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Building Design and Construction Handbook - Merritt - Ventech!

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5.32 SECTION FIVE<br />

5.5.2 Reactions<br />

As an example of the application of the equations of equilibrium (Art. 5.2.1) to the<br />

determination of the reactions of a statically determinate beam, we shall compute<br />

the reactions of the 60-ft-long beam<br />

with overhangs in Fig. 5.17. This beam<br />

carries a uniform load of 200 lb/lin ft<br />

over its entire length <strong>and</strong> several concentrated<br />

loads. The supports are 36 ft<br />

apart.<br />

To find reaction R1, we take moments<br />

about R2 <strong>and</strong> equate the sum of the mo-<br />

FIGURE 5.17 Beam with overhangs loaded<br />

with both uniform <strong>and</strong> concentrated loads.<br />

ments to zero (clockwise rotation is considered<br />

positive, counterclockwise, negative):<br />

�2000 � 48 � 36R1 � 4000 � 30 � 6000 � 18 � 3000 � 12<br />

�200 � 60 � 18 � 0<br />

R1 � 14,000 lb<br />

In this calculation, the moment of the uniform load was found by taking the moment<br />

of its resultant, which acts at the center of the beam.<br />

To find R 2, we can either take moments about R 1 or use the equation �V � 0.<br />

It is generally preferable to apply the moment equation <strong>and</strong> use the other equation<br />

as a check.<br />

3000 � 48 � 36R2 � 6000 � 18 � 4000 � 6 � 2000 � 12<br />

� 200 � 60 � 18 � 0<br />

R2 � 13,000 lb<br />

As a check, we note that the sum of the reactions must equal the total applied<br />

load:<br />

14,000 � 13,000 � 2000 � 4000 � 6000 � 3000 � 12,000<br />

27,000 � 27,000<br />

5.5.3 Internal Forces<br />

Since a beam is in equilibrium under the forces applied to it, it is evident that at<br />

every section internal forces are acting to prevent motion. For example, suppose<br />

we cut the beam in Fig. 5.17 vertically just to the right of its center. If we total<br />

the external forces, including the reaction, to the left of this cut (see Fig. 5.18a),<br />

we find there is an unbalanced downward load of 4000 lb. Evidently, at the cut<br />

section, an upward-acting internal force of 4000 lb must be present to maintain<br />

equilibrium. Again, if we take moments of the external forces about the section,<br />

we find an unbalanced moment of 54,000 ft-lb. So there must be an internal moment<br />

of 54,000 ft-lb acting to maintain equilibrium.<br />

This internal, or resisting, moment is produced by a couple consisting of a force<br />

C acting on the top part of the beam <strong>and</strong> an equal but opposite force T acting on

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