Structures

Chapter 4

ME 108 - Statics

Outline

• Applications

• Simple truss

• Method of joints

• Method of section

Germany

Tacoma Narrows Bridge http://video.google.com/videoplay?docid=-323172185412005564&q=bruce+lee&pl=true

UK

Applications

Trusses are commonly used to

support a roof.

For a given truss geometry and

load, how can we determine the

forces in the truss members and

select their sizes?

A more challenging question is

that for a given load, how can

we design the trusses’ geometry

to minimize cost?

Applications

Trusses are also used in a variety

of structures like cranes and the

frames of aircraft or space

stations.

How can we design a light weight

structure that will meet load,

safety, and cost specifications?

Simple Trusses

• A truss is a structure composed of slender members

joined together at their end points.

• Joint connections are formed by bolting or welding the

ends of the members to a common plate, called a gusset

plate, or by simply passing a large bolt or pin through

each of the members

Simple Trusses

Planar Trusses

• Planar trusses lie on a single plane and are used to support

roofs and bridges

• The truss ABCD shows a typical roof-supporting truss

• Roof load is transmitted to the truss at joints by means of a

series of purlins, such as DD’

• The analysis of the forces developed in the truss members is 2D

Assumptions for Design

• When designing both the member and the joints of a

truss, first it is necessary to determine the forces in each

truss member. This is called the force analysis of a truss.

When doing this, two assumptions are made:

1.“All loadings are applied at the joint”

– Assumption true for most applications of bridge and

roof trusses

– Weight of the members neglected since forces

supported by the members are large in comparison

– If member’s weight is considered, apply it as a

vertical force, half of the magnitude applied at each

end of the member

Assumptions for Design

2.“The members are joined together by smooth pins”

– Assumption true when bolted or welded joints are

used, provided the center lines of the joining

members are concurrent

Assumptions for Design

• Each truss member acts as a two force member,

therefore the forces at the ends must be directed

along the axis of the member

• If the force tends to elongate the member, it is a

tensile force

• If the force tends to shorten the member, it is a

compressive force

• Important to state the nature of the force in the

actual design of a truss – tensile or compressive

• Compression members must be made thicker

than tensile member to account for the buckling

or column effect during compression

Simple Truss

• To prevent collapse, the form of a truss

must be rigid

• The four bar shape ABCD will collapse

unless a diagonal member AC is added

for support

• The simplest form that is rigid or stable is

a triangle

• A simple truss is constructed starting with

a basic triangular element such as ABC

and connecting two members (AD and

BD) to form an additional element. For

these trusses, the number of members

(M) and the number of joints (J) are

related by the equation M = 2 J – 3.

Method of Joints

Procedure for Analysis

Truss analysis by the method of joints: 1) draw FBDs for all joints in the structure.

2) apply equations of equilibrium to each joint.

3) solve for unknowns.

Method of Joints

• Before beginning, it is usually necessary to draw a

free-body diagram of the entire truss (i.e. treat the

truss as a single object) & determine the reactions

at its supports:

– E.g. consider the Warren truss which has

members 2 m in length & support loads at B & D

Method of Joints

– From the equilibrium equations:

Σ Fx = Ax = 0

Σ Fy = Ay + E 400 N 800 N = 0

Σ Mpoint A = (1 m)(400 N) (3 m)(800 N) + (4 m)E = 0

We obtain the reactions:

Ax = 0, Ay = 500 N & E = 700 N

• The next step is to choose a joint & draw its free-

body diagram:

– Isolate joint A by cutting members AB & AC

Method of Joints

Method of Joints

– The terms TAB & TAC are the axial forces in

members AB & AC, respectively

– Although the directions of the arrows

representing the unknown axial forces can be

chosen arbitrarily, notice that we have chosen

them so that a member is in tension if we

obtain a positive value for the axial force

– Consistently choosing the directions in this

way helps avoid errors

Method of Joints

– The equilibrium equations for joint A are:

Σ Fx = TAC + TAB cos 60°= 0

Σ Fy = TAB sin 60° + 500 N = 0

Solving these equations, we obtain the axial

force TAB = 577 N & TAC = 289 N

– Member AB is in compression & member AC

is in tension

Method of Joints

• Next, obtain a free-body diagram of joint

B by cutting members AB, BC & BD:

– From the equilibrium equations for joint B:

Σ Fx = TBD + TBC cos 60° + 577 cos 60° = 0

Σ Fy = 400 N + 577 sin 60° TBC sin 60° = 0

We obtain TBC = 115 N & TBD = 346 N

– Member BC is in tension & member BD is in compression

Method of Joints

• By continuing to draw free-body diagrams of the

joints, we can determine the axial forces of all

the members

• In 2 dimensions, you can obtain only 2

independent equilibrium equations from the free-

body diagram of a joint

• Summing the moments about a point does not

result in an additional independent equation

because the forces are concurrent

Method of Joints

• Therefore when applying the method of joints, you should choose joints to analyze that are subjected to no more than 2 unknown forces:

– In our example, we analyzed joint A first because it was subjected to the known reaction exerted by the pin support & 2 unknown forces, the axial forces TAB & TAC

– We could then analyze joint B because it was subjected to 2 known forces & 2 unknown forces, TBC & TBD

– If we had attempted to analyze joint B first, there would have been 3 unknown forces

Method of Joints

• When determining the axial forces in the

members of a truss, it will be simpler if you are

familiar with 3 particular types of joints:

1.Truss joints with 2 collinear members & no

load: the sum of the forces must equal zero,

T1 = T2. The axial forces are equal.

