METHOD OF SECTIONS

The method of sections is useful if one needs to know the forces of a small number of interior members of a truss. It eliminates the tedium of working joint by joint across the truss. It is also advantageous in that answers are found directly without dependency on the results of a previous calculation. This means a mistake in one part of the solution does not necessarily affect the rest of the problem

The general approach of this method involves cutting the truss into two sections by drawing a line through not more than three members of interest. Select one side of the truss and draw a FBD. Either side can be selected. The resulting FBD of the truss section is a general force problem solvable using basic equations of equilibrium

There are two variations of the method of sections: the component method and the shear method. Both methods are quite similar with only the geometry of the truss affecting which method one uses

Method of Sections

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The component method is used when a truss does not have parallel chords

Begin by finding support reactions. Section the truss and choose the section with which you wish to work. Your section will have three non parallel unknowns. Any pair of unknowns will have a point of ‑concurrency. These points of concurrency may or may not lie on the truss. If you sum moments about that point of concurrency, the resulting moment equation will have only a single unknown variable.

The remaining unkowns can be solved similarly or one can sum of forces in the x and y directions‑ ‑

Method of Sections

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For example, assume one wishes to find forces in members BC, BH, and JH Solve for reactions Section the truss Determine the point of

intersection of each pair of unknowns

Sum moments about point 'O' to find the load in BH

Sum moments about joint 'H' to find the load in BC

Sum moments about joint 'B' to find the load in JH

Method of Sections

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400 lbf

BC

D

EGHJA

400 lbf

B

JA

BC

BH

JH HO

Alternatively, one can write a single sum of moments about any one of joints 'O', 'H', or 'B' plus two sums of forces

The shear method is used when the top and bottom chords of the truss are parallel. Because they are parallel, there is no point of concurrency for those two members; a sum of moments will not work

However, the third member will lie at an angle to these parallel members. As such, the force in that member can be found by summing forces in a direction orthognal to the parallel members. Once found, the other members can be solved using moment equations in a manner similar to that used in the component method

Method of Sections

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For example, assume one wishes to find the forces in members BC, BH, and JH Solve for reactions Section the truss

Use a ΣFyto determine the

load on member BH Sum moments about joint 'B'

to find the load in JH Sum moments about joint 'H'

to find the load in BC -or-Sum forces in the x direction ‑to find the load in BC

Method of Sections

6

400 lbf

B C D

EGHJA

400 lbf

B

JA

BC

BH

JH H

Consider the truss shown below. Find the forces in members BC, BH and JH

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Example 3 – Component Method

72 kN

BC

D

EGHJA 4m 4m

4m

4m 4m

3m 3m

Reactions

M A = 0 =−72 12E 16

E = 54 kN

F y = 0 = A−7254A=18 kN

Section the truss through members BC, BH, and JH. Draw a FBD of the left side making sure the reaction at 'A' is shown correctly. Locate the points of intersection of each pair of unknown members

To find the intersection of BC-JH, we only need extend BC along its slope until it intersect a line passing through JH. Member BC is at a slope of 1:4 and point 'B' is 3 meters above JH. This means point 'O' is located at the same elevation as JH but at a distance of 12 meters to the left of joint 'B'

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Example 3 – Component Method

BC

M H = 0 =−18 8 −4

17BC 3−

1

17BC 4

BC =−37.154BC = 37.154 kN C

BH

M O = 0 = 188−35

BH 12−45

BH 3

BH =15BH = 15 kN T

JHM B = 0 =−184HG 3

HG =24HG = 24 kN T

18 kN

B

HJA 4m

3m

O

41

43

BC

BG

JH

12 m

4m

The truss below is hinged at joint 'A' and is supported by a roller at joint 'F'. Determine the forces in members DE, DG and HG

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Example 4 – Shear Method

12 kipB

C

D E

GH

J

4'

6'

12 kip

A

F6'

6'8'8'8'

2 kip

2 kip

2 kip

ReactionsM A = 0 =−2 6−212−2 18−128−1216F 24

F = 15 kip

F x = 0 = 222−A xAx = 6 kip

F y = 0 = A y−12−1215A y = 9 kip

Use the right section of the truss to perform the member analysis.

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Example 4 – Shear Method

Since the top and bottom chords are parallel withno point of concurrency, solve for DG using F y

F y = 0 =35DG−1215

DG =−5DG = 5 kip C

M D = 0 =−GH 6−12 81516

GH =24GH = 24 kip T

F x = 0 =−24−DE−45

−5

DE =−20DE = 20 kip C

12 kip

E

G F

6'8'

DE

DG

GH

15 kip

43

D

When a student first learns truss analysis by the method of joints and the method of sections, they often consider them discrete. In other words, you must use one method or the other through out truss solution. This couldn't be further from the truth. It is quite common to use the method of sections to get inside a truss, then apply the method of joints to solve for the value one needs. Consider again the previous example. It is a simple matter to perform a joint analysis on joint 'E' to determine the load on member EG

These two methods of solution can compliment each other. Use them as necessary to efficiently accomplish the task at hand

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Joints and Sections

EDE=20 kip

EF

EG

43

F x = 0 = 20−45EF

EF = 25 kip C

F y = 0 =35

25−EG

EG = 15 kip T