# Statics_Part36.pdf

## Civil Engineering 2110 with Shana at Marquette University *

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Engineering Mechanics - Statics Chapter 7 a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F AB b b 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC − F− 0= d− b 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC F CD + 0= b b 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC P 1 − 0= F CD − f f 2 ab+() 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ F DE + 0= ab+ f 2 ab+() 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ F DE P 2 − 0= F AB F BC F CD F DE P 1 P 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F BC , F CD , F DE , P 1 , P 2 ,()= F AB F BC F CD F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 12.50 10.31 10.00 11.79 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= P 1 P 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.50 6.25 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN= T max max F AB F BC , F CD , F DE ,()= T max 12.50 kN= F max max F AB F BC , F CD , F DE ,= F max 12.50 kN= 736 © 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 The cable supports the loading shown. Determine the distance x B from the wall to point B. Given: W 1 8 lb= W 2 15 lb= a 5 ft= b 8 ft= c 2 ft= d 3 ft= Solution: Guesses T AB 1 lb= T BC 1 lb= T CD 1 lb= x B 1 ft= Given x B − a 2 x B 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ T AB x B d− b 2 x B d−() 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T BC − W 2 + 0= a a 2 x B 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB b b 2 x B d−() 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC − 0= x B d− b 2 x B d−() 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T BC d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T CD − 0= b b 2 x B d−() 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T CD − W 1 − 0= T AB T BC T CD x B ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find T AB T BC , T CD , x B ,()= T AB T BC T CD ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 15.49 10.82 4.09 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= x B 5.65 ft= Problem 7-97 Determine the maximum uniform loading w, measured in lb/ft, that the cable can support if it is 737 Problem 7-96 © 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 g pp capable of sustaining a maximum tension T max before it will break. Given: T max 3000 lb= a 50 ft= b 6 ft= Solution: y 1 F H xxw ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d= wx 2 2F H = y w 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = x a 2 = yb= F H wa 2 8b = dy dx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ tan θ max ()= w F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 4b a = θ max atan 4b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max 25.64 deg= T max F H cos θ max () = wa 2 8b cos θ max () = w T max 8b cos θ max () a 2 = w 51.93 lb ft = Problem 7-98 The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in the cable. Units Used: kip 10 3 lb= Given: w 250 lb ft = a 50 ft= b 6 ft= 738 © 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 Solution: y wx 2 2F H = b w 2F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = F H wa 2 8b = F H 13021 lb= tan θ max () x y d d x a 2 = ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max atan wa 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max 25.64 deg= T max F H cos θ max () = T max 14.44 kip= The minimum tension occurs at θ 0 deg= T min F H = T min 13.0 kip= Problem 7-99 The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine the equation of the curve y = f(x) which defines the cable shape AB, and the maximum tension developed in the cable. Units Used: kip 10 3 lb= Given: w 250 lb ft = a 20 ft= b 30 ft= Solution: y 1 F H xx wx b ⌠ ⎮ ⎮ ⌡ d ⌠ ⎮ ⎮ ⌡ d= y 1 F H wx 3 6b c 1 x+ c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 739 © 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 Apply boundary conditions yx= 0= and x y d d 0= x 0=, Thus C 1 C 2 = 0= y wx 3 6F H b = set ya= xb= a wb 3 6F H b = F H wb 2 6a = F H 1.875 kip= θ max atan wb 2 2F H b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max 63.43 deg= T max F H cos θ max () = T max 4.19 kip= Problem 7-100 The cable supports a girder which has weight density γ. Determine the tension in the cable at points A, B, and C. Units used: kip 10 3 lb= Given: γ 850 lb ft = a 40 ft= b 100 ft= c 20 ft= Solution: y 1 F H xxγ ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d= y γx 2 2F H = x y d d γx F H = Guesses x 1 1 ft= F H 1 lb= 740 © 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 c γx 1 2 2F H = a γ 2F H bx 1 −() 2 = x 1 F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find x 1 F H ,()= F H 36.46 kip= tan θ A () γ F H x 1 b−()= θ A atan γ F H x 1 b−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = θ A 53.79− deg= tan θ C () γ F H x 1 = θ C atan γ F H x 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ C 44.00 deg= T A F H cos θ A () = T B F H = T C F H cos θ C () = T A T B T C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 61.71 36.46 50.68 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kip= Problem 7-101 The cable is subjected to the triangular loading. If the slope of the cable at point O is zero, determine the equation of the curve y = f(x) which