,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: 𝐹 = (0.40)(39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
𝐹 ( 2)
1
Each Rear Wheel: 𝐹 = (0.60)(39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
𝑅 (2)
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
𝐹 245 N = 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
Δ𝐿 = =
𝐾 4500 N/m
𝑃 3200 N N
1.22 𝜎= = 3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
=
𝐴 (𝜋𝐷 2⁄4) [𝜋(10 mm) 2]⁄4 mm 2
N
𝑃 20×10 N
3
= 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
1.23 𝜎= =
𝐴 (10)(30) mm 2 mm 2
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 = 𝟕� � . 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4
, 𝑃 310×10 N
3
= 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 ∙ sin 35° = 1.428 𝐵𝐶
∑ 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
𝐴𝐵 = 33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
Stress in Rod AB: 𝜎𝐴𝐵 = 𝐴 [𝜋(20 mm) 2]/4
𝐵𝐶 = 23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
Stress in Rod BC: 𝜎𝐵𝐶 = 𝐴 [𝜋(20 mm) 2]/4
𝐵𝐷 = 41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
Stress in Rod BD: 𝜎𝐵𝐷 = 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=
4 = 201 mm2
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm 2
,1.32 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80)3 kN = 150 kN
150×10 N
𝐹𝐴𝐵
=
𝜎𝐴𝐵 = 𝐴 900 mm2
= 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
For BC: 𝐹𝐵𝐶 = 110 − 40 = 703 kN
𝐹𝐵𝐶
= 70×10 N = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝜎𝐵𝐶 = 𝐴 900 mm 2
For CD: 𝐹𝐶𝐷 = 110 kN 110×103 N
𝐹𝐶𝐷
=
𝜎𝐶𝐷 = 𝐴
900 mm2
= 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
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
3 = −17.52 kN
−17.52×10 N
𝐹𝐴𝐵
𝜎𝐴𝐵 = 𝐴1 = 491 mm2
= −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
−21.97×103 N
𝐹𝐵𝐶
𝜎𝐵𝐶 = 𝐴1 = = −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
491 mm2
For CD: 𝐹𝐶𝐷 = −9.65 kN−9.65×103 N
𝐹𝐶𝐷
𝜎𝐶𝐷 = 𝐴2 = 201 mm2
= −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
1.34 𝐴 = 515.8 mm2 [𝐷𝑁 40 Pipe-Appendix A-9(b)]
𝐹𝐵𝐶 11 000 N
= = 𝟐𝟏. 𝟑 𝐌𝐏𝐚 Tension
