determine the force in each member of the truss, Schemes and Mind Maps of Civil Engineering

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determine the force in each member of the truss

Free-body diagram of entire truss. Calculating the reactions is a good place to start because they are usually easy to compute, and they can be used in the equilibrium equations for the joints where the reactions act.

3 m

5 m

C

A B

10 kN

Equilibrium equations for entire truss

Fx = 0: Ax + 10 kN = 0 (1)

Fy = 0: Ay + By = 0 (2)

MA = 0: (10 kN)(3 m) + By(5 m) = 0 (3)

Solving these equations simultaneously gives

Ax = 10 kN, Ay = 6 kN, and By = 6 kN

Ax Ay (^) B y

3 m

5 m

C

A B

10 kN

1. Determine the force in each member of the truss and state whether the force is tension or compression.

By = 6 kN

B

FBC = 6 kN

FAB

A

C

B

10 kN

6.0 (C)

11.66 (T)

7 Free-body diagram of joint B

An "Answer diagram" summarizes the analysis of the entire truss (All forces are in kN).

(^9) Equilibrium equation for joint B 10

Fx^ = 0: FAB = 0

Solving gives

FAB = 0 Ans.

The force FBCis directed toward

the joint because member BC is

known to be in compression.

2. Determine the force in each member of the truss and state whether the force is tension or compression.

2 kip 4 kip 2 kip

14 ft

10 ft 10 ft 10 ft 10 ft

A

B C D

E

F G H

Free-body diagram of joint E. This joint is chosen because only two unknown forces are present. Thus we know that we can solve for these forces because two equations of equilibrium are available for the joint. Note also that we assume that both unknown forces are in tension (directed away from the joint).

Using = 54.46° in Eqs. 4 and 5 and solving simultaneously gives

FDE = 2.857 kip (T) Ans.

FEH = 4.916 kip = 4.916 kip (C) Ans.

We arbitrarily assumed member EH to be in tension. We then found that the member force was negative, so we know that our assumption was wrong. Member EH is in compression, and we show this by writing a positive "4.916" followed by "(C)".

Equilibrium equations for joint E

Fx = 0: FDE FEH cos = 0 (4)

Fy = 0: FEH sin 4 kip = 0 (5)

= tan-1( ) = 54.46°

Geometry

Ey = 4 kip

E

FEH

FDE

H

E

D

14 ft

10 ft

Use a free-body diagram of joint H next because only two member forces are unknown.

FEH = 4.916 kip (C)

FDE = 2.857 kip (T)

Equilibrium equations for joint H

Fx = 0: FGH (4.916 kip) cos 54.46° = 0 (6)

Fy = 0: FDH + (4.916 kip) sin 54.46° = 0 (7)

Solving simultaneously gives

FGH = 2.858 kip = 2.858 kip (C) Ans.

and

FDH = 4.0 kip (T) Ans.

As before, we assume that the unknown member forces (GH and DH in this instance) are tension, so are directed away from the joint. The force in member EH has already been found to be 4.916 kip compression, so it is directed towards the joint, not away from it.

Free-body diagram of joint H

F

A

G H

B C D

E

2 kip (^) 4 kip 2 kip

H

FEH = 4.916 kip (C)

FGH

FDH

FDE = 2.857 kip (T)

FEH = 4.916 kip (C)

F

A

G H

B C D

E

2 kip (^) 4 kip 2 kip

FGH = 2.857 kip(C)

FDG= 2.458 kip (C)

FDH = 4.0 kip (T)

FCD = 4.286 kip (T) Use a free-body diagram of joint C because only two member forces are unknown. Free-body diagram of joint C The unknown forces in members CG and BC are assumed to be tension, so are directed away from the joint. The force in member CD has already been found to be 4. kip (T).

Equilibrium equations for joint C.

Fx = 0: FBC + 4.286 kip = 0 (10)

Fy = 0: FCG 4 kip = 0 (11)

Solving gives

FBC = 4.286 kip (T) Ans.

and

FCG = 4 kip (T) Ans. 4 kip

C

FCG

FBC^ FCD = 4.286 kip (T)

F G H

E

B C D

A

All remaining bar forces follow from symmetry.

