1 VukazichCE160TrussAnalysisMethodofSections[3]

CE 160 Notes: Truss Method of Sections Example

A truss is pin supported at point A, roller supported at point L and subjected to the point loads shown. Determine the axial force in truss member FH.

Draw a Free-Body Diagram (F.B.D.) of the entire truss to find the support reactions

5 kN 5 kN 5 kN

1 kN

1 kN

1 kN

1 kN

1 kN

8 m

5 m 5 m 5 m 5 m 5 m 5 m

A

B

D

F

H

J

L

C E G I K

5 kN 5 kN 5 kN

1 kN

1 kN

1 kN

1 kN

1 kN

8 m

5 m 5 m 5 m 5 m 5 m 5 m

A

B

D

F

H

J

L

C E G I K

Ay

Ax

Ly

2 VukazichCE160TrussAnalysisMethodofSections[3]

Notes:

• Senses for the reaction force components Ax , Ay and Ly are assumed; • All appropriate dimensions and directions of forces are shown • Three unknown forces: Ax , Ay , and Ly.

Apply Equations of Equilibrium to find support reactions

Usually we can isolate an unknown by taking moment equilibrium about a convenient point Choose point A

𝑀! = 0

Counterclockwise moments about point A are positive

Note that unknown forces: Ax and Ay are concurrent at A and so their moment about A is zero.

− 1 𝑘𝑁 + 5 𝑘𝑁 5 𝑚 − 1 𝑘𝑁 + 5 𝑘𝑁 10 𝑚 − 1 𝑘𝑁 + 5 𝑘𝑁 15 𝑚 − 1 𝑘𝑁 )( 20 𝑚

− 1 𝑘𝑁 )(25 𝑚 + 𝐿! )( 30 𝑚 = 0

𝑳𝒚 = 𝟕.𝟓 𝒌𝑵 (Ly is positive so Ly acts upward as assumed)

Force Equilibrium

+→ 𝐹! = 0

forces positive to the right

𝑨𝒙 = 𝟎

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

𝐴! − 1 𝑘𝑁 − 5 𝑘𝑁 − 1𝑘 𝑁 − 5 𝑘𝑁 − 1 𝑘𝑁 − 5 𝑘𝑁 − 1 𝑘𝑁 − 1𝑘𝑁 + Ly 7.5 𝑘𝑁 = 0

𝑨𝒚 = 𝟏𝟐.𝟓 𝒌𝑵 (Ay is positive so Ay acts upward as assumed)

+

3 VukazichCE160TrussAnalysisMethodofSections[3]

F.B.D of entire truss in equilibrium with all support reaction forces shown

Draw a F.B.D. or a portion of the truss, cutting through the member whose force we are interested in finding F.B.D. of truss section to the right of cut a-a

Notes:

• Assume member forces in FH, GH, and GI are in tension (forces pulling away from joint);

• Moment equilibrium equation about point G will contain only unknown force FFH.

5 kN 5 kN 5 kN

1 kN

1 kN

1 kN

1 kN

1 kN

8 m

5 m 5 m 5 m 5 m 5 m 5 m

A

B

D

F

H

J

L

C E G I K

12.5 kN

7.5 kN

a

a

1 kN

1 kN

5 m 5 m

H

J

L

I K

7.5 kN

a

a

5 m

8 m

G

F FFH

FGH

FGI

4 VukazichCE160TrussAnalysisMethodofSections[3]

Principle of Transmissibility Advantageous to slide FFH to point F and break into components (note that it would also be advantageous to slide FFH to point L)

Moment Equilibrium of Truss Section about point G

𝑀! = 0

Counterclockwise moments about point G are positive

Note that unknown forces FGH and FGI are concurrent at G and so their moment about G is zero.

𝐹!"1517 8 𝑚 − 1 𝑘𝑁 )( 5 𝑚 − 1 𝑘𝑁)( 10 𝑚 + (7.5 𝑘𝑁 )( 15 𝑚) = 0

𝑭𝑭𝑯 = −𝟏𝟑.𝟖𝟏 𝒌𝑵 (FFH is negative so FH is in compression)

+

5 m 5 m 5 m

1 kN

1 kN

H

J

L

I K

7.5 kN

a

a

8 m

G

F

FFH (8/17)

FGH

FGI

FFH (15/17)

17 8

15