For BC: 𝜎𝐵𝐶 = 𝐴 515.8 mm 2
For AB: 𝐹𝐴𝐵 = 11𝐹000 + 2(36 000 cos 30°) = 73 354 N
𝐴𝐵 = 73 354 N = 𝟏𝟒𝟐. 𝟐 𝐌𝐏𝐚 Tension
𝜎𝐴𝐵 = 𝐴 515.8 mm
1.35 ∑ 𝑀𝐶 = 0 = 13 000 N(1.2 m) − 𝐹𝐵𝐷(0.8)
𝐹𝐵𝐷 = 19 500 N
𝐹𝐵𝐷
= 19 500 N = 𝟒𝟖. 𝟖 𝐌𝐏𝐚 Tension
𝜎𝐵𝐷 = 𝐴 (25)(16) mm 2
,1.36 𝐴𝐷 sin 30° = 5.25 kN
𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
Stresses: 9.09×103 N
𝐴𝐵, 𝐵𝐶: 𝜎 =𝜎 = = 𝟐𝟓. 𝟑 𝐌𝐏𝐚
𝐴𝐵 𝐵𝐶 (12)(30) mm 2
Tension
10.5×103 N
𝐵𝐷: 𝜎 = = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
𝐵𝐷 (2)(10)(30) mm 2
𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
𝜎𝐴𝐷 = 𝜎𝐶𝐷 = 500 mm2
= −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
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 𝜃31 200
𝐵𝐶
𝐶𝐹 = = = 52.0 kN Compression
cos 𝜃 0.6
𝐶𝐸 = 50 − 𝐶𝐹 sin 𝜃 = 50 − 52.0(0.8)
CE = 8.4 kN Compression
, 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 N
3
= 𝟓𝟎 𝐌𝐏𝐚
𝜎=
(20) 2 mm 2
𝜋(13)
1.39 𝐴= 2
4 = 132.7 mm2
𝐹 56 000 N = 𝟒𝟐𝟐 𝐌𝐏𝐚
𝜎= =
𝐴 132.7 mm 2
1.40 𝐴 = (68)(36) + 2[(36)(12)(1/2)] = 2880 mm2
𝐹 230 000 N
𝜎= = = 𝟕𝟗. 𝟗 𝐌𝐏𝐚
𝐴 2880 mm 2
𝜋(40)
1.41 𝐴 = (80)(40) − (60)(15) + 2
4 = 3557 mm2
N
𝐹 640×10
3
= 𝟏𝟖𝟎 𝐌𝐏𝐚
𝜎= =
𝐴 3557 mm 2
,1.42 Direct Shear – Single Shear
𝜋(12.0) 2 4
] mm3 2
𝐴𝑆 = [ = 113 mm2
𝐹 N
𝜏= 16.5×10 = 𝟏𝟒𝟔 𝐌𝐏𝐚
=
𝐴𝑆 113 mm 2 Pin is in single
shear
1.43 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃 (45)
𝐹𝑃 = 177 N
𝜋(3.0) 2 = 7.07 mm 2
𝐴𝑆 = 4
𝐹𝑃 177 N = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 7.07 mm 2
1.44 From Problem
𝜋(10) 2
1-46: 𝐹 = 23 695 N
] = 157 mm2 4
𝐴𝑆 = 𝐹2 [ 23 695 N Double Shear
𝜏= = = 𝟏𝟓𝟏 𝐌𝐏𝐚
𝐴𝑆 157 mm 2
1.45 𝐴𝑆 = 𝐹(75)(90) = 6750 mm2
𝜏 = = 1800 N
𝐴𝑆 6750 mm 2 = 𝟏. 𝟐 𝐌𝐏𝐚
1.46 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
3
𝐹 N
𝜏= 38.6×10 = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 475.7 mm 2
1.47 𝐿 = √152 + 102 = 18.03 mm
𝜋(20)
+ 2(18)] 5
𝐴𝑆 = [2(40) +
2
2
𝐴𝑆 = 𝐹737 200
mm000 N
𝜏= = = 𝟐𝟕𝟏 𝐌𝐏𝐚
𝐴𝑆 737 mm 2
1.48 𝑇 = 𝐹𝑆 ∙ 𝑅 95 N∙m 103 mm
𝑇
= 5429 N
𝐹𝑆 = = ∙
𝑅 35 mm/2 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝐹 5429 N = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜏= 𝑆 =
𝐴𝑆 220 mm 2
, 𝑇 900 N∙m
