Engineering Mechanics - Statics Chapter 7 V 2 x() B y − 1 2 w ab+ x− b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ x−()+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 lb = M 2 x() B y ab+ x−()M 0 − 1 2 w ab+ x− b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ x−() ab+ x− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x() 0= M 3 x() M 0 − 1 kip ft⋅ = 0 2 4 6 8 10121416 1500 1000 500 0 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 0 5 10 15 10 5 0 Distance (ft) Moment (kip-ft) M 1 x 1 () M 2 x 2 () M 3 x 3 () x 1 ft x 2 ft , x 3 ft , 721 © 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 Draw the shear and moment diagrams for the beam. Solution: 722 Problem 7-85 © 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 7-86 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given: w 2 kN m = a 3 m= b 3 m= Solution: x 1 0 0.01 a, a = V 1 x() 1 2 wb 1 2 w ax− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ax−()+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN = M 1 x() 1− 2 wb 2b 3 a+ x− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w ax− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ax−() ax− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x() 1 2 wb 1 2 w xa− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ xa−()− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN = M 2 x() 1− 2 wb a 2b 3 + x− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w xa− b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ xa−() xa− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN m⋅ = 0123456 5 0 5 10 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 723 © 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 0123456 20 10 0 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () x 1 x 2 , Problem 7-87 Draw the shear and moment diagrams for the beam. Units Used: kip 10 3 lb= Given: w 5 kip ft = M 1 15 kip ft⋅= M 2 15 kip ft⋅= a 6 ft= b 10 ft= c 6 ft= Solution: M 1 Ab− M 2 − w a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b a 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= AB+ wb− w ac+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= A M 1 M 2 − w a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b a 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w b 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w c 2 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − b = Bwb ac+ 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A−= A B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 40.00 40.00 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip= x 1 0 0.01 a, a = V 1 x() w− x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 1 kip = M 1p x() w− x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 x 3 M 1 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = 724 © 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 2 a 1.01a, ab+ = V 2 x() Aw a 2 − wx a−()− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip = M 2p x() M 1 − w a 2 x 2a 3 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − Ax a−()+ wx