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 7-72 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. Given: F 1 400 lb= F 2 800 lb= w 100 lb in = a 4 in= b 12 in= c 4 in= Solution: F 1 awb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − F 2 b− Bb c+()+ 0= B w b 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 b+ F 1 a− bc+ = B 950.00 lb= 701 © 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 = V 1 x() F 1 − 1 lb = M 1 x() F 1 − x 1 lb in⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () B− F 2 + wa b+ x 2 − () + ⎡ ⎣ ⎤ ⎦ 1 lb = M 2 x 2 () Ba b+ c+ x 2 − () F 2 ab+ x 2 − () − w ab+ x 2 − () 2 2 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 lb in⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x 3 () B− lb = M 3 x 3 () Ba b+ c+ x 3 − () 1 lb in⋅ = 0 5 10 15 20 1000 0 1000 2000 Distance (in) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 in x 2 in , x 3 in , 0 5 10 15 20 2000 0 2000 4000 Distance (ini) Moment (lb-in) M 1 x 1 () M 2 x 2 () M 3 x 3 () x 1 in x 2 in , x 3 in , 702 © 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 7-73 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given: F 1 10 kN= F 2 10 kN= M 0 12 kN m⋅= a 2 m= b 2 m= c 2 m= d 2 m= Solution: F 1 bc+ d+()M 0 − F 2 d+ A y ab+ c+ d+()− 0= A y F 1 bc+ d+()M 0 − F 2 d+ ab+ c+ d+ = A y 8.50 kN= A y B y + F 1 − F 2 − 0= B y F 1 F 2 + A y −= B y 11.50 kN= x 1 0 0.01 a, a = V 1 x() A y 1 kN = M 1 x() A y x 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x() A y F 1 − () 1 kN = M 2 x() A y xF 1 xa−()− ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x() A y F 1 − () 1 kN = M 3 x() A y xF 1 xa−()− M 0 + ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = x 4 ab+ c+ 1.01 ab+ c+(), ab+ c+ d+ = V 4 x() B y − 1 kN = M 4 x() B y ab+ c+ d+ x−() 1 kN m⋅ = 703 © 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 012345678 15 10 5 0 5 10 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () V 3 x 3 () V 4 x 4 () x 1 x 2 , x 3 , x 4 , 012345678 0 10 20 30 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () M 3 x 3 () M 4 x 4 () x 1 x 2 , x 3 , x 4 , Problem 7-74 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 © 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 F 200 lb= w 100 lb ft = M 300 lb ft⋅= a 1 ft= b 4 ft= c 1 ft= Solution: Fa b+()Ab− wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M− 0= A Fa b+() wb 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M− b = A 375.00 lb= x 1 0 0.01 a, a = V 1 x 1 () F− 1 lb = M 1 x 1 () F− x 1 1 lb ft⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () F− A+ wx 2 a− () − ⎡ ⎣ ⎤ ⎦ 1 lb = M 2 x 2 () F− x 2 Ax 2 a− () + w x 2 a− () 2 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 lb ft⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x 3 () 0 1 lb = M 3 x 3 () M− 1 lb ft⋅ = 0123456 400 200 0 200 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 , 705 Given: © 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 300 200 100 0 Distance (ft) Moment (lb-ft) M 1 x 1 () M 2 x 2 () M 3 x 3 () x 1 ft x 2 ft , x 3 ft , Problem 7-75 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given: F 8 kN= M 20 kN m⋅= w 15 kN m = a 2 m= b 1 m= c 2 m= d 3 m= Solution: A− ab+ c+()M− Fc+ wd d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= A Fc M− w d 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ab+ c+ = A 14.30− kN= x 1 0 0.01 a, a = V 1 x 1 () A 1 kN = M 1 x 1 () Ax 1 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () A 1 kN = M 2 x 2 () Ax 2 M+ () 1 kN m⋅ = x 3 ab+ 1.01 ab+(), ab+ c+ = V 3 x 3 () AF−() 1 kN = M 3 x 3 () Ax 3 M+ Fx 3 a− b− () − ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = 706 © 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 4 ab+ c+ 1.01 ab+ c+(), ab + c + d + = V 4 x 4 () wa b+ c+ d+ x 4 − () 1 kN = M 4 x 4 () w− ab+ c+ d+ x 4 − () 2 2 1 kN m⋅ = 012345678 40 20 0 20 40 60 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () V 3 x 3 () V 4 x 4 () x 1 x 2 , x 3 , x 4 , 012345678 80 60 40 20 0 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () M 3 x 3 () M 4 x 4 () x 1 x 2 , x 3 , x 4 , 707 © 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 7-76 Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing. Units Used: kN 10 3 N= Given: w 2 kN m = F 4 kN= a 0.8 m= b 0.2 m= Solution: Ba b+()Fa− wa a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= B Fa w a 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ab+ = B 3.84 kN= AB+ wa− F− 0= AwaF+ B−= A 1.76 kN= x 1 0 0.01 a, a = V 1 x 1 () Awx 1 − () 1 kN = M 1 x 1 () Ax 1 w x 1 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () B− 1 kN = M 2 x 2 () Ba b+ x 2 − () 1 kN m⋅ = 0 0.2 0.4 0.6 0.8 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 0 0.2 0.4 0.6 0.8 0.5 0 0.5 1 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () x 1 