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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 9, Problem 11P
Textbook Problem
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For the truss of Problem 8 .51, determine the maximum tensile and compressive axial forces in member DI due to a concentrated live load of 40 k, a uniformly distributed live load of 4 k/ft, and a uniformly distributed dead load of 2 k/ft.

Chapter 9, Problem 11P, For the truss of Problem 8 .51, determine the maximum tensile and compressive axial forces in member

FIG. P8.51

To determine

Find the maximum tensile and compressive axial forces in member DI.

Explanation of Solution

Given Information:

The concentrated live load (P) is 40 k.

The uniformly distributed live load (wL) is 4 k/ft.

The uniformly distributed dead load (wD) is 2 k/ft.

Calculation:

Find the support reactions.

Apply 1 k moving load from A to G in the bottom chord member.

Draw the free body diagram of the truss as in Figure 1.

Refer Figure 1,

Find the reaction at C and E when 1 k load placed from A to G. (0x96ft).

Apply moment equilibrium at C.

ΣMC=01(32x)Ey(32)=032Ey=(32x)32Ey=x32

Ey=x321

Apply force equilibrium equation along vertical.

Consider the upward force as positive (+) and downward force as negative ().

ΣFy=0Cy+Ey1=0Cy+x321=1Cy=2x32

Equation member force CD.

The expressions for the member force FCD can be determined by passing an imaginary section aa through the members HI, CI, and CD and then apply a moment equilibrium at I.

Draw the free body diagram of the section as shown in Figure 2.

Refer Figure 2.

Apply 1 k load just the left of C (0x32ft).

Find the equation of member force CD from A to C.

Consider the section DG.

Apply moment equilibrium equation at I.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMI=0

FCD(20)+Ey(32)=020FCD=32EyFCD=85Ey

Substitute x321 for Ey.

FCD=85(x321)=x2085

Apply 1 k load just the right of C (32ftx96ft).

Find the equation of member force CD from C to G.

Consider the section AC.

Apply moment equilibrium equation at I.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMI=0

FCD(20)=0FCD=0

Thus, the equation of force in the member CD,

FCD=x2085, 0x32ft        (1)

FCD=0, 32ftx96ft        (2)

Influence line for the force in member DI.

The expressions for the member force FDI can be determined by passing an imaginary section bb through the members IJ, DI and CD and then apply moment equilibrium at J.

Draw the free body diagram of the section bb as shown in Figure 3.

Refer Figure 3.

Apply 1 k load just the left of C (0x32ft).

Find the equation of member force DI from A to C.

Consider the section DG.

Apply moment equilibrium equation at J.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMJ=0Ey(16)FCD(24)(16162+202)FDI(24)=015FDI=16Ey24FCDFDI=1615Ey1.6FCD

Substitute x2085 for FCD and x321 for Ey.

FDI=1615(x321)1.6(x2085)=x3016152x25+6425=112757x150

Apply 1 k load just the right of C (32ftx96ft).

Find the equation of member force DI from C to G.

Consider the section AC.

Apply moment equilibrium equation at J.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMJ=0Cy(16)FCD(24)20162+202FDI(16)16162+202FDI(4)=012.5FDI+2.5FDI=16Cy24FCD15FDI=16Cy24FCD

FDI=1615Cy2415FCD

Substitute 2x32 for Cy and 0 for FCD.

FDI=1615(2x32)2415(0)=3215x30

Thus, the equation of force in the member DI,

FDI=112757x1500x32ft        (5)

FDI=3215x3032x96ft        (6)

Find the force in member DI using the Equation (5) and (6) and then summarize the value in Table 1

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