Axial Deformation

Axial deformation has been experimentally introduced in Lecture 2, when the tension test of a structural steel specimen was described. The case in point of this lecture is a systematic description of all aspects pertinent to axial deformation.

The three aspects discussed during Lecture 2, equilibrium equations, strain-displacement equations and the constitutive relation, are applied to the case of axial deformation. A local coordinate system , having axis coincident with the longitudinal axis of the member, is employed.

1.1 Strain-Displacement Equation

A direct consequence of part (b) of the definition 1 is the fact that axial deformation depends only on the variable, which represents the position of a particular cross-section on the longitudinal axis of the member.

The physical phenomenon described by definition 1 is illustrated in Figure 1, where the undeformed and deformed conditions of the linear member are presented.

Adjacent cross-sectionsandoriginally located at distance from each other, as shown in Figure 1.a, are found after the deformation to be located at distance. The change in the position of the two cross-sections are described by their displacements and, respectively.

Figure 1 Geometrical Aspects of the Axial Deformation

(a)    Undeformed Member and (b) Deformed Member

The extensional strain is expressed as:

(1)

Equation (1) is called the strain-displacement equation. The cross-section distribution of the elongation strain is shown in Figure 2. The elongation strain is a function only of the cross-section position described by the variable.

Figure 2 Extensional Strain Distribution

The rest of the generalized strain tensor components are:

(2)

(3)

(4)

From equations (2) through (4) is evident that a transversal reduction of the cross-section takes place concomitant with the axial deformation.

The displacement pertinent to a particular cross-section is obtained by integration from the equation (1):

(5)

where is the displacement at the beginning at the integration interval.

Consequently, the total elongation of the member is calculated from equation (5) as:

(6)

where is the total length of the bar.

Constitutive Equation

The constitutive equation reflects, as was describe in Lecture 2, the relation between the stress and the strain. If the linear elastic material behavior is considered, the relation between the normal stress and extension strain for the case of the axial deformation is written:

(7)

Equation (7) represents the application of Hooks Law for the case of axial deformation. The material constant, the modulus of elasticity, has a value unique to each specific material and is obtained from tensile tests. The cross-section of the bar is a small surface and the variation of the modulus of elasticity is negligible on this surface. In this case the constitutive equation (7) is expressed as:

(8)

Equation (8) implies that the normal stress varies only along the length of the member, but has a constant value on the entire cross-section. The representation of the normal stress is shown in Figure 3.

The rest of the stress tensor components are zero:

(9)

(10)

Figure 3 Cross-Section Normal Stress Distribution

Cross-Section Stress Resultants

Considering the stress distribution represented by equation (8) through (10) the cross-section stress resultants are obtained as:

(11)

(12)

(13)

The relation between the normal stress and the cross-section resultants, and is derived using the notation shown in Figure 4.

Figure 4 Normal Stress and Stress Resultants

(a)    Stress Resultants and (b) Normal Stress

If the axes and of the coordinate system intersect such that the axis passes through the cross-section centroid, the static moments and are zero and the axial force remains the only non-zero stress resultant. Then, from equation (11) the normal stressis calculated as:

(14)

Consequently, it is to be concluded that a beam made from a linear elastic material undergoes an axial deformation if the axial force passes through the cross-section centroid.

Equilibrium Equation

The equilibrium equation pertinent to the case of axial deformation was derived in Section 4.3 of Lecture 4. The detailed derivation is not repeated and only the final result expressed as the differential relation (4.7) is employed. Considering the exterior loading and stress resultants shown in Figure 5, acting on an infinitesimal volume of length separated from the body of the beam, the following differential equation is written:

(13)

where is the distributed loading parallel to the beam longitudinal axis.

Figure 5 Infinitesimal Volume Equilibrium

If the axial stress resultant is orientated as shown in Figure 5 and tends to lengthen the segment the internal force is called tension. The corresponding normal stress is called tension stress. In contrast, if the stress resultant force is orientated to shorten or compress the segment the corresponding internal force and normal stress are called compression and compression stress, respectively.

Integrating equation (15) the stress resultant force is calculated:

(16)

where is the value of the axial force at the origin of the integration interval.

1.5 Thermal Effects on Axial Deformation

The equations obtained in the previous sections are derived considering only the exterior load action and neglecting the change in temperature. In this section the effects of the thermal change are introduced.

