rezi_2corectat

7/31/2019 REZI_2corectat

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8. OBLIQUE BENDING {biaxial bending}8.1. Definition

A cross section is subjected to oblique bending if the normal stresses {direct stresses} x are reducedin the centroid G to the bending moments My and Mz.

!!!

- the vectorial senses of the My and Mz from the fig.,

are the positive ones.My and Mz are positive when their resultant Mi tensions

{stretch} the 1st

quadrant {of y and z positive}.

!!!

- the vectors My and Mz are always included in the

cross section plan.

- generally, oblique bending appears simultaneously with

oblique shering, produced by the shear forces Tz and Ty. If

the shear forces Tz=Ty=0 we discuse about pure obliquebending.

8.2. Mode of loading- oblique bending with shering appears in 2 main modes of loading:

a. the loads are applied perpendicular to the torsion axis X, and parallel to the principalplanes xGz and xGy.

- we assume that for a c.s. the significant system of axis of G and C are shown in figure:

!!!

- the forces lines must pass through the shere

center C to avoid the existence of a

supplementary solicitation, called torsion.

Otherwise, if the forces act to a certain

distance from C, they must be reduced in C,

and beside My, Mz, Tz and Ty a torsional

moment Mt will act.

- the vectorial forces Pz produce Tz and My. The

forces plan is xCz and the force line is Cz. Tz will

act in C, My act in G. The neutral axis for

bending is Gy.

- the forces Py produce Ty and Mz acting in C,

respectively, G. The forces line is Cy, while the

neutral axis is Gz.


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b. the forces act in a plan which pass through the shere centre C (contains x axis), but it isinclined with an angle with respect to the principal system of axis xGz and xGy.

- this case of loading can be reduced to

decomposing from the bending forces P to the

components with respect to Gy and Gz axis.

!!!

- in the 1st case of loading, the ration My/Mzisnt

constant along the bare, so the deformed axis will

be a skew {stramb} curve in space.

- in the 2nd

case of loading My/Mz is constant, in

any c.s. so, the deformed axis will be a plane

curve.

8.3. The normal stresses x in oblique bending- we consider the beam loaded by 2 equal forces situated to equal distances from supports and

individual with the angle with respect to z axis. For simplification we consider that the c.s. is double

symmetrical.

Pz = P* cos

Py= P* sin


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- similarly, Mz will produce bending

with the neutral axis Gz and x will becalculated also withNaviers Formula.

- using the hypothesis from strength of

materials regarding the smalldeformation of the construction

members and considering the materials

having a linear elastic behavior we may

superpose the effects of each straight

bending from My and Mz obtaining the

formula ofx for oblique bending.

- certainly, in strength calculation we are interested in calculation the max normal stress xmaxR.- xmax will be produced by Mymax and Mzmax, but taken from the same section.

ex:

- sect.2: dangerous sections for oblique bending

Mymax = -P*l

Mzmax = P*l

dangerous sections:

-sect D:8

*2

max

lqMy oblique

bending

-sect B: 0yafM straight

bending

2

* aPM

y

zaf

aPM yz *max


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- in case of oblique bending we have to determine the position of the neutral axis. This is calculated

equalizing x to 0.

0** yI

Mz

I

M

z

z

y

y

x

yI

I

M

Mz

z

y

y

z ** - n.a. equation.

- the neutral axis is a straight line which pass through G and it is inclined with an angle with respectto Gy axis.

z = -m*y

m = arctg = arctg

z

y

y

z

I

I

M

M*

!!! The resultant moment Mi(the vector) dosent act anymore along the neutral axis.

111 ** yI

Mz

I

M

z

z

y

y

x

222 ** yI

Mz

I

M

z

z

y

y

x


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9. Oblique shering9.1.Definition

We have seen in the previous chapter that oblique bending is usually accompanied by oblique

shering which is a solicitation produced by 2 shere forces Tz and Ty. Applying the shere forces in the shere

center C, their positive convention is:

- if in oblique bending we can use the super position of

both terms from My and Mz, in oblique shering its more

difficult to make this super position because any time

we have to write a static moment with respect to n.a. (an

inclined straight line) the distances which appear in

there static moment , are difficult to be calculated.

- easier, we shall calculate separate from Tz,

respectively Ty, and finally we shall add this values to find in a

certain point.