Method of Joints

2.Truss joints with 2 noncollinear members

& no load: because the sum of the forces in

the x direction must equal zero, T2 = 0.

therefore T1 must also equal zero. The axial

forces are zero.

Method of Joints

3.Truss joints with 3 members, 2 of which

are collinear & no load: because the sum of

the forces in the x direction must equal zero,

T3 = 0. The sum of the forces in the y direction

must equal zero, so T1 = T2. The axial forces in

the collinear members are equal & the axial

force in the 3rd member is zero.

A member that supports no force is called a zero-

force member.

Analysis of trusses is simplified if we can easily

identify (such as by inspection) any zero force

members.

Zero force members

zero force member

not a zero force

member

example: identify any zero force members.

A

B

D

F

H

J

L

C E G I K

A

B

D

F

H

J

L

C E G I K

example: identify any zero force members.

Example 1

Determine the axial forces in the members of the truss in Figure.

Strategy

1st, draw a free-body diagram of the entire truss, treating it as a single object & determine the reactions at the supports. Then apply the method of joints, simplifying the task by identifying any special joints

Example 1

Solution

Determine the Reactions at the Supports:

Draw the free-body diagram of the entire truss:

Example 1

Solution

From the equilibrium equations:

Σ Fx = Ax + B = 0

Σ Fy = Ay 2 kN = 0

Σ Mpoint B = (6 m) Ax (10 m)(2 kN) = 0

We obtain the reactions Ax = 3.33 kN, Ay = 2 kN &

B = 3.33 kN.

Example 1

Solution

Identify Special Joints:

Because joint C has 3 members, 2 of which are

collinear & no load, the axial force in member BC is

zero, TBC = 0 & the axial forces in the collinear

members AC & CD are equal, TAC = TCD.

Draw Free-Body Diagrams of the Joints:

We know the reaction exerted on joint A by the

support & joint A is subjected to only 2 unknown

forces, the axial forces in members AB & AC.

Example 1

Solution

Example 1

Solution

The angle = arctan (5/3) = 59.0°

The equilibrium equations for joint A are:

Σ Fx = TAC sin 3.33 kN = 0

Σ Fy = 2 kN TAB TAC cos = 0

Solving these equations, we obtain TAB = 0 &

TAC = 3.89 kN.

Example 1

Solution

Now draw the free-body diagram of joint B:

Example 1

Solution From the equilibrium equation:

Σ Fx = TBD + 3.33 kN = 0

We obtain TBD = 3.33 kN. The negative sign

indicates that member BD is in compression.

The axial forces in the members are:

AB: 0

AC: 3.89 kN in tension (T)

BC: 0

BD: 3.33 kN in compression (C)

CD: 3.89 kN in tension (T)

Example 1

Critical Thinking

• Observe how our solution was simplified by

recognizing that joint C is the type of special joint

with 3 members, 2 of which are collinear & no

load:

– This allowed us to determine the axial forces

in all members of the truss by analyzing only

2 joints

Example 2

Each member of the truss in Figure will safely

support a tensile force of 10 kN & a compressive

force of 2 kN. What is the largest downward load F

that the truss will safely support?

Example 2

Strategy

This truss is identical to the one we analyzed in

Example 1. By applying the method of joints in the

same way, the axial forces in the members can be

determined in terms of the load F. The smallest

value of F that will cause a tensile force of 10 kN or

a compressive force of 2 kN in any of the members

is the largest value of F that the truss will support.

Example 2

Solution By using the method of joints in the same way as

in Example 1, we obtain the axial forces:

AB: 0

AC: 1.94F (T)

BC: 0

BD: 1.67F (C)

CD: 1.94F (T)

Example 2

Solution

For a given load F, the largest tensile force is 1.94F

(in members AC & CD) & the largest compressive

force is 1.67F (in member BD).

The largest safe tensile force would occur when

1.94F = 10 kN or when F = 5.14 kN.

The largest safe compressive force would occur

when 1.67F = 2 kN or when F = 1.20 kN.

Therefore, the largest load F that the truss will safely

support is 1.20 kN.

