Bending of Beams Sollero

8/18/2019 Bending of Beams Sollero

1/34

Bending Stress

Cantilever of Galileo


2/34

Introduction

• When a slender member is subjected

to transverse loading, we say it actsas a beam. Examples of beam

actions are:

– The horizontal members in

buildings.

– The leaf springs of an automobile

suspension.

– The wings of an airplane.


3/34

Introduction

• If we sectioned a transversely

loaded member a shear force and a

bending moment would in general

have to act on the cross section in

order to maintain equilibrium.

• Our aim is to determine the

distributions of stresses which

have the shear force V and the

bending moment Mb as their

resultant.

• We shall also obtain an exact

solution within the theory of elasticity

for the special case of a beam

subjected to pure bending.


4/34

Symmetrical Beam Subjected to Pure

Bending

• In this moment we shall restrict our analysisto beams with the next characteristics:

– Originally straight beam which is uniformalong its length.

– The cross section is symmetrical aboutthe plane of loading.

– The material properties are constant

along the length and symmetrical withrespect to the plane of loading.

– Only Constant bending moment (purebending ).


5/34

Plane surfaces perpendicular to the

plane of symmetry of the beam


6/34

Symmetrical Beam Subjected to Pure Bending

M dA y M

dA z M

dA F

x z

x y

x x

0

0

• These requirements may be applied to the sumsof the components and moments of the statically

indeterminate elementary internal forces.

• Internal forces in any cross section are equivalent

to a couple. The moment of the couple is thesection bending moment .

• From statics, a couple M consists of two equal

and opposite forces.

• The sum of the components of the forces in any

direction is zero.

• The moment is the same about any axis

perpendicular to the plane of the couple and

zero about any axis contained in the plane.


7/34


Beam with a plane of symmetry in pure

bending:• member remains symmetric

• bends uniformly to form a circular arc

• cross-sectional plane passes through arc center

and remains planar

• length of top decreases and length of bottom

increases

• a neutral surface must exist that is parallel to theupper and lower surfaces and for which the length

does not change

• stresses and strains are negative (compressive)

above the neutral plane and positive (tension)

below it


8/34


Consider a beam segment of length L.After deformation, the length of the neutral

surface remains L. At other sections,

m x

mm

x

c

y

c ρ

c

y y

L

y y L L

y L

or

linearly)ries(strain va


9/34


• For a linearly elastic material,

linearly)varies(stressm

m x x

c

y

E c y E

• For static equilibrium,

dA yc

dAc

ydA F

m

m x x

0

0

First moment with respect to neutral plane is zero. Therefore, the

neutral surface must pass through

the section centroid.

• For static equilibrium,

I

My

c

y

S M

I Mc

c

I dA y

c M

dAc

y ydA y M

x

m x

m

mm

m x

ngSubstituti

2


10/34

Stress and deformation in symmetrical elastic beams

• Deformation due to bending moment M is

quantified by the curvature of the neutral surface

EI

M

I

Mc

Ec Ecc

mm

11

• Although cross sectional planes remain planar

when subjected to bending moments, in-plane

deformations are nonzero,

y y x z x y

• Expansion above the neutral surface and

contraction below it cause an in-plane curvature,

curvaturecanticlasti1


11/34

Stress and deformation in symmetrical elastic

beams


12/34


13/34

Example 1.

SOLUTION:

Based on the cross section geometry, calculate

the location of the section centroid and

moment of inertia.

mm38

3000

10114 3

A

A yY

3

3

3

32

101143000

104220120030402109050180090201

mm,mm,mmArea,

A y A

A y y

49-3

23

12123

121

23

1212

m10868mm10868

18120040301218002090

I

d Abhd A I I x


14/34

Example 1.

• Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

49

49

mm10868

m038.0mkN3

mm10868

m022.0mkN3

I

c M

I

c M

I

Mc

B B

A A

m

MPa0.76 A

MPa3.131 B

• Calculate the curvature

49- m10868GPa165mkN3

1

EI

M

m7.47

m1095.201 1-3


15/34

Example 2

• Problem. A steel beam 25mm. Wide and 75mm. Deep is pinned to supports

at points A and B, where the suuport B is on rollers and free to movehorizontally. When the ends of the beam area loaded with 5kN loads, find

the maximum bending stress at the mid-span of the beam and also the angle

ΔΦ0 subtended by the cross sections at A and B in the deformed beam.


16/34

Example 2

• Solution.

1. Determine the bending moment.

From this diagram we see that

the portion AB is one of constant

bending moment (state of pure

bending).

2. To calculate the bending stress,

we must locate the axes and

calculate the moment of intertia:


17/34

Example 2

• The maximum bending stress occurs at the farthest from the neutral

surface. At the mid-span the bending stress at the top of the beam isfound to be:

• If we use y=-37.5mm., we obtain a numerically equal compressive

stress at the bottom of the beam.

• To obtain the angle change ΔΦ0 we use the moment-curvature

relation (curvature definition from calculus):


18/34

Example 2

• Then, the total angle change between A and B is found byintegration:


19/34

Example 3.

For the timber beam and loading

shown, draw the shear and bend-

moment diagrams and determine the

maximum normal stress due to

bending.

SOLUTION:

• Treating the entire beam as a rigid body, determine the reaction forces

• Identify the maximum shear and

bending-moment from plots of theirdistributions.

• Apply the elastic flexure formulas to

determine the corresponding

maximum normal stress.

• Section the beam at points near

supports and load application points.

Apply equilibrium analyses on

resulting free-bodies to determine

internal shear forces and bending

couples


20/34

Example 3.

• Identify the maximum shear and bending-

moment from plots of their distributions.

mkN50kN26 Bmm M M V


determine the correspondingmaximum normal stress.

