User:Stephen Mckelvey/sandbox

When a fastener is tightened, a tension preload is develops in the bolt, while an equal compressive preload forms in the clamped parts. This system can be modeled as a spring-like assembly, where the clamped parts experience compressive strain, and the bolt tensile strain. When an external tensile load is applied, it reduces the compressive strain in the clamped parts and increases the tensile strain in the bolt. The load carried by the bolt and the clamped parts is in proportion to their stiffness because they both experience the same induced strain. As a result, the external load is shared across the joint rather than being solely carried by the bolt. In a well-designed joint, about 10-20% of the applied tensile load is carried by the bolt with the greater part transferred through the clamped parts because they are much stiffer than it. This reduction in the proportion of load transferred to the bolt is important in applications with cyclic loading, as bolts have low fatigue strength due to the stress concentration in their threads.

The accompanying graph and table illustrate how the relative stiffness of the clamped parts and the bolt affects the portion of applied load carried by it. For example, when the stiffness of the clamped parts equals that of the bolt (the blue curve), an external load in the range from minus to plus twice the preload results in only 50% of the applied load being transferred to the bolt, as the total load in the bolt only varies by twice the preload. If the tensile applied load exceeds twice the preload, the clamped parts separate, and the bolt carries the entire load. Conversely, if the compressive load lower than twice the preload, separation at the bolt head occurs, and the force in the bolt is zero. The curve representing a clamped parts-to-bolt stiffness ratio of 0.01 shows that when the relative stiffness of the clamped parts is very low, almost all of the load is transferred to the bolt, down to the point where a compressive load equals the preload, and separation at the bolt head occurs, reducing the force in the bolt to zero.

When a fastener is torqued, a tension preload develops in the bolt and an equal compressive preload develops in the parts being fastened. This can be modeled as a spring-like assembly that has some assumed distribution of compressive strain in the clamped joint components. When an external tension load is applied, it relieves the compressive strains induced by the preload in the clamped components, hence the preload acting on the compressed joint components provides the external tension load with a path (through the joint) other than through the bolt. In a well designed joint, perhaps 80-90% of the externally applied tension load will pass through the joint and the remainder through the bolt.

When the fastened parts are less stiff than the fastener (those that use soft, compressed gaskets for example), this model breaks down and the fastener is subjected to a tension load that is the sum of the tension preload and the external tension load.

Effect of Clamped Parts to Bolt Stiffness Ratio

Clamped Parts Stiffness Bolt Stiffness	Portion of Applied Load Transferred to Bolt	Range of Joint Integrity Applied Load Bolt Preload
		Separation at Bolt Head	Separation of Clamped Parts
0.01	99 %	-1.0	100
0.5	66 %	-1.5	3.0
1.0	50 %	-2.0	2.0
3.0	25 %	-4.0	1.3

Deviation of engineering stress from true stress

When a uniaxial compressive load is applied to an object it will become shorter and spread laterally so its original cross sectional area (${\textstyle A_{o}}$) increases to the loaded area (${\textstyle A}$) . Thus the true stress (${\acute {\sigma }}=F/A$) deviates from engineering stress ($\sigma _{e}=F/A_{o}$). Tests that measure the engineering stress at the point of failure in a material are often sufficient for many routine applications, such as quality control in concrete production. However, determining the true stress in materials under compressive loads is important for research focused on the properties on new materials and their processing.

The geometry of test specimens and friction can significantly influence the results of compressive stress tests. Friction at the contact points between the testing machine and the specimen can restrict the lateral expansion at its ends (also known as 'barreling') leading to non-uniform stress distribution. This is discussed in section on contact with friction.

