Engineering Mechanics Statics - Examples Part 10 ppsx

Engineering Mechanics - Statics Chapter 7 V C 6 kip= Problem 7-11 Determine the internal normal force, shear force, and moment at points E and D of the compound beam. Given: M 200 Nm⋅= c 4 m= F 800 N= d 2 m= a 2 m= e 2 m= b 2 m= Solution: Segment BC : M− C y de+()+ 0= C y M de+ = B y − C y + 0= B y C y = Segment EC : N E − 0= N E 0 N= N E 0.00 = V E C y + 0= V E C y −= V E 50.00 − N= M E − M− C y e+ 0= M E C y eM−= M E 100.00 − Nm⋅= Segment DB : N D − 0= N D 0 N= N D 0.00 = V D F− B y + 0= V D FB y −= V D 750.00 N= M D − Fb− B y bc+()+ 0= M D F− bB y bc+()+= M D 1300 − Nm⋅= 631 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Problem 7-12 The boom DF of the jib crane and the column DE have a uniform weight density γ . If the hoist and load have weight W, determine the normal force, shear force, and moment in the crane at sections passing through points A, B, and C. Treat the boom tip, beyond the hoist, as weightless. Given: W 300 lb= γ 50 lb ft = a 7 ft= b 5 ft= c 2 ft= d 8 ft= e 3 ft= Solution: + → Σ F x = 0; N A − 0= N A 0 lb= N A 0.00 lb= + ↑ Σ F y = 0; V A W− γ e− 0= V A W γ e+= V A 450 lb= Σ Μ Α = 0; M A γ e e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − We− 0= M A γ e 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ We+= M A 1125.00 lb ft⋅= + → Σ F x = 0; N B − 0= N B 0 lb= N B 0.00 lb= + ↑ V B γ de+()− W− 0= V B γ de+()W+= V B 850 lb= Σ F y = 0; Σ Μ Β = 0; M B γ de+() de+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − Wd e+()− 0= M B 1 2 γ de+() 2 Wd e+()+= M B 6325.00 lb ft⋅= 632 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 + → Σ F x = 0; V C 0 = V C 0 lb= V C 0.00 lb= + ↑ Σ F y = 0; N C cd+ e+() γ − W− γ b()− 0= N C γ cd+ e+ b+()W+= N C 1200.00 lb= M C cd+ e+() γ cd+ e+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − Wc d+ e+()− 0= Σ Μ C = 0; M C cd+ e+() γ cd+ e+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Wc d+ e+()+= M C 8125.00 lb ft⋅= Problem 7-13 Determine the internal normal force, shear force, and moment at point C. Units Used: kip 10 3 lb= Given: a 0.5 ft= d 8ft= b 2 ft= e 4 ft= c 3 ft= w 150 lb ft = Solution: Entire beam: Σ M A = 0; w− de+() de+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Ta b+()+ 0= T wd e+() 2 2 ab+() = T 4.32 kip= Σ F x = 0; A x T− 0= A x T= A x 4.32 kip= Σ F y = 0; A y wd e+()− 0= A y wd e+()= A y 1.80 kip= Segment AC: 633 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Σ F x = 0; A x N C + 0= N C A x −= N C 4.32 − kip= Σ F y = 0; A y wc− V C − 0= V C A y wc−= V C 1.35 kip= Σ M C = 0; A y − cwc c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M C + 0= M C A y cw c 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= M C 4.72 kip ft⋅= Problem 7-14 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN 10 3 N= Given: w 400 N m = a 2.5 m= b 3 m= c 6 m= Solution: Σ M A = 0; 1− 2 wc 2 3 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c+ 0= F BC 1 3 wc a 2 c 2 + a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F BC 2080 N= + → c a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC A x − 0= Σ F x = 0; A x c a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC = A x 1920 N= + ↑ A y 1 2 wc− a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC + 0= Σ F y = 0; 634 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 A y 1 2 wc a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC −= A y 400 N= + → Σ F x = 0; N D A x − 0= N D A x = N D 1.920 kN= + ↑ F y = 0; A y 1 2 w b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b− V D − 0= Σ V D A y 1 2 w b 2 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= V D 100 N= Σ F y = 0; A y − b 1 2 w b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M D + 0= M D A y b 1 6 w b 3 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= M D 900 Nm= Problem 7-15 The beam has weight density γ . Determine the internal normal force, shear force, and moment at point C. Units Used: kip 10 3 lb= Given: γ 280 lb ft = a 3 ft= b 7 ft= c 8 ft= d 6 ft= 635 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 θ atan c d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = W γ ab+()= Guesses A x 1 lb= A y 1 lb= B x 1 lb= N C 1 lb= V C 1 lb= M C 1 lb