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
SOLUTIONS MANUAL FOR 𝑾 = 𝟏3. 𝟕 𝐤𝐍
1.13 Total Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN
APPLIED STRENGTH
1
Each Front Wheel: 𝐹𝐹 = (2) (0.40)(34.34 kN) = 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = (2) (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
OF MATERIALS 1.14 Loading = Total Force / Area
Total Force = 𝑚𝑔 = 5900 kg ∙ 9.81 m/s2 = 57.9 kN
Area = (4.5 m)(3.5 m) = 15.8 m2
Loading = 57.9 kN⁄15.8 m2 = 3.66 kN⁄m2 = 𝟑.66 𝐤𝐏𝐚
7th Edition 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 = 𝐾 = 4800 N/m
= 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
Complete Chapter Solutions Manual
are included (Ch 1 to 14)
by
Robert L. Mott
Joseph A. Untener
** Immediate Download 1.16 𝑚=
𝑤
𝑔
3250 lb 2
= 32.2 (ft/s2 ) = 101 lb∙sft = 101 𝐬𝐥𝐮𝐠𝐬
** Swift Response 𝑤 11 600 lb 2
𝑚= = 32.2 (ft/s2 ) = 360 lb∙sft = 𝟑60 𝐬𝐥𝐮𝐠𝐬
** All Chapters included 1.17 𝑔
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 = 14 000 psi ∙ 6.895 (kPa⁄psi) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚 1.33 𝜎 =
𝑃
=
(29500 lb)/3
= 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2
𝑠𝑢 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚 3500 N
𝑃
2π rad 1 min 1.34 𝜎 = 𝐴
= (8.0 mm)2 = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
3600 rev 𝐫𝐚𝐝
1.22 𝑛= min
× rev
× 60s = 377 𝐬
2
1.35 𝑊 = 𝑚𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
(25.4mm) 𝟐
1.23 𝐴 = 26.1 in × 2
in
2
= 16 839 𝐦𝐦 𝐴𝐵𝑋 = 𝐴𝐵 sin 35°
1.24 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦 𝐴𝐵𝑌 = 𝐴𝐵 cos 35°
1.25 Dimensions: 18 in × 25.4 (mm/in) = 457 mm 𝐵𝐶𝑋 = 𝐵𝐶 sin 55°
12 in × 25.4 (mm/in) = 305 mm 𝐵𝐶𝑌 = 𝐵𝐶 cos 55°
Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐 ∑ 𝐹𝑋 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
Area = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐 0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55°
Volume = 𝑉 = Area × Height sin 55°
𝐴𝐵 = 𝐵𝐶 ∙ = 1.428 𝐵𝐶
𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑 sin 35°
∑ 𝐹𝑉 = 0 = 𝐴𝐵𝑌 + 𝐵𝐶𝑌 − 41.2 kN = 𝐴𝐵 cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭 𝟑
0 = (1.428 𝐵𝐶) cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
