• Author / Uploaded
• Harish

Citation preview

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16–47. The W14 * 43 simply supported beam is made of A992 steel and is subjected to the loading shown. Determine the deflection at its center C.

3 kip/ft

60 kip!ft

B

A

C x 12 ft

12 ft

SOLUTION Elastic Curve: The elastic curves for the partial uniform distributed load and couple moment are drawn separately as shown in Fig. a. Method of Superposition: Using the table in Appendix C, the required displacements shown in Fig. a are

( vC ) 1 =

5(3) ( 244 ) 6480 kip # ft 3 5wL4 = = T 768EI 768EI EI

( vC ) 2 =

MO x 2 ( x - 3Lx + 2L2 ) 6EIL

=

60(12) 6EI(24)

= -

3 122

- 3(24)(12) + 2 ( 122 ) 4

2160 kip # ft 3 EI

vC = ( vC ) 1 + ( vC ) 2 =

=

2160 kip # ft 3 EI

6480 kip # ft 3 EI

+

T

2160 kip # ft 3 EI

=

8640 kip # ft 3 EI

T

For W14 * 43 wide-flange section, I = 428 in4. Also, E = 29 ( 103 ) ksi for A992 steel. Then vC =

8640 ( 123 ) 29 ( 103 ) (428)

Ans.

= 1.203 in. = 1.20 in.T

Ans: vC = 1.20 in.T 1268

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16–50. The W14 * 43 simply supported beam is made of A-36 steel and is subjected to the loading shown. Determine the slope at A and B.

2 kip/ft 40 kip!ft A

B

C 10 ft

10 ft

SOLUTION ! uA ! = uA1 + uA2 =

=

7wL3 ML + 384EI 6EI 7(2) 12

( 2403 )

384EI

+

40(12)(240) 6EI

=

61,200 29 ( 103 ) (428) Ans.

uA = -0.00493 rad = - 0.283° uB = uB1 + uB2 =

=

3wL3 ML + 128EI 3EI 3(2) 12

( 2403 )

128EI

+

40(12)(240) 3EI

=

92,400 29 ( 103 ) (428) Ans.

= 0.007444 rad = 0.427°

Ans: uA = -0.283°, uB = 0.427° 1271