PROBE 5. 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. Reactions: From A to B: 0 < x < a Pb Σ = 0: A bp = 0 A = Pa Σ A = 0: ap = 0 = Pb Σ F = 0: = 0 Pb Σ J = 0: x = 0 = Pb = Pbx From B to : a < x < Pa Σ F = 0: + = 0 Pa Σ K = 0: + ( x) = 0 Pa = Pa( x) = Pab At section B: = PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5. For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bending-moment curves. From A to B (0 < x < a) : F = 0: wx = 0 x J = 0: ( wx) + = 0 From B to ( a < x < ) : = wx wx = F = 0: wa = 0 = wa a a J = 0: ( wa) x + = 0 = wax PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.4 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. At x =, wx 0 Σ F = 0: x = 0 wx 0 x Σ J = 0: x + = 0 wx 0 = wx 0 = 6 w w = 0 0 = 0 6 w = w 0 = 6 PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.7 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the imum absolute value (a) of the shear, (b) of the bending moment. Reactions: = 0: (00)(4) (40)() (60)(7) + B = 0 B = 70lb F = 0: 00+ 40 60+ 70= 0 = 70lb From A to : F = 0: 00 = 0 = 00lb = 0: (00)( x) + = 0 = 00x From to D: F = 0: 00+ 70 = 0 = + 40lb = 0: (00) x (70)( x 4) + = 0 From D to E: = 90 + 40x F = 0: 60+ 70= 0 = + 90lb = 0: (70)(6 x) (60)( x) = 0 From E to B: = 40 + 90x F = 0: + 70= 0 = 70lb 4 = 0: (70)(6 x) = 0 = 70 70x (a) = 40 lb (b) = 00 lb in PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5. Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the imum absolute value (a) of the shear, (b) of the bending moment. Reactions: = 0: F (8)(60) (4)(60) = 0 A EF F EF = 640 kips F = 0: A 640= 0 A = 640 kips x x x F = 0: A 60 60= 0 A = 0 kips From A to : (0 < x < 8 in.) F = 0: 0 = 0 = 0 kips = 0: 0x = 0 = 0 x kip in J From to D: (8 in. < x < 6 in.) F = 0: 0 60 = 0 = 60 kips Y = 0: 0x + 60( x 8) = 0 J From D to B: (6 in. < x < 4 in.) F = 0: 60= 0 = 60 kips = 0: 60(4 x) = 0 J = (60x + 480)kips in = (60x 440) kip in (a) = 0.0 kips (b) = 440 kip in = 0.0 kip ft PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5. Assuming that the reaction of the ground is uniforml distributed, draw the shear and bending-moment diagrams for the beam AB and determine the imum absolute value (a) of the shear, (b) of the bending moment. Over the whole beam, A to : (0 x < ft) Σ F = 0: w ()() 4 ()() = 0 w = kips/ft Σ F = 0: x x = 0 = ( x) kips At, x x +Σ J = 0: ( x) + ( x) + = 0 x = ft = kips, = 4.5 kip ft = (0.5 x ) kip ft to D: ( ft x < 6 ft) Σ F = 0: x ()() = 0 = (x 6) kips x Σ K = 0: ( x) + ()() x + = 0 = (.5x 6x + 9) kip ft At D, x = 6 ft = kips, = 7 kip ft D to B: Use smmetr to evaluate. (a) (b) =.00 kips = 7.0 kip ft PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.5 For the beam and loading shown, determine the imum normal stress due to bending on a transverse section at. Reaction at A: Use A as free bod. B = 0: 4.5 A + (.0)() + (.5)() + (.8)(4.5)(.5) = 0 A = 7.05 kn Σ = 0: (7.05)(.5) + (.8)(.5)(0.75) = 0 = 8.55 kn m = 8.55 0 N m I c = bh = (80)(00) = 80 0 mm 6 4 6 4 = 80 0 m = (00) = 50 mm = 0.50 m c (8.55 0 )(0.50) 6 