consent of McGraw–Hill Education.
SOLUTION Continued
consent of McGraw–Hill Education.
PROBLEM 11.86
Solve Problem 11.85, assuming that the magnitude of the
force applied at G is increased from 250 lb to 400 lb.
PROBLEM 11.85 For the loading shown, determine (a) the
stress at points A and B, (b) the point where the neutral axis
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
consent of McGraw–Hill Education.
PROBLEM 11.87
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
SOLUTION
Add y– and z–axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
3 3 6 4 64
3 3 3 4 64
3 2 32
11
(75)(125) (51)(101) 7.8283 10 mm 7.8283 10 m
12 12
11
(125)(75) (101)(51) 3.2781 10 mm 3.2781 10 m
12 12
(75)(125) (51)(101) 4.224 10 mm 4.224 10 m
z
y
I
I
A
−
−
−
= − =×=×
= − =×=×
= − =×=×
Resultant force and bending couples:
3
14 28 28 70 kN 70 10 N
(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN) 2625 N m
(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN) 525 N m
z
y
P
M
M
=++= =×
=−++ =⋅
=− + + =−⋅
(a)
3
36 6
70 10 (2625)( 0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yA
zA
Azy
Mz
My
P
AI I
σ
−− −
× −−
=−+= − +
×× ×
6
31.524 10 Pa
= ×
31.5 MPa
A
σ
=
3
36 6
70 10 (2625)(0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yB
zB
Bzy
Mz
My
P
AI I
σ
−− −
×−
=−+= − +
×× ×
6
10.39 10 Pa=−×
10.39 MPa
B
σ
= −
63
36
0.0375 m, ?, 0
0
7.8283 10 70 10 ( 525)(0.0375)
2625 4.224 10 3.2781 10
0.03151 m 31.51 mm
σ
−
−−
= = =
=−+
× ×−
= += +
××
= =
H HH
yH
zH
zy
zH
Hzy
zy
Mz
My
P
AI I
I Mz
P
yMA I
31.51 62.5 94.0 mm+=
Answer: 94.0 mm above point A.
consent of McGraw–Hill Education.
SOLUTION
Add y– and z–axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
3 3 6 4 64
3 3 6 4 64
3 2 32
11
(75)(125) (51)(101) 7.8283 10 mm 7.8283 10 m
12 12
11
(125)(75) (101)(51) 3.2781 10 mm 3.2781 10 m
12 12
(75)(125) (51)(101) 4.224 10 mm 4.224 10 m
z
y
I
I
A
−
−
−
= − =×=×
= − =×=×
= − =×=×
Resultant force and bending couples:
3
14 28 42 kN 42 10 N
(62.5 mm)(14 kN) (62.5 mm)(28 kN) 875 N m
(37.5 mm)(14 kN) (37.5 mm)(28 kN) 525 N m
z
y
P
M
M
=+= =×
=−+ =⋅
=−+ =⋅
(a)
3
36 6
42 10 (875)( 0.0625) (525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yA
zA
Azy
Mz
My
P
AI I
σ
−− −
×−
=−+= − +
×××
6
22.935 10 Pa= ×
22.9 MPa
A
σ
=
3
36 6
42 10 (875)(0.0625) (525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yB
zB
Bzy
Mz
My
P
AI I
σ
−− −
×
=−+= − +
×××
6
8.9631 10 Pa= ×
8.96 MPa
B
σ
=
(b) Let point K be the point where the neutral axis intersects BD.
63
63
?, 0.0625 m, 0
0
3.2781 10 (875)(0.0625) 42 10
525 7.8283 10 4.224 10
0.018465 m 18.465 mm
σ
−
−−
= = =
=−+
××
= −= −
××
=−=−
KK H
yH
zH
zy
yzH
Hyz
zy
Mz
My
P
AI I
IMy P
zMI A
37.5 18.465 56.0 mm+=
PROBLEM 11.89
Knowing that P = 90 kips, determine the largest distance a for
which the maximum compressive stress does not exceed 18 ksi.
SOLUTION
2
3 34
3 34
(5 in.)(6 in.) 2(2 in.)(4 in.) 14 in
11
(5in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
11
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
=−=
=−=
= +=
Force–couple system at C:
(2.5 in.)
xz
P P M P M Pa= = =
For P = 90 kips:
90 kips (90 kips)(2.5 in.) 225 kip in. (90kips)
xz
PM M a= = =⋅=
Maximum compressive stress at B:
18 ksi
B
s
= −
(3 in.) (2.5 in.)
xz
Bxz
PM M
AI I
s
=−− −
24 4
90 kips (225 kip in.)(3 in.) (90 kips) (2.5 in.)
