chapter vector mechanics for engineers: …engr.uconn.edu/.../ce2110/other/beervm11e_ppt_ch06.pdfh...

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eleventh Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

David F. Mazurek

CHAPTER

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

6Analysis of Structures


Vector Mechanics for Engineers: Statics

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Contents

6 - 2

Application

Introduction

Definition of a Truss

Simple Trusses

Method of Joints

Joints Under Special Loading

Conditions

Space Trusses

Sample Problem 6.1

Method of Sections

Trusses Made of Several Simple

Trusses

Sample Problem 6.3

Analysis of a Frame

Frames That Collapse Without

Supports

Sample Problem 6.4

Machines



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Application

6 - 3



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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 those 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 of

supporting large lateral loads. Loads must be applied at

the joints.



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6 - 7



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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 triangular

truss.



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Method of Joints

6 - 9

• Dismember the truss and create a freebody

diagram for each member and pin.

• The two forces exerted on each member are

equal, have the same line of action, and

opposite 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.



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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 the

number 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 and

6 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.

STRATEGY:

• What’s the first step to solving this

problem? Think, then discuss this with a

neighbor.

• DRAW THE FREE BODY DIAGRAM

FOR 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.



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Sample Problem 6.1

6 - 13

MODELING and ANALYSIS:

• Based on a free body diagram of the entire truss,

solve the 3 equilibrium equations for the reactions

at E and C.

ft 6ft 12lb 1000ft 24lb 2000

0

E

MC

lb 000,10E

xx CF 0 0xC

yy CF lb 10,000 lb 1000 - lb 20000

lb 7000yC

• 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.



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Sample Problem 6.1

6 - 14

534

lb 2000 ADAB FF

CF

TF

AD

AB

lb 2500

lb 1500

FDB FDA

FDE 2 35 FDA CF

TF

DE

DB

lb 3000

lb 2500

• 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.



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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.

lb 3750

25001000054

54

BE

BEy

F

FF

CFBE lb 3750

lb 5250

375025001500053

53

BC

BCx

F

FF

TFBC lb 5250

• There is one remaining unknown member

force at joint E (or C). Use joint E and

assume the member is in tension.

lb 8750

37503000053

53

EC

ECx

F

FF

CFEC lb 8750



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Sample Problem 6.1

6 - 16

checks 087507000

checks 087505250

54

53

y

x

F

F

REFLECT and THINK: Using the computed

values of FCB and FCE, you can determine the

reactions Cx and Cy by considering the equilibrium

of Joint C (Fig. 6). Since these reactions have

already been determined from the equilibrium of

the entire truss, this provides two checks of your

computations. You can also simply use the

computed values of all forces acting on the joint

(forces in members and reactions) and check that

the joint is in equilibrium.



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Method of Sections

6 - 17

• When the force in only one member or the

forces in a very few members are desired, the

method 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 FBD.

• An 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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Method of Sections

6 - 18

• Using the left-side FBD, write one

equilibrium equation that can be solved to

find FBD. 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, so

long as the cut creates a separated section.

k

k

pp



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Trusses Made of Several Simple Trusses

6 - 19

• 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 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.

STRATEGY:

• 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 and

solve 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


• Take the entire truss as a free body.

Apply the conditions for static equilib-

rium to solve for the reactions at A and L.

MA 0 5 m 6 kN 10 m 6 kN 15 m 6 kN

20 m 1 kN 25 m 1 kN 25 m L

L 7.5 kN

Fy 0 20 kN L Ay

Ay 12.5 kN

Fx 0 Ax

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.

MH 0

7.50 kN 10 m 1 kN 5 m FGI 5.33 m 0

FGI 13.13 kN

• Sum of the moments about point H:

TFGI kN 13.13

• 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

tan FG

GL

8 m

15 m 0.5333 28.07

MG 0

7.5 kN 15 m 1 kN 10 m 1 kN 5 m

FFH cos 8 m 0

FFH 13.82 kN CFFH kN 82.13

tan GI

HI

5 m23

8 m 0.9375 43.15

ML 0

1 kN 10 m 1 kN 5 m FGH cos 10 m 0

FGH 1.371 kN CFGH kN 371.1

• FFH is shown as its components. What one

equilibrium equation will determine FFH?

• There are many options for finding FGH at this

point (e.g., SFx=0, SFy=0). Here is one more:



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Sample Problem 6.3

2 - 24

REFLECT and THINK:

Sometimes you should resolve a force into

components to include it in the equilibrium

equations. By first sliding this force along its

line of action to a more strategic point, you

might eliminate one of its components from a

moment equilibrium equation.



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Analysis of a Frame

6 - 25

• 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 magnitude

and line of action. They must be represented with two

unknown components.



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Frames That Collapse Without Supports

6 - 26

• Some frames may collapse if removed from

their supports. Such frames can not be treated

as rigid 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-body

diagrams 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 - 27

Members ACE and BCD are

connected by a pin at C and by the

link DE. For the loading shown,

determine the force in link DE and the

components of the force exerted at C

on member BCD.

STRATEGY:

Follow the general procedure discussed in

this section. First treat the entire frame as a

free body, which will enable you to find

the reactions at A and B. Then dismember

the frame and treat each member as a

free body, which will give you the

equations needed to find the force at C.



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Sample Problem 6.4

6 - 28


Create a free-body diagram for the complete frame

and solve for the support reactions.

Fy 0 Ay 480 N

Ay 480 N

MA 0 480 N 100 mm B 160 mm

B300 N

Fx 0 B Ax

Ax 300 N

Ax 300 N

07.28tan150801

Note:



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Sample Problem 6.4

6 - 29

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 - 30

N 561

mm 100N 480mm 06N 300mm 250sin0

DE

DEC

F

FM

CFDE N 561

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

components at C.

N 300cosN 561 0

N 300cos0

x

DExx

C

FCF

N 795xC

N 480sinN 5610

N 480sin0

y

DEyy

C

FCF

N 216yC

Using the best FBD for member BCD, what is

the one equilibrium equation that can directly

find FDE? Please discuss.

07.28tan150801



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Sample Problem 6.4

6 - 31

REFLECT and THINK:

Check the computations by

considering the free body ACE .

0mm 220795mm 100sin561mm 300cos561

mm 220mm 100sinmm 300cos

xDEDEA CFFM

(checks)



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Machines

6 - 32

• 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 complete

machine, 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,

Pb

aQbQaPM A 0

chapter vector mechanics for engineers: …engr.uconn.edu/.../ce2110/other/beervm11e_ppt_ch06.pdfh...

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