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 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 333 471− 333 0 66 7 66 7 471 943− 66 7− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... publisher. Engineering Mechanics - Statics Chapter 7 + ↑ Σ F y = 0; V C wc− B y + F 2 − 0= V C wc B y − F 2 +...

Ngày tải lên : 21/07/2014, 17:20

40 774 1

Engineering Mechanics - Statics Episode 2 Part 2 potx

... publisher. Engineering Mechanics - Statics Chapter 6 F AB F AF F BC F BF F FC F FE F ED F EC F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 377.1− 189.7 26 6. 7− 26 6. 7 188 .6 56. 4 56. 4 0.0 188 .6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... publisher. Engineering Mechanics - Statics Chapter 6 Solution: Initial Gues...

Ngày tải lên : 21/07/2014, 17:20

40 322 0

Process Engineering Equipment Handbook Episode 2 Part 6 potx

... possible 11. Handles 2- phase mixtures, liquids and solids mixed 12. Suitable for high-temperature (20 00°F/1093°C), low-temperature (-3 25 °F/ -1 98°C), and high-pressure (up to 5000 psi/3 52 kg/cm 2 ) operating ... the other axis (see Fig. L-3). Third peak cyclic usage factor 93 2 = - - È Î Í ˘ ˚ ˙ 0 100 0 2 2 3 Second peak cyclic usage factor 97 2 = - - ()...

Ngày tải lên : 21/07/2014, 16:22

50 351 0

Engineering Mechanics - Statics Episode 3 Part 6 pps

... publisher. Engineering Mechanics - Statics Chapter 10 Solution: y c 2ab b 2 2cb b c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 2ab 2cb+ = y c 12. 50 mm= I x 1 12 2ab 3 2ab y c b 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 2 1 12 bc 3 bc b c 2 + y c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ += I x 908.3 ... Given: a 25 mm= b 25 0 mm= c 50 mm= d 150 mm= Solutuion: y c b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b2ab c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2dc+ b2ac2d+...

Ngày tải lên : 21/07/2014, 17:20

40 506 1

Engineering Mechanics - Statics Episode 3 Part 4 potx

... publisher. Engineering Mechanics - Statics Chapter 9 Solution: Lab+ a 2 c 2 ++ b 2 c 2 ++= x c 1 L ab+() ab+ 2 a 2 c 2 + a 2 + b 2 c 2 + a b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = x c 6. 50 in= y c 1 L a 2 c 2 + c 2 b 2 c 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ... Engineering Mechanics - Statics Chapter 9 Solution: A 2a 3 2 a 2 2 π = A 3 π a 2 = V 2 1...

Ngày tải lên : 21/07/2014, 17:20

40 1.8K 1

Engineering Mechanics - Statics Episode 2 Part 10 ppt

... 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⋅ = 024 68 10 10 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 024 68 10 60 40 20 0 20 Distance ... publisher. Engineering Mechanics - Statics Chapter 8 T N A F A F N ground ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟...

Ngày tải lên : 21/07/2014, 17:20

40 1.3K 1

Engineering Mechanics - Statics Episode 2 Part 9 ppsx

... y−()= Σ M = 0; Mwy y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wa a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wby− 0= My() wby 1 2 wy 2 − 1 2 wb 2 − 1 2 wa 2 −= My() 16 lb y 2 lb ft y 2 − 40 lb ft⋅−= Problem 7-1 21 Determine the ... y w 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = dy dx w F H x= h w 2F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = F H w 2h a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = tan θ max () wa 2F H = cos θ max () 2F H 4 F H 2 wa...

Ngày tải lên : 21/07/2014, 17:20

40 559 0

Engineering Mechanics - Statics Episode 2 Part 8 pdf

... publisher. Engineering Mechanics - Statics Chapter 7 0 2 4 6 8 10 121 4 161 820 1000 500 0 500 1000 1500 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 024 68 10 121 4 161 820 1 . 10 4 5000 0 5000 Distance ... publisher. Engineering Mechanics - Statics Chapter 7 024 68 10 121 4 30 20 10 0 10 20 30 Distance (m) Moment...

Ngày tải lên : 21/07/2014, 17:20

40 400 0

Engineering Mechanics - Statics Episode 2 Part 7 pdf

... publisher. Engineering Mechanics - Statics Chapter 7 0 123 4 567 89 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 , x 3 , 0 123 4 567 89 0 20 0 400 60 0 Distance ... ft⋅ = 024 68 10 12 1000 500 0 500 1000 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 024 68 10 12 0 20 00 4000 Distance (...

Ngày tải lên : 21/07/2014, 17:20

40 507 2

Engineering Mechanics - Statics Episode 2 Part 5 ppsx

... E y , () = C x F CD d c 2 d 2 + = C y F CD c c 2 d 2 + = D x C x −= D y C y −= A x B x B y C x C y D x D y E x E y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 160 80 26 .66 7 160 1 06. 667 160 − 1 06. 667 − 8 .69 4− ... from the publisher. Engineering Mechanics - Statics Chapter 6 F W 1 W 2 + 2 = F 102lb= Man: + ↑ Σ F y = 0; N C W 1 − 2 F 2...

Ngày tải lên : 21/07/2014, 17:20

40 661 2