Solution Manual for Applied Strength of Materials (7th Edition, 2022) by Mott and Untener

Covers All 13 Chapters




SOLUTIONS MANUAL

,Chapter 1 Answers
1.1 to 1.15 Basic Concepts in Strength of Materials
in text.

1.16 𝑊 = 𝑚 ∙ 𝑔 = 1800 kg ∙ 9.81 m/s2 = 17 658 (kg ∙ m)/s2 = 17 × 103 N
𝑾 = 𝟏𝟕. 𝟕 𝐤𝐍

1.17 Total Weight = 𝑚 𝑔 = 4000
1
kg ∙ 9.81 m/s2 = 39.24 kN
Each Front Wheel: 𝐹 = ( ) (0.40)(39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
𝐹 2
1
Each Rear Wheel: 𝐹 = ( ) (0.60)(39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
𝑅 2

1.18 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 6800 kg ∙ 9.81 m/s2 = 66.7 kN
Area = (5.0 m)(3.5 m) = 17.5 m2
Loading = 66.7 kN⁄17.5 m2 = 3.81 kN⁄m2 = 𝟑. 𝟖𝟏 𝐤𝐏𝐚
1.19 Force = Weight = 𝑚 𝑔 = 25 kg ∙ 9.81 m/s2 = 245 N

K = Spring Scale = 4500 N⁄m = 𝐹/Δ𝐿
= 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
Δ𝐿 = =
245 N
𝐾 4500 N/m
1.22 𝑊 = 17.7 kN = 17 700 N ∙ 0.2248 (lb⁄N) = 𝟑𝟗𝟖𝟎 𝐥𝐛
1.23 𝐹 = 7.85 kN = 7850 N ∙ 0.2248 (lb⁄N) = 𝟏𝟕𝟔𝟓 𝐥𝐛

𝐹 = 11.77 kN = 11 770 N ∙ 0.2248 (lb⁄N) = 𝟐𝟔𝟒𝟔 𝐥𝐛
3.81×10 𝐥𝐛
1.24 Loading = 3.81 kPa = 3
N 0.2248 lb 1m
2
= 𝟕𝟗
× ×
m2 N (3.28 ft)2 𝐟𝐭𝟐

1.25 𝐹 = 245 N ∙ 0.2248 (lb⁄N) = 𝟓𝟓. 𝟏 𝐥𝐛
𝐥𝐛
4500 N 0.2248 lb 1m = 𝟐𝟓. 𝟕
𝐾= × ×
m N 39.37 in 𝐢𝐧
𝐹 = 𝟐. 𝟏𝟒 𝐢𝐧
Δ𝐿 = =
55.1 lb
𝐾 25.7 (lb⁄in)
lb∙s 2
𝑤 2750 lb = 85.4 = 𝟖𝟓. 𝟒 𝐬𝐥𝐮𝐠𝐬
1.26 𝑚= =
𝑔 32.2 (ft/s2) ft
𝑤 12800 lb lb∙s
2
= 𝟑𝟗𝟖 𝐬𝐥𝐮𝐠𝐬
1.27 𝑚= = = 398
𝑔 32.2 (ft/s2) ft
1.29 𝑝 = 1200 psi ∙ 6.895 (kPa⁄psi) = 𝟖𝟐𝟕𝟒 𝐤𝐏𝐚


1.30 𝜎 = 21 600 psi ∙ 6.895 (kPa⁄psi) = 149 000 kPa = 𝟏𝟒𝟗 𝐌𝐏𝐚




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,1.31 𝑠 = 14 000 psi ∙ 6.895 (kPa⁄psi) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚
𝑠 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
1750 rev 2π rad 1 min 𝐫𝐚𝐝
1.32 𝑛= × × =60s𝟏𝟖𝟑 𝐬
min (25.4rev
mm) 2
1.33 𝐴 = 14.1 in × 2

= 𝟗𝟎𝟗𝟕 𝐦𝐦𝟐
in2
1.34 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
1.35 Dimensions: 18 in × 25.4 (mm/in) = 457 mm
12 in × 25.4 (mm/in) = 305 mm
Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐
Area = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
Volume = 𝑉 = Area × Height
𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉 = (209 × 103 mm2) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑

