PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 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.

SOLUTION

�

�

Reactions:

0: 0CPb

M LA bP AL

! = " = =

0: 0APa

M LC aP CL

! = " = =

From A to B: 0 x a< <

0: 0yPb

F VL

! = " =

PbV

L= W

0: 0JPb

M M xL

! = " =

Pbx

ML

= W�

From B to C: a x L< <

0: 0yPa

F VL

! = + =

PaV

L= " W

0: ( ) 0KPa

M M L xL

! = " + " =

( )Pa L x

ML

"= W

At section B: 2Pab

ML

= W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.3

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.

SOLUTION

From A to B (0 )x a< < :

0 : 0yF wx V¦ = " " =

V wx= " W

0 : ( ) 02Jx

M wx M¦ = + =

2

2wx

M = " W

From B to C ( ) :a x L< <

0 : 0yF wa V¦ = " " = V wa= " W

0 : ( ) 02Ja

M wa x M§ ·¦ = " + =¨ ¸© ¹

2a

M wa x§ ·= " "¨ ¸© ¹

W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 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.

SOLUTION

�

010: 02y

w xF x V

L! = " # " =

2

0

2w x

VL

= " W

010: 0

2 3Jw x x

M x ML

! = # # + =

3

0

6w x

ML

= " W�

At ,x L=

0

2w L

V = " 0max| |

2w L

V = W

2

0

6w L

M = " 2

0max| |

6w L

M = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.7

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION Reactions: 0 : (300)(4) (240)(3) (360)(7) 12 0 170 lbCM B¦ = " " + = = $B

0 : 300 240 360 170 0 730 lbyF C¦ = " + " " + = = $C

From A to C:

0 : 300 0 300 lbyF V V¦ = " " = = "

1 0 : (300)( ) 0 300M x M M x¦ = + = = "

From C to D:

0 : 300 730 0 430 lbyF V V¦ = " + " = = +

2 0 : (300) (730)( 4) 02920 430

M x x M

M x

¦ = " " + == " +

From D to E:

0 : 360 170 0 190 lbyF V V¦ = " + = = +

3 0 : (170)(16 ) (360)(11 ) 01240 190

M x x M

M x

¦ = " " " " == " +

From E to B:

0 : 170 0 170lbyF V V¦ = + = = "

4 0 : (170)(16 ) 02720 170

M x M

M x

¦ = " " == "

� (a) max 430 lbV = W

(b) max 1200 lb inM = # W�

�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.11

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Reactions:

0 : 3 (8)(60) (24)(60) 0A EFM F¦ = " " =

640 kipsEFF =

0 : 640 0 640 kipsx x xF A A¦ = " = = %

0 : 60 60 0 120 kipsy y yF A A¦ = " " = = $

From A to C: (0 8 in.)x< <

0 :yF¦ =

120 0

120 kipsV

V

" ==

0 : 120 0 120 kip inJM M x M x¦ = " = = # From C to D: (8 in. 16 in.)x< <

0 : 120 60 0 60 kipsYF V V¦ = " " = =

0 : 120 60( 8) 0(60 480)kips in

JM M x x

M x

¦ = " + " == + #

From D to B: (16 in. 24 in.)x< <

0 : 60 0 60 kipsyF V V¦ = " = =

0 : 60(24 ) 0(60 1440) kip in

JM M x

M x

¦ = " " " == " #

(a) max 120.0 kipsV = W�

(b) max 1440 kip in 120.0 kip ftM = # = # W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.13

Assuming that the reaction of the ground is uniformly distributed, draw the shear and bending-moment diagrams for the beam AB and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Over the whole beam,

0: 12 (3)(2) 24 (3)(2) 0yF w! = " " " = 3 kips/ftw =

A to C : (0 3 ft)x& <

0: 3 2 0yF x x V! = " " = ( ) kipsV x=

0: (3 ) (2 ) 02 2Jx x

M x x M+! = " + + = 2(0.5 ) kip ftM x= #

At C, 3 ftx =

3 kips, 4.5 kip ftV M= = #

C to D : (3 ft 6 ft)x& <

0: 3 (2)(3) 0yF x V! = " " = (3 6) kipsV x= "

30: (3 ) (2)(3) 02 2x

MK x x M§ · § ·! = " + " + =¨ ¸ ¨ ¸© ¹ © ¹

2(1.5 6 9) kip ftM x x= " + #

At ,D" 6 ftx =

12 kips, 27 kip ftV M= = #

D to B: Use symmetry to evaluate.

(a) max| | 12.00 kipsV = W

(b) max| | 27.0 kip ftM = # W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.15

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Reaction at A:

0: 4.5 (3.0)(3) (1.5)(3) (1.8)(4.5)(2.25) 0BM A= " + + + = 7.05 kN= $A

Use AC as free body.

