complex stresses (2nd year)

Complex Stresses

Dr Alessandro Palmeri <[email protected]>

Teaching schedule Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff 1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- --- 2 Shear centres A P Basic Concepts J E-R Shear Centre A P 3 Principle of Virtual

forces J E-R Indeterminate Structures J E-R Virtual Forces J E-R

4 The Compatibility Method

J E-R Examples J E-R Virtual Forces J E-R

5 Examples J E-R Moment Distribution -Basics

J E-R Comp. Method J E-R

6 The Hardy Cross Method

J E-R Fixed End Moments J E-R Comp. Method J E-R

7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R 8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R 9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P 10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric

Bending A P

11 Complex Stress/Strain A P Complex Stress/Strain A P Complex Stress/Strain

A P

Christmas Holiday

12 Revision 13 14 Exams 15 2

MoAvaAons (1/5) •  Failure of structures is oJen the result of different

stresses acAng together at the same locaAon

•  This is the case, for instance, of: –  Welded connec)ons, in which there are in general two direct

stresses to consider (σ normal and σ parallel to the weld axis) and two shear stresses (τ along and τ normal to the weld axis)

3 Weld cross section. International Journal of Fatigue, Volume 68, 2014, 178 - 185

MoAvaAons (2/5) –  Reinforced concrete

beams, in which different inclinaAons of the cracks appear, depending on the prevalent shear or bending acAon, as well as on the amount and distribuAon of reinforcement

4

RA=qL/2 RB=RA

B

q

A

L

V

M

prevalent shear force V

prevalent bending moment M

MoAvaAons (3/5) –  Cracking of concrete is

always due to the tensile stress σ exceeding the tensile strength (fct) of the material, which produces a rupture (ideally controlled by the steel reinforcement)

–  However this means that the orientaAon of the tensile stress is different for flexural cracks and shear cracks

5

–  Also, how do we get tensile stresses close to the beam supports, where the prevalent acAon is the shear force V?

•  And we know (Zhuravskii’s formula) that V induces τ, not σ in the cross secAon!

σ > fct

MoAvaAons (4/5)

–  Another example: While tesAng a concrete specimen in compression, fricAon sets up shear stresses at the base, whose effect is to change the cracking pa\ern, i.e. the direcAon of tensile stresses

6

Damage modes observed in cylinder compression tests as a funcAon of the boundary condiAons

Typical hourglass failure mode of a concrete cylinder

MoAvaAons (5/5)

–  Cylinder spli;ng test (also known as ‘Brazilian test’) is used to determine the tensile strength of concrete

–  We use a compression force (in the verAcal direcAon) to set up tensile stresses in the horizontal direcAon (verAcal fracture)… How?

7

Learning Outcomes When we have completed this unit (2 lectures + 1 tutorial), you should be able to:

•  Use the Mohr’s circle to determine: –  principal stresses, and their direcAons; – maximum shear stress, and the inclinaAon of the planes where it occurs;

–  normal stress and shear stress in any inclined plane

•  Only the case of plane stress will be considered, i.e. no out-‐of-‐plane stresses

8

Further reading

•  R C Hibbeler, “Mechanics of Materials”, 8th Ed, PrenAce Hall – Chapter 9 on “Stress TransformaAon”

•  T H G Megson, “Structural and Stress Analysis”, 2nd Ed, Elsevier – Chapter 14 on “Complex Stress and Strain” (eBook)

9

Stresses in Beams and Columns (1/3)

Within the limits of the Saint-‐Venant’s principle, stresses in beams and columns can be considered to be in plane state:

10

The difference between the effects of two

different but sta6cally equivalent loads becomes very small at sufficiently large distances from load

Adhémar Jean Claude Barré de Saint-‐Venant (1797-‐1886) was a French mechanician and mathemaAcian

•  i.e. normal stresses σ (sigma) and shear stresses τ (tau), all lie in the same plane


11

Axial (Normal) Force, N Shear Force, V

Bending Moment, M Twis)ng Moment, T

σ = NA

N N x

yz

τ = V ′QI b

V

y

z

σ = M dI

Mx

yz

M

τ = T rJ

T

y

z


12

•  There oJen situaAons where two or more internal forces act simultaneously

•  In a curved composite bridge, for instance, N, V, M and T induce complex stresses in the deck secAon

•  How the stresses combine?

Plane Stresses (1/2)

13

•  Let’s consider a material element subjected to plane stresses, such as…

–  In a reinforced concrete shear wall, with combined compressive and lateral forces (e.g. due to a seismic event)

V

P

xyz

σ z

τ xz

Plane Stresses (2/2)

14

–  In the the web and in the flanges of a thin-‐walled steel cross secAon

•  Without loss of generality, the plane {x,z} will be considered in our analyses

x

y

z

τ xz σ x

ConstrucAon of the Mohr’s Circle (1/7)

1.  The values of the three stresses σx, σz and τxz for a given material element are known

15

x

z

σ xσ x

σ z

σ z

τ xz

τ xz In this example: •  σx= 14 MPa (tension) •  σz= -‐6 MPa (compression) •  τxz= 8 MPa

ConstrucAon of the Mohr’s Circle (2/7) 2.  Draw the references axes in the Mohr’s plane

–  Normal stresses σ in the horizontal axis •  Posi2ve if in tension

–  Shear stresses τ in the verAcal axis •  Posi2ve if inducing a clockwise rota2on of the material element

16

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

σ

τ

tension compression

clockwise

an2clockwise

σ

τ

tension'compression'

clockwise'

an0clockwise'


