Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Question
Chapter 6.1, Problem 6.36P
To determine

The force in each of the members of the truss for P=(2184 N)j^ and Q=0 .

Expert Solution & Answer
Check Mark

Answer to Problem 6.36P

The force in member AC is 676 N and the member AC is in compression, the force in members AB and AD is 861 N and these members are in compression, the force in members BC and CD is 162.5 N and these members are in tension, the force in member BD is 244 N and it is in tension.

Explanation of Solution

The free-body diagram of the entire truss is shown in figure 1.

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>Vector</x-custom-btb-me> Mechanics for Engineers: Statics and Dynamics, Chapter 6.1, Problem 6.36P , additional homework tip  1

Refer to figure 1 and use symmetry.

Dx=BxDy=By (I)

Here, Dx is the x component of the reaction at the point D, Dy is the y component of the reaction at the point D, Bx is the x component of the reaction at the point B and By is the y component of the reaction at the point B.

The x component of the net force must be equal to zero.

ΣFx=0 (II)

Here, ΣFx is the x component of the net force.

Write the expression for ΣFx .

ΣFx=2Bxi^

Put the above equation in equation (II).

2Bxi^=0Bx=0

Put equation (I) in the above equation.

Dx=0

The z component of the net force must be equal to zero.

ΣFz=0 (III)

Here, ΣFz is the z component of the net force.

Write the expression for ΣFz .

ΣFz=Bz

Here, Bz is the z component of the reaction at the point B.

Put the above equation in equation (III).

Bz=0

Write the equilibrium equations taking the moments about the point C in the z direction.

ΣMcz=0 (IV)

Here, ΣMcz is the sum of the moments about the point C in the z direction.

Write the equation for ΣMcz .

ΣMcz=2By(2.8 m)+(2184 N)(2 m)

Put the above equation in equation (IV).

2By(2.8 m)+(2184 N)(2 m)=0By=780 N

Write the expression for the reaction at the point B.

B=Byj^

Here B is the reaction at the point B.

Substitute 780 N for By in the above equation to find B .

B=(780 N)j^

Consider the free-body joint A. The free-body diagram of joint A is shown in figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 6.1, Problem 6.36P , additional homework tip  2

Refer to figure (2) and write the expression for the forces.

FAB=FABABAB (V)

Here, FAB is the force exerted by member AB and AB is the magnitude of the vector AB .

Write the expression for AB .

AB=0.8i^4.8j^+2.1k^

Find the magnitude of AB .

AB=(0.8)2+(4.8)2+(2.1)2=5.30

Substitute 0.8i^4.8j^+2.1k^  for AB and 5.30 for AB in equation (V) to find the expression for FCA .

FAB=FAB(0.8i^4.8j^+2.1k^)5.30 (VI)

Write the expression for FAC .

FAC=FACACAC

Here, FAC is the force exerted by member AC and AC is the magnitude of the vector AC .

Substitute 2i^4.8j^ for AC and 5.20 for AC in the above equation to find the expression for FAC .

FAC=FAC(2i^4.8j^)5.20 (VII)

Write the expression for FAD .

FAD=FADADAD

Here, FAD is the force exerted by member AD and AD is the magnitude of the vector AD .

Substitute 0.8i^4.8j^2.1k^ for AD and 5.30 for AD in the above equation to find the expression for FAD .

FAD=FAD(0.8i^4.8j^2.1k^)5.30 (VIII)

The net force must be equal to zero.

ΣF=0 (IX)

Here, ΣF is the net force.

Write the expression for ΣF .

ΣF=FAB+FAC+FAD(2184 N)j^

Put the above equation in equation (IX).

FAB+FAC+FAD(2184 N)j^=0

Put equations (VI), (VII) and (VIII) in the above equation.

FAB(0.8i^4.8j^+2.1k^)5.30+FAC(2i^4.8j^)5.20+FAD(0.8i^4.8j^2.1k^)5.30(2184 N)j^=0 (X)

Equate the coefficient of i^ in equation (X) to zero.

0.85.30(FAB+FAD)+25.20FAC=0 (XI)

Equate the coefficient of j^ in equation (X) to zero.

4.85.30(FAB+FAD)4.85.20FAC2184 N=0 (XII)

Equate the coefficient of k^ in equation (X) to zero.

2.15.30(FABFAD)=0FAD=FAB (XIII)

Multiply equation (XI) by 6 and add equation (XII).

(16.85.20)FAC2184 N=0FAC=676 NFAC=676 N, compression

Put equation (XIII) in equation (XI).

0.85.30(FAB+FAB)+25.20FAC=0(0.85.30)2FAB+25.20FAC=0

Substitute 676 N for FAC in the above equation.

(0.85.30)2FAB+25.20(676 N)=0FAB=861.25 NFAB=861 N, compression

Put the above equation in equation (XIII).

FAD=861 N, compression

Consider the free-body joint B. The free-body diagram of joint B is shown in figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 6.1, Problem 6.36P , additional homework tip  3

Refer to figure (3) and write the expression for the forces.

FAB=FABABAB

Substitute 861.25 N for FAB , 0.8i^4.8j^+2.1k^ for AB and 5.30 for AB in the above equation to find the expression for FAB .

FAB=(861.25 N)(0.8i^4.8j^+2.1k^)5.30=(130 N)i^(780 N)j^+(341.25 N)k^ (XIV)

Write the expression for FBC .

FBC=FBCBCBC

Here, FBC is the force exerted by member BC and BC is the magnitude of the vector BC .

Substitute 2.8i^2.1k^ for BC and 3.5 for BC in the above equation to find the expression for FBC .

FBC=FBC(2.8i^2.1k^)3.5=FBC(0.8i^0.6k^) (XV)

Write the expression for FBD .

FBD=FBDk^ (XVI)

Here, FBD is the force exerted by member BD

Write the expression for ΣF .

ΣF=FAB+FBC+FBD+B

Put the above equation in equation (IX).

FAB+FBC+FBD+B=0

Put equations (XIV), (XV) and (XVI) in the above equation.

(130 N)i^(780 N)j^+(341.25 N)k^+FBC(0.8i^0.6k^)FBDk^+B=0

Substitute (780 N)j^ for B in the above equation.

(130 N)i^(780 N)j^+(341.25 N)k^+FBC(0.8i^0.6k^)FBDk^+(780 N)j^=0 (XVII)

Equate the coefficient of i^ in equation (XVII) to zero.

130 N+0.8FBC=0FBC=162.5 N, tension

Equate the coefficient of k^ in equation (XVII) to zero.

341.25 N0.6FBCFBD=0FBD=341.25 N0.6FBC

Substitute 162.5 N for FBC in the above equation.

FBD=341.25 N0.6(162.5 N)=244 N

From symmetry,

FCD=FBC

Here, FCD is the force exerted by CD.

Substitute 162.5 N, tension for FBC in the above equation to find FCD .

FCD=162.5 N, tension

Conclusion:

Thus, the force in member AC is 676 N and the member AC is in compression, the force in members AB and AD is 861 N and these members are in compression, the force in members BC and CD is 162.5 N and these members are in tension, the force in member BD is 244 N and it is in tension.

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Chapter 6 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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