Engineering Mechanics - Statics Chapter 7 Support Reactions: From FBD (b), wL 2 L 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ B y L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= B y wL 4 = From FBD (a), A y wL 2 − B y − 0= A y 3 wL 4 = B y − L 2 w L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − M A + 0= M A w L 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Shear and Moment Functions: From FBD (c) For 0 x≤ L≤ A y wx− V− 0= V w 4 3L 4x−()= M A A y x− wx x 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M+ 0= M w 4 3Lx 2x 2 − L 2 − () = 681 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 The beam has depth a and is subjected to a uniform distributed loading w which acts at an angle θ from the vertical as shown. Determine the internal normal force, shear force, and moment in the beam as a function of x. Hint:The moment loading is to be determined from a point along the centerline of the beam (x axis). Given: a 2 ft= L 10 ft= θ 30 deg= w 50 lb ft = Solution: 0 x≤ L≤ Σ F x = 0; Nwsin θ () x+ 0= Nx() w− sin θ () x= Σ F y = 0; V− wcos θ () x− 0= Vw− cos θ () x= Σ M = 0; wcos θ () x x 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ wsin θ () x a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − M+ 0= Mx() w− cos θ () x 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ wsin θ () xa 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += Problem 7-56 Draw the shear and moment diagrams for the beam. Given: w 250 lb ft = L 12 ft= 682 Problem 7-55 © 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 y = 0; V− 1 2 x wx L − 0= Vx() wx 2 2L − 1 lb = Σ M = 0; M x 2 wx L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= Mx() w− x 3 6L 1 lb ft⋅ = 024681012 2000 1000 0 Distance (ft) Force (lb) Vx() x ft 024681012 1 . 10 4 5000 0 Distance (ft) Moment (lb-ft) Mx() x ft Problem 7-57 The beam will fail when the maximum shear force is V max or the maximum moment is M max . Determine the largest intensity w of the distributed loading it will support. 683 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 Given: L 18 ft= V max 800 lb= M max 1200 lb ft⋅= Solution: For 0 x≤ L≤ V w− x 2 2L = M w− x 3 6L = V max wL 2 = w 1 2 V max L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w 1 88.9 lb ft = M max w 2 L 2 6 = w 2 6 M max L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w 2 22.2 lb ft = Now choose the critical case w min w 1 w 2 , () = w 22.22 lb ft = Problem 7-58 The beam will fail when the maximum internal moment is M max . Determine the position x of the concentrated force P and its smallest magnitude that will cause failure. 