재료역학 7판 대학교재솔루션 Gere Up
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재료역학 7판 대학교재솔루션 Gere Up
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재료역학 7판 대학교재솔루션 Gere
재료역학 7판 솔루션 Gere
재료역학 7판 솔루션 GereDraft solutions manual for 7th edition of Mechanics of Materials, J. Gere & B. Goodno, 2009 -- see also updated Answers to Problems -- see also errata table NOTE: This file contains only solutions for those problems which are either new or revised versions of those in the 6e of this text; please consult the 6e ISM for any problem solutions not shown here
========================================================================================= SOLUTION Part (a) P1 = 1700 dAB = 1.25 tAB = 0.5 dBC = 2.25 tBC = 0.375
2 2 π ? ? dAB ? dAB ? 2? tAB ? ? ? AAB = 4
(
)
AAB = 1.178
σ AB =
P1 AAB
σ AB = 1.443 × 10
1443 psi
3
psi
Part (b)
2 2 π ? ? dBC ? dBC ? 2? tBC ? ? ? ABC = 4
(
)
ABC = 2.209
P2 = σ AB? ABC ? P1 P1 + P2
P2 = 1.488 × 10
1488 lbs 3
3
lbs
CHECK: Part (c) P2 = 2260 P1 + P2 σ AB
2
ABC
= 1.443 × 10
= ABC
P1 + P2 σ AB
= 2.744
green box contains answer in accordance with significant digits rule (Sec. B. 4)
(dBC ? 2? tBC)
2 4 ? P1 + P2 ? ? = dBC ? ? ? π ? σ AB ?
?
?
dBC ? 2? tBC =
dBC ?
2
4 ? P1 + P2 ? ? ?? π ? σ AB ?
?
?
dBC ? tBC =
dBC ? 2
2
? 4 ? P1 + P2 ? ?? π ? σ AB ? ? ?
tBC = 0.499
inches
======================================================================================== SOLUTION P = 70 N δ = 0.214 mm P Ae δ L N mm
?4 2
Ae = 1.075 mm
2
L = 460 mm
σ =
σ = 65.116
65.1 MPa
MPa =
ε = σ ε
ε = 4.652 × 10
4.65 x 10^-4 5
= 1.4 × 10
MPa
σ ε ? ( 1000)
= 139.97
GPa
^ 6e, p. 71 - E for cables is about 140 GPa or 20,000 ksi
========================================================================================= SOLUTION
2 2
T = 45 lbs
Apad = 0.625 in
Acable = 0.00167
in
(a) cantilever brakes - braking force RB & pad pressure
∑ MA = 0
RB = 4 2 RB Apad T ?T
RB? ( 1) = TDCh ( 3) + TDCv? ( 1) RB = 127.279
127.3 lbs
TDCh = TDCv 4
TDCh =
T 2
lbs
= 2.828 2 ^ vs 4.25 for V brakes 4.25 = 1.503 2.828 < same for V-brakes (below)
σ pad =
σ pad = 203.647
204 psi
psi
4
σ cable =
Acable
σ cable = 2.695 × 10
26,946 psi
psi
(b) V brakes - braking force RB & pad pressure
∑ MA = 0
RB = 4.25? T
RB = 191.25
191.3 lbs
lbs
< 50% more effective
σ pad =
RB Apad
σ pad = 306
psi
========================================================================================= SOLUTION P = 3200 kips 1 2? 2 A = ( 24 + 20) ? ( 20 + 16 + 8) ? ? ? 8 ? ? 20 ? 2 ? ? A = 1.504 × 10 P A
3
in
2
show result as 2.13 ksi (i.e., 3 significant figures)
σc =
σ c = 2.128 ksi
( 24) ? ( 20 + 16) ? 12 + ( 24 ? 8) ? ( 8) ? ? 8 + xc = xc = 19.22 inches
? ?
1 2 ?2 ? 24 ? 8 ? ? + ( 20) ? ( 16 + 8) ? ( 24 + 10) + ? 8 ? ? ? 8? 2 ? 2 ?3 ? A
( )
( 24) ? ( 20 + 16) ? ? 8 + ? yc =
?
20 + 16 ? ? 16 + 8 ? + ( 24 ? 8) ? ( 8) ? ( 4) + 1 ? 82 ? ? 2 ? 8? ? + ( 20) ? ( 16 + 8) ? ? ? ? ? 2 ? 2 ? 2 ? ?3 ? A
(
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[문서정보]
문서분량 : 520 Page
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자료제목 :
파일이름 : 솔루션 물리 재료역학 7판 솔루션 Gere.pdf
키워드 : 재료역학,7판,솔루션,Gere,대학교재솔루션
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