Solutions Manual for Applied Strength of Materials, 7th Edition by Mott & Untener (All Chapters)

SolutionsgManualgforgAppliedgStrengthgofgMaterials,g7egbygRobert

Mott,gJosephgUntenerg(AllgChapters)
Chapterg1 BasicgConceptsgingStrengthgofgMaterials
1.1gtog1.11gAnswersgingtext.
1.12 𝑊g=g𝑚g∙g𝑔g=g1400gkgg∙g9.81gm/s2g=g13g734g(kgg∙gm)/s2g=g14g×g103gN
𝑾g=g𝟏3.g𝟕g𝐤𝐍
1.13 TotalgWeightg=g𝑚 𝑔g=g3500gkgg∙g9.81gm/s2g=g34.34gkN
1
EachgFrontgWheel:g𝐹𝐹g =g ( )g(0.40)(34.34gkN)g=g6.87g𝐤𝐍
2
1
EachgReargWheel:g𝐹𝑅g =g ( )g(0.60)(34.34gkN)g=g𝟏0.32g𝐤𝐍
2
1.14 Loadingg=gTotalgForceg/gArea
TotalgForceg=g𝑚 𝑔g=g5900gkgg∙g9.81gm/s2g=g57.9gkN
Areag=g (4.5gm)(3.5gm)g =g 15.8gm2
Loadingg=g57.9gkN⁄15.8gm2g=g3.66gkN⁄m2g=g𝟑.66g𝐤𝐏𝐚
1.15 Forceg = 𝑚g 𝑔g=g35gkgg∙g9.81gm/s2g=g343gN
Kg =g Springg Scaleg =4800gN⁄mg=g𝐹/Δ𝐿

Δ𝐿g= 𝐹g=g343gN =g0.0715gmg=g71.5g×g10−3 mg=g71.g𝟓g𝐦𝐦
𝐾 4800g N/m




𝑤g lb∙s2 =g101g𝐬𝐥𝐮𝐠𝐬
1.16 𝑚g =g = 3250gl =g101g
b
𝑔 32.2g(ft/s2) ft

𝑤g 11g600glb lb∙s2 =g𝟑60g𝐬𝐥𝐮𝐠𝐬
1.17 𝑚g = g =g
𝑔 32.2g(ft/s2)=g360g ft
1.19 𝑝g=g1700gpsig∙g6.895g(kPa⁄psi)g=g11g722g𝐤𝐏𝐚
1.20 𝜎g =g 24g300gpsig∙g6.895g(kPa⁄psi)g =g 167g549gkPag =g 𝟏68g 𝐌𝐏𝐚

, 1.21 𝑠𝑢g =g 14g000gpsig∙g6.895g(kPa⁄psi)g =g 96g500gkPag =g 𝟗𝟔.g𝟓g𝐌𝐏𝐚
𝑠𝑢g =g 76g000gpsig∙g6.895g(kPa⁄psi)g =g 524g000gkPag =g 𝟓𝟐𝟒g𝐌𝐏𝐚

3600grevg 2πgradg 1gming 𝐫𝐚𝐝
1.22 𝑛g=g ×g ×g
min rev 60s =g377g 𝐬
2
(25.4mm)
𝐴g=g26.1gin2g×g
g
1.23 =g16g839g𝐦𝐦𝟐
i2n
1.24 𝑦g =g 0.08ging∙g25.4g(mm⁄in)g =g 𝟐.g𝟎𝟑g𝐦𝐦
1.25 Dimensions:g18ging ×g 25.4g(mm/in)g =g 457gmm
12ging ×g 25.4g(mm/in)g =g 305gmm

Areag=g(18gin)2g =g 𝟑𝟐𝟒g𝐢𝐧𝟐

Areag=g(457gmm)2g =g 𝟐.g𝟎𝟗g×g𝟏𝟎𝟓g 𝐦𝐦𝟐
Volumeg=g𝑉g =gAreag×gHeight

𝑉g =g 324gin2g ×g12ging =g 𝟑𝟖𝟖𝟖g𝐢𝐧𝟑

𝑉g =g (1.5gft)2g ×g1.0gftg =g 𝟐.g𝟐𝟓g𝐟𝐭𝟑
𝑉g =g (209g×g103g mm2)g×g305gmmg =g 𝟔.g𝟑𝟕g×g𝟏𝟎𝟕g 𝐦𝐦𝟑

