Engineering Mechanics - Statics Episode 3 Part 6 pps

... publisher. Engineering Mechanics - Statics Chapter 10 Solution: I y ab 3 3 1 36 ac 3 + 1 2 ac b c 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 36 db c+() 3 + 1 2 db c+() 2 bc+() 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 += I y 1971in 4 = Problem 1 0-4 5 Locate ... publisher. Engineering Mechanics - Statics Chapter 10 Solution: I x' 1 36 bh 3 = x c 2 3 a 1 2 ha a ba− 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 hb...

Ngày tải lên: 21/07/2014, 17:20

... Engineering Mechanics - Statics Chapter 9 Units Used: kip 10 3 lb= Given: γ 56 lb ft 3 = c 8ft= w 6ft= d 4ft= a 10 ft= e 3ft= b 2ft= f 4ft= Solution: w B w γ b= w B 67 2 lb ft = w C w γ bc+()= ... the publisher. Engineering Mechanics - Statics Chapter 9 Units Used: Mg 10 3 kg= kN 10 3 N= Given: L 8m= ρ w 1.0 Mg m 3 = a 3m= b 2m= g 9.81 m s 2 = Solution: F 3...

Ngày tải lên: 21/07/2014, 17:20

... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 F 2 8N= e 20 mm= F 3 6N= f 35 mm= a 0.2 m= g 15 mm= b 0 .35 m= θ 1 30 deg= c 0.25 m= θ 2 15 deg= Solution: Positive ... from the publisher. Engineering Mechanics - Statics Chapter 4 Solution: r AC a r− b− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AC 1 3 2− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= F v F r AC r AC = F v 21 .38 1...

Ngày tải lên: 21/07/2014, 17:20

... LL v v 4 43 4 1 2 3 3 2 2 1 2 21 31 1 2 31 2 21 2 1 248 4 12 3 = - Ê Ë ˆ ¯ +-+ () + () {} -+ - () () {} K LL kD EI LL LL kD EI LL LL v v 3 32 1 2 22 2 31 1 2 31 2 21 2 4 12 3 = -+ - () -+ + () ... LL 2 3 2 1 2 31 2 2 1 2 21 31 2 21 2 4 = + () + () -+ () + () - () () {} K LL LLLL LL 1 32 3 3 131 2 21 2 = - () - () - (...

Ngày tải lên: 21/07/2014, 16:21

... publisher. Engineering Mechanics - Statics Chapter 11 a b c 60 (+ 60 (- 60 ( Solution: a sin 60 deg() b sin 60 deg θ − () = ba sin 60 deg θ − () sin 60 deg() = VW 2 3 dsin 60 deg( ) cos θ () bcos 30 ... publisher. Engineering Mechanics - Statics Chapter 11 V γπ a 2 h 2 2 W3a 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ () = θ V d d γπ a 2 h 2 2 W3a 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − sin θ () = 2 θ...

Ngày tải lên: 21/07/2014, 17:20

... γ . Given: a 3in= b 3in= γ 38 0 lb ft 3 = Solution: m 0 b y γπ a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= m 2 .66 lb= I y 0 b y γπ a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 1 2 a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= I y 6. 46 ... from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: Moment of inertia I x and I y : I x 1 12 ba 3 = I x 90 10 3...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 9 Solution: V 2 3 π a 3 π 3 a 2 a−= π 3 a 3 = z c 1 V 5a 8 2 3 π a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 4 a π 3 a 3 ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = z c a 2 = Problem 9-7 6 Determine ... Engineering Mechanics - Statics Chapter 9 Solution: A 2a 3 2 a 2 2 π = A 3 π a 2 = V 2 1 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 2 a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 6 a2...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 9 Given: a 2ft= b 2ft= Solution: V 0 b z π a 2 z b ⌠ ⎮ ⎮ ⌡ d= V 12. 566 ft 3 = z c 1 V 0 b zz π a 2 z b ⌠ ⎮ ⎮ ⌡ d= z c 1 .33 3 ft= Problem 9 -3 5 Locate ... publisher. Engineering Mechanics - Statics Chapter 9 Given: a 4in= b 2in= c 3in= Solution: A b ab+ y bc y ⌠ ⎮ ⎮ ⌡ d= A 6. 592 in 2 = y c 1 A b ab+ yy bc y ⎛ ⎜...

Ngày tải lên: 21/07/2014, 17:20

... from the publisher. Engineering Mechanics - Statics Chapter 8 M 0 R r μ p 0 p 0 R r− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 ⌠ ⎮ ⎮ ⌡ d 0 2 π θ 1 ⌠ ⎮ ⌡ d= π 6 μ p 0 R 3 = π 6 μ 3P π R 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ R 3 = μ PR 2 = M μ PR 2 = Thus, Problem ... Engineering Mechanics - Statics Chapter 8 Given: a 2in= b 3in= P 500 lb= M 3lbft= Solution: M a b μ k P b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − b 2...

Ngày tải lên: 21/07/2014, 17:20

... x, () = P N B F B T N D F D ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 13. 33 33. 33 6. 67 6. 67 30 .00 6. 67 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= x 0 .67 ft= Now checke the assumptions F Dmax μ D N D = Since F D 6. 67 lb= < F Dmax 9.00 ... publisher. Engineering Mechanics - Statics Chapter 8 Since x 0 .67 ft= < b 2 0.75 ft= then the block does not tip. So our origi...

Ngày tải lên: 21/07/2014, 17:20