Solution Manual for Applied Strength of Materials, 7th Edition by Robert L. Mott | All Chapters 1-14| Latest Edition 2025

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SOLUTIONS MANUAL
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,Chapter 1 df Basic Concepts in Strength of Materialsdf df df df df




1.1 to 1.15 Answers in text.
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1.16 𝑊 = 𝑚 ∙ 𝑔 = 1800 kg ∙ 9.81 m/s2 = 17 658 (kg ∙ m)/s2 = 17 × 103 N
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𝑾 = 𝟏𝟕. 𝟕 𝐤𝐍
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1.17 Total Weight = 𝑚𝑔 = 4000 kg ∙ 9.81 m/s2 = 39.24 kN
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Each Front Wheel: 𝐹𝐹 = (1) (0.40)(39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
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2

Each Rear Wheel: 𝐹𝑅 = (1) (0.60)(39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
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2

1.18 Loading = Total Force / Area df df df df df




Total Force = 𝑚𝑔 = df df df df d f 6800 kg ∙ 9.81 m/s2 = 66.7 kN df df df df df df df




Area = (5.0 m)(3.5 m) = 17.5 m2
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Loading = 66.7 kN⁄17.5 m2 = 3.81 kN⁄m2 = 𝟑. 𝟖𝟏 𝐤𝐏𝐚
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1.19 Force = Weight = 𝑚𝑔 = 25 kg ∙ 9.81 m/s2 = 245 N
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K = Spring Scale = 4500 N⁄m = 𝐹/Δ𝐿
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𝐹 d f
= 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
Δ𝐿 = df df =245 df
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N
𝐾 4500 dfN/m

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

1.25 𝐹 = 245 N ∙ 0.2248 (lb⁄N) = 𝟓𝟓. 𝟏 𝐥𝐛
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4500 dfN d f 0.2248 df lb df = 𝟐𝟓. 𝟕 𝐥𝐛
𝐾 = df df × df
1× df
df df d f

m
m N 39.37 d f in 𝐢𝐧
𝐹 d f
= 𝟐. 𝟏𝟒 𝐢𝐧
Δ𝐿 = df df =55.1 df
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lb
𝐾 25.7 df(lb⁄in)
2 d f

1.26 𝑤 df = 85.4 df d f
lb∙s = 𝟖𝟓. 𝟒 𝐬𝐥𝐮𝐠𝐬
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𝑚
lb
= df df =2750 df


𝑔 32.2 df(ft/s2) ft
2 d f

1.27 𝑤 df = df 398 d f
lb∙s = 𝟑𝟗𝟖 𝐬𝐥𝐮𝐠𝐬df df
𝑚 df =dfdlbf
12800 = df
𝑔 32.2 df(ft/s2) ft

1.29 𝑝 = 1200 psi ∙ 6.895 (kPa⁄psi) = 𝟖𝟐𝟕𝟒 𝐤𝐏𝐚
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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
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𝑠𝑢 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
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1750 df rev d f 2π df rad d f 1 df min df 𝐫𝐚𝐝
1.32 𝑛 = df df × df × df = 𝟏𝟖𝟑 df d f
min rev 60s 𝐬
2 d f

1.33 𝐴 = 14.1 in2 ×
df df df df df
(25.4 dfmm) = 𝟗𝟎𝟗𝟕 𝐦𝐦𝟐 d f df

in2

1.34 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
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1.35 Dimensions: 18 in df df d f × d f 25.4 (mm/in) = 457 mm df df df df




12 in df d f × d f 25.4 (mm/in) = 305 mmdf df df df




Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐
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Area = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
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Volume = 𝑉 = Area × Height df df df df df df




𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
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𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
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𝑉 = (209 × 103 mm2) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
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𝑉 = (0.457 m)2
df df df d f × d f 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
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1.36 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
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2 d f

𝐴 = 0.200 in2 ×
df df df df df
(25.4 dfmm) = 𝟏𝟐𝟗 𝐦𝐦𝟐 df df

in2
1.37 𝜎 = 𝑃 = 3200 df = 40.7 N = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
df
3200 N df
df df d f
=
N
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𝐴 (𝜋𝐷2 ⁄4) [𝜋(10 df mm2
mm) 2]⁄4


1.38 𝜎 = df df
𝑃 df
=20×103 df
= 66.7 df
N = 𝟔𝟔. 𝟕 𝐌𝐏𝐚df df df

N
𝐴 (10)(30) d f mm2 mm2

1.39 𝜎 = df
860 lb
df
𝑃 df = d f d f = 𝟓𝟑𝟕𝟓 𝐩𝐬𝐢df df

df

𝐴 (0.40 df in)2

𝑃 df 1850 df = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢df df df
1.40 𝜎
lb = df df =
𝐴 [𝜋(0.375 dfin)2]⁄4

1.41 Load on Shelf df df d f = 𝑊 = 𝑚𝑔 = 1840 kg ∙ 9.81 m⁄s2 = 18 050 N
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𝑊/2 = 9025 N On each side
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∑ 𝑀𝐴 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
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𝐶𝑉 = 4512 N
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𝐶 = 𝐶𝑉/ sin 30° = 9025 N
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𝑃 df 𝐶 d f
= 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎
N
= df df = df = df
9025 df df df



𝐴 𝐴 [𝜋(12 dfmm)2]⁄4
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, 𝑃 df = d f 𝟏𝟑𝟗𝟑 df 𝐩𝐬𝐢
1.42 𝜎 d f = df
70000 df lb = df df
𝐴 [𝜋(8 df in)2]/4




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