Beam theory 1

8/9/2019 Beam theory 1

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9966001305 Aeronautical Structures and Vibrations

Pablo Morata, [email protected]

February 17, 2014

Structural Analysis.

Beam Theory (I).


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Internal Forces

A deformable body is in equilibrium with the

externally applied loads (free-body diagram)

Cut by a section will create 2 solids each of one has

to be in equilibrium (external loads and

internal loads)Internal loads can be reduce to a force and a

moment applied at some point O

These internal forces have components in a given

coordinate system: consider the cut plane and

its normal vector to define that coordinate

system


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Internal Forces in a prismatic member

We define a prismatic member as the solid(elastic) created by the translation along a

curve known as directrix of a cross section

The cross section may vary along the directrix,

however in most of our applications will not

change

The directrix curvature should change smoothly

for this theory to be applicable. In most of

our applications the directrix will be a

straight line

We choose a coordinate system whosexaxis isalong directrix andyzplane is the cross

section

y

x

z

directrixCross

section


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Internal Forces in a prismatic member

At any section of the prismatic memberwe can reduce the internal forces to

a force and a moment:

Axial force, Shear forces

Torsional moment (torque, twisting

moment), bending moments

Most of our problems deal with prismatic members in

2D, that means only QyandMzare considered

zy

zy

MMT

QQN

,,

,,

y

x

z

yQ

N

zQ

yM

T

zM yQzM


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Sign ConventionsFor the external loads we consider them positive along the

positive sense of the coordinate axes defined.

For the internal forces the sign convention is the following:

Positive sense according to positive sense of axes defined in the

left hand side of the section. this means:

Mzis positive when it creates tension in the lower fiberand compression in the upper fiber

Qyis positive when it acts counterclockwise against the

material

yQyQ

z z


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Support reactionsBoundary conditions

The surface forces that develop at the supports or points of

contact between bodies are called reactions.

As a general rule, if the support prevents translationin a given

direction, then aforcemust be developed on the member in

that direction. Likewise, if rotationis prevented, a couple

momentmust be exerted on the member.This is what we call boundary conditions. For the usual 2D cases:


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Externally applied loadsBoundary conditions

Surface forces are caused by the direct contact of one body with the surface of

another.In all cases these forces are distributed over the area of contact between the

bodies.

If this area is small in comparison with the total surface area of the body, then

the surface force can be idealized as a single concentrated force, which is

applied to a point on the body,P. For example, the force of the ground on the wheels of a bicycle canbe considered as a concentrated force.If the surface loading is applied along a narrow strip of area, the loading can

be idealized as a linear distributed load, q

The resultant force of q is equivalent to the

area under the distributed loading curve, and

this resultant acts through the centroid Corgeometric center of this area. The loading along thelength of a beam is a typical example of where this idealization is often

applied.

A body force is developed when one body exerts a

force on another body without direct physical

contact between the bodies. Examples include the effectscaused by the earths gravitation or its electromagnetic field

q

RF


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Externally applied loadsBeams

Types of loading)(xq

)(xm

P

M

can be constant, linear or

any shape

Point force

Point Moment

10)( xxxxfq

Nm/mN][][

N/m]/[][][

Fm

LFq

Distributed force or moment Units:

Units:

Units:

N][][ FP

mN][][][ LFM


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Analysis of Statically determinate beams

First step is to determine the reactions. If the beam is statically determinate

there will be as many equations (3 equations in 2D analysis) as unknowns(reactions at supports)

P

l

a

Ox

y

SIMPLY SUPPORTED BEAM

Reactions in blue

ByROyR

OxR


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Analysis of Statically determinate beamsReactions

We consider the free-body diagram to set the equilibrium equations

P

l

a

Ox

y

EQUILIBRIUM EQUATIONS: 3EQ., 3 UNKNOWNSSTATICALLY

DETERMINATE

0:0 ByOyy RRPF

ByROyR

OxR

0:0 Oxx RF

0:0 lRPaM ByO

l

aPRBy

l

bP

l

alP

l

aPROy

)1(

b


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Analysis of Statically determinate beamsShear Forces and Bending Moments

We consider the free-body diagram to set the equilibrium equations after

applying the method of sections cutting at a generic location x

P

la

Ox

y

EQUILIBRIUM EQUATIONS APPLIED TO ANY OF THE 2 PORTIONS.

