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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual 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:

0:0

C

P

b

M LA bP A

L

!

=

"

= =

0:0

A

P

a

M LC aP C

L

!

=

"

= =

From

A

to

B

:

0

x

a

< <

0:0

y

Pb

F V

L

!

=

"

=

P

b

V

L

=

W

0:0

J

Pb

M M x

L

!

=

"

=

P

bx

M

L

=

W

From

B

to

C

:

a x L

< <

0:0

y

Pa

F V

L

!

= + =

P

a

V

L

=

"

W

0:( ) 0

K

Pa

M M L x

L

!

=

"

+

"

=

( )

P

a L x

M

L

"

=

W

At section

B

:

2

P

ab

M

L

=

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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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

< <

:

0:0

y

F wx V

¦

=

""

=

V wx

=

"

W

0:( ) 0

2

J

x

M wx M

¦

= + =

2

2

wx

M

=

"

W

From

B

to

C

( ):

a x L

< <

0:0

y

F wa V

¦

=

""

=

V wa

=

"

W

0:( ) 0

2

J

a

M wa x M

§ ·

¦

=

"

+ =

¨ ¸

© ¹

2

a

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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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

0

1

0:0

2

y

w x

F x V

L

!

=

"#"

=

2

0

2

w x

V

L

=

"

W

0

1

0:0

2 3

J

w x x

M x M

L

!

=

##

+ =

3

0

6

w x

M

L

=

"

W

At

,

x

L

=

0

2

w L

V

=

"

0

max

| |

2

w L

V

=

W

2

0

6

w L

M

=

"

2

0

max

| |

6

w 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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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 lb

C

M B

¦

=

""

+ = =

$

B

0:300 240 360 170 0 730 lb

y

F C

¦

=

"

+

""

+ = =

$

C

From

A

to

C

:

0:300 0 300 lb

y

F V V

¦

=

""

= =

"

1

0:(300)( ) 0 300

M

x M M x

¦

= + = =

"

From

C

to

D

:

0:300 730 0 430 lb

y

F V V

¦

=

"

+

"

= = +

2

0:(300) (730)( 4) 0

2920 430

M x x M

M

x

¦

=

""

+ =

=

"

+

From

D

to

E

:

0:360 170 0 190 lb

y

F V V

¦

=

"

+ = = +

3

0:(170)(16 ) (360)(11 ) 0

1240 190

M x x M

M

x

¦

=

""""

=

=

"

+

From

E

to

B

:

0:170 0 170lb

y

F V V

¦

= + = =

"

4

0:(170)(16 ) 0

2720 170

M x M

M

x

¦

=

""

=

=

"

(

a

)

max

430 lb

V

=

W

(

b

)

max

1200 lb in

M

=

#

W

PROPRIETARY MATERIAL.

© 2012 The McGraw-Hill Companies, Inc.

All rights reserved. No part of this Manual may be displayed,

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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) 0

A EF

M F

¦

=

""

=

640 kips

EF

F

=

0:640 0 640 kips

x x x

F A A

¦

=

"

= =

%

0:60 60 0 120 kips

y y y

F A A

¦

=

""

= =

$

From

A

to

C

:

(0 8 in.)

x

< <

0:

y

F

¦

=

120 0

120 kips

V

V

"

=

=

0:120 0 120 kip in

J

M M x M x

¦

=

"

= =

#

From

C

to

D

:

(8 in.16 in.)

x

< <

0:120 60 0 60 kips

Y

F V V

¦

=

""

= =

0:120 60( 8) 0

(60 480) kips in

J

M M x x

M x

¦

=

"

+

"

=

= +

#

From

D

to

B

:

(16 in.24 in.)

x

< <

0:60 0 60 kips

y

F V V

¦

=

"

= =

0:60(24 ) 0

(60 1440) kip in

J

M M x

M x

¦

=

"""

=

=

"#

(

a

)

max

120.0 kips

V

=

W

(

b

)

max

1440 kip in 120.0 kip ft

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 t

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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) 0

y

F w

!

=

"""

=

3 kips/ft

w

=

A

to

C

:

(0 3 ft)

x

&

<

0:3 2 0

y

F x x V

!

=

""

=

( ) kips

V x

=

0:(3 ) (2 ) 0

2 2

J

x x

M x x M

+

!

=

"

+ + =

2

(0.5 ) kip ft

M x

=

#

At

C

,

3 ft

x

=

3 kips,4.5 kip ft

V M

= =

#

C

to

D

:

(3 ft 6 ft)

x

&

<

0:3 (2)(3) 0

y

F x V

!

=

""

=

(3 6) kips

V x

=

"

3

0:(3 ) (2)(3) 0

2 2

x

MK x x M

§ · § ·

!

