© 2016 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 7–1 Determine the shear force and moment at points C and D 500 lb 300 lb 200 lb B A C ft ft E D ft ft ft SOLUTION Support Reactions: FBD (a) a + ©MB = 0; 500(8) - 300(8) -Ay (14) = Ay = 114.29 lb Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = : x + c ©Fy = 0; NC = 114.29 - 500 - VC = a + ©MC = 0; Ans VC = -386 lb Ans MC + 500(4) - 114.29 (10) = MC = -857 lb # ft Ans Applying the equations of equilibrium to segment ED [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0; ND = VD - 300 = -MD - 300 (2) = Ans VD = 300 lb Ans MD = -600 lb # ft Ans Ans: NC = VC = - 386 lb MC = - 857 lb # ft ND = VD = 300 lb MD = - 600 lb # ft 611 © 2016 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 7–2 Determine the internal normal force and shear force, and the bending moment in the beam at points C and D Assume the support at B is a roller Point C is located just to the right of the 8-kip load kip 40 kip ft A C ft D ft B ft SOLUTION Support Reactions: FBD (a) a + ©MA = 0; + c ©Fy = 0; By (24) + 40 - 8(8) = Ay + 1.00 - = + ©F = : x By = 1.00 kip Ay = 7.00 kip Ax = Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = : x + c ©Fy = 0; a + ©MC = 0; NC = 7.00 - - VC = MC - 7.00(8) = Ans VC = -1.00 kip MC = 56.0 kip # ft Ans Ans Applying the equations of equilibrium to segment BD [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0; VD + 1.00 = ND = Ans VD = -1.00 kip Ans 1.00(8) + 40 - MD = MD = 48.0 kip # ft Ans Ans: NC = VC = -1.00 kip MC = 56.0 kip # ft ND = VD = -1.00 kip MD = 48.0 kip # ft 612 © 2016 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 7–3 Two beams are attached to the column such that structural connections transmit the loads shown Determine the internal normal force, shear force, and moment acting in the column at a section passing horizontally through point A 30 mm 40 mm 250 mm 16 kN kN 23 kN 185 mm kN Solution A + ΣFx = 0; 6 - - VA = S Ans VA = 0 + c ΣFy = 0; - NA - 16 - 23 = Ans NA = - 39 kN a + ΣMA = 0; 125 mm - MA + 16(0.155) - 23(0.165) - 6(0.185) = MA = - 2.42 kN # m Ans Ans: VA = NA = -39 kN MA = -2.425 kN # m 613 © 2016 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 *7–4 The beam weighs 280 lb>ft Determine the internal normal force, shear force, and moment at point C B ft ft ft Solution A Entire beam : a + ΣMA = 0; C ft - 2.8 (3) + Bx (8) = Bx = 1.05 kip + ΣFx = 0; Ax - 1.05 = S Ax = 1.05 kip + c ΣFy = 0; Ay - 2.8 = Ay = 2.80 kip Segment AC : + Q ΣFx = 0; - NC + 1.05 cos 53.13° + 2.80 sin 53.13° - 0.84 sin 53.13° = Ans NC = 2.20 kip a + ΣFy = 0; - VC - 0.84 cos 53.13° + 2.80 cos 53.13° - 1.05 sin 53.13° = Ans VC = 0.336 kip a + ΣMC = 0; - 2.80 cos 53.13° (3) + 1.05 sin 53.13° (3) + 0.84 cos 53.13° (1.5) + MC = MC = 1.76 kip # ft Ans Ans: NC = 2.20 kip VC = 0.336 kip MC = 1.76 kip # ft 614 © 2016 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 7–5 The pliers are used to grip the tube at B If a force of 20 lb is applied to the handles,determine the internal shear force and moment at point C Assume the jaws of the pliers exert only normal forces on the tube 20 lb 40Њ 10 in 0.5 in in C B A SOLUTION + ©MA = 0; 20 lb -20(10) + RB (1.5) = RB = 133.3 lb Segment BC: +Q ©Fy = 0; VC + 133.3 = Ans VC = - 133 lb a + ©MC = 0; - MC + 133.3 (1) = MC = 133 lb # in Ans Ans: VC = - 133 lb MC = 133 lb # in 615 © 2016 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 7–6 Determine the distance a as a fraction of the beam’s length L for locating the roller support so that the moment in the beam at B