Answer diagram

4.92 (C)

2.86 (C) 2.86 (C)

4.00 (T)

4.0 (T)

4.0 (T)

4.92 (C) 2.46 (C)

4.29 (T) 4.29 (T)

2.46 (C)

2.86 (T) 2.86 (T)

All forces in kips.

900 lb

400 lb

10 ft

12 ft

B C

D

E

Ex Ey

Because AB is a two-force member, the line of action of FAB must pass through A and B.

Free body-diagram of entire truss

Equilibrium equations for entire truss

Fx = 0: FAB sin 60° + Ex + (900 lb) cos 30° = 0 (1)

Fy = 0: FAB cos 60° + Ey + (900 lb) sin 30° 400 lb = 0 (2)

MC = 0: (400 lb)(10 ft + FAB cos 60°(10 ft) + Ex(12 ft) = 0 (3)

Solving simultaneously gives

FAB = 347.8 lb

Ex = 478.2 lb,

Ey = 123.9 lb

FAB

Free-body diagram of joint D. Joint D is chosen because only two member forces are unknown there.

Equilibrium equations for joint D

Fx = 0: FDE = 0 (4)

Fy = 0: FBD 400 lb = 0 (5)

Solving gives

FBD = 400 lb (T) Ans.

FDE = 0 Ans.

Free-body diagram of joint C. Joint C is chosen because only two member forces are unknown there.

Equilibrium equations for joint C

Fx = 0: FBC +779.4 lb = 0 (6)

Fy = 0: FCE + 450 lb = 0 (7)

Solving gives

FBC = 779.4 lb (T) Ans.

FCE = 450.0 lb (T) Ans.

D

400 lb

FDE

FBD

FCE

C

FBC

(900 lb) sin 30° = 450.0 lb

(900 lb) cos 30°= 779.4 lb

The unknown forces have been assumed to be tension. The unknown forces have been assumed to be tension.

4. Determine the force in each member of the truss and state whether the force is tension or compression. The truss is symmetric.

2 kip

2 kip

2 kip

2 kip

2 kip

6 ft 6 ft 6 ft 6 ft 6 ft 6 ft

A

B C

D

E F G H

I J

K

K

I J

E F G H

D

B C

A

6 ft 6 ft 6 ft 6 ft 6 ft 6 ft

2 kip

2 kip

2 kip

2 kip

2 kip

Ax Ay (^) D y

Free-body diagram of entire truss

Equilibrium equations for entire truss

Fx = 0: Ax = 0

Fy = 0: Ay + Dy 2 kip 2 kip 2 kip 2 kip 2 kip = 0

MA = 0: 2(kip)(6 ft) 2 kip(12 ft) 2(kip)(18 ft) 2 kip(24 ft) 2(kip)(30 ft) + Dy(36 ft) = 0

Solving simultaneously gives

Ax = 0

Ay = 5 kip

Dy = 5 kip

Three unknown member forces are present at joint I, but two of them, FEI and FIK, are collinear, so summing forces perpendicular to FEI and FIK would give an equation with FFI as the only unknown.

Free-body diagram of joint I

Geometry of members at joint I

Equilibrium equations for joint I

Fy = 0: FFI sin 90° (2 kip) sin 60° = 0

Solving gives

FFI = 1.732 = 1.732 kip (C) Ans.

So member FI is perpendicular to the x axis. Thus the member force FFI lies on the y axis.

K

I J

E F G H

D

B C

A

2 kip

2 kip

2 kip

2 kip

2 kip

Ax = 0 Ay = 5 kip Dy = 5 kip

2 kip

I

y

x

FIK

FEI

FFI

E 60°

I

A

F

x

y

Use the same technique at joint F as was used at joint I: sum forces perpendicular to collinear members BF and FK.

Geometry of members at joint F

Free-body diagram of joint F

Equilibrium equations for joint F

Fy = 0: FEF sin 60° (1.732 kip) sin 60° = 0

Solving simultaneously gives

FEF = 1.732 kip (T) Ans.

K

I J

E F G H

D

B C

A

2 kip

2 kip

2 kip

2 kip

2 kip

Ax = 0 Ay = 5 kip Dy = 5 kip

E F

I

B

FFI = 1.732 kip (C) (already known)

F

y 60°

x

FFI = 1.732 kip (C)

FBF

FEF

FFK