6.7 Analysis of trusses By the method of sections

The method of joints is most effective when the forces in all the members of a truss are to be determined. If, however, the force in only one member of the forces in a very few members are desired,

an other method, the method of sections, is more efficient.

Assume for example, that we want to determine the force in member BD of the truss shown in figure 1(a).To do this, we must determine the forces with which member BD acts on either joint B or joint D. If we were to use the method of joints, we would choose either joint B or joint D as a free body.However, we can also choose as a free body a larger portion of the truss, composed of several joints and members,provided that the desired force is one of the external forces acting on that portion. If, in addition, the portion of the trussis chosen so that there is a total of only three unknown forces acting upon it, the desired force can be obtained by solvingthe equations of equilibrium for this portion of the truss.In practice, the portion of the truss to be utilized is obtained by passing a section through three members of the truss, one of which is the desired member, that is, by drawing a line which divides the truss into two completely separate parts but does not intersect more than three members.Either of the two portions of the truss obtained after the intersected members have been removed can then be used as afree body.In figure 1 (a), the section nn has been passed through members BD, BE, and CE, and the portion ABC of the truss is chosen as the free body ( figure 1b). The forces acting on the free body are the loads P1 and P2 at points A and B and the three unknown forces FBD, FBE, and FCE. Since it is not known whether the members removed were in tension or compression, the three forces have been arbitrarily drawn away from the free body as if the members were in tension.

Sample problems 6.2

Determine the force in members EF and GI of the truss shown.

Free body : Entire Truss. A free body diagram of the entire truss is drawn; external forces acting on this free body consist of the applied loads and the reactions at B and J. We write the following equilibrium equations.

= 0

= 0

Force in member EF : Section nn is passed through the truss so that it intersects member EFand only two additional members. After the intersected members have been removed, the the left handportion of the truss is chosen as a free body. Three unknown are involved; to eliminate the two horizontalforces, we write

The sense of FEF was chosen assuming member EF to be in tension ; the negative sign obtained indicates that the number is in compression.

FEF = 22.5 kN

Force in member GI . Section mm is passed through the truss so that it intersects member GI and only two additional members. After the intersected members have been removed, we choose the right-hand portion of the truss as a free body. Three unknown forces are again involved;to eliminate the forces passing through point H, we write

= 0

FGI = 46.0 kN

Sample problems 6.3

Determine the force in members FH, GH, and GI of the roof truss shown

SOLUTION:

• Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

kN 5.12

kN 200

kN 5.7

m 25kN 1m 25kN 1m 20

kN 6m 15kN 6m 10kN 6m 50

A

ALF

L

L

M

y

A

tan α = GL 15 m

8 mFG= = 0.5333 α = 28.07º

30 m

• Pass a section through members FH, GH, and GI and take the right-hand section as a free body.

kN 13.13

0m 33.5m 5kN 1m 10kN 7.50

0

GI

GI

H

F

F

M

• Apply the conditions for static equilibrium to determine the desired member forces.

TFGI kN 13.13

Force in member GI

kN 82.13

0m 8cos

m 5kN 1m 10kN 1m 15kN 7.5

0

07.285333.0m 15

m 8tan

FH

FH

G

F

F

MGL

FG

CFFH kN 82.13

kN 371.1

0m 10cosm 5kN 1m 10kN 1

0

15.439375.0m 8

m 5tan

32

GH

GH

L

F

F

M

HI

GI

CFGH kN 371.1

Force in member FH

Force in member GH

15

Problems

A Warren bridge truss is loaded as shown.Determine the force in members CE, DE and DF

Problems 6.42

Ky

Kx

Ay 3kN 3kN

FBD Truss

Ky

Kx

Ay 3kN 3kN

A

B

C

D

E

FDE

FDF5m

5m

3kN4.2kN

FCE

6m 12

5

13

Ky

Kx

Ay 3kN 3kN

A

B

C

D

E

FDE

FDF5m

5m

3kN4.2kN

FCE

6m

6.10 Analysis of a Frame

• Frames and machines are structures with at least one multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces.

• A free body diagram of the complete frame is used to determine the external forces acting on the frame.

• Internal forces are determined by dismembering the frame and creating free-body diagrams for each component.

• Forces between connected components are equal, have the same line of action, and opposite sense.

• Forces on two force members have known lines of action but unknown magnitude and sense.

• Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components.

• Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies.

• A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions.

• The frame must be considered as two distinct, but related, rigid bodies.

• With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components.

• Equilibrium requirements for the two rigid bodies yield 6 independent equations.

6.11 Frames which cease to be rigid when detached from their support

Sample problem 6.4

In the frame shown, members ACE and BCD are connected by a pinat C and by the link DE. For the loading shown, Determine the force in link DE and the components of the force exerted at C on member BCD

SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

Members. We now dismember the frame. Since only two members are connected at C, the components of the unknown forces acting on ACEand BCD are, respectively, equal and opposite and are assumed directedas shown. We assume that link DE is in tension and exerts equal and opposite forces at D and E, directed as shown.