𝑉g =g (0.457gm)2g ×g 0.305gmg =g 0.0637gm3g =g 𝟔.g𝟑𝟕g×g𝟏𝟎−𝟐g 𝐦𝟑

1.26 𝐴g=g𝜋𝐷2⁄4g=g𝜋(0.505gin)2⁄4g=g𝟎.g𝟐𝟎𝟎g𝐢𝐧𝟐
(25.4gmm)2g
𝐴g =g 0.200g in2g ×g =g 𝟏𝟐𝟗g𝐦𝐦𝟐
in2
𝑃 2800gN 2800gN N
1.27 𝜎g =g = =g 35.7 =g35.g𝟕g𝐌𝐏𝐚
𝐴=
(𝜋𝐷2g⁄4) 2
[𝜋(10gmm) ]⁄4 mm2
𝑃g 3g N N
1.28 𝜎g =g = 18×10 =g 50.7 =g50.g𝟕g𝐌𝐏𝐚
𝐴 (12)(30)g mm2 mm2
𝑃g 1150glbg
1.29 𝜎g =g =g g =g7188g𝐩𝐬𝐢
𝐴 (0.40g in)2

𝑃g =g 𝟏𝟔g𝟕𝟓𝟎g𝐩𝐬𝐢
1.30 𝜎g = g = 1850gl
b
𝐴 [𝜋(0.375gin)2]⁄4
1.31 LoadgongShelfg =g𝑊g=g𝑚𝑔g=g1650gkgg∙g9.81gm⁄s2g=g16g187gN
𝑊/2g=g8093gNgOngeachgside
∑g𝑀𝐴g=g0g=g(8093gN)(600gmm)g−g𝐶𝑉(1200gmm)
𝐶𝑉g=g4047gN
𝐶g=g𝐶𝑉/gsing30°g=g8093gN
𝑃g 𝐶 g g 9025gN
𝜎g=g =g =
𝐴g g 𝐴 [𝜋(12g mm)2]⁄4g g =g 71.6g 𝐌𝐏𝐚
𝑃 70000glbg g
1.32 𝜎g=g =g g = g891g𝐩𝐬𝐢
𝐴 [𝜋(10gin)2]/4

, 𝑃 (29500glb)/3g
1.33 𝜎g=g =g =g 𝟖𝟎𝟑g 𝐩𝐬𝐢
𝐴 (3.5gin)2

𝑃 3500gN =g 𝟓𝟒.g𝟕g𝐌𝐏𝐚
1.34 𝜎g=g =g g
𝐴
(8.0gmm)
2

1.35 𝑊g=g𝑚 𝑔g=g4200gkgg∙g9.81gm/s2g=g41.2gkN
𝐴𝐵𝑋g =g 𝐴𝐵gsing35°
𝐴𝐵𝑌g =g 𝐴𝐵gcosg35°
𝐵𝐶𝑋g =g 𝐵𝐶gsing55°
𝐵𝐶𝑌g =g 𝐵𝐶gcosg55°
∑g𝐹𝑋g=g0g=g𝐴𝐵𝑋g−g𝐵𝐶𝑋

0g =g 𝐴𝐵gsing35°g−g𝐵𝐶gsing55°
sing55°
𝐴𝐵g=g𝐵𝐶g∙
=g1.428g𝐵𝐶gsing
35°
∑g𝐹𝑉g =g0g =g𝐴𝐵𝑌g+g𝐵𝐶𝑌g−g41.2gkNg=g 𝐴𝐵gcosg35°g+g𝐵𝐶gcosg55°g−g41.2gkNg0g=g(1.428g𝐵𝐶)gcosg3
5°g+g𝐵𝐶gcosg55°g−g41.2gkN
41.2gkNg =g 𝐵𝐶[1.170g+g0.574]g =g 1.743g𝐵𝐶
41.2gkNg
𝐵𝐶g =g =g 23.63gkN
1.743
𝐴𝐵g =g 1.428g𝐵𝐶g =g 33.75gkN
𝐴𝐵g 33.75×10g g3Ng g
StressgingRodgAB:g𝜎 =g =g g =g 𝟏𝟎𝟕.g𝟒g𝐌𝐏𝐚
𝐴𝐵
𝐴 [𝜋(20gmm)2]/4
𝐵𝐶g 23.63×10g 3Ng g
StressgingRodgBC:g𝜎 =g =g g =g 𝟕𝟓.g𝟐g𝐌𝐏𝐚
𝐵𝐶 [𝜋(20gmm)2]/4
𝐴
𝐵𝐷g 41.2×10g3g N
StressgingRodgBD:g𝜎 =g = =g 𝟏𝟑𝟏.g𝟏g𝐌𝐏𝐚
𝐵𝐷 2
[𝜋(20gmm) ]/4
𝐴
1.36 𝐹g=g0.01097g𝑚 𝑅 𝑛2g=g(0.01097)(0.40)(0.60)(3000)2gN
𝐹g =g 23g695gN
𝜋(16gmm)2g 2
𝐴g =g =g 201g mm
4
𝐹g 23695gNg
𝜎g = g = g =g 𝟏𝟏𝟖g𝐌𝐏𝐚
𝐴 201gmm2

