Kekuatan Material

MECHANICS - CASE STUDY SOLUTION

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Cylindrical Gas Storage Tank

 

A gas storage tank needs to be designed to hold pressurized gas at 10 MPa. The tank inside diameter is set at 34 cm due to tank stacking system on a rail car. For safety reasons, a factor of safety of 2.0 is required. The material is steel with a yield stress of 250 MPa. The thickness of the material needs to be determined.

To account for the stress interaction between the hoop and axial directions, the maximum distortion energy theory (von Mises' Yield Criterion) will be used to predict failure.

It is assumed that the end caps will not fail and only the cylinder middle section will be considered (end cap stresses are complex and not studied in this eBook).

          Hoop and Axial Stresses

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Functions for the hoop and axial stress can be determined for a cylindrical pressure vessel. These are

     σ_h = Pr/t = (10 MPa)(0.17 m)/t = 1.7/t

     σ_h = Pr/(2t) = (10 MPa)(0.17 m)/(2t) = 0.85/t

Both the stresses are functions of t.

          Failure Criteria

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Stress Element at Cylinder Section  

The maximum distortion energy criteria takes into consideration stresses in multiple directions. The equation is

     

or for this case,

     

The yield stress is given as 250 MPa. However, to account for a factor of safety of 2.0, the actual yield stress is reduced in half. Substituting into the failure equation gives,

     

     2.890 + 0.7225 - 1.445 = 15,625 t²

     t = 0.1178 m = 11.78 mm

MECHANICS - THEORY

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    Failure Thoeries

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Uniaxial Stress-Strain Curve with
Yield Stress and Ultimate Stress

 

If all structures where loaded in only one direction, it would be easy to predict failure. All that would be needed was a single uniaxial test to find the yield stress and ultimate stresslevels. If it is a brittle material, then the ultimate stress will determine failure. For ductile material, failure is assumed to be when the material starts to yield and permanently deform.
However, when a structure has multiple stresses at a given local (σ_x, σ_y and τ_xy for 2D as discussed in Stresses at a Point section), then the interaction between those stresses may effect the final failure. This section presents three basic failure theories that can be used for different types of materials to help predict failure when multiple stresses are applied.
For simplification, all theories are based on principal stresses (σ₁, σ₂) which can be determined from any (σ_x, σ_y and τ_xy) stress state. This removes the shear stress terms since the shear stress is zero at the principal directions. Using principal stresses does not change the results from the failure theories.

          Maximum Normal Stress Theory

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Maximum Normal Stress Criterion
(Blue Regoin is Safe)  

The simpliest theory ignores any interaction between the normal principal stresses, and assumes that failure occurs when either of the normal stresses exceed the ultimate stress. This is written as
This can be visualized on a plot by normalizing the two normal stress as σ₁/σ_ult and σ₂/σ_ult. This gives a square region where the stress state is safe. Outside the region is failure.
This failure criteria is really good for brittle materials and should not used for ductile material like steel, aluminum, and plastics.

          Maximum Shear Stress Theory
(Tresca's Yield Criterion)

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Maximum Shear Stress Criterion
or Tresca's Yield Criterion
(Blue Regoin is Safe)  

The maximum shear stress assumes failure occurs when the maximum shear stress exceeds the shear stress in a simple uniaxial test. In a unixial test, the principal stresses are σ₁ = σ_x (axial direction) and σ₂ = 0 (transverse to axial direction). Using the stress transformation equations, the maximum shear stress for this stress state is
     τ_max = (σ_x - σ_y)/2 = σ_x/2 = σ_yld/2
Thus, for any combination of loading for σ₁ and σ₂, the shear stress cannot exceed = σ_yld/2. This condition gives three seperate possible situations that need to checked,
This criteria is actually fairly accurate for ductile materials like steel, alumunim and plastics. The difficulty is that three conditions need to be checked.

          Maximum Distortion Energy Theory
(von Mises' Yield Criterion)

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Maximum Distortion Energy Criterion
or von Mises' Yield Criterion
(Blue Regoin is Safe)  

The third theory looks at the total energy at failure and compares that with the total energy in a unixial test at failure. Any elastic member under load acts like a spring and stores energy. This is commonly called distortational energy and can be calculated as
     
The G is the shear modulus. The distortion energy for a general stress state can be compared to distortion energy for a uniaxial test that fails at σ_x = σ₁ = σ_yld. This gives,
     
Thus the general criterion for failure would be
This criteria is especially useful since it is a single equation. It is also accurate for ductile materials. The shape of the region is an ellispe that is rotated 45 degrees.

      

MECHANICS - THEORY

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  Maximum and Minimum Normal Stress

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Rotating Stresses from x-y Coordinate
System to new x'-y' Coordinate System

 

Rotating the stress state of a stress element can give stresses for any angle. But usually, the maximum normal or shear stresses are the most important. Thus, this section will find the angle which will give the maximum (or minimum) normal stress.
Start with the basic stress transformation equation for the x or y direction.
     
To maximize (or minimize) the stress, the derivative of σ_x′ with respective to the rotation angle θ is equated to zero. This gives,
     dσ_x′ / dθ = 0 - (σ_x - σ_x) sin2θ_p + 2τ_xy cos2θ_p = 0
where subscript p represents the principal angle that produces the maximum or minimum. Rearranging gives,

     
Principal Stresses, σ₁ and σ₂,
at Principal Angle, θ_p 

The angle θ_p can be substituted back into the rotation stress equation to give the actual maximum and minimum stress values. These stresses are commonly referred to as σ₁ (maximum) and σ₂ (minimum),
For certain stress configurations, the absolute value of σ₂ (minimum) may actually be be larger than σ₁ (maximum).
For convenience, the principal stresses, σ₁ and σ₂, are generally written as,
where the +/- is the only difference between the two stress equations.
It is interesting to note that the shear stress, τ_x′y′ will go to zero when the stress element is rotated θ_p.

