Solid Mechanics 1 (ENG1066) - Multi-axial Stress Analysis Lecture Notes PDF

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All of the lecture notes typed up for one of the three sections for the solid mechanics 1 module in semester 2 of first year...

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Stress Analysis Yield stress is the stress such that a material begins to undergo plastic deformation Proof stress is the stress such that a small plastic strain is attained, often 0.2%.

The modulus of rigidity, G, is the ratio of shear stress to shear strain, and is given by the slope of the initial linear part of the stress-strain plot:

And in an isotropic material:

G=

G=

τ γ

E 2(1+v )

Multi-axial stress and strain relationship

Normal stresses acting on each face

Shear stresses on each face

Tri-axial stresses Hooke’s law in tri-axial stress:

ε x=

σx v − ( σ +σ ) E E y z

where a similar expression can be found for strains in another direction

The above strain equation(s) can be rearranged to find the following stress equation(s):

Bulk modulus, K, is a measure of how incompressible a material is and is defined by:

σm E = ε v 3 (1−2 v )

And if a material is incompressible, v = 0.5 so K is infinity. If it is very compressible, it will have a Poisson’s ratio of 0, giving K = E/3 The rate of change of volume per unit vol, or volumetric strain, is: ε v =

(1−2 v ) [ σ x +σ y +σ z ] E

Plane (bi-axial) stresses Commonly encountered with thin plates of material, and stress in z-axis (through thickness) is assumed to be 0. Therefore, we neglect the out-of-plane shear stresses τ zx and τ yz . Therefore, Hooke’s law in a plane:

ε x=

1 ( σ −v σ y) where σ z=0 E x

But for the z direction, the strain is instead:

ε z=

−v (σ + σ ) E x y

Plane strain Where a long item is being stretched, and strain in z is neglected, which also implies any out of plane shear strains are zero:

Spherical pressure vessel

A state of plane stress, meaning radial stress is negligible is assumed. But this is also only valid for positive internal pressure and when points in the shell are away from the supports, openings, etc.

Planes of failure

The planes on which stresses are calculated and on which failure occurs are not always the same. Example, brittle material under the action of torsion. Where theta is the angle between the shear S N and the vertical axis. Also, noting that the area on the incline plane is A’ = A/cos(theta) P

Therefore, the normal stress and shear stress acting on the inclined plane are:

σθ=

N P⋅ cos θ P = ⋅cos2 θ=σ x cos 2 θ = ' ' A A A

τθ=

S −P⋅ sin θ −P = = ⋅sin θ cos θ=−σ x ⋅ sin θ cos θ A A' A'

We can plot a Mohr’s Circle of Stress, which is a type of plot where the above normal stress and shear stress are plotted as loci of points where theta varies from 0 to 180. Shear stress left is positive, and normal stress is always positive. An example of shear failure in a timber under compression produces a 45-degree angle of the failure plane. As shown in the Mohr Stress Circle, maximum shear stress is at 45 degrees so this is expected. However, the equations from above are only applicable if there is one stress, if there are applied normal stress in x and y: Equilibrium of forces normal to the inclined face:

σ θ =σ x cos2 θ+σ y sin2 θ +2 τ xy sin θ cos θ And again, but parallel to the inclined face:

τ θ =− ( σ x −σ y ) cos θ sin θ−τ xy (sin2 θ−cos 2 θ) And then substituting standard trigonometric identities will give us the final equations for the stresses acting at angle θ are:

These equations therefore enable us to calculate the stress in any direction at a point in the x-y plane if we know the state of about any other direction, e.g. x and y. These are the Stress Transformation Equations in Plane Stress [a full derivation is on SurreyLearn]. Construction of the Mohr Circle of Stress If the stresses on mutually perpendicular planes is known, a circle can be extrapolated. Simply put, plot points ( σ x , τ xy ¿ versus ( σ y ,−τ xy ¿ Principal Stress and Maximum Shear Stress From the Mohr Circle of Stress, we see that: 1. The maximum and minimum normal stresses occur on the planes where the shear stress is zero – these are known as the Principal Stresses, σ 1 and s 2 and act in the Principal Directions 2. The maximum shear stress, T max , occurs on a plane at 45 degrees to the Principal Planes

Geometry of the Mohr Stress Circle

Example Application – Torsion Torsion is a form of bending that occurs when the element is twisted, it is a twisting moment. A twisting moment, Mx, or a torque, T, is applied about the longitudinal axis of the shaft.

torque acting on cross-section

Critical Failure Plastic yielding – when a yielding ductile material undergoes a permanent strain. The stress which causes plastic yield is the limiting stress.

Fracture – in brittle materials, failure is often from crack growth (tensile) or crushing (compressive). The limiting stress is the maximum normal tensile or compressive stress When there is a general stress state acting at a point, the stress at which failure occurs is not obvious and so we must consider potential failure criteria by taking the multi axial stress into consideration. We will consider three criteria: 1) Maximum Principal Stress (Rankine) 2) Maximum Shear Stress (Tresca) 3) Maximum Shear Strain Energy (von Mises) 1. Maximum Principal Stress (or Rankine) Failure Criterion Failure occurs when the greatest principal stress reaches the failure stress (yield or fracture). Therefore, the Rankine Failure Criterion is simply:

|σ 1|=σ failure We can then plot a failure envelope for, say, plane stress. This would be a region where σ 1 on the x axis and σ 2 on the y axis such that they are both lower than the magnitude of the failure stress. This is only applicable for most brittle materials 2. Maximum Shear Stress (or Tresca) Failure Criterion Failure in ductile materials can often be seen to be related to a shearing action of shear stress. We can use the maximum shear stress, τ max , if we know the principal stresses ( σ 1 , σ 2 and σ 3 ¿ in a 3D environment. It is calculated as being numerically equal to the radius of the Mohr Stress Circle. For example:

|

τ max 12 =

|

σ 1−σ 2 2

So, if we say the above difference in stress is the yield stress, we can say:

σ yield=2 τ max The failure envelope for plane stress Tresca Criterion is:

3. Maximum Shear Strain Energy (von Mises) Failure Criterion Yielding will occur when the shear strain energy reaches a critical value. Using the Conservation of Energy Principle:

WORK DONE ,W =ELASTIC STRAIN ENERGY , U The work done, W, and shear strain energy stored, U, are given by the area the load-displacement curve.