𝑉 = (209 × 103 mm2 ) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
41.2 kN = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
41.2 kN
𝐵𝐶 = = 23.63 kN
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐 1.743
(25.4 mm)2 𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
𝐴 = 0.200 in2 × in2
= 𝟏𝟐𝟗 𝐦𝐦𝟐
𝐴𝐵 33.75×103 N
𝑃2800 N 2800 N N Stress in Rod AB: 𝜎𝐴𝐵 = = [𝜋(20 = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
1.27 𝜎 = 𝐴 = (𝜋𝐷 = [𝜋(10 mm)2 ]⁄4 = 35.7 = 35. 𝟕 𝐌𝐏𝐚 𝐴 mm)2 ]/4
2 ⁄4) mm2
𝐵𝐶 23.63×103 N
𝑃 18×10 3 N N Stress in Rod BC: 𝜎𝐵𝐶 = = [𝜋(20 = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
1.28 𝜎 = 𝐴 = (12)(30) = 50.7 = 50. 𝟕 𝐌𝐏𝐚 𝐴 mm)2 ]/4
mm2 mm2
𝐵𝐷 41.2×103 N
𝑃 1150 lb Stress in Rod BD: 𝜎𝐵𝐷 = = [𝜋(20 mm)2 ]/4 = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
1.29 𝜎 = 𝐴 = (0.40 in)2 = 7188 𝐩𝐬𝐢 𝐴
𝑃 1850 lb 1.36 𝐹 = 0.01097 𝑚𝑅𝑛2 = (0.01097)(0.40)(0.60)(3000)2 N
1.30 𝜎= = [𝜋(0.375 in)2 ]⁄4 = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
𝐴
𝐹 = 23 695 N
1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N 𝜋(16 mm)2
𝐴= 4
= 201 mm2
𝑊/2 = 8093 N On each side
𝐹 23695 N
∑ 𝑀𝐴 = 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm) 𝜎 = 𝐴 = 201 mm2 = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝐶𝑉 = 4047 N
𝐶 = 𝐶𝑉 / sin 30° = 8093 N
𝑃 𝐶 9025 N
𝜎 = 𝐴 ==𝐴 [𝜋(12 mm)2 ]⁄4 = 71.6 𝐌𝐏𝐚
𝑃 70000 lb
1.32 𝜎 = = [𝜋(10 = 891 𝐩𝐬𝐢
𝐴 in)2]/4
,1.37 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80) kN = 150 kN 1.41 𝐴𝐷 sin 30° = 5.25 kN
𝐹𝐴𝐵 150×103 N 𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝜎𝐴𝐵 = 𝐴
= 900 mm2
= 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
For BC: 𝐹𝐵𝐶 = 110 − 40 = 70 kN
Stresses:
𝐹𝐵𝐶 70×103 N
𝜎𝐵𝐶 = 𝐴
= 900 mm2
= 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension 9.09×103 N
𝐴𝐵, 𝐵𝐶: 𝜎𝐴𝐵 = 𝜎𝐵𝐶 = (12)(30) mm2 = 𝟐𝟓. 𝟑 𝐌𝐏𝐚 Tension
For CD: 𝐹𝐶𝐷 = 110 kN
10.5×103 N
𝐹𝐶𝐷 110×103 N 𝐵𝐷: 𝜎𝐵𝐷 = (2)(10)(30) mm2 = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
𝜎𝐶𝐷 = 𝐴
= 900 mm2
= 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
1.38 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2 𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