18 ksi 14 in 68.67 in 21.17 in
18 6.429 9.830 10.628
1.741 10.628
a
a
a
⋅
−=− − −
−=− − −
−=−
0.1638 in.a=
PROBLEM 11.90
Knowing that
1.25 in.,a=
determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
ten
10 ksi
s
=
comp
18 ksi
s
=
SOLUTION
2
3 34
3 34
(5in.)(6 in.) (2)(2 in.)(4 in.) 14 in
11
(5in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
11
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
=−=
=−=
= +=
Force–couple system at C: For
1.25 in.,a=
(2.5 in.)
x
P PM P= =
(1.25 in.)
y
M Pa= =
Maximum compressive stress at B:
18 ksi
B
s
= −
24 4
(3 in.) (2.5 in.)
(2.5 in.)(3in.) (1.25 in.)(2.5 in.)
18 ksi 14 in 68.67 in 21.17 in
18 0.0714 0.1092 0.1476
18 0.3282 54.8 kips
xz
Bxz
PM M
AI I
PP P
PPP
PP
=−− −
−=− − −
−=− − −
−= =
s
Maximum tensile stress at D:
10 ksi
D
s
= +
(3 in.) (2.5 in.)
10 ksi 0.0714 0.1092 0.1476
10 0.1854 53.9 kips
xz
Dxz
PM M
AI I
PPP
PP
s
=−+ +
+=− + +
= =
The smaller value of P is the largest allowable value.
53.9 kipsP=
PROBLEM 11.91
A horizontal load P is applied to the beam shown. Knowing that
a=20 mm and that the tensile stress in the beam is not to exceed
75 MPa, determine the largest permissible load P.
SOLUTION
Locate the centroid.
3
Y
8 10
3200
2.5 mm
Ay
A
=
×
=
=
∑
∑
PROBLEM 11.92
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
SOLUTION
SOLUTION Continued
63 33
6
36 6
663
3
For point , 50 mm, 2.5 mm
2.0267 10 100 10 ( 2.5)(100 10 )( 2.5 10 ) 75 10
50 10 3200 10 0.40667 10
2.0267 10 {31.25 1.537 75} 10 1.7111 10 N m
50 10
yx
yx
y
IP My
M Ax y
xA I
M
−−
−− −
−
−
= +− = =−
× × − × −×
= + −×
×× ×
×
= + −×=− × ⋅
×
σ
33
3
(1.7111 10 ) 17.11 10 m
100 10
y
M
aP
×
=−=− = ×
×
17.11 mma=
consent of McGraw–Hill Education.
PROBLEM 11.86
Solve Problem 11.85, assuming that the magnitude of the
force applied at G is increased from 250 lb to 400 lb.
PROBLEM 11.85 For the loading shown, determine (a) the
stress at points A and B, (b) the point where the neutral axis
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
consent of McGraw–Hill Education.
PROBLEM 11.87
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
SOLUTION
Add y– and z–axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
3 3 6 4 64
3 3 3 4 64
3 2 32
11
(75)(125) (51)(101) 7.8283 10 mm 7.8283 10 m
12 12
11
(125)(75) (101)(51) 3.2781 10 mm 3.2781 10 m
12 12
(75)(125) (51)(101) 4.224 10 mm 4.224 10 m
z
y
I
I
A
−
−
−
= − =×=×
= − =×=×
= − =×=×
Resultant force and bending couples:
3
14 28 28 70 kN 70 10 N
(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN) 2625 N m
(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN) 525 N m
z
y
P
M
M
=++= =×
=−++ =⋅
=− + + =−⋅
(a)
3
36 6
70 10 (2625)( 0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yA
zA
Azy
Mz
My
P
AI I
σ
−− −
× −−
=−+= − +
×× ×
6
31.524 10 Pa
= ×
31.5 MPa
A
σ
=
3
36 6
70 10 (2625)(0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yB
zB
Bzy
Mz
My
P
AI I
σ
−− −
×−
=−+= − +
×× ×
6
10.39 10 Pa=−×
10.39 MPa
B
σ
= −
63
36
0.0375 m, ?, 0
0
7.8283 10 70 10 ( 525)(0.0375)
2625 4.224 10 3.2781 10
0.03151 m 31.51 mm
σ
−
−−
= = =
=−+
× ×−
= += +
××
= =
H HH
yH
zH
zy
zH
Hzy
zy
Mz
My
P
AI I
I Mz
P
yMA I
31.51 62.5 94.0 mm+=
Answer: 94.0 mm above point A.
consent of McGraw–Hill Education.