          Maximum Shear Stress

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Maximum Shear Stresses, τ_max,
at Angle, θ_τ-max

 

Like the normal stress, the shear stress will also have a maximum at a given angle, θ_τ-max. This angle can be determined by taking a derivative of the shear stress rotation equation with respect to the angle and set equate to zero.
      
When the angle is substituted back into the shear stress transformation equation, the shear stress maximum is
The minimum shear stress will be the same absolute value as the maximum, but in the opposite direction. The maximum shear stress can also be found from the principal stresses, σ₁ and σ₂, as
     

          Plotting Stresses vs Angle

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Stresses as a Function of Angle
 

The relationships between principal normal stresses and maximum shear stress can be better understood by examining a plot of the stresses as a function of the rotation angle.
Notice that there are multiple θ_p and θ_τ-max angles because of the periodical nature of the equations. However, they will give the same absolute values.
At the principal stress angle, θ_p, the shear stress will always be zero, as shown in the diagram. And the maximum shear stress will occur when the two principal normal stresses, σ₁ and σ₂, are equal.

     

MECHANICS - THEORY

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    Combined Stress (or Loads)

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Stresses at a point (Stress Element)
for a Cantilever Beam
(Element on Beam is Movable)  

In the previous sections, both the bending and shear stresses were presented for beams. Recall, the bending stress will cause a normal stress (either tension or compression, depending on vertical location) and the shear stress will cause a tearing stress. Both of the stresses can act at the same point and should be considered at the same time. Since only linear elastic materials are considered in this eBook, both stresses can be added together using the principle of superposition. This is shown at the left on a square element called a 'stress element'. This element is really just a point, but to see the stress direction, the point is shown as a square element.

     

  Stress Element

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Stresses at a point (Stress Element)
for Pressurized Pipe

 

The beam above is just one possible configuration for multiple stresses acting at a point. Another possibility is a pipe that is pulled, pressurized and twisted at the same time. These three loads on the pipe will cause tension normal stress in both directions (axial and circumferential) and cause a twisting or shear stress. All three loads and their associate stresses can be combined together to give a total stress state at any point.
The stress element for this example is shown at the left. This section will examine a stress element to better understand stresses at a point and how they can be analyzed.

          Sign Convention for Stress Element

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Sign Convention for Stress Element
Positive Directions  

The sign convention for stresses at a point is similar to other stresses. Normal tension stress in both the x and y direction are assumed positive. The shear stress is assumed positive as shown in the diagram at the left. Shear stress act on four sides of the stress element, causing a pinching or shear action. All shear stresses on all four sides are the same, thus
     τ_xy = τ_yx

        Stress Rotation

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Stress on an Inclined Plane

 

In the Normal Stress section, stress on an inclined plane was presented. It was noted that stresses are not vector quantities, and are not rotated by just using a single sin or cos function. For the normal and shear stress on an inclined plane, the rotated stresses were found to be

     
Rotated Stress Element  

Similarly, the stresses at a point (stress element) can also be rotated to give a new stress state at any particular angle. The rotation angle, θ, is assumed positive using the right hand rule (counter-clockwise in the x-y plane is positive). The new coordinate system is labeled as x' and y'. The new rotated stresses are shown in the diagram at the left. The shear stresses, τ_x'y' and τ_y'x, are still equal.

     

Plane Section through Stress Element

Surface Area on Stress Element Plane

 

The objective is to relate the new stress in x' and y' coordinate system to the original stresses in the x and y coordinate. To do this, the original stress element is sliced at an angle θ, as shown in the diagram at the left. The stresses on the cut plane must be in equilibrium with the stresses on the outside surfaces of the stress element. Remember, nothing is moving, so all stresses and their associated forces must obey static equilibrium equations, ΣF = 0 and ΣM = 0.
Before the stresses are actually summed, the area on each surface needs to be defined. The plane section at the angle θ is assumed to have a basic area of dA. The stress element is really just a point, so the area is infinitesimal, or just dA. The other two surfaces are based on dA. The bottom surface will be 'sinθ dA' and the left surface will be 'cosθ dA', which are shown in the diagram at the left.
Summing the forces in each direction gives
     ΣF_x = 0 = (σ_x´ dA) cosθ - (τ_x´y´ dA) sinθ
                           - σ_x (cosθ dA) - τ_xy (sinθ dA)
     ΣF_y = 0 = (σ_x´ dA) sinθ - (τ_x´y´ dA) cosθ
                           - σ_y (sinθ dA) - τ_xy (cosθ dA)
There are two unknowns, σ_x´ and τ_x´y´ and two equations, so they can be determined, giving
     σ_x´ = σ_x cos²θ + σ_y sin²θ + 2 τ_xy sinθ cosθ
     τ_x´y´ = - (σ_x - σ_y ) sinθ cosθ +τ_xy (cos²θ - sin²θ)
The y' direction can be developed in the same way, but the section plane is 90^o offset. The final equation is
     σ_y´ = σ_x sin²θ + σ_y cos²θ - 2 τ_xy sinθ cosθ

     
Rotated Stress Element  

Using double angle trigonometry identities, these three equations can be simplified to

          Stress Rotation Plot

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Stress Rotation Tool   To help visualize how the stress changes when the stress element is rotated, the simulation at the left plots σ_x´σ_y´ and τ _x´ y´ as a function of the angle. Notice, the period is 180^o. The initial stress state can be changed.