3

0: (7.05)(1.5) (1.8)(1.5)(0.75) 0

8.55 kN m 8.55 10 N mC C

C

M M

M

! = " + =

= # = ' #

3 3 6 4

6 4

1 1 (80)(300) 180 10 mm12 12180 10 m1 (300) 150 mm 0.150 m2

I bh

c

"

= = = '

= '

= = =

3

66

(8.55 10 )(0.150) 7.125 10 Pa180 10

Mc

I( "

'= = = ''

7.13 MPa( = W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.19

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Use entire beam as free body.

0 :BM¦ =

90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0A" + + + + + =

9.5 kipsA =

Use portion AC as free body.

0 : (15)(9.5) 0142.5 kip in

CM M

M

¦ = " == #

For 38 18.4, 14.4 inS S' =

Normal stress: 142.514.4

M

S( = =

9.90 ksi( = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.44

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Reaction at A:

0: 3.0 (1.5)(3.0)(3.5) (1.5)(3) 0

6.75 kNBM A! = " + + =

= #A

Reaction at B: 6.75 kN= #B

Beam ACB and loading: (See sketch.)

Areas of load diagram:

A to C : (2.4)(3.5) 8.4 kN=

C to B: (0.6)(3.5) 2.1 kN=

Shear diagram:

6.75 kN6.75 8.4 1.65 kN

1.65 3 4.65 kN

4.65 2.1 6.75 kN

A

C

C

B

V

V

V

V

"

+

== " = "

= " " = "

= " " = "

Over A to C, 6.75 3.5V x= "

At G, 6.75 3.5 0 1.9286 mG GV x x= " = =

Areas of shear diagram:

A to G : 1 (1.9286)(6.75) 6.5089 kN m2

= $

G to C : 1 (0.4714)( 1.65) 0.3889 kN m2

" = " $

C to B: 1 (0.6)( 4.65 6.75) 3.42 kN m2

" " = " $

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.44 (Continued)

Bending moments: 00 6.5089 6.5089 kN m6.5089 0.3889 6.12 kN m

6.12 2.7 3.42 kN m

3.42 3.42 0

A

G

C

C

B

M

M

M

M

M

"

+

== + = #= " = #

= " = #

= " =

(a) max| | 6.75 kNV = W

(b) max| | 6.51 kN mM = # W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.45

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

0:3 (1)(4) (0.5)(4) 0

2 kN

BM

A

¦ =" + + =

= $A

0: 3 (2)(4) (2.5)(4) 0

6 kNAM B¦ = " " =

= $B

Shear diagram:

A to C: 2 kNV =

C to D: 2 4 2 kNV = " = "

D to B: 2 4 6 kNV = " " = "

Areas of shear diagram:

A to C: (1)(2) 2 kN mVdx = = #³

C to D: (1)( 2) 2 kN mVdx = " = " #³

D to E: (1)( 6) 6 kN mVdx = " = " #³

Bending moments:

0AM =

0 2 2 kN m

2 4 6 kN m

6 2 4 kN m

4 2 6kN m

6 6 0

C

C

D

D

B

M

M

M

M

M

"

+

"

+

= + = #

= + = #

= " = #

= + = #

= " =

(a) max 6.00 kNV = W�

(b) max 6.00 kN mM = # W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.50

For the beam and loading shown, determine the equations of the shear and bending-moment curves, and the maximum absolute value of the bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.

SOLUTION

0 0 0

00

20

0 1

12

00

( )(1 ) .

(1 )

(1 )2

0 at 0 0

(1 )2

w x kw L x w xw k kw

L L Lw xdV

w kw kdx L

w xV kw x k C

LV x C

w xdMV kw x k

d x L

"= " = + "

= " = " +

= " + +

= = =

= = " +

2 30 0

2

22 3

0 0

(1 )2 6

0 at 0 0

(1 )2 6

kw x w xM k C

LM x C

kw x k w xM

L

= " + +

= = =

+= "

(a) 1.k = 2

00

w xV w x

L= " W

2 3

0 0

2 3w x w x

ML

= " W

Maximum M occurs at .x L= 2

0max 6

w LM = W

(b) 1 .2

k = 2

0 032 4

w x w xV

L= " W

2 3

0 0

4 4w x w x

ML

= " W�

�20 at3

V x L= =

At ( ) ( )2 32 2 2

0 03 3 200

2 , 0.03704 3 4 4 27

w L w L w Lx L M w L

L= = " = =

At , 0x L M= =

2

0max| |

27w L

M = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.55

Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending.

SOLUTION

0 : (2)(1) (3)(4)(2) 4 05.5 kN

CM B

B

¦ = " + ==

0 : (5)(2) (3)(4)(2) 4 08.5 kN

BM C

C

¦ = + " ==

Shear:

A to C: 2 kNV = "

:C+ 2 8.5 6.5 kNV = " + =

B: 6.5 (3)(4) 5.5 kNV = " = "

Locate point D where 0.V =

4 12 266.5 5.5d d

d"= =

2.1667 m 4 3.8333 md d= " =

Areas of the shear diagram:

