Applied Strength of Materials – Seventh Edition by Robert L. Mott & Joseph A. Untener – Complete Solution Manual (Chapters 1-14) provides detailed, step-by-step solutions aligned with the main text, covering direct and shear stresses, torsion, bending, de

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SOLUTIONS MANUAL

,Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.15 Answers 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 kg ∙ 9.81 m/s2 = 39.24 kN

Each Front Wheel: 𝐹𝐹 = (1)2(0.40)(39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍

Each Rear Wheel: 𝐹𝑅 = (1) (0.60)(39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
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
Δ𝐿 =
𝐹
=
245 N = 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
𝐾 4500 N/m

1.22 𝑊 = 17.7 kN = 17 700 N ∙ 0.2248 (lb⁄N) = 𝟑𝟗𝟖𝟎 𝐥𝐛
1.23 𝐹𝐹 = 7.85 kN = 7850 N ∙ 0.2248 (lb⁄N) = 𝟏𝟕𝟔𝟓 𝐥𝐛
𝐹𝑅 = 11.77 kN = 11 770 N ∙ 0.2248 (lb⁄N) = 𝟐𝟔𝟒𝟔 𝐥𝐛
3 2
3.81×10 N
1.24 Loading = 3.81 kPa = × 0.2248 lb × 1m
= 𝟕𝟗 𝐥𝐛
m2 N (3.28 ft)2 𝐟𝐭𝟐

1.25 𝐹 = 245 N ∙ 0.2248 (lb⁄N) = 𝟓𝟓. 𝟏 𝐥𝐛
4500 N 0.2248 lb 1m = 𝟐𝟓. 𝟕 𝐥𝐛
𝐾= × ×
m N 39.37 in 𝐢𝐧
𝐹 55.1 lb = 𝟐. 𝟏𝟒 𝐢𝐧
Δ𝐿 = =
𝐾 25.7 (lb⁄in)
2
lb∙s
1.26 𝑚=
𝑤
=
2750 lb = 85.4 = 𝟖𝟓. 𝟒 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 (ft/s2) ft
2
lb∙s
1.27 𝑚=
𝑤
=
12800 lb = 398 = 𝟑𝟗𝟖 𝐬𝐥𝐮𝐠𝐬
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 𝑠𝑢 = 14 000 psi ∙ 6.895 (kPa⁄psi) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚
𝑠𝑢 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
1750 rev 2π rad 1 min 𝐫𝐚𝐝
1.32 𝑛= × × = 𝟏𝟖𝟑
min rev 60s 𝐬
2
1.33 𝐴 = 14.1 in2 × (25.4 mm) = 𝟗𝟎𝟗𝟕 𝐦𝐦𝟐
in 2

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 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
2
𝐴 = 0.200 in2 × (25.4 mm) = 𝟏𝟐𝟗 𝐦𝐦𝟐
in 2
3200 N N
1.37 𝜎=𝑃= =
3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
𝐴 (𝜋𝐷2⁄4) [𝜋(10 mm) 2]⁄4 mm2

𝑃 20×10 3 N = 66.7 N
= 𝟔𝟔. 𝟕 𝐌𝐏𝐚
1.38 𝜎= =
𝐴 (10)(30) mm2 mm2
860 lb
1.39 𝜎=𝑃= = 𝟓𝟑𝟕𝟓 𝐩𝐬𝐢
𝐴 (0.40 in)2
𝑃 1850 lb = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.40 𝜎= =
𝐴 [𝜋(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
∑ 𝑀𝐴 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4512 N
𝐶 = 𝐶𝑉/ sin 30° = 9025 N
𝑃 𝐶 9025 N = 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4
𝑃 70000 lb = 𝟏𝟑𝟗𝟑 𝐩𝐬𝐢
1.42 𝜎= =
𝐴 [𝜋(8 in)2]/4

, (29500 lb)/3
1.43 𝜎= 𝑃 = = 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2
3500 N
1.44 𝜎=𝑃= = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 mm)2

1.45 𝑊 = 𝑚 𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
𝐴𝐵𝑋 = 𝐴𝐵 sin 35°
𝐴𝐵𝑌 = 𝐴𝐵 cos 35°
𝐵𝐶𝑋 = 𝐵𝐶 sin 55°
𝐵𝐶𝑌 = 𝐵𝐶 cos 55°
∑ 𝐹𝑋 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55°
sin 55°
𝐴𝐵 = 𝐵𝐶 ∙ = 1.428 𝐵𝐶 sin
35°
∑ 𝐹𝑉 = 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 = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
41.2 kN
𝐵𝐶 = = 23.63 kN
1.743

𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
Stress in Rod AB: 𝜎 =
𝐴𝐵
=
33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴𝐵 𝐴 [𝜋(20 mm) 2]/4

Stress in Rod BC: 𝜎 =
𝐵𝐶
=
23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐶 𝐴 [𝜋(20 mm) 2]/4

Stress in Rod BD: 𝜎 =
𝐵𝐷
=
41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐵𝐷 𝐴 [𝜋(20 mm) 2]/4

1.46 𝐹 = 0.01097 𝑚 𝑅 𝑛2 = (0.01097)(0.40)(0.60)(3000)2 N
𝐹 = 23 695 N
𝜋(16 mm)2
𝐴= = 201 mm2
4
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm2