SOLUTION MANUAL
Applied Statics and Strength of Materials, 7th
i i
Edition
i
By George Limbrunner (All Chapters 1 – 20)
i i i i i
SOLUTION MANUAL i
, Contents
Chapteri1 1
Chapteri2 12
Chapteri3 20
Chapteri4 49
Chapteri5 71
Chapteri6 115
Chapteri7 158
Chapteri8 177
Chapteri9 202
Chapteri10 221
Chapteri11 232
Chapteri12 267
Chapteri13 285
Chapteri14 330
Chapteri15 361
Chapteri16 398
Chapteri17 428
Chapteri18 474
Chapteri19 502
Chapteri20 513
iii
, Solution Manual i
FOR
APPLIED STATICS AND STRENGTH OF MATERIALS
i i i i i
SeṿenthiEdition
GeorgeiF.iLimbrunner,iP.E.,i(INACTIṾE)iC
raigiT.iD’Allaird,iP.E.
NOTES:
1. Theisolutionsipresentedihereiniare,iinigeneral,isomewhatiabbreṿiateditoico
nserṿeispace.iṾeryilittleiexplanationiisifurnished.iSketchesiareikeptitoiaimini
mum.iFewichecksiareishown.
2. Theisolutionsifollowitheiproceduresideṿelopediinitheiexamplesiinitheitext.
3. Theisolutionsiareibasedionitheilimiteditablesifurnishediinitheitextiand/orith
eiappendices.iTheitablesifurnishediareiforitheipurposesiofithisitextionlyiand
ishouldinotibeiusediforidesign.
4. Theisolutionsiforitheidesigniproblemsiareigenerallyinotitheionlyisoluti
oninoriareitheyinecessarilyitheimostieconomicalisolutions.
5. Pleaseinoteithatiprobleminumbersiinitheisolutionimanualiareidepictediwit
hibothidashesi(‐
)iandiperiodsi(.)ibetweenitheichapteriandiprobleminumbers.iTheseiareiinte
rchangeable.
6. Itishouldibeinotedithatitheipreṿiousieditionsiofitheitextiusedilowercaseisitoi
denoteistresses.iThisihasichangediinithei7thiedition,ibutidueitoitimeiconstra
intsithisisolutionimanualihasinotibeenicompletelyiupdateditoireflectisuch.
7. Ifiyouifindierrorsiinithisimanualioriinitheitext,ipleaseiforwardit
hemitoimeiatic.dallaird@hṿcc.edu.
, Prob.i1.1
(a) 𝑐i =i √102i +i72i =i 12.21ift
(b)i 𝑏i =i √202i −i162i =i 12.00im
Prob.i1.2
(a) 𝑎i =i 25isini48°i =i 18.58ift
(b) 𝑏i =i √252i −i18.582i =i 16.73ift
(c) ℎi =i bisini48°i =i 16.73i sini48°i =i 12.43ift
Prob.i1.3
𝑎i =i √722i −i67.32i =i 25.59ift
Ai=icos–1(67.3/72)i=i20.8
Bi=isin–1(67.3/72)i=i69.2°
Prob.i1.4
ABi=i28isini70°i=i26.3ift
Prob.i1.5
𝑐i =i √102i +i62i =i 11.6ift
𝜃i =i tan−1(6⁄10)i =i 31.0°
Prob.i1.6
𝐴𝐵i=i √122i+i162i =i20ift
𝐵𝐶i =i √122i +i322i =i 34.2ift
𝐴i =i tan−1(12⁄16)i =i 36.9°
𝐶i =i tan−1(12⁄32)i =i 20.6°
Prob.i1.7
θi =i sin−1(5⁄6)i =i 56.4°
𝑥i =i √122i +i102i =i 6.63ift
1-1
Applied Statics and Strength of Materials, 7th
i i
Edition
i
By George Limbrunner (All Chapters 1 – 20)
i i i i i
SOLUTION MANUAL i
, Contents
Chapteri1 1
Chapteri2 12
Chapteri3 20
Chapteri4 49
Chapteri5 71
Chapteri6 115
Chapteri7 158
Chapteri8 177
Chapteri9 202
Chapteri10 221
Chapteri11 232
Chapteri12 267
Chapteri13 285
Chapteri14 330
Chapteri15 361
Chapteri16 398
Chapteri17 428
Chapteri18 474
Chapteri19 502
Chapteri20 513
iii
, Solution Manual i
FOR
APPLIED STATICS AND STRENGTH OF MATERIALS
i i i i i
SeṿenthiEdition
GeorgeiF.iLimbrunner,iP.E.,i(INACTIṾE)iC
raigiT.iD’Allaird,iP.E.
NOTES:
1. Theisolutionsipresentedihereiniare,iinigeneral,isomewhatiabbreṿiateditoico
nserṿeispace.iṾeryilittleiexplanationiisifurnished.iSketchesiareikeptitoiaimini
mum.iFewichecksiareishown.
2. Theisolutionsifollowitheiproceduresideṿelopediinitheiexamplesiinitheitext.
3. Theisolutionsiareibasedionitheilimiteditablesifurnishediinitheitextiand/orith
eiappendices.iTheitablesifurnishediareiforitheipurposesiofithisitextionlyiand
ishouldinotibeiusediforidesign.
4. Theisolutionsiforitheidesigniproblemsiareigenerallyinotitheionlyisoluti
oninoriareitheyinecessarilyitheimostieconomicalisolutions.
5. Pleaseinoteithatiprobleminumbersiinitheisolutionimanualiareidepictediwit
hibothidashesi(‐
)iandiperiodsi(.)ibetweenitheichapteriandiprobleminumbers.iTheseiareiinte
rchangeable.
6. Itishouldibeinotedithatitheipreṿiousieditionsiofitheitextiusedilowercaseisitoi
denoteistresses.iThisihasichangediinithei7thiedition,ibutidueitoitimeiconstra
intsithisisolutionimanualihasinotibeenicompletelyiupdateditoireflectisuch.
7. Ifiyouifindierrorsiinithisimanualioriinitheitext,ipleaseiforwardit
hemitoimeiatic.dallaird@hṿcc.edu.
, Prob.i1.1
(a) 𝑐i =i √102i +i72i =i 12.21ift
(b)i 𝑏i =i √202i −i162i =i 12.00im
Prob.i1.2
(a) 𝑎i =i 25isini48°i =i 18.58ift
(b) 𝑏i =i √252i −i18.582i =i 16.73ift
(c) ℎi =i bisini48°i =i 16.73i sini48°i =i 12.43ift
Prob.i1.3
𝑎i =i √722i −i67.32i =i 25.59ift
Ai=icos–1(67.3/72)i=i20.8
Bi=isin–1(67.3/72)i=i69.2°
Prob.i1.4
ABi=i28isini70°i=i26.3ift
Prob.i1.5
𝑐i =i √102i +i62i =i 11.6ift
𝜃i =i tan−1(6⁄10)i =i 31.0°
Prob.i1.6
𝐴𝐵i=i √122i+i162i =i20ift
𝐵𝐶i =i √122i +i322i =i 34.2ift
𝐴i =i tan−1(12⁄16)i =i 36.9°
𝐶i =i tan−1(12⁄32)i =i 20.6°
Prob.i1.7
θi =i sin−1(5⁄6)i =i 56.4°
𝑥i =i √122i +i102i =i 6.63ift
1-1
