Engineering Mechanics - Statics Episode 3 Part 7 pptx

Engineering Mechanics - Statics Episode 3 Part 7 pptx

Engineering Mechanics - Statics Episode 3 Part 7 pptx

... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: I xy 0 b y a 2 y b ya y b ⌠ ⎮ ⎮ ⌡ d= I xy 0.6 67 in 4 = 10 73 © 20 07 R. C. Hibbeler. Published by Pearson ... 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 × mm 4 = I y 1 12 ab 3 =...
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Engineering Mechanics - Statics Episode 1 Part 7 pptx

Engineering Mechanics - Statics Episode 1 Part 7 pptx

... F v F cos φ () sin θ () cos φ () cos θ () sin φ () − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = M A r AC F v ×= M A 5 .38 5− 13. 0 93 11 . 37 7 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-4 5 The pipe assembly is subjected to the force F . Determine the moment of this force about point B. 2 43 © 20 07 R. C. Hibbeler. ... publisher. Engineering Mechanics - Statics Chapter 4 c 1m= Solution: r AB 0 0 ab+ ⎛ ⎜ ⎜...
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Engineering Mechanics - Statics Episode 3 Part 8 pdf

Engineering Mechanics - Statics Episode 3 Part 8 pdf

... Engineering Mechanics - Statics Chapter 11 Problem 1 1-1 7 Each member of the pin-connected mechanism has a mass m 1 . If the spring is unstretched ... 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 V d d 2 γπ a 2 h 2 2 W3a 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ... publisher. En...
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Engineering Mechanics - Statics Episode 3 Part 6 pps

Engineering Mechanics - Statics Episode 3 Part 6 pps

... publisher. Engineering Mechanics - Statics Chapter 10 the x and y axes. Given: a 30 mm= b 170 mm= c 30 mm= d 140 mm= e 30 mm= f 30 mm= g 70 mm= Solution: I x 1 3 ac d+ e+() 3 1 3 bc 3 + 1 12 ge 3 + ... 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+()...
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Engineering Mechanics - Statics Episode 3 Part 5 pps

Engineering Mechanics - Statics Episode 3 Part 5 pps

... 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 ρ w gabL= F 3 470 .88 kN= F 2 ρ w gabL= ... writing from the publisher. Engineering Mechanics - Statics Chapter 9 F 7. 62 10 3 ×= N x c 1 F 0 5 xx 240− x 1+ 34 0+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 6 ⌠ ⎮ ⎮ ⌡ d= x c 2 .74...
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Engineering Mechanics - Statics Episode 3 Part 4 potx

Engineering Mechanics - Statics Episode 3 Part 4 potx

... 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 π ⎛...
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Engineering Mechanics - Statics Episode 3 Part 3 docx

Engineering Mechanics - Statics Episode 3 Part 3 docx

... 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 ... writing from the publisher. Engineering Mechanics - Statics Chapter 9 Problem 9 -3 6 Locate the centroid of the quarter-cone. Solution: r a h hz−()= z c z= x c y...
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Engineering Mechanics - Statics Episode 3 Part 2 pdf

Engineering Mechanics - Statics Episode 3 Part 2 pdf

... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Given: M 75 kg= a 70 0 mm= b 25 mm= c 30 0 mm= d 200 mm= e 1 60 mm= μ s 0 .3= e 2 .71 8= Solution: Initial guesses: T 1 1N= ... publisher. Engineering Mechanics - Statics Chapter 8 F E Pcb b 2 c 2 + cos β () b sin β () a+ () = F E 72 .7 N= The equilibrium of clamped block requires tha...
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Engineering Mechanics - Statics Episode 3 Part 1 doc

Engineering Mechanics - Statics Episode 3 Part 1 doc

... 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 or...
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Engineering Mechanics - Statics Episode 2 Part 7 pdf

Engineering Mechanics - Statics Episode 2 Part 7 pdf

... from the publisher. Engineering Mechanics - Statics Chapter 7 01 234 5 678 9 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 , 01 234 5 678 9 0 200 400 600 Distance ... L 3 x≤ 2L 3 ≤ + ↑ Σ F y = 0; V 2 0 = Σ M x = 0; M 2 M 0 = For 2L 3 x≤ L≤ + ↑ Σ F y = 0; V 3 0 = Σ M x = 0; M 3 0 = b() x 1 0 0.01 L, L 3 = x 2 L...
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