trusses method of sections

6.3 Trusses: Method of Sections

6.3 Trusses: Method of Sections Example 1, page 1 of 2

5 m

C D

EF

G H

5 m

5 m

B

3 m

A

1. Determine the force in members AC, CD, and DF, and

state whether the force is tension or compression.

1 Pass a section through the three

members whose forces are to be

determined.

A B

C D

EF

G H

4 kN

2 kN

6 kN 6 kN

2 kN

4 kN


F x = 0: 2 kN + 4 kN + F

CD = 0

F y = 0: F

AC F DF = 0

M D = 0: F

AC(3 m) (2 kN)(5 m + 5 m)

(4 kN)(5 m) = 0

Solving simultaneously gives

F AC = 13.33 kN (T) Ans.

F CD = 6.0 kN = 6.0 kN (C) Ans.

F DF = 13.33 kN = 13.33 kN (C) Ans.

+

++

We had assumed member CD to be in tension.

Calculations showed that F CD is negative, so our

assumption was wrong: CD must be in compression.

Similarly DF must be in compression.

5

2 kN

4 kN

C

Equations of equilibrium for the portion of the

truss (Note that moments are summed about

point D, even though point D is not part of the

free body):

4

F AC

F CD

F DF

At each cut through a

member, a force is shown

to represent the effect of

the portion of the member

on one side of the section

pulling on the portion on

the other side. It is

convenient to always

assume the force to be

tension.

3

Free-body diagram of portion of truss above the section2

3 m

5 m

HG

FE

D

5 m


2. Determine the force in members CD, CH, and GH, and state whether the force is tension or compression.

12 ft 12 ft 12 ft 12 ft

800 lb 800 lb 800 lb

A

F G H

B C D

E

800 lb

A

B

800 lb

C

F G

D

800 lb

H

E

1 Pass a section through the three


determined.

8 ft


D

800 lb

H

E

F GH

F CH

F CD

E y

2 Free-body diagram of portion of truss to right of section

At each cut through a member, a force is shown.3

4 Equations of equilibrium for the portion of the

truss:

F x = 0: F

GH F CH sin F

CD = 0 (1)

F y = 0: F

CH cos 800 lb + E y = 0 (2)

M H = 0: F

CD (8 ft) + E y (12 ft) = 0 (3)

12 ft

+

++

5

Three equations but four unknowns, so another

equation is needed.

8 ft

12 ft

6

Geometry

= tan-1 = 56.31°

H

DC

8 ft

12 ft8 ft


800 lb

12 ft12 ft

7

A y

B

800 lb

C

Free-body diagram of entire truss.

F G

12 ft

D

800 lb

H

E y

EA x A

12 ft

8

+

9

Equilibrium equation for entire truss. This will give the needed fourth equation.

M A = 0: (800 lb)(12 ft) (800 lb)(2 12 ft) (800 lb)(3 12 ft) + E

y(4 12 ft) = 0

Solving gives E y = 1,200 lb.

Substituting E y = 1,200 lb into Eqs. 1, 2, and 3 and solving simultaneously gives

F CD = 1,800 lb (T) Ans.

F CH = 721 lb (T) Ans.

F GH = 2,400 lb = 2,400 lb (C) Ans.

8 ft


4 m 4 m 4 m

3. The diagonal members are not connected to each other where they cross. Determine the force

in members BG, CF, and FG, and state whether the force is tension or compression.

E

3 kN

F G H

D

CBA

3 kN 3 kN

BA

3 kN 3 kN

E F

C

3 kN

G

D

H

1 Pass a section through the three members

whose forces are to be determined.

2.5 m


F FG

F BG

F CF

2

3 At each cut through a member, a force is shown

4 m

4 Equations of equilibrium for the portion of the truss:

F x = 0: F

FG F BG sin F

CF sin = 0 (1)

F y = 0: F

BG cos + F CF cos + D

y 3 kN = 0 (2)

M G = 0: F

CF sin (2.5 m) + D y(4 m) = 0 (3)

+

5

Three equations but four unknowns, so

another equation is needed.

4 m

C

GF

6

= tan-1 = 58.0°

Geometry

D y

++

2.5 m

2.5 m

2.5 m4 m

G

3 kN

C

Free-body diagram of portion of truss to right of section

D

H


Free-body diagram of entire truss (This will give the needed fourth equation).

3 kN3 kN

E x

7

FE

A B

E y

D y

3 kN

HG

C

D

8

+

Equilibrium equation for entire truss.

