Solution Manual for Applied Strength of Materials, 7th Edition by Robert L. Mott & Joseph A. Untener | Verified Solutions | Latest 2025/2026 Update | A+ Graded

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 N = N 14 N000 Npsi N∙ N6.895 N(kPa⁄psi) N= N 96 N500 NkPa N = N 𝟗𝟔. N𝟓 N N 𝐌𝐏𝐚

𝑠𝑢 N N= N 76 N000 Npsi N∙ N6.895 N(kPa⁄psi) N = N 524 N000 NkPa N = N 𝟓𝟐𝟒 N N𝐌𝐏𝐚
1.22
N 3600 N 𝑛 N= N
rev × N 2π N rad N × N 1 Nmin N N= N377 N N 𝐫𝐚𝐝
1.23 rev 60s
2 𝐬
min (25.4mm) 𝟐

𝐴 N= N26.1
i2n = N16 N839 N𝐦𝐦
N in2
×
1.24 𝑦 N = N 0.08 Nin N∙ N25.4 N(mm⁄in) N = N 𝟐. N𝟎𝟑 N𝐦𝐦
1.25 Dimensions: N18 Nin N × N 25.4 N(mm/in) N= N 457 Nmm
12 Nin N × N2 N25.4
Nmm N Area N= N(18 N in)
N (mm/in)
= NN𝟑𝟐𝟒 N𝐢𝐧
𝟐 N = N 305

Area N= N(457 Nmm)2 N N= N 𝟐. N𝟎𝟗 N× N𝟏𝟎𝟓 N N 𝐦𝐦𝟐
Volume N= N𝑉 N= NArea N× NHeight
𝑉 N = N 324 Nin2 N × N12 Nin N = N𝟑𝟖𝟖𝟖 N𝐢𝐧𝟑
𝑉 N = N (1.5 Nft)2 N × N1.0 Nft N = N 𝟐. N𝟐𝟓 N𝐟𝐭𝟑
𝑉 N = N (209 N× N103 N mm2) N× N305 Nmm N = N 𝟔. N𝟑𝟕 N× N𝟏𝟎𝟕 N N 𝐦𝐦𝟑

𝑉 N = N (0.457 Nm)2 N N N × N 0.305 Nm N = N 0.0637 Nm3 N N= N 𝟔. N𝟑𝟕 N× N𝟏𝟎−𝟐 N N 𝐦𝟑
1.26 N𝐴 N= N𝜋𝐷2⁄4 N= N𝜋(0.505 Nin)2⁄4 N= N𝟎. N𝟐𝟎𝟎 N𝐢𝐧𝟐
2
𝐴 N = N 0.200 Nin2 N × = N 𝟏𝟐𝟗 N𝐦𝐦𝟐
(25.4 N mm) N
N

in2

1𝑃N.27 𝜎 2800 N N N N N N 2800 N N N = N35.7 = N35. N𝟕 N𝐌𝐏𝐚
=
N =
𝐴 N
= [𝜋(10 N mm2
N(
𝜋𝐷2⁄4) N mm)2]⁄4
𝑃NN
1.28 𝜎 N= N =
18×103 N
= N50. N𝟕 N𝐌𝐏𝐚
N
= N50.7
N
𝐴 (12)(30) Nmm2 mm2

lb 𝑃NN
1.29 𝜎 N= N = = N7188 N𝐩𝐬𝐢

1150
N
𝐴
N in)2
(0.40


lb 𝑃 N = N 𝟏𝟔 N𝟕𝟓𝟎 N𝐩𝐬𝐢
1.30 𝜎 N= N =

1850
N
𝐴 [𝜋(0.375 Nin)2]⁄4

1.31 Load Non NShelf N = N𝑊 N= N𝑚𝑔 N= N1650 Nkg N∙ N9.81 Nm⁄s2 N= N16 N187 NN

, 𝑊/2 N= N8093 NN NOn Neach Nside
∑ N𝑀𝐴N= N0 N= N(8093 NN)(600 Nmm) N− N𝐶𝑉(1200 Nmm)

𝐶𝑉 N= N4047 NN
𝐶 N= N𝐶𝑉/ Nsin N30° N= N8093 NN
𝑃 𝐶 9025 NN N
𝜎 N= N 𝐴 N N =
=𝐴 [𝜋(12 Nmm) 2]⁄4 N N N N = N71.6 N 𝐌𝐏𝐚

1.32 N𝜎 N = NN NN = N N N 70000 Nlb



𝑃
N