- from:yz

yzT

xzIb

STT z

*

*:

- from:zy

zyT

xyIb

STT y

*

*:

yzT

x

T

xz !!! with algebric signs

9.2.Oblique shering for rectangular c.s.From straight shering (Tz)we know that has a parabolicvariation along the side which is parallel to Tz, having the

max value in the n.a.:

A

Tz 5.1max

- in a similar manner, from Ty we shall have also a

parabolic variation of along the side parallel to Ty, with

the max value A

Ty 5.1

max


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- lets represent these variations in a perspective view:

- the max. value of is in the centroid G, but as from Tyis xy, we can only compose these max.

values.

()

9.3.Double T (I) section made from narrow rectangles


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9.4.Symmetrical c.s. with 2 webs (inima)

9.5.The shear center positionThe shear center C also named the center of bendingtorsion and it is the 2

ndsignificant point of the

c.s. with respect to C the shear unit stresses are reduced, obtaining the main stresses: the sh ear forces Tz

and Ty, respectively the torsional moment Mt.

- for a double symmetrical c.s., C=G, while for a unit symmetrical c.s., C is on the symmetry axis,

9.5.1. The channel (U) cross section


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- we admit that a shear force Tz pass through the shere center C, and from this shere force we make the

corresponding diagrams, xz in the web with linear distribution.

- we calculated all significant values of

2

**

2

***

*

2

***

*

8

*

2

**

*

4**

22***

**

*

ma x

1

ma x

2

hb

I

Thtb

It

T

htb

Id

T

hdhtd

Id

T

hd

hhtb

Id

T

Ib

ST

y

z

y

z

xy

y

z

xz

y

z

y

z

yz

yz

xz

- we calculate the resultant R of these diagrams, representing the va l. of each diagram R=A*t ( is

constant on the thickness t of the thin profile)

Rproportional in the medium line of each rectangle

2

**

12

*2*

4*

12

*

12

*

4

****

2*

2

***

2*

12

*

2

****

2

**

**

3

2*

2

***

*

**3

2*

23233

2

2

322

111

ma x

ma x

htbhdht

tbhdI

thb

I

Tt

bhb

I

Tt

bR

hdhtb

I

Idh

hd

Id

Ih

htb

Id

T

dhR

y

y

z

y

z

xy

y

z

y

z

y

z

xzxzxz

- we determine the shear center position from the condition that the torsional moment produced by these

resultants, written with respect to C is O.

(Mt)c=0: -R1*+R2*h=0

yI

thb

R

hR

*4

***2

1

2


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9.5.2. The calculation steps in finding the shear center positionWe admit that for a unit symmetrical c.s. we can calculate the shere center position following the

steps:

1) We position C on the symmetry axis2) In C we apply a shear force perpendicular to the symmetry axis.3) From this shear force we draw all diagrams.4) We calculate all significant value function the constant I=k=const.5) We calculate the resultant of all diagrams R=ct *T and we position this resultant in the

medium line of each corresponding narrow rectangle taking into account the shear stress

senses.

6) We write the condition that the torsional moment given by these resultants, with respect to C is0.

(Mt)c=0 in order to avoid a supplement solicitation: the torsion.


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10.Bending and axial action(exccentric tension or compression)10.1.Definition

A c.s. is subjected to bending with axial force if the interior unit stresses are reduced to: an axialforce N and 2 bending moments My and Mz.

- the stresses from this fig. will produce oblique bending with

axial force. This case of solicitation can appear when certainforce F parallel to the bare axis Gx acts in a certain point Q(y0; z0):

N = F

My = F*Z0 = N*Z0

Mz = F*y0 = N*y0

Oblique

bending +

axial for

A particular case of solicitation is the one when 1 bending moment is 0, obtaining straight bending with

axial force.

N = F N = F

My = F*Z0 My = 0

Mz = 0 Mz=F*y0

straight

bending+N

- all these stresses, M, My, Mz produce normal stress , which in general appears simultaneously with theshear stress , produced by Tz and Ty. is calculated from oblique shering.


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10.2. The calculation of the normal stresses from oblique bending with axial forceWe admit a c.s. is subjected by 2 bending moments and an axial force, acting positive.

From statically point of view the same solicitation can be represented assuming the internal stresses N, My,

Mz are reduced in Q to a axial force N applied to the exentricities y0, z0 (b), distance which must be:

Similar to oblique bending we may superpose these 3 relations, taking into account the hypothesis of the

small deformations and admitting a linear elastic behavior of the material:

yI

Mz

I

M

A

N

z

z

y

y

x **

A

I

yy

A

I

zz

A

N

yI

yNz

I

zN

A

N

yNMzNM

zyx

zy

x

zy

**1*

**

**

*;*

00

00

;00

In the above relation : ; , representing the square of the gyration{inertia} radius:

2

0

2

0**

1*zy

xi

yy

i

zz

A

N

From the above relation, we observe that if y=z=0 {in the centroid G}, , so in the centroid thenormal stress appears only due to the axial force.


