Solutions Manual for Applied Strength of Materials 7th Edition by Robert L. Mott and Joseph A. Untener

SOLUTIONS MANUAL FOR
APPLIED STRENGTH
OF MATERIALS

7th Eḍ ition
Complete Chapter Solutions
Manual
are incluḍeḍ (Ch 1 to 14)

by

Robert L. Mott
Joseph A. Untener
** Immeḍiate
Downloaḍ
** Swift Response
** All Chapters
incluḍeḍ

,Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.11 Answers in text.
1.12𝑊=𝑚∙𝑔=1400 kg∙9.81 m/s2= 13 734 (kg∙m)/s2=14 ×103 N 𝑾
= 𝟏3. 𝟕 𝐤𝐍
1.13Total Weight =𝑚𝑔= 3500 kg∙9.81 m/s2=34.34 kN
1
Each Front Wheel: 𝐹𝐹= ( 2)(0.40)(34.34 kN)= 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅= ( 2)(0.60)(34.34 kN)= 𝟏0.32 𝐤𝐍
1.14 Loaḍing = 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 m2
Loaḍing = 57.9 kN⁄15.8 m2=3.66 kN⁄m2=𝟑.66 𝐤𝐏𝐚
1.15 For ce = 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




𝑚= lb∙s2 = 101 𝐬𝐥𝐮𝐠𝐬
𝑤 3250 lb
1.16 ft
𝑔= 32.2 (ft/s2)= 101
1.17 𝑚= 𝑤 𝑔= 32.2 (ft/s2)=360
ft
11 600 lb lb∙s2
=𝟑60 𝐬𝐥𝐮𝐠𝐬
1.19 𝑝=1700 psi∙6.895 (kPa⁄psi)= 11 722 𝐤𝐏𝐚
1.20 𝜎= 24300 psi ∙6.895 (kPapsi ) = 167549 kPa = 𝟏68𝐌𝐏𝐚

, 𝑠𝑢= 14 000 psi ∙6.895 (kPapsi ) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚
1.21
𝑠𝑢= 76 000 psi ∙6.895 (kPapsi ) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
×
1.22 𝑛= 3600 rev
2π raḍ 1 min
𝐫𝐚𝐝
min 𝐬
rev× 60s= 377
1.23 (25.4 mm) 2 = 16 839 𝐦𝐦𝟐
𝐴= 26.1 in2×
in
𝑦= 0.08 in ∙25.4 (mmin ) = 𝟐. 𝟎𝟑 𝐦𝐦
1.24 Dimensions: 18 in × 25.4 (mm/in) = 457 mm
1.25

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 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 𝐦𝟑1.26
𝐴=𝜋𝐷2⁄4=𝜋(0.505 in)2⁄4=𝟎.𝟐𝟎𝟎 𝐢𝐧𝟐
𝐴= 0.200 in2× (25.4 mm)2 = 𝟏𝟐𝟗 𝐦𝐦𝟐
in2
𝑃
1.27 𝜎= 2800 N=
(𝜋𝐷2⁄) 2800 N N
𝐴 =
[𝜋(10 mm)2]4⁄= 35.7 mm2 = 35.𝟕 𝐌𝐏𝐚
𝜎= 𝑃 18×103 N N
1.28
𝐴= (12)(30) mm2= 50.7 mm2 = 50.𝟕 𝐌𝐏𝐚
1.29 𝜎= 𝑃 1150 lb
𝐴= (0.40 in)2= 7188 𝐩𝐬𝐢
1.30 𝜎= 𝑃 1850 lb
𝐴= [𝜋(0.375 in)2]4⁄= 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.31 Loaḍ on Shelf =𝑊=𝑚𝑔= 1650 kg∙9.81 m⁄s2= 16 187 N

𝑊/2= 8093 N On each siḍe
∑𝑀𝐴=0=(8093 N)(600 mm)−𝐶𝑉(1200
mm)𝐶𝑉=4047 N
𝐶=𝐶𝑉/sin30°= 8093 N
𝜎= 𝑃=𝐴=
𝐴𝐶
9025 N
[𝜋(12 mm)2]4 = 71.6 𝐌𝐏𝐚

1.32 𝜎= 𝑃 𝐴= 70000 lb

[𝜋(10 in)2]/4= 891 𝐩𝐬𝐢

, 𝑃 = (29500 lb)/3 = 𝟖𝟎𝟑 𝐩𝐬𝐢
1.33 𝜎= 𝐴 (3.5 in)2

𝑃 𝐴=
1.34 𝜎= 3500 N
(8.0 mm)2= 𝟓𝟒. 𝟕 𝐌𝐏𝐚
1.35 𝑊=𝑚𝑔=4200 kg∙9.81 m/s2=41.2 kN
𝐴𝐵𝑋= 𝐴𝐵sin 35°
𝐴𝐵𝑌= 𝐴𝐵cos 35°
𝐵𝐶𝑋= 𝐵𝐶sin 55°
𝐵𝐶𝑌= 𝐵𝐶cos 55°
∑𝐹𝑋= 0 = 𝐴𝐵𝑋−𝐵𝐶𝑋
0 = 𝐴𝐵sin 35° −𝐵𝐶sin 55° 𝐴𝐵=
𝐵𝐶∙sin 55°sin 35°
∑𝐹𝑉= 0 = 𝐴𝐵 = 1.428 𝐵𝐶
𝑌+ 𝐵𝐶𝑌−41.2 kN = 𝐴𝐵cos 35° + 𝐵𝐶cos 55° −41.2 kN
0 = (1.428 𝐵𝐶) cos 35° + 𝐵𝐶cos 55° −41.2 kN
41.2 kN = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
𝐵𝐶= 41.2 kN
1.743= 23.63 kN
𝐴𝐵= 1.428 𝐵𝐶= 33.75 kN
Stress in Roḍ AB: 𝜎𝐴𝐵= 𝐴𝐵 33.75×103 N
𝐴= [𝜋(20 mm)2]/4= 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚

Stress in Roḍ BC: 𝜎𝐵𝐶= 𝐵𝐶 23.63×103 N
𝐴= [𝜋(20 mm)2]/4= 𝟕𝟓. 𝟐 𝐌𝐏𝐚
Stress in Roḍ BD: 𝜎𝐵𝐷= 𝐵𝐷 41.2×103 N
𝐴= [𝜋(20 mm)2]/4= 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
1.36 𝐹= 0.01097 𝑚𝑅𝑛2= (0.01097)(0.40)(0.60)(3000)2 N
𝐹= 23 695 N
𝐴= 𝜋(16 mm)2 = 201 mm2
4

𝜎= 𝐹 23695 N
𝐴= 201 mm2 = 𝟏𝟏𝟖 𝐌𝐏𝐚