Therefore, we can express the Strain Energy Density in Shear, u, as U per Unit Volume, u

u=

τ2 ΔU = ΔV 2G

And substituting the maximum shear stresses gives:

u=

1 2 2 2 σ 1−σ 2) +( σ 2 −σ 3 ) +( σ 1−σ 3) ] ( [ 8G Failure will occur when u reaches a critical value. Using a uniaxial tensile test, we find:

σ 1=σ yield

and

σ2 = σ3 = 0 Therefore,

2

ν=

σ 1 ( 2 σ 2yield) = yield 8G 4G

Finally, we can say the von Mises Failure Criterion can be expressed as: 2 2 2 z ( σ 1 −σ 2 ) +( σ 2−σ 3 ) + ( σ 1− σ 3) =2 σ yⅈeld

If we looked at the usual conditions of plane-stress, to avoid failure the von Mises Criterion states:

And the failure envelope for Plane Stress with the von Mises Criterion is:

Comparison of failure criteria If we plot a graph with all failure criterion and the yielding stress for different materials, we find:   

Tresca failure stress is lower than the von Mises stress, apart from particular points on the corners of the Tresca plot Brittle fracture of cast iron governed by Rankine criterion Ductile plastic yield failure for steel, copper indicated by von Mises + Tresca criterion

Beams under shear stress and loading 

Recall that bending stress distribution,

σ max =

My I NAx

, has it maximum value furthest from

neutral axis. How might a beam fail?  

Shear deflection – if we neglect bending deflections and just look at that due to shear force, we can see potential tensile or compressive shearing failure that will result. The shear deflection can either cause buckling from compression, or cracks from tension.

If a beam is split into two, with no shear connection on the separate beams, the neutral axis will change:

B D3 I NAx= 12

will go to

I NAx=2

( ) D 2 12

3

[ ] B

=

B D3 48

A bi-metallic strip has two metals with different thermal properties bonded together, an example of shear connection. A change in temperature causes the strip to bend, useful for thermal switches.

If we consider the distribution of shear stresses in a rectangular beam section, and then the equilibrium of the horizontal forces acting on a beam section, abfe, we can say produce the following:

This equation is finding a shear stress acting on a plane at distance y1 from the NAX, or a shear stress distribution. Shear stress has it’s maximum value on the neutral axis. Maximum shear stress in beams calculation For a rectangular cross section:

τ m ax=

3 × Smax 1 ⋅5 S m ax = 2 × ( A rea of section ) BD

I-section beam with thin web section:

τ m ax ≈

S Smax = m ax Area of web D ×t

Strain measurement As it is sometimes not possible to we must instead measure

measure stresses directly, deformations.

Instrumentation for monitoring

strain

1. Electrical Resistance – thin filament of wire

Strain Gauges (ERS gauges) within a polymer based is

glued on. Wire strains with the surfaces, and a Wheatstone Bridge circuit is used to monitor change in resistance. R1 is the active gauge, R2 being the dummy. When balanced, V BD will be zero, but if R1 changes due to strain, current flows across BS, which is calibrated for strain.

2. Demec Gauge (extensometer) – these have a known length which is set with a template and by adhering studs into which the gauge fits. Subsequent readings if different from the first indicate strain change, ΔL/L. Useful for small strains. 3. Vibrating Wire Gauges – A thin wire with natural frequency proportional to wire’s tension. A change in frequency of oscillation will be a result of change in tensile force and therefore change in distance between the two ends, which is proportional to the strain. Strain Transformation Equations With Hooke’s law, we can express the direct normal and shear strains, acting on an inclined plane in terms of normal and shear stresses.

ε and γ respectively,

γ 1 1 ε θ = ( ε x + ε y )+ ( ε x −ε y) cos 2 θ+ xy sin 2 θ 2 2 2 γ θ=− ( ε x −ε y ) sin 2 θ+γ xy cos 2 θ

We can also construct a Mohr Circle for Strain, by replacing σ x

γ xy : 2

with

ε x and T xy with

And similar to stresses, the Principle Strains, ε 1 and can be found by looking at the geometry of the circle:

ε 2 , and Maximum Shear Strain, γ max ,

Strain Gauge Rosette Analysis It is very difficult to measure shear strain directly in practice. Therefore, if the principal directions are not known, it is normal to use a stain gauge rosette to monitor the normal strains in 3 or more independent directions. Common configurations include rectangular 45 degrees or delta 45 degrees. Considering the rectangular rosette: Using the plane strain transformation equations, we can use the normal strain data for the 3 directions to calculate the shear strain, γ0,90.

And considering a 60-degree delta rosette: Use the plane strain transformation equations.

These simultaneous equations can be solved. Substituting for the principle strains and resolving yields:...