C-D; 𝐴2 = 𝜋(16)2 /4 = 201 mm2 𝜎𝐴𝐷 = 𝜎𝐶𝐷 = 500 mm2
= −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
For AB: 𝐹𝐴𝐵 = −9.65 − 12.32 + 4.45 = −17.52 kN 1.42 ∑ 𝑀𝐴 = 0 = 6000(6) + 12 000(12) − 𝑅𝐹 (18)
𝐹𝐴𝐵 −17.52×103 N
𝜎𝐴𝐵 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression 𝑅𝐹 = 10 000 lb
𝐴1 491 mm2
∑ 𝑀𝐹 = 0 = 12 000(6) + 6000(12) − 𝑅𝐴 (18)
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
𝐹𝐵𝐶 −21.97×103 N
𝑅𝐴 = 8000 lb
𝜎𝐵𝐶 = 𝐴1
= 491 mm2
= −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
For CD: 𝐹𝐶𝐷 = −9.65 kN
𝐹𝐶𝐷 −9.65×103 N
𝜎𝐶𝐷 = 𝐴2
= 201 mm2
= −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
𝜋[(1.90)2−(1.61)2] 1
1.39 𝐴 = 4
= 0.799 in2 [1 2 in Pipe-Appendix A-9(a)]
𝐹𝐵𝐶 2500 lb
For BC: 𝜎𝐵𝐶 = = = 𝟑𝟏𝟐𝟗 𝐩𝐬𝐢 Tension
𝐴 0.799 in2 𝑅𝐴 = 𝐴𝐵 sin 𝜃 = 𝐴𝐵(0.8)
For AB: 𝐹𝐴𝐵 = 2500 + 2(8000 cos 30°) = 16 356 lb 𝑅 8000
𝐴𝐵 = 0.8𝐴 = 0.8
= 10 000 lb Compression
𝐹𝐴𝐵 16 356 lb
𝜎𝐴𝐵 = = = 𝟐𝟎 𝟒𝟕𝟏 𝐩𝐬𝐢 Tension
𝐴 0.799 in2 𝐴𝐷 = 𝐴𝐵 cos 𝜃 = 10 000(0.6) = 6000 lb Tension
1.40 ∑ 𝑀𝐶 = 0 = 2800(45) − 𝐹𝐵𝐷 (30)
𝐹𝐵𝐷 = 4200 lb
𝐹𝐵𝐷 4200 lb
𝜎𝐵𝐷 = 𝐴
= (2.0)(0.65) in2 = 𝟑𝟐𝟑𝟏 𝐩𝐬𝐢 Tension
𝐵𝐸 sin 𝜃 + 6000 − 𝐴𝐵 sin 𝜃 = 0
𝐴𝐵 sin 𝜃−6000 10 000(0.8)−6000
𝐵𝐸 = sin 𝜃
= 0.8
= 2500 lb Tension
𝐵𝐶 = 𝐴𝐵 cos 𝜃 + 𝐵𝐸 cos 𝜃 = 10 000(0.6) + 2500(0.6)
[Continued on next page]
, 𝐵𝐶 = 7500 lb Compression 1.47 Direct Shear – Single Shear
𝜋(12.0)2
𝐴𝑆 = [ 4
] mm2 = 113 mm2
𝐵𝐶 = 𝐶𝐹 cos 𝜃
𝐹 16.5×103 N
𝐵𝐶 7500 𝜏=𝐴 = 113 mm2
= 𝟏𝟒𝟔 𝐌𝐏𝐚
𝐶𝐹 = = = 12 500 lb Compression 𝑆
cos 𝜃 0.6
1.48 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃 (45)
𝐶𝐸 = 12 000 − 𝐶𝐹 sin 𝜃 = 12 000 − 12 500(0.8)
𝐹𝑃 = 177 N
CE = 2000 lb Compression
𝜋(3.0)2
EF = CF cos = 12 500 lb(0.6) = 7500 lb Tension 𝐴𝑆 = 4
= 7.07 mm2
𝐹𝑃 177 N
Areas of members: Appendixes A-5(a) and A-6(a) 𝜏= = = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝐴𝑆 7.07 mm2 Pin is in single
AD, DE, EF – 2(0.484 in2) = 0.968 in2 shear
BD, BE, CE – 0.484 in2 1.49 From Problem 1-46: 𝐹 = 23 695 N
AB, BC, CF – 2(1.21 in2) = 2.42 in2 𝜋(10)2
𝐴𝑆 = 2 [ 4
] = 157 mm2 Double Shear
Stresses: 𝐹 23 695 N
𝜏 = 𝐴 = 157 mm2 = 𝟏𝟓𝟏 𝐌𝐏𝐚
AD = DE = 6000/0.968 = +6198 psi 𝑆
EF = 7500/0.968 = +7748 psi 1.50 𝐴𝑆 = (3.0)(3.5) = 10.5 in2
BD = 0
𝐹 1650 lb