3.  Locate point X≡{σx,τxz}, representaAve of the face of the material element orthogonal to the x axis

17

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

X ≡ {14,8}

σ x

τ xz

σ

τ


clockwise'

an0clockwise'


4.  Locate point Z≡{σz,-‐τxz}, representaAve of the face of the material element orthogonal to the z axis

18

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

X

Z ≡ {-6,-8}

σ z

−τ xz


5.  Locate the centre of the circle, Cσ≡{σave,0}, as the centre of the diameter XZ, at the average stress between σx and σz

σ

τ


clockwise'

an0clockwise'

Z

19

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

X

Cσ

σ ave

σ ave =σ x +σ z

2= 4MPa


6.  Use the Pythagoras’ Theorem to calculate the radius Rσ of the circle

σ

τ


clockwise'

an0clockwise'

Z

20

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

X

Cσ

σ ave

Rσ = 12

σ x −σ z( )2 + 2τ xz( )2

= 12.81MPa

`

Rσ


7.  Draw the Mohr’s circle, with centre , Cσ≡{σave,0} and radius Rσ

σ

τ


clockwise'

an0clockwise'

Z

21

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

X

Cσ

σ ave

Rσ

ProperAes of the Mohr’s Circle (1/4) •  Each point of the Mohr’s circle is representaAve of a the stresses (σ and τ) experienced by a face in the material element of a given inclinaAon

σ

τ


clockwise'

an0clockwise'

Z

X

Cσ

σ ave

Rσ

22

compression*

ProperAes of the Mohr’s Circle (2/4) •  RotaAng the faces of the material element, different stresses will be seen, and the representaAve points will move along the Mohr’s circle

σ

τ


clockwise'

an0clockwise'

Z

X

Cσ

σ ave

23

compression*Rσ

ProperAes of the Mohr’s Circle (3/4) •  Extremes points of each diameter are associated with stress condiAons on orthogonal faces, such as points X and point Z

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

24

compression*

ProperAes of the Mohr’s Circle (4/4) •  A rotaAon α of the faces in the material element corresponds to an angle 2α in the Mohr’s circle (in the same direcAon, e.g. both counterclockwise)

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

25

compression*

α=0°

α=22.5°

α=45° 2α

=90°

α=67.5°

Principal Stresses (1/4) •  It is always possible to find two orthogonal faces of the

material element in which there are no shear stresses, but normal stresses only

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

26

compression*

•  These are the “principal stresses”, i.e. maximum and minimum values of the normal stress for the varying inclinaAon of the element’s face

Principal Stresses (2/4) •  The principal stresses σp and σq are represented in the Mohr’s

circle by the extreme points P and Q of the diameter on the horizontal axis (where τ=0)

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

27

P Qσ p =σ ave − Rσ = −8.81MPaσ q =σ ave + Rσ = 16.81MPaτ pq = 0

⎧

⎨⎪

⎩⎪

Rσ

x

z

σ xσ x

σ z

σ z

τ xz

τ xz

x

zσ q

σ q

σ p

σ p

Reference element

Rotated element

Principal stresses

Principal Stresses (3/4) •  The principal stresses occur along the

principal direcAons of the stress, p and q; •  They are orthogonal each other, and can

be determined considering the corresponding angles in the Mohr’s circle

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

28

P Q

Rσ

x

zσ q

σ q

σ p

σ pRotated element

Principal stresses

α xq =α zp = − 12sin−1 τ xz

Rσ

⎛⎝⎜

⎞⎠⎟

= − 38.62

= −19.3

α xq

2α xqτ xz

•  Important: It is assumed here that angles α are posi2ve if an2clockwise

Principal Stresses (4/4) •  In case of bri\le material, such as concrete,

cracks may appear orthogonally to the direcAon of the maximum tensile stresses

σ

τ


clockwise'

an0clockwise'

Cσ

σ ave

Z

X

29

P Q

Rσ

x

zσ q

σ q

σ p

σ pCracked element

α xq

2α xqτ xz

Maximum Shear Stress •  The maximum value of the shear stress is τmax=Rσ and

happens in two mutually orthogonally faces, which are inclined by 45° with respect to the principal direcAons of the stress (represented by points R and S in the Mohr’s circle)

σ

τ


clockwise'

an0clockwise'

Z

X

30

P Q

τmax = RσCσ

σ ave

R

S

x

z

σ ave

σ ave

σ ave

σ ave

τmax

τmax

Stresses on an Arbitrary Inclined Face •  Having drawn the

Mohr’s circle, it is possible to evaluate the normal stress σm and the shear stress τmn in any generic face, inclined by αxm with respect to the horizontal axis

σ

τ


clockwise'

an0clockwise'

Z

X

31

P Q

Rσ

2α xq

τ mnCσ

σ ave

M

N

σm

2α xm

σm =σ ave + Rσ cos 2α xm − 2α xq( )τmn = Rσ sin 2α xm − 2α xq( )⎧⎨⎪

⎩⎪

•  Important: It is assumed here that angles α are posi2ve if an2clockwise

Key Learning Points 1.  Normal stresses and shear stresses acAng on a given

material element change their values depending on the inclinaAon of the elementary area being considered

2.  The Mohr’s circle allows evaluaAng –  The extreme values of the normal stress σp and σq –  The extreme value of the shear stress τmax

–  The inclinaAon of the faces where such values are seen –  The stresses σm and τmn for an arbitrary inclinaAon

32

complex stresses (2nd year)

Education