defines the cable shape OB, and the maximum tension developed in the cable. Units used: kip 10 3 lb= Given: w 500 lb ft = b 8 ft= a 15 ft= 741 © 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 y 1 F H xx wx a ⌠ ⎮ ⎮ ⌡ d ⌠ ⎮ ⎮ ⌡ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = y 1 F H w a x 3 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ C 1 x+ C 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = x y d d 1 F H wx 2 2a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ C 1 F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += At x = 0, x y d d 0= , C 1 0= At x = 0, y = 0, C 2 0= y wx 3 6aF H = x y d d wx 2 2aF H = At x = a, y = b b wa 3 6aF H = F H 1 6 w a 2 b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H 2343.75 lb= 742 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 x y d d tan θ max ()= wa 2 2aF H = θ max ()atan wa 2 2aF H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max ()57.99 deg= T max F H cos θ max () = T max 4.42 kip= Problem 7-102 The cable is subjected to the parabolic loading w = w 0 (1− (2x/a) 2 ). Determine the equation y = f(x) which defines the cable shape AB and the maximum tension in the cable. Units Used: kip 10 3 lb= Given: ww 0 1 2x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = a 100 ft= w 0 150 lb ft = b 20 ft= Solution: y 1 F H xxwx() ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d= y 1 F H xw 0 x 4x 3 3a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ C 1 + ⌠ ⎮ ⎮ ⎮ ⌡ d= y 1 F H w 0 x 2 2 x 4 w 0 3a 2 − C 1 x+ C 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = dy dx 1 F H w 0 x 4w 0 x 3 3a 2 − C 1 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = 743 © 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 At x 0= dy dx 0= C 1 0= At x 0= y 0= C 2 0= Thus y 1 F H w 0 x 2 2 x 4 w 0 3a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = dy dx 1 F H w 0 x 4w 0 x 3 3a 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = At x a 2 = we have b 1 F H w 0 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 w 0 3a 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 − ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = 5 48 w 0 a 2 F H ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F H 5w 0 a 2 48b = F H 7812.50 lb= tan θ max () x y d d a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 1 F H w 0 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4w 0 3a 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = w 0 a 3F H = θ max atan w 0 a 3F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max 32.62 deg= T max F H cos θ max () = T max 9.28 kip= Problem 7-103 The cable will break when the maximum tension reaches T max . Determine the minimum sag h if it supports the uniform distributed load w. Given: kN 10 3 N= T max 10 kN= w 600 N m = a 25 m= Solution: The equation of the cable: y 1 F H xxw ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d= y 1 F H wx 2 2 C 1 x+ C 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = dy dx 1 F H wx C 1 +()= 744 © 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 Boundary Conditions: y = 0 at x = 0, then from Eq.[1] 0 1 F H C 2 ()= C 2 0= x y d d = 0 at x = 0, then from Eq.[2] 0 1 F H C 1 ()= C 1 0= Thus, y w 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = dy dx w F H x= h w 2F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = F H w 2h a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = tan θ max () wa 2F H = cos θ max () 2F H 4F H 2 wa() 2 + = T max F H cos θ max () = F H 2 wa() 2 4 += wa 2 a 2 16h 2 1+= Guess h 1 m= Given T max wa 2 a 2 16h 2 1+= h Find h()= h 7.09 m= Problem 7-104 A fiber optic cable is suspended over the poles so that the angle at the supports is θ. Determine the minimum tension in the cable and the sag. The cable has a mass density ρ and the supports are at the same elevation. Given: θ 22 deg= ρ 0.9 kg m = a 30 m= g 9.81 m s 2 = Solution: θ max θ= 745 © 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 w 0 ρg= x y d d tan θ()= sinh w 0 a 2 F H ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F H w 0 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ asinh tan θ()() = F H 336 N= T max F H cos θ() = T max 363 N= h F H w 0 cosh w 0 a 2 F H ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = h 2.99 m= Problem 7-105 A cable has a weight density γ and is supported at points that are a distance d apart and at the same elevation. If it has a length L, determine the sag. Given: γ 3 lb ft = d 500 ft= L 600 ft= Solution: Guess F H 100 lb= Given L 2 F H γ sinh γ F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 0= F H Find F H ()= F H 704.3 lb= h F H γ cosh 1 2 γ F H d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h 146 ft= Problem 7-106 Show that the deflection curve of the cable discussed in Example 7.15 reduces to Eq. (4) in Example 7.14 when the hyperbolic cosine function is expanded in terms of a series and only the 746 © 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 p yp f py first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) Solution: cosh x() 1 x 2 2! + ..+= Substituting into y F H w 0 cosh w 0 F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H w 0 1 w 0 2 x 2 2F H 2 + ..+ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = w 0 x 2 2F H = Using the boundary conditions yh= at x L 2 = h w 0 2F H L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = F H w 0 L 2 8h = We get y 4h L 2 x 2 = Problem 7-107 A uniform cord is suspended between two points having the same elevation. Determine the sag-to-span ratio so that the maximum tension in the cord equals the cord's total weight. Solution: s F H w 0 sinh w 0 F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = y F H w 0 cosh w 0 F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = At x L 2 = x y d d tan θ max ()= sinh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = max cos θ max () 1 cosh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 747 © 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 T max F H cos θ max () = w 0 2sF H cosh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 2F H sinh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F H cosh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = tanh w 0 L 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 = k 1 atanh 0.5()= k 1 0.55= w 0 L 2F H k 1 = when x L 2 = yh= h F H W 0 cosh k 1 () 1−()= k 2 cosh k 1 () 1−= k 2 0.15= hk 2 F H w 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Lk 1 2F h w 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ratio h L = k 2 2k 1 = ratio k 2 2k 1 = ratio 0.14= Problem 7-108 A cable has a weight denisty γ. If it can span a distance L and has a sag h determine the length of the cable. The ends of the cable are supported from the same elevation. Given: γ 2 lb ft = L 100 ft= h 12 ft= Solution: From Eq. (5) of Example 7-15 : h F H γ γ 2 L F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 2 ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ = F H 1 8 γ L 2 h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H 208.33 lb= From Eq. (3) of Example 7-15: l 2 F H γ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh γ F H L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = l 2 F H γ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh 1 2 γ L F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = l 104 ft= 748 © 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 7-109 The transmission cable having a weight density γ is strung across the river as shown. Determine the required force that must be applied to the cable at its points of attachment to the towers at B and C. Units Used: kip 10 3 lb= Given: γ 20 lb ft = b 75 ft= a 50 ft= c 10 ft= Solution: From Example 7-15, y F H γ cosh γ F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = dy dx sinh γx F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guess F H 1000 lb= Given At B: c F H γ cosh γa F H − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H Find F H ()= F H 2.53 kip= tan θ B ()sinh γa F H − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ B atan sinh γ− a F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ B 22.06− deg= tan θ C ()sinh γb F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ C atan sinh γb F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ C 32.11 deg= T B F H cos θ B () = T C F H cos θ C () = T B T C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.73 2.99 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip= Problem 7-110 Determine the maximum tension developed in the cable if it is subjected to a uniform load w. 749 © 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 MN 10 6 N= Given: w 600 N m = a 100 m= b 20 m= θ 10 deg= Solution: The Equation of the Cable: y 1 F H xxwx() ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d= 1 F H wx 2 2 C 1 x+ C 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = dy dx 1 F H wx C 1 +()= Initial Guesses: C 1 1 N= C 2 1 Nm⋅= F H 1 N= Given Boundary Conditions: x = 0 0 1 F H C 2 = tan θ() 1 F H C 1 ()= y = b at x = a b w 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 C 1 F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a+= C 1 C 2 F H ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find C 1 C 2 , F H ,()= C 1 0.22 MN= C 2 0.00 Nm⋅= F H 1.27 MN= tan θ max () 1 F H wa C 1 +()= θ max atan wa C 1 + F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ max 12.61 deg= T max F H cos θ max () = T max 1.30 MN= 750 Units Used: © 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 chain of length L has a total mass M and is suspended between two points a distance d apart. Determine the maximum tension and the sag in the chain. Given: L 40 m= M 100 kg= d 10 m= g 9.81 m s 2 = Solution: wM g L = s F H w ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh w F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = y F H w ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cosh w F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = x y d d sinh w F H x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses F H 10 N= h 10 m= Given L 2 F H w ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh w F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h F H w cosh w F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find F H h,()= θ max atan sinh w F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T max F H cos θ max () = F H 37.57 N= h 18.53 m= T max 492 N= Problem 7-112 The cable has a mass density ρ and has length L. Determine the vertical and horizontal components of force it exerts on the top of the tower. Given: ρ 0.5 kg m = L 25 m= θ 30 deg= d 15 m= g 9.81 m s 2 = 751 Problem 7-111 © 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 Solution: xs 1 1 1 F H 2 sρg ⌠ ⎮ ⌡ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 2 + ⌠ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⌡ d= Performing the integration yields: x F H ρg asinh ρgs C 1 + F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ C 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = dy dx 1 F H sρg ⌠ ⎮ ⌡ d= 1 F H ρgs C 1 +()= At s = 0; dy dx tan θ()= Hence C 1 F H tan θ()= dy dx ρgs F H tan θ()+= Applying boundary conditions at x = 0; s = 0 to Eq.