During Lecture 2, the thermal strain along the axis was introduced as:

(17)

where is the thermal expansion coefficient and is the change in the member temperature.

The total elongation strain is the sum of the elongation strain induced by the exterior load action and thermal effects and is expressed as:

(18)

By substitution of equation (14) into (18) the elongation strain is obtained as:

(19)

Then, using equation (19) in equation (6) the total elongation of the member is written as:

(20)

A special case of axial deformation frequently encountered in structural engineering is the case of uniform-axial deformation shown in Figure 6. Systems composed of many members subjected to uniform-axial deformation are also commonly used in structural practice. A typical example is the plane truss, where each individual member is subjected to uniform-axial deformation under the action of to tension or compression forces.

The formulation related with the definition of the uniform-axial member and its application in the investigation of the statically determinate and indeterminate structures is presented in this section.

2.1 Members Subjected to Uniform-Axial Deformation

Definition 2 is mathematically transcribed as:

(21)

(22)

(23)

.

Figure 6 Member Exhibiting Uniform Axial-Deformation

Note: Equation (23) implies the absence of the distributed load in equation (15)

Rewriting equations (6), (8) and (14) obtained in the previous section for the case in point of a member with uniform axial-deformation the following equations are obtained:

(24)

(25)

(26)

In the computer applications equation (26) takes the following form:

(27)

where is called the axial stiffness coefficient.

The axial stiffness coefficient represents the force applied to the member ends when the elongation is equal with the unit length. The unit is [F/L]. The reciprocal value of the axial stiffness is called the axial flexibility coefficient. The flexibility coefficient is calculated:

(28)

Using the flexibility coefficient expression equation (28) is cast in a new format:

(29)

If the change in temperature is also constant along the entire length of the member the formula (20) are amended as follows:

(30)

Using the flexibility coefficient the elongation is calculated as:

(31)

Rewriting the equation (31) the force is obtained:

(32)

Statically Determinate Structure

An example of a statically determinate structure is shown in Figure 7. A rigid, weightless beam AC is supported at end A by a column and at end C by a vertical rod CD. The rod is attached at point D to a leveling jack, a component which permits a limited vertical movement and has the mission to keep the beam AC leveled. The beam AC is loaded with a vertical force P located at distance from the left end A. To calculate the axial forces acting on the column and rod the system is decomposed into three components: the beam, the column and the rod.

Figure 7 Statically Determinate Structure

The free-body diagram of the beam is depicted in Figure 8 and is used to calculate the reaction forces and. The connection between the column and the beam requires in general, two reaction forces. In the absence of any horizontal force the horizontal component is null and is not shown

Figure 8 Free-Body Diagram

The reaction forces can be calculated using the following equilibrium equations:

(33)

(34)

Solving equations (33) and (34) the reaction forces are obtained:

Compression in column (35)

Tension in rod (36)

Note: The column and the rod are loaded with forces having opposite directions than the reaction forces calculated in equations (35) and (36). This became obvious if one attempt to draw the free-body diagram for the rod and column.

From equations (35) and (36) it is evident that the column and the rod are loaded with compression force and tension force , respectively. These two elements are characterized, using definition 2, as members with uniform axial-deformation.

Using the formulae (24) through (26) the following related column values are calculated:

(37)

(38)

(39)

Note: The force is compression and the column is shortened by the loading action.

Related quantities for the rod CD are similarly calculated:

(40)

(41)

(42)

Note: The force is tension and the rod is elongated by the loading action.

For the beam to stay in perfect horizontal balance the displacement at point A and C should be equal and manifest in the same direction:

(41)

The expressions of the displacements at points A and C are calculated using the formula (5).The displacement at point A is equal with the column elongation, because the initial displacement at ground connecting point is considered zero:

(42)

The displacement at point C is calculated as:

(43)

where is the displacement allowed by the leveling jack.

The necessary displacement allowed by the leveling device is calculated from equation (41):

(44)

Note: The displacement has to be positive for the device to work properly (the device works only in tension). This means that the deformation of the column has to be larger than the rod deformation.

The numerical application for this example is presented in section 6.1.

