Explanation Calculation: Find the support reactions. Apply 1 k moving load from A to G in the top chord member. Find the reaction at A and G when 1 k load placed from A to G . ( 0 ≤ x ≤ 180 ft ) . Apply moment equilibrium at A . Consider clockwise moment as positive and anticlockwise moment as negative. Σ M A = 0 1 ( x ) − G y ( 180 ) = 0 180 G y = x G y = x 180 Apply force equilibrium equation along vertical. Consider the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . Σ F y = 0 A y + G y − 1 = 0 A y + x 180 = 1 A y = 1 − x 180 Influence line for the force in member CD . The expressions for the member force F C D can be determined by passing an imaginary section aa through the members MN , IN , DI , and CD and then apply a vertical force equilibrium. Draw the free body diagram as shown in Figure 2. Refer Figure 2. The force in the member DI and IN is equal and opposite nature F I N = − F D I . Find the equation of member force DI . Apply a 1 k load at just left of C ( 0 ≤ x ≤ 60 ft ) . Consider the section DG . Apply force equilibrium equation along vertical in the section DG . Consider the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . Σ F y = 0 G y + F D I ( 1 2 2 + 1 2 ) − F I N ( 1 2 2 + 1 2 ) = 0 1 5 F D I − 1 5 F I N = − G y Substitute − F D I for F I N and x 180 for G y . 1 5 F D I − 1 5 ( − F D I ) = − ( x 180 ) 2 5 F D I = − 5 180 x F D I = − 5 360 x Apply a 1 k load at just right of D ( 90 ft ≤ x ≤ 180 ft ) . Consider the section AD . Apply force equilibrium equation along vertical in the section AD . Consider the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . Σ F y = 0 A y − F D I ( 1 2 2 + 1 2 ) + F I N ( 1 2 2 + 1 2 ) = 0 − 1 5 F D I + 1 5 F I N = − A y Substitute − F D I for F I N and 1 − x 180 for A y . − 1 5 F D I + 1 5 ( − F D I ) = − ( 1 − x 180 ) − 2 5 F D I = − 1 + x 180 F D I = 5 2 − 5 360 x Thus, the equation of force in the member DI , F D I = − 5 360 x , 0 ≤ x ≤ 60 ft (1) F D I = 5 2 − 5 360 x , 90 ft ≤ x ≤ 180 ft (2) Find the force in member DI using the Equation (1) and (2) and then summarize the value in Table 1. x (ft) Apply 1 k load Force in member AB (k) Influence line ordinate for the force in member DI (k/k) 0 A − 5 360 ( 0 ) = 0 0 30 B − 5 360 ( 30 ) = − 0.186 − 0.186 60 C − 5 360 ( 60 ) = − 0.373 − 0.373 90 D 5 2 − 5 360 ( 90 ) = 0

6th Edition

ISBN: 9780357030981

Author: Aslam Kassimali

Publisher: Cengage Learning US

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Influence Lines

The graphs that show the variation of a function at a specific point on a beam or truss due to a unit load placed at any point along the structure are known as influence lines in engineering. Some common functions that are studied with influence lines ar…

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Chapter 8, Problem 52P

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Draw the influence lines for the force in member CD and DI.

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Chapter 8, Problem 51P

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Draw the influence lines for the force in member CD and DI .

Solution Summary: The author illustrates how the influence lines for the force in member CD and DI can be determined by passing an imaginary section through the members MN, IN, DI, and CD.

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Draw the influence lines for the force in member CD and DI.

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Chapter 8 Solutions