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Solutions for Applied Strength of Materials, 7th Edition Mott (All Chapters included)
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Complete Solutions Manual for Applied Strength of Materials, 7th Edition by Robert L. Mott, Joseph A. Untener ; ISBN13: 9781032002224. (Full Chapters included Chapter 1 to 14).... 1. Basic Concepts in Strength of Materials. 2. Design Properties of Materials. 3. Direct Stress, Deformation, and De...

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  • December 23, 2023
  • 377
  • 2021/2022
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  • ABA - Applied Behavior Analysis
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SOLUTIONS MANUAL FOR
APPLIED STRENGTH
OF MATERIALS

7th Edition
Complete Chapter Solutions Manual
are included (Ch 1 to 14)

by

Robert L. Mott
Joseph A. Untener
** Immediate Download
** Swift Response
** All Chapters included

,Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.11 Answers in text.
1.12 𝑊 = 𝑚 ∙ 𝑔 = 1400 kg ∙ 9.81 m/s2 = 13 734 (kg ∙ m)/s2 = 14 × 103 N
𝑾 = 𝟏3. 𝟕 𝐤𝐍
1.13 Total Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN
1
Each Front Wheel: 𝐹𝐹 = (2) (0.40)(34.34 kN) = 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = (2) (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍

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




𝑤 3250 lb 2
1.16 𝑚= = 32.2 (ft/s2 ) = 101 lb∙sft = 101 𝐬𝐥𝐮𝐠𝐬
𝑔

𝑤 11 600 lb 2
1.17 𝑚= = 32.2 (ft/s2 ) = 360 lb∙sft = 𝟑60 𝐬𝐥𝐮𝐠𝐬
𝑔

1.19 𝑝 = 1700 psi ∙ 6.895 (kPa⁄psi) = 11 722 𝐤𝐏𝐚
1.20 𝜎 = 24 300 psi ∙ 6.895 (kPa⁄psi) = 167 549 kPa = 𝟏68 𝐌𝐏𝐚

, 1.21 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
3600 rev 2π rad 1 min 𝐫𝐚𝐝
1.22 𝑛= × × = 377
min rev 60s 𝐬
2
(25.4mm)
1.23 𝐴 = 26.1 in × 2
in
2
= 16 839 𝐦𝐦𝟐

1.24 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
1.25 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 𝐟𝐭 𝟑
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
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 mm)2
𝐴 = 0.200 in2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2
𝑃2800 N 2800 N N
1.27 𝜎 = 𝐴 = (𝜋𝐷 = [𝜋(10 mm)2 ]⁄4 = 35.7 = 35. 𝟕 𝐌𝐏𝐚
2 ⁄4) mm2

𝑃 18×10 3 N N
1.28 𝜎 = 𝐴 = (12)(30) = 50.7 = 50. 𝟕 𝐌𝐏𝐚
mm2 mm2
𝑃 1150 lb
1.29 𝜎 = 𝐴 = (0.40 in)2 = 7188 𝐩𝐬𝐢
𝑃 1850 lb
1.30 𝜎 = 𝐴 = [𝜋(0.375 in)2 ]⁄4 = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢

1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N
𝑊/2 = 8093 N On each side
∑ 𝑀𝐴 = 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4047 N
𝐶 = 𝐶𝑉 / sin 30° = 8093 N
𝑃 𝐶 9025 N
𝜎 = 𝐴 ==𝐴 [𝜋(12 mm)2 ]⁄4 = 71.6 𝐌𝐏𝐚
𝑃 70000 lb
1.32 𝜎 = = [𝜋(10 = 891 𝐩𝐬𝐢
𝐴 in)2]/4

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