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 = 𝑚𝑔 = 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 = 𝑚𝑔 =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 = 𝑚𝑔 = 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
= 101 𝐬𝐥𝐮𝐠𝐬
𝑚=𝑤= 3250 lb
1.16 = 101
lb∙s2
𝑔 32.2 (ft/s 2) ft
𝑤 11 600 lb∙s
1lb.17 𝑚= = = 360 2
𝑔 32.2 (ft/s 2)
= 𝟑60 𝐬𝐥𝐮𝐠𝐬
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 3600
𝑛=
× 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. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
2
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) mm 2 mm 2
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
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 = 𝑚𝑔 = 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 = 𝑚𝑔 =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 = 𝑚𝑔 = 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
= 101 𝐬𝐥𝐮𝐠𝐬
𝑚=𝑤= 3250 lb
1.16 = 101
lb∙s2
𝑔 32.2 (ft/s 2) ft
𝑤 11 600 lb∙s
1lb.17 𝑚= = = 360 2
𝑔 32.2 (ft/s 2)
= 𝟑60 𝐬𝐥𝐮𝐠𝐬
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 3600
𝑛=
× 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. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
2
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) mm 2 mm 2
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