is zero P P C SOLUTION a + ©MA = 0; a -P a Cy = a + ©M = 0; B A L/3 L 2L - a b + Cy1L - a2 + Pa = 2P A L3 - a B L - a M = 2P A L3 - a B L - a 2PL a a = a L b = L - ab = L Ans Ans: a = 616 L © 2016 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 7–7 Determine the internal shear force and moment acting at point C in the beam kip/ft A B C ft ft Solution Support Reactions Referring to the FBD of the entire beam shown in Fig a, a+ ΣMA = 0; By(12) - (4)(6)(4) = 0 By = 4.00 kip Internal Loadings Referring to the FBD of the right segment of the beam sectioned through C, Fig b, + c ΣFy = 0; VC + 4.00 = 0 VC = - 4.00 kip Ans a+ ΣMC = 0; 4.00(6) - MC = Ans MC = 24.0 kip # ft Ans: VC = -4.00 kip MC = 24.0 kip # ft 617 © 2016 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 *7–8 Determine the internal shear force and moment acting at point C in the beam 500 lb/ ft 900 lb и ft A ft 900 lb и ft C ft B ft ft Solution Support Reactions Referring to the FBD of the entire beam shown in Fig a a+ ΣMB = 0; 500(12)(6) + 900 - 900 - Ay(12) = Ay = 3000 lb + ΣFx = 0; S Ax = Internal Loadings Referring to the FBD of the left segment of beam sectioned through C, Fig b, + c ΣFy = 0; 3000 - 500(6) - VC = 0 VC = Ans a+ ΣMC = 0; MC + 500(6)(3) + 900 - 3000(6) = MC = 8100 lb # ft = 8.10 kip # ft Ans Ans: VC = MC = 8.10 kip # ft 618 © 2016 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 7–9 Determine the normal force, shear force, and moment at a section passing through point C Take P = kN B 0.1 m 0.5 m C 0.75 m 0.75 m P SOLUTION a + ©MA = 0; 0.75 m A -T(0.6) + 8(2.25) = T = 30 kN + ©F = 0; : x Ax = 30 kN + c ©Fy = 0; Ay = kN + ©F = 0; : x -NC - 30 = Ans NC = - 30 kN + c ©Fy = 0; VC + = Ans VC = - kN a + ©MC = 0; - MC + 8(0.75) = MC = kN # m Ans Ans: NC = - 30 kN VC = - kN MC = kN # m 619 © 2016 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 7–10 The cable will fail when subjected to a tension of kN Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading B 0.1 m 0.5 m C 0.75 m -2(0.6) + P(2.25) = Ans P = 0.533 kN + ©F = 0; : x Ax = kN + c ©Fy = 0; Ay = 0.533 kN + ©F = 0; : x -NC - = Ans NC = - kN + c ©Fy = 0; VC - 0.533 = Ans VC = -0.533 kN a + ©M C = 0; 0.75 m P SOLUTION a + ©MA = 0; 0.75 m A - MC + 0.533(0.75) = MC = 0.400 kN # m Ans Ans: P = 0.533 kN NC = -2 kN VC = -0.533 kN MC = 0.400 kN # m 620 © 2016 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 7–111 Determine the maximum tension developed in the cable if it is subjected to the triangular distributed load y B 20 ft 15Њ A Solution The Equation of The Cable Here, w(x) = 15x 1 w(x)dx dx y = FH 1 y = y = y = 300 lb/ft 20 ft 1 15x dx dx FH 1 15 a x2 + C1 bdx FH a x3 + C1x + C2 b FH (1) dy 15 = a x + C1 b dx FH x (2) Boundary Conditions At x = 0, y = Then Eq (1) gives = (0 + + C2) C2 = FH Also, at x = 0, dy = tan 15° Then Eq (2) gives dx (0 + C1) C1 = FH tan 15° tan 15° = FH Thus, the equation of the cable becomes a x + FH tan 15°xb y = FH y = x + tan 15° x 2FH (3) And the slope of the cable is dy 15 = x + tan 15° dx 2FH Also, at x = 20 ft, y = 20 ft.Then using Eq 3, 20 = a (4) b(203) + tan 15°(20) 2FH FH = 1366.03 lb umax occurs at x = 20 ft Then Eq (4) gives tan umax = Thus Tmax = dy 15 ` = c d (202) + tan 15° umax = 67.91° dx x = 20 ft 2(1366.02) FH 1366.02 = = 3632.65 lb = 3.63 kip cos umax cos 67.91° Ans 733 Ans: Tmax = 3.63 kip © 2016 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 *7–112 The cable will break when the maximum tension reaches Tmax = 10 kN Determine the minimum sag h if it supports the uniform distributed load of w = 600 N>m 25 m h 600 N/m SOLUTION The Equation of The Cable: y = = ( w(x)dx)dx FH L L w0 ¢ x + C1x + C2 ≤ FH [1] dy (w x + C1) = dx FH [2] Boundary Conditions: y = at x = 0, then from Eq.