1.49 𝐹𝑆 = 𝑅
= 0.025 m
= 36 000 N
𝐴𝑆 = 𝑏 ∙ 𝐿 = (12)(60) = 720 mm2
𝐹 36 000 N
𝜏= 𝑆= = 𝟓𝟎 𝐌𝐏𝐚
𝐴𝑆 720 mm 2
1.50 Pin: Double
𝐹 Shear; 𝐴𝑆 = 2[𝜋(13)2/4] = 265.5 mm2
𝜏 = = 000 N90
𝐴𝑆 265.5 mm 2 = 𝟑𝟑𝟗 𝐌𝐏𝐚
Collar: Shear Collar from Connector Body
𝐴𝑆 = 𝐹𝜋𝑑𝑡 = 𝜋(22)(5) = 345.6 mm2
𝜏 = = 90 000 N
𝐴𝑆 345.6 mm 2 = 𝟐𝟔𝟎. 𝟒 𝐌𝐏𝐚
1.51 ∑ 𝑀𝐴 = 0 = 3600(2) − 𝐵𝑉(0.2)
𝐵𝑉 = 36 000
𝐵
N
𝐵 = cos 20°𝑉
= 38 310 N
𝜋(10) 2
] = 157 mm2 4
𝐴𝑆 = 𝐵2 [ 38 310 N
𝜏= = = 𝟐𝟒𝟒 𝐌𝐏𝐚
𝐴𝑆 157 mm 2
1.52 𝐴𝑆 = (40)(12) = 480 mm2
3
𝐹 N
𝜏= 88×10 = 𝟏𝟖𝟑 𝐌𝐏𝐚
=
𝐴𝑆 480 mm 2
1.53 𝐴𝑆 = (40)(120) = 4800 mm2
3
𝐹 N
𝜏= 88.2×10 = 𝟏𝟖. 𝟒 𝐌𝐏𝐚
=
𝐴𝑆 4800 mm 2
1.54 𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(12)(8) = 301.6 mm2
3
𝐹 N
𝜏= 22.3×10 = 𝟕𝟑. 𝟗 𝐌𝐏𝐚
=
𝐴𝑆 301.6 mm 2
1.55 𝐴𝑆 = 2[𝜋(12) /4] = 226.2 mm2 Two Rivets – Single Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟒𝟓. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 226.2 mm 2
1.56 𝐴𝑆 = 4[𝜋(12) /4] = 452.4 mm2 Two Rivets – Double Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟐𝟐. 𝟓𝟓 𝐌𝐏𝐚
=
𝐴𝑆 452.4 mm 2
, Structural Shapes 2, 6.075 × 104 mm3 with long side vertical 1-58
1-57 [Appendix
[Appendix A-4(b)] A = 4050
A-5(c)] 20.719 x 2mm
= 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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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: 𝐹 = (0.40)(39.24 kN) = 𝟕. 𝟖𝟓 𝐤𝐍
𝐹 ( 2)
1
Each Rear Wheel: 𝐹 = (0.60)(39.24 kN) = 𝟏𝟏. 𝟕𝟕 𝐤𝐍
𝑅 (2)
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
𝐹 245 N = 0.0545 m = 54.5 × 10−3 m = 𝟓𝟒. 𝟓 𝐦𝐦
Δ𝐿 = =
𝐾 4500 N/m
𝑃 3200 N N
1.22 𝜎= = 3200 N = 40.7 = 𝟒𝟎. 𝟕 𝐌𝐏𝐚
=
𝐴 (𝜋𝐷 2⁄4) [𝜋(10 mm) 2]⁄4 mm 2
N
𝑃 20×10 N
3
= 66.7 = 𝟔𝟔. 𝟕 𝐌𝐏𝐚
1.23 𝜎= =
𝐴 (10)(30) mm 2 mm 2
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 = 𝟕� � . 𝟖 𝐌𝐏𝐚
𝜎= = =
𝐴 𝐴 [𝜋(12 mm) 2]⁄4
, 𝑃 310×10 N
3
= 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 ∙ sin 35° = 1.428 𝐵𝐶
∑ 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
𝐴𝐵 = 33.75×10 3 N = 𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚
Stress in Rod AB: 𝜎𝐴𝐵 = 𝐴 [𝜋(20 mm) 2]/4
𝐵𝐶 = 23.63×10 3 N = 𝟕𝟓. 𝟐 𝐌𝐏𝐚
Stress in Rod BC: 𝜎𝐵𝐶 = 𝐴 [𝜋(20 mm) 2]/4