, 1.37 𝐴g=g(30gmm)2g=g900gmm2
ForgAB:g𝐹𝐴𝐵g =g (110g−g40g+g80)gkNg =g 150gkN
𝐹𝐴𝐵g 150×103g Ng
𝜎 =g =g =g𝟏𝟔𝟕g𝐌𝐏𝐚gTension
𝐴𝐵 𝐴 900gmm2
ForgBC:g𝐹𝐵𝐶g =g 110g−g40g =g 70gkN
𝐹𝐵𝐶g 70×103g Ng
𝜎 =g =g =g 𝟕𝟕.g𝟖g 𝐌𝐏𝐚g Tension
𝐵𝐶 𝐴 900gmm2
ForgCD:g𝐹𝐶𝐷g =g 110gkN
𝐹𝐶𝐷g 110×103g Ng
𝜎 =g =g =g 𝟏𝟐𝟐g 𝐌𝐏𝐚g Tension
𝐶𝐷 𝐴 900gmm2
1.38 Areas:gA-C;g𝐴1g=g𝜋(25)2/4g=g491gmm2
C-D;g 𝐴2g =g 𝜋(16)2/4g =g 201g mm2
ForgAB:g 𝐹𝐴𝐵g =g −9.65g−g12.32g+g4.45g =g −17.52gkN
𝐹𝐴𝐵g −17.52×103g Ng
𝜎 =g =g =g −𝟑𝟓.g𝟕g 𝐌𝐏𝐚g Compression
𝐴𝐵 𝐴1 491gmm2
ForgBC:g 𝐹𝐵𝐶g =g −9.65g−g12.32g =g −21.97gkN
𝐹𝐵𝐶g −21.97×103g Ng
𝜎 =g =g =g −𝟒𝟒.g𝟕g𝐌𝐏𝐚g Compression
𝐵𝐶 𝐴1 2
491gmm
ForgCD:g𝐹𝐶𝐷g =g −9.65gkN
𝐹𝐶𝐷g −9.65×103g Ng
𝜎 =g =g =g −𝟒𝟖.g𝟎g𝐌𝐏𝐚g Compression
𝐶𝐷 2
𝐴 201gmm
2
2 2 1
1.39 𝐴g= 𝜋[(1.90) −(1.61) ] =g 0.799gin2 [1g g ingPipe-AppendixgA-9(a)]
4 2
𝐹 𝐵𝐶 g 2500glbg
ForgBC:g𝜎 =g =g =g 𝟑𝟏𝟐𝟗g𝐩𝐬𝐢g Tension
𝐵𝐶
𝐴 0.799gin2
ForgAB:g 𝐹𝐴𝐵g =g 2500g+g2(8000gcosg30°)g =g 16g356glb
𝐹𝐴𝐵g 16g 356glbg
𝜎 =g =g =g 𝟐𝟎g 𝟒𝟕𝟏g𝐩𝐬𝐢g Tension
𝐴𝐵 2
𝐴 0.799g in
1.40 ∑g𝑀𝐶g=g0g=g2800(45)g−g𝐹𝐵𝐷(30)
𝐹𝐵𝐷g =g 4200glb

𝐹𝐵𝐷g =g 𝟑𝟐𝟑𝟏g𝐩𝐬𝐢g Tension
𝜎 4200
=g glb =
𝐵𝐷 (2.0)(0.65)g in2
𝐴