75 k 40 k Hinge 250 k-ft E A B D F12 ft 12 ft 12 f- -12 ft- El = constant E = 29,000 ksi I= 6,000 in.4 FIG. P6.33, P6.59 2.5 k/ft Hinge 35 k A B -8 ft 8 ft- 16 ft I= 4,000 in.4 I= 3,000 in.4 E = constant = 30,000 ksi FIG. P6.34, P6.60

75 k 40 k Hinge 250 k-ft E A B D F12 ft 12 ft 12 f- -12 ft- El = constant E = 29,000 ksi I= 6,000 in.4 FIG. P6.33, P6.59 2.5 k/ft Hinge 35 k A B -8 ft 8 ft- 16 ft I= 4,000 in.4 I= 3,000 in.4 E = constant = 30,000 ksi FIG. P6.34, P6.60

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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Concept explainers

Beams

In structural engineering, a beam is a component made of a variety of materials (including steel alloys, and wood) that can resist loads applied laterally to the beam axis. Members, elements, rafters, shafts, and purlins are all terms used to describe beams.

Question

Use the conjugate-beam method to determine the slopes and deflections at points B and D of the beams shown in Figs. P6.33 and P6.34.

75 k
40 k
Hinge
250 k-ft
E
A
B
D
F12 ft 12 ft 12 f-
-12 ft-
El = constant
E = 29,000 ksi
I= 6,000 in.4
FIG. P6.33, P6.59
2.5 k/ft
Hinge
35 k
A
B
-8 ft 8 ft-
16 ft
I= 4,000 in.4
I= 3,000 in.4
E = constant = 30,000 ksi
FIG. P6.34, P6.60

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