Statically Indeterminate Structure

A typical example of a statically indeterminate structure is shown in Figure 9. This type of structures requires a more involved methodology for in order to calculate the stress and strain distributions.

Figure 9 Statically Indeterminate Structure

The system shown in Figure 9 is composed of a rigid member AD, pinned into the wall at point A, and two unequal linear elastic rods, BE and CF. The rods, BE and CF, are attached to the ceiling at points E and F, respectively. The system is loaded with a concentrated vertical force at point D. The free-body diagram used to write the equilibrium equations is shown in Figure 10.

Figure 10 Free-Body Diagram

The following equilibrium equations are written:

(47)

(48)

(49)

The equilibrium equations (48) and (49) contain three unknown reaction forces, and . In order to solve these unknown quantities one additional equation is necessary. This equation is obtained from the deformation compatibility condition schematically described in Figure 11. Because the beam AD is rigid, purely geometric relations between the rod elongations, and, and the rotation angle are written as:

(50)

(51)

Figure 11 Deformation Notation

Using equation (29) and the relations (50) and (51) the forces in the rods are expressed as:

(52)

(53)

The stiffness coefficients, and, are calculated from the geometrical and material properties characteristics of the rods:

(54)

(55)

Substituting equations (52) and (53) into the equilibrium equations (48) and (49), and are eliminated from the system leaving only two unknowns, and:

(56)

(57)

Solving the algebraic system, the two unknowns are found as:

(58)

where is the rotation angle corresponding to

(59)

Introducing the rotation angle into equations (52) and (53) the rod forces are calculated:

(60)

(61)

where and are the forces in the rods in

The rod stresses are also calculated:

(62)

(63)

The rod elongations are obtained using equations (50) and (51) as:

(64)

(65)

If the allowable vertical force is required, the stress in the rods must be compared against the allowable stress:

(66)

(67)

At limit, the relations (66) and (67) are rewritten as:

(68)

(69)

Introducing the rod forces, equations (60) and (61), into equations (68) and (69) the allowable vertical force admitted by the system is calculated as:

(70)

The numerical application for this example is presented in section 6.2.

The definition of a member with uniform-axial deformation is specified in Section 3. If any one of the assumptions contained in definition 2 is violated the axial deformation is called nonuniform-axial deformation. The most common cases of nonuniform-axial deformation are treated in the following subsections.

Non-homogeneous Cross-Section Members

The theory developed in the previous sections assumed that the cross-section is made from a homogeneous material described by its modulus of elasticity. In structural engineering practice it is not uncommon to have a case in which a member constructed from two different materials bounded together at their interface is forced to undergo an axial deformation. The members made from different materials but behaving together as a single member are called composite sections.

For the purpose of analysis it is assumed that the member is made from two materials, each being characterized by a specific modulus of elasticity ( and) and area ( and). The strain distribution is, as before, assumed to be a function of only the variable, the position of the particular cross-section of interest:

(71)

Considering that both materials are homogeneous linear elastic materials the Hooks law may be written for each material as:

(72)

(73)

The total axial force , as expressed in equations (11), is divided into two axial forces, each acting at the centroid of the corresponding bar cross-section.

(74)

where

(75)

(76)

Substituting the axial stresses expressed in equations (72) and (73) into equations (75) and (76) the equilibrium equation (74) is written as:

(77)

Consequently, the elongation strain is obtained:

(78)

Accordingly, the normal stresses are calculated employing the equations (72) and (73) as:

(79)

(80)

In general, the stress resultant moments, as expressed in equations (12) and (13), do not vanish and additional restrictions regarding the geometrical characteristics of the cross-section must be imposed. Nullification of the stress resultant moments is obtained by using symmetric cross-section about one or both centroidal axes.

The frequently encountered practical case when the areas andare constants along the entire length and the member is subjected to a constant force is considered below. It is additionally assumed that the cross-section is symmetrical only about the vertical axis . Then, the normal stresses developed in each material zone are calculated, following the equations (79) and (80), as:

(81)

(82)

The assumption regarding the symmetrical aspect of the cross-section about the vertical axis nullify the stress resultant moment when the bar local coordinate system is centroidal. The second stress resultant moment can be made zero by manipulating the position of the application of the force . It was previously argued that the individual cross-section resultants and must be located at the centroids of the cross-sections and. This raises the question about the location of the applied force in order that only axial forces are induced in each of the individual bars composing the cross-section. This situation is solved by replacing the axial forces and with a resultant force and a null moment written about a horizontal axis. The stress resultant moment about the vertical axis has a zero value due to the imposed symmetry of the cross-section about that axis. The moment equation, written about an horizontal axis passing through the application point of the force , is:

(83)

where is the distance between the centroids of cross-sections and, while is the distance between the application point of the force and the centroid of the area .

