Engineering Mechanics - Statics Chapter 6 W 1 γ L L 2 a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a = W 1 3lb= + ↑ Σ F y = 0; R A W 1 − γ L− 0= R A W 1 γ L+= R A 9lb= Σ M B = 0; W 2 acos φ () γ Lcos φ () L 2 a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − R A La−( ) cos φ () − 0= W 2 γ L L 2 a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ R A La−()+ a = W 2 21lb= + ↑ R B W 2 − R A − γ L− 0= Σ F y = 0; R B W 2 R A + γ L+= R B 36lb= Σ M C = 0; R B La−( ) cos φ () γ L L 2 a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos φ () + W 3 acos φ()− 0= W 3 R B La−() γ L L 2 a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + a = W 3 75lb= Problem 6-124 The three-member frame is connected at its ends using ball-and-socket joints. Determine the x, y, z components of reaction at B and the tension in member ED. The force acting at D is F . Given: F 135 200 180− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 6ft= e 3ft= b 4ft= f 1ft= 601 © 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 6 d 6ft= g 2ft= cgf+= Solution: AC and DE are two-force members. Define some vectors r DE e− b− g− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u DE r DE r DE = r AC d− e− b− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u AC r AC r AC = r BD e f− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BA ed+ c− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses B x 1lb= B y 1lb= B z 1lb= F DE 1lb= F AC 1lb= Given B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F DE u DE + F AC u AC + F+ 0= r BD F DE u DE F+ () × r BA F AC u AC () ×+ 0= B x B y B z F DE F AC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find B x B y , B z , F DE , F AC , () = B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 30− 13.333− 3.039 10 12− × ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ lb= F DE F AC ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 270 16.415 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= 602 © 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 6 The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is F max , determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pon at G only exert force components on the frame. Given: F max 800 N= a 300 mm= b 600 mm= c 600 mm= Solution: Σ M x = 0; P− 2 c b b 2 c 2 + F max c+ 0= P F max b 2 b 2 c 2 + = P 282.843 N= B z D z + F max c b 2 c 2 + − 0= D z B z = B z F max c 2 b 2 c 2 + = D z B z = B z 283 N= D z 283 N= B y D y + F max b b 2 c 2 + − 0= D y B y = B y F max b 2 b 2 c 2 + = D y B y = B y 283 N= D y 283 N= B x D x = 0= Problem 6-126 The structure is subjected to the loading shown. Member AD is supported by a cable AB and a roller at C and fits throu g h a smooth circular hole at D . Member ED is su pp orted b y a roller at 603 Problem 6-125 © 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 6 g pp y D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB. Units Used: kN 10 3 N= Given: F 0 0 2.5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= a 0.5 m= d 0.3 m= b 0.4 m= e 0.8 m= c 0.3 m= Solution: AB c− d− 0 e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AB 1kN= D x 1kN= D z 1kN= D z2 1kN= E x 1kN= E y 1kN= M Dx 1kNm⋅= M Dz 1kNm⋅= C x 1kN= M Ex 1kNm⋅= M Ey 1kN⋅ m= Given F F AB AB AB + C x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + D x 0 D z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= M Dx 0 M Dz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ C x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ d b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F×+ cd+ b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F AB AB AB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×+ 0= D x − 0 D z2 D z − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ E x E y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= M Dx − 0 M Dz − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ M Ex M Ey 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ D x 0 D z D z2 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= 604 © 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 6 C x D x D z D z2 E x E y F AB M Dx M Dz M Ex M Ey ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find C x D x , D z , D z2 , E x , E y , F AB , M Dx , M Dz , M Ex , M Ey , () = C x D x D z D z2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0.937 0 1.25 1.25 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= M Dx M Dz ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0.5 