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Solution Manual For Applied Strength Of
Materials, 7th Edition By Robert L. Mott | All
Chapters 1-14| Latest Edition 2026
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Table Of Contents
1. 1. Basic Concepts In Strength Of Materials
2. 2. Design Properties Of Materials
3. 3. Direct Stress, Deformation, And Design
4. 4. Design For Direct Shear, Torsional Shear, And Torsional Deformation
5. 5. Shearing Forces And Bending Moments In Beams
6. 6. Centroids And Moments Of Inertia Of Areas
7. 7. Stress Due To Bending
8. 8. Shearing Stresses In Beams
9. 9. Deflection Of Beams
10. 10. Combined Stresses
11. 11. Columns
12. 12. Pressure Vessels
13. 13. Connections
14. 14. Thermal Effects And Elements Of More Than One Material
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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 each
front wheel: 𝐹𝐹 = (1 ) (0.40)(34.34 kn ) = 6.87 𝐤𝐍
2
each rear wheel: 𝐹𝑅 = (1 ) (0.60)(34.34 kn ) = 𝟏 0.32 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 m 2
loading = 57.9 kn ⁄15.8 m 2 = 3.66 kn ⁄m 2 = 𝟑.66 𝐤𝐏𝐚
1.15 force = 𝑚 𝑔 = 35 kg ∙ 9.81 m/s2 = 343 n
k = spring scale =4800 n⁄m = 𝐹/δ𝐿
δ𝐿 = 𝐹 = 343 n = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
𝐾 4800 n/m
1.16 𝑚= 𝑤
= 3250 lb lb∙s2 = 101 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 (ft/s2 )
= 101 ft
1.17 𝑚=
𝑤
=
11 600 lb
= 360
lb∙s2 = 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
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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
(25.4mm)
1.23 𝐴 = 26.1 in 2 × I2n
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 in 2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉 = (209 × 103 mm 2 ) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m 3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
1.26 𝐴 = 𝜋𝐷2 ⁄4 = 𝜋(0.505 in )2 ⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
2
𝐴 = 0.200 in 2 × (25.4 mm) = 𝟏𝟐𝟗 𝐦𝐦𝟐
in 2
𝑃 2800 n
1.27 𝜎= 2800 n
= = 35.7 n
= 35. 𝟕 𝐌𝐏𝐚
𝐴= (𝜋𝐷 2 ⁄4 ) [𝜋 (10 mm )2 ] ⁄4 mm 2
3
n
1.28 𝜎= 𝑃 = 18×10 = 50.7 n = 50. 𝟕 𝐌𝐏𝐚
𝐴 (12 )(30 ) mm 2 mm 2
1.29 𝜎= 𝑃
= 1150 lb = 7188 𝐩𝐬𝐢
𝐴 (0.40 in)2
1.30 𝜎=
𝑃
=
1850 lb = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
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
𝜎 = 𝑃 == 𝐶 9025 n 2 = 71.6 𝐌𝐏𝐚
𝐴 𝐴 [ 𝜋 (12 mm ) ]⁄4
1.32 𝜎 = 𝑃 70000 lb = 891 𝐩𝐬𝐢
=
𝐴 [ 𝜋 (10 in)2 ] /4
Solution Manual For Applied Strength Of
Materials, 7th Edition By Robert L. Mott | All
Chapters 1-14| Latest Edition 2026
, Stuvia.com - The Marketplace to Buy and Sell your Study Material
Table Of Contents
1. 1. Basic Concepts In Strength Of Materials
2. 2. Design Properties Of Materials
3. 3. Direct Stress, Deformation, And Design
4. 4. Design For Direct Shear, Torsional Shear, And Torsional Deformation
5. 5. Shearing Forces And Bending Moments In Beams
6. 6. Centroids And Moments Of Inertia Of Areas
7. 7. Stress Due To Bending
8. 8. Shearing Stresses In Beams
9. 9. Deflection Of Beams
10. 10. Combined Stresses
11. 11. Columns
12. 12. Pressure Vessels
13. 13. Connections
14. 14. Thermal Effects And Elements Of More Than One Material
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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 each
front wheel: 𝐹𝐹 = (1 ) (0.40)(34.34 kn ) = 6.87 𝐤𝐍
2
each rear wheel: 𝐹𝑅 = (1 ) (0.60)(34.34 kn ) = 𝟏 0.32 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 m 2
loading = 57.9 kn ⁄15.8 m 2 = 3.66 kn ⁄m 2 = 𝟑.66 𝐤𝐏𝐚
1.15 force = 𝑚 𝑔 = 35 kg ∙ 9.81 m/s2 = 343 n
k = spring scale =4800 n⁄m = 𝐹/δ𝐿
δ𝐿 = 𝐹 = 343 n = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
𝐾 4800 n/m
1.16 𝑚= 𝑤
= 3250 lb lb∙s2 = 101 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 (ft/s2 )
= 101 ft
1.17 𝑚=
𝑤
=
11 600 lb
= 360
lb∙s2 = 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
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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
(25.4mm)
1.23 𝐴 = 26.1 in 2 × I2n
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 in 2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉 = (209 × 103 mm 2 ) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m 3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
1.26 𝐴 = 𝜋𝐷2 ⁄4 = 𝜋(0.505 in )2 ⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
2
𝐴 = 0.200 in 2 × (25.4 mm) = 𝟏𝟐𝟗 𝐦𝐦𝟐
in 2
𝑃 2800 n
1.27 𝜎= 2800 n
= = 35.7 n
= 35. 𝟕 𝐌𝐏𝐚
𝐴= (𝜋𝐷 2 ⁄4 ) [𝜋 (10 mm )2 ] ⁄4 mm 2
3
n
1.28 𝜎= 𝑃 = 18×10 = 50.7 n = 50. 𝟕 𝐌𝐏𝐚
𝐴 (12 )(30 ) mm 2 mm 2
1.29 𝜎= 𝑃
= 1150 lb = 7188 𝐩𝐬𝐢
𝐴 (0.40 in)2
1.30 𝜎=
𝑃
=
1850 lb = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
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
𝜎 = 𝑃 == 𝐶 9025 n 2 = 71.6 𝐌𝐏𝐚
𝐴 𝐴 [ 𝜋 (12 mm ) ]⁄4
1.32 𝜎 = 𝑃 70000 lb = 891 𝐩𝐬𝐢
=
𝐴 [ 𝜋 (10 in)2 ] /4
