Problem 6.37 The truss supports loads at A and H.Use the method of sections to determine the axial forcesin members CE, BE, and BD.

A

B

C

D

E

F

G H

300 mm

400 mm

18 kN 24 kN

400 mm 400 mm 400 mm

Solution: First find the external support loads on the truss

18 kN24 kN

C

D

E G

BX

BYFY

H

0.8 m

1.2 m

0.4 m

∑Fx : Bx D 0

∑Fy : By C Fy % 18 % 24 D 0 (kN)

∑MB: 0.8Fy % .1.2-.24- C .0.4-18 D 0

Solving: Bx D 0

By D 15 kN

Fy D 27 kN

Method of sections:

18 kN

C CE E

BD

BY

0.4 m

0.3 mBE

θ θ

tan & D 3

4

& D 36.87°

By D 15 kN∑Fx : CE C BE cos & C BD D 0

∑Fy : By % 18 C BE sin & D 0

C∑

MB: C .0.4-.18- % .0.3-.CE- D 0

Solving,CE D 24 kN .T-BE D 5 kN .T-BD D %28 kN .C-

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Problem 6.49 Use the method of sections to determinethe axial force in member EF.

4 ft

4 ft

4 ft

4 ft

A

C

E

G

I

12 ft

D

B

F

H

10 kip

10 kip

Solution: The included angle at the apex BAC is

˛ D tan%1

(12

16

)D 36.87°.

The interior angles BCA, DEC, FGE, HIG are % D 90° % ˛ D 53.13°.The length of the member ED is LED D 8 tan ˛ D 6 ft. The interiorangle DEF is

ˇ D tan%1

(4

LED

)D 33.69°.

The complete structure as a free body: The moment about H is MH D%105124 % 105164 C I5124 D 0, from which I D 280

12D 23.33 kip.

The sum of forces:∑Fy D Hy C I D 0,

from which Hy D %I D %23.33 kip.

∑Fx D Hx C 20 D 0,

from which Hx D %20 kip. Make the cut through EG, EF, and DE.Consider the upper section only. Denote the axial force in a memberjoining I, K by IK. The section as a free body: The sum of the momentsabout E is ME D %10544 % 10584 C DF5LED4 D 0, from which DF D120

6D 20 kip.

The sum of forces:∑Fy D %EF sin ˇ % EG sin % % DF D 0,

∑Fx D %EF cos ˇ C EG cos % C 20 D 0,

from which the two simultaneous equations: 0.5547EF C 0.8EG D%20, and 0.8320EF % 0.6EG D 20. Solve: EF D 4.0 kip (T)

F = 10 kip

F = 10 kip

DF EF EG

Eβ γ

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Problem 6.79 The frame supports a 6-kN load at C.Determine the reactions on the frame at A and D.

0.4 m0.8 m

A B

D EF

C

0.5 m

0.4 m 1.0 m6 kN

Solution: Note that members BE and CF are two force members.Consider the 6 kN load as being applied to member ABC.

Ax

Ay

FCFFBE

B C

0.4 m

1.0 m6 kN

θ φ

tan & D 0.5

0.4& D 51.34°

tan * D 0.5

0.2* D 68.20°

Member DEF

Dx

Dy

FCFFBE

FE

0.8 m 0.4 m

θφ

Equations of equilibrium:

Member ABC:∑Fx : Ax C FBE cos & % FCF cos * D 0

∑Fy : Ay % FBE sin & % FCF sin * % 6 D 0

C∑

MA: % /0.4.FBE sin & % /1.4.FCF sin * % 1.4/6. D 0

Member DEF:∑Fx : Dx % FBE cos & C FCF cos * D 0

∑Fy : Dy C FBE sin & C FCF sin * D 0

C∑

MD: /0.8./FBE sin &. C 1.2FCF sin * D 0

Unknowns Ax, Ay, Dx, Dy, FBE, FCF we have 6 eqns in 6unknowns.

