5-5

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For 4 ft < x < 13 ft

2

0 ( ) 10 5

( ) 5 10 kips

(13) 55 kips

( ) 52

5( ) +

Σ

Σ 0

10( 4) 40 kip2

(13) 332.5 ki

10( 4)

ft

ftp

y

C

C

z

V x x

V x x

V VxM x x

M x x x

M M

F

M x

= =- + -

=- +

= =-æ ö÷ç- - ÷ç ÷ç ÷è ø

=

= = + -

- -

= =- ⋅

- ⋅

P5.4. Write the equations for shear V and

moment M between points B and C. Take the

origin at point A. Evaluate V and M at point C

using the equations. B

AC D

w = 5 kips/ft

P = 10 kips

4ʹ9ʹ4ʹ

P5.4

5 kips/ft

B

x

10 kips4 ft

AM(x)

V(x)

z

5-23

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FBD “AB”

2

2

/1 2

/1

5 (20 )Σ 0; 12 (20 ) (20 ) 0

2

62

Σ 0; (20 ) 62 12 0

50

kk

A B

kB

k k ky y

ky

M V

V

F A F

A

+

+

¢ ¢ ¢= + - =

=

¢ = - + - =

=

FBD “BCD”

/1 25 (30 )Σ 0; 62 (30 ) (24 ) 0

2

171.25

Σ 0; 62 5 (30 ) 171.25 0

40.75

kk

D y

ky

k k ky y

ky

M C

C

F C

D

+

+

¢¢ ¢= - - + =

= ¢ = - - + + =

=

P5.21. For each beam, draw the shear and

moment curves label the maximum values

of shear and moment, locate points of inflection,

and sketch the deflected shape.

P = 12 kips

w = 5 kips/ft

B C DA

20ʹ 24ʹ6ʹ

hinge

P5.21

5-49

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P5.47. For the frame in Figure P5.47, draw the

shear and moment curves for all members. Next

sketch the deflected shape of the frame. Show

all forces acting on a free-body diagram of

joint C.

A

B

D

C E

20 kips

w = 5 kips/ft

4ʹ

hinge

4ʹ

4ʹ

6ʹ

6ʹ

P5.47

M Diagrams Deflected Share

Joint C