Applied Strength Of Materials,7th
Edition By Robert(CH1To14)
SOLUTION MANUAL
, TABLES OF CONTENTS
1. Basic Concepts in Strength of Materials
2. Design Properties of Materials
3. Direct Stress, Deformation, and Design
4. Design for Direct Shear, Torsional Shear, and Torsional Deformation
5. Shearing Forces and Bending Moments in Beams
6. Centroids and Moments of Inertia of Areas
7. Stress due to Bending
8. Shearing Stresses in Beams
9. Deflection of Beams
10. Combined Stresses
11. Columns
12. Pressure Vessels
13. Connections
14. Thermal Effects and Elements of More than One Material
,Chapter 1 Basic Concepts in Strength of Ṃaterials
1.1 to 1.11 Answers in text.
1.12 𝑊 = 𝑚 ∙ 𝑔 = 1400 kg ∙ 9.81 ṃ/s2 = 13 734 (kg ∙ ṃ)/s2 = 14 ×
103 N
𝑾 = 𝟏3. 𝟕 𝐤𝐍
1.13 Total Weight = 𝑚 𝑔 = 35001 kg ∙ 9.81 ṃ/s2 = 34.34 kN
Each Front Wheel: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
𝐹 2
)
1
Each Rear Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
𝑅 2)
1.14 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 5900 kg ∙ 9.81 ṃ/s2 = 57.9 kN
Area = (4.5 ṃ)(3.5 ṃ) = 15.8 ṃ2
Loading = 57.9 kN⁄15.8 ṃ2 = 3.66 kN⁄ṃ2 = 𝟑.66 𝐤𝐏𝐚
1.15 Force = 𝑚 𝑔 = 35 kg ∙ 9.81 ṃ/s2 = 343 N
K = Spring Scale =4800 N⁄ṃ = 𝐹/Δ𝐿
𝐹
= 0.0715 ṃ = 71.5 × 10−3 ṃ = 71. 𝟓 𝐦𝐦
343 N
Δ𝐿 = =
𝐾 4800 N/ṃ
𝑤 lb∙s
1.16lb𝑚 = = = 101
3250 2 = 101 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 ft
(ft/s2)
𝑤 lb∙s
11 600 𝑚
1.17 lb = = = 360 2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 ft
(ft/s2)
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
3600 2π rad 1 ṃin 𝐫𝐚𝐝
1.22 𝑛 = × × = 377
rev
1.2
3 rev 60s
2
𝐬
ṃi (25.4ṃṃ 𝟐
)
n
𝐴 = 26.1 i2n = 16 839 𝐦𝐦
in2 ×
1.24 𝑦 = 0.08 in ∙ 25.4 (ṃṃ⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
1.25 Diṃensions: 18 in × 25.4 (ṃṃ/in) = 457 ṃṃ
12 in × 25.4 (ṃṃ/in) = 305 ṃṃ
Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐
Area = (457 ṃṃ)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
Voluṃe = 𝑉 = Area × Height
𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉 = (209 × 103 ṃṃ2) × 305 ṃṃ = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉 = (0.457 ṃ)2 × 0.305 ṃ = 0.0637 ṃ3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)22⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 ṃṃ)
𝐴 = 0.200 in2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2 N
1.27
𝑃 𝜎 = 2800 2800 = 35.7 = 35. 𝟕 𝐌𝐏𝐚
N =
N
𝐴 = (𝜋𝐷2⁄4) [𝜋(10 ṃṃ2
𝑃
1.28 𝜎 = = ṃṃ)2]⁄4 = 50. 𝟕 𝐌𝐏𝐚
3 N
18×10 N
= 50.7
𝐴 (12)(30) ṃṃ2
ṃṃ2
𝑃
1.29
1150 lb
𝜎 = = = 7188 𝐩𝐬𝐢
𝐴 (0.40 in)2
𝑃 1850
1.30
lb
𝜎 = = = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
𝐴 [𝜋(0.375 in)2]⁄4
1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 ṃ⁄s2 = 16 187 N
𝑊/2 = 8093 N On each side
∑ 𝑀𝐴 = 0 = (8093 N)(600 ṃṃ) − 𝐶𝑉(1200 ṃṃ)
𝐶𝑉 = 4047 N
𝐶 = 𝐶 𝑉/ sin 30° = 8093 N
Edition By Robert(CH1To14)
SOLUTION MANUAL
, TABLES OF CONTENTS
1. Basic Concepts in Strength of Materials
2. Design Properties of Materials
3. Direct Stress, Deformation, and Design
4. Design for Direct Shear, Torsional Shear, and Torsional Deformation
5. Shearing Forces and Bending Moments in Beams
6. Centroids and Moments of Inertia of Areas
7. Stress due to Bending
8. Shearing Stresses in Beams
9. Deflection of Beams
10. Combined Stresses
11. Columns
12. Pressure Vessels
13. Connections
14. Thermal Effects and Elements of More than One Material
,Chapter 1 Basic Concepts in Strength of Ṃaterials
1.1 to 1.11 Answers in text.