Frictionless contact

With a compressive load on a test specimen it will become shorter and spread laterally so its cross sectional area increases and the true compressive stress is$${\acute {\sigma }}=F/A$$and the engineering stress is$${\sigma _{e}}=F/A_{o}$$The cross sectional area (${\textstyle A}$) and consequently the stress ( ${\textstyle {\acute {\sigma }}}$) are uniform along the length of the specimen because there are no external lateral constraints. This condition represents an ideal test condition. For all practical purposes the volume of a high bulk modulus material (e.g. solid metals) is not changed by uniaxial compression . So$$Al=A_{o}l_{o}$$Using the strain equation from above$$A=A_{o}/(1+\epsilon _{e})$$and$${\acute {\sigma }}=\sigma _{e}(1+\epsilon _{e})$$Note that compressive strain is negative, so the true stress (${\acute {\sigma }}$ ) is less than the engineering stress (${\textstyle \sigma _{e}}$). The true strain (${\acute {\epsilon }}$) can be used in these formulas instead of engineering strain (${\textstyle \epsilon _{e}}$) when the deformation is large.

Contact with friction

As the load is applied, friction at the interface between the specimen and the test machine restricts the lateral expansion at its ends. This has two effects:

• It can cause non-uniform stress distribution across the specimen, with higher stress at the centre and lower stress at the edges, which affects the accuracy of the result.

• It causes a barreling effect (bulging at the centre) in ductile materials. This changes the specimen’s geometry and affects its load-bearing capacity, leading to a higher apparent compressive strength.

Various methods can be used to reduce the friction according to the application:

• Applying a suitable lubricant, such as MoS2, oil or grease; however, care must be taken not to affect the material properties with the lubricant used.

• Use of PTFE or other low-friction sheets between the test machine and specimen.
• A spherical or self-aligning test fixture, which can minimize friction by applying the load more evenly across the specimen's surface.

Three methods can be used to compensate for the effects of friction on the test result:

1. Correction formulas
2. Geometric extrapolation
3. Finite element analysis

Correction formulas

Round test specimens made from ductile materials with a high bulk modulus, such as metals, tend to form a barrel shape under axial compressive loading due to frictional contact at the ends. For this case the equivalent true compressive stress for this condition can be calculated using$${\acute {\sigma }}=C\sigma _{a}$$where

$C={(1-2R/d_{2})\ln(1-d_{2})/(2R))}^{-1}$

$R=(l^{2}+(d_{2}-d_{1})^{2})/(4(d_{2}-d_{1}))$

$\sigma _{a}=4F/(\pi d_{2}^{2})$

$l$ is the loaded length of the test specimen,

$d_{1}$is the loaded diameter of the test specimen at its ends, and

$d_{2}$is the maximum loaded diameter of the test specimen.

Note that if there is frictionless contact between the ends of the specimen and the test machine, the bulge radius becomes infinite (${\textstyle R=\infty }$) and ${\textstyle C=1}$ . In this case, the formulas yield the same result as ${\textstyle {\acute {\sigma }}=\sigma _{e}(1+\epsilon _{e})}$ because ${\textstyle \sigma _{a}}$ changes according to the ratio $(d_{o}/d_{2})^{2}$.

The parameters (${\textstyle F,d_{1},d_{2},l}$) obtained from a test result can be used with these formulas to calculate the equivalent true stress ${\textstyle {\acute {\sigma }}}$ at failure.

Specimen shape effect

The graph of specimen shape effect shows how the ratio of true stress to engineering stress (σ´/σ_e) varies with the aspect ratio of the test specimen (${\textstyle d_{o}/l_{o}}$). The curves were calculated using the formulas provided above, based on the specific values presented in the table for specimen shape effect calculations. For the curves where end restraint is applied to the specimens, they are assumed to be fully laterally restrained, meaning that the coefficient of friction at the contact points between the specimen and the testing machine is greater than or equal to one (μ ⩾ 1). As shown in the graph, as the relative length of the specimen increases (${\textstyle d_{o}/l_{o}\rightarrow 0}$), the ratio of true to engineering stress (${\acute {\sigma }}/\sigma _{e}$) approaches the value corresponding to frictionless contact between the specimen and the machine, which is the ideal test condition.