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10.3. The neutral axis positionIt will be determined from the condition that in the n.a., x=0 => 0

**1

2

0

2

0 zy i

yy

i

zz

- the equation give the n.a. position, representing the equation of a straight line which doesnt pass

anymore{like in straight or oblique bending} through the centroid G.Thats why the n.a. may intersect the

c.s. (a) dividing it into a tension and compression part, or it may be tangent (b), or outside completely from

the c.s. (c). In these 2 final cases, the normal stresses will have a unique sign (+ or -) depending on theaxial force sign. This situation appears in case of big axial forces.

The n.a. position can be determined from the above equation if we make an turn y=0, respectively

z=0, obtaining the n.a. cuts{taieturi}:

;0z 0

2

z

iz

y

n

;0y 0

2

y

iy z

n

- this means that the n.a. will pass through the points A(y0,0) and B(0,z0).

A

Ii

N

Mz

A

Ii

N

My

z

z

y

y

y

z

2

0

2

0

;

;

- observing the relation of the n.a. cuts, we may conclude that the point of application of the axial excentric

force N, Q(yo,zo) and of the n.a. given by the cuts yn,zn are always situated on one part and the other with

respect to the c.s. centroid G(yn and zo, respectively zn and zo) have always opposite signs.

- also, the n.a. position is given {concerning its inclination} by the bending moment M y and Mz {like in thecase of oblique bending}. The influence of N on the n.a. position is produced bt the displacement of n.a.,

from the position given onley by oblique bending {for N=0}


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10.4. Geometrical correlations between the excentric point f application of N and n.a.The excentric axial force acts in the point Q(yo,zo), where ; yo and zo appear in the relationof yn and zn, so, the following correlations may be written:

a) the n.a and the axial force N will be situated always in

opposite quadrant with respect to centroid G, obtaining in

consequence max. normal stress in the extreme point close

to N position.

b) as far is the point Q with respect to the centroid G, as close

is the n.a. with respect to G, and inversely.

- if yo = zo = N yn = zn = 0(the case of oblique bending My, Mz- if yo = zo = 0 (My = Mz yn = zn (the case of axial solicitation, N c) the n.a. inclination is due only to the bending moment M y

and Mz, the axial force presence having the effect of

displacement the n.a. in the opposite reaction with respect to

Q.

d) if the point Q is situated on a principal inertia axis, n.a. will

be perpendicular to that axis.

*zI

M

A

N

0M0z

z

y

iy

z

z

x

y0

n

0

2

z

n

e) if the point Q(yo,zo) is displaced on a straight line d-d, the n.a.

is rotated around a point Q from the n.a.


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R

yI

Mz

I

M

A

N

yI

Mz

I

M

A

N

xxx

z

z

y

y

x

z

z

y

y

x

minmaxmax

min

max

,max||

**

**

10.5. The strength verification in case of bending with axial force

xmin

xmax

- for the rectangular c.s. the general situation is for oblique

bending with axial force:

6

*

6

*

2

12

*

2

2

3

min

max

hbW

hb

h

hb

z

IW

W

M

W

M

A

N

W

M

W

M

A

N

z

y

y

z

z

y

y

x

z

z

y

y

x

- in this case, if the excentric axial force N is situated on a principal inertia

axis {ex.:Gy}, we shall have the partial case of straight bending with axialforce.


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We note the excentricity yo with e Mz = N*e.

b

e

A

N

W

eN

A

N

z

x

*61

* - the normal stress x for the rectangular c.s.

- due to the presence of the excentricity e of the axial

force, we may have 3 different situations:

a) ifx

be

6will have 2 different extreme values: one

of compression, one of tension.

b) if

0

*2

6

min

max

x

xA

Nbe

c) if 6

be : both extreme values will have the same

sign (+ or -)


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10.6. Excentric compression in case of materials which cant undertake tension (materials withweak tensile strength)

From previous paragraph we have seen that the position of the excentric axial force N which gives implicit

the n.a. position, give different distribution of the normal stresses .

- there are elements of construction made from materials which

have a very weak tensile strength, practically neglected: simple

concrete, brick(stone) masonry, the foundation soil. For themits recomanded to work mainly in compression, thats why the

elements made from this material (mainly the foundations) must

have such dimension in order to be subjected only to

compression (or a small part may be subjected to tension),

solicitation undertaken by the reinforcement.