[1] = (C ) FH dy = at x = 0, then from Eq.[2] = (C ) dx FH C2 = C1 = Thus, w0 x 2FH [3] dy w0 x = dx FH [4] y = y = h, at x = 12.5m, then from Eq.[3] h = w0 (12.52) 2FH FH = 78.125 w0 h u = umax at x = 12.5 m and the maximum tension occurs when u = umax From Eq.[4] tan umax = Thus, cos umax = dy = dx x - 12.5m w0 x 18.125 h w0 = 0.0128h(12.5) = 0.160h 20.0256h2 + The maximum tension in the cable is Tmax = 10 = FH cos umax 18.125 h (0.6) 20.0256h2 + Ans h = 7.09 m 734 Ans: h = 7.09 m © 2016 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 7–113 The cable is subjected to the parabolic loading w = 15011 - 1x>50222 lb>ft, where x is in ft Determine the equation y = f1x2 which defines the cable shape AB and the maximum tension in the cable 100 ft y A B 20 ft x SOLUTION y = ( w(x)dx)dx FH L L 150 lb/ft x3 [150(x ) + C1]dx FH L 3(50)2 y = y = x4 (75x2 + C1x + C2) FH 200 dy C1 150x = x3 + dx FH 50FH FH At x = , dy = C1 = dx At x = , y = C2 = y = At x = 50 ft , y = 20 ft x2 b a 75x2 FH 200 FH = 7813 lb y = x2 x2 a 75 b ft 7813 200 Ans dy 4x3 = a 150x b = tan umax dx 7813 200 x = 50 ft umax = 32.62° Tmax = FH 7813 = = 9275.9 lb cos umax cos 32.62° Ans Tmax = 9.28 kip Ans: x2 x2 a75 b 7813 200 Tmax = 9.28 kip y = 735 © 2016 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 7–114 The power transmission cable weighs 10 lb>ft If the resultant horizontal force on tower BD is required to be zero, determine the sag h of cable BC 300 ft A SOLUTION 200 ft B 10 ft h C D The origin of the x, y coordinate system is set at the lowest point of the cables Here, w0 = 10 lb>ft Using Eq of Example 7–13, y = FH w0 Bcosh ¢ x ≤ - R w0 FH y = FH 10 Bcosh ¢ x ≤ - R ft 10 FH Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft, 10 = 10(150) (FH)AB Bcosh ¢ ≤ - 1R 10 (FH)AB Solving by trial and error yields (FH)AB = 11266.63 lb Since the resultant horizontal force at B is required to be zero, (FH)BC = (FH)AB = 11266.62 lb Applying the boundary condition of cable BC y = h at x = -100 ft to Eq (1), we obtain h = 10(- 100) 11266.62 c cosh B R - 1s 10 11266.62 Ans = 4.44 ft Ans: h = 4.44 ft 736 © 2016 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 7–115 The power transmission cable weighs 10 lb>ft If h = 10 ft, determine the resultant horizontal and vertical forces the cables exert on tower BD 300 ft A SOLUTION 200 ft B 10 ft h C D The origin of the x, y coordinate system is set at the lowest point of the cables Here, w0 = 10 lb>ft Using Eq of Example 7–13, y = y = FH w0 Bcosh ¢ w0 x≤ - 1R FH FH 10 Bcosh ¢ x ≤ - R ft 10 FH Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft, 10 = 10(150) (FH)AB Bcosh ¢ ≤ - 1R 10 (FH)AB Solving by trial and error yields (FH)AB = 11266.63 lb Applying the boundary condition of cable BC, y = 10 ft at x = - 100 ft to Eq (2), we have 10 = 10(100) (FH)BC Bcosh ¢ ≤ - 1R 10 (FH)BC Solving by trial and error yields (FH)BC = 5016.58 lb Thus, the resultant horizontal force at B is (FH)R = (FH)AB - (FH)BC = 11266.63 - 5016.58 = 6250 lb = 6.25 kip Using Eq (1), tan (uB)AB = sin h B sin h B 10(150) R = 0.13353 and 11266.63 Ans tan (uB)BC = 10( - 100) R = 0.20066 Thus, the vertical force of cables AB and BC acting 5016.58 on point B are (Fv)AB = (FH)AB tan (uB)AB = 11266.63(0.13353) = 1504.44 lb (Fv)BC = (FH)BC tan (uB)BC = 5016.58(0.20066) = 1006.64 lb The resultant vertical force at B is