truss example in two-dimensions ambar k. mitra · whole truss free-body-diagram (fbd) ... stick to...

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Truss Example in Two-Dimensions

Ambar K. Mitra

This document contains screen-shots from the software Statics-Power.

Visit www.actuspotentia.com for details.

What is a truss?

A truss is an assembly of two force members (rods or cables) that are joined at their ends with other members by pins. The entire assembly is then supported with

either two pins or with one roller and one pin.

What is truss analysis? • Determine the support forces from the pins/roller.

• Determine the force on each two-force member.

• Determine the nature of the force, i.e. tension/compression, on each two-force member.

Whole Truss Free-Body-Diagram (FBD)

Pin-Roller Supported Truss

Draw FBD of whole truss. Write three equilibrium equations and determine Ax, Ay,

Cy (or Cx).


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Pin-Pin Supported Truss

In general, the four unknowns, Ax, Ay, Bx, By, cannot be determined from three

equilibrium equations. However, in some special situations, you can determine two out of the four unknowns.

Joint FBD

A pin or joint is a point mass; therefore, we can write two force balance equations

for equilibrium. Find a pin or joint that has two unknown forces. Determine the two unknown forces from the two equilibrium equations. Determine the forces in

all the members by enforcing equilibrium conditions at a series of pins that have two unknown forces.

Sign Convention for the Forces

Consider that three members meet at a joint.

The FBD of the members and the pin are


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Note that we assumed that all the members are in tension. It is a good idea to

stick to this convention. When we know that a force is compression, while writing

the equilibrium equations, we insert a negative numerical value for this force.

Visual Clue Tensile forces in members show up as arrows pointing outward from a joint.

Zero-Force Member

• No force is acting on the joint P.

• Three members (PQ, PR, PS) meet at the joint. • Two out of three members (PQ and PR) are aligned with one straight line.

• Force on the third member (PS), F(PS) = 0.

• F(RP) = F(PQ)

Example


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Figure-1a

A truss is supported with two pins at A and F and is loaded as shown. Determine the support forces at the pins and the force on each member. Identify the forces in

the members as tension or compression.

• No force is acting on joint B. • Three members (AB, BC, and BG) meet at the joint.

• Two out of three members (AB and BC) are aligned with one straight line. • Force on the third member, F(BG) = 0.

• F(AB) = F(BC)

By removing the zero-force member BG from the truss, we arrive at the truss of

Figure-1b.


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Figure-1b

• No force is acting on joint G. • Three members (AG, GF, and GC) meet at the joint.

• Two out of three members (AG and GF) are aligned with one straight line. • Force on the third member, F(GC) = 0.

• F(AG) = F(GF)

By removing the zero-force member GC from the truss, we arrive at the truss of Figure-1c.


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Figure-1c

• No force is acting on joint C.

• Three members (BC, CD, and CF) meet at the joint. • Two out of three members (BC and CD) are aligned with one straight line.

• Force on the third member, F(CF) = 0. • F(BC) = F(CD)

By removing the zero-force member CF from the truss, we arrive at the truss of

Figure-1d.


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Figure-1d

Note: By identifying and removing one zero-force member in a truss, you may

start a chain reaction that creates other zero-force members and removal of new zero-force members greatly simplifies the analysis of the truss.

Problem Solution

We choose to enforce the equilibrium conditions at joint E.


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Figure-1e

• Joint label: E

• Joint coordinate: (7.5,6) • Joint force: (0,-2000)

• Pin/roller at joint: None • Members meet at this joint: 2 (ED and EF)

• Members with unknown forces: 2 (ED and EF) • Member data:

o Member ED

� Label: D � Coordinate of D: (6,6)

� F(ED): unknown o Member EF


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� Label: F

� Coordinate of F: (6,4) � F(EF): unknown

The equilibrium equations for joint E are:

Figure-1f

By solving the equations, we find:

F(EF) = -2500lb (2500lb compression) F(ED) = 1500lb (1500lb tension)

Next, we choose to enforce the equilibrium conditions at joint F.


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Figure-1g

• Joint label: F

• Joint coordinate: (6,4) • Joint force: (0,0)

• Pin/roller at joint: Pin/roller • Fx at pin: unknown

• Fy at pin: unknown • Members meet at this joint: 3 (EF, DF, and FG)

• Members with unknown forces: 2 (DF and FG)


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Figure-1h

There are four unknowns at this joint, namely, Fx, Fy, F(DF), and F(FG), and two equilibrium equations. Therefore, this joint is not solvable.

Next, we choose to enforce the equilibrium conditions at joint D.


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Figure-1i

• Joint label: D • Joint coordinate: (6,6)

• Joint force: (0,-1200) • Pin/roller at joint: None

• Members meet at this joint: 3 (DC, DF, and DE) • Members with unknown forces: 2 (DC and DF)

• Member data: o Member DC

� Label: C � Coordinate of D: (5,5)

� F(DC): unknown

o Member DF � Label: F


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� Coordinate of F: (6,4)

� F(DF): unknown o Member DE

� Label: E � Coordinate of F: (7.5,6)

� F(DE): known = 1500lb

The equilibrium equations for joint D are:

Figure-1j

By solving the equations we find:

F(DC) = 2121lb = 2121lb (tension)

F(DF) = -2700lb = 2700lb (compression)

truss example in two-dimensions ambar k. mitra · whole truss free-body-diagram (fbd) ... stick to...

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