1.36 𝐴 = 𝜋𝐷2⁄4 = (0.505 in)22⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 mm)
𝐴 = 0.200 in2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2
𝑃 3200 N N
1.37 𝜎= = 3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
=
𝐴 (𝜋𝐷2⁄4) [(10 mm)2]⁄4 mm2
N
𝑃
3
20×10 N = 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
1.38 𝜎= =
𝐴 (10)(30) mm2 mm2
860 lb
1.39 𝜎= = = 𝟓𝟑𝟕𝟓 𝐩𝐬𝐢
𝐴 (0.40 in)2
𝑃 1850 lb = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.40 𝜎= =
𝐴 [(0.375 in)2]⁄4
1.41 Load on Shelf = 𝑊 = 𝑚𝑔 = 1840 kg ∙ 9.81 m⁄s2 = 18 050 N
𝑊/2 = 9025 N On each side
∑ 𝑀 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶 = 4512 N

𝐶 = 𝐶𝑉/ sin 30° = 9025 N
𝐶 9025 N = 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [(12 mm)2]⁄4
𝑃 70000 lb = 𝟏𝟑𝟗𝟑 𝐩𝐬𝐢
1.42 𝜎= =
𝐴 [(8 in)2]/4


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, (29500 lb)/3
1.43 𝜎= = = 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2
𝑃 3500 N
1.44 𝜎= = = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 mm)2
1.45 𝑊 = 𝑚 𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
𝐴𝐵 = 𝐴𝐵 sin 35°
𝐴𝐵 = 𝐴𝐵 cos 35°
𝐵𝐶 = 𝐵𝐶 sin 55°
𝐵𝐶 = 𝐵𝐶 cos 55°
∑ 𝐹 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55°
sin 55°

𝐴𝐵 = 𝐵𝐶 ∙ sin 35° = 1.428 𝐵𝐶
∑ 𝐹 = 0 = 𝐴𝐵𝑌 + 𝐵𝐶𝑌 − 41.2 kN = 𝐴𝐵 cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
0 = (1.428 𝐵𝐶) cos 35° + 𝐵𝐶 cos 55° − 41.2 kN

41.2 kN = 𝐵[1.170 + 0.574] = 1.743 𝐵𝐶
41.2 kN
𝐵𝐶 = = 23.63 kN
1.743

𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
𝐴𝐵 = 33.75×103 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
Stress in Rod AB: 𝜎𝐴 = 𝐴 [(20 mm)2]/4
𝐵𝐶 = 23.63×103 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
Stress in Rod BC: 𝜎𝐵 = 𝐴 [(20 mm)2]/4
𝐵𝐷 = 41.2×103 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
Stress in Rod BD: 𝜎𝐵 = 𝐴 2
[(20 mm) ]/4
1.46 𝐹 = 0.01097 𝑚 𝑅 𝑛 =2 (0.01097)(0.40)(0.60)(3000)2
N

𝐹 = 23 695 N
(16 mm) 2
𝐴=
4 = 201 mm2
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm2




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,1.47 𝐴 = (30 mm)2 = 900 mm2

For AB: 𝐹𝐴 = (110 − 40 + 80) 3kN = 150 kN
150×10 N
𝐹𝐴𝐵
=
𝜎𝐴𝐵 = 𝐴 900 mm2
= 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
For BC: 𝐹𝐵 = 110 − 40 = 70 3kN
𝐹𝐵𝐶
= 70×10 N = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝜎𝐵𝐶 = 𝐴 900 mm2

For CD: 𝐹𝐶 = 110 kN 110×103 N
𝐹𝐶𝐷
=
𝜎𝐶𝐷 = 𝐴 900 mm2
= 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
1.48 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2
C-D; 𝐴2 = 𝜋(16)2/4 = 201 mm2

For AB: 𝐹𝐴 = −9.65 − 12.32 + 4.45 = −17.52 kN
−17.52×103 N
𝐹𝐴𝐵

𝜎𝐴𝐵 = 𝐴1 = 491 mm2
= −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
For BC: 𝐹𝐵 = −9.65 − 12.32 = −21.97 kN
−21.97×103 N
𝐹𝐵𝐶

𝜎𝐵𝐶 = 𝐴1 = 491 mm2
= −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
For CD: 𝐹𝐶 = −9.65 kN −9.65×103 N
𝐹𝐶𝐷
𝜎𝐶𝐷 =
2
𝐴2
2
= 201 mm2
= −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
[(1.90) −(1.61) ] 1
1.49 𝐴= = 0.799 in2 [1 in Pipe-Appendix A-9(a)]
4 2
𝐹𝐵𝐶
= 2500 lb
= 𝟑𝟏𝟐𝟗 𝐩𝐬𝐢 Tension
For BC: 𝜎𝐵 = 𝐴 0.799 in2

