Engineering Mechanics Statics - Examples Part 2 pps

... deg= Solution: cos α () 2 cos β () 2 + cos γ () 2 + 1= F x F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 cos β () 2 + F z F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1= F F x 2 F z 2 + 1 cos β () 2 − = F 2. 02 kN= F y F cos β () = F y 0.5 kN= Problem 2- 7 4 The eye ... F 1v 175 175 24 7.5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 2h F 2 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2y F 2 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2v F 2h cos θ () F 2h − sin θ () F...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 11 Solution: b 2 a 2 x 2 + 2axcos θ () −= Solving xacos θ () a 2 cos θ () 2 b 2 + a 2 ++= Virtual displacements b 2 a 2 x 2 + 2axcos θ () −= 02x δ x 2acos θ () δ x− ... m= I G 1 12 2a ρ r 2a() 2 2a ρ r y c 2 + 1 12 b ρ r b 2 + b ρ r b 2 y c − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 2 π d 2 ρ s d 2 + π d 2 ρ s bd+...

Ngày tải lên: 11/08/2014, 02:21

... publisher. Engineering Mechanics - Statics Chapter 7 024 6810 20 0 0 20 0 Distance in m Force in kN V 1 x 1 () V 2 x 2 () x 1 x 2 , 024 6810 400 20 0 0 20 0 400 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () x 1 x 2 , Problem ... xwxa−() xa− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M 2 x()+ 0= M 2 x() A y xw xa−() 2 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN m⋅ = 0 0.5 1 1.5 2 2.5 3 3.5 4 4 2 0 2...

Ngày tải lên: 11/08/2014, 03:20

... publisher. Engineering Mechanics - Statics Chapter 1 Given: a 1 0.631 Mm= b 1 8.60 kg= a 2 35 mm= b 2 48 kg= Solution: a() a 1 b 1 2 8.5 32 km kg 2 = b() a 2 2 b 2 3 135.48 kg 3 m 2 ⋅= 12 © 20 07 R. ... publisher. Engineering Mechanics - Statics Chapter 2 F 3 50 N= θ 20 deg= c 3= d 4= Solution: F Rx F 1 − d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 sin θ ()() −...

Ngày tải lên: 11/08/2014, 02:22

... without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 FM A c 2 d 2 + da ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 35 .2 N= Problem 4 -2 2 Determine the clockwise direction θ 0deg θ≤ ... Lab+ d e c− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e e 2 d 2 + = M A F 1 − a( )cos θ 2 () ⎡ ⎣ ⎤ ⎦ F 2 ab+( ) sin θ 1 () − F 3 L−= M A 2. 5 32 kN m⋅= Solution Using Principle of Moments: M A F 1...

Ngày tải lên: 11/08/2014, 02:22

... 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 = d 11.3 ft= Problem 4-1 50 The ... 10 3 N= Given: w 1 800 N m = w 2 200 N m = a 2m= b 3m= Solution: F R w 2 bw 1 a+ 1 2 w 1 w 2 − () b+= F R 3.10 kN= xF R w 1 a a 2 1...

Ngày tải lên: 11/08/2014, 02:22

... ab+() = F 2 724 N= F 1 2 F 2 = F 1 1.448 kN= + → Σ F x = 0; A x d c 2 d 2 + T max − 0= A x d c 2 d 2 + T max = A x 1 .20 kN= + ↑ Σ F y = 0; A y F 2 − F 1 − c c 2 d 2 + T max + 0= 3 82 © 20 07 R. ... 4= θ 30deg= Solution: Σ M A = 0 ; F 1 2 F 2 = 2 F 2 acos θ () F 2 ab+( ) cos θ () − d c 2 d 2 + T max ab+( ) sin θ () + c c 2 d 2 + T max ab+( ) cos θ ()...

Ngày tải lên: 11/08/2014, 02:22

... F CD 2 a 2 a() 2 b 2 + − 0= F 2 − F CD b 2 a() 2 b 2 + − 0= Joint B F BA − F CB + 0= F 1 − F BD − 0= Joint D F CD F DA − F DE − () 2 a 2 a() 2 b 2 + 0= F BD F CD F DA + F DE − () b 2 a() 2 b 2 + + ... publisher. Engineering Mechanics - Statics Chapter 5 w 1 ab+() 1 2 w 2 w 1 − () ab+()+ F− W− 0= w 1 ab+() ab+ 2 1 2 w 2 w 1 − () ab+() 2 3 ab+...

Ngày tải lên: 11/08/2014, 02:22

... the publisher. Engineering Mechanics - Statics Chapter 6 F CD − b a 2 b 2 + F BD − b− a 2 b 2 + c 2 + F AD b b 2 c 2 + F DF −+ 0= F 3 − F DE − c b 2 c 2 + F DF − c− a 2 b 2 + c 2 + F AD + 0= Joint ... without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 A y F 1 − F 2 − F OE a a 2 b 2 + − 0= F LK − 2a()F...

Ngày tải lên: 11/08/2014, 02:22

... 2 c b b 2 c 2 + F max c+ 0= P F max b 2 b 2 c 2 + = P 28 2.843 N= B z D z + F max c b 2 c 2 + − 0= D z B z = B z F max c 2 b 2 c 2 + = D z B z = B z 28 3 N= D z 28 3 N= B y D y + F max b b 2 c 2 + − ... from the publisher. Engineering Mechanics - Statics Chapter 6 Given M− 2 gcos θ 1 () F AB cos θ 2 () − F CB b a 2 b 2 + + 0= M− 2 gsin θ 1 () F...

Ngày tải lên: 11/08/2014, 02:22