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Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.15 Answers in text.
1.16 𝑊 = 𝑚 ∙ 𝑔 = 1800 kg ∙ 9.81 m/s2 = 17 658 (kg ∙ m)/s2 = 17 × 103 N
𝑾 = 𝟏𝟕. 𝟕 𝐤𝐍
1.17 Total Weight = 𝑚 𝑔 = 4000 kg ∙ 9.81 m/s2 = 39.24 kN
1
Each Front Wheel: 𝐹𝐹 = ( )2 (0.40)( 39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = ( )2 (0.60)( 39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
1.18 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 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 = 𝟑. 𝟖𝟏 𝐤𝐏𝐚
1.19 Force = Weight = 𝑚 𝑔 = 25 kg ∙ 9.81 m/s2 = 245 N
K = Spring Scale = 4500 N⁄m = 𝐹/Δ𝐿
Δ𝐿 =
𝐹
=
245 N
= 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
𝐾 4500 N/m
𝑃 3200 N N
1.22 𝜎= = =
3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
𝐴 (𝜋𝐷2⁄4) [𝜋(10 mm) 2]⁄4 mm2
𝑃 20×10 3 N N
1.23 𝜎= = = 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
𝐴 (10)(30) mm2 mm2
𝑃 3500 N
1.24 𝜎= = = 𝟑𝟓. 𝟎 𝐌𝐏𝐚
𝐴 (0.010 m)2
𝑃 8300 N
1.25 𝜎= = = 𝟏𝟑𝟎. 𝟓 𝐌𝐏𝐚
𝐴 [𝜋(9.0 mm)2]⁄4
1.26 Load on Shelf = 𝑊 = 𝑚𝑔 = 1840 kg ∙ 9.81 m⁄s2 = 18 050 N
𝑊/2 = 9025 N On each side
∑ 𝑀𝐴 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4512 N
𝐶 = 𝐶𝑉/ sin 30° = 9025 N
𝑃 𝐶 9025 N = 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4
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𝑃 310×10 3 N = 𝟗. 𝟖𝟕 𝐌𝐏𝐚
1.27 𝜎= =
𝐴 [𝜋(0.2 m) 2]/4
𝑃 (132 000 N)/3
1.28 𝜎= = = 𝟔. 𝟏 𝐌𝐏𝐚
𝐴 (85 mm)2
𝑃 3500 N
1.29 𝜎= = = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 mm)2
1.30 𝑊 = 𝑚 𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
𝐴𝐵𝑋 = 𝐴𝐵 sin 35°
𝐴𝐵𝑌 = 𝐴𝐵 cos 35°
𝐵𝐶𝑋 = 𝐵𝐶 sin 55°
𝐵𝐶𝑌 = 𝐵𝐶 cos 55°
∑ 𝐹𝑋 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55° sin
55°
𝐴𝐵 = 𝐵𝐶 ∙ = 1.428 𝐵𝐶
sin 35°
∑ 𝐹𝑉 = 0 = 𝐴𝐵𝑌 + 𝐵𝐶𝑌 − 41.2 kN = 𝐴𝐵 cos 35° + 𝐵𝐶 cos 55° − 41.2 kN 0
= (1.428 𝐵𝐶) cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
41.2 kN = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
41.2 kN
𝐵𝐶 = = 23.63 kN
1.743
𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
Stress in Rod AB: 𝜎 𝐴𝐵 33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴𝐵 = =
𝐴 [𝜋(20 mm) 2]/4
Stress in Rod BC: 𝜎 𝐵𝐶 23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐶 = =
𝐴 [𝜋(20 mm) 2]/4
Stress in Rod BD: 𝜎 𝐵𝐷 41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐵𝐷 = =
𝐴 [𝜋(20 mm) 2]/4
1.31 𝐹 = 0.01097 𝑚 𝑅 𝑛2 = (0.01097)(0.40)(0.60)(3000)2 N
𝐹 = 23 695 N
𝜋(16 mm)2
𝐴= = 201 mm2
4
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm2
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1.32 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80) kN = 150 kN