- in order to see if a construction element is subjected only to

compression, we are interested in finding a domain from the

member c.s. where if the excentric force N is applied only

compressive stresses x will exist. This domain is called central

core (sambure central) or kern.

10.6.1.The central coreIt is a ficticous notion representing the geometric place of all points of application of the excentric axial

force N, corresponding to the n.a. tg to the c.s. always is in a central zone, situated around the centroid G.

- if the excentric force N is inside the central core or at limit on the

central core content the entire c.s. will be subjected only in

compression (for N>0).

- if the excentric force N is outside the central core n.a. will

intersect the c.s. having both tensile and compressive normalstresses.


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The central core properties:

a) The c.s. and the c.c. are reciprocal figures meaning that for every side of the c.s. it willcorrespond a c.c. vertex.

b) The central core of a convex polygonal section is also a convex polygon, having the nr. of vertexequal to the nr. of sides of the polygon in which the c.s. is inscribed.

c) If the c.s. is symmetric with respect to 1 or 2 axes, the central core vertex will be alsosymmetric.

d) The n.a. tg to the c.s. contour, should not intersect the c.s.

e) The coordinates of the central core vertex are calculated in the principal inertia system of axes.

In order to represent the central core of a c.s. we represent on turn n.a. obtaining coordinates ofcorresponding vertex Pi(yvi,zvi), calculated with the relations:

,


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10.6.2.The central core for some usual c.s.a) The rectangular c.s ,

!!!For the rectangular c.s. the c.c. is a rhomb with the semi diagonal , respectively - the same central core have the c.s. which may be inscribed into a rectangle.


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b) The circular c.s.

- the central core is also a circle of radius

.

10.6.3.The active zone in case of materials with weak tensile strengthWe have seen in the chapter of centric compression that the most dangerous in the case of foundation

is the one of contact between foundation and soil where we had to check if xmaxRt.Now, in case of excentric compression we have to take into account the possibility that a part of thiscontact surface is subjected to tension, producing tensile stresses that cant be overtaken by foundation and

soil.

For any c.s. there are possible 2 situations:

1. The axial compressive stresses N acts in the central core (a) or at the limit of the centralcore (b), situation when the entire c.s. will be compressed.

2. The compressive force N acts outside the core the n.a. cutting now the c.s. dividing itinto a compressed zone, called active zone and a tension zone which will be completely neglected.


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The excentric force N is carried out only by the compressed part on the c.s. (active zone). The

corresponding axis which cuts the c.s. is no longer the n.a., but an axis called zero axis (zero-because x=0

in this axis). It has a different position from n.a., being close to the most compressed fibre of the c.s.We are interested in finding the position of this 0 axis(dist. o).

Assuming that x has a linear variation we may write the normal stress at a certain level ,

We write the stresses acting in c.s. with respect to 0 axis, from a strength calculation.

- the axial excentric force:

- the bending moment:

Iothe moment of inertia of the active area with respect to 0 axis.

- dividing relation (b) to relation (a) we obtain 0 related to the compressive force N position.

cdistance from the excenric force position to the most compressed fibre: la=0+c

- replacing la in the relation (a) we obtain:

- when N is outside the central core

)(*0

0

maxc

S

N


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For the rectangle c.s.

h

The max normal stress:

!!! When N acts outside the central core and only for rectangular c.s.


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11.Energetical methods for calculating the bar disolacements11.1. The strain energy (energia de deformatie)

Using energetical methods, the total potential strain energy Ud, has the general formula:

Applying this formula for the simple case of solicitation: axial solicitation, straight bending, straight

shering and torsion, we get the following expression for the strain energy.

The terms EA, EI, GA and GIt are modulus of rigidity corresponding to axial solicitation straight

bending, straight shering and torsion.

Gshear modulus (transversal modulus of elasticity)

- for mild steel: G=8.1x10

5daN/cm

2

Atransformed area of the c.s.

, kshape factork = 1.2:

k = 22.4:

Itmoment of inertia in torsion.

- based on the method of virtual displacement Castigliono demonstrated two theorems very useful in

displacement calculations.

- the 1st

theorem is :

For a linear structure the partial derivative of the strain energy with respect to P i is equal to

the corresponding displacement di considering that the strain energy is expressed as a function of

loads.

- we apply this theorem for the strain energy written before:

[

]

- but in case of a linear elastic structure, the 1st

derivative of any stresses can be noted with:

,

, , ,- n, m, t, mtinfluence coefficients representing internal stresses provided by a virtual unique force

Pi=l

- as the diagrams n, m, t and mt may have maximum linear variation, the integrals, which are infinite

sums may be transformed into finite sums.

dxI*Em*N

- the most important term


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- the displacement di can be calculated using thise finite sum applying the rule of integration ofVeresciaghin,

which introduces a new form for di:

IE

d iii*

*

- where: - ithe area of the real diagram of moment.