[1] and using the result C 1 F H tan θ()= yields C 2 asinh tan θ()()−= . Hence Guess F H 1 N= Given d F H ρg ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ asinh 1 F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ρgL F H tan θ()+() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ asinh tan θ()()()− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = F H Find F H ()= F H 73.94 N= At A tan θ A () ρgL F H tan θ()+= θ A atan ρgL F H tan θ()+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ A 65.90 deg= F A F H cos θ A () = F Ax F A cos θ A ()= F Ay F A sin θ A ()= F Ax F Ay ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 73.94 165.31 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= 752 © 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 7-113 A cable of length L is suspended between two points a distance d apart and at the same elevation. If the minimum tension in the cable is T min , determine the total weight of the cable and the maximum tension developed in the cable. Units Used: kip 10 3 lb= Given: L 50 ft= d 15 ft= T min 200 lb= Solution: T min F H = F H T min = F H 200 lb= From Example 7-15: s F H w 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh w 0 x F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guess w 0 1 lb ft = Given L 2 F H w 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh w 0 F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w 0 Find w 0 ()= w 0 79.93 lb ft = Totalweight w 0 L= Totalweight 4.00 kip= tan θ max () w 0 F H L 2 = θ max atan w 0 L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F H ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = θ max 84.28 deg= Then, T max F H cos θ max () = T max 2.01 kip= 753 © 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 7-114 The chain of length L is fixed at its ends and hoisted at its midpoint B using a crane. If the chain has a weight density w, determine the minimum height h of the hook in order to lift the chain completely off the ground. What is the horizontal force at pin A or C when the chain is in this position? Hint: When h is a minimum, the slope at A and C is zero. Given: L 80 ft= d 60 ft= w 0.5 lb ft = Solution: Guesses F H 10 lb= h 1 ft= Given h F H w cosh w F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = L 2 F H w ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh w F H d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find hF H ,()= F A F H = F C F H = F A F C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 11.1 11.1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= h 23.5 ft= Problem 7-115 A steel tape used for measurement in surveying has a length L and a total weight W. How much horizontal tension must be applied to the tape so that the distance marked on the ground is a? In practice the calculation should also include the effects of elastic stretching and temperature changes on the tape’s length. Given: L 100 ft= W 2 lb= a 99.90 ft= 754 © 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 Solution: w 0 W L = w 0 0.02 lb ft = Guess F H 10 lb= Given L 2 F H w 0 sinh w 0 F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= F H Find F H ()= F H 12.9 lb= Problem 7-116 A cable of weight W is attached between two points that are a distance d apart, having equal elevations. If the maximum tension developed in the cable is T max determine the length L of the cable and the sag h. Given: W 100 lb= d 50 ft= T max 75 lb= Solution: Guesses F H 20 lb= L 20 ft= θ max 20 deg= h 2 ft= Given h F H L W cosh W F H L d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = tan θ max ()sinh W F H L d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T max F H cos θ max () = L 2 F H L W ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sinh W F H L d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H L θ max h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F H L, θ max , h,()= F H 55.90 lb= L h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 55.57 10.61 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft= θ max 41.81 deg= Problem 7-117 Determine the distance a between the supports in terms of the beam's length L so that the moment in the symmetric beam is zero at the beam's center. 755 © 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 Solution: Support Reactions: ΣM D = 0; w 2 La+() a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ B y a()− 0= B y w 4 La+()= Internal Forces: ΣM C = 0; w a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w La− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2aL+ 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 4 La+() a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= 2a 2 2aL+ L 2 − 0= b 2− 12+ 4 = b 0.366= abL= 756 © 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. dhanesh_h Mathcad - CombinedMathcads

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