𝐹𝐴𝐵 150×103 N
𝜎 = = = 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
𝐴𝐵 𝐴 900 mm2
For BC: 𝐹𝐵𝐶 = 110 − 40 = 70 kN
𝐹𝐵𝐶 70×103 N
𝜎 = = = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝐵𝐶 𝐴 900 mm2
For CD: 𝐹𝐶𝐷 = 110 kN
𝐹𝐶𝐷 110×103 N
𝜎 = = = 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
𝐶𝐷 𝐴 900 mm2
1.33 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2
C-D; 𝐴2 = 𝜋(16)2/4 = 201 mm2
For AB: 𝐹𝐴𝐵 = −9.65 − 12.32 + 4.45 = −17.52 kN
𝐹𝐴𝐵 −17.52×103 N
𝜎 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
𝐴𝐵 𝐴1 491 mm2
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
𝐹𝐵𝐶 −21.97×103 N
𝜎 = = = −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
𝐵𝐶 𝐴1 491 mm2
For CD: 𝐹𝐶𝐷 = −9.65 kN
𝐹𝐶𝐷 −9.65×103 N
𝜎 = = = −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
𝐶𝐷 𝐴2 201 mm2
1.34 𝐴 = 515.8 mm2 [𝐷𝑁 40 Pipe-Appendix A-9(b)]
For BC: 𝜎 =
𝐹𝐵𝐶
=
11 000 N = 𝟐𝟏. 𝟑 𝐌𝐏𝐚 Tension
𝐵𝐶 𝐴 515.8 mm2
For AB: 𝐹𝐴𝐵 = 11 000 + 2(36 000 cos 30°) = 73 354 N
𝜎𝐴𝐵 =
𝐹𝐴𝐵
=
73 354 N = 𝟏𝟒𝟐. 𝟐 𝐌𝐏𝐚 Tension
𝐴 515.8 mm
1.35 ∑ 𝑀𝐶 = 0 = 13 000 N(1.2 m) − 𝐹𝐵𝐷(0.8)
𝐹𝐵𝐷 = 19 500 N
𝜎𝐵𝐷 =
𝐹𝐵𝐷
=
19 500 N = 𝟒𝟖. 𝟖 𝐌𝐏𝐚 Tension
𝐴 (25)(16) mm2
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1.36 𝐴𝐷 sin 30° = 5.25 kN
𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
Stresses:
9.09×10 3 N
𝐴𝐵, 𝐵𝐶: = 𝜎𝐵𝐶 = = 𝟐𝟓. 𝟑 𝐌𝐏𝐚
(12)(30) mm2
𝜎𝐴𝐵
Tension
10.5×10 3 N
𝐵𝐷: 𝜎𝐵𝐷 = = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
(2)(10)(30) mm2
𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
𝜎𝐴𝐷 = 𝜎𝐶𝐷 = = −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
500 mm2
1.37 ∑ 𝑀𝐴 = 0 = 25(1.5) + 50(3) − 𝑅𝐹(4.5)
𝑅𝐹 = 41.7 kN
∑ 𝑀𝐹 = 0 = 25(3) + 50(1.5) − 𝑅𝐴(4.5)
𝑅𝐴 = 33.3 kN
𝑅𝐴 = 𝐴𝐵 sin 𝜃 = 𝐴𝐵(0.8)
𝐴𝐵 = 41.6 kN Compression
𝐴𝐷 = 𝐴𝐵 cos 𝜃 = 25 kN Tension
𝐵𝐷 = 0
𝐵𝐸 sin 𝜃 + 25 − 𝐴𝐵 sin 𝜃 = 0
𝐴𝐵 sin 𝛩−25 41.63(0.8)−25
𝐵𝐸 = = = 10.4 kN Tension
sin 𝛩 0.8
𝐵𝐶 = 𝐴𝐵 cos 𝜃 + 𝐵𝐸 cos 𝜃 = 41.63(0.6) + 10.4(0.6)
𝐵𝐶 = 31.2 kN Compression
𝐵𝐶 = 𝐶𝐹 cos 𝜃
𝐵𝐶 31 200
𝐶𝐹 = = = 52.0 kN Compression
cos 𝛩 0.6
𝐶𝐸 = 50 − 𝐶𝐹 sin 𝜃 = 50 − 52.0(0.8)
CE = 8.4 kN Compression
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EF = CF cos = 52.0 N(0.6) = 31.2 kN Tension
Areas of members: Appendixes A-5(c) and A-6(d)
AD, DE, EF – 2(475 mm2) = 950 mm2
BD, BE, CE = 475 mm2
AB, BC, CF – 2(608 mm2) = 1216 mm2
Stresses:
AD = DE = 25 000/950 = +26.3 MPa
EF = 31 200/950 = +32.8 MPa
BD = 0
BE = 10 400/475 = +21.9 MPa
CE = -8400/475 = -17.7 MPa [NOTE: Compression members must be
AB = -41 600/1216 = -34.2 MPa checked for column buckling.]