BE = 2500/0.484 = +5165 psi 𝜏 = 𝐴 = 10.5 in2
= 157 𝐩𝐬𝐢
𝑆
CE = -2000/0.484 = -4132 psi [NOTE: Compression members must be
AB = -10 000/2.42 = -4132 psi checked for column buckling.] 1.51 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
BC = -7500/2.42 = -3099 psi 𝐹 38.6×103 N
𝜏=𝐴 = = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
CF = -12 500/2.42 = -5165 psi 𝑆 475.7 mm2
1.52 𝐿 = √0.42 + 0.62 = 0.721 in
1.43 ∑ 𝑀𝐶 = 0 = (12.5)(4.0) − 𝐴𝐵(2.5)
𝜋(0.8)
𝐴𝐵 = 20 kN 𝐴𝑆 = [2(1.60) + 2
+ 2(0.721)] 0.194
20×103 2
N
𝜎 = (20)2 mm2 = 𝟓𝟎 𝐌𝐏𝐚 𝐴𝑆 = 1.144 in
𝐹 45 000 lb
2 𝜏 = 𝐴 = 1.144 in2 = 𝟑𝟗 𝟑𝟐𝟒 𝐩𝐬𝐢
𝜋(0.505)
1.44 𝐴 = 4
= 0.200 in2 𝑆
𝐹 12 600 lb
1.53 𝑇 = 𝐹𝑆 ∙ 𝑅
𝜎 = 𝐴 = 0.200 in2 = 𝟔𝟑 𝟎𝟎𝟎 𝐩𝐬𝐢
𝑇 95 N∙m 103 mm
𝐹𝑆 = 𝑅 = 35 mm/2 ∙ m
= 5429 N
1.45 𝐴 = (2.65)(1.40) + 2[(1.40)(0.5)(𝑡)] = 4.41 in2
𝐹 52 000 lb 𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝜎=𝐴= 4.41 in2
= 𝟏𝟏 𝟕𝟗𝟏 𝐩𝐬𝐢
𝐹 5429 N
2
𝜏 = 𝐴𝑆 = 220 mm2 = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜋(40) 𝑆
1.46 𝐴 = (80)(40) − (60)(15) + 4
= 3557 mm2
𝐹 640×103 N
𝜎=𝐴= 3557 mm2
= 𝟏𝟖𝟎 𝐌𝐏𝐚
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
SOLUTIONS MANUAL FOR 𝑾 = 𝟏3. 𝟕 𝐤𝐍
1.13 Total Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN
APPLIED STRENGTH
1
Each Front Wheel: 𝐹𝐹 = (2) (0.40)(34.34 kN) = 6.87 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = (2) (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍
OF MATERIALS 1.14 Loading = Total Force / Area
Total Force = 𝑚𝑔 = 5900 kg ∙ 9.81 m/s2 = 57.9 kN
Area = (4.5 m)(3.5 m) = 15.8 m2
Loading = 57.9 kN⁄15.8 m2 = 3.66 kN⁄m2 = 𝟑.66 𝐤𝐏𝐚
7th Edition 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 = 𝐾 = 4800 N/m
= 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦𝐦
Complete Chapter Solutions Manual
are included (Ch 1 to 14)
by
Robert L. Mott
Joseph A. Untener
** Immediate Download 1.16 𝑚=
𝑤
𝑔
3250 lb 2
= 32.2 (ft/s2 ) = 101 lb∙sft = 101 𝐬𝐥𝐮𝐠𝐬
** Swift Response 𝑤 11 600 lb 2
𝑚= = 32.2 (ft/s2 ) = 360 lb∙sft = 𝟑60 𝐬𝐥𝐮𝐠𝐬
** All Chapters included 1.17 𝑔
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 = 14 000 psi ∙ 6.895 (kPa⁄psi) = 96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚 1.33 𝜎 =
𝑃
=
(29500 lb)/3
= 𝟖𝟎𝟑 𝐩𝐬𝐢
𝐴 (3.5 in)2
