PROBLEM 7.37 SOLUTION

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1 PROLEM 7.37 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maimum absolute values of the shear and bending moment. Free bod: Entire beam Σ M = 0: E(6 ft) (6 kips)( ft) (1 kips)(4 ft) (4.5 kips)(8 ft) = 0 Σ F = 0: = 0 E =+ 16 kips E = 16 kips Σ F = 0: + 16 kips 6 kips 1 kips 4.5 kips = 0 = kips = 6.50 kips (a) Shear and bending moment Just to the right of : V = kips M = Just to the right of C: Σ F = 0: 6.50 kips 6 kips V = 0 V = kips Σ M = 0: M (6.50 kips)( ft) = 0 M =+ 13 kip ft Just to the right of D: Σ F = 0: V = 0 V 3 = kips Σ M = 0: M (6.50)(4) (6)() = 0 M 3 =+ 14 kip ft PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 1051

2 PROLEM 7.37 (Continued) Just to the right of E: Σ F = 0: V 4.5 = 0 V 4 =+ 4.5 kips 4 Σ M = 0: M (4.5) = 0 M 4 = 9kip ft 4 4 t : V = M = 0 (b) V ma = 11.50kips M ma = 14.00kip ft PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 105

PROLEM 7.39 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maimum absolute values of the shear and bending moment.

0 kn (a) Shear and bending-moment diagrams.

m: M C =+ 9.0kN m For = 3m: M D =+ 78.0kN m PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved.

3 PROLEM 7.39 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maimum absolute values of the shear and bending moment. Free bod: Entire beam Σ M = 0: (5 m) (60 kn)( m) (50 kn)(4 m) = 0 Σ F = 0: = 0 Σ F = 0: kn 6.0 kn 50 kn = 0 = kn = 64.0 kn = kn = 46.0 kn (a) Shear and bending-moment diagrams. From to C: Σ F = 0: 46 V = 0 V =+ 46 kn Σ M = 0: M 46= 0 M = (46 )kn m From C to D: Σ F = 0: V = 0 V = 14 kn Σ M = 0: M ( ) = 0 j M = (10 14 )kn m For = m: M C =+ 9.0kN m For = 3m: M D =+ 78.0kN m PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 1055

4 PROLEM 7.6* In order to reduce the bending moment in the cantilever beam, a cable and counterweight are permanentl attached at end. Determine the magnitude of the counterweight for which the maimum absolute value of the bending moment in the beam is as small as possible and the corresponding value of M ma. Consider (a) the case when the distributed load is permanentl applied to the beam, (b) the more general case when the distributed load ma either be applied or removed. M due to distributed load: M due to counter weight: Σ MJ = 0: M w = 0 1 M = w Σ M = 0: M + w= 0 J M = w (a) oth applied: w M = W dm W = W w= 0at = d w W nd here M = > 0 so M ma ; M min must be at = L w 1 So Mmin = WL wl. For minimum M ma set ma min M = M, so W 1 WL wl or w = + W + wlw w L = 0 W = wl± w L (need + ) W = ( 1) wl= wl PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 1093

5 PROLEM 7.6* (Continued) (b) w ma be removed M ma W ( 1) = = w wl Mma = wL Without w, M = W M = WLat ma With w (see Part a) w M = W W W Mma = at = w w 1 Mmin = WL wl at = L For minimum Mma, set Mma (no w) = Mmin (with w) With 1 1 WL = WL + wl W = wl 4 M ma 1 = wl 4 1 W = wl 4 PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 1094

6 PROLEM 7.68 Using the method of Section 7.6, solve Problem PROLEM 7.34 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maimum absolute values of the shear and bending moment. Free bod: Entire beam Σ F = 0: P = 0 = P Σ M = 0: M M + PL = 0 0 Shear diagram t : V = P M = 0 V ma = P ending-moment diagram t : M = M0 = PL M ma = PL PROPRIETRY MTERIL. 013 The McGraw-Hill Companies, Inc. ll rights reserved. No part of this Manual ma be displaed, reproduced or distributed in an form or b an means, without the prior written permission of the publisher, or used beond the limited distribution to teachers and educators permitted b McGraw-Hill for their individual course preparation. If ou are a student using this Manual, ou are using it without permission. 1100

PROBLEM 5.1 SOLUTION. Reactions: Pb L Pa L. From A to B: 0 < x < a. Pb L Pb L Pb L Pbx L. From B to C: a < x < L Pa L. Pa L. L Pab At section B: M = L

PROBLEM 5.1 SOLUTION. Reactions: Pb L Pa L. From A to B: 0 < x < a. Pb L Pb L Pb L Pbx L. From B to C: a < x < L Pa L. Pa L. L Pab At section B: M = L PROBEM 5.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOUTION Reactions: From A to B: 0 < x < a