A to C: ( 2.0)(1) 2.0 kN mVdx³ = " = " #

C to D: 1 (2.16667)(6.5) 7.0417 kN m2

Vdx³ = = #

D to B: 1 (3.83333)( 5.5) 5.0417 kN m2

Vdx³ = " = " #

Bending moments: 0AM =

0 2.0 2.0 kN mCM = " = " #

2.0 7.0417 5.0417 kN mDM = " + = #

5.0417 5.0417 0BM = " =

Maximum 35.0417 kN m 5.0417 10 N mM = # = ' #

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.55 (Continued)

For pipe: 1 1 1 1(160) 80 mm, (140) 70 mm2 2 2 2o o i ic d c d= = = = = =

( )4 4 4 4 6 4(80) (70) 13.3125 10 mm4 4o iI c c) ) ª º= " = " = '¬ ¼

6

3 3 6 313.3125 10 166.406 10 mm 166.406 10 m80o

IS

c"'= = = ' = '

Normal stress: 3

66

5.0417 10 30.3 10 Pa166.406 10

M

S( "

'= = = ''

30.3 MPa( = W

�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 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 1750 psi.

SOLUTION

Reactions: By symmetry, A = D.

1 10 : (3)(1.5) (6)(1.5) (3)(1.5) 02 2

6.75 kips

yF A D

A D

¦ = " " " " =

= = $

Shear diagram: 6.75 kipsAV =

16.75 (3)(1.5) 4.5 kips2BV = " =

4.5 (6)(1.5) 4.5 kipsCV = " = "

14.5 (3)(1.5) 6.75 kips2DV = " " = "

Locate point E where 0V = :

By symmetry, E is the midpoint of BC.

Areas of the shear diagram:

2to : (3)(4.5) (3)(2.25) 18 kip ft3

A B + = #

1to : (3)(4.5) 6.75 kip ft2

B E = #

1to : (3)( 4.5) 6.75 kip ft2

E C " = " #

to : Byantisymmetry, 18 kip ftC D " #

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.67 (Continued)

Bending moments: 0AM =

0 18 18 kip ftBM = + = #

18 6.75 24.75 kip ftEM = + = #

24.75 6.75 18 kip ftCM = " = #

18 18 0DM = " =

3max maxmax

max

(24.75 kip ft)(12 in/ft) 169.714 in1.750 ksi

M MS

S(

(#= = = =

For a rectangular section, 216

S bh=

6 6(169.714) 14.27 in.5

Sh

b= = = 14.27 in.h = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.73

Knowing that the allowable stress for the steel used is 160 MPa, select the most economical wide-flange beam to support the loading shown.

SOLUTION

21

1

2

2 32

2

2 3

18 66 (6 2 ) kN/m6

6 2

60 at 0, 0

6

133

0 at 0, 0133

w x x

dVw x

dx

V x x C

V x C

dMV x x

dx

M x x C

M x C

M x x

"§ ·= + = +¨ ¸© ¹

= " = " "

= " " += = =

= = " "

= " " +

= = =

= " "

max occurs at 6 m.M x =

2 3 3max

1= (3)(6) (6) 80 kN m 180 10 N m3

M§ ·" " = # = ' #¨ ¸© ¹

6all 160 MPa 160 10 Pa( = = '

3

3 3 3 3min 6

all

180 10 1.125 10 m 1125 10 mm160 10

MS

("'= = = ' = '

'

Shape S, ( 3 310 mm )

W530 66' 1340 * W460 74' 1460 Lightest acceptable wide flange beam: W530 66' W

W410 85' 1510 W360 79' 1270 W310 107' 1600

W250 101' 1240

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.74

Knowing that the allowable stress for the steel used is 160 MPa, select the most economical wide-flange beam to support the loading shown.

SOLUTION

all

Section modulus160 Mpa! =

6 3max

minall

3 3

286 kN m 1787 10 m160 MPa

1787 10 mm

MS

!"#= = = $

= $

Use W530 92$ W

Shape S, ( 3 310 mm )

W610 101$ 2520

W530 92$ 2080 % W

W460 113$ 2390

W410 114$ 2200

W360 122$ 2020

W310 143$ 2150

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 5.89

A 54-kip is load is to be supported at the center of the 16-ft span shown. Knowing that the allowable normal stress for the steel used is 24 ksi, determine (a) the smallest allowable length l of beam CD if the W12 50! beam AB is not to be overstressed, (b) the most economical W shape that can be used for beam CD. Neglect the weight of both beams.

SOLUTION

(a)

8ft 16ft 22l

d l d= " = " (1)

Beam AB (Portion AC):

For W12 50,! 364.2 inxS = all 24 ksi# =

all all (24)(64.2) 1540.8 kip in 128.4 kip ft27 128.4 kip ft 4.7556 ft

x

C

M S

M d d

#= = = $ = $= = $ =

Using (1), 16 2 16 2(4.7556) 6.4888 ftl d= " = " = 6.49 ftl = W �

(b) Beam CD: all6.4888 ft 24 ksil #= =

max

minall

3

(87.599 12) kip in24 ksi

43.800 in

MS

#! $= =

=

Shape 3(in )S

W18 35! 57.6 W16 31! 47.2 ĸ W14 38! 54.6 W12 35! 45.6 W10 45! 49.1

W16 31.! W