σ = = = 7.5 0 Pa 6 I 80 0 σ = 7. Pa PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.9 For the beam and loading shown, determine the imum normal stress due to bending on a transverse section at. Use entire beam as free bod. B = 0: 90 A + (75)(5) + (60)(5) + (45)() + (0)() + (5)() = 0 A = 9.5 kips Use portion A as free bod. = 0: (5)(9.5) = 0 = 4.5 kip in For S8 8.4, S = 4.4 in Normal stress: σ = = S 4.5 4.4 σ = 9.90 ksi PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.44 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the imum absolute value (a) of the shear, (b) of the bending moment. Reaction at A: Reaction at B: Σ = 0:.0 A + (.5)(.0)(.5) + (.5)() = 0 B A = 6.75 kn B = 6.75 kn Beam AB and loading: (See sketch.) Areas of load diagram: A to : (.4)(.5) = 8.4 kn to B: (0.6)(.5) =. kn Shear diagram: A + B = 6.75 kn Over A to, = 6.75.5x = 6.75 8.4 =.65 kn =.65 = 4.65 kn = 4.65. = 6.75 kn At G, = 6.75.5x = 0 x =.986 m Areas of shear diagram: G G A to G: G to : to B: (.986)(6.75) = 6.5089 kn m (0.474)(.65) = 0.889 kn m (0.6)( 4.65 6.75) =.4 kn m PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.44 (ontinued) Bending moments: = 0 A G + B = 0 + 6.5089 = 6.5089 kn m = 6.5089 0.889 = 6. kn m = 6..7 =.4 kn m =.4.4 = 0 (a) (b) = 6.75 kn = 6.5 kn m PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.45 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the imum absolute value (a) of the shear, (b) of the bending moment. = 0: B A + ()(4) + (0.5)(4) = 0 A = kn = 0: B ()(4) (.5)(4) = 0 A Shear diagram: A to : = kn to D: = 4 = kn D to B: = 4 = 6kN Areas of shear diagram: B = 6kN A to : dx = ()() = kn m to D: dx = ()( ) = kn m D to E: dx = ()( 6) = 6 kn m Bending moments: A = + D D + B 0 = 0 + = kn m = + 4 = 6kN m = 6 = 4kN m = 4 + = 6kN m = 6 6 = 0 (a) (b) = 6.00 kn = 6.00 kn m PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.50 For the beam and loading shown, determine the equations of the shear and bending-moment curves, and the imum absolute value of the bending moment in the beam, knowing that (a) k =, (b) k = 0.5. (a) k =. wx 0 kw0( x) wx 0 w= = ( + k) kw. d wx 0 = w= kw0 ( + k) dx wx 0 0 = kw x ( + k) + = 0 at x= 0 = 0 d wx 0 = = kw 0 x ( + k ) dx 0 w0x kw x = ( + k) + 6 = 0 at x = 0 = 0 0 ( + k) w0x kw x = 6 aximum occurs at x =. wx 0 = w0 x wx 0 wx 0 = w 0 6 = (b) k =. At = 0 at x = ( ) ( ) 0 0 0 0 w w w x =, = = = 0.0704 w 4 4 7 At x =, = 0 wx 0 wx 0 = 4 0 wx 0 wx = 4 4 w 0 = 7 PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.55 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the imum normal stress due to bending. = 0:()() ()(4)() + 4B = 0 B = 5.5 kn = 0:(5)() + ()(4)() 4 = 0 B = 8.5 kn Shear: A to : = kn + : = + 8.5 = 6.5kN B: = 6.5 ()(4) = 5.5 kn ocate point D where = 0. Areas of the shear diagram: d 4 d = d = 6 6.5 5.5 d =.667 m 4 d =.8 m A to : dx = (.0)() =.0kN m to D: dx = (.6667)(6.5) = 7.047 kn m D to B: Bending moments: A = 0 dx = (.8)( 5.5) = 5.047 kn m = 0.0 =.0kN m =.0 + 7.047 = 5.047 kn m D = 5.047 5.047 = 0 B aximum = 5.047 kn m = 5.047 0 N m PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.55 (ontinued) For pipe: co = do = (60) = 80 mm, ci = di = (40) = 70 mm ( o i ) π π I = c c = (80) (70) =.5 0 mm 4 4 4 4 4 4 6 4 S 6 I.5 0 = = = 66.406 0 mm = 66.406 0 m c 80 o 6 Normal stress: 5.047 0 6 σ = = = 0. 