
dD D D=−=
2
d
b=
(b)
EI
M
ρ
=
ρ
is maximum when I is maximum,
3
1
12 bd
is maximum, or
26
bd
is maximum.
2 26
()D dd−
is maximum.
25 7
6 80Dd d−=
3
2
dD=
22
31
42
bD D D=−=
3
d
b=
PROBLEM 11.103
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
SOLUTION Continued
P Pec
This image cannot currently be displayed.
consent of McGraw–Hill Education.
PROBLEM 11.104
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress
σ
m will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
SOLUTION
2 2 2 2 22
32
1 11
12 2 6
Dbd d Db
I
I bd c d bd
c
=+=−
= = =
(a)
σ
m is the minimum when
I
c
is maximum.
22 2 3
22
1 11
()
6 66
13 1
0
66 3
IbD b Db b
c
dI Db b D
db c
= −= −
= −= =
22
12
33
dD D D=−=
2
d
b=
(b)
EI
M
ρ
=
ρ
is maximum when I is maximum,
3
1
12 bd
is maximum, or
26
bd
is maximum.
2 26
()D dd−
is maximum.
25 7
6 80Dd d−=
3
2
dD=
22
31
42
bD D D=−=
3
d
b=