Applied Strength of Materials (6th Edition) SI Units Version – Solution Manual by Robert L. Mott & Joseph A. Untener (2018) – A complete companion manual providing fully-worked solutions for each chapter covering basic concepts, design properties of mater

,Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.15 Answers in text.
1.16 𝑊 = 𝑚 ∙ 𝑔 = 1800 kg ∙ 9.81 m/s2 = 17 658 (kg ∙ m)/s2 = 17 × 103 N
𝑾 = 𝟏𝟕. 𝟕 𝐤𝐍
1.17 Total Weight = 𝑚 𝑔 = 4000 kg ∙ 9.81 m/s2 = 39.24 kN
1
Each Front Wheel: 𝐹𝐹 = ( )2 (0.40)( 39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = ( )2 (0.60)( 39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍

1.18 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 6800 kg ∙ 9.81 m/s2 = 66.7 kN
Area = (5.0 m)(3.5 m) = 17.5 m2
Loading = 66.7 kN⁄17.5 m2 = 3.81 kN⁄m2 = 𝟑. 𝟖𝟏 𝐤𝐏𝐚
1.19 Force = Weight = 𝑚 𝑔 = 25 kg ∙ 9.81 m/s2 = 245 N
K = Spring Scale = 4500 N⁄m = 𝐹/Δ𝐿
Δ𝐿 =
𝐹
=
245 N
= 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
𝐾 4500 N/m
𝑃 3200 N N
1.22 𝜎= = =
3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
𝐴 (𝜋𝐷2⁄4) [𝜋(10 mm) 2]⁄4 mm2

𝑃 20×10 3 N N
1.23 𝜎= = = 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
𝐴 (10)(30) mm2 mm2


𝑃 3500 N
1.24 𝜎= = = 𝟑𝟓. 𝟎 𝐌𝐏𝐚
𝐴 (0.010 m)2
𝑃 8300 N
1.25 𝜎= = = 𝟏𝟑𝟎. 𝟓 𝐌𝐏𝐚
𝐴 [𝜋(9.0 mm)2]⁄4



1.26 Load on Shelf = 𝑊 = 𝑚𝑔 = 1840 kg ∙ 9.81 m⁄s2 = 18 050 N
𝑊/2 = 9025 N On each side
∑ 𝑀𝐴 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4512 N
𝐶 = 𝐶𝑉/ sin 30° = 9025 N
𝑃 𝐶 9025 N = 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4




@SOLUTIONSSTUDY

, 𝑃 310×10 3 N = 𝟗. 𝟖𝟕 𝐌𝐏𝐚
1.27 𝜎= =
𝐴 [𝜋(0.2 m) 2]/4


𝑃 (132 000 N)/3
1.28 𝜎= = = 𝟔. 𝟏 𝐌𝐏𝐚
𝐴 (85 mm)2
𝑃 3500 N
1.29 𝜎= = = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 mm)2

1.30 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
Stress in Rod AB: 𝜎 𝐴𝐵 33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴𝐵 = =
𝐴 [𝜋(20 mm) 2]/4

Stress in Rod BC: 𝜎 𝐵𝐶 23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐶 = =
𝐴 [𝜋(20 mm) 2]/4

Stress in Rod BD: 𝜎 𝐵𝐷 41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐵𝐷 = =
𝐴 [𝜋(20 mm) 2]/4



1.31 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 = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm2

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

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

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



1.33 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
𝜎 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
𝐴𝐵 𝐴1 491 mm2

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

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



1.34 𝐴 = 515.8 mm2 [𝐷𝑁 40 Pipe-Appendix A-9(b)]
For BC: 𝜎 =
𝐹𝐵𝐶
=
11 000 N = 𝟐𝟏. 𝟑 𝐌𝐏𝐚 Tension
𝐵𝐶 𝐴 515.8 mm2

For AB: 𝐹𝐴𝐵 = 11 000 + 2(36 000 cos 30°) = 73 354 N
𝜎𝐴𝐵 =
𝐹𝐴𝐵
=
73 354 N = 𝟏𝟒𝟐. 𝟐 𝐌𝐏𝐚 Tension
𝐴 515.8 mm



1.35 ∑ 𝑀𝐶 = 0 = 13 000 N(1.2 m) − 𝐹𝐵𝐷(0.8)
𝐹𝐵𝐷 = 19 500 N
𝜎𝐵𝐷 =
𝐹𝐵𝐷
=
19 500 N = 𝟒𝟖. 𝟖 𝐌𝐏𝐚 Tension
𝐴 (25)(16) mm2




@SOLUTIONSSTUDY