=

"

+

"

+ =

¨ ¸ ¨ ¸

© ¹ © ¹

2

(1.5 6 9) kip ft

M x x

=

"

+

#

At

,

D

"

6 ft

x

=

12 kips,27 kip ft

V M

= =

#

D

to

B

: Use symmetry to evaluate.

(

a

)

max

| | 12.00 kips

V

=

W

(

b

)

max

| | 27.0 kip ft

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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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) 0

B

M 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 m

C C

C

M M

M

!

=

"

+ =

=

#

=

'#

3 3 6 4

6 4

1 1

(80)(300) 180 10 mm

12 12

180 10 m

1

(300) 150 mm 0.150 m

2

I bh

c

"

= = =

'

=

'

= = =

3

6

6

(8.55 10 )(0.150)

7.125 10 Pa

180 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,

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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:

B

M

¦

=

90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0

A

"

+ + + + + =

9.5 kips

A

=

Use portion AC as free body.

0:(15)(9.5) 0

142.5 kip in

C

M M

M

¦

=

"

=

=

#

For

3

8 18.4,14.4 in

S S

'

=

Normal stress:

142.5

14.4

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,

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he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual 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 kN

B

M 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 kN

6.75 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 3.5

V x

=

"

At

G

,

6.75 3.5 0 1.9286 m

G G

V x x

=

"

= =

Areas of shear diagram:

A

to

G

:

1

(1.9286)(6.75) 6.5089 kN m

2

=

$

G

to

C

:

1

(0.4714)( 1.65) 0.3889 kN m

2

"

=

"$

C

to

B

:

1

(0.6)( 4.65 6.75) 3.42 kN m

2

""

=

"$

PROPRIETARY MATERIAL.

© 2012 The McGraw-Hill Companies, Inc.

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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PROBLEM 5.44

(Continued)

Bending moments:

0

0 6.5089 6.5089 kN m

6.5089 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 kN

V

=

W

(

b

)

max

| | 6.51 kN m

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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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

B

M

A

¦

=

"

+ + =

=

$

A

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

6 kN

A

M B

¦

=

""

=

=

$

B

Shear diagram:

A

to

C

:

2 kN

V

=

C

to

D

:

2 4 2 kN

V

=

"

=

"

D

to

B

:

2 4 6 kN

V

=

""

=

"

Areas of shear diagram:

A

to

C

:

(1)(2) 2 kN m

Vdx

= =

#

³

C

to

D

:

(1)( 2) 2 kN m

Vdx

=

"

=

"#

³

D

to

E

:

(1)( 6) 6 kN m

Vdx

=

"

=

"#

³

Bending moments:

0

A

M

=

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 kN

V

=

W

(

b

)

max

6.00 kN m

M

=

#

W

PROPRIETARY MATERIAL.

© 2012 The McGraw-Hill Companies, Inc.

All rights reserved. No part of this Manual may be displayed,

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he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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

0

0

2

0

0 1

1

2

0

0

( )

(1 ).

(1 )

(1 )

2

0 at 0 0

(1 )

2

w x kw L x w x

w k kw

L L L

w x

dV

w kw k

dx L

w x

V kw x k C

L

V x C

w x

dM

V kw x k

d x L

"

=

"

= +

"

=

"

=

"

+

=

"

+ +

= = =

= =

"

+

2 3

0 0

2

2

2 3

0 0

(1 )

2 6

0 at 0 0

(1 )

2 6

kw x w x

M

k C

L

M x C

kw x k w x

M

L

=

"

+ +

= = =

+

=

"

(

a

)

1.

k

=

2

0

0

w x

V w x

L

=

"

W

2 3

0 0

2 3

w x w x

M

L

=

"

W

Maximum

M

occurs at

.

x

L

=

2

0

max

6

w L

M

=

W

(

b

)

1

.

2

k

=

2

0 0

3

2 4

w x w x

V

L

=

"

W

2 3

0 0

4 4

w x w x

M

L

=

"

W

2

0 at

3

V x L

= =

At

( )

( )

2 3

2 2

2

0 0

3 3

2

0

0

2

,0.03704

3 4 4 27

w L w L

w L

x

L M w L

L

= =

"

= =

At

,0

x L M

= =

2

0

max

| |

27

w L

M

=

W

PROPRIETARY MATERIAL.

© 2012 The McGraw-Hill Companies, Inc.

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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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 0

5.5 kN

C

M B

B

¦

=

"

+ =

=

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

8.5 kN

B

M C

C

¦

= +

"

=

=

Shear:

A

to

C

:

2 kN

V

=

"

:

C

+

2 8.5 6.5 kN

V

=

"

+ =

B

:

6.5 (3)(4) 5.5 kN

V

=

"

=

"

Locate point

D

where

0.

