Solutions Manual for Applied Strength of Materials, 7e by Robert Mott, Joseph Untener (All Chapters)

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
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∙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




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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= × × = 377
min rev 60s 𝐬
2
(25.4mm)
1.23 𝐴 = 26.1 in2 × 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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, 𝑃 (29500 lb)/3
1.33 𝜎 = = = 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2

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

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 33.75×103 N
Stress in Rod AB: 𝜎𝐴𝐵 = = [𝜋(20 = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴 mm)2 ]/4

𝐵𝐶 23.63×103 N
Stress in Rod BC: 𝜎𝐵𝐶 = = [𝜋(20 = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐴 mm)2 ]/4

𝐵𝐷 41.2×103 N
Stress in Rod BD: 𝜎𝐵𝐷 = = [𝜋(20 = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐴 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 = 𝐴 = 201 mm2 = 𝟏𝟏𝟖 𝐌𝐏𝐚




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, 1.37 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80) kN = 150 kN
𝐹𝐴𝐵 150×103 N
𝜎𝐴𝐵 = = = 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
𝐴 900 mm2

For BC: 𝐹𝐵𝐶 = 110 − 40 = 70 kN
𝐹𝐵𝐶 70×103 N
𝜎𝐵𝐶 = = = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝐴 900 mm2

For CD: 𝐹𝐶𝐷 = 110 kN
𝐹𝐶𝐷 110×103 N
𝜎𝐶𝐷 = = = 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
𝐴 900 mm2

1.38 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2
C-D; 𝐴2 = 𝜋(16)2 /4 = 201 mm2
For AB: 𝐹𝐴𝐵 = −9.65 − 12.32 + 4.45 = −17.52 kN
𝐹𝐴𝐵 −17.52×103 N
𝜎𝐴𝐵 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
𝐴1 491 mm2

For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
𝐹𝐵𝐶 −21.97×103 N
𝜎𝐵𝐶 = = = −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
𝐴1 491 mm2

For CD: 𝐹𝐶𝐷 = −9.65 kN
𝐹𝐶𝐷 −9.65×103 N
𝜎𝐶𝐷 = = = −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
𝐴2 201 mm2

𝜋[(1.90)2−(1.61)2] 1
1.39 𝐴 = 4
= 0.799 in2 [1 2 in Pipe-Appendix A-9(a)]
𝐹𝐵𝐶 2500 lb
For BC: 𝜎𝐵𝐶 = = 0.799 in2 = 𝟑𝟏𝟐𝟗 𝐩𝐬𝐢 Tension
𝐴

For AB: 𝐹𝐴𝐵 = 2500 + 2(8000 cos 30°) = 16 356 lb
𝐹𝐴𝐵 16 356 lb
𝜎𝐴𝐵 = = = 𝟐𝟎 𝟒𝟕𝟏 𝐩𝐬𝐢 Tension
𝐴 0.799 in2

1.40 ∑ 𝑀𝐶 = 0 = 2800(45) − 𝐹𝐵𝐷 (30)
𝐹𝐵𝐷 = 4200 lb
𝐹𝐵𝐷 4200 lb
𝜎𝐵𝐷 = = (2.0)(0.65) in2 = 𝟑𝟐𝟑𝟏 𝐩𝐬𝐢 Tension
𝐴




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