- ithe ordinate corresponding to the area.

- ithe number of intervals on which Mi and mi have continuous variation.- i, measured in midiagram on the line corresponding to the centroid of i.!!! iand i are taken with algebraic signs.

- neglecting the term (which takes account the static nature of

displacement) from the above relation, we may write Mohr-

Maxwell formula for the displacement calculation (its also

called the unit load method for linearly elastic structures).

- the formula was demonstrated 1

stby Maxwell in 1874, but it was 1

stused by the german engineer in 1864.

- supplementary, the formula has 2 other terms regarding the displacement produced by temperature

variation and from supports.

- in the above formula N, M, T and M t are internal stresses obtained from real loads; n, m, t, mt are

internal stresses obtained from a virtual action.

- a virtual unit force applied in the section where the deflection v and w, respectively the elongation l,

should be calculated.

- a virtual unit moment 1 applied in the section where the rotation should be calculated.


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12.Pure torsionA member is subjected to pure torsion if in any c.s. the single stress is the moment of torsion Mt(torque).

---

---

---

- generally, construction members are subjected to torsion, but associated with other solicitation: bending,

axial solicitation, shering.

- the study of torsion is a complex problem, because only for circular or ring-shaped sections the

fundamental hypothesis used in strength of materials, can be used. For other type of c.s., rectangular c.s.,

section made from narrow rectangles (open or closed) these hypothesis are no longer valid, imposing the

using ofTheorem of Elasticity. This is a consequence of the fact that durin the member torsion, the c.s. aredistorted (every point from a c.s. has different deformations). Only for circular sections, due to the symmetry

of solicitation and geometry, this distortion doesnt appear (the c.s. remain plane after deformation).

- a rectangular section becomes after twisting approximately a hyperbolic paraboloid.

!!! If the c.s. distorsion is freely produced, the torsion is called free or uniform. If the distorsion is

prevented, we discuss about prevented or non-uniform torsion.

- in this chapter we discuss about free torsion.


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12.1. Torsion of bars with circular (or ring-shape) sections12.1.1.Geometrical aspect

We consider a model with a circular section, rectangular shered by parallel circles angle equidistant

generatrics.

- the bar is subjected at both ends by 2 equals moments of torsion.

- after torsion, the circular sections remain plane and equidistant, the single modification being the

inclination of longitudinal generatrics, all with the angle . So, the single specific def. will be the angular

deformation, .

- as the initial plane c.s. remain plane after deformation. Bernoullis hypothesis is still valid.

- we isolate a differential element from this model, of length dx. To understand better the phenomena we

consider that this differential element is fixed at one end (the bottom bar) which the other end is free.


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- the c.s. from the top part (free end) will be rotated with a differential angle d.

- the angle between the twisted generatrics BD and the initial generatrics BD represents the modificationof straight angle, being in fact the specific deformation

- due to the symmetry of deformation, the relative displacement of B, BB is perpendicular to

- to express the specific deformation we write this displacement from 2D: DOBB and DBBD.

Where the specific rotation (twist), representing the angle of rotation per unit length

* + 12.1.2.Physical aspect

- admitting the bar has a linear elastic behavior, Hooks low is valid, so: - so, the single distinct stress is the tangential stress which is proportional with the radius .

- will have a linear variation with zero value in the centroid and max. value on the circle contour, for = R

The tangential stress sense is given by the torsional moment sense.

12.1.3.Statical aspect- from statical point of view Nt=Mto.

- this stress, Nt can be written from interior from a strength calculation.

- the specific angle - the term GIp is the modulus of rigidity of free torsion.

- we replace in formula


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- we are interested in the max value of which corresponds to =R

- radius

- to use a general formula for the entire solicitation of free torsion we shall replace Ip and Wp with It, Wt (for

circular c.s. Wt = Wp and Wt=It).

- we may calculate also the total rotation : - observing the relation of we may conclude that its similar to Naviers formula, but only from

mathematical point of view. From mechanical point of view its completely different because it can beapplied only to circular sections.

12.2. Torsion of beams with non-circular sections- as we explained before the relation from the previous paragraph are no longer valid, because different

points from a c.s. have different displacements, the c.s. being distorted.

