Elementary Geometrical Statics; An Introduction to Graphic Statics (Paperback)


This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1897 Excerpt: ...ad, obtained by the above process, must pass through the intersection of de and ef. Thus, for equilibrium, in constructing the force diagram, the end of the line representing the last force must coincide with the beginning of the line representing the first force (this is expressed by saying the force polygon closes), and in drawing the dotted lines of the space diagram, we start with the point of intersection of the first two lines of action, and end with the point of intersection of the last two. 94. The method of Art. 92 may be still more abbreviated in simple cases, as, for instance, in the following: It is required to find the resultant of four given forces, P, Q, R, S, acting along the given lines indicated in the space diagram. Draw straight lines A B, BC, CD, DE, to represent the forces P, Q, R, S respectively in magnitude and direction. Let the lines of action of P, Q, R, S be now marked ah, bc, cd, de respectively, as in the figure. Join AC, AE, CE. Through the intersection of ab and bc draw ac parallel to AC; and through the intersection of cd and de draw ce parallel to CE. Then we shall show that the straight line ae, drawn through the intersection of ac and ce parallel to AE, is the line of action of the resultant, and that AE represents the resultant in magnitude and direction. For, the forces P and Q are equivalent to a force represented by A C acting along ac; also, the forces R and S are equivalent to a force represented by CE acting along ce; and, if P, Q, R, S be replaced by these two forces, they in turn may be replaced by a force represented by AE, and acting along ae. For the system to be in equilibrium, it is necessary and sufficient that A and E should coincide, and that the lines ac, ce should form one continuous straight line. Thus, .

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1897 Excerpt: ...ad, obtained by the above process, must pass through the intersection of de and ef. Thus, for equilibrium, in constructing the force diagram, the end of the line representing the last force must coincide with the beginning of the line representing the first force (this is expressed by saying the force polygon closes), and in drawing the dotted lines of the space diagram, we start with the point of intersection of the first two lines of action, and end with the point of intersection of the last two. 94. The method of Art. 92 may be still more abbreviated in simple cases, as, for instance, in the following: It is required to find the resultant of four given forces, P, Q, R, S, acting along the given lines indicated in the space diagram. Draw straight lines A B, BC, CD, DE, to represent the forces P, Q, R, S respectively in magnitude and direction. Let the lines of action of P, Q, R, S be now marked ah, bc, cd, de respectively, as in the figure. Join AC, AE, CE. Through the intersection of ab and bc draw ac parallel to AC; and through the intersection of cd and de draw ce parallel to CE. Then we shall show that the straight line ae, drawn through the intersection of ac and ce parallel to AE, is the line of action of the resultant, and that AE represents the resultant in magnitude and direction. For, the forces P and Q are equivalent to a force represented by A C acting along ac; also, the forces R and S are equivalent to a force represented by CE acting along ce; and, if P, Q, R, S be replaced by these two forces, they in turn may be replaced by a force represented by AE, and acting along ae. For the system to be in equilibrium, it is necessary and sufficient that A and E should coincide, and that the lines ac, ce should form one continuous straight line. Thus, .

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Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

May 2012

Availability

Supplier out of stock. If you add this item to your wish list we will let you know when it becomes available.

First published

May 2012

Authors

Dimensions

246 x 189 x 4mm (L x W x T)

Format

Paperback - Trade

Pages

70

ISBN-13

978-1-236-01458-0

Barcode

9781236014580

Categories

LSN

1-236-01458-8



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