𝑠𝑢 = 76 000 psi ∙ 6.895 (kPa⁄psi) = 524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚 3500 N
𝑃
2π rad 1 min 1.34 𝜎 = 𝐴
= (8.0 mm)2 = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
3600 rev 𝐫𝐚𝐝
1.22 𝑛= min
× rev
× 60s = 377 𝐬
2
1.35 𝑊 = 𝑚𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
(25.4mm) 𝟐
1.23 𝐴 = 26.1 in × 2
in
2
= 16 839 𝐦𝐦 𝐴𝐵𝑋 = 𝐴𝐵 sin 35°
1.24 𝑦 = 0.08 in ∙ 25.4 (mm⁄in) = 𝟐. 𝟎𝟑 𝐦𝐦 𝐴𝐵𝑌 = 𝐴𝐵 cos 35°
1.25 Dimensions: 18 in × 25.4 (mm/in) = 457 mm 𝐵𝐶𝑋 = 𝐵𝐶 sin 55°
12 in × 25.4 (mm/in) = 305 mm 𝐵𝐶𝑌 = 𝐵𝐶 cos 55°
Area = (18 in)2 = 𝟑𝟐𝟒 𝐢𝐧𝟐 ∑ 𝐹𝑋 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
Area = (457 mm)2 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟓 𝐦𝐦𝟐 0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55°
Volume = 𝑉 = Area × Height sin 55°
𝐴𝐵 = 𝐵𝐶 ∙ = 1.428 𝐵𝐶
𝑉 = 324 in2 × 12 in = 𝟑𝟖𝟖𝟖 𝐢𝐧𝟑 sin 35°
∑ 𝐹𝑉 = 0 = 𝐴𝐵𝑌 + 𝐵𝐶𝑌 − 41.2 kN = 𝐴𝐵 cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
𝑉 = (1.5 ft)2 × 1.0 ft = 𝟐. 𝟐𝟓 𝐟𝐭 𝟑
0 = (1.428 𝐵𝐶) cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
𝑉 = (209 × 103 mm2 ) × 305 mm = 𝟔. 𝟑𝟕 × 𝟏𝟎𝟕 𝐦𝐦𝟑
41.2 kN = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
𝑉 = (0.457 m)2 × 0.305 m = 0.0637 m3 = 𝟔. 𝟑𝟕 × 𝟏𝟎−𝟐 𝐦𝟑
41.2 kN
𝐵𝐶 = = 23.63 kN
1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋(0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢𝐧𝟐 1.743
(25.4 mm)2 𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
𝐴 = 0.200 in2 × in2
= 𝟏𝟐𝟗 𝐦𝐦𝟐
𝐴𝐵 33.75×103 N
𝑃2800 N 2800 N N Stress in Rod AB: 𝜎𝐴𝐵 = = [𝜋(20 = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
1.27 𝜎 = 𝐴 = (𝜋𝐷 = [𝜋(10 mm)2 ]⁄4 = 35.7 = 35. 𝟕 𝐌𝐏𝐚 𝐴 mm)2 ]/4
2 ⁄4) mm2
𝐵𝐶 23.63×103 N
𝑃 18×10 3 N N Stress in Rod BC: 𝜎𝐵𝐶 = = [𝜋(20 = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
1.28 𝜎 = 𝐴 = (12)(30) = 50.7 = 50. 𝟕 𝐌𝐏𝐚 𝐴 mm)2 ]/4
mm2 mm2
𝐵𝐷 41.2×103 N
𝑃 1150 lb Stress in Rod BD: 𝜎𝐵𝐷 = = [𝜋(20 mm)2 ]/4 = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
1.29 𝜎 = 𝐴 = (0.40 in)2 = 7188 𝐩𝐬𝐢 𝐴
𝑃 1850 lb 1.36 𝐹 = 0.01097 𝑚𝑅𝑛2 = (0.01097)(0.40)(0.60)(3000)2 N
1.30 𝜎= = [𝜋(0.375 in)2 ]⁄4 = 𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢
𝐴
𝐹 = 23 695 N
