Solution Manual of Applied Strength of Materials of Mott 6th edition pdf

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APPLIED STRENGTH OF MATERIALS
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1.1 to 1.15 Answers in text.
1.16 = = 1800 kg 9.81 m/s2 = 17 658 (kg m)/s2 = 17 × 103 N
= .
1.17 Total Weight = = 4000 kg 9.81 m/s2 = 39.24 kN
Each Front Wheel: = ( 1
2 ) (0.40)(39.24 kN) = .
Each Rear Wheel: = ( 1
2 ) (0.60)(39.24 kN) = .
1.18 Loading = Total Force / Area
Total Force = = 6800 kg 9.81 m/s2 = 66.7 kN
Area = (5.0 m)(3.5 m) = 17.5 m2
Loading = 66.7 kN 17.5 m2⁄ = 3.81 kN m2⁄ = .
1.19 Force = Weight = = 25 kg 9.81 m/s2 = 245 N
K = Spring Scale = 4500 N m⁄ = /Δ
Δ =
245 N
4500 N/m = 0.0545 m = 54.5 × 10−3 m = .
1.22 = 17.7 kN = 17 700 N 0.2248 (lb N⁄ ) =
1.23 = 7.85 kN = 7850 N 0.2248 (lb N⁄ ) =
= 11.77 kN = 11 770 N 0.2248 (lb N⁄ ) =
1.24 Loading = 3.81 kPa = 3.81×103 N
m2 × 0.2248 lb
1.25 = 245 N 0.2248 (lb N⁄ ) = .
= 4500 N
1.26 =
1.29 = 1200 psi 6.895 (kPa psi⁄ ) =
1.30 = 21 600 psi 6.895 (kPa psi⁄ ) = 149 000 kPa =
1.31 = 14 000 psi 6.895 (kPa psi⁄ ) = 96 500 kPa = .
= 76 000 psi 6.895 (kPa psi⁄ ) = 524 000 kPa =
1.32 = 1750 rev
in2 =
1.34 = 0.08 in 25.4 (mm in⁄ ) = .
1.35 Dimensions: 18 in × 25.4 (mm/in) = 457 mm
12 in × 25.4 (mm/in) = 305 mm
Area = (18 in)2 =
Area = (457 mm)2 = . ×
Volume = = Area × Height
= (1.5 ft)2 × 1.0 ft = .
= (209 × 103 mm2) × 305 mm = . ×
= (0.457 m)2 × 0.305 m = 0.0637 m3 = . × −
1.36 = 2 4⁄ = (0.505 in)2 4⁄ = .
= 0.200 in2 × (25.4 mm)2
in2 =
1.37 =
[(0.375 in)2] 4⁄ =
1.41 Load on Shelf = = = 1840 kg 9.81 m s2⁄ = 18 050 N
/2 = 9025 N On each side
∑ = 0 = (9025 N)(600 mm) − (1200 mm)
= 4512 N
=
1.42 =
1.45 = = 4200 kg 9.81 m/s2 = 41.2 kN
= sin 35°
= cos 35°
= sin 55°
= cos 55°
∑ = 0 = −
= sin 55°
sin 35° = 1.428
∑ = 0 = + − 41.2 kN = cos 35° + cos 55° − 41.2 kN
0 = (1.428 ) cos 35° + cos 55° − 41.2 kN
41.2 kN = [1.170 + 0.574] = 1.743
= 41.2 kN
1.743 = 23.63 kN
Stress in Rod AB: =
=
=
=
1.46 = 0.01097 2 = (0.01097)(0.40)(0.60)(3000)2 N
= 23 695 N
For AB: = (110 − 40 + 80) kN = 150 kN
=
For BC: = 110 − 40 = 70 kN
=
For CD: = 110 kN
=
1.48 Areas: A-C; 1 = (25)2/4 = 491 mm2
C-D; 2 = (16)2/4 = 201 mm2
For AB: = −9.65 − 12.32 + 4.45 = −17.52 kN
=
For BC: = −9.65 − 12.32 = −21.97 kN
=
For CD: = −9.65 kN
=
1.49 = [(1.90)2−(1.61)2]
4 = 0.799 in2 [1
For AB: = 2500 + 2(8000 cos 30°) = 16 356 lb
=
1.50 ∑ = 0 = 2800(45) − (30)
= 4200 lb
= 10.5 kN =
Stresses:
(12)(30) mm2 = . Tension
: = 10.5×103 N
(2)(10)(30) mm2 = . Tension
, : = (30)2 − (20)2 = 500 mm2
= = −10.5×103 N
500 mm2 = −. Compression
1.52 ∑ = 0 = 6000(6) + 12 000(12) − (18)
= 10 000 lb
= 8000 lb
=
= cos = 10 000(0.6) = 6000 lb Tension
sin + 6000 − sin = 0
= sin −6000
sin =
= cos + cos = 10 000(0.6) + 2500(0.6)
[Continued on next page]
= 12 000 − sin = 12 000 − 12 500(0.8)
CE = 2000 lb Compression
EF = CF cos = 12 500 lb(0.6) = 7500 lb Tension
Areas of members: Appendixes A-5(a) and A-6(a) AD, DE, EF – 2(0.484 in2) = 0.968 in2
BD, BE, CE – 0.484 in2
AB, BC, CF – 2(1.21 in2) = 2.42 in2
Stresses: AD = DE = 6000/0.968 = +6198 psi EF = 7500/0.968 = +7748 psi BD = 0 BE = 2500/0.484 = +5165 psi CE = -2000/0.484 = -4132 psi [NOTE: Compression members must be AB = -10 000/2.42 = -4132 psi checked for column buckling.] BC = -7500/2.42 = -3099 psi CF = -12 500/2.42 = -5165 psi
1.53 ∑ = 0 = (12.5)(4.0) − (2.5)
= 20 kN
1.55 = (2.65)(1.40) + 2[(1.40)(0.5)()] = 4.41 in2
=
4 = 3557 mm2
= [ (12.0)2
= 177 N
1.59 From Problem 1-46: = 23 695 N
= 2 [ (10)2
=
=
=
1.62 = √0.42 + 0.62 = 0.721 in
= [2(1.60) + (0.8)
2 + 2(0.721)] 0.194
= 1.144 in2
1.63 =
=