Solving, we get

Ax D %16.8 kNAy D 11.25 kNDx D 16.3 kNDy D %5.25 kN

Also, FBE D 20.2 kN /T.

FCF D %11.3 kN /C.

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Problem 6.85 Determine the forces on member ABC.

1 m

D C

E

A B

1 m

2 m 2 m 1 m

6 kN

Solution: The frame as a whole: The equations of equilibrium are

∑FX D AX C EX D 0,

∑FY D AY C EY % 6000 N D 0,

and, with moments about E,∑ME D 2AX % .5-6000 D 0.

Solving for the support reactions, we get AX D 15,000 N and EX D%15,000 N. We cannot yet solve for the forces in the y direction at Aand E.

Member ABC: The equations of equilibrium are

∑FX D AX % BX D 0,

∑FY D AY % BY % CY D 0,

and summing moments about A,

∑MA D %2BY % 4CY D 0.

Member BDE: The equations of equilibrium are

∑FX D EX C DX C BX D 0,

∑FY D EY C DY C BY D 0,

and, summing moments about E,

∑ME D .1-DY C .1-DX C .2-BY C .2-BX D 0.

Member CD: The equations of equilibrium are

∑FX D %DX D 0,

∑FY D %DY C CY % 6000 D 0,

and summing moments about D,

∑MD D %.4-6000 C 3CY D 0.

Solving these equations simultaneously gives values for all of theforces in the frame. The values are AX D 15,000 N, AY D %8,000 N,BX D 15,000 N, BY D %16,000 N, CY D 8,000 N, DX D 0, andDY D 2,000 N.

EYEX E DY

DX

DX D

BXBY

DY

BX

AXAY BY

CY

CY

D

B

BA

C

C

6 kN

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Problem 6.95 Determine the forces on member AD.

400 N

200 N

C

D

130 mm

400 mm

400 mm 400 mm

BA

Solution: Denote the reactions of the support by Rx and Ry . Thecomplete structure as a free body:

∑Fx D Rx % 400 D 0,

from which Rx D 400 N. The sum of moments:∑MA D 800C % 400+930* C 400+530* % 400+200* D 0,

from which C D 300 N.∑Fy D C C Ry % 400 % 200 D 0,

from which Ry D 300 N. Element ABC : The sum of the moments:

∑MA D %4By C 8C D 0,

from which By D 600 N. Element BD : The sum of the forces:

∑Fy D By % Dy % 400 D 0,

from which Dy D 200 N.

Element AD: The sum of the forces:∑Fy D Ay C Dy % 200 D 0,

from which Ay D 0: Element AD: The sum of the forces:

∑Fx D Ax C Dx D 0

and∑

MA D %400+200* C 800Dy % 400Dx D 0

Ax D %200 N, and Dx D 200 N.

Element BD: The sum of forces:∑Fx D Bx % Dx % 400 D 0

from which Bx D 600 N. This completes the solution of the nineequations in nine unknowns, of which Ax , Ay , Dx , and Dy are thevalues required by the Problem.

200 N

400 N

400 N

Dy

Dy

Bx

BxRx

Ax

AxAy

Ay

Ry By

By

Dx

Dx

C

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Problem 6.106 The woman exerts 20-N forces on thehandles of the shears. Determine the magnitude of theforces exerted on the branch at A.

BAC

D

E

25 mm25 mm36 mm

65 mm 20 N

20 N

Solution: Assume that the shears are symmetrical.

Consider the 2 pieces CD and CE∑Fx D 0 ) Dx D Ex

∑Fy D 0 ) Dy D Ey

∑MC D 0 ) Dx D Ex D 0

20 N

20 NEy

Ex

C

Dx

Dy

Now examine CD by itself∑MC D %*20 N)*90 mm) C Dy*25 mm) D 0 ) Dy D 72 N

20 N

Cy

Cx

Dx = 0

Dy

Finally examine DBA∑MB : A*36 mm) % Dy*50 mm) D 0

Bx

Dx = 0By

Dy

A

Solving we find A D 100 N

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