684 © 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: For ξ < x, M 1 P ξ Lx−() L = For ξ > x, M 2 Px L ξ − () L = Note that M 1 = M 2 when x = ξ M max M 1 = M 2 = Px L x−() L = P L Lx x 2 − () = x Lx x 2 − () d d L 2x−= x L 2 = Thus, M max P L L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 2 L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 4M max L = Problem 7-59 Draw the shear and moment diagrams for the beam. Given: w 30 lb ft = M C 180 lb ft⋅= a 9 ft= b 4.5 ft= Solution: A y − a 1 2 wa a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M C − 0= 685 © 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 wa 6 M C a −= A y B y + 1 2 wa− 0= B y wa 2 A y −= x 1 0 0.01 a, a = A y 1 2 w x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x− V 1 x()− 0= V 1 x() A y wx 2 2a − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 lb = A y − x 1 2 w x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M 1 x()+ 0= M 1 x() A y x wx 3 6a − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 lb ft⋅ = x 2 a 1.01a, ab+ = A y 1 2 wa− B y + V 2 x()− 0= V 2 x() A y B y + wa 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 lb = A y − x 1 2 wa x 2a 3 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + B y xa−()− M 2 x()+ 0= M 2 x() A y xB y xa−()+ wa 2 x 2a 3 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 lb ft⋅ = 024681012 200 100 0 100 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () x 1 ft x 2 ft , 686 © 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 024681012 200 100 0 100 Diastance (ft) Moment (lb-ft) M 1 x 1 () M 2 x 2 () x 1 ft x 2 ft , Problem 7-60 The cantilevered beam is made of material having a specific weight γ . Determine the shear and moment in the beam as a function of x. Solution: By similar triangles y x h d = y h d x= W γ V= γ 1 2 yxt ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = γ 1 2 h d x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ xt ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = γ ht 2d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = Σ F y = 0; V γ ht 2d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 − 0= V γ ht 2d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = Σ M = 0; M− γ ht 2d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 x 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= 687 © 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 M γ ht 6d − x 3 = Problem 7-61 Draw the shear and moment diagrams for the beam. Given: kip 10 3 lb= w 1 30 lb ft = w 2 120 lb ft = L 12 ft= Solution: w 1 L L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w 2 w 1 − () L L 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + A y L− 0= A y w 1 L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w 2 w 1 − 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += A y w 1 x− 1 2 w 2 w 1 − () x L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x− Vx()− 0= Vx() A y w 1 x− 1 2 w 2 w 1 − () x 2 L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 lb = A y − xw 1 x x 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 1 2 w 2 w 1 − () x L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + Mx()+ 0= Mx() A y xw 1 x 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − w 2 w 1 − () x 3 6L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = 688 © 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 024681012 500 0 500 Distance (ft) Force (lb) Vx() x ft 024681012 0 1 2 Distance (ft) Moment (kip-ft) Mx() x ft Problem 7-62 Draw the shear and moment diagrams for the beam. Units Used: kip 10 3 lb= Given: F 1 20 kip= F 2 20 kip= w 4 kip ft = a 15 ft= b 30 ft= c 15 ft= Solution: F 1 ab+()wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + F 2 c− A y b− 0= A y F 1 ab+ b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ wb 2 + F 2 c b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= A y B y + F 1 − F 2 − wb− 0= B y F 1 F 2 + wb+ A y −= 689 © 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 1 0 0.01 a, a = F 1 − V 1 x()− 0= V 1 x() F 1 − 1 kip = F 1 xM 1 x()+ 0= M 1 x() F 1 − x 1 kip ft⋅ = x 2 a 1.01a, ab+ = F 1 − wx a−()− A y + V 2 x()− 0= V 2 x() F 1 − wx a−()− A y + ⎡ ⎣ ⎤ ⎦ 1 kip = F 1 xA y xa−()− wx a−() xa− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M 2 x()+ 0= M 2 x() F 1 − xA y xa−()+ w xa−() 2 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kip ft⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x() F 2 − 0= V 3 x() F 2 1 kip = M 3 x()− F 2 ab+ c+ x−()− 0= M 3 x() F 2 − ab+ c+ x−() 1 kip ft⋅ = 0 102030405060 100 50 0 50 100 Distance (ft) Force (kip) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 690 © 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 Force (kN) 2 V1( x1) 0 V2( x2) 2 4 0 0.5 1 1.5 2 2.5 3 2. 5 3 3.5 4 x1 , x2 Distance (m) Moment (kN-m) 2 M1( x1) 1 M2( x2) 0 1 0 0.5 1 1.5 2 3.5 4 x1 , x2 Distance (m) Problem 7-7 0 Draw the shear and moment diagrams for the beam Given: lb ft w = 30 MC = 180 lb⋅ ft a = 9 ft b = 4.5 ft Solution: − Ay a + Ay = 1 2 wa 6 ⎛ a⎞ − M = ⎟ C ⎝ 3⎠ w a⎜ − 0 MC a 6 98 © 20 07... without permission in writing from the publisher 1 lb⋅ ft Engineering Mechanics - Statics Chapter 7 Force (lb) 1500 V1( x1) 1000 V2( x2) 500 V3( x3) 0 500 1000 0 2 4 6 8 10 12 14 16 18 20 x1 x2 x3 , , ft ft ft Distance (ft) Moment (lb-ft) 5000 M1( x1) 0 M2( x2) M3( x3) 5000 1 10 4 0 2 4 6 8 10 12 14 16 18 x1 x2 x3 , , ft ft ft Distance (ft) Problem 7 -8 0 The beam consists of three segments pin connected... the publisher 6 Engineering Mechanics - Statics Chapter 7 x4 = a + b + c , 1.01( a + b + c) a + b + c + d V 4 ( x4 ) = w( a + b + c + d − x4 ) M4 ( x4 ) = −w 1 kN (a + b + c + d − x4 )2 1 2 kN⋅ m 60 Force (kN) V1( x1) 40 V2( x2) 20 V3( x3) 0 V4( x4) 20 40 0 1 2 3 4 5 6 7 8 5 6 7 8 x1 , x2 , x3 , x4 Distance (m) Moment (kN-m) 0 M1( x1) 20 M2( x2) M3( x3) 40 M4( x4) 60 80 0 1 2 3 4 x1 , x2 , x3 , x4 Distance... publisher Engineering Mechanics - Statics Ay + By − By = wa 2 1 Chapter 7 wa = 0 2 − Ay x1 = 0 , 0.01a a 2 ⎛ wx ⎞ 1 ⎜ Ay − ⎟ 2 a ⎠ lb ⎝ V 1 ( x) = M1 ( x) = 3 ⎛ wx ⎞ 1 ⎜ Ay x − ⎟ 6 a ⎠ lb⋅ ft ⎝ x2 = a , 1.01a a + b ⎛A + B − wa⎞ 1 ⎜ y y 2 ⎟ ⎝ ⎠ lb V 2 ( x) = ⎡ ⎣ M2 ( x) = ⎢Ay x + B y( x − a) − wa⎛ 2 ⎜x − ⎝ 2 a ⎞⎤ 1 3 ⎟⎥ ⎠⎦ lb⋅ ft Force (lb) 100 V1( x1) V2( x2) 0 100 20 0 0 2 4 6 8 10 12 x1 x2 , ft ft... publisher Engineering Mechanics - Statics x1 = 0 , 0.01a a Chapter 7 V 1 ( x) = −F 1 x2 = a , 1.01a a + b M2 ( x2 ) 1 M1 ( x) = −F1 x lb V 2 ( x2 ) = ⎡−B + F2 + w( a + b − x2 )⎤ ⎣ ⎦ 1 lb⋅ in 1 lb ⎡ (a + b − x2 )2 1 ⎥ = ⎢B( a + b + c − x2 ) − F2 ( a + b − x2 ) − w⋅ 2 ⎣ ⎦ lb⋅ in V 3 ( x3 ) = x3 = a + b , 1.01( a + b) a + b + c −B lb M3 ( x3 ) = B ( a + b + c − x3 ) 1 lb⋅ in Force (lb) 20 00 V1( x1) V2( x2) 1000... publisher Engineering Mechanics - Statics Chapter 7 10 Force (kN) V1( x1) 5 V2( x2) 0 V3( x3) 5 V4( x4) 10 15 0 1 2 3 4 5 6 7 5 6 7 8 x1 , x2 , x3 , x4 Distance (m) Moment (kN-m) 30 M1( x1) M2( x2) 20 M3( x3) M4( x4) 10 0 0 1 2 3 4 8 x1 , x2 , x3 , x4 Distance (m) Problem 7-7 4 Draw the shear and moment diagrams for the shaft The support at A is a journal bearing and at B it is a thrust bearing 704 © 20 07... Engineering Mechanics - Statics Chapter 7 Given: F = 20 0 lb w = 100 lb M = 300 lb⋅ ft ft Solution: F ( a + b) − A b + w b⎛ ⎜ F( a + b) + b⎞ ⎟−M= ⎝ 2 x1 = 0 , 0.01a a a = 1 ft A = 0 V 1 ( x1 ) = −F ⎛ w b2 ⎞ ⎜ ⎟−M ⎝ 2 ⎠ M1 ( x1 ) = −F x1 1 lb M2 ( x2 ) c = 1 ft A = 375.00 lb b V 2 ( x2 ) = ⎡−F + A − w( x2 − a)⎤ ⎣ ⎦ x2 = a , 1.01a a + b b = 4 ft 1 lb⋅ ft 1 lb ⎡ ⎤ (x − a )2 1 ⎢−F x2 + A( x2 − a) − w 2. .. 2 4 6 8 10 12 x1 x2 , ft ft Diastance (ft) Problem 7-7 1 Draw the shear and moment diagrams for the beam 3 kip = 10 lb Given: w1 = 30 lb ft w2 = 120 lb ft L = 12 ft Solution: ⎛ L⎞ + ⎟ 2 w1 L⎜ (w2 − w1)L⎛ 3 ⎞ − Ay L = 0 ⎜ ⎟ 2 L 1 ⎝ ⎠ ⎛ L ⎞ + ⎛ w2 − w1 ⎞ ⎛ L ⎞ ⎟⎜ ⎟ ⎟ ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 3 ⎠ Ay = w1 ⎜ ⎡ ⎣ ⎦ ⎡ x ⎣ 2 M ( x) = ⎢A y x − w1 2 x 1 (w2 − w1) L ⎥ lb 2 1 V ( x) = ⎢Ay − w1 x − 2 − ( w2 − w1 ) x ⎤... w = 2 kN m F = 4 kN ⎛ a⎞ = ⎟ 2 B ( a + b ) − F a − w a⎜ Solution: a = 0 .8 m x2 = a , 1.01a a + b B = 0 a+b A = wa + F − B V 1 ( x1 ) = ( A − w x1 ) V 2 ( x2 ) = −B ⎛ a2 ⎞ ⎟ 2 F a + w⎜ A + B − wa − F = 0 x1 = 0 , 0.01a a b = 0 .2 m M1 ( x1 ) 1 kN B = 3 .84 kN A = 1.76 kN ⎡ ⎛ x 2 ⎞⎤ ⎢A x1 − w⎜ 1 ⎟⎥ 1 = ⎣ ⎝ 2 ⎠⎦ kN⋅ m M2 ( x2 ) = B ( a + b − x2 ) 1 kN 1 kN⋅ m Force (kN) 5 V1( x1) V2( x2) 0 5 0 0 .2 0.4... beam Given: MB = 80 0 lb⋅ ft a = 5 ft F = 100 lb b = 5 ft Solution: V 1 ( x) = F x2 = a , 1.01a a + b V 2 ( x) = F Force (lb) x1 = 0 , 0.01a a 1 M1 ( x) = ⎡−F ( a + b − x) − MB⎤ ⎣ ⎦ lb 1 M2 ( x) = −F( a + b − x) lb 1 lb⋅ ft 1 lb⋅ ft V1( x1) 100 V2( x2) 99.9 99 .8 0 2 4 6 8 10 x1 x2 , ft ft Distance (ft) Moment (lb-ft) 0 M1( x1) M2( x2) 1000 20 00 0 2 4 6 8 10 x1 x2 , ft ft Distane (ft) 694 © 20 07 R C Hibbeler . x 3 6L = V max wL 2 = w 1 2 V max L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w 1 88 .9 lb ft = M max w 2 L 2 6 = w 2 6 M max L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = w 2 22. 2 lb ft = Now choose the critical case w min w 1 w 2 , () = w 22 .22 lb ft = Problem 7-5 8 The beam will fail when the. publisher. Engineering Mechanics - Statics Chapter 7 0 0.5 1 1.5 2 2.5 3 3.5 4 4 2 0 2 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 0 0.5 1 1.5 2 2.5 3 3.5 4 1 0 1 2 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () x 1 x 2 , Problem. the publisher. Engineering Mechanics - Statics Chapter 7 024 681 0 12 500 0 500 Distance (ft) Force (lb) Vx() x ft 024 681 0 12 0 1 2 Distance (ft) Moment (kip-ft) Mx() x ft Problem 7-6 2 Draw the shear