VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 6.1, Problem 6.39P

The truss shown consists of nine members and is support by a ball-and-socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that completely constrained, and that the reactions at its suppo are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.

Chapter 6.1, Problem 6.39P, The truss shown consists of nine members and is support by a ball-and-socket at B, a short link at

Fig. P6.39

(a)

Expert Solution
Check Mark
To determine

Verify that the truss is a simple truss, completely constrained, and the reactions at the supports are statically determinate.

Answer to Problem 6.39P

The reactions at supports B, C, and D is B=(800N)j;C=(100N)j;D=(300N)j_.

Explanation of Solution

Given information:

The value of the force P is P=(1200N)j.

The value of the force Q is zero.

Calculation:

Show the free-body diagram of the truss as in Figure 1.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 6.1, Problem 6.39P , additional homework tip  1

Find the vector coordinates by taking moment about point B.

MB=01.8i×Cyj+(1.8i3k)×(Dyj+Dzk)+(0.6i0.75k)×(1200j)=01.8Cyk+1.8Dyk1.8Dzj3Dyi720k+900i=0

Equate the coefficients of i to zero.

3Dy900=0Dy=300ND=(300N)j

Equate the coefficients of j to zero.

Dz=0

Equate the coefficients of k to zero.

1.8Cy+1.8(Dy)720=0

Substitute 300 N for Dy.

1.8Cy+1.8(300)720=01.8Cy+540720=0Cy=100NC=(100N)j

Resolve the force components in y-axis.

Fy=0Byj+Dyj+Cyj1200j=0

Substitute 300 N for Dy and 100 N for Cy.

Byj+300j+100j1200j=0By=800NB=(800N)j

The unknown reactions can be calculated with the equilibrium equations. Therefore, the truss is statically determinate, completely constrained and simple truss.

Thus, the reactions at supports B, C, and D is B=(800N)j;C=(100N)j;D=(300N)j_.

(b)

Expert Solution
Check Mark
To determine

Find the force in each member of the truss.

Answer to Problem 6.39P

The force in the member AB is 840N(C)_.

The force in the member BC is 160N(T)_.

The force in the member BE is 200N(T)_.

The force in the member AC is 110.6N(C)_.

The force in the member CE is 233N(C)_.

The force in the member CD is 225N(T)_.

The force in the member AD is 394N(C)_.

The force in the member DE is 120N(T)_.

The force in the member AE is zero_.

Explanation of Solution

Show the free-body diagram of the joint B as in Figure 2.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 6.1, Problem 6.39P , additional homework tip  2

Resolve the force components as follows;

F=0;FBA+FBC+FBE+(800N)j=0

Write the vector value of BA as follows;

BA=0.6i+3j0.75k

Find the scalar quantity of BA using the relation.

BA=0.62+32+(0.75)2=3.15m

Find the force in the member AB as follows;

FBA=FABBABA

Substitute (0.6i+3j0.75k) for BA and 3.15 m for BA.

FBA=FAB(0.6i+3j0.75k)3.15=FAB(0.19048i+0.95238j0.23810k)

Find the force in the member BC as follows;

FBC=FBCi

Find the force in the member BE as follows;

FBE=FBEk

Equate the coefficients of j to zero.

0.95238FAB+800=0FAB=840N=840N(C)

Equate the coefficients of i to zero.

0.19048FAB+FBC=0

Substitute –840 N for FAB.

0.19048(840)+FBC=0FBC=160N(T)

Equate the coefficients of k to zero.

(0.23810)FABFBE=0

Substitute –840 N for FAB.

(0.23810)(840)FBE=0FBE=200N(T)

Therefore,

The force in the member AB is 840N(C)_.

The force in the member BC is 160N(T)_.

The force in the member BE is 200N(T)_.

Show the free-body diagram of the joint C as in Figure 3.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 6.1, Problem 6.39P , additional homework tip  3

Resolve the force components as follows;

F=0;FCA+FCB+FCD+FCE+(100N)j=0

Write the vector value of CA as follows;

CA=1.20i+3j0.75k

Find the scalar quantity of CA using the relation.

CA=(1.20)2+32+(0.75)2=3.317m

Find the force in the member AC as follows;

FCA=FACCACA

Substitute (1.20i+3j0.75k) for CA and 3.317 m for CA.

FCA=FAC(1.20i+3j0.75k)3.317=FAC(0.36177i+0.90443j0.22611k)

Find the force in the member CB as follows;

FCB=FBC=(160N)i

Find the force in the member CD as follows;

FCD=FCDk

Write the vector value of CE as follows;

CE=1.80i3k

Find the scalar quantity of CE using the relation.

CE=(1.80)2+(3)2=3.499m

Find the force in the member CE as follows;

FCE=FCECECE

Substitute (1.80i3k) for CE and 3.499 m for CE.

FCE=FCE(1.80i3k)3.499=FCE(0.51443i0.85739k)

Equate the coefficients of j to zero.

0.90443FAC+100=0FAC=110.6N=110.6N(C)

Equate the coefficients of i to zero.

0.36177FAC1600.51443FCE=0

Substitute –110.6 N for FAC.

0.36177(110.6)1600.51443FCE=0FCE=233N=233N(C)

Equate the coefficients of k to zero.

0.22611FACFCD0.85739FCE=0

Substitute –110.6 N for FAC and –233 N for FCE.

0.22611(110.6)FCD0.85739(233)=0FCD=225N(T)

Therefore,

The force in the member AC is 110.6N(C)_.

The force in the member CE is 233N(C)_.

The force in the member CD is 225N(T)_.

Show the free-body diagram of the joint D as in Figure 4.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 6.1, Problem 6.39P , additional homework tip  4

Resolve the force components as follows;

F=0;FDA+FDC+FDE+(300N)j=0

Write the vector value of DA as follows;

DA=1.20i+3j+2.25k

Find the scalar quantity of DA using the relation.

DA=(1.20)2+32+2.252=3.937m

Find the force in the member AD as follows;

FDA=FADDADA

Substitute (1.20i+3j+2.25k) for DA and 3.937 m for DA.

FDA=FAD(1.20i+3j+2.25k)3.937=FAD(0.30480i+0.7620j0.57150k)

Find the force in the member DC as follows;

FDC=FCDk=(225N)k

Find the force in the member DE as follows;

FDE=FDEi

Equate the coefficients of j to zero.

0.7620FAD+300=0FAD=394N=394N(C)

Equate the coefficients of i to zero.

0.30480FADFDE=0

Substitute –394 N for FAD.

0.30480(394)FDE=0FDE=120N(T)

Therefore,

The force in the member AD is 394N(C)_.

The force in the member DE is 120N(T)_.

Show the free-body diagram of the joint E as in figure 5.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 6.1, Problem 6.39P , additional homework tip  5

The member AE is not in the xz plane.

Therefore, the force in the member AE is zero_.

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Chapter 6 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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