SOLUTION MANUAL
For Applied Strength of Materials 7th Edition by
Robert Mott, Joseph Untener (All Chapters, 100% Original
Verified, A+ Grade)
, 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 = 𝑚𝑔 = 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 = 𝑚𝑔 = 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
Δ𝐿 = 𝐹 = 343 N = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
𝐾 4800 N/m
𝑤 3250 lb lb∙s2 = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚= = = 101
𝑔 32.2 (ft/s2) ft
𝑤 11 600 lb lb∙s2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
1.17 𝑚= = = 360
𝑔 32.2 (ft/s2) ft
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
For Applied Strength of Materials 7th Edition by
Robert Mott, Joseph Untener (All Chapters, 100% Original
Verified, A+ Grade)
, 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 = 𝑚𝑔 = 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 = 𝑚𝑔 = 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
Δ𝐿 = 𝐹 = 343 N = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
𝐾 4800 N/m
𝑤 3250 lb lb∙s2 = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚= = = 101
𝑔 32.2 (ft/s2) ft
𝑤 11 600 lb lb∙s2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
1.17 𝑚= = = 360
𝑔 32.2 (ft/s2) ft
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
