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Engineering Mechanics - Statics Episode 3 Part 3 docx

Engineering Mechanics - Statics Episode 2 Part 3 ppsx

Engineering Mechanics - Statics Episode 2 Part 3 ppsx

... publisher. Engineering Mechanics - Statics Chapter 6 Given: F 1 30 kN= F 2 20 kN= F 3 20 kN= F 4 40 kN= a 4m= b 4m= Solution: F 2 − aF 3 2a()− F 4 3a()− G y 4a()+ 0= G y F 2 2F 3 + 3F 4 + 4 = G y 45kN= Guesses ... writing from the publisher. Engineering Mechanics - Statics Chapter 6 compression. Units Used: kN 10 3 N= Given: F 1 2kN= F 4 5kN= a 5m= F 2 4kN= F 5 3k...
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Engineering Mechanics - Statics Episode 1 Part 3 pptx

Engineering Mechanics - Statics Episode 1 Part 3 pptx

... publisher. Engineering Mechanics - Statics Chapter 2 Problem 2-9 3 The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. Given: F BA 35 0 ... writing from the publisher. Engineering Mechanics - Statics Chapter 2 b 8m= c 3m= d 4m= Solution: r c− d− b− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r 3 12− 4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m=...
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Engineering Mechanics - Statics Episode 1 Part 8 docx

Engineering Mechanics - Statics Episode 1 Part 8 docx

... the publisher. Engineering Mechanics - Statics Chapter 4 Given: F 1 450 N= a 2m= F 2 30 0 N= b 4m= F 3 700 N= c 3m= θ 60 deg= M 1500 N m⋅= φ 30 deg= Solution: F Rx F 1 cos θ () F 3 sin φ () −= F Rx 125− ... publisher. Engineering Mechanics - Statics Chapter 4 Given: F 1 30 lb= F 2 40 lb= F 3 60 lb= a 1ft= b 3ft= c 2ft= d 3= e 4= Solution: F R F 1 0 1− 0 ⎛ ⎜ ⎜ ⎝...
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Engineering Mechanics - Statics Episode 2 Part 10 ppt

Engineering Mechanics - Statics Episode 2 Part 10 ppt

... Engineering Mechanics - Statics Chapter 8 μ s 0 .3= M 5Nm⋅= a 50 mm= b 200 mm= c 400 mm= r 150 mm= P 1 30 N= P 2 70 N= Solution: To hold lever: Σ M O = 0; F B rM− 0= F B M r = F B 33 .33 3 N= Require N B F B μ s = ... the publisher. Engineering Mechanics - Statics Chapter 8 a 3ft= Solution: The brace is a two-force member. μ s N N L 2 a 2 − a = μ s aL 2 a 2 −= La1 μ s...
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Engineering Mechanics - Statics Episode 2 Part 9 ppsx

Engineering Mechanics - Statics Episode 2 Part 9 ppsx

... publisher. Engineering Mechanics - Statics Chapter 7 0 5 10 15 20 60 40 20 0 20 Distance (ft) Moment (kip-ft) M 1p x 1 () M 2p x 2 () M 3p x 3 () x 1 ft x 2 ft , x 3 ft , Problem 7-8 8 Draw the ... from the publisher. Engineering Mechanics - Statics Chapter 7 Problem 7-8 6 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given: w 2 kN m = a...
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Engineering Mechanics - Statics Episode 2 Part 8 pdf

Engineering Mechanics - Statics Episode 2 Part 8 pdf

... writing from the publisher. Engineering Mechanics - Statics Chapter 7 01 234 5678 5 0 5 10 15 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () M 3 x 3 () x 1 x 2 , x 3 , Problem 7-6 9 Draw the shear and ... from the publisher. Engineering Mechanics - Statics Chapter 7 Problem 7-8 3 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given:...
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Engineering Mechanics - Statics Episode 2 Part 7 pdf

Engineering Mechanics - Statics Episode 2 Part 7 pdf

... L 3 x≤ 2L 3 ≤ + ↑ Σ F y = 0; V 2 0 = Σ M x = 0; M 2 M 0 = For 2L 3 x≤ L≤ + ↑ Σ F y = 0; V 3 0 = Σ M x = 0; M 3 0 = b() x 1 0 0.01 L, L 3 = x 2 L 3 L 3 1.01, 2L 3 = x 3 2 L 3 2L 3 1.01, L ... from the publisher. Engineering Mechanics - Statics Chapter 7 01 234 56789 1 0.5 0 0.5 1 Distance in m Shear force in N V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 x 2...
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Engineering Mechanics - Statics Episode 2 Part 6 pot

Engineering Mechanics - Statics Episode 2 Part 6 pot

... Engineering Mechanics - Statics Chapter 6 F AB F AG F BC F BG F CD F CE F CG F DE F EG ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 33 3 471− 33 3 0 667 667 471 9 43 667− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... publisher. Engineering Mechanics - Statics Chapter 7 N C F 3 cos θ () = N C 21.7 kN= Σ F y = 0; V C F 2 − F 3 si...
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Engineering Mechanics - Statics Episode 2 Part 5 ppsx

Engineering Mechanics - Statics Episode 2 Part 5 ppsx

... 0= D y W 1 − W 3 − C y + 0= W 3 xW 1 e+ D y cd+ e+()− 0= B y C y D y x ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find B y C y , D y , x, () = A y B y C y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 5.5 10 3 × 6 .33 3 10 3 × 3. 167 10 3 × ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= ... without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Given: w 1 500 N m = w 2 400 N m = w 3 600 N m =...
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Engineering Mechanics - Statics Episode 2 Part 4 pdf

Engineering Mechanics - Statics Episode 2 Part 4 pdf

... from the publisher. Engineering Mechanics - Statics Chapter 6 Positive (T) Negative (C) F AB F AC F AD F AE F BC F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 30 0− 5 83. 095 33 3 .33 3 666.667− 0 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= F BF F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 30 0− 30 0− 0 424.264 30 0− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb=...
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