ft= Given Entire beam: A x B x − 0= A y W− 0= B x cW d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= Bottom Section A x N C cos θ () − V C sin θ () + 0= A y W a ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − N C sin θ () − V C cos θ () − 0= M C V C a− W a ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ () − 0= A x A y B x N C V C M C ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , B x , N C , V C , M C , () = A x A y B x ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1.05 2.80 1.05 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kip= N C V C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.20 0.34 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip= M C 1.76 kip ft⋅= Problem 7-16 Determine the internal normal force, shear force, and moment at points C and D of the beam. 636 Solution: © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Units Used: kip 10 3 lb= Given: w 1 60 lb ft = a 12 ft= b 15 ft= w 2 40 lb ft = c 10 ft= d 5 ft= F 690 lb= e 12= f 5= Solution: θ atan e f ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses B y 1 lb= N C 1 lb= V C 1 lb= M C 1 lb ft⋅= N D 1 lb= V D 1 lb= M D 1 lb ft⋅= Given B y bFsin θ () bc+()− w 2 b b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1− 2 w 1 w 2 − () b b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= N C − F cos θ () − 0= V C 1 2 w 1 w 2 − () ba− b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ba−()− B y w 2 ba−()− F sin θ () −+ 0= N D − F cos θ () − 0= V D F sin θ () − 0= M D − F sin θ () d− 0= M C − w 2 ba−() ba− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 w 1 w 2 − () ba− b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ba−() ba− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − B y ba−()F sin θ () cb+ a−()−+ 0= 637 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 B y N C V C M C N D V D M D ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find B y N C , V C , M C , N D , V D , M D , () = N C V C N D V D ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 265− 649− 265− 637 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= M C M D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.23− 3.18− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip ft⋅= Problem 7-17 Determine the normal force, shear force, and moment acting at a section passing through point C. Units Used: kip 10 3 lb= Given: F 1 800 lb= F 2 700 lb= F 3 600 lb= θ 30 deg= a 1.5 ft= b 1.5 ft= c 3 ft= d 2 ft= e 1 ft= fab+ c+ d− e−= Solution: Guesses B y 1 lb= A x 1 lb= A y 1 lb= N C 1 lb= V C 1 lb= M C 1 lb ft⋅= 638 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Given A x − V C sin θ () + N C cos θ () + 0= A y V C cos θ () − N C sin θ () + 0= M C A x a( ) sin θ () − A y a( ) cos θ () − 0= A x − F 1 sin θ () + F 3 sin θ () − 0= A y B y + F 2 − F 1 cos θ () − F 3 cos θ () − 0= F 1 − ab+()F 2 ab+ c+( ) cos θ () − F 3 cos θ () ab+ c+ d+ e+( ) cos θ () − F 3 sin θ () fsin θ () B y 2 ab+ c+( ) cos θ () ++ 0= A x A y B y N C V C M C ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , B y , N C , V C , M C , () = A x A y B y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 100 985 927 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= N C V C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 406− 903 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= M C 1.355 kip ft⋅= Problem 7-18 Determine the normal force, shear force, and moment acting at a section passing through point D. Units Used: kip 10 3 lb= 639 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 F 1 800 lb= F 2 700 lb= F 3 600 lb= θ 30 deg= a 1.5 ft= b 1.5 ft= c 3 ft= d 2 ft= e 1 ft= fab+ c+ d− e−= Solution: Guesses B y 1 lb= A x 1 lb= A y 1 lb= N D 1 lb= V D 1 lb= M D 1 lb ft⋅= Given V D sin θ () N D cos θ () − F 3 sin θ () − 0= B y V D cos θ () + N D sin θ () + F 3 cos θ () − 0= M D − F 3 e− B y ef+( ) cos θ () + 0= A y B y + F 2 − F 1 cos θ () − F 3 cos θ () − 0= A x − F 1 sin θ () + F 3 sin θ () − 0= F 1 − ab+()F 2 ab+ c+( ) cos θ () − F 3 cos θ () ab+ c+ d+ e+( ) cos θ () − F 3 sin θ () fsin θ () B y 2 ab+ c+( ) cos θ () ++ 0= A x A y B y N D V D M D ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , B y , N D , V D , M D , () = A x A y B y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 100 985 927 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= N D V D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 464− 203− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= M D 2.61 kip ft⋅= 640 Given: © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. [...]