- a good solution for these twisted bars was given by the french scientist Barrie de Saint-Verne. Anyway, he

used the methods from theory of elasticity. After that, he obtained very good results making an analogy

between the phenomena of torsion and that of deformation of an elastic membrane. The procedure the

analogy of the elastic membrane, applying the observation that both phenomenon have an identical structure

with different equation containing partial derivatives.

- explanations of the membrane theory:-we assume that a box is provided with a lid which has a hole of the same form and dimension as the

bar c.s.

- over this hole an elastic membrane (has no rigidity in bending) is tensioned by a constant tension

force on contour.

- the box is provided with another hole on a lateral surface, where a gas is introduced under a

pressure p which will act on the box walls.

- due to these 2 forces the membrane is deformed becoming a curved surface, the tension force from

the membrane equilibrating the exterior forces. It was demonstrated that the different equation of the

deflected surface of the membrane has the same form as the equation which govern the stress distribution in

a c.s. of a torsion bar.- these equations may be written for the elastic membrane:


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- we shall use 3 similitudes:1) the tangent to a counter line in any point of the deflected membrane corresponding to the direction ofx stress in the c.s. point of the twisted bar.2) the moment of torsion Mt is twice the value included by the surface of the deflected membrane and

the horizontal plane.

3) the max slope of the membrane in any point is even the tangential stress in the corespomding pointin the twisted bar.

12.2.1.The narrow rectangular cross sections- a rectangle is considered to be narrow if

. for this section, the influence of the short side is

neglected, so in the membrane analogy, the deflected shape of the membrane may be generatrics parallel to

the long side.

- we consider a strip having a unit width and we represent in section the arc corresponding to this strip with

the forces acting on it.

- as the membrane has no rigidity in bending we write the bending moment and we make it equal to zero.

From symmetry condition we may write only for a half of arc.

- replacing the expression of from the member analogy we have x=Gy2

- this equation is the equation of a parabola of 2nd

degree, representing even the equation of the deflected

membrane.

- thats why well similitudes

1) the direction of is || to the long sides of the narrow rectangle.2) the slope of the membrane is even the - the above relation proved the linear variation of with value for y=0 and max. value for y = 3)Mt is 2x val.

- replacing this in the formula of :


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- the max value of is for

12.2.2.The broad rectangular c.s.

A rectangle is considered to be broad if

.

If for the narrow rectangle has the same distribution alongthe long side of the rectangle for the broad rectangle is max incentral line and 0 in corner. Its max on the rectangle counter and 0 in

the centroid, but it has a parabolic variation.

, , numerical factors function the ratio

h/t 1 1.5 1.75 2 2.5 3 4 5

0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299

0.141 0.196 0.214 0.229 0.249 0.263 0.281 0.292

1 0.252 0.82 0.795 0.766 0.753 0.745 0.744


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12.2.3.Sections made from narrow rectangles, having an open contour(simple conex sections)

- for the simple conex sections we generalize the formulas for 1 narrow rectangle;

- at the 3rd power will be always the thickness- Wt corresponds to the rectangle having the greatest thickness, tmax;

; ; ;- the stress distribution in each narrow rectangle is identical to the one of a narrow rectangle ( || to the longsides).

- if the c.s. is made only from 1 laminated profile (I, U), or if its connected by b olds or rivets, it is affected

by a coefficient becoming : =1.2

=1.12

=1

=1 {welded}

=0.5 {riveted{bolded}}


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12.2.4.The c.s. made from narrow rectangles, but with a closed contour(duble conex sections)

- the surface is considered as thearea closed by the medium line of

each rectangle. indicates that maxwill be in the rectangle of the

minimum thickness;

- the stress distribution on eachrectangles thickness .


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13.Torsion of thin-walls members which have the cross-sections prevented from working(preventedor non-uniform torsion)

13.1. Generalities. The phenomena of working torsion- the members studied in pure torsion have the c.s. completely free to work(torsion). In reality, the

construction members are supported, so the c.s. are prevented from working.

- we have seen the case of a torsion rectangular c.s., which after torsion gecame a hyperbolic paraboloid.

- besides the c.s. distortion, a distortion of the median line will exist. As the median line Gx axis remains in

the initial plane the c.s. median plane is zero .- lets consider now a bar having an I c.s. subjected to torsion by two equal moments acting in the end ofbars. We represent, after torsion a top view of the member, observing the flanges positions.

- as the above bar is free to work(torsion), it is subjected to free uniform torsion.- now lets consider the same bar, but having an end fixed(built in).

- now that end cant be distorted.