1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 kg ∙ 9.81 m⁄s2 = 16 187 N 𝜋(16 mm)2
𝐴= 4
= 201 mm2
𝑊/2 = 8093 N On each side
𝐹 23695 N
∑ 𝑀𝐴 = 0 = (8093 N)(600 mm) − 𝐶𝑉(1200 mm) 𝜎 = 𝐴 = 201 mm2 = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝐶𝑉 = 4047 N
𝐶 = 𝐶𝑉 / sin 30° = 8093 N
𝑃 𝐶 9025 N
𝜎 = 𝐴 ==𝐴 [𝜋(12 mm)2 ]⁄4 = 71.6 𝐌𝐏𝐚
𝑃 70000 lb
1.32 𝜎 = = [𝜋(10 = 891 𝐩𝐬𝐢
𝐴 in)2]/4
,1.37 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80) kN = 150 kN 1.41 𝐴𝐷 sin 30° = 5.25 kN
𝐹𝐴𝐵 150×103 N 𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝜎𝐴𝐵 = 𝐴
= 900 mm2
= 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
For BC: 𝐹𝐵𝐶 = 110 − 40 = 70 kN
Stresses:
𝐹𝐵𝐶 70×103 N
𝜎𝐵𝐶 = 𝐴
= 900 mm2
= 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension 9.09×103 N
𝐴𝐵, 𝐵𝐶: 𝜎𝐴𝐵 = 𝜎𝐵𝐶 = (12)(30) mm2 = 𝟐𝟓. 𝟑 𝐌𝐏𝐚 Tension
For CD: 𝐹𝐶𝐷 = 110 kN
10.5×103 N
𝐹𝐶𝐷 110×103 N 𝐵𝐷: 𝜎𝐵𝐷 = (2)(10)(30) mm2 = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
𝜎𝐶𝐷 = 𝐴
= 900 mm2
= 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
1.38 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2 𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
C-D; 𝐴2 = 𝜋(16)2 /4 = 201 mm2 𝜎𝐴𝐷 = 𝜎𝐶𝐷 = 500 mm2
= −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
For AB: 𝐹𝐴𝐵 = −9.65 − 12.32 + 4.45 = −17.52 kN 1.42 ∑ 𝑀𝐴 = 0 = 6000(6) + 12 000(12) − 𝑅𝐹 (18)
𝐹𝐴𝐵 −17.52×103 N
𝜎𝐴𝐵 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression 𝑅𝐹 = 10 000 lb
𝐴1 491 mm2
∑ 𝑀𝐹 = 0 = 12 000(6) + 6000(12) − 𝑅𝐴 (18)
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
𝐹𝐵𝐶 −21.97×103 N
𝑅𝐴 = 8000 lb
𝜎𝐵𝐶 = 𝐴1
= 491 mm2
= −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
For CD: 𝐹𝐶𝐷 = −9.65 kN
𝐹𝐶𝐷 −9.65×103 N
𝜎𝐶𝐷 = 𝐴2
= 201 mm2
= −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
𝜋[(1.90)2−(1.61)2] 1
1.39 𝐴 = 4
= 0.799 in2 [1 2 in Pipe-Appendix A-9(a)]
𝐹𝐵𝐶 2500 lb
For BC: 𝜎𝐵𝐶 = = = 𝟑𝟏𝟐𝟗 𝐩𝐬𝐢 Tension
𝐴 0.799 in2 𝑅𝐴 = 𝐴𝐵 sin 𝜃 = 𝐴𝐵(0.8)
For AB: 𝐹𝐴𝐵 = 2500 + 2(8000 cos 30°) = 16 356 lb 𝑅 8000
𝐴𝐵 = 0.8𝐴 = 0.8
= 10 000 lb Compression
𝐹𝐴𝐵 16 356 lb
𝜎𝐴𝐵 = = = 𝟐𝟎 𝟒𝟕𝟏 𝐩𝐬𝐢 Tension
𝐴 0.799 in2 𝐴𝐷 = 𝐴𝐵 cos 𝜃 = 10 000(0.6) = 6000 lb Tension
1.40 ∑ 𝑀𝐶 = 0 = 2800(45) − 𝐹𝐵𝐷 (30)
𝐹𝐵𝐷 = 4200 lb
𝐹𝐵𝐷 4200 lb
𝜎𝐵𝐷 = 𝐴