... in writing from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7-2 4 The jack AB is used to straighten the bent beam DE using the arrangement shown If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam Neglect the weight of the beams Units Used: kip = 103 lb Given: P = 5000 lb a = 2 ft b = 10 ft Solution: Segment: ΣMC =... this material may be reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7-2 7 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame Units Used: kN = 103 N Given: w1 = 200 N w2 = 400 N m m a = 2.5 m b = 3m c = 6m Solution: Σ MA = 0; F BC + → Σ Fx = 0; + ↑Σ Fy... material may be reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7-3 2 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame Units Used: 3 kN = 10 N Given: kN m w = 0.75 F = 4 kN a = 1.5 m d = 1.5 m b = 1.5 m e = 3 c = 2.5 m f = 4 Solution: Σ MC = 0;.. .Engineering Mechanics - Statics Chapter 7 Problem 7-1 9 Determine the normal force, shear force, and moment at a section passing through point C Units Used: 3 kN = 10 N Given: P = 8 kN c = 0.75 m a = 0.75m d = 0.5 m b = 0.75m r = 0.1 m Solution: ΣMA = 0; −T ( d + r) + P( a + b + c)... any means, without permission in writing from the publisher Engineering Mechanics - Statics γ = 150 Chapter 7 lb ft a = 2 ft b = 10 ft Solution: Beam: + ↑Σ Fy = 0; P − 2 γ ( a + b) − 2 R = 0 R = P 2 − γ ( a + b) R = 700 lb Segment: ΣMC = 0; M C + R b + γ ( a + b) MC = −R b − γ ⎛ a + b⎞ = ⎜ 2 ⎟ ⎝ ⎠ ( a + b) 0 2 2 MC = −17.8 kip⋅ ft Problem 7-2 6 Determine the normal force, shear force, and moment in the... permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 Solution: ΣF x = 0; NC = 0 ΣF y = 0; wa ΣMC = 0; MC − 2 NC = 0 wa − − VC = 0 2 ⎛ w a⎞a + ⎛ w a ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ VC = 0 ⎛ a⎞ = ⎜3⎟ ⎝ ⎠ 0 2 MC = wa MC = 54.00 kip⋅ ft 3 Problem 7-2 2 Determine the internal shear force and moment acting at point D of the beam Units Used: kip = 103 lb Given: w = 2 kip ft a = 6 ft b =... or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7-2 8 Determine the normal force, shear force, and moment at sections passing through points E and F Member BC is pinned at B and there is a smooth slot in it at C The pin at C is fixed to member CD Units Used: 3 kip = 10 lb Given: M = 350 lb⋅ ft w = 80 lb ft c = 2 ft F = 500 lb d = 3... writing from the publisher Engineering Mechanics - Statics Chapter 7 MC = −MA − Ax r sin ( φ ) + Ay r( 1 − cos ( φ ) ) MC = −1.59 kip⋅ ft Problem 7-3 1 The cantilevered rack is used to support each end of a smooth pipe that has total weight W Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section Units Used: 3 kip = 10 lb Given: W = 300 lb... permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7-2 3 The shaft is supported by a journal bearing at A and a thrust bearing at B Determine the internal normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the F2 force Units Used: kip = 103 lb Given: F 1 = 2500 lb a = 6 ft... permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 b = −0.866 Σ Mo = 0; N = w0 ⋅ r⋅ b −M + r w0 ( θ ) + b r w0 r = 0 2 M = w0 r ( θ + b) 2 c = θ+b c = 1.2284 2 M = c r w0 Problem 7-3 8 Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly Neglect the weight of the pipe Units Used: kip = 103 lb Given: ⎛ 0 ⎞ F 1 = ⎜ 350 ⎟ . permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Units Used: kip 10 3 lb= Given: w 1 60 lb ft = a 12 ft= b 15 ft= w 2 40 lb ft = c 10 ft= d 5 ft= F 690 lb= e 12= f 5= Solution: θ atan e f ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses B y 1 lb=. the publisher. Engineering Mechanics - Statics Chapter 7 Problem 7-1 9 Determine the normal force, shear force, and moment at a section passing through point C. Units Used: kN 10 3 N= Given: P. publisher. Engineering Mechanics - Statics Chapter 7 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN 10 3 N= Given: w 1 200 N m = w 2 400 N m = a