[email protected] engr-36_lec-15_trusses-2.pptx 1 bruce mayer, pe engineering-36: engineering...

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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 6:Trusses-2



Introduction: MultiPiece Structures For the equilibrium of structures made of several

connected parts, the internal forces as well the external forces are considered.

In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

The Major Categories of Engineering Structures• Frames: contain at least one multi-force member,

i.e., a member acted upon by 3 or more forces• Trusses: formed from two-force members, i.e.,

straight members with end point connections• Machines: structures containing moving parts

designed to transmit and modify forces



Definition of a Truss A truss consists of straight members

connected at joints. No member is continuous through a joint.

Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and NO couple. Only two-force members are considered LoA CoIncident with Geometry

A truss carries ONLY those loads which act in its plane, allowing the truss to be treated as a two-dimensional structure.

When forces tend to pull the member apart, it is in tension. When the forces tend to push together the member, it is in compression.

Tens

ion

Com

pres

sion



Truss Defined Members of a truss are SLENDER and NOT

capable of supporting large LATERAL loads• i.e.; IN-Plane, or 2D, loading only• Members are of NEGLIBLE Weight

Loads MUST be applied at the JOINTS to Ensure AXIAL-ONLY Loads on Members.• Mid-Member Loads Produce BENDING-Loads

which Truss Members are NOT Designed to Support

Beams Apply RoadBed Load at

JOINTS Only



Trusses Made of Simple Trusses• Compound trusses are statically

determinant, rigid, and completely constrained.

32 nm

• Truss contains a redundant member and is statically indeterminate. 32 nm

• Necessary but INsufficient condition for a compound truss to be statically determinant, rigid, and completely constrained,

nrm 2

non-rigid

32 nm

• Additional reaction forces may be necessary for a nonrigid truss.

rigid

42 nm



Method of Sections

To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side.

With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD

When the force in only one member or the forces in a very few members are desired, the method of sections works well.



Example Method of Sections

Given the Truss with Loading and Geometry Shown

Use the Method of Sections to Determine the Force in Member FD




Take Section to Expose FFD

Now Take ΣME = 0

''' 1020kip1510kip150 FDF

kip45

10ft

ftkip300150

FDF nCompressio kip45FDF




Determine the force in members just right of Center: • FH • GH • GI

SOLUTION PLAN• Take the entire truss as a free

body. Apply the conditions for static equilibrium to solve for the reactions at A and L.

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

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



Example Method of Sections SOLUTION PLAN

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

kN 5.12

kN 200

kN 5.7

m 30kN 1m 25kN 1m 20

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

y

yy

A

A

ALF

L

L

M



Example Method of Sections Pass a section (n-n) through

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

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

kN 13.13

0m 33.5m 5kN 1m 10kN 7.50

0

GI

GI

H

F

F

M

TFGI kN 13.13




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 15cosm 5kN 1m 10kN 1

0

15.439375.0m 8

m 5tan

32

GH

GH

L

F

F

M

HI

GI

CFGH kN 371.1



Method of Sections - Summary1. If needed Determine Support Reactions

2. Decide on How to CUT the Truss into Sections and draw the Corresponding Free Body Diagrams

3. Try to Apply the Eqns of Equilibrium to avoid generation of simultaneous Eqns

• Moments should be Summed about points that lie at the intersection of the LoA’s of 2+Forces, making simpler the solution for the remaining forces



Pick: Pivot & PoA

When doing Sections Recall that the LoA for Truss Members are defined by the Member Geometry

Use Force Transmissibility → Forces are SLIDING Vectors• Pick a Pivot Point, on or Off the Body,

where the LoA’s of many Force LoA’s Cross• Apply the Force of interest so that ONE of

its X-Y Components passes Thru the Pivot



Pick: Pivot & PoA Example

After Finding support RCNs find force in Member ED → Use Section a-a

Pick Pt-B as Pivot to Eliminate from Moment Calc FAB, FFB, FEB, 1000N



Pick: Pivot & PoA Example

Pick Pt-C as the Point of Appliction (PoA) for FED

Using Pt-C as the PoA permits using F•d to find the Moment about Pivot-B



WhiteBoard Work

Let’s WorkSome TrussProblems

Find Forces in EL & LM

Find Forces in EL & LM



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix

[email protected] engr-36_lec-15_trusses-2.pptx 1 bruce mayer, pe engineering-36: engineering...

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