BC = -31 200/1216 = -25.7 MPa
CF = -52 000/1216 = -42.8 MPa
1.38 ∑ 𝑀𝐶 = 0 = (12.5)(4.0) − 𝐴𝐵(2.5)
𝐴𝐵 = 20 kN
20×10 3 N
𝜎= = 𝟓𝟎 𝐌𝐏𝐚
(20) 2 mm2
𝜋(13)2
1.39 𝐴= = 132.7 mm2
4
𝐹 56 000 N = 𝟒𝟐𝟐 𝐌𝐏𝐚
𝜎= =
𝐴 132.7 mm2
1.40 𝐴 = (68)(36) + 2[(36)(12)(1/2)] = 2880 mm2
𝐹 230 000 N
𝜎= = = 𝟕𝟗. 𝟗 𝐌𝐏𝐚
𝐴 2880 mm2
𝜋(40)2
1.41 𝐴 = (80)(40) − (60)(15) + = 3557 mm2
4
𝐹 640×10 3 N
𝜎= = = 𝟏𝟖𝟎 𝐌𝐏𝐚
𝐴 3557 mm2
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1.42 Direct Shear – Single Shear
𝜋(12.0)2
𝐴𝑆 = [ ] mm2 = 113 mm2
4
𝐹 16.5×10 3 N
𝜏= = = 𝟏𝟒𝟔 𝐌𝐏𝐚
𝐴𝑆 113 mm2
1.43 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃(45)
𝐹𝑃 = 177 N
𝜋(3.0)2
𝐴𝑆 = = 7.07 mm2
4
𝐹𝑃 177 N = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 7.07 mm2 Pin is in single
shear
1.44 From Problem 1-46: 𝐹 = 23 695 N
𝜋(10)2
𝐴𝑆 = 2 [ ] = 157 mm2 Double Shear
4
𝐹 23 695 N
𝜏= = = 𝟏𝟓𝟏 𝐌𝐏𝐚
𝐴𝑆 157 mm2
1.45 𝐴𝑆 = (75)(90) = 6750 mm2
𝐹 1800 N
𝜏= = = 𝟏. 𝟐 𝐌𝐏𝐚
𝐴𝑆 6750 mm2
1.46 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
𝐹 38.6×10 3 N
𝜏= = = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
𝐴𝑆 475.7 mm2
1.47 𝐿 = √152 + 102 = 18.03 mm
𝜋(20)
𝐴𝑆 = [2(40) + + 2(18)] 5
2
𝐴𝑆 = 737 mm2
𝐹 200 000 N
𝜏= = = 𝟐𝟕𝟏 𝐌𝐏𝐚
𝐴𝑆 737 mm2
1.48 𝑇 = 𝐹𝑆 ∙ 𝑅
103 mm
𝐹𝑆 =
𝑇
=
95 N∙m ∙ = 5429 N
𝑅 35 mm/2 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝐹𝑆 5429 N = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 220 mm2
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𝑇 900 N∙m
1.49 𝐹 = = = 36 000 N
𝑆 𝑅 0.025 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (12)(60) = 720 mm2
𝐹𝑆 36 000 N
𝜏= = = 𝟓𝟎 𝐌𝐏𝐚
𝐴𝑆 720 mm2
1.50 Pin: Double Shear; 𝐴𝑆 = 2[𝜋(13)2/4] = 265.5 mm2
𝐹 90 000 N
𝜏= = = 𝟑𝟑𝟗 𝐌𝐏𝐚
𝐴𝑆 265.5 mm2
Collar: Shear Collar from Connector Body
𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(22)(5) = 345.6 mm2
𝐹 90 000 N
𝜏= = = 𝟐𝟔𝟎. 𝟒 𝐌𝐏𝐚
𝐴𝑆 345.6 mm2
1.51 ∑ 𝑀𝐴 = 0 = 3600(2) − 𝐵𝑉(0.2)
𝐵𝑉 = 36 000 N
𝐵𝑉
𝐵= = 38 310 N
cos 20°
𝜋(10)2
𝐴𝑆 = 2 [ ] = 157 mm2
4
𝐵 38 310 N
𝜏= = = 𝟐𝟒𝟒 𝐌𝐏𝐚
𝐴𝑆 157 mm2
1.52 𝐴𝑆 = (40)(12) = 480 mm2
𝐹 88×10 3 N
𝜏= = = 𝟏𝟖𝟑 𝐌𝐏𝐚
𝐴𝑆 480 mm2
1.53 𝐴𝑆 = (40)(120) = 4800 mm2
𝐹 88.2×10 3 N
𝜏= = = 𝟏𝟖. 𝟒 𝐌𝐏𝐚
𝐴𝑆 4800 mm2
1.54 𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(12)(8) = 301.6 mm2
𝐹 22.3×10 3 N
𝜏= = = 𝟕𝟑. 𝟗 𝐌𝐏𝐚
𝐴𝑆 301.6 mm2
1.55 𝐴𝑆 = 2[𝜋(12)2/4] = 226.2 mm2 Two Rivets – Single Shear
𝐹 10.2×10 3 N
𝜏= = = 𝟒𝟓. 𝟏 𝐌𝐏𝐚
