Solutions Manual For Applied Strength of Materials 7th Edition By Robert Mott, Joseph Untener (All Chapters) (1)

Applied Strengthof Materials,
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7th edition
By Mott, Joseph Untener (All Chapters)
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Solution matual gh

,Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.11 Answers in text.
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1.12 𝑊 = 𝑚 ∙ 𝑔 = 1400 kg ∙ 9.81 m/s2 = 13 734 (kg ∙ m)/s2 = 14 × 103 N
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𝑾 = 𝟏3. 𝟕 𝐤𝐍 gh gh gh gh




1.13 Total Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN
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1
Each Front Wheel: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
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𝐹 )
12
Each Rear Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
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𝑅 2)
1.14 Loading = Total Force / Area gh gh gh gh gh




Total
Area Force
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𝑚𝑔 = m)
= (4.5=m)(3.5 5900= kg ∙ 9.81
15.8
2
m2 m/s = 57.9 kN
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Loading = 57.9 kN⁄15.8 m2 = 3.66 kN⁄m2 = 𝟑.66 𝐤𝐏𝐚 gh gh gh gh
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1.15 Force = 𝑚𝑔 = 35 kg ∙ 9.81 m/s2 = 343 N gh g h gh gh gh gh gh gh
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K = Spring Scale =4800 N⁄m = 𝐹/Δ𝐿
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𝐹 343 N
Δ𝐿 = = = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
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𝐾 4800 N/m g h




𝑤 3250 lb = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚 = = = 101
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gh lb∙s2
𝑔 32.2 (ft/s2) gh ft
𝑤 11 600
1lb.17 𝑚 = = = 360
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= 𝟑60 𝐬𝐥𝐮𝐠𝐬
32.2 (ft/s2)
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𝑔 gh lb∙s
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1.19 𝑝 = 1700 psi ∙ 6.895 (kPa⁄psi) = 11 722 𝐤𝐏𝐚
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1.20 𝜎 = 24 300 psi ∙ 6.895 (kPa⁄psi) = 167 549 kPa = 𝟏68 𝐌𝐏𝐚
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,1.21 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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1.22
3600 𝑛=
× 2π rad × 1 min = 377 𝐫𝐚𝐝
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1.23 rev 60s
2 𝐬
min (25.4mm) 𝟐

𝐴 = 26.1 in2
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× i2n = 16 839 𝐦𝐦 gh gh gh




1.24 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐.𝟎𝟑 𝐦𝐦
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1.25 Dimensions: 18 in × 25.4 (mm/in) = 457 mm gh gh g h g h gh gh gh gh




mm Area = 12
gh (18inin)×2 =25.4
𝟑𝟐𝟒 (mm/in)
𝐢𝐧𝟐
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= 305 gh gh
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Area = (457 mm) = 𝟐. 𝟎𝟗 × 𝟏𝟎 𝐦𝐦𝟐
2
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𝟓
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Volume = 𝑉 = Area × Height gh gh gh gh gh gh




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 × 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐
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𝐦𝟑
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
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2
2 (25.4
𝐴 = 0.200 in ×
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g h mm) N

in2

1𝑃. 27 𝜎 2800 N 2800 N = 35.7 = 35. 𝟕 𝐌𝐏𝐚
=
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g h =
𝐴 g h
= (𝜋𝐷2⁄4) gh
[𝜋(10 mm)2]⁄4 gh gh mm2
𝑃
1.28 𝜎 = =
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18×103
N
gh

= 50.7 gh = 50. 𝟕 𝐌𝐏𝐚 gh gh gh




N
𝐴 (12)(30)mm2 h
g mm2

lb 𝑃
1.29 𝜎 = = = 7188 𝐩𝐬𝐢
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1150
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𝐴in)2 (0.40
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lb 𝑃 = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.30 𝜎= =
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1850
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𝐴 [𝜋(0.375 in)2]⁄4 gh




1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N
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𝑊/2 = 8093 N On each side gh gh gh gh gh gh




∑ 𝑀𝐴= 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm)
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, 𝐶𝑉 = 4047 N gh
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𝐶 = 𝐶𝑉/ sin 30° = 8093 N
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𝑃 𝐶 9025 N gh gh




𝜎= gh g h
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=
=𝐴 [𝜋(12 mm) 2]⁄4
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1.32 𝜎 = 70000 lb
=
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𝑃
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