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MODULE 3

METHOD OF SECTIONS

INTRODUCTION

The method of sections is another method of determining forces in members of a truss.

Its use permits us to determine directly the forces in almost any member instead of

proceeding to that member by a joint to joint analysis. In the method of sections, a

cutting plane is passed through the entire truss, separating it into two parts without

cutting more than three members. If this plane actually severed the members, the truss

would collapse. But suppose, at the instant of severing the members, that an external

force is applied to each side of the cut members exactly equivalent to the load being

transmitted by the members. We shall then have two parts of the truss, each

constituting a non concurrent system of forces in equlibrium under the action of the

known loads that act on each part and the unknown forces (stresses) that the members

of one part exert on the other.

SAMPLE PROBLEM 3.1

Using Method of Section determine the forces of member BC and FE

A

B C

DEF

Ra RD

12 KN

3 m. 3 m. 3 m.

4 m. 4 m.

53.13 53.13

Figure 3.1a

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SOLUTION

Determine first the reaction of the two support RAand RD

MD= 0; RA(9)12(3) = 0; RA= 4 KN

Fy = 0; RA + RD -12 = 0; RD = 8 KN

Next step is to pass a plane a-a to section BC and FE and consider only the left hand

portion of the truss

A

B C

DEF

Ra RD

12 KN

3 m. 3 m. 3 m.

4 m. 4 m.

53.13 53.13

a

a

Figure 3.1b

53.13A

B

F

BC

FE

RA= 4 KN

4 m

3 m

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Solve for BC and FE by taking moments at point F and at point B respectively

MF = 0; BC(4) 4(3) = 0; BC = 3 KN

MB = 0; FE(4) 4(3) = 0; FE = 3 KN

In our example unlike the method of joints you have to go through joint A, B and F

before you determine member BC and FE.

SAMPLE PROBLEM 3.2

Using Method of Section determine member CD, HI and HD

Cutting Planea

a

A

B C D E F

G H I J KL

RA RL

1 m 1 m 1 m 1 m 1 m 1 m

2 m

10 KN

Figure 3.2a

SOLUTION

Solve for the two reaction RAand RL

ML = 0; RA(6) 10(2) = 0; RA = 3.33 KN

Fy = 0; RA + RL 10 = 0; RL = 6.67 KN

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Next is to pass a plane a-a at member CD HI and HD cutting the truss into two halves

and consider only the left hand portion of the truss.

A

B C

G H

RA= 3.33 KN

1 m 1 m

63.43

CD

HD

HI

2 m

Figure 3.2b

Solve for member CD by taking moments at point H

MH = 0; 3.33(2) CD(2) = 0; CD = 3.33 KN

Solve for member HD and HI by taking moments at point C

MC = 0; 3.33(2) HI(2) + HD(2)COS63.43 = 0 equation 1

Fy = 0; 3.33 + HD(SIN63.43) = 0; HD = -3.72 KN

Negative sign indicates that our assume direction is wrong, HD is compression with

respect to point H

Substitute the value of HD = -3.72 KN in equation 1, we get

6.66HI(2) + (-3.72)COS63.43 = 0; HI = 2.5 KN

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PROBLEM

Using Method of Section Determine the forces in the lettered bars of the trusses shown

1.

10 KN

RARC

6 m

2 m.

a

b

2.

Ra RD

3 m. 3 m. 3 m.

4 m. 4 m.

53.13 53.13

a

b

12 KN

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3.

a

b

c

10 KN

2 m 2 m 2 m

3 m

4.

6 @ 3 m

3 m

3 m

a

b

20 KN

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7.

6 @ 2 m

2 m3 m

4 m

a

b

c

d

e

f

20 KN

8.

10 KN

10 KN

10 KNa

b

c

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9.

1.2 KN

1.2 KN2.4 KN

a

10.

3 m

3 @ 2 m

10 KN

a

b

c