M E = 0: (3 kN)(4 m) (3 kN)(2 4 m) + D

y(3 4 m) = 0

Solving gives D y = 3.0 kN. Then substituting = 58.0° and D

y = 3.0 kN into Eqs. 1, 2, and 3 and solving

simultaneously gives

F BG = 5.66 kN (T) Ans.

F CF = 5.66 kN (T) Ans.

F FG = 9.6 kN = 9.6 kN (C) Ans.

2.5 m

4 m4 m4 m


3 m1.5 m

3 m

6 m

3 m

3 m

3 m

4. Determine the force in members CE, EF, HF, and CF, and state whether the force is tension or compression.

I

G

E

C

A B

D

F

H

J

6 kN

4 kN

6 m

4 kN

6 kN

Pass a section through at

least some of the members

whose forces are to be

determined. The general

idea is to choose as few

members as possible --three

in this instance-- because

each time a member is cut

by a section, an additional

unknown is introduced into

the equilibrium equations.

1

I J

HG

E F

CD

BA


3 m

3 m

I

G

E F

H

J 6 kN

4 kN

L GH

L EF

2 Free-body diagram of portion of truss above section (Using

the upper portion of the truss rather than the lower

eliminates the need to calculate the reactions at the bottom

of the truss).


MG = 0: (6 kN)(3 m) + FEF(3 m) F FH cos (L

GH) = 0 (1)

M F = 0: (6 kN)(2 3 m) 4 kN)(3 m) + F

CE cos (L EF) = 0 (2)

F x = 0: F

CE sin + F EF + F

FH sin + 4 kN + 6 kN = 0 (3)+

++

F EF

F CE

F FH


E

CD

G

I

F

H

J

BA

4 Geometry5

3 m

3 m

3 m

3 m

3 m

3 m

3 m

3 m

3 m 3 m

3 m

3 m

3 m

(6 m) tan

(3 m) tan

= tan-1 = tan-1 = 7.125°

L GH = 3 m + (3 m) tan + (3 m) tan = 3.75 m

L EF = 3 m + (6 m) tan + (6 m) tan = 4.50 m

6 Substituting these values for , LGH, and LEF into Eqs. 1, 2, and

3 and solving simultaneously gives:

F CE = 10.75 kN Ans.

F EF = kN = 7.33 kN (C) Ans.

F FH = 10.75 kN = 10.75 kN (C) Ans.

B'1.5 m 1.5 m

1.5 m

4 3 mJB'

BB'


8 Equilibrium equations for joint F

F x = 0: F

CF cos + F DF sin 7.125° + (10.75 kN)(sin 7.125°) 7.33 kN = 0 (4)

F y = 0: F

CF sin F DF cos 7.125° (10.75 kN)(cos 7.125°) = 0 (5)

+

+

L EF = 4.50 m

(3 m) tan

E F

DC

9 Geometry

= tan-1 = 31.61°

10 Substituting = 31.608° into Eqs. 4 and 5 and solving simultaneously gives:

F DF = 14.97 kN = 14.97 kN (C)

F CF = 7.99 kN (T) Ans.

F EF = 7.33 kN (C)

F CF

F DF

F FH = 10.75 kN (C)

3 m

(3 m) tan 7.125° + 4.50 m

3 m = 7.125°

7 Free-body diagram of joint F.

This free body will enable us to

calculate the remaining unknown

force the force in member CF.

F


G

M

1 Pass a section through the four


determined. It does not appear

possible to find a section that cuts

only three of these members.

5. Determine the force in members RS, LS, FL, and EF, and state whether the force is tension or compression.

N O P Q R S T

H I J K L M

AB C D E F

G

2 m

3 m

2 m

3 m3 m3 m3 m3 m

4 kN 4 kN 4 kN

C

N O

A

H

B

I L

P Q R S

4 kN 4 kN

D

J

4 kN

E F

K

T


3 m

F

S

2 mG

T

M2 m

2 Free body diagram of truss portion to right of section line

F RS

F LS

F FL

F EF

G y


M S = 0: F

EF(2 2 m) + G y(3 m) = 0 (1)

M F = 0: F

RS(2 2 m) + G y(3 m) = 0 (2)

F y = 0: F

FL F LS + G

y = 0 (3)++

+

Three equations with five unknowns so two more

equations are needed.4


4 kN

3 m

P

3 m

A

3 m

B C

N

H

O

I J

3 m3 m

D

4 kN

3 m

4 kN

E F

Q SR

K L

2 m

G

T

M2 m

5 Free-body diagram of entire truss (This free body will enable us to calculate the reaction at G).

6 Equation of equilibrium for the entire truss.

M A = 0: (4 kN)(2 3 m) (4 kN)(3 3 m) (4 kN)(4 3 m) + G

y(18 m) = 0 (4)

Solving gives

G y = 6 kN

Substituting G y = 6 kN into Eqs. 1 and 2 and solving gives:

F EF = 4.5 kN (T) Ans.

