Use the conjugate-beam method to determine the slopes at ends A and D and the deflections at points B and C of the beam shown in Fig. 6.16(a) 60 k40 kBC20 ft10 ft 10 ft-El = constantE=1,800 ksiI= 46,000 in.4(a) Real Beam800El600EIA.D.BM(b)ElDiagram ()k-ftEl800El600EIDBAy20 ft10 ft10 ft-(c) Conjugate BeamFIG. 6.16

Solution: Given that; EI is constant E=1800 ksiI=46,000 in4EI=1800×46000=82.8×106 k-in2EI=575×103…

60 k 40 k B C 20 ft 10 ft 10 ft- El = constant E=1,800 ksi I= 46,000 in.4 (a) Real Beam 800 El 600 EI A. D. B M (b) El Diagram () k-ft El 800 El 600 EI D B Ay 20 ft 10 ft 10 ft- (c) Conjugate Beam FIG. 6.16

60 k 40 k B C 20 ft 10 ft 10 ft- El = constant E=1,800 ksi I= 46,000 in.4 (a) Real Beam 800 El 600 EI A. D. B M (b) El Diagram () k-ft El 800 El 600 EI D B Ay 20 ft 10 ft 10 ft- (c) Conjugate Beam FIG. 6.16

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Use the conjugate-beam method to determine the slopes at ends A and D and the deflections at points B and C of the beam shown in Fig. 6.16(a)

60 k
40 k
B
C
20 ft
10 ft 10 ft-
El = constant
E=1,800 ksi
I= 46,000 in.4
(a) Real Beam
800
El
600
EI
A.
D.
B
M
(b)
El
Diagram ()
k-ft
El
800
El
600
EI
D
B
Ay
20 ft
10 ft
10 ft-
(c) Conjugate Beam
FIG. 6.16

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