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

Solutions Manual For
Applied Strength Of Materials
Author: Robert L. Mott


7th Edition

,Solutions Manual For Applied Strength Of Materials, 7e By Robert
Mott, Joseph Untener (All Chapters)
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
Δ𝐿 = 𝐹 = 343 N = 0.0715 M = 71.5 × 10−3 M = 71. 𝟓 𝐦𝐦
𝐾 4800 N/M




𝑤 3250 Lb∙S101
= 2 = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚
Lb = =
𝑔 32.2 (Ft/S2) Ft

𝑤 11 600 = 360
2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
1.17 𝑚
Lb = = Lb∙S
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

,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 In2 × i2n
= 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 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
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 In)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 Mm)2
𝐴 = 0.200 In2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
In2
𝑃 2800 N 2800 N N
1.27 𝜎= = = = 35.7 = 35. 𝟕 𝐌𝐏𝐚
𝐴 (𝜋𝐷2 ⁄4) [𝜋(10 Mm)2]⁄4 Mm2
𝑃 3 N N
1.28 𝜎= = 18×10 = 50.7 = 50. 𝟕 𝐌𝐏𝐚
𝐴 (12)(30) Mm2 Mm2
𝑃 1150 Lb
1.29 𝜎= = = 7188 𝐩𝐬𝐢
𝐴 (0.40 In)2
𝑃 1850 = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.30 𝜎
Lb = =
𝐴 [𝜋(0.375 In)2]⁄4

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= == = 71.6 𝐌𝐏𝐚
𝐴 𝐴 [𝜋(12 Mm)2]⁄4
𝑃 70000 Lb
1.32 𝜎 = = = 891 𝐩𝐬𝐢
𝐴 [𝜋(10 In)2]/4

, 𝑃 (29500 Lb)/3
1.33 𝜎 = = = 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 In)2

𝑃 3500 N
1.34 𝜎 = = = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 Mm)2

1.35 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
𝐴𝐵 33.75×10 3 N
Stress In Rod AB: 𝜎 = = = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴𝐵 𝐴 [𝜋(20 Mm)2]/4
𝐵𝐶 23.63×10 3 N
Stress In Rod BC: 𝜎 = = = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐶 𝐴 [𝜋(20 Mm)2]/4
𝐵𝐷 41.2×10 3 N
Stress In Rod BD: 𝜎 = = = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐵𝐷 𝐴 [𝜋(20 Mm)2]/4

1.36 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
𝜎= = = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝐴 201 Mm2