KEY

P12.36. Draw the influence lines for the vertical and horizontal reactions, AX and AY, at support A B and the bar forces in members AD, CD, and BC. If the truss is loaded by a uniform dead load of 4 kips/ft over the entire length of the top chord, determine the magnitude of the bar forces in members AD and CD.

hinge

C

D

E

A

G

4 @ 20 = 80

P12.36

F 20

r

KEY IS

ZFM

?- otto

M?0

AT T

' Ho

'

Ay

F

DG

EM A = o

fi 160) - Fps (407=0

Fog = E Z

FBC

= FCD

=

wA

=

4

k

/

ft

?????

1 2

?????

(1)20

= 40 kips

t EF?=o F%

Ax t?Fpg=o

FA*D

=

1 2

(40)??????-2 2

??????? 4

kips/ ft

= -40 2 kips

-

IAx=/

Note* Positive and negative areas on ends cancel out.

T EFy=o

Ah

faffs Ay - It

-

/Ay=h5I 12-39

Copyright ? 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

P7.1. Derive the equations for slope and

deflection for the beam in Figure P7.1. Compare

the deflection at B with the deflection at

w

midspan.

A

B

L

P7.1

Analysis by Double Integration

w(L - x)2 2

=

M

d2y dx 2

=

M EI

=

-

w 2EI

(L2

-2xL +

x2)

2EI

d2y dx2

=

-lL2

+

2wxL

-

wx 2

2EI

dy dx

= -wL2 x + 2wL

x2 8

- wx3 3

+ C1?0

=0

(1)

At

x = 0,

dy dx

=

0

\

C1

=

0

2

EIy

=

-

wL2 2

x

2

+ wLx3 3

- wx4 12

+ C2?0

=0

(2)

At x = 0, y = 0 \ C2 = 0

Compute

D B

;

Set x = L

in Eq(2)

D B

=

y

=

1 2EI

??????-

wL4 2

+

wL4 3

-

wL4 12

??????

D = - wL4

B

BEI

Compute B ; Set x = L in Eq1

B

=

1 2EI

??????-wL3

+ wL3

-

wL3 3

??????

B

=

wL3 2EI

?????-1+1- 13????? =

- wL3 6EI

B

= - wL3 6EI

D at

L 2

;

Set x =

L 2

in Eq2

D = -17wL4 = - wL4

384EI

22.59

D Compare D

at at

L

2 B

=

1 22.59

1

=

0.35

8

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P7.3. Derive the equations for slope and

deflection for the beam in Figure P7.3. Compute

the maximum deflection. Hint: Maximum

M

deflection occurs at point of zero slope.

M 2

A

B

L

P7.3

+

Compute RA : MB = 0

0

=

M

-

M 2

-

RA

L

RA

=

M 2L

Ecaluate Mx : M

Mx = M - RAx

=

M

-

M 2L

x

d2y dx 2

=

Mx EI

-?????M

-

Mx 2L

?????

1 EI

(1)

EI

dy dx

=

Mx

-

Mx 2 2L 2

+ C1

(2)

EIy

=

Mx 2 2

-

Mx3 12L

+ C1x

+ C2

(3)

Substi y = 0 @ x = 0 in Eq (3) 0 = 0 - 0 + 0 + C2

\ C2 = 0

Substi y = 0 @ x - L in Eq (3)

0

=

ML2 2

-

ML2 12

+ C1L

C1 = -152 ML

EI

dy dx

=

Mx

-

Mx 2 4L

-5 12

ML

(2a)

EIy

=

Mx 2 2

-

Mx3 12L

-

5 12

MLx

(3a)

Compute Dmax :

Set

dy dx

=

0

in

E

q

2a

to locate position Dmax

0

=

Mx

-

Mx 2 4L

-5 12

ML

x

2

-

4

Lx

=

-

5L 12

(4

L2

)

(x

- 2L)2

=

7 3

L2

x = 0.4725L

Substi x = 0.4725L into (Eq.3a)

Dmaxy

=

M EI

????

(0.4725L 2

)2

-

(0.4725L)3 12L

-

5 L(0.142725L ) ????

= -0.094ML2

EI

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