Chapter06-Analysis of Structures

7/24/2019 Chapter06-Analysis of Structures

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VECTOR MECHANICS FOR ENGINEERS:

STATICS

Tenth Editionin SI Units

Ferdinand P. Beer

E. Russell Johnston, Jr.David F. Mazurek

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

6 Analysis of Structures




Vector Mechanics for Engineers: StaticsT en t h

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Contents

6 - 2

Introduction

Definition of a Truss

Simple Trusses

Analysis of Trusses by the Method

of Joints Joints Under Special Loading

Conditions

Space Trusses

Sample Problem 6.1

Analysis of Trusses by the Methodof Sections

Trusses Made of Several SimpleTrusses

Sample Problem 6.3

Analysis of a Frame

Frames Which Cease to be RigidWhen Detached From Their

Supports

Sample Problem 6.4

Machines





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Application

6 - 3

Design of support structures requiresknowing the loads, or forces, that each

member of the structure will

experience.

Functional elements, such as the holding

force of this pliers, can be determined

from concepts in this section.

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Introduction

6 - 4

• 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.

• Three categories of engineering structures are considered:

a) Trusses : formed from two-force members, i.e.,

straight members with end point connections and

forces that act only at these end points.b) Frames : contain at least one multi-force member,

i.e., member acted upon by 3 or more forces.

c) Machines : structures containing moving parts

designed to transmit and modify forces.

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6 - 5

• 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.

• Most structures are made of several trusses joined

together to form a space framework. Each truss

carries loads which act in its plane and may be

treated as a two-dimensional structure.

• When forces tend to pull the member apart, it is in

tension. When the forces tend to compress the

member, it is in compression.

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6 - 6

•

Members of a truss are slender and not capable ofsupporting large lateral loads. Loads must be applied

at the joints.

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6 - 7 T I




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Simple Trusses

6 - 8

• A rigid truss will not collapse under

the application of a load.

• A simple truss is constructed by

successively adding two members and

one connection to the basic triangulartruss.

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Analysis of Trusses by the Method of Joints

6 - 9

• Dismember the truss and create a free body

diagram for each member and pin.

• The two forces exerted on each member are

equal, have the same line of action, andopposite sense.

• Forces exerted by a member on the pins or

joints at its ends are directed along the member

and equal and opposite.

• Conditions of equilibrium are used to solve for

2 unknown forces at each pin (or joint), giving a

total of 2n solutions, where n=number of joints.

Forces are found by solving for unknown forces

while moving from joint to joint sequentially.

• Conditions for equilibrium for the entire truss

can be used to solve for 3 support reactions.

V t M h i f E i St tiT I n




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Joints Under Special Loading Conditions

6 - 10

• Forces in opposite members intersecting in

two straight lines at a joint are equal.• The forces in two opposite members are

equal when a load is aligned with a third

member. The third member force is equal

to the load (including zero load).

• The forces in two members connected at a joint are equal if the members are aligned

and zero otherwise.

• Recognition of joints under special loading

conditions simplifies a truss analysis.





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Space Trusses

6 - 11

• An elementary space truss consists of 6 members

connected at 4 joints to form a tetrahedron.

• A simple space truss is formed and can be

extended when 3 new members and 1 joint are

added at the same time.

• Equilibrium for the entire truss provides 6

additional equations which are not independent of

the joint equations.

• In a simple space truss, m = 3n - 6 where m is thenumber of members and n is the number of joints.

• Conditions of equilibrium for the joints provide 3n

equations. For a simple truss, 3n = m + 6 and the

equations can be solved for m member forces and6 support reactions.





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Sample Problem 6.1

6 - 12

Using the method of joints, determine

the force in each member of the truss.

SOLUTION:

• What’s the first step to solving this

problem? Think, then discuss this with a

neighbor.

• DRAW THE FREE BODY DIAGRAMFOR THE ENTIRE TRUSS (always first)

and solve for the 3 support reactions

• Draw this FBD and compare your sketch

with a neighbor. Discuss with each other

any differences.

V t M h i f E i St tiT e

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Vector Mechanics for Engineers: StaticsTen t h

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Sample Problem 6.1

6 - 13

SOLUTION:

• Based on a free body diagram of the entire truss,solve the 3 equilibrium equations for the reactions

at E and C .

• Looking at the FBD, which “sum of moments”

equation could you apply in order to find one of

the unknown reactions with just this one equation?