x 2 , 708 © 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 7-77 Draw the shear and moment diagrams for the beam. Given: w 20 lb ft = M B 160 lb ft⋅= a 20 ft= b 20 ft= Solution: w− a a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ M B − C y ab+()+ 0= C y w a 2 2 M B + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = C y 104.00 lb= A y wa− C y + 0= A y wa C y −= A y 296.00 lb= x 1 0 0.01 a, a = V 1 x() A y wx− () 1 lb = M 1 x() A y xw x 2 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 lb ft⋅ = x 2 a 1.01a, ab+ = V 2 x() C y − 1 lb = M 2 x() C y ab+ x−() 1 lb ft⋅ = 0 5 10 15 20 25 30 35 40 200 0 200 400 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () x 1 ft x 2 ft , 709 © 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 10203040 0 1000 2000 3000 Distance (ft) Moment (lb-ft) M 1 x 1 () M 2 x 2 () x 1 ft x 2 ft , Problem 7-78 The beam will fail when the maximum moment is M max or the maximum shear is V max . Determine the largest distributed load w the beam will support. Units used: kip 10 3 lb= Given: M max 30 kip ft⋅= V max 8 kip= a 6 ft= b 6 ft= Solution: Set w 1 kip ft = and then scale the answer at the end A− bw a 2 a 3 b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wb b 2 + 0= A w a 2 a 3 b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w b 2 2 + b = A 7.00 kip= AB+ wb− w a 2 − 0= Bwb a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A−= B 2.00 kip= Shear limit - check critical points to the left and right of A and at B V big max B w a 2 , w a 2 A−, ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = V big 4.00 kip= w shear V max V big ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w= w shear 2.00 kip ft = 710 © 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. [...]... material may be reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 7 30 M1( x1) 20 Moment (kN-m) M2( x2) M3( x3) 10 M4( x4) 0 M5( x5) M6( x6) 10 20 30 0 2 4 6 8 10 12 14 x1 , x2 , x3 , x4 , x5 , x6 Distance (m) Problem 7-8 1 Draw the shear and moment diagrams for the beam Solutions: Support Reactions: ΣMx = 0; ⎛ L ⎞ − w0 L... 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 Moment (kN-m) 10 M1( x1) 5 M2( x2) 0 5 0 1 2 3 4 5 6 x1 , x2 Distance (m) Problem 7-8 4 Draw the shear and moment diagrams for the beam Units Used: 3 kip = 10 lb Given: w = 100 Guesses: Given lb ft M0 = 9 kip⋅ ft Ay = 1 lb... 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 Moment (kN-m) 0 M1( x1) M2( x2) 10 20 0 1 2 3 4 5 6 x1 , x2 Distance (m) Problem 7-8 7 Draw the shear and moment diagrams for the beam Units Used: 3 kip = 10 lb Given: w = 5 kip ft M1 = 15 kip⋅ ft M2 = 15 kip⋅ ft a = 6 ft... 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 20 Moment (kip-ft) M1p( x1) M2p( x2) M3p( x3) 0 20 40 60 0 5 10 15 20 x1 x2 x3 , , ft ft ft Distance (ft) Problem 7-8 8 Draw the shear and moment diagrams for the beam Units Used: 3 kip = 10 lb Given: w1 = 2 kip ft w2 = 1 kip ft a = 15 ft... 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 (kip) 4 V( x ) 2 0 0 2 4 6 8 10 12 14 x ft Distance (ft) Moment (kip-ft) 40 M( x) 20 0 0 2 4 6 8 10 12 14 x ft Distance (ft) Problem 7-8 9 Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal... 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 7-8 6 Draw the shear and moment diagrams for the beam Units Used: 3 kN = 10 N Given: kN m w = 2 a = 3m b = 3m Solution: x1 = 0 , 0.01a a V 1 ( x) = ⎡ 1 w b + 1 w⎛ a − x ⎞ ( a −.. .Engineering Mechanics - Statics Chapter 7 Moment limit - check critical points at A and betwen A and B B ⎛ a⎞⎛ a ⎞ MA = −w⎜ ⎟ ⎜ ⎟ x = w ⎝ 2 ⎠⎝ 3 ⎠ Mbig = max ( MA , MAB wmoment = MAB ) ⎛ x2 ⎞ = B x − w⎜ ⎟ ⎝2⎠ ⎛ MA ⎞ ⎛ −6.00 ⎞ ⎜... 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 7-8 2 Draw the shear and moment diagrams for the beam Units Used: 3 kip = 10 lb Given: F = 2000 lb a = 9 ft lb ft b = 9 ft w = 500 Solution: ⎛ 2b ⎞ ⎛ a⎞ w b⎜ ⎟ + F b + w a⎜ b +... + c − x3 ) − w −M 2 ⎣ ⎦ lb⋅ ft 711 © 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 1 lb⋅ ft Engineering Mechanics - Statics Chapter 7 Force (lb) 1500... 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 Tmax = max ( TAB , TBC , TCD , TDE) yB = 3.53 m P = 800.00 N Tmax = 8.17 kN Problem 7-9 0 Cable ABCD supports the lamp of mass M1 and the lamp of mass M2 Determine the maximum tension in the cable and the sag of point B Given: . 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 7-7 2 Draw the shear. the publisher. Engineering Mechanics - Statics Chapter 7 0123456 300 200 100 0 Distance (ft) Moment (lb-ft) M 1 x 1 () M 2 x 2 () M 3 x 3 () x 1 ft x 2 ft , x 3 ft , Problem 7-7 5 Draw the shear. writing from the publisher. Engineering Mechanics - Statics Chapter 7 0 10203040 0 1000 2000 3000 Distance (ft) Moment (lb-ft) M 1 x 1 () M 2 x 2 () x 1 ft x 2 ft , Problem 7-7 8 The beam will fail