FRS = 4.5 kN = 4.5 kN (C) Ans.

A x

A y G

y

+


S

7 Free-body diagram of joint S. This free body will

enable us to calculate the force in member LS.

F LS

F MS

F ST

F RS = 4.5 kN (C)

8 Equations of equilibrium for joint S. Note that there

are three unknowns but only two equations.

F x = 0: 4.5 kN + F

ST + F MS cos = 0 (5)

F y = 0: F

LS F MS sin = 0 (6)

+

+

9 Geometry

S T

M

2 m

3 m

= tan-1 = 33.69°

10 Free body diagram of joint T

F ST

F MT

T

11 Two members meet at joint T, they are not collinear and no

external force acts at joint T, so members ST and MT are zero-

force members.

Substituting F ST = 0 in Eq. 5 and solving Eqs. 5 and 6

simultaneously gives:

F MS = 5.41 kN = 5.41 kN (C)

F LS = 3.0 kN (T) Ans.

12 Substituting F LS = 3.0 kN and G

y = 6 kN into Eq. 3 and

solving gives:

F FL = 3.0 kN = 3.0 kN (C) Ans.

2 m3 m


5 m

5 m

T

N

A

B C D E F G H I J K L

M

O P Q R S

U V WX

10 kN12 panels @ 4 m each

6. Determine the force in members TU, EF, and EU. State whether the force is tension or compression.

SQ RPON

A

CB ED

10 kN

HGF JI

M

LK

T U V W X

1 Even though we were not asked to determine the force in member EP,

we have to pass the section through it because we must make the

section go completely through the truss.


T

F EF

F EP

F EU

F TU

A y

A x

5 m

5 m

4 m 4 m 4 m 4 m

2 Free-body diagram of portion of the truss to the left of the section


F x = 0: A

x + F TU + F

EP cos + F EF = 0 (1)

F y = 0: A

y + F EU + F

EP sin = 0 (2)

M E = 0: A

y (4 4 m) F TU(2 5 m) = 0 (3)

+

++

4 Three equations with six unknowns so three more

equations are needed.

5 Geometry

P

F

E

4 m

5 m 4 m5 m = tan-1 = 51.34°

ON

A

B C D E


12 panels @ 4 m each

CB

A

N

T

RQPO

10 kN

GFED JIH

S

LK

M

XU V W

6 Free-body diagram of entire truss (This free body will enable us to calculate the reactions at support A).

A x

A y

M y

7 Equations of equilibrium for entire truss. Note that we

only write two equations because we only need to

calculate Ax and Ay, since only Ax and Ay appear in

Eqs. 1, 2, and 3.

F x = 0: A

x = 0 (4)

M M = 0: (10 kN)(6 4 m) A

y(12 4 m) = 0 (5)

Solving gives A x = 0 and A

y = 5 kN.

Consideration of joint F shows that member FP is a zero-force

member, so F FP = 0.

But if member FP is removed (because it is a zero-force

member), consideration of joint P shows that member EP is also

a zero-force member, so F EP = 0.

8

+

+


9 Substituting

= 51.34°,

A x = 0,

A y = 5 kN,

and

F EP = 0

into Eqs. 1, 2, and 3, and solving gives:

F TU = 8 kN = 8 kN (C) Ans.

F EF = 8 kN (T) Ans.

F EU = 5 kN = 5 kN (C) Ans.


7. Determine the force in

members KM, LM, and

DK. State whether the

force is tension or

compression.

1 We choose a section that cuts at least some of the

members whose forces are to be determined. But the

section should cut as few other members as possible,

since each time a member is cut, an additional

unknown appears in the equilibrium equations.

L

2 kN

2 kN

A

G

2 kN

2 kN

B C

I

D

KJ

M

E

H

2 kN

2 kN

F

2 kN

I

2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m

I

6 m2 kN

2 kN

2 kN

2 kN

2 kN

2 kN

2 kN

H

L

M

KJI

G

FEDCB

A


ED

K L

2 kN

F

H

2 kN

2 kN

F KM

F ML

F CD

F y

2 Free-body diagram of portion of truss to right of section.

It is not essential but we can save some work if we use

the principle of transmissibility as shown in Step 3.

F CD

D

K

F y

2 kN

E

H

L

2 kN

F

2 kN

F KM cos

F KM sin

F ML cos

F ML sin

3 Same free body as in Step 2, but now the force F KM has been

moved along its line of action to joint D (principle of

transmissibility) and then expressed in terms of vertical and

horizontal components. Similarly F ML is moved to joint F.

