Disc

• Plane structure

• Loads only in disc plane!

• Thickness h << b,l

• Coordinate system located in middle plane

• Plane state of stress (Transformation to plane state of strain possible)

• Stresses are constant over the disc thickness

• Geometric linerarity

• Material linearity (Hookes law applies)

Integration of the respective stress over the thickness, due to assumption of constant stress, just stress * h:

Due to inplane loads only u and v in x and respectivly y

• epsilon-xx

• epsilon-yy

• gamma-xy

• (epsilon-zz, occurs but not relevant for calculation, dependent)

• (gamma-xz = 0 = gamma-yz)

The Disc Stiffness A devolops arising force flows from (easy if uniaxial) elongation

8 unknown quantities

• 3 stresses: sigma-xx; -yy; tau-xy (3 according force flows)

• 3 strains: epsilon-xx; -yy; gamma-xy

• 2 displacements: u; v

force flows, strains and displacements are always refered to the middle plane. (Stresses are constant over the thickness, so no variation from middle plane stress expected)

8 equations are necessary

• 2 equilibrium conditions

• 3 kinematic equations

• 3 constitutive equations

• 1 compatibility equation

The main issue is, that the problem is statically indeterminate. There are only two equilibrium conditions to develop three stresses.

Workaround: utilize the “Force method” (refers to the equilibrium of forces) to work out another equilibrium from compatibility, the so called disc equation

Starting point are the two equilibrium conditions and the compatibility

Substitution of epsilon-xx, -yy and gamma-xy in the compatibility equation via Hookes law

Leads to a compatibility in stresses:

To match, 2nd partial derrivation of equilibrium condition yields

—> Both equilibrium conditions equate to 0, adding them is permissable and leads to:

Adding compatibility in stresses and combined equilibrium conditions leads to the disc equation

Airy´s stress function - F - is introduced.

Expanded form:

Condensed form:

In general any arbitrary function of F can be used that fulfills the boundary conditions. However it must fullfill

In practice polynomial series expansions are used. However for most problems, Formulation for Airy´s stress function are given in literature

1) Find a Ansatz for the behaviour of Airy´s stress function —> utilize known symmetries and patterns, in this case

2) Perform disc equation

3) Choose appropriate Ansatz for the arising differential equation (in this case 4th order)

4) Use Boundary conitions to find Fn-bar —> F given.

Arising stresses can now be determined from Airy´s stress function.

—> Key TakeAway: Differentiate Airy´s stress function into known behaviour (e.g. Periodicity) and unkown behaviour (decay)

In relativly compact I-beams (large width compared to length) The flanges will not be fully utilized. Due to load input from web to flange in the middle, stresses will decay towards the outside. For easier calculation, an effective width is introduced.

(notice limit of ~0,2 after that, bad utilization of the flanges)

Analysis of

• Holes (Screws/ Rivets)

• Point load introductions

In a rotationaly symmetric situation:

• Derivations w.r.t to phi vanish (just dependend on r)

• tau-r-phi = 0

—> decay by 1/r^2

Analysis in polar coordinates advantegous because cartesian stresses are fully populated

Resultant stresses:

sigma-phi-phi and tau-r-phi equate to 0

due to the dependency 1/r —> Singularity at load introduction

Analysis in polar coordinates advantegous because cartesian stresses are fully populated

Resultant stresses:

sigma-phi-phi and tau-r-phi equate to 0

due to the dependency 1/r —> Singularity at load introduction

The given load introduction can be understood as a superposition of the individual cases of a Vertical and a Horizontal load introduction. Airy´s stress function represents this and is also a superposition of the base cases (Airy´s stress function just added)

• Radial stresses vanish at the inner and outer edge (sigma-rr = 0 @ r=Ri and r=Ra)

• The integral of the circumferencial stresses times their respective lever arm must result in the applied moment

(rotationally symmetric —> tau-r-phi = 0)

Stresses depend on

• Inner Radius

• Outer Radius

• coefficient of outer and inner radius

• Applied Moment

• Radial position in the disc

Stress concentration factor: 3

Tensile stresses at top and bottom of the hole

Compressive stresses at the sides towards the load

The Ansatz choosen is again superposition. For the case of additional tensile loads, calculated stresses can just be added with a shift of 90°.

The result for additional tensile loads is a stress concentration factor of 2 (remember compressive stresses vertical to load introduction - further remember that the hole is “supported” by add. tensile loads)

The result for additional compressive loads is a stress concentration factor of 4 (compressive stresses from earlier example turn their presign —> Add. tensile loads - further remember that the hole is additionally squashed by add. compressive loads)

Change perspective to the principal axis system —> Remember load case with tensile and compressive loads orthogonal to each other —> Stress concentration factor of 4

Isotropy - 1

Low shear number, fast decay (high longitudinal stiffness, low shear stiffness/ transverse stiffness)

High shear number, slow decay

Remembering the shear number

The stiffness in circumferencial direction must be increased. This is the case in e.g. laminates that utilize fibres in the circumferencial direction

The higher the shear number, the sharper the rise of the radial stresses and with a high shear number the max. stresses are reduced and more uniformly distributed.

For laminates this is advantegeous because radial stresses will generally lead to delamination —> Risk of delamination is reduced.

A higher shear number leads to:

• higher maximum stresses

• sharper rise of the stresses

• a worse utilization of the laminate middle plane