Complete Solutions Manual for Applied Strength of Materials 7th Edition (2026/2027) by Robert L. Mott & Joseph A. Untener – Step‑by‑Step Worked Solutions for All Chapters (1–14) with Full Calculations & Engineering Problem‑Solving Support

,SOLUTIONS MANUAL FOR
APPLIED STRENGTH
OF MATERIALS

7th Edition
Complete Chapter Sol𝓊tions
Man𝓊al
are incl𝓊ded (Ch 1 to 14)

by

Robert L. Mott
Joseph A. Untener
** Immediate
Download
** Swift Response
** All Chapters
incl𝓊ded

,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 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 m2
Loading = 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 (kPa psi ) = 167549 kPa = 𝟏68 𝐌𝐏𝐚

, 𝑠𝑢= 14 000 psi ∙6.895 (kPa psi ) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚
1.21
𝑠𝑢= 76 000 psi ∙6.895 (kPa psi ) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
×
1.22 𝑛= 3600 rev
2π rad 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 (mm in ) = 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
(𝜋𝐷2⁄N)= 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 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/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