Solution Manual for Applied Strength of Materials 7th Edition by Robert L. Mott & Joseph A. Untener – Latest Update 2025/2026

SOLUTION MANUAL FOR APPLIED
STRENGTH OF MATERIALS 7TH EDITION
BY ROBERT L.MOTT , JOSEPH
A.UNTENER




Solution matual

,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
1
Each Front Wheel: 𝐹 = ( (0.40)(34.34 kN) = 6.87 𝐤𝐍
𝐹 )
12
Each Rear Wheel: 𝐹 = ( (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
𝑅 2)
1.14 Loading = Total Force / Area
Total
Area =Force
(4.5 = 𝑚𝑔 =m)
m)(3.5 5900 kg ∙ 9.81
= 15.8
2
m2 m/s = 57.9 kN
Loading = 57.9 kN⁄15.8 m2 = 3.66 kN⁄m2 = 𝟑.66 𝐤𝐏𝐚
1.15 Force = 𝑚𝑔 = 35 kg ∙ 9.81 m/s2 = 343 N
K = Spring Scale =4800 N⁄m = 𝐹/Δ𝐿
𝐹 343 N
Δ𝐿 = = = 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦

𝐾 4800 N/m


𝑤 3250 lb = 101 𝐬𝐥𝐮𝐠𝐬
1.16 𝑚= = = 101
lb∙s2
𝑔 32.2 (ft/s2) ft
𝑤 11 600 lb∙s
1lb.17 𝑚= = = 360 2
= 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

,1.21 𝑠𝑢 = 14 000 psi ∙ 6.895 (kPa⁄psi) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚
P P P P P P P P P P P P P P P P




𝑠𝑢 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚
PP P P P P P P P P P P P P PP


1.22
3600 𝑛=
× 2π rad × 1 min = 377 𝐫𝐚𝐝
P P
P P P P P PP
P P
rev P P P

1.23 rev 60s
2 𝐬
min (25.4mm) 𝟐

𝐴 = 26.1
P P



i2n = 16 839 𝐦𝐦
in2
P P P
P




×
1.24 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦
P P P P P P P P P P



1.25 Dimensions: 18 in × 25.4 (mm/in) = 457 mm P P P P P P P P



Pmm Area =12 (18inin)×2 25.4
P = 𝟑𝟐𝟒(mm/in)
𝐢𝐧𝟐 = 305 P P
P
P
P PP P
P
P
P
P P



Area = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐
P P P
PP
P P P P
P P




Volume = 𝑉 = Area × Height P P P P P P




𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑
P P P
P
P P P P P




𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭𝟑
P P P
P
P P P P P P




𝑉 = (209 × 103 mm2) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
P P P P
P
P P P P P P P P
P P




𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
P P P
PP P
P P P P P
PP
P P P P
P P




1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐
P P P P P P P P P P
2
(25.4
𝐴 = 0.200 in2 ×
P P P
P
P = 𝟏𝟐𝟗 𝐦𝐦𝟐 P P

P mm) N

in2

1𝑃 .27 𝜎 2800 P N P P P 2800 P P N = 35.7 = 35. 𝟕 𝐌𝐏𝐚
=
P P P P P P




P =
𝐴 P = P(𝜋𝐷2⁄4) [𝜋(10 P Pmm)2]⁄4 mm2
𝑃PP
1.28 𝜎 = P P =
18×103
N
= 50.7 = 50. 𝟕 𝐌𝐏𝐚
P
P P P P



N
𝐴 (12)(30)Pmm2 mm2

lb 𝑃PP
1.29 𝜎 = P P = = 7188 𝐩𝐬𝐢 P P




1150
P

𝐴
P in)
2 (0.40


lb 𝑃P = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
1.30 𝜎= P P = P P P




1850
P

𝐴 [𝜋(0.375 Pin)2]⁄4

1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N
P P P P P P P P P P P P
P
P P P

, 𝑊/2 = 8093 N On each side
P P P P P P




∑ 𝑀𝐴 = 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm)
P
P P P P P P P P P




𝐶𝑉 = 4047 N
P P P



𝐶 = 𝐶𝑉/ sin 30° = 8093 N
P P P P P P P


𝑃 𝐶 9025 PN P
𝜎= P P
𝐴P P=
= 𝐴 [𝜋(12 Pmm) 2]⁄4 P P P P = 71.6 𝐌𝐏𝐚
P P




1.32 𝜎 = 70000 Plb
=
P P PP PP
PP P




𝑃
P