therefore (Fv)R = (Fv)AB + (Fv)BC = 1504.44 + 1006.64 Ans = 2511.07 lb = 2.51 kip 737 Ans: (Fh)R = 6.25 kip (Fv)R = 2.51 kip © 2016 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 *7–116 The man picks up the 52-ft chain and holds it just high enough so it is completely off the ground The chain has points of attachment A and B that are 50 ft apart If the chain has a weight of lb/ft, and the man weighs 150 lb, determine the force he exerts on the ground Also, how high h must he lift the chain? Hint: The slopes at A and B are zero h A 25 ft SOLUTION Deflection Curve of The Cable: x = L 31 + ds 11>F H 21 w0 where w0 = lb>ft ds2242 Performing the integration yields x = FH b sinh-1 B 13s + C12 R + C2 r FH (1) From Eq 7–14 dy 1 13s + C12 w0 ds = = dx FH L FH (2) Boundary Conditions: dy = at s = From Eq (2) dx = 10 + C12 FH C1 = Then, Eq (2) becomes dy 3s = tan u = dx FH (3) s = at x = and use the result C1 = From Eq (1) x = FH b sinh-1 B 10 + 02 R + C2 r FH C2 = Rearranging Eq (1), we have s = FH x≤ sinh ¢ FH (4) Substituting Eq (4) into (3) yields dy x≤ = sinh ¢ dx FH Performing the integration y = FH x cosh FH y = at x = From Eq (5) = B (5) + C3 FH FH cosh + C3 ,thus, C3 = 3 738 25 ft © 2016 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 *7–116. Continued Then, Eq (5) becomes 26 ft at cosh (6) 25 ft From Eq (4) 26 sinh 25 154.003 lb By trial and error at 25 ft From Eq (6) 154.003 25 cosh 154.003 Ans 6.21 ft From Eq (3) tan 26 ft The vertical force 26 26.86° 0.5065 154.003 that each chain exerts on the man is tan 154.003 tan 26.86° 78.00 lb Equation of Equilibrium: By considering the equilibrium of the man, 0; 150 78.00 Ans 306 lb Ans: h = 6.21 ft Nm = 306 lb 739 © 2016 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 7–117 The cable has a mass of 0.5 kg>m, and is 25 m long Determine the vertical and horizontal components of force it exerts on the top of the tower A SOLUTION B ds x = L b1 + 30 (w0 ds)2 r F2H 15 m Performing the integration yields: x = FH b sinh - c (4.905s + C1) d + C2 r 4.905 FH (1) rom Eq 7-13 dy = w ds dx FH L dy = (4.905s + C1) dx FH At s = 0; dy = tan 30° Hence C1 = FH tan 30° dx dy dx = 4.905s + tan 30° FH (2) Applying boundary conditions at x = 0; s = to Eq.(1) and using the result C1 = FH tan 30° yields C2 = - sinh - 1(tan 30°) Hence x = FH b sinh - c (4.905s+FH tan 30°) d - sinh - 1(tan 30°)r 4.905 FH (3) At x = 15 m; s = 25 m From Eq.(3) 15 = FH b sinh - c (4.905(25) + FH tan 30°) R - sinh - 1(tan 30°) r 4.905 FH By trial and error FH = 73.94 N At point A, s = 25 m From Eq.(2) tan uA = dy 4.905(25) + tan 30° = dx s = 25 m 73.94 uA = 65.90° (Fv )A = FH tan uA = 73.94 tan 65.90° = 165 N Ans (FH)A = FH = 73.9 N Ans Ans: (Fv) A = 165 N (Fh)A = 73.9 N 740 © 2016 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 7–118 A 50-ft cable is suspended between two points a distance of 15 ft apart and at the same elevation If the minimum tension in the cable is 200 lb, determine the total weight of the cable and the maximum tension developed in the cable SOLUTION Tmin = FH = 200 lb From Example 7–13: s = FH w0 x sinh a b w0 FH w0 15 200 50 sinh a = a bb w0 200 Solving, w0 = 79.9 lb>ft Ans Total weight = w0 l = 79.9 (50) = 4.00 kip dy w0 s = tan umax = ` dx max FH umax = tan - B 79.9 (25) R = 84.3° 200 Then, Tmax = FH 200 = 2.01 kip = cos umax cos 84.3° Ans Ans: W = 4.00 kip Tmax = 2.01 kip 741 © 2016 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 7–119 Show that the deflection curve of the cable discussed in Example 7–13 reduces to Eq in Example 7–12 