= (2.0)(0.65) in2 = 𝟑𝟐𝟑𝟏 𝐩𝐬𝐢 Tension
𝐵𝐸 sin 𝜃 + 6000 − 𝐴𝐵 sin 𝜃 = 0
𝐴𝐵 sin 𝜃−6000 10 000(0.8)−6000
𝐵𝐸 = sin 𝜃
= 0.8
= 2500 lb Tension
𝐵𝐶 = 𝐴𝐵 cos 𝜃 + 𝐵𝐸 cos 𝜃 = 10 000(0.6) + 2500(0.6)
[Continued on next page]
, 𝐵𝐶 = 7500 lb Compression 1.47 Direct Shear – Single Shear
𝜋(12.0)2
𝐴𝑆 = [ 4
] mm2 = 113 mm2
𝐵𝐶 = 𝐶𝐹 cos 𝜃
𝐹 16.5×103 N
𝐵𝐶 7500 𝜏=𝐴 = 113 mm2
= 𝟏𝟒𝟔 𝐌𝐏𝐚
𝐶𝐹 = = = 12 500 lb Compression 𝑆
cos 𝜃 0.6
1.48 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃 (45)
𝐶𝐸 = 12 000 − 𝐶𝐹 sin 𝜃 = 12 000 − 12 500(0.8)
𝐹𝑃 = 177 N
CE = 2000 lb Compression
𝜋(3.0)2
EF = CF cos = 12 500 lb(0.6) = 7500 lb Tension 𝐴𝑆 = 4
= 7.07 mm2
𝐹𝑃 177 N
Areas of members: Appendixes A-5(a) and A-6(a) 𝜏= = = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝐴𝑆 7.07 mm2 Pin is in single
AD, DE, EF – 2(0.484 in2) = 0.968 in2 shear
BD, BE, CE – 0.484 in2 1.49 From Problem 1-46: 𝐹 = 23 695 N
AB, BC, CF – 2(1.21 in2) = 2.42 in2 𝜋(10)2
𝐴𝑆 = 2 [ 4
] = 157 mm2 Double Shear
Stresses: 𝐹 23 695 N
𝜏 = 𝐴 = 157 mm2 = 𝟏𝟓𝟏 𝐌𝐏𝐚
AD = DE = 6000/0.968 = +6198 psi 𝑆
EF = 7500/0.968 = +7748 psi 1.50 𝐴𝑆 = (3.0)(3.5) = 10.5 in2
BD = 0
𝐹 1650 lb
BE = 2500/0.484 = +5165 psi 𝜏 = 𝐴 = 10.5 in2
= 157 𝐩𝐬𝐢
𝑆
CE = -2000/0.484 = -4132 psi [NOTE: Compression members must be
AB = -10 000/2.42 = -4132 psi checked for column buckling.] 1.51 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
BC = -7500/2.42 = -3099 psi 𝐹 38.6×103 N
𝜏=𝐴 = = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
CF = -12 500/2.42 = -5165 psi 𝑆 475.7 mm2
1.52 𝐿 = √0.42 + 0.62 = 0.721 in
1.43 ∑ 𝑀𝐶 = 0 = (12.5)(4.0) − 𝐴𝐵(2.5)
𝜋(0.8)
𝐴𝐵 = 20 kN 𝐴𝑆 = [2(1.60) + 2
+ 2(0.721)] 0.194
20×103 2
N
𝜎 = (20)2 mm2 = 𝟓𝟎 𝐌𝐏𝐚 𝐴𝑆 = 1.144 in
𝐹 45 000 lb
2 𝜏 = 𝐴 = 1.144 in2 = 𝟑𝟗 𝟑𝟐𝟒 𝐩𝐬𝐢
𝜋(0.505)
1.44 𝐴 = 4
= 0.200 in2 𝑆
𝐹 12 600 lb
1.53 𝑇 = 𝐹𝑆 ∙ 𝑅
𝜎 = 𝐴 = 0.200 in2 = 𝟔𝟑 𝟎𝟎𝟎 𝐩𝐬𝐢
𝑇 95 N∙m 103 mm
𝐹𝑆 = 𝑅 = 35 mm/2 ∙ m
= 5429 N
1.45 𝐴 = (2.65)(1.40) + 2[(1.40)(0.5)(𝑡)] = 4.41 in2
𝐹 52 000 lb 𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝜎=𝐴= 4.41 in2
= 𝟏𝟏 𝟕𝟗𝟏 𝐩𝐬𝐢
𝐹 5429 N
2
𝜏 = 𝐴𝑆 = 220 mm2 = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜋(40) 𝑆
1.46 𝐴 = (80)(40) − (60)(15) + 4
= 3557 mm2
𝐹 640×103 N
𝜎=𝐴= 3557 mm2
= 𝟏𝟖𝟎 𝐌𝐏𝐚
