Engineering Mechanics - Statics Episode 1 Part 10 potx

... the publisher. Engineering Mechanics - Statics Chapter 5 Given: W 850 lb= a 0.5 ft= b 1ft= c 1. 5 ft= θ 1 10 deg= θ 2 30 deg= θ 3 10 deg= Solution: Guesses F A 1lb= F B 1lb= F C 1lb= Given Σ M B ... publisher. Engineering Mechanics - Statics Chapter 5 Units Used: kN 10 3 N= Given: F 500 N= a 0 .15 m= L 3m= Solution: The initial guesses w 1 1 kN m = w 2 1 kN...

Ngày tải lên: 21/07/2014, 17:20

... .4 .25 .25 .1 .05 Horizontal centrifugal compressors .25 .2 .15 .15 .1 .05 Barrel compressor .15 .1 .1 .1 .05 .025 Gears Parallel Shaft .25 .2 .15 .15 .1 .05 Epicyclic .15 .1 .1 .1 .1 .05 Steam turbines ... get (C -1 0 ) which, if we solve for t (C -1 1 ) or in more general terms (C -1 2 ) xe ce ce cmt cm km t cm km t =+ [] - ( ) ( ) - ( ) [] - ( ) - ( ) [] 2 1...

Ngày tải lên: 21/07/2014, 16:22

... the publisher. Engineering Mechanics - Statics Chapter 9 Given: a 12 0 mm= b 240 mm= c 12 0 mm= Solution: Σ M B = 0; Wx c Pl 1 − W 1 l 1 l 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= x c Pl 1 W 1 l 1 l 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − W = Put ... publisher. Engineering Mechanics - Statics Chapter 9 b 4in= c 6in= Solution: x c a− b b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab 1 2 ac+ = x c 1. 143− in= y c ab a 2 ⎛...

Ngày tải lên: 21/07/2014, 17:20

... M 1 x 1 () Ax 1 wx 1 x 1 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 kN m⋅ = x 2 a 1. 01a, ab+ = V 2 x 2 () C− 1 kN = M 2 x 2 () M− Ca b+ x 2 − () + ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = 0246 810 10 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 0246 810 60 40 20 0 20 Distance ... publisher. Engineering Mechanics - Statics Chapter 7 Given wa b a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Aa b+()− M− 0= AC+ wa−...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 6 P 2 10 00 lb= a 10 ft= b 10 ft= Solution: θ atan b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses: F AB 1lb= F AG 1lb= F BG 1lb= F BC 1lb= F DC 1lb= F DE 1lb= F EG 1lb= ... publisher. Engineering Mechanics - Statics Chapter 6 Solution: θ atan e a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses: F AB 1kN= F AG 1kN= F CF 1kN= F BC 1kN= F BG 1kN= F D...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: w 1 1.5 kN m = a 3m= w 2 1 kN m = b 3m= w 3 2.5 kN m = c 1. 5 m= Solution: F R w 1 aw 2 b+ w 3 c+= F R 11 .25 kN= M A w 1 a a 2 w 2 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ... the publisher. Engineering Mechanics - Statics Chapter 4 kip 10 3 lb= Given: w 1 50 lb ft = w 2 300 lb ft = w 3 1...

Ngày tải lên: 21/07/2014, 17:20

... N= F Ry F 1 − sin θ () F 3 cos φ () − F 2 −= F Ry 1. 296− 10 3 × N= FF Rx 2 F Ry 2 += F 1. 302 10 3 × N= θ 1 atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 84.5 deg= F Ry xF 1 sin θ () bF 3 cos φ () c− M−= x F 1 sin θ () bF 3 cos φ () c− ... deg= Solution: F Rx F 1 cos θ () F 3 sin φ () −= F Rx 12 5− N= F Ry F 1 − sin θ () F 3 cos φ () − F 2 −= F Ry 1. 296− 10 3 × N= FF Rx 2 F Ry...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: x 1m= F 6 2− 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= M O 4 5 14 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Solution: Initial Guesses: y 1m= z 1m= Given ... publisher. Engineering Mechanics - Statics Chapter 4 Initial guesses: θ 10 deg= M 3 10 N m⋅= Given + → Σ M x = 0; M 1 M 3 cos θ () − M 2 cos θ 1 () − 0= + ↑...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kip 10 00 lb= Given: F 275 lb= a 85 ft= θ 30 deg= Solution: M A1 F sin θ () a= M A1 11 .7 kip ft⋅= M A2 F sin θ () a= M A2 11 .7 kip ... without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 3 Solution: Initial Guesses F OA 10 lb= F OC 10 lb= F OD 10 lb= Given c a...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 3 kN 10 3 N= Given: T 15 kN= e 4m= a 2m= f 6m= b 10 m= g 6m= c 12 m= h 6m= d 2m= i 2m= gravity 9. 81 m s 2 = Solution: F B T= F C T= F D T= M 1kg= The ... publisher. Engineering Mechanics - Statics Chapter 3 Given: F 0 60 lb= k 40 lb ft = d 2ft= y 1 2ft= Solution: Initial spring stretch: s 1 F 0 k = s 1 1.50...

Ngày tải lên: 21/07/2014, 17:20