4 Equations of equilibrium for free body in Step 3. Note that

because we were not asked to determine F CD, we choose two

moment equations in which F CD does not appear.

M F = 0: (2 kN)(2 m) + (2 kN)(3 2 m) F

KM sin (4 2 m) = 0 (1)

M D = 0: (2 kN)(2 m) (2 kN)(3 2 m) (2kN)(4 2 m) + F

y(4 2 m) + F ML sin (4 2 m) = 0 (2)

5 Two equations but three unknown forces, so another

equilibrium equation is needed.

++

2 m2 m 2 m 2 m


6 m

6 Geometry

= tan-1 ( 2 m ) = 71.56°

= tan-1 ( 2 m + 8 m

) = 30.96°

2 m

8 m

6 m

6 m

M

DF


C D

I

2 kN

LJ K

M

2 kN

FE

F y

2 kN

H

2 kN

8 Equilibrium equation for entire truss

M A = 0: (2 kN)(2 m) (2 kN)(3 2 m) (2 kN)(5 2 m)

(2 kN)(7 2 m) (2 kN)(9 2 m) (2 kN)(10 2 m) + F y(10 2 m) = 0 (3)

9 Solving gives F y = 7 kN. Substituting F

y = 7 kN, = 71.56°, and

= 30.96° in Eqs. 1 and 2, and solving simultaneously gives:

F KM = 2.11 kN (T) Ans.

F ML = 5.83 kN = 5.83 kN (C) Ans.

+

I

2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m

Free-body diagram of entire

truss. This free body will

enable us to calculate the

reaction at support F.

7

2 kN

2 kN

AA x

A y

2 kN

G

B


F DK

10 Free-body diagram of joint K (This free body will enable

us to calculate the force in member DK).

11 Since there are only two unknown forces, F KL and F

DK,

we could write force-equilibrium equations in the x

and y directions and then solve them simultaneously.

However, we can save work by noticing that a

zero-force member is present.

2 kN

L

2 kN

A

G

2 kN

B C

I

D

KJ

2 kN

2 kN M

E

H

2 kN

2 kN

FA

x

A y

F y

12 Free- body diagram of entire truss13 Consideration of joint K shows that KL must

be a zero-force member, so F KL = 0.

I

K

y

x

F KM = 2.11 kN (T)

F KL


14 Free-body diagram of joint K (repeated)

F y = 2.11 kN F

DK = 0 (4)

Solving gives

F DK = 2.11 kN (T) Ans.

+

F KM = 2.11 kN (T)

y

F KL

F DK

K

x


A

B

F

C D E

G H I

Cable

J30°

1 The section must pass through the

cable. Otherwise the portion of the

truss to the left of the section could

not be isolated as a free body.

5 ft 5 ft 5 ft 5 ft 5 ft

JIHGF

EDCB

A

Cable

8. Determine the force in members GH,

CD, and CH. State whether the force is

tension or compression. Also, find the

tension in the cable.

5 kip

3 ft

5 kip

30°


5 kip

A

B

F

C

G

2 Free-body diagram of portion of truss to left of section

T

F GH

F CH

F CD

3 The tension in the cable is

one of the unknowns.


M C = 0: T cos 30°(3 ft) F

GH(3 ft) + (5 kip)(2 5 ft) = 0 (1)

M G = 0: (5 kip)(2 5 ft) + F

CD(3 ft) + F CH sin (3 ft) = 0 (2)

F y = 0: 5 kip + T sin 30° + F

CH cos 0 (3)++

+

C

G H

5 Geometry

3 ft

5 ft

= tan-1 ( 3 ft

) = 59.04°

6 Three equations but four

unknown forces, so another

equilibrium equation is needed.

3 ft

5 ft 5 ft

5 ft

30°


J x

8

5 kip

7

Equation of equilibrium for the entire truss. Only one equation is

used because we need to calculate T only; the reactions at J are not

needed.

M J = 0: (5 kip)(5 5 ft) T sin 30°(3 5 ft) = 0 (4)

T = 16.67 kip Ans.

Substituting = 59.04° and T = 16.67 kip into Eqs. 1, 2, and 3 and

solving simultaneously gives:

F GH = 2.23 kip (T) Ans.

F CD = 11.11 kip = 11.11 kip (C) Ans.

F CH = 6.48 kip = 6.48 kip (C) Ans.

+

9

5 ft5 ft5 ft5 ft5 ft

3 ft

30°

E

Free-body diagram of entire truss

A

B

F

C D

G

T

H

J y

I J

trusses method of sections

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