a−() xa− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x() w ab+ c+ x− c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ c+ x− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 kip = M 3p x() w− ab+ c+ x− c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ c+ x− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ c+ x− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ M 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = 0 5 10 15 20 40 20 0 20 40 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 725 © 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 0 5 10 15 20 60 40 20 0 20 Distance (ft) Moment (kip-ft) M 1p x 1 () M 2p x 2 () M 3p x 3 () x 1 ft x 2 ft , x 3 ft , Problem 7-88 Draw the shear and moment diagrams for the beam. Units Used: kip 10 3 lb= Given: w 1 2 kip ft = w 2 1 kip ft = a 15 ft= Solution: x 0 0.01a, a = Vx() w 2 xw 1 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip = Mx() w 2 x x 2 w 1 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 x 3 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = 726 © 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 0 2 4 6 8 101214 0 2 4 Distance (ft) Force (kip) Vx() x ft 0 2 4 6 8 101214 0 20 40 Distance (ft) Moment (kip-ft) Mx() x ft Problem 7-89 Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal. Also, compute the sag y B and the maximum tension in the cable. Units Used: kN 10 3 N= 727 © 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 4 m= F 1 4 kN= b 6 m= F 2 6 kN= c 3 m= d 2 m= e 3 m= Solution: Initial guesses: y B 1 m= P 1 kN= T AB 1 kN= T BC 1 kN= T CD 1 kN= T DE 1 kN= Given a− a 2 y B 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB T BC + 0= y B a 2 y B 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ T AB F 1 − 0= T BC − c c 2 y B e− () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T CD + 0= y B e− c 2 y B e− () 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T CD P− 0= c− c 2 y B e− () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T CD d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T DE + 0= y B e− () − c 2 y B e− () 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T CD e e 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T DE + F 2 − 0= y B P T AB T BC T CD T DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find y B P, T AB , T BC , T CD , T DE , () = 728 © 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 max T AB T BC , T CD , T DE , () = y B 3.53 m = P 800.00 N= T max 8.17 kN= Problem 7-90 Cable ABCD supports the lamp of mass M 1 and the lamp of mass M 2 . Determine the maximum tension in the cable and the sag of point B. Given: M 1 10 kg= M 2 15 kg= a 1 m= b 3 m= c 0.5 m= d 2 m= Solution: Guesses y B 1 m= T AB 1 N= T BC 1 N= T CD 1 N= Given a− a 2 y B 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB b b 2 y B d− () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC + 0= y B a 2 y B 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ T AB y B d− b 2 y B d− () 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T BC + M 1 g− 0= b− b 2 y