Solutions for Applied Strength of Materials 7th Edition by Robert L. Mott

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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




Solutions for Applied Strength of
Solutions
Materials,
for7th
Applied
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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 = ( ) (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
2

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 = 4800 N/m
= 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦




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

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




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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= min
× rev
×
60s = 377 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 × in2
= 𝟏𝟐𝟗 𝐦𝐦𝟐
𝑃 2800 N 2800 N N
1.27 𝜎 = 𝐴 = (𝜋𝐷 2 ⁄4)
= [𝜋(10 mm)2 ]⁄4 = 35.7
mm2
= 35. 𝟕 𝐌𝐏𝐚
𝑃 18×10 3 N N
1.28 𝜎 = 𝐴 = (12)(30) = 50.7 mm2
= 50. 𝟕 𝐌𝐏𝐚
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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Applied
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𝑃 (29500 lb)/3
1.33 𝜎 = = = 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2

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

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 = 1.743
= 23.63 kN

𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
𝐴𝐵 33.75×103 N
Stress in Rod AB: 𝜎𝐴𝐵 = 𝐴
= [𝜋(20 mm)2 ]/4 = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐵𝐶 23.63×103 N
Stress in Rod BC: 𝜎𝐵𝐶 = 𝐴
= [𝜋(20 mm)2 ]/4 = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐷 41.2×103 N
Stress in Rod BD: 𝜎𝐵𝐷 = 𝐴
= [𝜋(20 mm)2 ]/4 = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚

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 = 𝐴 = 201 mm2 = 𝟏𝟏𝟖 𝐌𝐏𝐚




Solutions for Applied Strength of
Solutions
Materials,
for7th
Applied
Edition
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Strength
by
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Robert
of Materials,
L. Mott.(exthttps://online2pdf.com/remove-pdf-header-footer)
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