𝐵𝐷 = 41.2×10 3 N = 𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚
Stress in Rod BD: 𝜎𝐵𝐷 = 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=
4 = 201 mm2
𝐹 23695 N = 𝟏𝟏𝟖 𝐌𝐏𝐚
𝜎= =
𝐴 201 mm 2
,1.32 𝐴 = (30 mm)2 = 900 mm2
For AB: 𝐹𝐴𝐵 = (110 − 40 + 80)3 kN = 150 kN
150×10 N
𝐹𝐴𝐵
=
𝜎𝐴𝐵 = 𝐴 900 mm2
= 𝟏𝟔𝟕 𝐌𝐏𝐚 Tension
For BC: 𝐹𝐵𝐶 = 110 − 40 = 703 kN
𝐹𝐵𝐶
= 70×10 N = 𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension
𝜎𝐵𝐶 = 𝐴 900 mm 2
For CD: 𝐹𝐶𝐷 = 110 kN 110×103 N
𝐹𝐶𝐷
=
𝜎𝐶𝐷 = 𝐴
900 mm2
= 𝟏𝟐𝟐 𝐌𝐏𝐚 Tension
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
3 = −17.52 kN
−17.52×10 N
𝐹𝐴𝐵
𝜎𝐴𝐵 = 𝐴1 = 491 mm2
= −𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression
For BC: 𝐹𝐵𝐶 = −9.65 − 12.32 = −21.97 kN
−21.97×103 N
𝐹𝐵𝐶
𝜎𝐵𝐶 = 𝐴1 = = −𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression
491 mm2
For CD: 𝐹𝐶𝐷 = −9.65 kN−9.65×103 N
𝐹𝐶𝐷
𝜎𝐶𝐷 = 𝐴2 = 201 mm2
= −𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression
1.34 𝐴 = 515.8 mm2 [𝐷𝑁 40 Pipe-Appendix A-9(b)]
𝐹𝐵𝐶 11 000 N
= = 𝟐𝟏. 𝟑 𝐌𝐏𝐚 Tension
For BC: 𝜎𝐵𝐶 = 𝐴 515.8 mm 2
For AB: 𝐹𝐴𝐵 = 11𝐹000 + 2(36 000 cos 30°) = 73 354 N
𝐴𝐵 = 73 354 N = 𝟏𝟒𝟐. 𝟐 𝐌𝐏𝐚 Tension
𝜎𝐴𝐵 = 𝐴 515.8 mm
1.35 ∑ 𝑀𝐶 = 0 = 13 000 N(1.2 m) − 𝐹𝐵𝐷(0.8)
𝐹𝐵𝐷 = 19 500 N
𝐹𝐵𝐷
= 19 500 N = 𝟒𝟖. 𝟖 𝐌𝐏𝐚 Tension
𝜎𝐵𝐷 = 𝐴 (25)(16) mm 2
,1.36 𝐴𝐷 sin 30° = 5.25 kN
𝐴𝐷 = 10.5 kN = 𝐶𝐷
𝐴𝐵 = 𝐴𝐷 cos 30° = 9.09 kN = 𝐵𝐶
Stresses: 9.09×103 N
𝐴𝐵, 𝐵𝐶: 𝜎 =𝜎 = = 𝟐𝟓. 𝟑 𝐌𝐏𝐚
𝐴𝐵 𝐵𝐶 (12)(30) mm 2
Tension
10.5×103 N
𝐵𝐷: 𝜎 = = 𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension
𝐵𝐷 (2)(10)(30) mm 2
𝐴𝐷, 𝐶𝐷: 𝐴 = (30)2 − (20)2 = 500 mm2
−10.5×103 N
𝜎𝐴𝐷 = 𝜎𝐶𝐷 = 500 mm2
= −𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression
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 𝜃31 200
𝐵𝐶
𝐶𝐹 = = = 52.0 kN Compression
cos 𝜃 0.6
𝐶𝐸 = 50 − 𝐶𝐹 sin 𝜃 = 50 − 52.0(0.8)
CE = 8.4 kN Compression
, 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 N
3
= 𝟓𝟎 𝐌𝐏𝐚
𝜎=
(20) 2 mm 2
𝜋(13)
1.39 𝐴= 2
4 = 132.7 mm2
𝐹 56 000 N = 𝟒𝟐𝟐 𝐌𝐏𝐚
𝜎= =
𝐴 132.7 mm 2
1.40 𝐴 = (68)(36) + 2[(36)(12)(1/2)] = 2880 mm2
𝐹 230 000 N
𝜎= = = 𝟕𝟗. 𝟗 𝐌𝐏𝐚
𝐴 2880 mm 2