B d− () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T CD + 0= y B d− () − b 2 y B d− () 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T BC d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T CD + M 2 g− 0= 729 © 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 B T AB T BC T CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find y B T AB , T BC , T CD , () = T AB T BC T CD ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 100.163 38.524 157.243 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= T max max T AB T BC , T CD , () = T max 157.2 N= y B 2.43 m= Problem 7-91 The cable supports the three loads shown. Determine the sags y B and y D of points B and D. Given: a 4 ft= e 12 ft= b 12 ft= f 14 ft= c 20 ft= P 1 400 lb= d 15 ft= P 2 250 lb= Solution: Guesses y B 1 ft= y D 1 ft= T AB 1 lb= T BC 1 lb= T CD 1 lb= T DE 1 lb= Given b− b 2 y B 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB c c 2 fy B − () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC + 0= y B b 2 y B 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ T AB fy B − c 2 fy B − () 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ T BC − P 2 − 0= c− c 2 fy B − () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T BC d d 2 fy D − () 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ T CD + 0= 730 © 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. [...]... = 0 at x = 0, then from Eq. [2] 0 = w ⎞ 2 y=⎛ ⎜ 2F ⎟ x ⎝ H⎠ w dx Thus, tan ( θ max) = Tmax = dy dx 2FH cos ( θ max) Guess Tmax = 1 FH (C2) C2 = 0 (C1) C1 = 0 ⎛ a⎞ h= ⎜ ⎟ 2FH ⎝ 2 ⎠ w x = FH + ( w a) 4 2 = 2 a wa 2 ⎛ a⎞ FH = ⎜ ⎟ 2h ⎝ 2 ⎠ w 2 2 FH 4 F H + ( w a) 2 2 2 2 2 +1 16h h = 1m Given FH FH cos ( θ max) = wa FH = 1 wa 2 2 a 16h 2 +1 h = Find ( h) h = 7. 09 m Problem 7-1 04 A fiber optic cable is suspended... TAB = 10 lb TCD = 30 lb TBC = 20 lb xB = 5 ft Given xB − d ⎛ −xB ⎟ ⎡ ⎞ ⎤ ⎜ ⎥ TBC + P = TAB − ⎢ ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB ⎢ BC ⎜ 2 2⎟ 2 2⎥ xB + a ⎠ (xB − d) + b ⎦ ⎝ ⎣ 0 0 xB − d ⎡ ⎤ f ⎞ ⎢ ⎥ TBC − ⎛ d ⎞ TCD + ⎛ ⎜ 2 2⎟ ⎜ 2 2 ⎟F = ⎢ ( xB − d) 2 + b2⎥ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ ⎦ 0 b ⎡ ⎤T − ⎛ c ⎞T − ⎛ e ⎞F = ⎢ ⎥ BC ⎜ 2 2 ⎟ CD ⎜ 2 2 ⎟ (xB − d )2 + b2⎦ ⎝ c +d ⎠ ⎝ e +f ⎠ ⎣ 0 ⎛ TAB... c ⎛ −b ⎞ T + ⎡ ⎤T = BC ⎜ 2 ⎟ AB ⎢ 2 2 2 b + yB ⎠ c + ( f − yB) ⎦ ⎝ ⎣ 0 yB f − yB ⎛ ⎞ ⎤ ⎜ ⎟ TAB − ⎡ ⎢ ⎥ TBC − P2 = ⎜ b2 + yB2 ⎟ ⎢ c2 + ( f − yB) 2 ⎝ ⎠ ⎣ ⎦ 0 −c d ⎡ ⎤T + ⎡ ⎤T = BC ⎢ CD ⎢ 2 2⎥ 2 2⎥ c + ( f − yB) ⎦ d + ( f − yD) ⎦ ⎣ ⎣ 0 f − yB f − yD ⎡ ⎤ ⎤ ⎢ ⎥ TBC + ⎡ ⎢ ⎥ TCD − P1 = ⎢ c2 + ( f − yB) 2 ⎢ d2 + ( f − yD) 2 ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎤T + ⎡ ⎤T = CD ⎢ DE ⎢ 2 2⎥ 2 2⎥ d + ( f − yD) ⎦ e + ( a + yD) ⎦... Engineering Mechanics - Statics ⎛ w0 L ⎞ ⎟ ⎝ 2 FH ⎠ FH Tmax = Chapter 7 w0 2 s = F H cosh ⎜ cos ( θ max) ⎛ w0 L ⎞ ⎛ w0 L ⎞ ⎟ = FH cosh ⎜ ⎟ ⎝ 2 FH ⎠ ⎝ 2FH ⎠ ⎛ w0 L ⎞ ⎟= ⎝ 2 FH ⎠ 2 F H sinh ⎜ k1 = atanh ( 0.5) when x = L ⎛ FH ⎞ ⎟ ⎝ w0 ⎠ h= ⎛ 2 Fh ⎞ ⎟ ⎝ w0 ⎠ h = k2 ⎜ w0 L k1 = 0.55 y=h 2 tanh ⎜ L = k1 ⎜ W0 (cosh (k1) − 1) ratio = 2 = k1 2 FH FH 1 k2 h = 2 k1 L k2 = cosh ( k1 ) − 1 