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN m⋅= E x E y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN= M Ex M Ey ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0.5 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN m⋅= F AB 1.562 kN= Problem 6-127 The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C. 605 © 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 6 Given: a 2m= M 800 N m⋅= b 1.5 m= F 250 N= c 3m= θ 1 60 deg= d 4m= θ 2 45 deg= θ 3 60 deg= Solution: Guesses B x 1N= B y 1N= A x 1N= A y 1N= A z 1N= C x 1N= C y 1N= C z 1N= M Bx 1Nm⋅= M By 1Nm⋅= M Cy 1Nm⋅= M Cz 1Nm⋅= Given A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x B y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= c a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x B y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × M 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + M Bx − M By − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= F cos θ 1 () cos θ 2 () cos θ 3 () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x − B y − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + C x C y C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= 0 0 bd+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F cos θ 1 () cos θ 2 () cos θ 3 () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × 0 0 b ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x − B y − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ M Bx M By 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 M Cy M Cz ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= 606 © 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 6 A x A y A z C x C y C z B x B y M Bx M By M Cy M Cz ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , A z , C x , C y , C z , B x , B y , M Bx , M By , M Cy , M Cz , () = A x A y A z C x C y C z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 172.3− 114.8− 0 47.3 61.9− 125− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N= M Cy M Cz ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 429− 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Nm⋅= Problem 6-128 Determine the resultant forces at pins B and C on member ABC of the four-member frame. Given: w 150 lb ft = a 5ft= b 2ft= 607 © 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 6 c 2ft= e 4ft= dab+ c−= Solution: The initial guesses are F CD 20 lb= F BE 40 lb= Given F CD cd+()F BE ec db−() 2 e 2 + − 0= w− ab+() ab+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BE e db−() 2 e 2 + a+ F CD ab+()− 0= F CD F BE ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find F CD F BE , () = F CD F BE ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 350 1531 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= Problem 6-129 The mechanism consists of identical meshed gears A and B and arms which are fixed to the gears. The spring attached to the ends of the arms has an unstretched length δ and a stiffness k. If a torque M is applied to gear A, determine the angle θ through which each arm rotates. The gears are each pinned to fixed supports at their centers. 608 © 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 6 Given: δ 100 mm= k 250 N m = M 6Nm⋅= r δ 2 = a 150 mm= Solution: Σ M A = 0; F− rPacos θ () − M+ 0= Σ M B = 0; Pacos θ () Fr− 0= 2Pacos θ () M= 2k 2a( ) sin θ () acos θ () M= 2ka 2 sin 2 θ () M= θ 1 2 asin M 2ka 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 16.1 deg= Problem 6-130 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN 1000 N= 609 © 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 6 Given: F 1 20 kN= F 2 10 kN= a 1.5 m= b 2m= Solution: θ atan b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guesses F AG 1kN= F BG 1kN= F GC 1kN= F GF 1kN= F AB 1kN= F BC 1kN= F CD 1kN= F CF 1kN= F DF 1kN= F DE 1kN= F EF 1kN= Given F AB − cos θ () F BC + 0= F AB − sin θ () F BG − 0= F GC cos θ () F GF + F AG − 0= F GC sin θ () F BG + F 1 − 0= F BC − F CD + F GC cos θ () − F CF cos θ () + 0= F GC − sin θ () F CF sin θ () − 0= F CD − F DE cos θ () + 0= F DF − F DE sin θ () − 0= F GF − F CF cos θ () − F EF + 0= 610 © 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. [...]... Units Used: 3 kN = 10 N Given: w = 400 N m a = 2. 