0 Pa 6 S 66.406 0 σ = 0. Pa PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.67 For the beam and loading shown, design the cross section of the beam, knowing that the grade of timber used has an allowable normal stress of 750 psi. Reactions: B smmetr, A = D. F = 0: A ()(.5) (6)(.5) ()(.5) D = 0 Shear diagram: A = D = 6.75 kips A = 6.75 kips B = 6.75 ()(.5) = 4.5 kips = 4.5 (6)(.5) = 4.5 kips D = 4.5 ()(.5) = 6.75 kips ocate point E where = 0 : B smmetr, E is the midpoint of B. Areas of the shear diagram: A to B : ()(4.5) + ()(.5) = 8 kip ft B to E : ()(4.5) = 6.75 kip ft E to : ()( 4.5) = 6.75 kip ft to D : B antismmetr, 8 kip ft PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.67 (ontinued) Bending moments: A = 0 = 0 + 8 = 8 kip ft B = 8 + 6.75 = 4.75 kip ft E = 4.75 6.75 = 8 kip ft = 8 8 = 0 D σ (4.75 kip ft)( in/ft) = S = = = 69.74 in S σ.750 ksi For a rectangular section, S = bh 6 6S 6(69.74) h = = = 4.7 in. h = 4.7 in. b 5 PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.7 Knowing that the allowable stress for the steel used is 60 Pa, select the most economical wide-flange beam to support the loading shown. Shape S, ( 0 mm ) W50 66 40 8 6 w = 6 + x = (6 + x)kn/m 6 d = w = 6 x dx = 6x x + = 0 at x = 0, = 0 d = = 6 x x dx = x x + = 0 at x = 0, = 0 = x x occurs at x = 6 m. = ()(6) (6) = 80kN m = 80 0 N m 6 σ all = 60 Pa = 60 0 Pa S 80 0 = = =.5 0 m = 5 0 mm min 6 σ all 60 0 W460 74 460 ightest acceptable wide flange beam: W50 66 W40 85 50 W60 79 70 W0 07 600 W50 0 40 PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.74 Knowing that the allowable stress for the steel used is 60 Pa, select the most economical wide-flange beam to support the loading shown. S min Section modulus σ = 60 pa all 86 kn m = = = 787 0 m σ 60 Pa all 6 = 787 0 mm Shape S, ( 0 mm ) W60 0 50 W50 9 080 W460 90 W40 4 00 W60 00 W0 4 50 Use W50 9 PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,
PROBE 5.89 A 54-kip is load is to be supported at the center of the 6-ft span shown. Knowing that the allowable normal stress for the steel used is 4 ksi, determine (a) the smallest allowable length l of beam D if the W 50 beam AB is not to be overstressed, (b) the most economical W shape that can be used for beam D. Neglect the weight of both beams. (a) l d = 8ft l = 6ft d () Beam AB (Portion A): For W 50, S = 64. in σ all = 4 ksi x all = σ all Sx = (4)(64.) = 540.8 kip in = 8.4 kip ft = 7d = 8.4 kip ft d = 4.7556 ft Using (), l = 6 d = 6 (4.7556) = 6.4888 ft l = 6.49 ft (b) Beam D: S min = = σ all l = 6.4888 ft σ = 4 ksi all (87.599 ) kip in 4 ksi = 4.800 in Shape S (in ) W8 5 57.6 W6 47. W4 8 54.6 W 5 45.6 W0 45 49. W6. PROPRIETARY ATERIA. 0 The cgraw-hill ompanies, Inc. All rights reserved. No part of this anual ma be displaed,