Engineering Mechanics - Statics Episode 3 Part 2 pdf

... 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 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = a μ k P 2b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 3 a 3 − b 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = μ k 2Mb ... publisher. Engineering Mechanics - Statics Chapter 8 P 0 2 π θ R 1 R...

Ngày tải lên: 21/07/2014, 17:20

... c 2 ce ea− = y c d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 1 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 c− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 1 3c+ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 2 c− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 2 3c+ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 c− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠...

Ngày tải lên: 21/07/2014, 17:20

... Solution: m 0 a x ρπ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 2 3 a ρπ b 2 = ρ 3m 2 π ab 2 = I x 0 a x 1 2 3m 2 π ab 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ b 2 1 x 2 a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 2 5 mb 2 = ... t 3 8 in= Solution: I x 2 1 12 at 3 at b t 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 t 2b 2t−() 3 += I x 55.55 in 4...

Ngày tải lên: 21/07/2014, 17:20

... 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+ b...

Ngày tải lên: 21/07/2014, 17:20

... 6in= Solution: A 4a 2 a 2 2 − π a 2 4 −= A 97.7 in 2 = x c 1 A a 2 − 2 2− 3 a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π a 2 4 a 4a 3 π − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = x c 0 .26 2− in= y c 1 A a 2 − 2 2a 3 π a 2 4 4a 3 π a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = ... p A 88 .29 kN m 2 = w A p A w= w A 706 . 32 kN m = p B ρ w ga= p B 49.05 kN m 2 = w B p B w= w B 39 2. 4 kN m = Equilibrium w B b 2...

Ngày tải lên: 21/07/2014, 17:20

... 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 π ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = V π 4 a 3 = Problem 9-8 9 Sand is ... 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...

Ngày tải lên: 21/07/2014, 17:20

... revolution. Solution: dV π z 2 dy= z 2 a 2 1 y 2 b 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = V 0 b y π a 2 1 y 2 b 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 1 3 b 3 b 2 b 2 − b 2 a 2 π = y c 3 2ba 2 π 0 b yy π a 2 1 y 2 b 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⌠ ⎮ ⎮ ⎮ ⌡ d= 3 8 ... h 2 π a h h 2 a 2 + h 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = z c 1 π aa 2 h 2 + 0 h zz2 π az h 1 a h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⌠ ⎮ ⎮ ⌡ d...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 8 of the wedge and the thickness of the beam. Units Used: kN 10 3 N= Given: F 1 2kN= a 3m= F 2 4kN= b 2m= F 3 4kN= c 3m= F 4 2kN= θ 15 deg= μ s 0 .25 = Solution: Guesses N 1 1kN= ... kg= a 2m= μ B 0.4= b 400 mm= μ C 0 .2= c 30 0 mm= w 800 N m = d 3= g 9.81 m s 2 = e 4= Solution: Member AB: Σ M A = 0; 1 2 wa ⎛ ⎜ ⎝ ⎞ ⎟...

Ngày tải lên: 21/07/2014, 17:20

... w 3 b+= F R 3. 9 kip= F R dw 1 a a 2 1 2 w 2 w 1 − () a 2a 3 + 1 2 w 2 w 3 − () ba b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 3 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += d 3 w 3 ba 2 w 3 b 2 + w 1 a 2 + 2 a 2 w 2 + 3 bw 2 a+ w 2 b 2 + 6F R = ... F Ry 900− N= x− F Ry w 3 − a a 2 w 2 b b 2 + 1 2 w 1 w 2 − () b 2b 3 += x w 3 − a a 2 w 2 b b 2 + 1 2 w 1 w 2 − (...

Ngày tải lên: 21/07/2014, 17:20

... F R cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 1 0 0 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ += F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 2x F 2y , F 2z , () = F 2 F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 2 17.1− 8.7 26 .2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 2 32 .4 lb= α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 2 F 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 12...

Ngày tải lên: 21/07/2014, 17:20