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- due to the built-in support a new deformation will appear,

the distortion of the c.s. median line varying along the bar

from zero in the fixed support to a max. value in the free

end. In this case we discuss about prevented torsion.

- the preventing of the medium line distortion represents the

preventing of the elastic deformation along the bar axis Gx.

For this reason a new normal stress will appear

As this normal stress varies along the bar axis, it must

be equilibrated by a new tangential stress .

- from these observation we may conclude that due to this prevented deformation, 3 distinct stresses will

appear in the c.s. (only from torsion):

1)A normal(direct) stress produced by the flanges bending:

- this normal stress doesntintroduce any known internal stress

because the moment M which

produce the flanges bending are

equilibrated. Any way it must be

introduced a new internalstress(fictious) as an measure of the

resultant of these normal stresses.

The new stress is called bimoment

B.

- so, the bimoment will produce the

normal stress .


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2)As we explained before a tangential stress produced by the distortion of the median linealong the bar axis will equilibrate x.

Prevented torsion Free torsion

- the resultant of this tangential stress, reduced in the shere center define a moment of torsion M calledmoment of prevented torsion.

3)As the moment of torsion which produce this solicitation isnt overtaken completely by

M, a free moment of torsion M will produce also a tangential stress calculated from pure(free)torsion. -from these three stresses:, and only the one is known from pure torsion.

13.2. The normal stress and the tangential stress As we have seen,

is produced by the new stress bimoment B, being demonstrated that :

Where: - - the sectorial coordinate (sectorial surface) [cm2];

- Ithe sectorial moment of inertia [cm6];

- the bimoment b is calculated function the total rotation angle of the deformed bar. - the tangential stress is produced by the prevented moment of torsion , being calculated:

Where: - S sectorial static moment [cm5];


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- between the bimoment B and the prevented moment of torsion M , a differential relation exists (similarto the one between the bending moment and the sher force).

- the stresses and are yet unknown because the total angle of rotation is unknown.13.3. The differential equation of the rotation angle

Where: - kthe bar characteristic at flexuranl torsion;

- k2the ration between the rigidity in pure torsion and in prevented torsion;

- as the differential equation of the rotation angle is unknown, the solution

is the sum of a general and a

particular solution; - finally, the solution of the differential equation is:

Where: , , - are called parameters in origin for x=0; they are calculated from

boundary condition written in supports:- for a simple supported or hinged support.

;x = 0 x =

- for a fixed support:

x = 0


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- Pthe particular solution, the solution of unknown differential equation depending on the type ofloading:

*

+


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14.Computing of section in elasto-plastic domain14.1. Generalities- all strength calculations considered in all previous chapters it was used the hypothesis of the linear elastic

material, where Hooks law is valid (=E*). We considered in all previous calculations that the maxnormal stress xmax is limited by the design strength: maxR, but Hooks law can be written until the yieldlimit, when

reaches the yield value

c.

- but, for a ductile material (mild steel) the reaching of c in a point doesnt produce the failure of thatelement of the construction. After producing the plastic deformation in one point, the strength capacity isntconsumed. This means that we can stil increase the load until the reaching of the complete limit state.

- to study an element beyond the elasticity limit, the material must be homogen and isotropic. The material

behavior can be illustrated in the ideal elasto-plastic curve (Prontdls curve) forductile material with yield

plateau.

- the yield limit for common steel:

- in Romania standards: Rc=2400 daN/cm

2

- in Eurocode 3: fy=235 N/mm2

14.2. Pure bending in elasto-plastic domain- we consider a bar having a symmetrical c.s. subjected to pure bending.


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- in diagram (a) My produces a normal stress x characteristic to a linear elastic material, the stressdistribution being linear with zero value in the n.a.

- the max normal stress xmaxdoesnt reach the yield limit.

- we continue to increase the load until in the most solicitated fibre the normal stress reach c (diagram(b)). In this situation we compute the limit elastic moment.

- if the load continues to increase, the specific deformations are increased in the plastified zone exceedingthe yield specific deformation c according to Parnells diagram, if >c, the normal stress remainconst. and equal to c. in the reman elastc zone Hooks low s stll vald being proportional to .- the stress distribution from diagram (c) shows that in the extreme zone reach c these zones beingcompletely plastified, while a central zone in still in the elastic zone, with the characteristic limit

variation.

- if the load is increased, the elastic zone is reduced and the limit situation is the one when the entirec.s. is plastified. We say that in that zone it was produced a plastic hinge (d).