ratio = k2 = 0.15 k2... writing from the publisher Engineering Mechanics - Statics Chapter 7 Solution: 2 y= ⎛ a⎞ b= ⎜ ⎟ 2FH ⎝ 2 ⎠ wx w 2FH tan ( θ max) = Tmax = d dx y ⎛x = ⎜ ⎝ 2 FH = w ⎛ a⎞ ⎟= ⎜ ⎟ 2 FH ⎝ 2 ⎠ a⎞ FH wa 2 F H = 13 021 lb 8b ⎞ θ max = atan ⎛ ⎜ 2F ⎟ ⎝ H⎠ wa θ max = 25 .64 deg Tmax = 14.44 kip cos ( θ max) The minimum tension occurs at θ = 0 deg Tmin = FH Tmin = 13.0 kip Problem 7 -9 9 The cable is subjected to... in writing from the publisher Engineering Mechanics - Statics dy x=0 At dx At x = 0 =0 y=0 Chapter 7 C1 = 0 C2 = 0 Thus ⎛ w0 x2 x4w0 ⎞ ⎜ ⎟ y= − ⎜ 2 2 ⎟ FH 3a ⎠ ⎝ 3⎞ ⎛ ⎜ w x − 4w0 x ⎟ = 0 2 ⎟ FH ⎜ dx 3a ⎠ ⎝ dy 1 x= At a we have 2 ⎡w0 ⎛ a ⎞ 2 w0 ⎛ a ⎞ 4⎥ ⎤ ⎢ ⎜ ⎟ − b= 2 = 2 FH ⎢ 2 ⎝ 2 ⎠ ⎥ 3a ⎝ ⎠ ⎦ ⎣ 1 tan ( θ max) ⎞ ⎛ w0 a2 ⎟ ⎜ 48 ⎜ F H ⎟ ⎝ ⎠ 5 FH = 5 w0 a 2 F H = 78 12. 50 lb 48b 3 ⎤ ⎛ a ⎞ = 1 ⎡w ⎛... ⎜ 2 2 ⎟ AB ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ⎝ b +d ⎠ ⎛ −d ⎞ F + F = ⎜ 2 2 ⎟ BC CD ⎝ b +d ⎠ ⎛ b ⎞F − P = ⎜ 2 2 ⎟ BC 1 ⎝ b +d ⎠ −F CD + 0 0 0 f ⎡ ⎤F = DE ⎢ 2 2⎥ ⎣ f + ( a + b) ⎦ a+b ⎡ ⎤F − P = DE 2 ⎢ 2 2⎥ f + ( a + b) ⎦ ⎣ 0 0 ⎞ ⎛ FAB ⎟ ⎜ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ = Find ( F , F , F , F , P , P ) AB BC CD DE 1 2 ⎜ FDE ⎟ ⎜ ⎟ ⎜ P1 ⎟ ⎜ P2 ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎛ 12. 50 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FBC ⎟ = ⎜ 10.31 ⎟ kN ⎜ FCD ⎟ ⎜ 10.00 ⎟ ⎜ ⎟ ⎜ 11. 79. .. publisher Engineering Mechanics - Statics Chapter 7 Problem 7 -9 6 The cable supports the loading shown Determine the distance xB from the wall to point B Given: W1 = 8 lb W2 = 15 lb a = 5 ft b = 8 ft c = 2 ft d = 3 ft Solution: Guesses TAB = 1 lb TBC = 1 lb TCD = 1 lb xB = 1 ft Given xB − d ⎛ −xB ⎟ ⎡ ⎞ ⎤ ⎜ ⎥ TBC + W2 = TAB − ⎢ ⎜ a2 + xB2 ⎟ ⎢ b2 + ( xB − d) 2 ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎡ ⎞ ⎤T = BC ⎜ 2 ⎟ TAB − ⎢ 2 2 2 .. .Engineering Mechanics - Statics Chapter 7 f − yB f − yD ⎡ ⎤ ⎤ ⎢ ⎥ TBC + ⎡ ⎢ ⎥ TCD − P1 = ⎢ c2 + ( f − yB) 2 ⎢ d2 + ( f − yD) 2 ⎣ ⎦ ⎣ ⎦ −d e ⎡ ⎡ ⎤ ⎤ ⎢ 2 ⎥ TCD + ⎢ 2 ⎥ TDE = 2 2 ⎣ d + ( f − yD) ⎦ ⎣ e + ( a + yD) ⎦ 0 a + yD ⎡ −( f − yD) ⎥ ⎡ ⎤ ⎤ ⎢ ⎥ TDE − P2 = TCD + ⎢ ⎢ d2 + ( f − yD) 2 ⎢ e2 + ( a + yD) 2 ⎣ ⎦ ⎣ ⎦ ⎛ TAB ⎞ ⎜ ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TCD ⎟ = Find... Engineering Mechanics - Statics Chapter 7 Given: F = 30 lb c = 2 ft xB = 6 ft d = 3 ft a = 5 ft e = 3 b = 8 ft f = 4 Solution: The initial guesses: TAB = 10 lb TCD = 30 lb TBC = 20 lb P = 10 lb Given xB − d ⎛ −xB ⎟ ⎡ ⎞ ⎤ ⎜ ⎥ TBC + P = TAB − ⎢ ⎜ xB2 + a2 ⎟ ⎢ ( xB − d) 2 + b2⎥ ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎞T − ⎡ ⎤T = AB ⎢ BC ⎜ 2 2⎟ 2 2⎥ a + xB ⎠ b + ( xB − d) ⎦ ⎝ ⎣ 0 0 xB − d ⎤ f ⎛ −d ⎞ T + ⎡ ⎞ ⎥ TBC + ⎛ CD ⎢ ⎜ 2 2⎟ . publisher. Engineering Mechanics - Statics Chapter 7 0 2 4 6 8 10 121 4 0 2 4 Distance (ft) Force (kip) Vx() x ft 0 2 4 6 8 10 121 4 0 20 40 Distance (ft) Moment (kip-ft) Mx() x ft Problem 7-8 9 Determine. 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 13 021 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 () =. from the publisher. 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 −