For AB: 𝐹𝐴 = 2500 + 2(8000 cos 30°) = 16 356 lb
𝐹𝐴𝐵
= 16 356 lb = 𝟐𝟎 𝟒𝟕𝟏 𝐩𝐬𝐢 Tension
𝜎𝐴𝐵 = 𝐴 0.799 in2
1.50 ∑ 𝑀 = 0 = 2800(45) − 𝐹𝐵𝐷(30)

𝐹𝐵 = 4200 lb
𝐹𝐵𝐷
= 4200 lb = 𝟑𝟐𝟑𝟏 𝐩𝐬𝐢 Tension

𝜎𝐵𝐷 = 𝐴 (2.0)(0.65) in2




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,1.51 𝐴𝐷 sin 30° = 5.25 kN
𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶

Stresses:
9.09×103 N
𝐴𝐵, 𝐵𝐶: 𝜎 =𝜎 = = 𝟐𝟓. 𝟑 𝐌𝐏𝐚 Tension
𝐴𝐵 𝐵𝐶 (12)(30) mm 2
10.5×103 N
𝐵𝐷: 𝜎 = = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
𝐵𝐷 (2)(10)(30) mm 2

𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N

𝜎𝐴𝐷 = 𝜎𝐶𝐷 = 500 mm2 = −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
1.52 ∑ 𝑀 = 0 = 6000(6) + 12 000(12) − 𝑅𝐹(18)
𝑅 = 10 000 lb
∑ 𝑀 = 0 = 12 000(6) + 6000(12) − 𝑅𝐴(18)


𝑅 = 8000 lb



𝑅 = 𝐴𝐵 sin 𝜃 = 𝐴𝐵(0.8)
𝑅 8000
𝐴𝐵 = = = 10 000 lb Compression
0.8 0.8




𝐴𝐷 = 𝐴𝐵 cos 𝜃 = 10 000(0.6) = 6000 lb Tension


𝐵𝐸 sin 𝜃 + 6000 − 𝐴𝐵 sin 𝜃 = 0
𝐴𝐵 sin 𝜃−6000 10 000(0.8)−6000
𝐵𝐸 = = = 2500 lb Tension
sin 𝜃 0.8
𝐵𝐶 = 𝐴𝐵 cos 𝜃 + 𝐵𝐸 cos 𝜃 = 10 000(0.6) + 2500(0.6)
[Continued on next page]




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, 𝐵𝐶 = 7500 lb Compression

𝐵𝐶 = 𝐶𝐹
𝐵𝐶
cos 𝜃7500
𝐶𝐹 = = = 12 500 lb Compression
cos 𝜃 0.6
𝐶𝐸 = 12 000 − 𝐶𝐹 sin 𝜃 = 12 000 − 12 500(0.8)
CE = 2000 lb Compression
EF = CF cos
members: = 12 500A-5(a)
Appendixes lb(0.6)and
= 7500 lb Tension Areas of
A-6(a)
AD, DE, EF – 2(0.484 in2) = 0.968 in2

BD, BE, CE – 0.484 in2
AB, BC, CF – 2(1.21 in2) = 2.42 in2
Stresses:
AD = DE = 6000/0.968 = +6198 psi

EF = 7500/0.968 = +7748 psi

BD = 0

BE = 2500/0.484 = +5165 psi

CE = -2000/0.484 = -4132 psi [NOTE: Compression members must be
AB = -10 000/2.42 = -4132 psi checked for column buckling.]
BC = -7500/2.42 = -3099 psi
CF = -12 500/2.42 = -5165 psi


1.53 𝐴𝐵 ∑= 𝑀
20 =kN03 = (12.5)(4.0) − 𝐴𝐵(2.5)
20×10 N = 𝟓𝟎 𝐌𝐏𝐚
𝜎= 2 2
(20) mm2
(0.505)
1.54 𝐴= = 0.200 in2
4
12 600 lb
𝜎= = = 𝟔𝟑 𝟎𝟎𝟎 𝐩𝐬𝐢
𝐴 0.200 in2

1.55 𝐴 = (2.65)(1.40) + 2[(1.40)(0.5)(𝑡)] = 4.41 in2
52 000 lb
𝜎= = = 𝟏𝟏 𝟕𝟗𝟏 𝐩𝐬𝐢
𝐴 4.41 in2 𝜋(40)2
1.56 𝐴 = (80)(40) − (60)(15) + = 3557 mm2
4
N
640×10
3
= 𝟏𝟖𝟎 𝐌𝐏𝐚
𝜎= =
𝐴 3557 mm2