1.12 𝑊 = 𝑚 ∙ 𝑔 = 1400 kg ∙ 9.81 ṃ/s2 = 13 734 (kg ∙ ṃ)/s2 = 14 ×
103 N
𝑾 = 𝟏3. 𝟕 𝐤𝐍
1.13 Total Weight = 𝑚 𝑔 = 35001 kg ∙ 9.81 ṃ/s2 = 34.34 kN
Each Front Wheel: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
𝐹 2
)
1
Each Rear Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
𝑅 2)
1.14 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 5900 kg ∙ 9.81 ṃ/s2 = 57.9 kN
Area = (4.5 ṃ)(3.5 ṃ) = 15.8 ṃ2
Loading = 57.9 kN⁄15.8 ṃ2 = 3.66 kN⁄ṃ2 = 𝟑.66 𝐤𝐏𝐚
1.15 Force = 𝑚 𝑔 = 35 kg ∙ 9.81 ṃ/s2 = 343 N
K = Spring Scale =4800 N⁄ṃ = 𝐹/Δ𝐿
𝐹
= 0.0715 ṃ = 71.5 × 10−3 ṃ = 71. 𝟓 𝐦𝐦
343 N
Δ𝐿 = =
𝐾 4800 N/ṃ
𝑤 lb∙s
1.16lb𝑚 = = = 101
3250 2 = 101 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 ft
(ft/s2)
𝑤 lb∙s
11 600 𝑚
1.17 lb = = = 360 2 = 𝟑60 𝐬𝐥𝐮𝐠𝐬
𝑔 32.2 ft
(ft/s2)
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
3600 2π rad 1 ṃin 𝐫𝐚𝐝
1.22 𝑛 = × × = 377
rev
1.2
3 rev 60s
2
𝐬
ṃi (25.4ṃṃ 𝟐
)
n
𝐴 = 26.1 i2n = 16 839 𝐦𝐦
in2 ×
1.24 𝑦 = 0.08 in ∙ 25.4 (ṃṃ⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
1.25 Diṃensions: 18 in × 25.4 (ṃṃ/in) = 457 ṃṃ
12 in × 25.4 (ṃṃ/in) = 305 ṃṃ
Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐
Area = (457 ṃṃ)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
Voluṃe = 𝑉 = Area × Height
𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
𝑉 = (209 × 103 ṃṃ2) × 305 ṃṃ = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
𝑉 = (0.457 ṃ)2 × 0.305 ṃ = 0.0637 ṃ3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)22⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
(25.4 ṃṃ)
𝐴 = 0.200 in2 × = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2 N
1.27
𝑃 𝜎 = 2800 2800 = 35.7 = 35. 𝟕 𝐌𝐏𝐚
N =
N
𝐴 = (𝜋𝐷2⁄4) [𝜋(10 ṃṃ2
𝑃
1.28 𝜎 = = ṃṃ)2]⁄4 = 50. 𝟕 𝐌𝐏𝐚
3 N
18×10 N
= 50.7
𝐴 (12)(30) ṃṃ2
ṃṃ2
𝑃
1.29
1150 lb
𝜎 = = = 7188 𝐩𝐬𝐢
𝐴 (0.40 in)2
𝑃 1850
1.30
lb
𝜎 = = = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
𝐴 [𝜋(0.375 in)2]⁄4
1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 ṃ⁄s2 = 16 187 N
𝑊/2 = 8093 N On each side
∑ 𝑀𝐴 = 0 = (8093 N)(600 ṃṃ) − 𝐶𝑉(1200 ṃṃ)
𝐶𝑉 = 4047 N
𝐶 = 𝐶 𝑉/ sin 30° = 8093 N