V

=

4

12 26

6.5 5.5

d d

d

"

= =

2.1667m 4 3.8333 m

d d

=

"

=

Areas of the shear diagram:

A

to

C:

( 2.0)(1) 2.0 kN m

Vdx

³

=

"

=

"#

C

to

D

:

1

(2.16667)(6.5) 7.0417 kN m

2

Vdx

³

= =

#

D

to

B

:

1

(3.83333)( 5.5) 5.0417 kN m

2

Vdx

³

=

"

=

"#

Bending moments:

0

A

M

=

0 2.0 2.0 kN m

C

M

=

"

=

"#

2.0 7.0417 5.0417 kN m

D

M

=

"

+ =

#

5.0417 5.0417 0

B

M

=

"

=

Maximum

3

5.0417 kN m 5.0417 10 N m

M

=

#

=

'#

PROPRIETARY MATERIAL.

© 2012 The McGraw-Hill Companies, Inc.

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he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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PROBLEM 5.55

(Continued)

For pipe:

1 1 1 1

(160) 80 mm,(140) 70 mm

2 2 2 2

o o i i

c d c d

= = = = = =

( )

4 4 4 4 6 4

(80) (70) 13.3125 10 mm

4 4

o i

I c c

) )

ª º

=

"

=

"

=

'

¬ ¼

6

3 3 6 3

13.3125 10

166.406 10 mm 166.406 10 m

80

o

I

S

c

"

'

= = =

'

=

'

Normal stress:

3

6

6

5.0417 10

30.3 10 Pa

166.406 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,

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he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

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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 1

0:(3)(1.5) (6)(1.5) (3)(1.5) 0

2 2

6.75 kips

y

F A D

A D

¦

=

""""

=

= =

$

Shear diagram:

6.75 kips

A

V

=

1

6.75 (3)(1.5) 4.5 kips

2

B

V

=

"

=

4.5 (6)(1.5) 4.5 kips

C

V

=

"

=

"

1

4.5 (3)(1.5) 6.75 kips

2

D

V

=

""

=

"

Locate point

E

where

0

V

=

:

By symmetry,

E

is the midpoint of

BC

.

Areas of the shear diagram:

2

to:(3)(4.5) (3)(2.25) 18 kip ft

3

A B

+ =

#

1

to:(3)(4.5) 6.75 kip ft

2

B E

=

#

1

to:(3)( 4.5) 6.75 kip ft

2

E C

"

=

"#

to:Byantisymmetry,18 kip ft

C 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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual is using it

without permission.

PROBLEM 5.67

(Continued)

Bending moments:

0

A

M

=

0 18 18 kip ft

B

M

= + =

#

18 6.75 24.75 kip ft

E

M

= + =

#

24.75 6.75 18 kip ft

C

M

=

"

=

#

18 18 0

D

M

=

"

=

3

max max

max

max

(24.75 kip ft)(12 in/ft)

169.714 in

1.750 ksi

M M

S

S

(

(

#

= = = =

For a rectangular section,

2

1

6

S bh

=

6 6(169.714)

14.27 in.

5

S

h

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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual 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

2

1

1

2

2 3

2

2

2 3

18 6

6 (6 2 ) kN/m

6

6 2

6

0 at 0,0

6

1

3

3

0 at 0,0

1

3

3

w x x

dV

w x

dx

V x x C

V x C

dM

V x x

dx

M x x C

M x C

M x x

"

§ ·

= + = +

¨ ¸

© ¹

=

"

=

""

=

""

+

= = =

= =

""

=

""

+

= = =

=

""

max

occurs at 6 m.

M x

=

2 3 3

max

1

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

3

M

§ ·

""

=

#

=

'#

¨ ¸

© ¹

6

all

160 MPa 160 10 Pa

(

= =

'

3

3 3 3 3

min

6

all

180 10

1.125 10 m 1125 10 mm

160 10

M

S

(

"

'

= = =

'

=

'

'

Shape

S

, (

3 3

10 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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual 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

Sectionmodulus

160 Mpa

!

=

6 3

max

min

all

3 3

286 kN m

1787 10 m

160 MPa

1787 10 mm

M

S

!

"

#

= = =

$

=

$

Use

W530 92

$

W

Shape

S

, (

3 3

10 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 t

he limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this m

anual 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 2

2

l

d l d

=

"

=

"

(1)

Beam

AB

(Portion

AC

):

For

W12 50,

!

3

64.2 in

x

S

=

all

24 ksi

#

=

all all

(24)(64.2) 1540.8 kip in 128.4 kip ft

27 128.4 kip ft 4.7556 ft

x

C

M S

M d d

#

= = =

$

=

$

= =

$

=

Using (1),

16 2 16 2(4.7556) 6.4888 ft

l d

=

"

=

"

=

6.49 ft

l

=

W

(

b

)

Beam

CD

:

all

6.4888 ft 24 ksi

l

#

= =

max

min

all

3

(87.599 12) kip in

24 ksi

43.800 in

M

S

#

!$

= =

=

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

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