• Next, apply the remaining

equilibrium conditions to

find the remaining 2 support

reactions.

m3m6kN5m12kN10

0

E

M C

kN50 E

x x C F 0 0 xC

y y C F kN50kN5-kN100

kN35 yC


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Sample Problem 6.1

6 - 14

DA DE

DA DB

F F

F F

5

32

• We now solve the problem by moving

sequentially from joint to joint and solving

the associated FBD for the unknown forces.

• Which joint should you start with, and why?

Think, then discuss with a neighbor.

• Joints A or C are equally good because each

has only 2 unknown forces. Use joint A and

draw its FBD and find the unknown forces.

• Which joint should you move to next, and why? Discuss.• Joint D, since it has 2 unknowns remaining

(joint B has 3). Draw the FBD and solve.

534

kN10 AD AB

F F

C . F

T . F

AD

AB

kN512

kN57

C F

T . F

DE

DB

kN15

kN512


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Sample Problem 6.1

6 - 15

• There are now only two unknown member

forces at joint B. Assume both are in tension.

• There is one remaining unknown member

force at joint E (or C ). Use joint E andassume the member is in tension.

kN7518

kN512kN505

4

5

4

. F

F . F

BE

BE y

C . F BE kN7518

kN2526

kN7518kN512kN5705

3

5

3

. F

... F F

BC

BC x

T . F BC kN2526

kN7543

kN7518kN1505

3

5

3

. F

. F F

EC

EC x

C . F EC kN7543


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Sample Problem 6.1

6 - 16

• All member forces and support reactions are

known at joint C . However, the joint equilibriumrequirements may be applied to check the results.

checks 0kN7543kN35

checks 0kN7543kN2526

5

4

5

3

. F

.. F

y

x


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Analysis of Trusses by the Method of Sections

6 - 17

• When the force in only one member or the

forces in a very few members are desired, themethod of sections works well.

• To determine the force in member BD, form a

section by “cutting” the truss at n-n 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 F BD.

• A FBD could have been created for the right

side, but why is this a less desirable choice?

Think and discuss.

• Notice that the exposed internal forces

are all assumed to be in tension.

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Analysis of Trusses by the Method of Sections

6 - 18

• Using the left-side FBD, write one

equilibrium equation that can be solved tofind F BD. Check your equation with a

neighbor; resolve any differences between

your answers if you can.

• So, for example, this cut with line p-p is

acceptable.

• Assume that the initial section cut was made

using line k-k . Why would this be a poor

choice? Think, then discuss with a neighbor.

• Notice that any cut may be chosen, solong as the cut creates a separated section.

k

k

p p

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Trusses Made of Several Simple Trusses

6 - 19

• Compound trusses are statically

determinate, rigid, and completelyconstrained.

32 nm

• Truss contains a redundant member

and is statically indeterminate. 32 nm

• Necessary but insufficient condition

for a compound truss to be statically

determinate, rigid, and completely

constrained,

nr m 2

non-rigid rigid

32 nm

• Additional reaction forces may be

necessary for a rigid truss.

42 nm





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Sample Problem 6.3

6 - 20

Determine the force in members FH ,

GH , and GI .

SOLUTION:

• List the steps for solving this problem.

Discuss your list with a neighbor.

1. Draw the FBD for the entire truss.

Apply the equilibrium conditions andsolve for the reactions at A and L.

3. Apply the conditions for static

equilibrium to determine the desired

member forces.

2. Make a cut through members FH ,

GH , and GI and take the right-hand

section as a free body (the left side

would also be good).





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Sample Problem 6.3

6 - 21

SOLUTION:

• Take the entire truss as a free body.

Apply the conditions for static

equilibrium to solve for the reactions at A

and L.

x x

y

y y

A

A F

. A

A L F

. L

L

M

0

kN512

kN200

kN57

m25kN1m25kN1m20

kN6m15kN6m10kN6m50

LAy

Ax




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Sample Problem 6.3

6 - 22

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

Draw this FBD.

kN1313

0m335m5kN1m10kN7.50

0

. F

. F

M

GI

GI

H

• Sum of the moments about point H :

T F GI kN13.13

• What is the one equilibrium equation that could be solved to find FGI ? Confirm your answer with

a neighbor.





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Sample Problem 6.3

6 - 23

kN8113

0m8cos

m5kN1m10kN1m15kN7.5

0

072853330 m15

m8tan

. F

F

M

..GL

FG

FH

FH

G

C . F FH kN8113

kN3711

0m15cosm5kN1m10kN1

0

154393750 m8

m5tan

3

2

. F

F

M

.. HI GI

GH

GH

L

C F GH kN371.1

• F FH is shown as its components. What one

equilibrium equation will determine F FH

?

