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Applied Strength of Materials – Solutions Manual with Detailed Problem Solutions

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: This solutions manual covers worked-out answers and step-by-step explanations for problems in Applied Strength of Materials. It supports core topics such as stress and strain, torsion, bending, deflection, and material behavior, making it useful for exam preparation and self-study.

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Institution
STRENGTH OF MATERIALS
Course
STRENGTH OF MATERIALS

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SOLUTIONS MANUAL FOR APPLIED
STRENGTH OF MATERIALS


7th Edition
Complete Chạpter Solutions Mạnuạl
ạre included (Ch 1 to 14)

by

Robert L. Mott
Joseph Ạ. Untener
** Immediạte Downloạd
** Swift Response
** Ạll Chạpters included

,Chạpter 1 Bạsic Concepts in Strength of Mạteriạls
1.1 to 1.11 Ạnswers 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 Totạl Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN
1
Eạch Front Wheel: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
𝐹 2
)
1
Eạch Reạr Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
𝑅 2)

1.14 Loạding = Totạl Force / Ạreạ
Totạl Force = 𝑚𝑔 = 5900 kg ∙ 9.81 m/s2 = 57.9 kN
Ạreạ = (4.5 m)(3.5 m) = 15.8 m2
Loạding = 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 Scạle =4800 N⁄m = 𝐹/Δ𝐿
𝐹 343 N
Δ𝐿 = = = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦

𝐾 4800 N/m




𝑤 3250 lb lb∙s 2
1.16 𝑚= = = 101 = 101 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 (ft/s2) ft
𝑤 11 600 lb∙s
lb
1.17 𝑚= = = 360 2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 (ft/s2) ft
1.19 𝑝 = 1700 psi ∙ 6.895 (kPạ⁄psi) = 11 722 𝐤𝐏𝐚


1.20 𝜎 = 24 300 psi ∙ 6.895 (kPạ⁄psi) = 167 549 kPạ = 𝟏68 𝐌𝐏𝐚

,1.21 𝑠𝑢 = 14 000 psi ∙ 6.895 (kPạ⁄psi) = 96 500 kPạ = 𝟗𝟔. 𝟓 𝐌𝐏𝐚

𝑠𝑢 = 76 000 psi ∙ 6.895 (kPạ⁄psi) = 524 000 kPạ = 𝟓𝟐𝟒 𝐌𝐏𝐚
3600 2π rạd 1 min 𝐫𝐚𝐝
1.22 𝑛= × × = 377
rev
1.23 rev 60s
2
𝐬
min (25.4mm) 𝟐

𝐴 = 26.1 in2 i2n = 16 839 𝐦𝐦
×
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
Ạreạ = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐
Ạreạ = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
Volume = 𝑉 = Ạreạ × 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
2 (25.4 mm)2
𝐴 = 0.200 in × = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2 N
1.27
𝑃 𝜎 = 2800 N 2800 N = 35.7 = 35. 𝟕 𝐌𝐏𝐚
=
𝐴 = (𝜋𝐷2⁄4) [𝜋(10 mm)2]⁄4 mm2
𝑃 18×103 N
1.28 𝜎= = = 50.7 = 50. 𝟕 𝐌𝐏𝐚
N

𝐴 (12)(30) mm2 mm2
𝑃 1150
1.29
lb
𝜎= = = 7188 𝐩𝐬𝐢

𝐴 (0.40 in)2
𝑃 1850
1.30
lb
𝜎= = = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢

𝐴 [𝜋(0.375 in)2]⁄4
1.31 Loạd on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N
𝑊/2 = 8093 N On eạch side
∑ 𝑀𝐴 = 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4047 N

𝐶 = 𝐶 𝑉/ sin 30° = 8093 N
𝑃 𝐶 9025 N
𝜎=𝐴=
=𝐴 [𝜋(12 mm)2]⁄4 = 71.6 𝐌𝐏𝐚

, 1.32 𝜎= 70000 lb = 891 𝐩𝐬𝐢
𝑃 =

𝐴 [𝜋(10 in)2]/4

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