𝜋(40)
1.41 𝐴 = (80)(40) − (60)(15) + 2
4 = 3557 mm2
N
𝐹 640×10
3
= 𝟏𝟖𝟎 𝐌𝐏𝐚
𝜎= =
𝐴 3557 mm 2
,1.42 Direct Shear – Single Shear
𝜋(12.0) 2 4
] mm3 2
𝐴𝑆 = [ = 113 mm2
𝐹 N
𝜏= 16.5×10 = 𝟏𝟒𝟔 𝐌𝐏𝐚
=
𝐴𝑆 113 mm 2 Pin is in single
shear
1.43 ∑ 𝐹𝐽 = 0 = 55(145) − 𝐹𝑃 (45)
𝐹𝑃 = 177 N
𝜋(3.0) 2 = 7.07 mm 2
𝐴𝑆 = 4
𝐹𝑃 177 N = 𝟐𝟓. 𝟏 𝐌𝐏𝐚
𝜏= =
𝐴𝑆 7.07 mm 2
1.44 From Problem
𝜋(10) 2
1-46: 𝐹 = 23 695 N
] = 157 mm2 4
𝐴𝑆 = 𝐹2 [ 23 695 N Double Shear
𝜏= = = 𝟏𝟓𝟏 𝐌𝐏𝐚
𝐴𝑆 157 mm 2
1.45 𝐴𝑆 = 𝐹(75)(90) = 6750 mm2
𝜏 = = 1800 N
𝐴𝑆 6750 mm 2 = 𝟏. 𝟐 𝐌𝐏𝐚
1.46 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 mm2
3
𝐹 N
𝜏= 38.6×10 = 𝟖𝟏. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 475.7 mm 2
1.47 𝐿 = √152 + 102 = 18.03 mm
𝜋(20)
+ 2(18)] 5
𝐴𝑆 = [2(40) +
2
2
𝐴𝑆 = 𝐹737 200
mm000 N
𝜏= = = 𝟐𝟕𝟏 𝐌𝐏𝐚
𝐴𝑆 737 mm 2
1.48 𝑇 = 𝐹𝑆 ∙ 𝑅 95 N∙m 103 mm
𝑇
= 5429 N
𝐹𝑆 = = ∙
𝑅 35 mm/2 m
𝐴𝑆 = 𝑏 ∙ 𝐿 = (10)(22) = 220 mm2
𝐹 5429 N = 𝟐𝟒. 𝟕 𝐌𝐏𝐚
𝜏= 𝑆 =
𝐴𝑆 220 mm 2
, 𝑇 900 N∙m
1.49 𝐹𝑆 = 𝑅
= 0.025 m
= 36 000 N
𝐴𝑆 = 𝑏 ∙ 𝐿 = (12)(60) = 720 mm2
𝐹 36 000 N
𝜏= 𝑆= = 𝟓𝟎 𝐌𝐏𝐚
𝐴𝑆 720 mm 2
1.50 Pin: Double
𝐹 Shear; 𝐴𝑆 = 2[𝜋(13)2/4] = 265.5 mm2
𝜏 = = 000 N90
𝐴𝑆 265.5 mm 2 = 𝟑𝟑𝟗 𝐌𝐏𝐚
Collar: Shear Collar from Connector Body
𝐴𝑆 = 𝐹𝜋𝑑𝑡 = 𝜋(22)(5) = 345.6 mm2
𝜏 = = 90 000 N
𝐴𝑆 345.6 mm 2 = 𝟐𝟔𝟎. 𝟒 𝐌𝐏𝐚
1.51 ∑ 𝑀𝐴 = 0 = 3600(2) − 𝐵𝑉(0.2)
𝐵𝑉 = 36 000
𝐵
N
𝐵 = cos 20°𝑉
= 38 310 N
𝜋(10) 2
] = 157 mm2 4
𝐴𝑆 = 𝐵2 [ 38 310 N
𝜏= = = 𝟐𝟒𝟒 𝐌𝐏𝐚
𝐴𝑆 157 mm 2
1.52 𝐴𝑆 = (40)(12) = 480 mm2
3
𝐹 N
𝜏= 88×10 = 𝟏𝟖𝟑 𝐌𝐏𝐚
=
𝐴𝑆 480 mm 2
1.53 𝐴𝑆 = (40)(120) = 4800 mm2
3
𝐹 N
𝜏= 88.2×10 = 𝟏𝟖. 𝟒 𝐌𝐏𝐚
=
𝐴𝑆 4800 mm 2
1.54 𝐴𝑆 = 𝜋𝑑𝑡 = 𝜋(12)(8) = 301.6 mm2
3
𝐹 N
𝜏= 22.3×10 = 𝟕𝟑. 𝟗 𝐌𝐏𝐚
=
𝐴𝑆 301.6 mm 2
1.55 𝐴𝑆 = 2[𝜋(12) /4] = 226.2 mm2 Two Rivets – Single Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟒𝟓. 𝟏 𝐌𝐏𝐚
=
𝐴𝑆 226.2 mm 2
1.56 𝐴𝑆 = 4[𝜋(12) /4] = 452.4 mm2 Two Rivets – Double Shear
2
3
𝐹 N
𝜏= 10.2×10 = 𝟐𝟐. 𝟓𝟓 𝐌𝐏𝐚
=
𝐴𝑆 452.4 mm 2
, Structural Shapes 2, 6.075 × 104 mm3 with long side vertical 1-58
1-57 [Appendix
[Appendix A-4(b)] A = 4050
A-5(c)] 20.719 x 2mm
= 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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