- in diagram (d) we observe a different distribution of the normal stress, with rectangular blocks in all

fibers =c). in this situation we compute the limit plastic moment Mlim.pl. ,still unknown because weobserve that now the passing from the tensional section of area Atto the compressed section of area Acis made at an unknown level, corresponding to the plastic n.a.- to determin the plastic n.a. position, we write the axial force from a strength calculus N

reswhich must be

equal to the one determined from static calculus Nres

= Nst

. But, from the bar loading the single distinct

stress is My so, Nres

= Nst

= 0.

- this final relation shows that plastic n.a. divides the c.s. into 2 equal areas(while elastic n.a. divided it

in 2 parts of equal static moments).

- to compute the limit plastic moment, we write My from a strength calculation:

- plastic strength modulus. , the static moment of the tensioned area respectively compressed area, written with respect toelastic n.a or plastic n.a..

- for the rectangular c.s.:

- we can introduce a characteristic notion of the plastic calculus, called index of efficiency:


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15.The buckling of compressed bars15.1. Generalities- the buckling problem specific only to axially compressed members.

- the phenomena of buckling represents a problem of stability in sense that any compressed element can lose

its stability and not a problem of resistance.

- for linear elements (bars) we discuss about buckling, but for plane elements (plates) we discuss about local

buckling.

- the problem of buckling is a very complex problem, which uses some simplifying hypotesys having a

linear elastic behavior.

- the axial compressive force is centric applied;

- the bar axis is considered to be perfectly straight;

- these hypothesis consider that the compressed member has an ideal behavior without imperfections. In

reality, the bars have geometrical and structural imperfections which makes the differences between the real

and ideal bar.

- the method which studies the real bar is the one which uses the notion of divergence of equilibrium

which consider the imperfections that affects the real bar.

- the ideal bars are studied in buckling using the concept ofbifurcation of equilibrium.- in what follows we shall study the ideal bar being intrusted in finding the 1st

force of bifurcation which

corresponds to the 1st

phase of losing the bar stability critical force of buckling.

15.2. The calculation of the critical force of buckling using the statical method15.2.1.The double hinged bar

- we consider a double hinged column axially compressed by the force F. if the load F is unstable, transverse

deflection will appear and finally the column fails. For slender columns, buckling occurs at a certain stress

beyond the yield limit and for this reason this critical stress isnt related to the strength of materials.

- the phenomenon elastic instability(buckling)


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15.2.2.The cantilever(grinda)

Boundary condition:

cos

c o s - the solution for this equation: The 1

st(min.) force of buckling correspond to:

n

- for cantilever

We write the bending moment at a certain level x and we introduce it into the differential equation of the

deformed axis.

The above equation is a differential homogen equation having only a general solution: w=Asin kx +Bcos kx

We observe that the general solution depends on 2 constants, A and B which will be calculated from

boundary condition written in supports.

Obviously , A

0

sin k=0the solution for this equation is : k = n


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For different values of n, we obtain different critical force of buckling. We are interested in finding the

1st(min) force of buckling, for n=1 - for the double hinged bar

For the bar hinged at one end and fixed at the other, 15.2.3.Eulers critical force of buckling

Observing the above relation for the critical force of buckling for several types of supporting, Euler

developed in 1744 the generalized Eulers formula:

- where:

Ei - modulus of rigidity in bending, which corresponds to the min. axis of inertia (EImin)f - the length of buckling: - the coefficient of buckling length depending on the type of support.

= 1 = 2 = 0.7 = 0.5 = 1 = 2f= f= 2 f= 0.7 f= 0.5 f= f= 2


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For the critical force of buckling we may calculate the corresponding critical buckling stress with the known

formula:

Where: - the slenderness coefficient

- the max. slenderness coefficient is limited for any material by an admissible slenderness coefficient, a(max a).- we observe that the critical force of buckling was calculated taking into account the influence of the

bending moment (the buckling is produced by bending).

- however, for composed columns (steel, wood), the sher force is imp. affecting .15.3. The calculation of compressed members using the practical method of the buckling coefficient With this method the verification of compressed members in buckling (stability) is made with the formula:

Where: the buckling coefficient which is taken with the minimum value between y and z

- i is taken from tables, function:- the slenderness coefficient;- the c.s. shape;

- the yield limit Rc = 2400 daN/cm2 (OL37);

- the buckling curves A, B, C;- min

- the above relation corresponds to the axially compressed member (the single stress acting in any c.s. is N)

- for excentric compressed members, the formula becomes:

;Where: - Cxcoefficient of reduction the bending moment, function on the bending moment

variation on the bare length;

- gcoefficient which takes into account the lateral buckling of the bent bar(coefficient dedeversare);

rezi_2corectat

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