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,1.57 Direct Shear – Single Shear
(12.0)2
] mm 2 2

𝐴𝑆 = [ 4
3
= 113 mm
𝐹 N
𝜏= 16.5×10 = 𝟏𝟒𝟔 𝐌𝐏𝐚
=
𝐴𝑆 113 mm2
1.58 ∑ 𝐹 = 0 = 55(145) − 𝐹𝑃 (45) Pin is in single
shear
𝐹 = 177 N
(3.0)2
= 7.07 mm 2

𝐴𝑆 = 4
𝐹 177 N = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 7.07 mm2

1.59 From Problem
(10)2
1-46: 𝐹 = 23 695 N
] = 157 mm2 4
𝐴 = 2𝐹[ 23 695 N Double Shear
𝜏= = = 𝟏𝟓𝟏 𝐌𝐏𝐚
𝐴𝑆 157 mm2

1.60 𝐴 = (3.0)(3.5)
𝐹 1800 lb= 10.5 in
2
𝜏= = = 𝟏𝟕𝟏 𝐩𝐬𝐢
𝐴𝑆 10.5 in2

1.61 𝐴 = [2(35) + 𝜋(8)](50) = 475.7 mm2
3
𝐹 N
𝜏= 38.6×10 = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 475.7 mm2
1.62 𝐿 = √0.42 + 0.62 = 0.721 in
(0.8)
+ 2(0.721)] 0.194
𝐴 = [2(1.60) +
2
𝐴 = 1.144
𝐹
in2
45 000 lb
𝜏= = = 𝟑𝟗 𝟑𝟐𝟒 𝐩𝐬𝐢
𝐴𝑆 1.144 in2

1.63 𝑇 = 𝐹𝑆 ∙ 𝑅
95 N∙m 103 mm
𝑇
= 5429 N
𝐹= 𝑅
= 35 mm/2
∙
m

𝐴 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝐹 5429 N = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 220 mm2




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, 𝑇
1.64 = = 8000 lb∙in
𝑆
= 8000 lb
𝑅 1.0 in

𝐴 = 𝑏 ∙ 𝐿 = (0.50)(2.25) = 1.125 in2
𝐹 8000 lb = 𝟕𝟏𝟏𝟏 𝐩𝐬𝐢
𝜏= =
𝐴𝑆 1.125 in2


1.65 Pin: Double
𝐹
Shear; 𝐴 = 2[𝜋(0.5)2/4] = 0.393 in2
20 000 lb
𝜏= = = 𝟓𝟎 𝟗𝟑𝟎 𝐩𝐬𝐢
𝐴𝑆 0.393 in2
Collar: Shear Collar from Connector Body

𝐴 = 𝜋𝑑𝑡
𝐹
= 𝜋(0.875)(0.1875) = 0.5154 in2
𝜏 = = 20 000 lb
𝐴𝑆 0.5154 in2 = 𝟑𝟖 𝟖𝟎𝟎 𝐩𝐬𝐢
1.66 ∑ 𝑀 = 0 = 800(80) − 𝐵𝑉(8)

𝐵 = 8000𝐵lb
𝐵 = cos 20°𝑉
= 8513 lb
(0.375)2
] = 0.221 in2 4
𝐴 = 2𝐵[
𝜏= = 8513 lb

𝐴𝑆 0.221 in2 = 𝟑𝟖 𝟓𝟒𝟎 𝐩𝐬𝐢
1.67 𝐴 = (40)(12) = 480 mm2
3
𝐹 N
𝜏= 88×10 = 𝟏𝟖𝟑 𝐌𝐏𝐚
=
𝐴𝑆 480 mm2
1.68 𝐴 = (40)(120) = 4800 mm2
3
𝐹 N
𝜏= 88.2×10 = 𝟏𝟖. 𝟒 𝐌𝐏𝐚
=
𝐴𝑆 4800 mm2
1.69 𝐴 = 𝜋𝑑𝑡 = 𝜋(12)(8) = 301.6 mm2
3
𝐹 N
𝜏= 22.3×10 = 𝟕𝟑. 𝟗 𝐌𝐏𝐚
=
𝐴𝑆 301.6 mm2
1.70 𝐴 = 2[𝜋(12) /4] = 226.2 mm2 Two Rivets – Single Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟒𝟓. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 226.2 mm2
1.71 𝐴 = 4[𝜋(12) /4] = 452.4 mm2 Two Rivets – Double Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟐𝟐. 𝟓𝟓 𝐌𝐏𝐚
=
𝐴𝑆 452.4 mm2




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