Example 2

Critical Thinking

• This example demonstrates why engineers

analyze structures:

– By doing so, they can determine the loads

that an existing structure will support or

design a structure to support given loads

– In this example, the tensile & compressive

loads the members of the truss will support

are given

Example 2

Critical Thinking

– Information of that kind must be obtained by

applying the methods of mechanics of

materials to the individual members

– Then statics can be used, as we have done in

this examples, to determine the axial loads in

the members in terms of the external loads on

the structure

Example 3

Determine the force in each member of the

truss and indicate whether the members are

in tension or compression.

Example 4

Determine the force in each member of the

truss and indicate whether the members are

in tension or compression.

Example 5

Example

The Method of Sections

• When we need to know the axial forces only in certain members of a truss, we often can determine them more quickly using the method of sections than the method of joints

• E.g. consider the Warren truss we used for the method of joints:

– It supports loads at B & D & each member is 2 m in length

– Suppose we need to determine only the axial force in member BC

The Method of Sections

– Just as in the method of joints, we begin by

drawing a free-body diagram of the entire

truss & determining the reactions at the

supports:

– The next step is to cut the members AC, BC &

BD to obtain a free-body diagram of a part, or

a section, of the truss:

The Method of Sections

– Summing moments about point B,

the equilibrium equations for the

section are:

Σ Fx = TAC + TBD + TBC cos 60° = 0

Σ Fy = 500 N 400 N TBC sin 60° = 0

Σ Mpoint B = (2 sin 60° m)TAC

(2 cos 60° m)(500 N) = 0

Solving them, we obtain TAC = 289 N, TBC = 115 N & TBD = 346 N.

The Method of Sections

• Notice how similar this method is to the method of joints:

– Both methods involve cutting members to obtain free-body diagrams of parts of a truss

– In the method of joints, we move from joint to joint, drawing free-body diagrams of the joints & determining the axial forces in the members as we go

– In the method of sections, we try to obtain a single free-body diagram that allows us to determine the axial forces in specific members

The Method of Sections

– In our example, we obtained a free-body diagram by cutting 3 members, including the 1 (member BC) whose axial force we wanted to determine

• In contrast to the method of joints, the forces on the free-body diagrams used in the method of sections are not usually concurrent:

– As in our example, we can obtain 3 independent equilibrium equations

– Although there are exceptions, it is usually necessary to choose a section that requires cutting no more than 3 members, or there will be more unknown axial forces than equilibrium equations

Example

The truss in Figure supports a 100-kN load. The

horizontal members are each 1 m in length.

Determine the axial force in member CJ & state

whether it is in tension or compression.

Example

Strategy

We need to obtain a section by cutting members

that include member CJ. By cutting members CD,

CJ & IJ, we will obtain a free-body diagram with 3

unknown axial forces.

Solution

To obtain a section, we cut members CD, CJ & IJ

& draw the free-body diagram of the part of the

truss on the right side of the truss

Example

Solution

From the equilibrium

equation:

Σ Fy = TCJ sin 45° 100 kN

= 0

We obtain TCJ = 141.4 kN.

The axial force in member

CJ is 141.4 kN (T).

Example

Critical Thinking

• We designed this example to demonstrate that

the method of sections can be very

advantageous when you only need to determine

the axial forces in particular members of a truss

– Imagine calculating the axial force in member

CJ using the method of joints

• But in engineering applications it is usually

necessary to know the axial forces in all the

members of a truss & in that case the 2 methods

are comparable

Example

Determine the axial forces in members DG & BE &

CG of the truss in Figure.

Example

Strategy

We can’t obtain a section that involves cutting

members DG & BE without cutting more than 3

members. However, cutting members DG, BE, CD

& BC results in a section with which we can

determine the axial forces in members DG & BE.

Example

Solution

Determine the Reactions at the Supports:

Draw the free-body diagram of the entire truss:

Example

Solution

From the equilibrium equations:

Σ Fx = Ax = 0

Σ Fy = Ay + K F 2F F = 0

Σ Mpoint A = LF (2L)(2F) (3L)F + (4L)K = 0

We obtain the reactions Ax = 0, Ay = 2F & K = 2F.

Example

Solution Choose a Section:

We obtain a section by cutting

members DG, CD, BC & BE.

Because the lines of action of

TBE, TBC & TCD pass through point

B, we can determine TDG by

summing moments about B:

Σ Mpoint B = L(2F) (2L)TDG = 0

Example

Solution

The axial force TDG = F.

Then from the equilibrium equation:

Σ Fx = TDG + TBE = 0

We see that TBE = TDG = F.

Member DG is in compression & member BE is in

tension.

Example

Critical Thinking

• This is a clever example but not 1 that is typical

of problems faced in practice:

– The section used to solve it might not be

obvious even to a person with experience

analyzing structures

– Notice that the free-body diagram of the

section of the truss is statically indeterminate,

although it can be sued to determine the axial

forces in members DG & BE

Example Problem

For the truss shown, determine

the forces in members DC and FG.