Free Body : member BCD Using the free body BCD, we write

FDE = 561 N C

From the signs obtained for Cx and Cy we conclude that the force components Cx and Cy exerted on memberBCD are directed, respectively, to the left and up. We have

Cx = 795 N Cy = 216 N

Free body : member ACE ( check )The computations are checked by considering the free body ACE. For example,

Check

Sample problem 6.5

Determine the components of the forces acting on each member of the frame shown.

FBD

Free body : Entire Frame. Since the external reactions involve only three unknowns,we compute the reactions by considering the free body diagram of the entire frame.

F = 1800 N

Ey = 600 N

Ex = 0

Members. The frame is now dismembered; since only two members are connected at each joint,equal and opposite components are shown on each member at each joint.

Free Body : Member BCD

Cy = + 3600 N

By = + 1200 N

We note that neither Bx nor Cx can be obtained by considering only member BCD. The positive values obtained for By and Cyindicate that the force components By and Cy are directed as assumed

Free Body : Member ABE

Bx = 0

Ax = 0

Ay = + 1800 N

Free Body : Member BCDReturning to member BCD, we write

Cx = 0

Free Body : Member ACF ( Check )All unknown components have now been found;to check the results, we verify that member ACF is in equilibrium.

Check

Sample problem 6.6

2.7 kN

0.75 m

0.75 m

0.75 m

0.75 m

1.8 m

A 2.7 kN horizontal force is applied to pin A of the frame shown. Determine the force acting on the two vertical members of the frame.

2.7 kN

1.8 m

3.0 m

2.7 kN

Free body: Entire FrameThe entire frame is chosen as a free body although the reaction involve four unknowns, Ey and Fy may be determined by writing

2.7 kN

0.75 m

1.5 m

4.5 kN 4.5 kN

0.75 m

2.25 m

2.7 kN

0.75 m

1.5 m

4.5 kN 4.5 kN

0.75 m

2.25 m

PROBLEMS

Problems 6.77

90 N 90 N

100mm250mm

50mm

200mm

Determine the force in member AC and the reaction at B when(a) = 30˚(b) = 60˚

90 N 90 N

100mm250mm

50mm

200mm

Solution

θ

90 N 90 N

100mm250mm

Bx

By

FAC

C D

90 N 90 N

100mm250mm

50mm

200mm

Solution

θ

90 N 90 N

100mm250mm

Bx

By

FAC

C D

90 N 90 N

100mm250mm

50mm

200mm

Solution

θ

90 N 90 N

100mm250mm

Bx

By

FAC

C D

6.12 Machines

Photo 1 : The lamp shown can be replaced in many positions.By considering various free bodies, the force in the springs and the internal forces at the joints can be determined.

• Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces.

• Given the magnitude of P, determine the magnitude of Q.

• Create a free-body diagram of the complete machine, including the reaction that the wire exerts.

• The machine is a nonrigid structure. Use one of the components as a free-body.

• Taking moments about A,

Pb

aQbQaPM A 0

6.12 Machines

Sample Problem 6.7

A hydraulic-lift table is used to raise a 1000 kg crate. It consists of a platform and two identical linkages on which hydraulic cylinders exert equal forces. (Only one linkage and one cylinder are shown. Members EDB and CG are each of length 2a, and member AD is pinned to the midpointof EDB. If the crate is placed on the table, so that half of its weight is supported by the systemshown,Determine the force exerted by each cylinder in raising the create for θ= 60˚ , a = 0.70 m, andL = 3.20 m.Show that the result obtained is independent of the distance d.

The machine considered consists of the platform and the linkage. Its free body diagram includes a force FDH exertedby the cylinder, the weight ½ W, and reactions at E ad G that we assume to be directed as shown. Since more than three unknowns are involved, this diagram will not be used. The mechanism is dismembered and a free body diagram is drawnfor each of its component parts. We note that AD, BC and CG are two force members. we already assumed member CGto be in compression. We now assume that AD and BC arein tension; the forces exerted on them are then directed as shown.Equal and opposite vectors will be used to represent the forces exerted by the two force members on the platform, on member BDE, and on roller C.

Solution

PROBLEMS

Problem 6.139

Determine the magnitude of the gripping forces exerted along line aa on the nut when two 240-N forcesare applied to the handles as shown.Assume that pins A and D slide freely in slots cut in the jaws.

Solution

0.01m0.03mA

By

Bx

G

240N

0.09m

0.015m

Cx

CyEx = 0

Ey = 4G 3

A = G 3

A

C

E

100mm

25mm

500mm

25mm

25mm

12.5mm

340N

340N

Problem 6.140

In using the bolt cutter shown, a worker applies two 340 N forces to the handles.Determine the magnitude of the forces exerted by the cutter on the bolt.

II

THE END