36

3

36

2

612

61

m1033.833

m N1050

m1033.833

m250.0m080.0

S

M

hbS

Bm

Pa100.60 6m


21/34

Example 4.

The structure shown is constructed of a

W10x112 rolled-steel beam. (a) Draw

the shear and bending-moment diagrams

for the beam and the given loading. (b)determine normal stress in sections just

to the right and left of point D.

SOLUTION:

• Section the beam at points near thesupport and load application points.

Apply equilibrium analyses on

resulting free-bodies to determine

internal shear forces and bending

couples.


determine the maximum normal

stress to the left and right of point D.


22/34

Example 4.


determine the maximum normal stress tothe left and right of point D.

3

3

in126

inkip1776

:

in126

inkip2016:

S

M

Dof right theTo

S

M Dof left theTo

m

m

ksi0.16m

ksi1.14m


23/34

Stresses in Symmetrical Elastic Beams

Transmitting Both Shear Force and Bending

Moment

• Pure bending is a relatively uncommon

type of loading for a beam. Instead it is

more common for a shear force to be

present.

• The presence of the shear force means

that the bending moment varies along the

beam (i.e. symmetry arguments are no

longer applicable).

• In Engineering applications we make the

assumption that the bending-stress

equation is valid even when a shear

force is present.


24/34

Shear Stresses in Beams

00

0

00

x z xz z

x y xy y

xy xz x x x

y M dA F

dA z M V dA F

dA z y M dA F

• Distribution of normal and shearing

stresses satisfies

• Transverse loading applied to a beam

results in normal and shearing stresses intransverse sections.

• When shearing stresses are exerted on the

vertical faces of an element, equal stressesmust be exerted on the horizontal faces

• Longitudinal shearing stresses must exist

in any member subjected to transverse

loading.


25/34


• Consider prismatic beam

• For equilibrium of beam element

A

C D

A D D x

dA y I

M M H

dA H F 0

xV xdx

dM M M

dA yQ

C D

A

• Note,

flow shear I

VQ

x

H q

x I

VQ H

• Substituting,


26/34


flow shear I VQ

x H q

• Shear flow,

• where

sectioncrossfullof momentsecond

aboveareaof momentfirst

'

21

A A

A

dA y I

y

dA yQ

• Same result found for lower area

H H

QQ

q I QV

x H q

axisneutralto

respecthmoment witfirst

0


27/34

Example 5

A beam is made of three planks,

nailed together. Knowing that the

spacing between nails is 25 mm andthat the vertical shear in the beam is

V = 500 N, determine the shear force

in each nail.

SOLUTION:

• Determine the horizontal force per

unit length or shear flow q on the

lower surface of the upper plank.

• Calculate the corresponding shear

force in each nail.


28/34

Example 5.

46

2

3121

3121

36

m1020.16

]m060.0m100.0m020.0

m020.0m100.0[2

m100.0m020.0

m10120

m060.0m100.0m020.0

I

y AQ

SOLUTION:

• Determine the horizontal force per

unit length or shear flow q on the

lower surface of the upper plank.

m N3704

m1016.20

)m10120)( N500(46-

36

I

VQq

• Calculate the corresponding shear

force in each nail for a nail spacing of25 mm.

m N q F 3704)(m025.0()m025.0(

N6.92 F


29/34

Determination of the Shearing Stress in a Beam

• The average shearing stress on the horizontal

face of the element is obtained by dividing theshearing force on the element by the area of

the face.

It

VQ

xt

x

I

VQ

A

xq

A

H ave

• On the upper and lower surfaces of the beam,

yx= 0. It follows that xy= 0 on the upper and

lower edges of the transverse sections.

• If the width of the beam is comparable or large

relative to its depth, the shearing stresses at D1

and D2 are significantly higher than at D.


30/34

Shearing Stresses τ xy in Common Types of Beams

• For a narrow rectangular beam,

A

V

c

y

A

V

Ib

VQ xy

2

3

12

3

max

2

2

• For American Standard (S-beam)

and wide-flange (W-beam) beams

web

ave

A

V

It VQ

max


31/34

Distribution of

Stresses in a Narrow Rectangular Beam

2

2

12

3

c

y

A

P xy

I

Pxy x

• Consider a narrow rectangular cantilever beam

subjected to load P at its free end:

• Shearing stresses are independent of the distance

from the point of application of the load.

• Normal strains and normal stresses are unaffected by

the shearing stresses.

• From Saint-Venant’s principle, effects of the load

application mode are negligible except in immediatevicinity of load application points.

• Stress/strain deviations for distributed loads are

negligible for typical beam sections of interest.


32/34

Example 6.

A timber beam is to support the three

concentrated loads shown. Knowing

that for the grade of timber used,

psi120 psi1800 all all

determine the minimum required depth

d of the beam.

SOLUTION:

• Develop shear and bending moment

diagrams. Identify the maximums.

• Determine the beam depth based onallowable normal stress.

• Determine the beam depth based on

allowable shear stress.

• Required beam depth is equal to thelarger of the two depths found.


33/34

Example 6.

SOLUTION:

Develop shear and bending moment

diagrams. Identify the maximums.

inkip90ftkip5.7

kips3

max

max

M

V


34/34

Example 6.

2

2

6

1

261

3121

in.5833.0

in.5.3

d

d

d bc

I S

d b I

• Determine the beam depth based on allowable

normal stress.

in.26.9

in.5833.0

in.lb1090 psi1800

2

3

max

d

d

S

M all

• Determine the beam depth based on allowable

shear stress.

in.71.10

in.3.5lb3000

23 psi120

2

3 max

d

d

A

V all

• Required beam depth is equal to the larger of the two.

in.71.10d

Documents

Bending of Beams Sollero