The distance is then calculated:

(84)

Note: In general the application point of the force does not correspond with the centroid of the composite cross-section.

3.2 Non-homogeneous Cross-Section Members Subjected to Thermal Changes

Consider the composite section described in Section 3.1 subjected to change in temperature. For generality, in this discussion, let be assumed that each individual bar undergoes a different change in temperature and, respectively. The notation employed in the Section 3.1 is maintained. In the absence of the exterior forces, the equation of equilibrium is:

(85)

The axial forces pertinent to each one of the bars is constant along their respective lengths and thus, the equation (85) becomes:

(86)

The equilibrium equation contains two unknown cross-sectional forces,and , and consequently the system is statically indeterminate. An additional equation is necessary to obtain these forces. This equation is derived from the condition of equality of the elongations for the two individual bars imposed by the existence of the rigid members attached at their ends. This equation is written as:

(87)

The elongations may be express using the formulation of equation (31):

(88)

(89)

Substituting equations (88) and (89) into equation (87) the second necessary equation is obtained:

(90)

Solving the algebraic equation system (86) and (90) the cross-section forces are calculated:

(91)

The axial stress in the bars is:

(92)

(93)

3.3 Heated Member with a Linear Temperature Variation

The slender beam shown in Figure 12.a is heated by a heating coil capable of producing a linearly varying temperature as shown in Figure 12.b. The beam has a constant cross-section area and is made from a linear elastic material. The beam is attached to rigid supports at ends A and B.

Figure 12 Heated Uniform Member

(a) Geometry and (b) Temperature Variation

In the absence of any applied forces between A and B, the equilibrium equation is written as:

(94)

The system is statically indeterminate containing two unknown reaction forces and. An additional equation is necessary. This equation is obtained by observing that the total elongation of the beam is null:

(95)

Following the equations (20) and (31) the total elongation is calculated:

(96)

where

Accordingly, with the temperature variation shown in Figure 12.b the thermal variation in a particular cross-section is:

(97)

The integral contained in the equation (96) is calculated using the expression (97):

(98)

Substituting (98) into (96) the total elongation is expressed as:

(99)

Imposing the condition (95) the force is found:

(100)

Using the equation (94) results that:

(101)

The normal stress is then calculated as:

(102)

4.1 Normal Stress in the Vicinity of the Load Application

A vertical prismatic bar characterized by a constant cross-section along its entire length is loaded at one end by a concentrated forceand supported at the other end as illustrated in Figure 13. Based on Newtons Law the constraint at the support generates a uniform distributed reaction opposed to the action. The free-body diagram is shown in Figure 13.a. The example considered falls in the category of members with uniform axial-deformation studied in Section 2. The exact determination of the distribution of normal stress along the length of the beam requires advanced methodologies employed in the Theory of Elasticity and for this reason only the results are presented. Analyzing the results illustrated in Figure 13 it is found that the formulae obtained in Section 2 are valid in the majority of the cross-sections except of those located in the vicinity of the ends. The perturbation zone has a length roughly equal to the width of the cross-section.

Figure 13 Normal Stress Distribution

Note: It can be concluded that for practical purposes the formulae obtained in Section 2 based on the assumptions contained in definition 1 are valid especially if the ratio. Special attention has to be given to the areas located near the point of load application or near abrupt changes in the cross-section. The application of Saint Venants Principle is valid for the case of the beam.

4.2 Stress Concentrations

In the theoretical development pertinent to the axial deformation of linear members, the area of the cross-section is considered as a smoothly varying function of the position . If discontinuities appear in the definition of the cross-sectional area, the formulae obtained in the preceding sections are invalid and the concept of stress concentration must be introduced.

Figure 14 Concentration of Stress

A typical case is shown in Figure 14 where a prismatic type linear member having a circular hole of diameter is subjected to a constant tension force . The member cross-section is described by the height and thickness.