𝐴𝑆 226.2 mm2
1.56 𝐴𝑆 = 4[𝜋(12)2/4] = 452.4 mm2 Two Rivets – Double Shear
𝐹 10.2×10 3 N
𝜏= = = 𝟐𝟐. 𝟓𝟓 𝐌𝐏𝐚
𝐴𝑆 452.4 mm2
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Structural Shapes
1-57 [Appendix A-4(b)] A = 4050 mm2, 6.075 × 104 mm3 with long side vertical 1-58
[Appendix A-5(c)] 20.719 x 2 = 41.438 N, 275 mm2
1-59 [Appendix A-6(d)] 16.51 x 3.8 = 62.74 N, A = 608 mm2
1-60 [Appendix A-7(e)] 471.7 x 2.55= 1203 N, A = 6261 mm2
1-61 [Appendix A-8(c)] 33.5 N/m, A = 444 mm2
1-62 [App. A-8(c)] A = 9344 mm2, 704 N/m
1-63 [App. A-9(b)] 73.03 mm, A = 1099 mm2
1-64 [App. A-9(b)] 219.1 mm, A = 5419 mm2
1-65 [Appendix A-9(d)] A = 1335 mm2, 100.6 N/m
1-66 [Appendix A-9(d)] 30 mm x 4 mm
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Chapter 2 Design Properties of Materials
Only those problems requiring numerical data are shown.
2-14 𝑠𝑢 = 621 MPa; 𝑠𝑦 = 414 MPa; 25% Elongation [Appendix A-10]
Because % Elongation >5%, it is ductile.
2-15 1020 HR: 36% Elongation – Greater ductility [Appendix A-10]
1040 HR: 25% Elongation
2-16 SAE 1141 OQT 700: High sulfur alloy steel with 0.41% carbon, quenched in oil,
tempered at 700℉. [Appendix A-10]
2-17 Yes, AISI 1141 steel has: 𝑠𝑦 = 1186 MPa @ OQT 700
[Appendix A-10]
2-18 𝐸 = 207 GPa for all carbon and alloy steels. [Appendix A-10]
kg
2-19 Mass = Density × Volume = (7680 ) (0.03)(0.1)(0.35)m3 = 𝟖. 𝟏 𝐤𝐠
m3
[Appendix A-10]
2-20 𝜋 2 5 3
Volume = Area × Length = (50) × 250 = 4.909 × 10 mm
4
7680 kg 4.909×10 5 mm3 1 m3 = 3.77 kg [Appendix A-10]
Steel Bar: Mass = × ×
m3 1 (10 3 mm)3
𝑊𝑡 = 𝑚 ∙ 𝑔 = 3.77 kg ∙ 9.81 m/s2 = 36.98 kg ∙ m/s2 = 𝟑𝟔. 𝟗𝟖 𝐍
2-21 Magnesium would stretch more because it has a lower 𝐸.
𝐸𝑀𝑎𝑔 = 45 GPa; 𝐸𝑇𝑖 = 114 GPa; Ti is stiffer. [Appendix A-11]
2-23 Alloy of aluminum with silicon and magnesium. Heat treated to T6 temper.
2-24
𝑠𝑢 𝑠𝑦 𝐸 Density
6061-0 124 MPa 55 MPa 69 𝐺𝑃𝑎 2770 km/m3
6061-T4 241 MPa 145 MPa 69 𝐺𝑃𝑎 2770 km/m3
6061-T6 310 MPa 276 MPa 69 GPa 2770 km/m3
[Appendix A-14]
2-29 𝑠𝑢𝑡 = 276 MPa; 𝑠𝑢𝑐 = 965 MPa [Appendix A-13]
2-31 Bending 𝜎𝑑 = 10.0 MPa; Tension 𝜎𝑑 = 5.9 MPa; Compression 6.9 MPa Parallel to
grain, 2.65 MPa Perpendicular to grain; Shear 𝜏𝑑 = 0.66 MPa [Appendix A-15]
2-32 14 - 48 MPa [Section 2-11]
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Chapter 1 Basic Concepts in Strength of Materials
1.1 to 1.15 Answers in text.
1.16 𝑊 = 𝑚 ∙ 𝑔 = 1800 kg ∙ 9.81 m/s2 = 17 658 (kg ∙ m)/s2 = 17 × 103 N
𝑾 = 𝟏𝟕. 𝟕 𝐤𝐍
1.17 Total Weight = 𝑚 𝑔 = 4000 kg ∙ 9.81 m/s2 = 39.24 kN
1
Each Front Wheel: 𝐹𝐹 = ( )2 (0.40)( 39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
1
Each Rear Wheel: 𝐹𝑅 = ( )2 (0.60)( 39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
1.18 Loading = Total Force / Area
Total Force = 𝑚 𝑔 = 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 = 𝟑. 𝟖𝟏 𝐤𝐏𝐚
1.19 Force = Weight = 𝑚 𝑔 = 25 kg ∙ 9.81 m/s2 = 245 N
K = Spring Scale = 4500 N⁄m = 𝐹/Δ𝐿
Δ𝐿 =
𝐹
=
245 N
= 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
𝐾 4500 N/m
𝑃 3200 N N
1.22 𝜎= = =
3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
𝐴 (𝜋𝐷2⁄4) [𝜋(10 mm) 2]⁄4 mm2
𝑃 20×10 3 N N
1.23 𝜎= = = 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
𝐴 (10)(30) mm2 mm2
𝑃 3500 N
1.24 𝜎= = = 𝟑𝟓. 𝟎 𝐌𝐏𝐚
𝐴 (0.010 m)2
𝑃 8300 N
1.25 𝜎= = = 𝟏𝟑𝟎. 𝟓 𝐌𝐏𝐚
𝐴 [𝜋(9.0 mm)2]⁄4
1.26 Load on Shelf = 𝑊 = 𝑚𝑔 = 1840 kg ∙ 9.81 m⁄s2 = 18 050 N
𝑊/2 = 9025 N On each side
∑ 𝑀𝐴 = 0 = (9025 N)(600 mm) − 𝐶𝑉(1200 mm)
𝐶𝑉 = 4512 N
𝐶 = 𝐶𝑉/ sin 30° = 9025 N
𝑃 𝐶 9025 N = 𝟕𝟗. 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4
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𝑃 310×10 3 N = 𝟗. 𝟖𝟕 𝐌𝐏𝐚
1.27 𝜎= =
𝐴 [𝜋(0.2 m) 2]/4
𝑃 (132 000 N)/3
1.28 𝜎= = = 𝟔. 𝟏 𝐌𝐏𝐚
𝐴 (85 mm)2
𝑃 3500 N
1.29 𝜎= = = 𝟓𝟒. 𝟕 𝐌𝐏𝐚
𝐴 (8.0 mm)2
1.30 𝑊 = 𝑚 𝑔 = 4200 kg ∙ 9.81 m/s2 = 41.2 kN
𝐴𝐵𝑋 = 𝐴𝐵 sin 35°
𝐴𝐵𝑌 = 𝐴𝐵 cos 35°
𝐵𝐶𝑋 = 𝐵𝐶 sin 55°
𝐵𝐶𝑌 = 𝐵𝐶 cos 55°
∑ 𝐹𝑋 = 0 = 𝐴𝐵𝑋 − 𝐵𝐶𝑋
0 = 𝐴𝐵 sin 35° − 𝐵𝐶 sin 55° sin
55°
𝐴𝐵 = 𝐵𝐶 ∙ = 1.428 𝐵𝐶
sin 35°
∑ 𝐹𝑉 = 0 = 𝐴𝐵𝑌 + 𝐵𝐶𝑌 − 41.2 kN = 𝐴𝐵 cos 35° + 𝐵𝐶 cos 55° − 41.2 kN 0
= (1.428 𝐵𝐶) cos 35° + 𝐵𝐶 cos 55° − 41.2 kN
41.2 kN = 𝐵𝐶[1.170 + 0.574] = 1.743 𝐵𝐶
41.2 kN
𝐵𝐶 = = 23.63 kN
1.743
𝐴𝐵 = 1.428 𝐵𝐶 = 33.75 kN
Stress in Rod AB: 𝜎 𝐴𝐵 33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
𝐴𝐵 = =
𝐴 [𝜋(20 mm) 2]/4
Stress in Rod BC: 𝜎 𝐵𝐶 23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
𝐵𝐶 = =
𝐴 [𝜋(20 mm) 2]/4
Stress in Rod BD: 𝜎 𝐵𝐷 41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
𝐵𝐷 = =
𝐴 [𝜋(20 mm) 2]/4
1.31 𝐹 = 0.01097 𝑚 𝑅 𝑛2 = (0.01097)(0.40)(0.60)(3000)2 N
𝐹 = 23 695 N
𝜋(16 mm)2
𝐴= = 201 mm2
4
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm2
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1.32 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80) kN = 150 kN
𝐹𝐴𝐵 150×103 N
𝜎 = = = 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
𝐴𝐵 𝐴 900 mm2
For BC: 𝐹𝐵𝐶 = 110 − 40 = 70 kN
𝐹𝐵𝐶 70×103 N
𝜎 = = = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝐵𝐶 𝐴 900 mm2
For CD: 𝐹𝐶𝐷 = 110 kN
𝐹𝐶𝐷 110×103 N
𝜎 = = = 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
𝐶𝐷 𝐴 900 mm2
1.33 Areas: A-C; 𝐴1 = 𝜋(25)2/4 = 491 mm2
C-D; 𝐴2 = 𝜋(16)2/4 = 201 mm2
For AB: 𝐹𝐴𝐵 = −9.65 − 12.32 + 4.45 = −17.52 kN
𝐹𝐴𝐵 −17.52×103 N
𝜎 = = = −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
𝐴𝐵 𝐴1 491 mm2
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
𝐹𝐵𝐶 −21.97×103 N
𝜎 = = = −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
𝐵𝐶 𝐴1 491 mm2
For CD: 𝐹𝐶𝐷 = −9.65 kN
𝐹𝐶𝐷 −9.65×103 N
𝜎 = = = −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
𝐶𝐷 𝐴2 201 mm2
1.34 𝐴 = 515.8 mm2 [𝐷𝑁 40 Pipe-Appendix A-9(b)]
For BC: 𝜎 =
𝐹𝐵𝐶
=
11 000 N = 𝟐𝟏. 𝟑 𝐌𝐏𝐚 Tension
𝐵𝐶 𝐴 515.8 mm2
For AB: 𝐹𝐴𝐵 = 11 000 + 2(36 000 cos 30°) = 73 354 N
𝜎𝐴𝐵 =
𝐹𝐴𝐵
=
73 354 N = 𝟏𝟒𝟐. 𝟐 𝐌𝐏𝐚 Tension
𝐴 515.8 mm
1.35 ∑ 𝑀𝐶 = 0 = 13 000 N(1.2 m) − 𝐹𝐵𝐷(0.8)
𝐹𝐵𝐷 = 19 500 N
𝜎𝐵𝐷 =
𝐹𝐵𝐷
=
19 500 N = 𝟒𝟖. 𝟖 𝐌𝐏𝐚 Tension
𝐴 (25)(16) mm2
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1.36 𝐴𝐷 sin 30° = 5.25 kN
𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
Stresses:
9.09×10 3 N
𝐴𝐵, 𝐵𝐶: = 𝜎𝐵𝐶 = = 𝟐𝟓. 𝟑 𝐌𝐏𝐚
(12)(30) mm2
𝜎𝐴𝐵
Tension
10.5×10 3 N
𝐵𝐷: 𝜎𝐵𝐷 = = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
(2)(10)(30) mm2
𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
𝜎𝐴𝐷 = 𝜎𝐶𝐷 = = −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
500 mm2
1.37 ∑ 𝑀𝐴 = 0 = 25(1.5) + 50(3) − 𝑅𝐹(4.5)
𝑅𝐹 = 41.7 kN
∑ 𝑀𝐹 = 0 = 25(3) + 50(1.5) − 𝑅𝐴(4.5)
𝑅𝐴 = 33.3 kN
𝑅𝐴 = 𝐴𝐵 sin 𝜃 = 𝐴𝐵(0.8)
𝐴𝐵 = 41.6 kN Compression
𝐴𝐷 = 𝐴𝐵 cos 𝜃 = 25 kN Tension
𝐵𝐷 = 0
𝐵𝐸 sin 𝜃 + 25 − 𝐴𝐵 sin 𝜃 = 0
𝐴𝐵 sin 𝛩−25 41.63(0.8)−25
𝐵𝐸 = = = 10.4 kN Tension
sin 𝛩 0.8
𝐵𝐶 = 𝐴𝐵 cos 𝜃 + 𝐵𝐸 cos 𝜃 = 41.63(0.6) + 10.4(0.6)
𝐵𝐶 = 31.2 kN Compression
𝐵𝐶 = 𝐶𝐹 cos 𝜃
𝐵𝐶 31 200
𝐶𝐹 = = = 52.0 kN Compression
cos 𝛩 0.6
𝐶𝐸 = 50 − 𝐶𝐹 sin 𝜃 = 50 − 52.0(0.8)
CE = 8.4 kN Compression
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EF = CF cos = 52.0 N(0.6) = 31.2 kN Tension
Areas of members: Appendixes A-5(c) and A-6(d)
AD, DE, EF – 2(475 mm2) = 950 mm2
BD, BE, CE = 475 mm2
AB, BC, CF – 2(608 mm2) = 1216 mm2
Stresses:
AD = DE = 25 000/950 = +26.3 MPa
EF = 31 200/950 = +32.8 MPa
BD = 0
BE = 10 400/475 = +21.9 MPa
CE = -8400/475 = -17.7 MPa [NOTE: Compression members must be
AB = -41 600/1216 = -34.2 MPa checked for column buckling.]
BC = -31 200/1216 = -25.7 MPa
CF = -52 000/1216 = -42.8 MPa
1.38 ∑ 𝑀𝐶 = 0 = (12.5)(4.0) − 𝐴𝐵(2.5)
𝐴𝐵 = 20 kN
20×10 3 N
𝜎= = 𝟓𝟎 𝐌𝐏𝐚
(20) 2 mm2
𝜋(13)2
1.39 𝐴= = 132.7 mm2
4
𝐹 56 000 N = 𝟒𝟐𝟐 𝐌𝐏𝐚
𝜎= =
𝐴 132.7 mm2
1.40 𝐴 = (68)(36) + 2[(36)(12)(1/2)] = 2880 mm2
𝐹 230 000 N
𝜎= = = 𝟕𝟗. 𝟗 𝐌𝐏𝐚
𝐴 2880 mm2
𝜋(40)2
1.41 𝐴 = (80)(40) − (60)(15) + = 3557 mm2
4
𝐹 640×10 3 N
𝜎= = = 𝟏𝟖𝟎 𝐌𝐏𝐚
𝐴 3557 mm2
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1.42 Direct Shear – Single Shear
𝜋(12.0)2
𝐴𝑆 = [ ] mm2 = 113 mm2
4
𝐹 16.5×10 3 N
𝜏= = = 𝟏𝟒𝟔 𝐌𝐏𝐚
𝐴𝑆 113 mm2
1.43 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃(45)
𝐹𝑃 = 177 N
𝜋(3.0)2
𝐴𝑆 = = 7.07 mm2
4
𝐹𝑃 177 N = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 7.07 mm2 Pin is in single
shear
1.44 From Problem 1-46: 𝐹 = 23 695 N
𝜋(10)2
𝐴𝑆 = 2 [ ] = 157 mm2 Double Shear
4
𝐹 23 695 N
𝜏= = = 𝟏𝟓𝟏 𝐌𝐏𝐚
𝐴𝑆 157 mm2
1.45 𝐴𝑆 = (75)(90) = 6750 mm2
𝐹 1800 N
𝜏= = = 𝟏. 𝟐 𝐌𝐏𝐚
𝐴𝑆 6750 mm2
1.46 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
𝐹 38.6×10 3 N
𝜏= = = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
𝐴𝑆 475.7 mm2
1.47 𝐿 = √152 + 102 = 18.03 mm
𝜋(20)
𝐴𝑆 = [2(40) + + 2(18)] 5
2
𝐴𝑆 = 737 mm2
𝐹 200 000 N
𝜏= = = 𝟐𝟕𝟏 𝐌𝐏𝐚
𝐴𝑆 737 mm2
1.48 𝑇 = 𝐹𝑆 ∙ 𝑅
103 mm
𝐹𝑆 =
𝑇
=
95 N∙m ∙ = 5429 N
𝑅 35 mm/2 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝐹𝑆 5429 N = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 220 mm2
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𝑇 900 N∙m
1.49 𝐹 = = = 36 000 N
𝑆 𝑅 0.025 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (12)(60) = 720 mm2
𝐹𝑆 36 000 N
𝜏= = = 𝟓𝟎 𝐌𝐏𝐚
𝐴𝑆 720 mm2
1.50 Pin: Double Shear; 𝐴𝑆 = 2[𝜋(13)2/4] = 265.5 mm2
𝐹 90 000 N
𝜏= = = 𝟑𝟑𝟗 𝐌𝐏𝐚
𝐴𝑆 265.5 mm2
Collar: Shear Collar from Connector Body
𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(22)(5) = 345.6 mm2
𝐹 90 000 N
𝜏= = = 𝟐𝟔𝟎. 𝟒 𝐌𝐏𝐚
𝐴𝑆 345.6 mm2
1.51 ∑ 𝑀𝐴 = 0 = 3600(2) − 𝐵𝑉(0.2)
𝐵𝑉 = 36 000 N
𝐵𝑉
𝐵= = 38 310 N
cos 20°
𝜋(10)2
𝐴𝑆 = 2 [ ] = 157 mm2
4
𝐵 38 310 N
𝜏= = = 𝟐𝟒𝟒 𝐌𝐏𝐚
𝐴𝑆 157 mm2
1.52 𝐴𝑆 = (40)(12) = 480 mm2
𝐹 88×10 3 N
𝜏= = = 𝟏𝟖𝟑 𝐌𝐏𝐚
𝐴𝑆 480 mm2
1.53 𝐴𝑆 = (40)(120) = 4800 mm2
𝐹 88.2×10 3 N
𝜏= = = 𝟏𝟖. 𝟒 𝐌𝐏𝐚
𝐴𝑆 4800 mm2
1.54 𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(12)(8) = 301.6 mm2
𝐹 22.3×10 3 N
𝜏= = = 𝟕𝟑. 𝟗 𝐌𝐏𝐚
𝐴𝑆 301.6 mm2
1.55 𝐴𝑆 = 2[𝜋(12)2/4] = 226.2 mm2 Two Rivets – Single Shear
𝐹 10.2×10 3 N
𝜏= = = 𝟒𝟓. 𝟏 𝐌𝐏𝐚
𝐴𝑆 226.2 mm2
1.56 𝐴𝑆 = 4[𝜋(12)2/4] = 452.4 mm2 Two Rivets – Double Shear
𝐹 10.2×10 3 N
𝜏= = = 𝟐𝟐. 𝟓𝟓 𝐌𝐏𝐚
𝐴𝑆 452.4 mm2
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Structural Shapes
1-57 [Appendix A-4(b)] A = 4050 mm2, 6.075 × 104 mm3 with long side vertical 1-58
[Appendix A-5(c)] 20.719 x 2 = 41.438 N, 275 mm2
1-59 [Appendix A-6(d)] 16.51 x 3.8 = 62.74 N, A = 608 mm2
1-60 [Appendix A-7(e)] 471.7 x 2.55= 1203 N, A = 6261 mm2
1-61 [Appendix A-8(c)] 33.5 N/m, A = 444 mm2
1-62 [App. A-8(c)] A = 9344 mm2, 704 N/m
1-63 [App. A-9(b)] 73.03 mm, A = 1099 mm2
1-64 [App. A-9(b)] 219.1 mm, A = 5419 mm2
1-65 [Appendix A-9(d)] A = 1335 mm2, 100.6 N/m
1-66 [Appendix A-9(d)] 30 mm x 4 mm
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Chapter 2 Design Properties of Materials
Only those problems requiring numerical data are shown.
2-14 𝑠𝑢 = 621 MPa; 𝑠𝑦 = 414 MPa; 25% Elongation [Appendix A-10]
Because % Elongation >5%, it is ductile.
2-15 1020 HR: 36% Elongation – Greater ductility [Appendix A-10]
1040 HR: 25% Elongation
2-16 SAE 1141 OQT 700: High sulfur alloy steel with 0.41% carbon, quenched in oil,
tempered at 700℉. [Appendix A-10]
2-17 Yes, AISI 1141 steel has: 𝑠𝑦 = 1186 MPa @ OQT 700
[Appendix A-10]
2-18 𝐸 = 207 GPa for all carbon and alloy steels. [Appendix A-10]
kg
2-19 Mass = Density × Volume = (7680 ) (0.03)(0.1)(0.35)m3 = 𝟖. 𝟏 𝐤𝐠
m3
[Appendix A-10]
2-20 𝜋 2 5 3
Volume = Area × Length = (50) × 250 = 4.909 × 10 mm
4
7680 kg 4.909×10 5 mm3 1 m3 = 3.77 kg [Appendix A-10]
Steel Bar: Mass = × ×
m3 1 (10 3 mm)3
𝑊𝑡 = 𝑚 ∙ 𝑔 = 3.77 kg ∙ 9.81 m/s2 = 36.98 kg ∙ m/s2 = 𝟑𝟔. 𝟗𝟖 𝐍
2-21 Magnesium would stretch more because it has a lower 𝐸.
𝐸𝑀𝑎𝑔 = 45 GPa; 𝐸𝑇𝑖 = 114 GPa; Ti is stiffer. [Appendix A-11]
2-23 Alloy of aluminum with silicon and magnesium. Heat treated to T6 temper.
2-24
𝑠𝑢 𝑠𝑦 𝐸 Density
6061-0 124 MPa 55 MPa 69 𝐺𝑃𝑎 2770 km/m3
6061-T4 241 MPa 145 MPa 69 𝐺𝑃𝑎 2770 km/m3
6061-T6 310 MPa 276 MPa 69 GPa 2770 km/m3
[Appendix A-14]
2-29 𝑠𝑢𝑡 = 276 MPa; 𝑠𝑢𝑐 = 965 MPa [Appendix A-13]
2-31 Bending 𝜎𝑑 = 10.0 MPa; Tension 𝜎𝑑 = 5.9 MPa; Compression 6.9 MPa Parallel to
grain, 2.65 MPa Perpendicular to grain; Shear 𝜏𝑑 = 0.66 MPa [Appendix A-15]
2-32 14 - 48 MPa [Section 2-11]