THE STANDARD IS TO KEEP LEFT HAND SIDE WHERE WE

DEFINED THE POSITIVE SIGN CONVENTION. Moments are taken

about the cut section, so for x>a

0:0 QRPF Oyy

ByR

OyR

OxR

0:0 Oxx RF

0)(:0 xRaxPMM Oyxat

b

z

M

yQ yQ

z

Mx


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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams

Now we have obtained M(x)and Q(x)we can draw these functions of x. This is

what we call shear forces and bending moment diagrams x

x

y




about the cut section

l

aPQ

xl

b

PaxPM

)(

a b

l

bPQ

xl

b

PM

ax ax


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021100:0 ByO RM


Example withP=-100 kN, a=1 m, b=1 m,l= 2 m

x

y

1 1

50OyR 50ByR

100

0100:0 ByOyy RRF

50ByR 5050100 OyR


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x

y

50 Q

1 1

50 Q 1x1x

50

50

050100:0 QFy050:0 QFy


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x

y

xM 50100

1 1

xM 50 1x1x

50

050)1(100:0 xxMM xat050:0 xMM xat


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Analysis of Statically determinate beams

First step is to determine the reactions. If the beam is statically determinate

there will be as many equations (3 equations in 2D analysis) as unknowns(reactions at supports)

P

l

a

Ox

y

CLAMPED BEAM

Reactions in blue

OyR

OxR

OzM


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Analysis of Statically determinate beamsReactions, clamped beam

We consider the free-body diagram to set the equilibrium equations

P

l

a

Ox

y

EQUILIBRIUM EQUATIONS: 3EQ., 3 UNKNOWNSSTATICALLY

DETERMINATE

0:0 Oyy RPF

OyR

OxR

0:0 Oxx RF

0:0 OzO MPaM

PROy PaMOz

OzM


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Analysis of Statically determinate beamsShear Forces and Bending Moments, clamped beam

We consider the free-body diagram to set the equilibrium equations after

applying the method of sections cutting at a generic location x

P

l

a

Ox

y




about the cut section, so for xa shear force and bending

moment are zero):

0:0 QRF Oyy

OyR

OxR

0:0 Oxx RF

0:0 xRMMM OyOzxat

z

M

yQ yQ

z

Mx

OzM


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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams, clamped beam

Now we have obtained M(x)and Q(x)we can draw these functions of x. This is

what we call shear forces and bending moment diagrams x

x

y




about the cut section

0 Q

0

M

a al

PQ

)( axPM

ax ax


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Relationships between Loads, Shear

Forces and Bending Moments

We consider a slice of a beam of length dx. We will see what is the relationshipbetween q, Q(x) and M(x).

)(xq

yy dQxQ )(

zz dMxM )(

)(xQy

)(xMz

Equilibrium (vertical forces):

0)())(()( dxxqdQxQxQ yyy

0)( dxxqdQy

)(xqdx

dQy


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We consider a slice of a beam of length dx. We will see what is the relationshipbetween q, Q(x) and M(x).

)(xq

yy dQxQ )(

zz dMxM )(

)(xQy

)(xMz

Equilibrium (of moments):

0))(()( kdxqdxdxQdMxMxM yzzz

0 kdxqdxdxQdM yz

kis a factor that identifies

the point of application

of the resultant of q(x)distribution. If q(x) is

constant k=1/2

As is a second order

differential we neglect it compared

to other terms resulting

yz Q

dx

dM

0 kdxqdx


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The signs in this formulas are due to our sign convention, a different signconvention would give different signs so be careful when using them.

yz Q

dx

dM)(xq

dx

dQy

These relationships are very helpful in drawing the shear force and bending

moment diagram:

because if the shear force iszero then the bending moment diagram

curve has a maximum or minimum (dM/dx=0) also the slopeof the bending moment diagram has the opposite sign(due

to sign convention) of the shear force diagram (dM/dx=-Q)

The curvature of the bending moment curve (convex or concave) is given

by the sign of qsince d2M/dx2=-dQ/dx=q (second derivative)


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Assignment #4aDraw the shear and bending moment diagrams of the following

beams

4a.1

4a.2


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Bibliography

Hibbeler, Mechanics of Materials

Ortiz Berrocal, Resistencia de Materiales

Timoshenko, Strength of Materials

Available at UEM Libraryhttp://biblioteca.uem.es/
http://biblioteca.uem.es/http://biblioteca.uem.es/

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Beam theory 1