SOLUTION
Add y– and z–axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
3 3 6 4 64
3 3 6 4 64
3 2 32
11
(75)(125) (51)(101) 7.8283 10 mm 7.8283 10 m
12 12
11
(125)(75) (101)(51) 3.2781 10 mm 3.2781 10 m
12 12
(75)(125) (51)(101) 4.224 10 mm 4.224 10 m
z
y
I
I
A
−
−
−
= − =×=×
= − =×=×
= − =×=×
Resultant force and bending couples:
3
14 28 42 kN 42 10 N
(62.5 mm)(14 kN) (62.5 mm)(28 kN) 875 N m
(37.5 mm)(14 kN) (37.5 mm)(28 kN) 525 N m
z
y
P
M
M
=+= =×
=−+ =⋅
=−+ =⋅
(a)
3
36 6
42 10 (875)( 0.0625) (525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yA
zA
Azy
Mz
My
P
AI I
σ
−− −
×−
=−+= − +
×××
6
22.935 10 Pa= ×
22.9 MPa
A
σ
=
3
36 6
42 10 (875)(0.0625) (525)(0.0375)
4.224 10 7.8283 10 3.2781 10
yB
zB
Bzy
Mz
My
P
AI I
σ
−− −
×
=−+= − +
×××
6
8.9631 10 Pa= ×
8.96 MPa
B
σ
=
(b) Let point K be the point where the neutral axis intersects BD.
63
63
?, 0.0625 m, 0
0
3.2781 10 (875)(0.0625) 42 10
525 7.8283 10 4.224 10
0.018465 m 18.465 mm
σ
−
−−
= = =
=−+
××
= −= −
××
=−=−
KK H
yH
zH
zy
yzH
Hyz
zy
Mz
My
P
AI I
IMy P
zMI A
37.5 18.465 56.0 mm+=
PROBLEM 11.89
Knowing that P = 90 kips, determine the largest distance a for
which the maximum compressive stress does not exceed 18 ksi.
SOLUTION
2
3 34
3 34
(5 in.)(6 in.) 2(2 in.)(4 in.) 14 in
11
(5in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
11
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
=−=
=−=
= +=
Force–couple system at C:
(2.5 in.)
xz
P P M P M Pa= = =
For P = 90 kips:
90 kips (90 kips)(2.5 in.) 225 kip in. (90kips)
xz
PM M a= = =⋅=
Maximum compressive stress at B:
18 ksi
B
s
= −
(3 in.) (2.5 in.)
xz
Bxz
PM M
AI I
s
=−− −
24 4
90 kips (225 kip in.)(3 in.) (90 kips) (2.5 in.)
18 ksi 14 in 68.67 in 21.17 in
18 6.429 9.830 10.628
1.741 10.628
a
a
a
⋅
−=− − −
−=− − −
−=−
0.1638 in.a=
PROBLEM 11.90
Knowing that
1.25 in.,a=
determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
ten
10 ksi
s
=
comp
18 ksi
s
=
SOLUTION
2
3 34
3 34
(5in.)(6 in.) (2)(2 in.)(4 in.) 14 in
11
(5in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
11
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
=−=
=−=
= +=
Force–couple system at C: For
1.25 in.,a=
(2.5 in.)
x
P PM P= =
(1.25 in.)
y
M Pa= =
Maximum compressive stress at B:
18 ksi
B
s
= −
24 4
(3 in.) (2.5 in.)
(2.5 in.)(3in.) (1.25 in.)(2.5 in.)
18 ksi 14 in 68.67 in 21.17 in
18 0.0714 0.1092 0.1476
18 0.3282 54.8 kips
xz
Bxz
PM M
AI I
PP P
PPP
PP
=−− −
−=− − −
−=− − −
−= =
s
Maximum tensile stress at D:
10 ksi
D
s
= +
(3 in.) (2.5 in.)
10 ksi 0.0714 0.1092 0.1476
10 0.1854 53.9 kips
xz
Dxz
PM M
AI I
PPP
PP
s
=−+ +
+=− + +
= =
The smaller value of P is the largest allowable value.
53.9 kipsP=
PROBLEM 11.91
A horizontal load P is applied to the beam shown. Knowing that
a=20 mm and that the tensile stress in the beam is not to exceed
75 MPa, determine the largest permissible load P.
SOLUTION
Locate the centroid.
3
Y
8 10
3200
2.5 mm
Ay
A
=
×
=
=
∑
∑
PROBLEM 11.92
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
SOLUTION
SOLUTION Continued
63 33
6
36 6
663
3
For point , 50 mm, 2.5 mm
2.0267 10 100 10 ( 2.5)(100 10 )( 2.5 10 ) 75 10
50 10 3200 10 0.40667 10
2.0267 10 {31.25 1.537 75} 10 1.7111 10 N m
50 10
yx
yx
y
IP My
M Ax y
xA I
M
−−
−− −
−
−
= +− = =−
× × − × −×
= + −×
×× ×
×
= + −×=− × ⋅
×
σ
33
3
(1.7111 10 ) 17.11 10 m
100 10
y
M
aP
×
=−=− = ×
×
17.11 mma=