Solutions Manual For Applied Strength of Materials 7th Edition By Robert Mott, Joseph Untener (All Chapters)

Applied Strength of Materials,
7th edition
By Mott, Joseph Untener (All Chapters)




Solution matual

,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: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
𝐹 )
12
Each Rear Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
𝑅 2)
1.14 Loading = Total Force / Area
Total
Area =Force
(4.5 = 𝑚𝑔 =m)
m)(3.5 5900 kg ∙ 9.81
= 15.8
2
m2 m/s = 57.9 kN
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 = = = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦

𝐾 4800 N/m


𝑤 3250 lb = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚= = = 101
lb∙s2
𝑔 32.2 (ft/s2) ft
𝑤 11 600 lb∙s
1lb.17 𝑚= = = 360 2
= 𝟑60 𝐬𝐥𝐮𝐠𝐬
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
1.22 𝑛 = 3600
× 2πrev
rad
× 160s
rev min 𝐫𝐚𝐝
1.23 2
= 377 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
122 in
Area = (18 in) =× 25.4
𝟑𝟐𝟒 𝐢𝐧𝟐(mm/in) = 305 mm
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
2
(25.4
𝐴 = 0.200 in2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
mm) N

in2

1𝑃 .27 𝜎= 2800 N 2800 N = 35.7 = 35. 𝟕 𝐌𝐏𝐚
𝐴 = (𝜋𝐷2⁄4) = [𝜋(10 mm)2]⁄4 mm2
𝑃
1.28 𝜎= =
18×103
N
= 50.7 = 50. 𝟕 𝐌𝐏𝐚
N
𝐴 (12)(30) mm2 mm2

lb 𝑃
1.29 𝜎= = = 7188 𝐩𝐬𝐢

1150
𝐴 (0.40 in)2

lb 𝑃
1.30 𝜎= = = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢

1850

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= 𝐴 =
=𝐴 [𝜋(12 mm) 2]⁄4 = 71.6 𝐌𝐏𝐚

1.32 𝜎 = 70000 lb
=
𝑃