Computer study—Influence of supports on frame behavior.

(a) Using the RISA-2D computer program, compute the initial elastic deflection at midspan of

the girder in Figure P8.44, given that the support at D is a roller. For the computer analysis,

replace the tapered members by 3-ft-long segments of constant depth whose properties are

based on each segment’s midspan dimensions; that is, there will be 9 members and 10 joints.

When you set up the problem, specify in GLOBAL that forces are to be computed at three

sections. This will produce values of forces at both ends and at the center of each segment.

To account for cracking of the reinforced concrete, assume for girder BCD that I e = 0.35I G ; for

column AB assume I e = 0.7I G (compression forces in columns reduce cracking). Since

deflections of beams and one-story rigid frames are due almost entirely to moment and not

significantly affected by the area of the member’s cross-section, substitute the gross area in

the Member Properties Table.

(b) Replace the roller at support D in Figure P8.44 by a pin to prevent horizontal displacement

of joint D, and repeat the analysis of the frame. The frame is now an indeterminate structure.

Compare your results with those in part (a), and briefly discuss differences in behavior with

respect to the magnitude of deflections and moments.

NOTE: Because reinforced concrete beams crack due to tensile stresses created by moment

and shear, initial elastic deflections are based on an empirical equation for moment of inertia

established from experimental studies of full-size beams (provided in the ACI Code). This

equation produces an effective moment of inertia I e that varies from about 0.35 to 0.5 of the

moment of inertia I G based on the gross area of the cross section. The additional deflection

due to creep and shrinkage that occurs over time, which can exceed the initial deflection, is

not considered.










When the support at D is a roller, the horizontal equilibrium of the entire structure requires

that the reaction at support A (i.e., the shear of column AB) be equal to zero, As a result,

column AB is subjected to an axial load only and beam BD acts as a simply supported beam.

Since end B of the beam rotates, column AB has to sway to the right in order to maintain the

90-degreee angle with the beam and, at the same time, the column has to remain straight

(i.e., no moment in the column.) Once the roller is replaced by a pin support, the horizontal

reaction at A produces a moment at the top end of the column, which restrains the beam and

reduces the vertical deflection at midspan by about a half; this restraining effect also reduces

the beam moment at midspan by 34%.

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