Specimen shape effect calculations

	Frictionless	Laterally Constrained
Constant volume	$\pi l_{o}d_{o}^{2}/4=\pi l(d_{2}^{2}+d_{1}^{2})/12$
Equal diameters	$d_{1}=d_{2}$	$d_{o}=d_{1}$
Solve for $d_{2}$	$\pi l_{o}d_{o}^{2}/4=\pi l(d_{2}^{2}+d_{2}^{2})/12$	$\pi l_{o}d_{o}^{2}/4=\pi l(d_{2}^{2}+d_{o}^{2})/12$
	$d_{2}=d_{o}{\sqrt {l_{o}/l}}$	$d_{2}=3d_{o}{\sqrt {(3l_{o}/l-1)/18}}$
Equivalent stress ratio	${\acute {\sigma }}/\sigma _{a}=1$	${\acute {\sigma }}/\sigma _{a}=C$
Engineering stress	$\sigma _{e}=4F/\pi d_{o}^{2}$
Average stress	$\sigma _{a}=4F/\pi d_{2}^{2}$
Average stress ratio	$\sigma _{a}/\sigma _{e}=(d_{o}/d_{2})^{2}$
True strain	${\acute {\epsilon }}=\ln(l/l_{o})$

Geometric extrapolation

As demonstrated in the section on correction formulas, as the length of test specimens increases and their aspect ratio approaches zero ($d_{o}/l_{o}\longrightarrow 0$), the compressive stress (σ) approaches the true value (σ′). However, conducting tests with excessively long specimens is impractical, as they would fail by buckling before reaching the material's true compressive strength. To overcome this, a series of tests can be conducted using specimens with varying aspect ratios, and the true compressive strength can then be determined through extrapolation.

Test Methods

Laboratory

Common Applications:

   Concrete Testing: To ensure the material meets required strength specifications for construction.

   Metal and Alloy Testing: For quality control in manufacturing, especially for components subject to high compressive forces.

   Ceramics and Plastics: To test the durability of materials used in packaging, automotive, and aerospace industries.

This type of testing machine is crucial in quality control, research, and development, ensuring materials meet safety and performance standards.

There are a number of standards with industry specific recommendations for specimen preparation, conduct of the tests and analysis of the results. Commonly used standards are:

• ASTM E9-89A, Standard Test Methods of Compression Testing of Metallic Materials at Room Temperature
• ASTM D575-91 Standard Test Methods for Rubber Properties in Compression
• ASTM D3410 Compression of Composites

Most commonly cylindrical specimens are used: either prepared specifically for the test or cut from an existing material or structure.

In-situ

With frictionless contact between the test machine and the ends of the test specimen, the cross-sectional area of the specimen remains uniform during compression. For uniaxially loaded homogeneous materials—those without internal voids, such as foams or granular materials—it can be demonstrated that:

$A_{l}=A_{o}/(1+h_{l}/h_{o})$

where:

$A_{l}$ is the loaded cross sectional area,

$A_{o}$ is the unloaded cross sectional area,

$h_{l}$ is the loaded height, and

$h_{o}$ is the unloaded height.

The engineering stress is

$\sigma _{e}=F/A_{o}$

and the true stress is

$\sigma _{t}=F/A_{l}$

With frictional contact between the test machine and the ends of the test specimen it becomes barrel shaped during compression and the volume of the

The axial strain is negative for an object under compression, meaning the object becomes shorter. As a result, the loaded area increases compared to the original area, causing the true stress to be lower than the engineering stress.

The volume of an unloaded specimen

$V_{o}=A_{o}\times l_{o}$

The volume of the loaded specimen

$V_{l}=A_{l}\times l_{l}$

The loaded length is the unloaded length minus the deflection

$l_{l}=l_{o}-\delta l$

and the strain in the specimen

$\epsilon =\delta l/l_{o}$Combining the three equations above gives

$V_{l}=A_{l}\times l_{o}\times (1+\epsilon )$For all practical purposes with uni axially loaded homogeneous materials (i.e those those that do not have voids such as foams and granular materials) the unloaded and loaded volumes are the same.

$V_{l}=V_{o}$