when the hyperbolic cosine function is expanded in terms of a series and only the 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 = + x2 + Á 21 Substituting into y = FH w0 B cosh ¢ x ≤ - R w0 FH = FH w 20x2 + Á - 1R B1 + w0 2F 2H = w0x2 2FH Using Eq (3) in Example 7–12, FH = We get y = w0L2 8h 4h x L2 QED 742 © 2016 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 *7–120 A telephone line (cable) stretches between two points which are 150 ft apart and at the same elevation The line sags ft and the cable has a weight of 0.3 lb> ft Determine the length of the cable and the maximum tension in the cable SOLUTION w = 0.3 lb>ft From Example 7–13, s = FH w sinh ¢ x≤ w FH y = FH w Bcosh ¢ x ≤ - R w FH At x = 75 ft, y = ft, w = 0.3 lb>ft = FH 75w Bcosh ¢ ≤ - 1R w FH FH = 169.0 lb dy w = tan umax = sinh ¢ x≤ ` ` dx max FH x = 75 ft umax = tan-1 csinh ¢ Tmax = s = 7510.32 169 ≤ d = 7.606° FH 169 = 170 lb = cos umax cos 7.606° Ans 169.0 0.3 sinh c 1752 d = 75.22 0.3 169.0 Ans L = 2s = 150 ft Ans: Tmax = 170 lb L = 150 ft 743 © 2016 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 7–121 A cable has a weight of lb> ft If it can span 100 ft and has a sag of 12 ft, determine the length of the cable The ends of the cable are supported from the same elevation SOLUTION From Eq (5) of Example 7–13: h = 12 = FH w0L Bcosh ¢ ≤ - 1R w0 2FH 211002 FH Bcosh ¢ ≤ - 1R 2FH 24 = FH B cosh ¢ 100 ≤ - 1R FH FH = 212.2 lb From Eq (3) of Example 7–13: s = FH w0 sinh ¢ x≤ w0 FH 21502 212.2 l = b sinh a 2 212.2 Ans l = 104 ft Ans: l = 104 ft 744 © 2016 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 7–122 A cable has a weight of lb ft and is supported at points that are 500 ft apart and at the same elevation If it has a length of 600 ft, determine the sag SOLUTION w0 = lb>ft From Example 7–15, s = w0 FH sinh a xb w0 FH At x = 250 ft, 300 = s = 300 ft 3(250) FH sinh a b FH FH = 704.3 lb y = h = w0 FH c cosh x - d w0 FH 3(250) 704.3 b - 1d c cosh a 704.3 Ans h = 146 ft Ans: h = 146 ft 745 © 2016 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 7–123 A cable has a weight of lb/ft If it can span 300 ft and has a sag of 15 ft, determine the length of the cable The ends of the cable are supported at the same elevation SOLUTION w0 = lb>ft From Example 7–15, y = w0 FH B cosh ¢ x ≤ - R w0 FH At x = 150 ft, y = 15 ft 150w0 15w0 = cosh ¢ ≤ - FH FH FH = 3762 lb s = w0 FH sinh x w0 FH s = 151.0 ft Ans L = 2s = 302 ft Ans: L = 302 ft 746 © 2016 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 *7–124 The 10 kg m cable is suspended between the supports A and B If the cable can sustain a maximum tension of 1.5 kN and the maximum sag is m, determine the maximum distance L between the supports L A SOLUTION The origin of the x, y coordinate system is set at the lowest point of the cable Here w0 = 10(9.81) N>m = 98.1 N>m Using Eq (4) of Example 7–13, y = FH w0 Bcosh ¢ x ≤ - R w0 FH y = FH 98.1x Bcosh ¢ ≤ - 1R 98.1 FH Applying the boundary equation y = m at x = = L , we have FH 49.05L Bcosh ¢ ≤ - 1R 98.1 FH The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal From Eq (1), tan umax = sinh ¢ 49.05L ≤ FH By referring to the geometry shown in Fig b, we have cos umax = A + sinh2 49.05L ¢ F ≤ H = cosh ¢ 49.05L ≤ FH hus, Tmax = FH cos umax 1500 = FH cosh ¢ 49.05L ≤ FH (3) Solving Eqs (2) and (3) yields Ans L = 16.8 m FH = 1205.7 N 747 3m B ... 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. .. 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. .. 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