5 m b = 3m c = 6m Solution: ΣMA = 0; −1 2 ⎛ 2 c⎞ + F ⎛ a ⎞ c = ⎜3 ⎟ BC ⎜ ⎝ ⎠ 2 2⎟ ⎝ a +c ⎠ wc F BC = + → Σ Fx = 0; ↑Σ Fy = 0; 3 wc ⎛ a2 + c2 ⎞ ⎜ ⎟ ⎝ a ⎠ ⎛ c ⎞F − A = x ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ Ax = + 1 Ay − ⎛ c ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ 1 2 wc + 0 F BC = 20 80 N 0 Ax = 1 920 N ⎛ a ⎞F = ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ 0 63 4 © 20 07 R C Hibbeler Published by Pearson Education,... publisher Engineering Mechanics - Statics Chapter 6 Given: mlink = 5 kg δ = 0.3 m N k = 400 m a = 0.1 m b = 0 .6 m g = 9.81 m 2 s Solution: Guesses θ = 10 deg F BD = 1 N Ex = 1 N Given mlink g a+b cos ( θ ) − FBD b cos ( θ ) + E x b sin ( θ ) = 0 2 2 mlink g a+b 2 cos ( θ ) + Ex 2 b sin ( θ ) = 0 F BD = k( 2 b sin ( θ ) − δ ) ⎛ FBD ⎞ ⎜ ⎟ ⎜ Ex ⎟ = Find ( FBD , Ex , θ ) ⎜ ⎟ ⎝ θ ⎠ θ = 21 .7 deg Problem 6- 1 32 The... the publisher Engineering Mechanics - Statics ΣMA = 0; Chapter 6 a+b ⎡Ey ( 2) ( a + b) sin ( θ ) − 2( M) ( g)⎤ ⎣ ⎦ 2 ( cos ( θ ) ) = 0 ⎡E ( 2) sin ( θ ) − 2 ( m) ⎢ y ⎣ g⎤ ⎥ ( cos ( θ ) = 0 2 ⇒ E y ( 2 sin ( θ ) ) = m ( g) ( cos ( θ ) ) Ey = ΣMC = 0; m( g) ( cos ( θ ) ) 2 sin ( θ ) m( g) ( cos ( θ ) ) 2 sin ( θ ) ( a + b) sin ( θ ) + m( g) ⎛ ⎜ a + b⎞ ⎟ cos ( θ ) − Fs( b cos ( θ ) ) = 0 ⎝ 2 ⎠ b m = Fs... form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 MD = Mmax 2 MD w1 = w1 = 100 w2 = 22 5 ab N m N m Assume that the maximum normal force in BC has been reached T = Pmax T2 d w2 = 2 ( a + b) c + d 2 w = min ( w1 , w2 ) Now choose the critical load w = 100 N m Problem 7-1 0 Determine the shear force and moment acting at a section passing through... V C = F2 + F3 sin ( θ ) V C = 62 . 50 kN ΣMC = 0; −MC − F2 b − F 3 sin ( θ ) 2 b = 0 MC = −F 2 b − F3 sin ( θ ) 2b MC = 22 5.00 kN⋅ m Problem 7-8 Determine the normal force, shear force, and moment at a section passing through point C Assume the support at A can be approximated by a pin and B as a roller Units used: 3 kip = 10 lb Given: F 1 = 10 kip a = 6 ft F 2 = 8 kip b = 12 ft kip ft c = 12 ft w =... writing from the publisher Engineering Mechanics - Statics Ay = + → Σ Fx = 0; + ↑Σ Fy = 0; Chapter 7 1 2 wc − ⎛ a ⎞F ⎜ 2 2 ⎟ BC ⎝ a +c ⎠ ND − Ax = 0 Ay − Ay = 400 N ND = A x b⎞ ⎜w ⎟ b − VD = c⎠ 1⎛ 2 0 2 1 ⎛b V D = A y − w⎜ 2 ⎝ c ΣF y = 0; − Ay b + ⎟ ⎠ V D = 100 N b⎞ ⎛ b⎞ ⎜ w ⎟ b⎜ 3 ⎟ + M D = 2 c⎠ ⎝ ⎠ 1⎛ MD = Ay b − 1 6 ND = 1. 920 kN 0 ⎛ b3 ⎞ w⎜ ⎟ ⎝c⎠ MD = 900 N m Problem 7-1 5 The beam has weight density... Solution: Section A: ΣF z = 0; F 2 − 2 F1 − NA = 0 NA = F 2 − 2 F 1 NA = 10.00 lb Section B: ΣF z = 0; F 2 − 2 F1 − NA + NB = 0 NB = −F 2 + 2 F1 + NA NB = 0.00 lb Problem 7-3 The shaft is supported by smooth bearings at A and B and subjected to the torques shown Determine the internal torque at points C, D, and E Given: M1 = 400 N⋅ m M2 = 150 N⋅ m M3 = 550 N⋅ m 62 3 © 20 07 R C Hibbeler Published by Pearson... F1 − F2 = 0 NA = F 1 + F 2 NA = 12. 0 kN F 1 a − F2 b − MA = 0 MA = F1 a − F 2 b MA = 0 kN⋅ m ↑Σ Fy = 0; ΣMA = 0; VA = 0 Applying equations of equillibrium to the top segment sectioned through point B, we have + → Σ Fx = 0; VB = 0 + NB − F1 − F2 − F3 = 0 NB = F 1 + F 2 + F 3 NB = 20 .0 kN F 1 a − F 2 b − F3 c + MB = 0 MB = −F1 a + F 2 b + F3 c MB = 1 .20 kN⋅ m ↑Σ Fy = 0; +ΣMB = 0; VB = 0 Problem 7 -2 The... in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics + ↑Σ Fy = 0; Chapter 7 VC − w c + By − F2 = 0 VC = w c − By + F2 V C = 0.5 kip ⎛ c ⎞ + B c − F ( c + d) = ⎟ y 2 ⎝ 2 −MC − w c⎜ ΣMC = 0; ⎛ c2 ⎞ ⎟ + B y c − F 2 ( c + d) 2 MC = −w⎜ 0 MC = 3 .6 kip⋅ ft Problem 7-9 The beam AB will fail if the maximum internal moment at D reaches Mmax or the... = 800 N⋅ m P max = 1500 N a = 4m b = 4m c = 4m d = 3m Solution: ⎛ a + b ⎞ − A ( a + b) = ⎟ y ⎝ 2 ⎠ w ( a + b) ⎜ Ay = w( a + b) 2 ⎛ a⎞ − A a + M = ⎟ y D 2 w a⎜ MD = w 0 ⎛ a b⎞ 2 ⎟ ⎝ ⎠ Ay − w( a + b) + T= 0 w( a + b) ⎛ d ⎞T = ⎜ 2 2⎟ ⎝ c +d ⎠ 2 c +d 0 2 2d Assume the maximum moment has been reached 62 9 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River, NJ All rights reserved . 300 mm= b 60 0 mm= c 60 0 mm= Solution: Σ M x = 0; P− 2 c b b 2 c 2 + F max c+ 0= P F max b 2 b 2 c 2 + = P 28 2.843 N= B z D z + F max c b 2 c 2 + − 0= D z B z = B z F max c 2 b 2 c 2 + = D z B z =. b 2 c 2 + = D z B z = B z 28 3 N= D z 28 3 N= B y D y + F max b b 2 c 2 + − 0= D y B y = B y F max b 2 b 2 c 2 + = D y B y = B y 28 3 N= D y 28 3 N= B x D x = 0= Problem 6- 1 26 The structure is subjected. publisher. Engineering Mechanics - Statics Chapter 6 Σ M A = 0; E y 2( )ab+( ) sin θ () 2 M()g()− ⎡ ⎣ ⎤ ⎦ ab + 2 cos θ ()() 0= E y 2( ) sin θ () 2 m() g 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ cos θ () 0= ( ⇒ E y 2 sin θ ()() mg(