• There are many options for finding FGH at this

point (e.g., S F x=0, S F y=0). Here is one more:

Vector Mechanics for Engineers: StaticsT en

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Analysis of a Frame

6 - 24

• Frames and machines are structures with at least one

multiforce (>2 forces) member. Frames are designed to

support loads and are usually stationary. Machines contain

moving parts and transmit and modify forces.

• A free body diagram of the complete frame is used to

determine the external forces acting on the frame.

• Internal forces are determined by dismembering the frame

and creating free-body diagrams for each component.

• Forces between connected components are equal, have the

same line of action, and opposite sense.

• Forces on two force members have known lines of action

but unknown magnitude and sense.

• Forces on multiforce members have unknown magnitudeand line of action. They must be represented with two

unknown components.


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Frames Which Cease To Be Rigid When Detached From Their Supports

6 - 25

• Some frames may collapse if removed from

their supports. Such frames cannot be treated asrigid bodies.

• A free-body diagram of the complete frame

indicates four unknown force components which

cannot be determined from the three equilibrium

conditions (statically indeterminate).

• The frame must be considered as two distinct, but

related, rigid bodies.

• With equal and opposite reactions at the contact

point between members, the two free-bodydiagrams show 6 unknown force components.

• Equilibrium requirements for the two rigid bodies

yield 6 independent equations. Thus, taking the

frame apart made the problem solvable.


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Sample Problem 6.4

6 - 26

Members ACE and BCD are connected

by a pin at C and by the link DE . For

the loading shown, determine the forcein link DE and the components of the

force exerted at C on member BCD.

SOLUTION:

1. Create a free body diagram for thecomplete frame and solve for the

support reactions.


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Sample Problem 6.4

6 - 27

SOLUTION:

1. Create a free-body diagram for the completeframe and solve for the support reactions.

N4800 y y A F N480 y A

mm160mm100 N4800 B M A

N300 B

N300

0

x

x x

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A B F

N300 x A


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Sample Problem 6.4

6 - 28

SOLUTION (cont.):

2. Create a free body diagram for member BCD (since the problem asked for forces on this

body). Choose the best FBD, then discuss your

choice with a neighbor. Justify your choice.

FDE,x

FDE,y

FDE

FDE

FDE,x

FDE,y


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Sample Problem 6.4

6 - 29

N561

mm100 N480mm06 N300mm250sin0

DE

DE C

F

F M

C F DE N561

• Sum of forces in the x and y directions may be used to find the force

components at C .

N300cos N5610

N300cos0

x

DE x xC

F C F N795 xC

N480sin N5610

N480sin0

y

DE y y

C

F C F

N216 yC

SOLUTION (cont.):

3. Using the best FBD for member BCD, what isthe one equilibrium equation that can directly

find F DE ? Please discuss.

07.28tan150

801


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Sample Problem 6.4

6 - 30

• With member ACE as a free body with no

additional unknown forces, check the

solution by summing moments about A.

0mm220795mm100sin561mm300cos561

mm220mm100sinmm300cos

x DE DE A C F F M

(checks)


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Machines

6 - 31

• Machines are structures designed to transmit

and modify forces. Typically they transform

input forces (P) into output forces (Q).

• Given the magnitude of P, determine the

magnitude of Q.

• Create a free-body diagram of the completemachine, including the reaction that the wire

exerts.

• The machine is a nonrigid structure. Use

one of the components as a free-body.

Discuss why the forces at A are such.

• Sum moments about A,

P b

aQbQaP M A 0


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Tutorials

Chapter 2

•

2.19 2.38 (2.F4) 2.67 2.70 2.121Chapter 3

• 3.4 3.9 3.10 3.49 3.97 3.101 3.114 3.148

Chapter 4

• (4.F1) (4.F4) 4.3 4.8 4.10 4.14 4.15 4.19 4.45

• (4.F7) 4.103 4.105 4.113 4.121

Chapter 5

• (Fig P5.1) (Fig P5.7) (Fig P5.15)

5.20 5.21 5.102 5.103 5.104

Chapter 6

• (Fig P6.2) (Fig P6.5)

Chapter 7• 7.29 7.35 7.36 7.37 7.58

Chapter 8

• (8.F1) (8.F2) (8.F3) (8.F4)

• 8.1 8.2 8.6 8.15 8.19

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Chapter06-Analysis of Structures

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