The normal stress around the hole can be significantly greater than and varies as a function of the ratio. For practical applications the coefficient, called the stress-concentration factor is introduced. This factor is defined as the ratio of the maximum normal stress around the hole to the normal stress calculated in the absence of the hole, called nominal stress.

(103)

. (104)

The average normal stress calculated with the formula (14) is:

. (105)

The variation of the stress-concentration factor , calculated using the Theory of Elasticity methods is illustrated in Figure 1 The stress concentration factor for the configuration under consideration varies from 2.3 to 3.0.

Figure 15 Variation of the Stress-Concentration Factor

Analyzing the variation, it is concluded that if the diameterdecreases the concentration factor also increases. This is somewhat misleading and is due to the way the chart in Figure 15 is constructed. Using the average normal stressin the definition of the stress concentration factor, equation (103) the following formula is obtained:

(106)

Considering the extreme cases shown in Figure 15 and applying these within equation (106) it can be concluded that the maximum normal stress increases with the increase in the diameter of the hole and varies as:

(107)

For small diameter holes the stress concentration disappears at relatively small distance from the hole. This is an example of the application of Saint Venants Principle.

4.3 Limits of Poissons Ratio

The volumetric strainwas calculated in Section 2.9.3 for the general case when the normal stresses, and are applied simultaneously. For the case of axial deformation only the normal stress has a non-zero value and, consequently, the equation (2.111) is written as:

(108)

Because the volume can not decrease during the tensioning of the axially deformed member the volumetric strain is a positive value. Mathematically this condition is enforced as:

(109)

Consequently, from physical reality and equation (108) it is seen that:

(110)

The limits established for Poissons ratio by the expression (110) are generally valid for all materials used in structural engineering. The cases representing the limiting values, and , are pertinent to cork and water, respectively. Structural steel has a Poissons ratio of 0.33. There are some cases when the material has a negative Poissons ratio. These materials are called swollen solids and this unusual behavior is characteristic of certain materials subjected to radiation.

The transversal contraction of the cross-section is similarly obtained as:

(111)

In the design of the members subjected to axial deformation two important factors, the load and the resistance , are considered. The load represents the maximum axial force that occurs in the specific member when subjected to the action of exterior forces or change in temperature. The most common definition for the resistance is the force which is developed in the member when the normal stress reaches the yielding value.

The design of the member subjected to axial deformation is conducted under the condition that the capacity of the member, represented by its resistance force, must always be greater or equal to the demand force . Mathematically, this assumption is expressed as:

(112)

In the calculation of the load and resistance forces there are typically a number of factors (load magnitudes and directions, material characterization, manufacturing tolerance, etc.) which are not known with absolute certainty. The degree of uncertainty may be treated using the concepts proper of Probability Theory. In order to circumvent the difficulties inherent in the rigorous application of probabilistic methods, a global factor , called the safety factor, encompassing all possible uncertainties is introduced as:

(113)

The allowable resistance is then used in place of the resistance in equation (112).

(114)

The safety factor is always greater than or equal to unity:

(115)

This is the approach used by the method called ultimate strength design method and was adopted by American Concrete Institute (ACI) and American Institute of Steel Construction (AISC).

If the relation between the stress and strain is linear, than a similar safety factor may be defined by limiting the value of the normal stress in the axially deformed member.

(116)

The design formula (114) is modified using the relationship between maximum normal stress and the allowable normal stress:

(117)

The formula (117) was used for a long period of time in a procedure known as the allowable-stress design. Due to the simplicity of application, this method is still commonly used in United States for the design of steel structures.

The application of the theoretical formulae developed in this lecture is illustrated in the following examples.

Statically Determinate Structure

The structure illustrated in Figure 7 was investigated in Section 2. The following numerical values are considered in the numerical application of the generic formulation:

The reaction forces are:

Compression in column

Tension in rod

The column related values are:

The rod related values are:

The displacement of the leveling device is calculated as:

Statically Indeterminate Structure

The statically indeterminate structure illustrated in Figure 9 is solved in Section 2.3. For the numerical application of the formulation developed the following data are employed:

The stiffness coefficients are:

The rotation angle is:

The rod forces and are calculated as:

The allowable vertical force for the system is: